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https://crypto.stackexchange.com/questions/86835/why-this-example-about-pseudorandom-generators-is-performed-like-that
[ "# Why this example about pseudorandom generators is performed like that?\n\nSuppose Alice wishes to authenticate herself to Bob, by proving she knows a secret that they share. With pseudorandom number generators (PRNGs) they could proceed as follows.\n\n1. They both seed a PRNG with the shared secret $$s$$.\n2. Bob picks sends Alice some random number $$i \\in \\mathbb{N}$$.\n3. Alice proves she knows the share secret by responding with the $$i$$-th random number generated by the PRNG.\n\nI am wondering why Bob sends a random number $$i$$ instead of just $$1$$. Then Alice just sends $$G(s)$$ (where $$G(\\cdot)$$ is the PRNG) to Bob. Only Alice and Bob should have access to the secret $$s$$, and hence this should also work. In the above example, Alice and Bob should both have to compute $$G(s)$$ $$i$$ times, making it much more inefficient.\n\nWhich is the difference between what I am proposing and the original example? Aren't them both equally secure?\n\n• Where is this scheme suggested? As it's not on the linked Wiki page. – Modal Nest Dec 11 '20 at 13:27" ]
[ null ]
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https://nhigham.com/2020/07/28/what-is-the-polar-decomposition/comment-page-1/
[ "# What is the Polar Decomposition?\n\nA polar decomposition of", null, "$A\\in\\mathbb{C}^{m \\times n}$ with", null, "$m \\ge n$ is a factorization", null, "$A = UH$, where", null, "$U\\in\\mathbb{C}^{m \\times n}$ has orthonormal columns and", null, "$H\\in\\mathbb{C}^{n \\times n}$ is Hermitian positive semidefinite. This decomposition is a generalization of the polar representation", null, "$z = r \\mathrm{e}^{\\mathrm{i}\\theta}$ of a complex number, where", null, "$H$ corresponds to", null, "$r\\ge 0$ and", null, "$U$ to", null, "$\\mathrm{e}^{\\mathrm{i}\\theta}$. When", null, "$A$ is real,", null, "$H$ is symmetric positive semidefinite. When", null, "$m = n$,", null, "$U$ is a square unitary matrix (orthogonal for real", null, "$A$).\n\nWe have", null, "$A^*A = H^*U^*UH = H^2$, so", null, "$H = (A^*\\!A)^{1/2}$, which is the unique positive semidefinite square root of", null, "$A^*A$. When", null, "$A$ has full rank,", null, "$H$ is nonsingular and", null, "$U = AH^{-1}$ is unique, and in this case", null, "$U$ can be expressed as", null, "$U = \\displaystyle\\frac{2}{\\pi} A \\displaystyle\\int_{0}^{\\infty} (t^2I+A^*\\!A)^{-1}\\, \\mathrm{d}t.$\n\nAn example of a polar decomposition is", null, "$A = \\left[\\begin{array}{@{}rr} 4 & 0\\\\ -5 & -3\\\\ 2 & 6 \\end{array}\\right] = \\sqrt{2}\\left[\\begin{array}{@{}rr} \\frac{1}{2} & -\\frac{1}{6}\\\\[\\smallskipamount] -\\frac{1}{2} & -\\frac{1}{6}\\\\[\\smallskipamount] 0 & \\frac{2}{3} \\end{array}\\right] \\cdot \\sqrt{2}\\left[\\begin{array}{@{\\,}rr@{}} \\frac{9}{2} & \\frac{3}{2}\\\\[\\smallskipamount] \\frac{3}{2} & \\frac{9}{2} \\end{array}\\right] \\equiv UH.$\n\nFor an example with a rank-deficient matrix consider", null, "$A = \\begin{bmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ 0 & 0 & 0 \\end{bmatrix},$\n\nfor which", null, "$A^*A = \\mathrm{diag}(0,1,1)$ and so", null, "$H = \\mathrm{diag}(0,1,1)$. The equation", null, "$A = UH$ then implies that", null, "$U = \\begin{bmatrix} 0 & 1 & 0 \\\\ 0 & 0 & 1 \\\\ \\theta & 0 & 0 \\end{bmatrix}, \\quad |\\theta| = 1,$\n\nso", null, "$U$ is not unique.\n\nThe polar factor", null, "$U$ has the important property that it is a closest matrix with orthonormal columns to", null, "$A$ in any unitarily invariant norm. Hence the polar decomposition provides an optimal way to orthogonalize a matrix. This method of orthogonalization is used in various applications, including in quantum chemistry, where it is called Löwdin orthogonalization. Most often, though, orthogonalization is done through QR factorization, trading optimality for a faster computation.\n\nAn important application of the polar decomposition is to the orthogonal Procrustes problem1", null, "$\\min \\bigl\\{\\, \\|A-BW\\|_F: W \\in \\mathbb{C}^{n\\times n},\\; W^*W = I \\,\\bigr\\},$\n\nwhere", null, "$A,B\\in\\mathbb{C}^{m\\times n}$ and the norm is the Frobenius norm", null, "$\\|A\\|_F^2 = \\sum_{i,j} |a_{ij}|^2$. This problem, which arises in factor analysis and in multidimensional scaling, asks how closely a unitary transformation of", null, "$B$ can reproduce", null, "$A$. Any solution is a unitary polar factor of", null, "$B^*\\!A$, and there is a unique solution if", null, "$B^*\\!A$ is nonsingular. Another application of the polar decomposition is in 3D graphics transformations. Here, the matrices are", null, "$3\\times 3$ and the polar decomposition can be computed by exploiting a relationship with quaternions.\n\nFor a square nonsingular matrix", null, "$A$, the unitary polar factor", null, "$U$ can be computed by a Newton iteration:", null, "$X_{k+1} = \\displaystyle\\frac{1}{2} (X_k + X_k^{-*}), \\quad X_0 = A.$\n\nThe iterates", null, "$X_k$ converge quadratically to", null, "$U$. This is just one of many iterations for computing", null, "$U$ and much work has been done on the efficient implementation of these iterations.\n\nIf", null, "$A = P \\Sigma Q^*$ is a singular value decomposition (SVD), where", null, "$P\\in\\mathbb{C}^{m\\times n}$ has orthonormal columns,", null, "$Q\\in\\mathbb{C}^{n\\times n}$ is unitary, and", null, "$\\Sigma$ is square and diagonal with nonnegative diagonal elements, then", null, "$A = PQ^* \\cdot Q \\Sigma Q^* \\equiv UH,$\n\nwhere", null, "$U$ has orthonormal columns and", null, "$H$ is Hermitian positive semidefinite. So a polar decomposition can be constructed from an SVD. The converse is true: if", null, "$A = UH$ is a polar decomposition and", null, "$H = Q\\Sigma Q^*$ is a spectral decomposition (", null, "$Q$ unitary,", null, "$D$ diagonal) then", null, "$A = (UQ)\\Sigma Q^* \\equiv P \\Sigma Q^*$ is an SVD. This latter relation is the basis of a method for computing the SVD that first computes the polar decomposition by a matrix iteration then computes the eigensystem of", null, "$H$, and which is extremely fast on distributed-memory manycore computers.\n\nThe nonuniqueness of the polar decomposition for rank deficient", null, "$A$, and the lack of a satisfactory definition of a polar decomposition for", null, "$m < n$, are overcome in the canonical polar decomposition, defined for any", null, "$m$ and", null, "$n$. Here,", null, "$A = UH$ with", null, "$U$ a partial isometry,", null, "$H$ is Hermitian positive semidefinite, and", null, "$U^*U = HH^+$. The superscript “", null, "$+$” denotes the Moore–Penrose pseudoinverse and a partial isometry can be characterized as a matrix", null, "$U$ for which", null, "$U^+ = U^*$.\n\nGeneralizations of the (canonical) polar decomposition have been investigated in which the properties of", null, "$U$ and", null, "$H$ are defined with respect to a general, possibly indefinite, scalar product.\n\n## References\n\nThis is a minimal set of references, which contain further useful references within.\n\n## Related Blog Posts\n\nThis article is part of the “What Is” series, available from https://nhigham.com/category/what-is and in PDF form from the GitHub repository https://github.com/higham/what-is.\n\n## Footnotes:\n\n1\n\nProcrustes: an ancient Greek robber who tied his victims to an iron bed, stretching their legs if too short for it, and lopping them if too long.\n\n## 2 thoughts on “What is the Polar Decomposition?”\n\n1.", null, "Michael Scharrer says:\n\nI’m very much enjoying this “What is” series.\n\nOne question: At the beginning of this post you write: “This decomposition is a generalization of the polar representation z = r*e^(i*theta) of a complex number”. Can this analogy be taken further using the matrix exponential, setting U=e^V for some other matrix V?\n\n2.", null, "Nick Higham says:\n\nYes – there are standard results on writing an orthogonal or unitary matrix as exp(S) for a skew-symmetric or skew-Hermitian matrix." ]
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https://ufdc.ufl.edu/AA00022793/00001
[ "Citation\n\n## Material Information\n\nTitle:\nA controller design method which applies to time varying linear systems\nCreator:\nKoenig, Kurt Walter, 1967-\nPublication Date:\n\n## Subjects\n\nSubjects / Keywords:\nAcceleration ( jstor )\nAircraft pitch ( jstor )\nAutomatic pilots ( jstor )\nEigenvalues ( jstor )\nLinear systems ( jstor )\nMissiles ( jstor )\nParametric models ( jstor )\nSimulations ( jstor )\nTrajectories ( jstor )\nVelocity ( jstor )\n\n## Record Information\n\nRights Management:\nCopyright [name of dissertation author]. Permission granted to the University of Florida to digitize, archive and distribute this item for non-profit research and educational purposes. Any reuse of this item in excess of fair use or other copyright exemptions requires permission of the copyright holder.\nResource Identifier:\n20381782 ( ALEPH )\n32490049 ( OCLC )\n\nFull Text\n\nA CONTROLLER DESIGN\n\nMETHOD WHICH APPLIES TO TIME VARYING\nLINEAR SYSTEMS\n\nBy\n\nKURT WALTER KOENIG\n\nA DISSERTATION PRESENTED TO THE GRADUATE SCHOOL\nOF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT\nOF THE REQUIREMENTS FOR THE DEGREE OF\nDOCTOR OF PHILOSOPHY\n\nUNIVERSITY OF FLORIDA\n\n1994\n\nA B ST R A CT . . . . . . . . . . . . . . . . . .\n\nCHAPTERS\n\n1 INTRODUCTION ...............................\n\n1.1 Earlier W orks . . . . . . . . . . . . . . . .\n1.2 P purpose . . . . . . . . . . . . . . . . .\n\n2 THEORY BEHIND THE DESIGN METHOD ..................\n\n2.1 A Geometric Inerpretation of Lyapunov's Linear Stability Theorem .\n2.2 Adding Control to the System ......................\n2.3 A Linear Feedback Set to Control xTp ................\n2.4 One Linear Feedback Matrix to Control xTpk . . . . . . .\n3 A TIME VARYING SECOND ORDER EXAMPLE . . . . . . .\n\n4 DERIVATION OF A MODEL OF THE EMRAAT MISSILE ........\n\n4.1 The Nonlinear Model ...........................\n4.2 The Linear M odel .............................\n\n5 THE DEPENDENCE OF GAINS ON FLIGHT PARAMETERS .....\n\nGenerating p and V from M and Q . .\nThe Flight Parameter Generator . . .\nInitializing the Iterative Lyapunov Design\nThe Iterative Lyapunov Design Method .\nFormulation of a State Tracker . . .\nComparing Gains with Flight Parameters\n\nMethod .......\n\n6 COMPUTING LOOK-UP TABLES ......................\n6.1 Determining a Grid of Points ......................\n6.2 Formulation of :he Design Constraints ...................\n6.3 Generating the Look-Up Table ......................\n\n7 GAIN SCHEDULING .....................\n\n7.1 C urve Fitting . . . . . . . . . . . . . . . 83\n7.2 Testing the Fit . . . . . . . . . . . . . . .. .. 83\n\n8 NONLINEAR SIMULATIONS . . . . . . . . . . ... .. 88\n\n8.1 The Nonlinear Simulation . . . . . . . . . . ... .. 88\n8.2 A Test of State Tracking . . . . . . . . . . . .. .. 90\n8.3 Simulation of Flight Scenarios ... ..... . . . . . . . . .. 93\n\n9 CONCLUSION . . . . . . . . . . . . . . . ... .. 104\n\nAPPENDICES\n\nA AERODYNAMIC DATA FOR THE EMRAAT AIRFRAME ........ 106\n\nB INERTIAL DATA FOR THE EMRAAT AIRFRAME ........... 117\n\nREFERENCES ................................... 118\n\nBIOGRAPHICAL SKETCH ............................. 120\n\n. . . 83\n\nAbstract of Dissertation Presented to the Graduate School\nof the University of Florida in Partial Fulfillment of the\nRequirements for the Degree of Doctor of Philosophy\nA CONTROLLER DESIGN METHOD WHICH APPLIES TO TIME VARYING\nLINEAR SYSTEMS\nBy\nKurt Walter Koenig\nAugust 1994\nChairman: Dr. Thomas E. Bullock\nMajor Department: Electrical Engineering\n\nA feedback controller design method has been formulated which applies to linear\n\ntime varying systems. The motivation behind this technique is the design of an\n\nautopilot for an air-to-air missile. The missile under study is the extended medium\n\n'range air-to-air technology (EMRAAT) missile, a theoretical bank-to-turn missile\n\nwhich is under study by the United States Air Force. Knowledge gained by this\n\nmissile will be applied to future bank-to-turn missiles.\n\nConventional autopilot design techniques use pole placement, linear quadratic\n\nregulators, linear quadratic Gaussian/loop transfer recovery techniques, or eigen-\n\nstructure assignment methods. These techniques are applied to a linear model which\n\ndepends on time varying flight parameters. In applying these methods, the false as-\n\nsumption is made that the linear model is slowly varying. Since the system is changing\n\nrapidly, no theoretical basis exists to support the success of the resulting controller.\n\nThe proposed design method finds state feedback gains which cause the closed loop\n\nlinearized system to be stable with respect to a specified Lyapunov function. Even\n\nthough flight parameters change rapidly, local stability around the operating point is\n\nachieved.\n\nThe design algorithm finds feedback gains which place the eigenvalues of the\n\nderivative of a given Lyapunov function. Limitations on placement of these eigenval-\n\nues are stated. This concept is applied to a second order linear time varying system\n\nwhere ordinary pole placement techniques fail. The design method is then applied\n\nto a linearized model of the EMRAAT missile which is a function of time varying\n\nflight parameters. Feedback gains are generated as a function of these flight param-\n\neters. It is discovered that the gains depend on dynamic pressure, mach number,\n\nand angle-of-attack. A combination of interpolation and polynomial fitting is used\n\nto create a look-up table for the gains. The resulting controller is programmed into\n\na nonlinear simulation which runs missile and target scenarios. Small miss distances\n\nare achieved.\n\nCHAPTER 1\nINTRODUCTION\n\nThe original goal of this dissertation was to formulate an autopilot design method\n\nwhich stabilizes the flight of a bank-to-turn air-to-air missile. The result was the de-\n\nvelopment of a controller design method which applies to time varying linear systems.\n\nWhen used on a nonlinear system, local stability is achieved. The remainder of this\n\nchapter gives a brief description of the missile used in this study, a summary of related\n\nworks preceding this study, and the purpose and outline of this dissertation.\n\nAn air-to-air missile is launched from a military aircraft with the intent of in-\n\ntercepting an enemy aircraft or target. Bank-to-turn (BTT) missiles have wings\n\ngiving them greater maneuverability than the conventional skid-to-turn (STT) mis-\n\nsiles which have fins. Thus, the target must work much harder to evade a BTT\n\nmissile. The dynamics of BTT airframes are highly unstable, making for a difficult\n\ncontrols problem.\n\nThe missile under study in this paper is the extended medium range air-to-air\n\ntechnology (EMRAAT) missile as shown in Figure 1.1 . The EMRAAT missile\n\nis a theoretical missile formulated by the Air Force with the intent of studying the\n\nfeasibility of the BTT concept. The knowledge gained from the study of this missile\n\nwill be used by the Air Force if it ever decides to make a real BTT missile.\n\nThe EMRAAT missile is equipped with a seeker. If the seeker is infrared based,\nthen it measures the line of sight angles to the target. In the case that the seeker is\n\nradar based, then, in addition to the line of sight angles, the range and range rate\n\nof the target are measured. This information is passed to the guidance law which\n\ndetermines the desired normal accelerations needed to intercept the target. The\n\ncross section, constant -\n\nStart -\nwedge\nfairings\n\n7.5 din\n\nAll dimensions are in inches\n\nV A-A\nView A-A\n\nFigure 1.1. The EMRAAT missile .\n\nZZ\n\nautopilot attempts to achieve these accelerations by applying the proper control to\n\nthe control surfaces.\n\n1.1 Earlier Works\n\nEarly autopilot designs separated the missile dynamics into two or three lower\n\norder models. These models were linearized, and classical techniques were used to\n\nstabilize them. Other designs applied more advanced techniques to higher order lin-\n\nearized models. Some of these techniques are pole placement, linear quadratic regula-\n\ntors, linear quadratic Gaussian/loop transfer recovery techniques and eigenstructure\n\nassignment methods Feedback controllers would be designed as a function of the\n\nlinearized models. Gains would then be scheduled against the flight parameters on\n\nwhich the models depend. Gain scheduling is a popular technique used in control\n\ndesigns and has been studied in depth by Shamma, Athans, and Cloutier [3,4,5,6].\n\nThe problem with these design methods is that they assume that the system is\ntime invariant or slowly varying. The linearized models depend on rapidly changing\n\nflight parameters. Although these methods often work, no theoretical basis exists\n\nto support the success of these designs. More recently, H, and yi synthesis design\n\ntechniques have been used to formulate one dynamic controller which would stabilize\n\nthe system for all modeled uncertainties [7,8]. A disadvantage of H.. controllers is\n\nthat they have very high orders. Also, the existence of a robust H\"\" controller which\n\nsatisfies the performance requirements is not guaranteed. The only way to determine\n\nthe stability of a given design is to test it using a nonlinear simulation.\n\nDesoer states if a given time varying system\n\nx= A(t)x (1.1)\n\nis stable for each t when t is frozen, then it is not necessarily stable when t is allowed\n\nto change. However, (1.1) is stable if A(t) changes slowly. A conservative upper\n\nAlpha vs. Time\n12, ,,\n12o ---i--\" --------'- --- ---------\n10-\n\n8\n\nS6\n12\n4\n\n2\n\n0-\n\n0 1 2 3 4 5 6 7 8\nSeconds\n\nFigure 1.2. Angle of attack from one trajectory of the EMRAAT missile.\n\nbound on supt>o IIA(t)lI is given which, if enforced, guarantees asymptotic stability\n\nof (1.1). As will now be shown, this limit is too restrictive for the EMRAAT missile.\n\nFigure 1.2 shows the angle of attack of the trajectory of a missile flown using an\n\nalready existing autopilot. Desoer proposes the Lyapunov function\n\nV(x, t) = xT(61I + P(t))x (1.2)\n\nwhere ei > 0 and P(t) is chosen so that\n\nAT(t)P(t) + P(t)A(t) = -31 (1.3)\n\nDesoer gives a bound on V,\n\nV < xTx[-3 + 2elaM + aM3m4 (1.4)\n\nwhere the following definitions are made:\n\naM :=sup IA(t)|| < c (1.5)\nt>0\n\naoo is positive and\n\nRe Ai[A(t)] < -2c7o Vi, Vt > 0 (1.6)\n\nm is a constant and depends only on ao and aM so that\n\n|exp(rA(t))|l mexp(-oCor), Vr > 0 Vt > 0 (1.7)\n\nAlso,\n\niaM:=sup IIA(t)l. (1.8)\nt>0\nIf c is allowed to be very small, then from (1.4), stability would result if\n4o.2\nM < -- (1.9)\n\nFor the linear pitch model of the given trajectory, o0 = 15.75. At .04 seconds into\nthe trajectory the closed loop matrix is\n\nA -3.2533 .8997 (1.10)\n-1678.2 -83.50\nAt this instant in time, m = 22.88 and aM = 2231. 42 = .0036. From this we see\n\n'that the linear system is changing much faster than the limit shown in (1.9). The\nsystem is not slowly varying. Note that m was only found for t = .04 sec. and not\nfor all time. If a greater m were found for the rest of the trajectory, then the limit in\n1.9 would be even more restrictive and the result would be the same. This test does\nnot show that the system is unstable. In fact, a Lyapunov function is known which\nshows that this linear system is stable but Desoer's theory can not support this fact.\nWilson, Cloutier, and Yedavalli give a stability condition for a constant linear\nsystem with time varying uncertainties. Given a system,\n\nx = [A+ E(t)]x (1.11)\n\nif a positive definite P can be found which solves the Lyapunov equation,\n\nPA + ATP +21 = 0\n\n(1.12)\n\nAlpha vs. Time\n12\n\n10\n\n8-\n\na)\n6-\n\n4-/\n0\n\n& 3\n\n0 1 2 3 4 5 6\nSeconds\n\nFigure 1.3. Angle of attack from one trajectory of the EMRAAT missile.\n\nthen the system will be asymptotically stable if\n\n,max[E(t)] <,, 1 Vt (1.13)\n7maz.(P)\n\nwhere 0,max is the maximum singular value. This idea is initially interesting because\n\nit requires the error, E(t), to be bounded, but does not constrain the time variation.\n\nThe above result can not be easily applied to the EMRAAT missile. This will be\n\nshown by applying the result to a trajectory of the EMRAAT missile whose angle of\n\nattack is shown in figure 1.3. In order to check the stability of the EMRAAT missile,\n\nA must be chosen since it may be any matrix which is stable. It is believed that the\n\nbest choice of A should come from a closed loop matrix in the trajectory. With this\n\nin mind, A was taken at t = 2 sec into the trajectory.\n\nA-B '-4.477 .9526 (1.14)\nA- BK = -1.684 -79.47 (\n\nSigma max[E(t)] vs. Time\n2 5 0,,\n\n200-\n\nFiso\n\nE\n150-\n*a-\nE\n05100-\n\n50\n\n0 1 2 3 4 5 6\nSeconds\n\nFigure 1.4. A plot of axB[E(t)] versus time.\n\nE(t) is the result of subtracting this matrix from the closed loop system from the\n\nrest of the trajectory. Solving for P in (1.12) gives\n\n[ 17.29 -.0454 (1.15)\nS -.0454 .012 (115\n\nComputing the upper bound in (1.13) gives\n\nS= .0578 (1.16)\nam)axj4P)\nFigure 1.4 shows the plot of crmax[E(t)]. Obviously since marn[E(t)] is greater than\n\n.0587 for most of the trajectory then (1.13) is not satisfied. This procedure was\n\nrepeated by taking the constant closed loop matrix from every point in the trajectory,\n\nand the result was the same. The time varying linear system from this trajectory is\n\nknown to be stable because a Lyapunov function exists which can show this. The\n\nupper bound in (1.13) can not be easily used if at all to support the claim of stability.\n\nDesoer gives a stability limit on the rate of variation of a given closed loop\n\ntime varying system. Wilson, Cloutier, and Yedavalli give a stability limit on\n\nthe size of a time varying uncertainty. Both of these tests were found to be too\n\nconservative when applied to an already existing stable autopilot, and the results\nwere inconclusive. In addition, these tests are analytic tools. As yet, no design\nprocedure exists which can give a closed loop linear system that will pass either\ntest. It would be desirable to formulate a design method that can yield a closed\nloop time varying linear system which satisfies a less conservative stability condition.\nVidyasagar gives two important results. The first can also be found in Hahn \n\nand says that given a time varying system\n\nS= A(t)x (1.17)\n\nif a positive definite matrix P can be found so that the matrix\n\nPA(t) + AT(t)P (1.18)\n\nis negative definite for all time, then the system is asymptotically stable.\nThe second result allows the first result to be applied to a nonlinear system. Given\n\na nonlinear system of the form\n\nS= f(t,x)\n\nwhere\n\nf(t,O) = 0\n\nand f is continuously differentiable, then let\n\nA(t) = 8f(t, x)\nox x=O\nand assume that\nIIf(t,x) A(t)xl\nlim sup = 0\nllxll-o t>0 ||x\n\nand A is bounded. If 0 is an asymptotically stable equilibrium point of i = A(t)z for\nall time, then 0 is a locally stable equilibrium point for the system\n\n5 = f(t,x)\n\nThese ideas can be applied in the following way. Given a nonlinear system\n\nx= f(x,w,u) (1.19)\n\ndefine\n\nA(x,w) = B(x,w)= (1.20)\nxw .(XW)nomna (X,W)nominai\nwhere x is a vector containing the states and w is a vector containing additional\nsystem parameters. At regions near the operating point, the system becomes close\nto\n\nAk = A(x, w)Ax + B(x, w)Au (1.21)\n\nwhere Ax and Au are small perturbations between the states and inputs and the\n\noperating point. We would like to find a feedback control law,\n\nu = -K(x,w)x (1.22)\n\nso that\n\nP[A(x, w) B(x, w)K(x, w)] + [A(x, w) B(x, w)K(x, w)]TP (1.23)\n\nis negative definite for all x and w where P is positive definite. If such a control law\nis found, then the perturbations from the nominal trajectory are locally stable for\n\nthe system\n\n= f(x,w, -K(x,w)x) (1.24)\n\nShahruz and Behtash give one control law which places some of the eigenvalues\nof (1.23) for the case where P = I. However unnecessary limitations on where the\n\neigenvalues can be placed exist. Also, while many feedback control laws make (1.23)\n\nnegative definite, Shahruz and Behtash give only a small subset of these control laws.\n\n1.2 Purpose\n\nThis paper gives an algorithm which can stabilize a linear time varying system by\n\nplacing the eigenvalues of (1.23) between desired bounds and has fewer limitations\n\nthan the control law in Shahruz and Behtash . Limitations on the placement of\n\nthese bounds are stated. This algorithm applies to all systems for which a constant\n\npositive definite P exists so that (1.23) can be made negative definite for all time.\n\nFirst, theory is presented leading to the formulation of this algorithm. Then, to\n\ndemonstrate the usefulness of this algorithm, it is applied to a linear time varying\n\nsystem where normal pole placement techniques fail. This algorithm is then applied\n\nto the EMRAAT missile. A nonlinear model is made, and from this, a time vary-\n\ning linearized model is generated which is a function of several flight parameters.\n\nGains are computed, and their dependence on flight parameters determined. A gain\n\nscheduling scheme is then implemented yielding local stability around the operating\n\npoint. The resulting design is tested in a nonlinear simulation. Finally, the results of\n\nthis test are given, and the usefulness of this design technique is evaluated.\n\nCHAPTER 2\nTHEORY BEHIND THE DESIGN METHOD\n\nThe material in this section gives the theory leading to the proposed controller\ndesign method. The first section presents a geometric interpretation of Lyapunov's\nlinear stability theorem. The effects of control on the velocity field of a system is\nstudied in the next section. The nonemptiness and convexity of the set of all feedback\ngains which stabilize a system with respect to a given Lyapunov function are then\ndiscussed. Finally, an iterative procedure that finds one element of this set is given.\n\n2.1 A Geometric Interpretation of Lvapunov's Linear Stability Theorem\n\nTo understand why time invariance is a necessary assumption for eigenstructure\ndesign methods the following second order linear time varying system was studied.\nThe example given now is from Vidyasagar's example 5.3,109 and can also be\nfound in Khalil . Given the following system\n\nS= A(t)x (2.1)\n\nwhere\n[ -1 + a cos2(t) 1 a sin(t)cos(t)\nA(t) -1 a sin(t) cost) -1 + a sin2(t) (2.2)\nVidyasagar notes that the transition matrix is given by\nP(t0) e(a-1)tcos(t) e-tsin(t) 1\nq^(t,0)-= --e(a-1)sin(t) e-* cos(t) (2.3)\n\nand the characteristic equation is\n\nA2+(2-a)A+(2-a)=0 (2.4)\n\nThe roots of (2.4) have negative real parts for 1 < a < 2. The exponents in the first\ncolumn of D(t, 0) indicate, however, that the system is unstable for these values of a.\n11\n\nb\n1.^ *- z Y \\\\\n\n1r\\\n.-0.5 _\\\n\n-10 0 5 10 41 -1 -. 0 0-5 1\nX- X1\n\nFigure 2.1. The trajectory of (a) the system (2.1) and (b) a frozen system.\n\nThis system was simulated using MATLAB with a = 1.5 and xo = [1, 0]T. The\n\neigenvalues of A(t) are -!+ s7j and -- j Figure 2.la shows the state trajectory\n\nof the system 2.1 where X, is assigned to the horizontal axis, and x2 is assigned to\n\nthe vertical axis. This plot demonstrates the instability of the system. If the system\n\nwere frozen, i.e. A(t) = A(0), then the stable trajectory of Figure 2.1b would result.\n\nThis trajectory is shown with the velocity vector field of the frozen system.\n\nTo explain the instability in Figure 2.la, we will now discuss the time varying\n\nvelocity field of equation (2.1). The plots in Figure 2.2 show the trajectory of the\n\ntime varying system in the state plane for eight instances in time. For each plot, the\n\nvelocity vector field for that instant in time is superimposed on to the trajectory. Each\n\nplot in Figure 2.2 shows the boundaries of four pie-shaped regions which this paper\n\nrefers to as positive and negative regions. Two positive regions exist that are defined\n\nas the set of all points whose velocities have an outer radial component, i.e. each of\n\nthese velocities have a component pointing directly away from the origin. Likewise,\n\nthere are two negative regions containing velocities with inner radial components.\n\nThe plots show that the boundaries separating the positive and negative regions\n\nrotate in a clockwise direction. Also, the current position in the trajectory remains\n\ninside one of the positive regions. Since the velocity of the system always has an\n\nouter radial component, the magnitude of the state vector is always increasing so\n\nthat the system (2.1) is unstable.\n\nA direct relationship exists between the positive and negative region boundaries\n\nand the eigenvectors of the system. In this case, the eigenvectors are rotating clock-\n\nwise at the same angular velocity as the region boundaries. It can be shown that if a\n\nsystem has constant eigenvectors with time varying eigenvalues which have negative\n\nreal parts for all time then the system is stable. The stability question, however, is\n\nnot as easy to answer for the case with moving eigenvectors.\n\nIf the system has no positive regions, then it will be stable regardless of how\n\nquickly it changes. However the converse is not true in general. Figure 2.3 shows a\n\nstate vector and its corresponding velocity for some dynamic system. The angle 9\n\ncan be computed in the following way.\n\nXTk\ncos(0) = xl (2.5)\n\nFor k to have an inner radial component then\n\nS< 0 < (2.6)\n22\n\nThis is true when\n\nxT < 0 (2.7)\n\nwhere\n\nk = Ax (2.8)\n\nSo if xTAx is negative everywhere for all time, then the system is stable. Since the\n\nsystem is linear, it is sufficient to check the sign of xTAx on the unit circle. If all\n\nvelocities on the unit spheroid point inside that boundary, then the same is true for\n\n10 S \\\n\n4 ) 4 ^ '\\\n\n0 F. -' >^ \\\n\n-5o\n-10 \\ \\ 0 10 is\n\n151\n-5 -10 .5 0 5 10 1I\n\nXI\n\nis 7 -A 1-\n107\n\n15 -1 0 5 1\n.S ,' />\n\n10 7 Ic j / /\n\n-1 -10 *I 0 5 10 15\nXI\n\n10 -, ^ '- N\n\n4 -4 -4 - ,- *-i *^\n\ns-10\n\n-s.. \" 4-^ f 4-\n\n-1 -10 -3 0 5 10 1S\n\n4\n15 -\n\n-1 -0 5 4 O 1\n10\n\n--^ -4 --\n5^ -^ 'p x S.\n\n0 '- 05 \\$ 0 i\\$\n\n,'F -~ F? 4\\ 4 ^\n\n-10\n4* 4- -4 - <\\ <- -' /\n-15 -\n-15 .10 -5 0 S 10 15\nxi\n\nis\n\n'F 4 ^\n\n-0' *, 4 ^i \\ \\ .\n\n\"' \\ ^ C\n\nS 4L\n-,.~I ^. >^-v\n\n-10. \\ I\\ t 4\n\\ \\ I\\ \\ I s\n\n-15 -10 .5 0 5 10 15\nXI\n\n4\n15 1\n\n10\n\n*^ ^ 4 4 4 ,\n\n.5-\nC*y 4- 4-^\n\n-151 is/\n\nXI\n\nFigure 2.2. The trajectory and velocity field of (2.1)\n\n15\n\n2\n\n1.5\n\n0.5\n0/\n\n-0.5\n-I\n\n-1.5\n\n-2 -1.5 -1 -0.5 0 0.5 1 1.5 2\nxl\n\nFigure 2.3. x and x\n\nall spheroids centered on the origin. This leads to the following stability condition\n\nwhich is given without proof.\n\nStability Condition 1: If xTic = xTAx < 0 Vx : XTX = 1,V t > 0 for\n\nthe system, 5 = A(t)x, then the origin is a global asymptotically stable\n\nequilibrium point.\n\nFigure 2.4 illustrates an example of a second order system which meets Stability\n\nCondition 1. Note, that while all figures given so far represent second order linear\n\nsystems, the above stability condition applies to linear systems of any order.\n\nThe above stability condition can be made less conservative by expanding the\n\nclass of shapes to ellipsoids defined in the following way.\n\nxTPx = 1 (2.9)\n\nwhere P is a symmetric, positive definite matrix. Figure 2.5 shows an ellipse of the\n\nform (2.9) defined for a second order system. The state vector, x, is drawn to some\n\n-1 -0.5 0 0.5 1\n\nFigure 2.4. An example of a\nponents\n\nsecond order system with negative radial velocity corn-\n\nFigure 2.5. One velocity vector on the ellipse xTpx = 1\n\n0\n\npoint on the ellipse, and its velocity, :, is also shown. The normal to the ellipse at x\nis Px. To ensure stability, it is sufficient to require that the projection of k onto the\nnormal of the ellipse, Px, is negative everywhere. The resulting term, xTpk is the\nnormal velocity component on the ellipse. The main thrust of this paper is to be able\n\nto control xTP* and to make this normal velocity component negative everywhere\nfor all time in order to achieve stability. As before, the linearity of the system allows\n\none to only check xTpk on the unit circle. This generalizes the previous stability\ncondition to\n\nStability Condition 2: For the system k = A(t)x, ifxTpk = xTPA(t)x <\n0 V x : XTX = 1, V t > 0 for some constant positive definite matrix P,\nthen the origin is a global asymptotically stable equilibrium point.\n\nThe above condition applies to linear systems of any order.\n\nLyapunov stated that if any positive definite function of the system states\nV(x) is always decreasing, i.e. V(x) < 0, then the system is stable asymptoticallyy\nstable for linear systems) . For the linear case, let\n\nV(x) = xTPx (2.10)\n\nV(x) is positive definite if and only if P is a positive definite matrix. Taking the\n\nderivative gives\n\nV(x) = xTPx + xTp (2.11)\n\n= xTATPx + xTPAx (2.12)\n\n= xT[ATp + PA]x (2.13)\n\nLyapunov's stability condition becomes\n\nxT[ATP+PA]x<0 Vx,Vt > 0\n\n(2.14)\n\nSince\n\nXTpx = XTPAx = xT[ATP + PA]x (2.15)\n2\nStability Condition 2 is equivalent to Lyapunov's linear stability condition.\n\nIt would be useful to compute the lower and upper bound of the rate of decay of\na given positive definite function,\n\nV(x) = xTPx (2.16)\n\nof the states of a time varying linear system by evaluating,\n\nc := min [AminI PA(t) + AT(t)P] (2.17)\n-- 1\n\nc := max [Ama [PA(t) + AT(t)P]] (2.18)\n\nSo,\n\nc < xT[PA(t) + AT(t)P]x < V x : xTx = 1, V t (2.19)\n_2\nThen c and Z are the lower and upper bounds respectively of the normal velocity\ncomponents, xTP*, on the unit spheroid. If c < 0, then the system is stable.\n\n2.2 Adding Control to the System\n\nThe result in section 2.1 is an analytical tool only. A design procedure is needed\nto control stability for the system\n\nx= A(t)x + B(t)u (2.20)\n\nwhere u is the control vector and x is the state vector. This section focuses on two\n\nquestions:\n\n1) What effect does u have on the velocity field of (2.20)?\n2) Can the normal velocity component xTP on the spheroid xTx = 1\nbe controlled?\n\nFigure 2.6. A velocity vector (a) without control and (b) with control.\n\nTo answer question 1), consider a second order single input system frozen at some\n\ninstant in time.\n\nS= Ax + Bu (2.21)\n\nA is a 2 x 2 matrix, and B is a two dimensional column vector, u is a scalar input\n\nand can take on any real value. If u = 0 then the velocity field of the system is\n\nk = Ax. Figure 2.6a shows the velocity, k = Ax, of some state vector, x, in the\n\nsystem. The control vector, B, is also shown. When u : 0, 5 has the extra term,\n\nBu. The direction of Bu is constant, but its magnitude is directly proportional to u.\n\nFigure 2.6b shows many possible values for by sweeping u through a wide range of\n\nvalues in small increments. As this diagram shows, the arrow head of each velocity\n\nvector can be placed on the line drawn parallel to B and intersecting the arrow head\n\nof Ax. This demonstrates that velocities can be controlled along the space spanned\n\nby the columns of B. The following theorem answers question 2).\n\nB\n0-5\n0.5\n\n-1\n1.5 -\n~2 -1.5 1 -0 .5 o o .5 1 '.5 2\n\nFigure 2.7. A case in which the normal velocity component cannot be controlled.\n\nTheorem 1\nConsider the system, k = A(t)x+B(t)u, where dim(x) = n, dim(u) = m\nand A(t) and B(t) have compatible dimensions. Let P be a constant\npositive definite matrix. The normal velocity component, xTpk, can be\narbitrarily set to any value with the right choice of u at a given time t,\nat any point x, on the spheroid, xTx = 1, unless x is contained in the set\n\nS(t) := P-'span(B_(t)) n {x: xTx = 1}\n\n(2.22)\n\nwhere Bj(t) is a basis of column vectors orthogonal to span(B(t)). If\nx E S(t) then\n\nxTpk = xTPA(t)x\n\n(2.23)\n\nand this velocity component cannot be controlled at time t.\n\nTheorem 1 gives the parts of the unit spheroid for which the velocity component\nxTP* is uncontrollable and can therefore be used to determine if a linear time varying\n\nsystem is stabilizable with an appropriate choice of constant positive definite P. If\nthere exists a constant positive definite P so that\n\nmax max xTPA(t)x < 0 (2.24)\nt xES(t)\nthe system is stabilizable. The above condition is equivalent to requiring the following\nmatrix to be negative definite for all time.\n\nIBT(t)[A(t)P-1 + P-1AT(t)]BL(t) (2.25)\n21\nThis fact was established by Fields and is true for the following reasons. Condi-\ntion (2.24) is true if and only if\n\n(P-l B (t),L)T(PA(t))(P-lB (t),t) <0VI' 0, Vt (2.26)\n(P-1B (t)'U)T(P- (B (t)j')\n\nSince the denominator of the above fraction is positive, with some simplification the\nabove statement is equivalent to\n\n21T BT(t)[A(t)P- -1 <0 # 0, + t (2.27)\n2 PA P- _L P^At)] y < 0 V IL k 0, V t (2.7\n\nwhich is equivalent to\n\nBT(t)[A(t)P-1 + P-lAT(t)]BI(t) <0 V t (2.28)\n\nIf a constant positive definite P exists satisfying condition (2.28), then the given time\nvarying system can be stabilized. This result is more general than a similar one given\nby Shahruz , and the two are equivalent when P = I.\nBefore giving a proof of Theorem 1, we give an intuitive explanation of Theorem\n1 for the second order case with a single input. Figure 2.7 shows the B vector for\nsuch a system along with some ellipsoid xTpx = 1. If a state vector, x, is drawn\nto some point on this ellipse whose normal is not perpendicular to B. then there is\n\nalways a control, u, which can place x inside the ellipsoid. This is true because the\nvelocity field can always be controlled in the B direction. This is not true, however,\nfor points on the ellipsoid whose normal is parallel to span(Bi.) as shown in figure\n2.7. These points on the ellipsoid can be found by computing\n\nP-'span(BI) (2.29)\n\nwhere BI is a basis of column vectors orthogonal to span(B). We now give the proof\nof Theorem 1.\n\nProof:\nCase 1: x E S(t)\n\nSubstituting 2.20 into xTp*, we have\n\nxTpk = xTPA(t)x + xTPB(t)u (2.30)\n\nSince x E S then\n\nx = P-'1(B.(t)) (2.31)\n\nwhere l(B-(t)) is any linear combination of Bj1(t). So\n\nxTPB(t)u = l(B_(t))TP-lPB(t)u (2.32)\n\n= l(B(t))TB(t)u (2.33)\n\n= 0 (2.34)\n\nand therefore from 2.30 we have,\n\nxTPk = xTPA(t)x\n\n(2.35)\n\nCase 2: x 0 S(t)\nSuppose we want to set xTpik = c where c is some arbitrarily chosen\nvalue. Then we want\n\nc = xTPA(t)x + xTPB(t)u (2.36)\n\nSince x S(t) then xTPB(t)u : 0 and there exists at least one u such\nthat 2.36 is true.\n\nOne could solve (2.36) for u to use as a control law, but this would not be practical\nto implement because A and B can not easily be computed. Linear state feedback is\nmore desirable. The next section will develop a procedure for computing the set of\nfeedback gains that will implement a feedback control law keeping the normal velocity\ncomponent, xTPk, within some specified limit.\n\n2.3 A Linear Feedback Set to Control xTP*\n\nRecall from Section 2.1 that the stability of a time varying linear system\n\nx = A(t)x (2.37)\n\ncould be analyzed by evaluating the bounds of xTPA(t)x on the spheroid\n\nxTx = 1 (2.38)\n\nNow we want to find the set of all linear state feedback gains for the control law,\n\nu = -K(t)x (2.39)\n\nfor the system\n\nx= A(t)x + B(t)u (2.40)\n\nso that condition (2.19) holds for the closed loop system where c and are now\nspecified. This section will give conditions for the nonemptiness of such a set followed\n\nby a discussion of its convexity. Finally this section will present an exhaustive search\n\nmethod for finding the boundary of this set.\n\nFor the remainder of this chapter, A and B will be frozen at one instant in\ntime. A discussion follows on how to find a controller or set of controllers which\n\nsatisfies (2.19) at one given instant. If a given system is time varying then the results\nwhich follow must be applied for all time with P being constant and positive definite.\n\nThese results apply to all systems which can be stabilized with respect to a Lyapunov\nfunction using a constant P. The success of these methods depends on the existence\nof a constant positive definite P so that\n\n2BT(tt)[A(t)P-1 + P-AT(t)]B.(t) < 0 V t (2.41)\n\nSince we require P to be zero, these results are conservative.\n\nLet KA and TC be the set of all K which satisfy the left and right inequalities\n\nrespectively of the following expression.\n\nc < xT[PA + ATP PBK KTBTPx V x : xTx = 1 (2.42)\n\nwhere P is positive definite. The objective is to find\n\n/C := k nC (2.43)\n\nThe following theorem gives conditions for the nonemptiness of _C and K?\n\nTheorem 2 Given the 'nth order m-input system,\n\nS= Ax+ Bu (2.44)\n\nk and KT are nonempty if and only if there exists a positive definite P so\nthat the following conditions are true:\n\nc < minxTPAx = min -xT[PA + ATP]x (2.45)\nX ES xES 2\n\nc> maxxT PAx = max 2xT[PA + ATp]x (2.46)\nxES xeS 2\n\nConditions (2.45) and (2.46) are respectively equivalent to\n\nc < A,.m[I(vH/)-'B [AP-1 + P-1AT]BH(v')1- (2.47)\n\nand\n\n>_ Anax[(V'-H)-'BT[AP-1 + P-1AT]B(-vH)1] (2.48)\n\nwhere\n\nH = (P-1B)Tp-1B (2.49)\n\nThe matrix H is square, full rank, and has dimension n m. The above is true for\nthe following reasons. Equation (2.46) can be rewritten as\n\nS> max 1xT[PA + ATp]x (2.50)\nx=P-lBBL,,xTx=l 2\n1 T T -1p1T1Rt.\n= max -yB[APl +P-lATBI (2.51)\nUTHL=I 2\n\nSince H is positive definite then ~ exists, is square, and has an inverse. By making\nthe substitution, y = (v/H)-lz, (2.51) becomes\n\nc > max zT(v/HB)-IBT[AP-1 + P-1A]B(v/H\")-1z (2.52)\n-zTz= 21\nwhich is equivalent to\n\nAm [vH)-1IT[AP-1 + P-1'A]B(vH)-] (2.53)\n\nEquation (2.45) is equivalent to (2.47) for similar reasons. We now give a proof of\nTheorem 2.\n\nProof:\n\nWe will show the nonemptiness of KX. KI can be shown to be nonempty in\n\na similar manner.\n\nCase 1:\n\nmaxxxTPAx < (2.54)\nxES\nthen\n\nxTPAx < V x E S (2.55)\n\nLet K = kBTP where k is a scalar value. Then\n\nx = lxT[PA + ATP PBK KTBTP]x (2.56)\n\nI XT[PA + ATP]x- k[xTPB][BTPx] (2.57)\n2i 2\n\nSince xTPB [BTPx]T then\n\nxTPBBTPx > 0 V x S, V xTx = 1 (2.58)\n\nIf (2.55) is true and if k is made large enough, then the second term in\n\n(2.57) will dominate V x S and x'p5c < V x : xTx = 1. Since\n\nkBTP E E then E is nonempty.\n\nCase 2:\n\nmaxxTPAx > (2.59)\nxr=S\nThen\n\nXTPAx > for some x E S. (2.60)\n\nBy Theorem 1, xPAx can not be controlled V x E S; so TC is empty.\n\nRemarks:\nThe nonemptiness of the intersection K: = 4c n is not guaranteed. If the designer\ndiscovers that no intersection exists, then the upper and lower velocity bounds will\n\nAnother useful property of CK: is convexity. This property is valuable in formulating\niterative search techniques to be described in the next section. K is convex if when\n\nx and y are elements of K, then cax + (1 a)y is also an element of )C for 0\n.\n\nTheorem 3 Let K: be the set of all K so that\n\nAmax [I[P(A BK) + (A BK)TP]] < (2.61)\n\nLet IC be the set of all K so that\n\nAmin[I[P(A BK) + (A BK)Tp]] > c (2.62)\n2\nThen /C, CK:, and : = K: fn kC are convex.\n\nProof:\n\nIf K: and K)C are convex, then the intersection K: is convex. Convexity of\nkC will be proven here. In a similar manner the proof for the convexity of\nKC can be written.\n\nLet K1 and K2 be elements of T. We must show that aK1 + (1 ca)K2\nis contained in KT when 0 < a < 1. We know that\n\nIxT[PA+ATP-PBK1 -K KTBTP]x < V xTx = 1 (2.63)\n\nand\n\nlxT[PA + ATP- PBK2 KTBTP]x < V x = 1 (2.64)\n2~ ~\n\nThen\n\nax T[PA + ATP PBK1 KBTP]x < acE V x = 1 (2.65)\n\nand\n\nI(1-aQ)xT[PA+ATP-PBK2-KTBTp]x < (1-_a) VxTx 1 (2.66)\n2\n\nwhere 0 < a < 1. Adding (2.65) to (2.66) gives\nxT[PA + ATP]x- _QXT[PBKl + KTBTP]x\n(2.67)\n-( a)xT[PBK2 + Kf2BTp]x < Z V xTx = 1\nwhich becomes\n\n1xT[PA+ATp-PB(aK+(1-a)K2)-(oaK+(1-a)K2)TBTP] < Vxx =1\n2\n(2.68)\nSo caK1 + (1 a)K2 E C for 0 < a c < 1. /C is convex.\n\nNow that we have conditions on the nonemptiness and convexity of KC and k_ it\nwould be desirable to find the boundary of these sets. We know that if one of the\neigenvalues of the square matrix Q is c then\n\ndet[Q cI] = 0. (2.69)\n\nThis is helpful in understanding the following exhaustive search method for comput-\ning the boundary of T. A similar procedure exists for finding Q9_.\n\n1) All eigenvalues of\n\nI-[PA + ATP- PBK KTBTP] -I (2.70)\n2\nneed to be less than or equal to zero. Fix all but one of the entries of K.\nLet the i,j'th entry be the free entry and be represented by k. Let Koi,j\n\ncontain all the fixed entries of K and\n\nSkl11\n\nbe defined as follows.\nklj kin 1\n\nkil 0 kin\n\nkml k77j km,\n\nThen\n\nK = kQij + Koi,\n\n1 j\n0 0\n\n0\n0 1\n0\n\n0 0 0\n\n(2.73)\n\n2) Find all values of the free entry which make\n\ndet[PA + ATP 2cI PBK KTBTp] = 0\n\n(2.74)\n\nThis problem reduces to finding the roots of a polynomial. After sub-\n\nstituting equation (2.72) into the argument of the above determinant, it\n\nbecomes\n\nk(-PBQi,j QJBTP) + (PA + ATP PBKoi,j KrjBTP) (2.75)\n\nLet\n\nPl(i,j) := -(PBQi,j + QT.B TP)\n\nPo(Koi,j) := PA + ATP PBKo,j KTojBTP\n\n(2.76)\n\n(2.77)\n\n(2.71)\n\nwhere\n\n(2.72)\n\nKoi,j :=\n\nQij \"=\n\nThen (2.74) becomes\n\ndet[kP1 + Po] = 0 (2.78)\n\nSolve (2.78) for k.\n\n3) Reject all complex roots. If all the roots are complex then skip the\n\nnext step.\n4) Test the intervals between the real roots by checking to see if\n\nAma[PA + ATp 2-cI PBK KTBTp] < 0 (2.79)\n\nThe K's that bound the interval which satisfies (2.79) lie on the boundary\n\nof T. Convexity of K implies that no more than two K's bound this\n\ninterval.\n\n5) Repeat the process for all possible values for the fixed entries in K.\n\nThe result is 9K.\n\n9KC can be computed in a similar way by reversing the inequalities in the above\nprocedure and by replacing max with Aiin. To find C the intersection of KC and TC\n\ncan be found in step 4).\n\nThe fixed entries in step 5) must be assigned to a finite number of grid points if\n\nthe above procedure is to be executed on a real system. The spacing of these grid\n\npoints must be smaller than the size of the feedback set. If the system has a high\n\norder or a multiple number of inputs, the number of grid points will become too large,\n\nand it will not be practical to implement this method. Understanding this procedure,\n\nhowever, leads to the formulation of an iterative method that can be used on high\n\norder, multiple input systems and will be presented in the following section. First,\n\nan example is given applying the exhaustive search method to a second order, single\n\ninput system.\n\nExample 1\n\nGiven the system\n\nx= Ax + Bu\n\n(2.80)\n\nwhere\n\nA '1 12 ]\n0 -1 1\nand\nB = cos(75) ]\nS sin(75o)\nlet\n\nP=I\n\nWe want to find all feedback gains which satisfy the following constraint.\n\nc\n\n(2.81)\n\nBefore specifying c and c, we must check the normal velocity components\nfor x E S as Theorem 1 requires. Let\n\nx=B = [ -sin75]\n\nxAxo = -12.4075\n\n1x TAxi =-12.40\n\n(2.82)\n\n(2.83)\n\nFrom Theorem 2 we must have\n\nc < -12.40 <\n\n(2.84)\n\nFrom this we choose\n\nc = -14\n\n-- -9\n\n(2.85)\n\n(2.86)\n\nThen\n\nIn this example n = 2 and m = 1. It was decided to set i = j = 1 so\nthat\n\nQ,j=Qi,i= [li 0] (2.87)\n\nKoj = K 01,1 =[ 0 /i2 ] (2.88)\n\nand\nK = kQi,i + Koi,i = k K2 ] (2.89)\n\nFrom equations (2.76) and (2.77) come\n\nPi_ = -[BQ,1 + Q,1B T] (2.90)\nPl = -[BQ1,, + QIBT] (2.91)\n\nPoC = A + AT 2cI BKo0,1 KTBT (2.92)\nPo- = A + AT 2I BKoi,1 KT' BT (2.93)\n\nThe roots of the following polynomials are computed in terms of k while\nincrementing K2 through a wide range of values.\n\ndet[kP + Po] = 0 (2.94)\n\ndet[kP +Po0] = 0 (2.95)\n\nRejecting complex roots and checking the regions separated by the real\nroots give\n\nk!(K2), kc(K2) (2.96)\nku(K2), kj(g2) (2.97)\n\nThe intersection of these regions are found.\n\nk(K2) = max[k,k-E] (2.98)\nk(K2) = mink[,K] (2.99)\n\n14\n\n13 -\n\n12\n\n11 .[7,11]\n\n10\n\n9\n\n81\n4 5 6 7 8 9 10 11 12 13 14\nki\n\nFigure 2.8. A plot of the boundary of IC which guarantees satisfaction of the design\nconstraints of Example 1.\n\nFinally,\n\nC C={[k(K2),K2] V K2 }u{[k(K2),K2] V K2} (2.100)\n\nA plot of 9aC is shown in Figure 2.8. Let K = [7 11]. We can check to see\n\nif K E KC by evaluating\n\nA1 = Amin,(A+AT-BK KTBT) (2.101)\n\n= -12.91 (2.102)\n\nA2 = Am(A + AT BK KTBT) (2.103)\n\n= -10.52 (2.104)\n\nAs the following shows,\n\nc < A1 (2.105)\n\nA2 < Z (2.106)\n\n2.4 One Linear Feedback Matrix to Control xTP*\n\nThe search procedure given in section 2.3 becomes impractical for high order\nsystems with multiple inputs because the number of grid points for the fixed entries\nof K becomes very large. This section discusses an iterative Lyapunov design method\nwhich saves on computations and finds one element K E K, if it exists, where KC =\nKC n KC. By applying this procedure at every instant in time to a time varying linear\nsystem, one can find a control law stabilizing the system if a constant positive definite\n\nP exists which satisfying the following condition.\n\n2BT(t) [A(t)P-1 + P-1AT(t)]B(t) <0 V t (2.107)\n\nBy Theorem 2, this condition guarantees the existence of a Z < 0 such that C(t)\nis nonempty for every instant in time. This procedure applies to all systems which\ncan be stabilized with respect to a Lyapunov function given by a constant positive\ndefinite P. The following is a discussion of the iterative Lyapunov method followed\nby the algorithm itself. An example is then given applying this procedure to one\n\noperating point of a fifth order linearized model of the EMRAAT missile.\nFigure 2.9 illustrates the iterative procedure. Kf is the feedback set which satisfies\nthe designer's predetermined constraints for some specified P. The constraints are\n\nCf < xT[PA + ATP PBK KTBTp]x < Zcf V x xTx = 1 (2.108)\n\nLet Ki be defined as the set of all K which satisfies the following constraints.\n\nci < xT[PA + ATp PBK TBTp]x < cZ V x : xTx = 1 (2.109)\n\nGiven Ki, we would like to find c, and ci so that if Ki 0 kf then Ki E MKi. We also\nrequire that Kf C Ki. If Ki E Kf, then we want K, = KC. The following definitions\nfor c, and ci meet these requirements. Let\n\ni := max xT[P(A BKi) + (A BK)TP]x (2.110)\nIlx||=1 2\n\nFigure 2.9. A geometric view of the iterative Lyapunov design method.\n\nand\n\nAi:= min Ix[P(A BKi) + (A BKi)Tp]x (2.111)\n||x||=i 2 'J'\nThen let\n\nS= max[cf, \\i (2.112)\n\nand\nci = min[cf,A ] (2.113)\n\nK0 is the initial guess in the search for Kf E kfA. Co and co are computed using\n(2.112) and (2.113) so that Ko is a member of 9ko and /Cf C PC0. Then a new\nfeedback matrix, K1 is found which lies inside of K0, but not on the boundary. New\nconstraining values are found in the same way as before so that the boundary of the\nnext feedback set contains K1. K2 is then found so that it lies inside of the present\nfeedback set, but not on its boundary. This process is continued until Ki = Kf e ACf.\n\nThe success of finding Kf depends on the following conditions.\n\n1. Given that Ki E a9)C we must be able to find Ki1 such that K++i G Ki.\n\n2. We must show that 4i+1 < c, when -, > Zf and ci+1 > ci when ci < cf.\n\n3. We must show that )Cf C kCi+l C Ki.\n\n4. 'Cf must be nonempty.\n\nWe now address these four points.\n\n1. Given Ki E aki, we need to find a second feedback matrix Ki2 C dKci where\nKi2 7 Ki. Then, due to convexity, 1 Ki + 1Ki2 is a member of kC{. Figure 2.10Oa shows\na second order example of a procedure for finding Ki+A The algorithm will be given\nshortly. The horizontal and vertical axis are assigned to k1l and k12 respectively. The\nregion PCi is enclosed by aK&L and 9Ci. Kj is known. 'i2 is found by searching along\n\nthe line that passes through K, and is parallel to the k1l axis. Point c is found by\ncomputing Ki + -Ki2. Since K;i is convex, then c E kC. This step is repeated again\nby searching along the k12 axis. Using similar arguments, point f is also in KCi. KJ+i\n\nis set equal to point f. For higher order systems, additional boundary points are\nfound and interior points computed by searching along directions which are parallel\nto the axis of the coordinate system.\n\nUnder normal conditions, this procedure works well. However numerical problems\ndo occur. These problems will now be discussed along with a cure. Figure 2.10b shows\na second order example of when the above method fails. The boundaries are shaped\n\nin such a way that searching along line 11 and 12 yields no new boundary points. A\nsolution to this problem is to relax one of the constraining values by setting, for ex-\nample, zi equal to + 6. 9Ad will then move so that point b can be found as shown in\nfigure 2.10c. Later, the relaxed constraining value can take on its original assignment.\n\n2. We need to show that c, > Zi+l when c, > 5f. In this case\n\nTi= Ai > Zf (2.114)\n\nSince Ki+j e KCU and Ki+j OQCi, then the following condition holds with strict\ninequality.\n\nxT[P(A BKi+1) + (A BKi+I)TP]x < a Vx : xTX = 1 (2.115)\n\nTaking (2.110) for i + 1 then\n\nIxT[P(A BKi+1) + (A BK<+)TP x V : xT = 1 (2.116)\n\nand there exists an x which satisfies the above condition for equality. Therefore\nZi > AX+i and Zi > maxc.f,\\i+l] = -i+n. Using similar arguments, it can be shown\n\n38\n\na b\n\nOKi\n\nS< \"/ /\n*// Ii0o-/ /\nK. a'ttily K ^2 / / /\n\n/ K,\na' l1\n/ a1 / h\n\nk12\nc\n\n--I-\n\n/\nf' /t9\n\n^i+i\nS b\n\nd ki\n\nFigure 2.10. Step 5 of the iterative Lyapunov design method for (a) a second order\nexample, (b) how it sometimes fails, and (c) how this problem is corrected.\n\nthat c9 < cj+1 when ci < cf.\n\n3. We now show that Cf C kCi+i C Ki. Since c, = max[f, Aj], then ci > c. We have\n\nalready seen that ci > >Ai+. So ci > max[f, AIi+] = ci+i. Similarly c ci+. Cki+l is\n\nthe set of all K so that\n\nc I+ < xT[P(A BK) + (A BK)TP]x < c6+1 V x xTx = 1 (2.117)\n\nSince T, > Ei+l and ci <_ 4i+1, then for every element of Kji+ the following holds.\n\nc < xT[P(A BK) + (A BK)TP]x < Z, V x: xTx = 1 (2.118)\n\nTherefore, .Ci+l C K;. Since c-i+ > cf and ci+l < cf, then using a similar argument\n\nkCf C kA+1.\n\n4. In using this design procedure, P is chosen so that the maximum uncontrol-\n\nlable normal velocity component is negative. Then from Theorem 2, cf can be made\n\nnegative in an attempt to achieve stability. -f must be greater than the maximum\n\nuncontrollable normal velocity component, and cf must be less than the minimum\n\nuncontrollable normal velocity component. From Theorem 2 this will guarantee the\n\nnonemptiness of Cf and kf. The nonemptiness of the intersection of these two sets,\n\nhowever, is unknown. If Cf is nonempty, then, as i becomes large, Ki G Kfj. If Cf\n\nis empty then ci and -. will converge to values which do not match the desired con-\n\nstraints and Ki will yield a closed loop system that meets the constraints given by cj\nand ci. The designer will either have to accept this result or try again with a different\n\nP or different constraining values or both. Since stability is desired, one approach\n\nwould be to keep P and Cf and lower Cf until Kf becomes large enough to intersect kf.\n\nThe outline of the iterative Lyapunov design method is as follows.\n\n1) Choose P, Cjf, and Zf. This selection must obey Theorem 2. It should\n\nbe noted that Theorem 2 guarantees the nonemptiness of Kf and )Cf, but\nnot their intersections. If the system is time varying, then, in order to\nuse this algorithm, P must be found so that\n\n2B T(t)[A(t)P-1 + P-lAT(t)]B(t) < 0 V t (2.119)\n\nOtherwise this algorithm cannot guarantee stability.\n\n2) Compute co and Co so that K0 = 0 E &aC0 and )Cf C Co. K0 will be\n\nthe initial guess in the search for Kf G ACf.\n\n3) Let i = 0\n\n4) Let i = Z+ 1\n\n5) Find Ki so that KA E KC1I and KJi 0 M9Ai-.\n\n6) Compute ci and c, so that Ki E 9C)i and KCf C /C, or so that Ki E kCf.\n\n7) Repeat steps 4) 5) and 6) until one of two events occur.\n\n1. ci = c and i = f .\n\n2. (ci ci-1) and (i c,_i) become very small.\n\nRemarks:\nIf event 1 occurs, then kCf is nonempty and Ki E Cf. If event 2 occurs, then kf is\n\nempty and Ki yields a closed loop system that meets the constraints corresponding\n\nto ci and ,i.\nWe now explain how to perform steps 2), 5), and 6).\n\nStep 2)\n\nThe lower and upper bound of xTP* for the open loop system is respec-\ntively,\n\nAo = Amird (PA + ATP)] (2.120)\n\nAo = Am,[ 2(PA + ATp)] (2.121)\n\nThen,\n\no= min[ff,Ao] (2.122)\n\no = max[[cf ,Ao] (2.123)\n\nStep 5)\n\nThe algorithm is now given.\n\nK = K,\nFor j = 1 to m\nFor k = 1 to n\nKoj,k = K kJ,kQj,k\nPi = -(PBQj,k + Q7kBTP)\n\nPo = P(A BKOj,k) + (A BKoj,k)TP 2cl\nSolve det[kP1 + Po0] = 0 in terms of k for c = ci, and Zi.\nReject all values which do not meet the constraints from\nsteps 2) and 6). The result is two intervals whose\nlower bound is k_ and k2 and whose upper bound is\nk and k2.\nFind the intersection of these intervals by evaluating\n\nk max[k, k2]\n\nk = mn[Ti, -k2.\nFind the midpoint of the interval by computing\n\nkj,k = (k +).\nLet K = ki,kQj,k + Koj,k\nNext k\n\nNext j\nKi+1 = K\n\nStep 6)\n\nLet\n\nAiA = min[(PA + ATP- PBKi KTBTP)] (2.124)\n\nSAmaxI(PA + ATp PBKi KTBTp)] (2.125)\n2\n\nSo,\n\n-* = min[-Cf,A] (2.126)\n\n\"i =max[-Cf, ] (2.127)\n\nThe following example illustrates the feasibility in applying this method to the\n\nEMRAAT missile.\n\nExample 2\nWe would like to apply the iterative Lyapunov design method to the EM-\n\nRAAT missile. The missile was flown in a simulation through a trajectory\n\nusing another autopilot design. The model was linearized, and the fol-\n\nlowing system was taken at 4.00 seconds into the flight.\n\nx: = Ax + Bu\n\n39 -.1\n13 -.4\n36 -1\n.2 -6.\n06 8;\n\nB=\n\n612 .0086\n100 .1079\n557 -2.078 -\n097 .0173 -\nL3.29 -.0261 -\n\n0.0 -.1296\n-.0149 0.0\n-1150 -31.06\n-4.494 -121.7\n1.234 -.2802\n\n(2.128)\n\n.9997\n0.0\n.1763\n.6403\n.1597\n\n0\n.10-\n-12:\n-4.7-\n-108\n\n0.0\n-.9997\n.0442\n.1611\n-.5701\n\n1.0\n14\n22\n47\n.6\n\n(2.129)\n\n(2.130)\n\nx = [a,0,p,q,r]T\n\n(2.131)\n\nStep 1)\n\nLet P = I. The minimum and maximum uncontrollable normal velocity\n\ncomponents were found by computing the limiting values in equations\n\n(2.47) and (2.48). Since P = I, these terms simplify and are evaluated as\n\nfollows:\n\nminlxT[PA+ATp]x= A'[mr[BT(A + AT)B.]\nXS .7263\n= -.7263\n\n(2.132)\n\n(2.133)\n\nmax xT [PA + ATp]x\nxEs 2\n\n= A [axc)I(A + A)B]\n\n= -.3017\n\nwhere\n\n-.99;\n.16\n-64.:\n-252\n-.584\n\nand\n\nand\n\n(2.134)\n\n(2.135)\n\nTheorem 2 guarantees the nonemptyness of T and K if f > -.3017 and\n\ncf < -.7263. With this in mind, we let Zf = -.3 and cf = -6000. Since\nthe intersection of T and K is not guaranteed, cf was chosen to be very\n\nnegative to increase the probability of getting an answer.\n\nStep 2)\n\nA = A/,,[(A+ AT)] (2.136)\n2\n= -781.4 (2.137)\n\n(2.138)\n\nA = Ama,,[(A + AT)] (2.139)\n2\n= 778.9 (2.140)\n\n(2.141)\n\nSo\n\nco = min(cf,A) (2.142)\n\n= -6000 (2.143)\n\nCo = max(Zf,A) (2.144)\n\n= 778.9 (2.145)\n\nSteps 3) 7)\nA program was written for MATLAB to carry out the iterations in steps\n\n3) 7). The program would terminate if Kf was found or when\n\n6 := I[i _i-1 < 6 = .0001 (2.146)\n\nkf was found before 6 < .0001. The program ran 16 iterations. The final\n\nresult is\n\nKf =\n\n.00046\n2.057\n-.00029\n\n2.125\n-.0012\n-.7259\n\n-.7974\n2.339\n.0263\n\n.0805\n-6.577\n-.1346\n\n.7535\n.2591\n-.7985\n\n(2.147)\n\nNow to check the result.\n\n1\nAf = A.n[(A+AT BKf KjBT)]\n2f\n= -818.0\n\nAf = Amax[(A + AT BKf KfTBT)]\n\n= -.3001\n\nThis meets the desired constraint.\n\naf < Af < Af < Cf\n\n(2.148)\n\n(2.149)\n\n(2.150)\n\n(2.151)\n\nCHAPTER 3\nA TIME VARYING SECOND ORDER EXAMPLE\n\nThe following problem gives a case when pole placement succeeds in giving eigen-\nvalues with negative real parts, but fails to stabilize the system. The Lyapunov\ndesign method is then employed, and the resulting closed loop system is shown to be\nasymptotically stable for all time.\nWe would like to find a feedback control law that stabilizes the system\n\nx= A(t)x + B(t)u (3.1)\n\nwhere\n\nA (t) -1 + 1.Scos(t)[cos(t) + cos(t + Ir/18)] 1 1.5sin(t)[cos(t) + cos(t + r/18)] 1\nA -1 1.5cos(t)[sin(t) + sin(t + 7r/18)] -1 + 1.5sin(t)[sin(t) + sin(t + 7r/18)]\n\n(3.2)\nand\nB(t)= [ cos(t + r/18) (33)\nsin(t + 7r/18) (\nThe eigenvalues of A(t) are -.4889 and 1.4661 for all time. The following control law\nis proposed.\n\nu = [ 1.5cos(t) -1.5sin(t) ]x (3.4)\n\nThe resulting closed loop system can be found in example 5.3,109 by Vidyasagar\n and also in Khalil . The eigenvalues for the resulting closed loop system are\nA = -.25 j.6614. Since the eigenvalues have negative real parts, one would expect\nthe closed loop system to be stable. However, Vidyasagar shows that the transition\nmatrix is\n(t,, 0) = e5tcos) esin(t) (3.5)\n1 _e.51sin^) e t cos(t)I\n\nTrajectory of a pole placement time varying system\n\n2501-\n\n200\n\n150\n\nV 100\n\n-50 -\n\n-n00 -100 0 100 200 300 400 500\nxl\n\nFigure 3.1. The trajectory of the above closed loop system based\nThe initial conditions are x0 = [1 O]T. The system is unstable.\n\non pole placement.\n\nIf the initial conditions of the system are xo = [1 0]T, the resulting trajectory is\n\nunstable as figure 3.1 shows.\n\nWe now turn to the Lyapunov design method. We choose P to be the identity\n\nmatrix. Before giving the design constraints, we need to check the the value of the\n\nuncontrollable normal velocity components for all time. At t = 0\n\nB = [.9848 .1736]T\n\n(3.6)\n\nSince P is the identity matrix, we are interested in the normal velocity component\n\nwhich is on the part of the unit circle whose tangent is parallel to B. So we let\n\nx = B = [.1736 .9848]T\n\nThe uncontrollable normal velocity component at t = 0 is\n\nxTp = =xT[PA(O) + AT(0)P]x = -.9548\nxT 2x\n\n(3.7)\n\n(3.8)\n\nThe uncontrollable velocity for this example is constant for all time. In selecting the\n\nconstraining values, we must have\n\nc < -.9548 < Z (3.9)\n\nc=-2 (3.10)\n\n= -.5 (3.11)\n\nSince the system is second order and has only one input it is possible to plot the set\nof all feedback gains which satisfy the following.\n\nC < lxT[P(A(t)-B(t)K(t))+(A(t)- B(t)K(t))TP]x < V x:xTx= 1, Vt (3.12)\n\nThis time varying feedback set is shown in figure 3.2. Since the time varying nature\nof the system is periodic, and from inspection of figure 3.2, the following control law\nis chosen.\n\nu = 3[cos(t) sin(t)]x (3.13)\n\nThe feedback matrix in (3.13) is shown to be inside the moving feedback set in figure\n\n3.2. The eigenvalues of the derivative of the resulting Lyapunov function is\n\nA[A(t) B(t)K(t) + AT(t) KT(t)BT(t)] = -1.1193, -0.8579 Vt (3.14)\n\nTherefore, the closed loop system is stable. Figure 3.3 shows a trajectory of the\nLyapunov based closed loop system where the initial condition is xO = [1 0]T.\n\n0\n\n4 -2 2 4 -\n\n____________ I) I p\n\n4-\n62 ---I -__________\n\n6 -2 a 2 4\nk11\n\n-2\n\n6 -4 -2 2 2 6\n611\n6 ____ d) t -3pir ___\n\n4\n\n2\n\n0-\n\n-6\n-6 -O2 2 4\n-8B 5 0 S ------ 4 ------\n\n-26 7 2 4\nk11 t.7p4\n\nFigure 3.2. The time varying feedback set which satisfies the design constraints.\n\n-0.05\n\n-0.1\n\n-0.15\n\n-0.2\n\n-0.25\n\nTrajectory of the Lyapunov based closed loop time varying system\n\n: r /\n\n-0.2 0 0.2 0.4 0.6 0.8 1\nxl\n\nFigure 3.3. A trajectory of the closed loop system using the Lyapunov design method.\nxo = [1 O]J. The system is stable.\n\nCHAPTER 4\nDERIVATION OF A MODEL OF THE EMRAAT MISSILE\n\nThe next section derives a nonlinear model of the EMRAAT missile. A linearized\n\nmodel is then generated in the following section and will be used for the design of the\n\nautopilot. Aerodynamic and inertial cross coupling are assumed negligible in order\n\nto reduce the order of the model. A linearized pitch model and a linearized roll-yaw\n\nmodel results. All assumptions are clearly stated.\n\n4.1 The Nonlinear Model\n\nBefore deriving the nonlinear model some variables and terms are defined. Figure\n\n4.1 shows the missile body coordinate frame of the EMRAAT missile . The three\n\naxes, x, y, and z, are fixed to the missile as shown. The velocity of the missile is\n\nrepresented by V. Angle of attack, a, is defined as the angle between x and the\n\nprojection of V onto the x-z plane. Sideslip, j, is the angle between x and the\n\nprojection of V onto the x-y plane. The velocity components, u, v, and w, are the\n\nprojections of V onto the x, y, and z axis respectively. The angle rates, p, q, and r,\n\nare named roll rate, pitch rate, and yaw rate respectively and are defined as the rate\n\nof rotation around the x, y, and z axis. Their directions obey the right hand rule.\n\nThese definitions are similar to those applied to aircraft as given by Etkin .\n\nThe following derivation of the nonlinear model is based on a similar model found\nin Smith and in Koenig . Assumptions are made here to separate the system\n\ninto two lower order models: the pitch model and the roll-yaw model. We begin the\n\nderivation by starting with the force equations. The forces acting on the missile are\n\nthrust, gravity, and aerodynamic forces. The autopilot design in this paper applies\n\nto the second phase of the flight when the engine is no longer burning so thrust is\n\np\n\nr^ \"- .- - ...q.\n\nz\n\nFigure 4.1. The missile body coordinate frame of the EMRAAT missile .\n\nzero. Also, the weight of the missile is small in comparison to the aerodynamic force\n\nso gravity is neglected. The aerodynamic force in the x direction is much smaller\n\nthan the aerodynamic force in the y and z directions, and therefore, will be ignored\n\nfor the rest of the paper. Newton's second law of motion implies the following:\n\nFy = m[i + ru pw] (4.1)\n\nFz = m[b + pv qu] (4.2)\n\nFy and Fz are the aerodynamic forces in the y and z directions. The quantities\n\ninside the brackets are the total accelerations in the y and z directions. We know\n\nthat\n\nV = [u2 + v2 + w2]1 (4.3)\n\nSince v and w are much smaller than u then\n\nV u (4.4)\n\nAngle of attack and sideslip are given by\n\na = arctan (-) (4.5)\n\n=3 arctan (-) (4.6)\n\nIf we assume that a and are small, then\n\nw\na w- (4.7)\nu\nU\n\nand\nV\n3-- (4.8)\nU\nEquations (4.1) and (4.2) can be rewritten as\n\nFy = mu + r p (4.9)\n\nFz = mu [+ p q]\n\nwhich simplifies to\n\nFy = mV [- +r pal\n\nFz =mV [ + po3- q]\n\nWe assume that the forward velocity changes slowly so that\n\n(4.13)\n\nThen\n\nw\nU\n\nV\n\nand\n\nSo (4.11) and (4.12) become\nFy\nmV\nFz\nmV\nThe aerodynamic forces are given by\n\n=O+r-pa (4.16)\n\n-& +p p -q (4.17)\n\nFy = QS [Cyfi + Cyp + Cyjr + Cy6,bp + Cyl\\6r]\n\nFz = -QS [CNQ + CNQ& + CNq + CNd6]q I\n\n(4.18)\n\n(4.19)\n\nQ is the dynamic pressure and is defined as\n\nQ= Ppv2\n\n(4.20)\n\nwhere p is the air density. S is the surface area of the wing. The aerodynamic\n\ncoefficients come from wind tunnel tests. They depend on mach number, and some\n\ndepend on angle of attack. The values of these coefficients have been put in tabular\n\n(4.10)\n\n(4.11)\n\n(4.12)\n\n(4.14)\n\n(4.15)\n\nform and are given in the first appendix along with other data pertaining to the\nEMRAAT missile.\nSubstituting (4.18) and (4.19) into (4.16) and (4.17) and solving for & and /3 gives\nQS\np= a r + -t [Cy + Cyp + Cyr + Cy,6/ + Cy,lr] (4.21)\n\nand\nS+ QSCN. q = QS [CN. + CNqq + CN,,6q] (4.22)\n\n= [ + QSC [q p (CN.aa + CNq + CN,) (4.23)\n\nEquations (4.21) and (4.23) are two nonlinear state equations. In order to separate\npitch dynamics from roll-yaw dynamics, /3 is assumed close to zero in (4.23). Thus,\n\n[I [1 i QSCN]1 \\ QSC \\ + (I-' CAq) q +--2S N6q]\n\n(4.24)\nEquation (4.21) can be written as\nS= ( y -\" \"-p(-1 -Cr) r-(--Cy6p) p-( P+ CY6,) 4,\n\\mV p/ V \\ mV ) \\mV p \\mV\n(4.25)\nWe now turn to the moment equations of the missile. Since thrust is zero and since\ngravity does not contribute any moment to the missile, then the moments around\nthe x, y, and z axis are given by 1, m, and n respectively, the moments due to\naerodynamic pressure. They are given by\n\nI = QSd [Ci3 + Cip + Cr + C,6p + Clfr\\] (4.26)\n\nm = QSd [Cma + Cm,& + Cmqq + C,6q 6,q] (4.27)\nand\n\nS= QSd [Co, + Cn,p + C1r + Cn,,p6 + Cn6r6r\n\n(4.28)\n\nwhere d is the missile diameter. Again, the aerodynamic coefficients come from\nwind tunnel testing and are tabulated in the first appendix. Euler's three moment\nequations can be written as follows.\n\nm =J c + H(p,q,r) (4.29)\nn r\nwhere\nIxx -IXY -Ixz\nJ= -IXY IYY -IYz (4.30)\n-Ixz -IYz Izz\nand\n-(Iyy Izz)qr + Iyz(r2 q2) Ixypq + Ixyrp\nH = -(Izz- Ixx)rp + Ixz(p2 r2) Ixyqr + Iyzpq (4.31)\n-(Ixx Iyy)pq + Ixy(q2 p2) Iyzrp + Ixzqr\nWe will assume that the inertial cross products, Ixy, Ixz, and Iyz are small. Also\ndue to symmetry of the airframe, Iyy and Izz are assumed to be equal. Inertial data\nfor the EMRAAT missile is given in the second appendix. Solving (4.29) for p, 4,\nand r gives\nS= --I (4.32)\n'xx\n= -- [m + (Izz Ixx)rp] (4.33)\nIYY\nand\nS= -- [n + (Ixx Iyy) pq] (4.34)\n'zz\nSubstituting (4.26), (4.27), and (4.28) into (4.32), (4.33), and (4.34) gives\n\nQSd [ + Cp + Cr + C1 6+ C,] (4.35)\n\niQSd Izzxx\nq QSd [Cmoa + Ca& + Cmq + Cm6qSq] + Izz Ixxrp (4.36)\nIYY i 1* YY\nand\n____ '~ xx Iyy\nS= [C/3 + Cp + Cnr + C 6,6p + C, 5] + 1- pq (4.37)\nIzz Izz\n\nIf gyroscope effects are considered small, then (4.24), (4.25), (4.35), (4.36), and (4.37)\ncan be written as two separate systems: the pitch dynamics, and the roll-yaw dy-\nnamics. The pitch dynamics are as follows.\n\n&=[ + QS\\ QNa l[(-2-. Qc e + (I ( -'CN,) q+ (+Q CN6) 6q]\nL rV J \\mV } \\ mV *} \\mV *\n(4.38)\nQSd [C.a + Cm& + Cqq + C,,,q (4.39)\n\nThe roll-yaw dynamic equations are\n\nQS (c + 2 Cy ))p (I + Q- (,) 2Cr+(Q2S C, ('S ) c r)\n\\mV */ \\ mV *) \\ mV mV p) \\mV\n(4.40)\nsd[C,3 + C1pp + C r + C6pSP + C \\r] (4.41)\nIx-x\n\n= Q-d- [C,3 + Cnp + Cn1r + C,6p6P + Cn,65,] (4.42)\n\nThe states of the pitch model are a and q with 6q as its input. For the roll-yaw\nmodel, the states are /3, p, and r, and the inputs are 6p and 64.\n\n4.2 The Linear Model\nThe previous section gave a model of the pitch dynamics and the roll-yaw dynam-\nics in the following form.\n[ = fq(a, q, q) (4.43)\n\n[/ ]T fpT(),p,r,6p,6r) (4.44)\n\nWe would like to have two linearized models for use with the proposed design method.\nThe result will be two systems in the following form.\n\nx = Ax + Bu (4.45)\n\nwhere x contains the states and u contains the inputs. A and B are matrices which\nare functions of several time varying flight parameters and are computed as follows.\n\nA = df (4.46)\ndx (X,W)nominal\n\ndB f (4.47)\nB=du\nd(X,W ).ominal\nwhere w contains additional flight parameters. Note that x and u are now pertur-\nbations from the point around which the linearization is taken.\nWe linearize the pitch model first. From inspection of (4.38) we see that\n\n=1 ( QSCNQ 1 (-QS CN\naqll = [1 +-'---- mI --- a\nM mV M mV\naq12 +( QSCN (1- QS CN,)\n\nbq= (I + QSCN) (- CN q)\nM n-V M \\mV q\nTo linearize (4.39), we must first substitute (4.38) in for &. Then we differentiate as\nin (4.46) and (4.47).\nQSd C + & QSCN (-QS \\\naq22 = 7y [cm + Cm ( + ~mV~) ()]\nh[U [( QSCNj1 (--QSCN ]\n\nbq21 = QSd Cm6q + Cm (11 + QSCN) (QS0)]\nIvy Irn V M\nThe resulting linearized pitch model is\nF 1 a=1, aq12 a I + bgil I 86 (4.48)\naq2l aq22 q bqg21\nThe same procedure is applied to the roll-yaw model. Assuming that a is constant\nin (4.40), and from inspection of equations (4.40) to (4.42) the following results.\nQQSCy QSCy, QSCyr\naprIl MV apr12 = a + ,-V apr13 = -1 + m\"V\n\n59\n\nbpr1i QSCYP bpr12 QSCy6,\nmV 'mV '\nQSdCi QSdCi, QSdCh,\napZ1 ~ xI apr22 ~ X ~,apr23 -\nQSdCj QSdCj ,\nbpr21 XX Q pr22 Ixx\nQSdC,, QSdCn, QSdCnr\napr31 = Z apr32 IZZ -I pr3 ~ IZZ '\n\n_Q~rdC'6^ QSdCn6T\nbpr3l QSdC,6 bpr32 QSdC\n\nThe linearized model is given by\n\napri apr12 apr13 + bpr11 bpr2 1 (p\np = apr21 apr22 apr23 p + bp21 bpr22 (449\n1 apr31 apr32 apr33 i \"r bpr31 bpr32\n\nCHAPTER 5\nTHE DEPENDENCE OF GAINS ON FLIGHT PARAMETERS\n\nThis chapter describes the procedure that was used to determine which flight\nparameters would be scheduled against the gains. Figure 5.1 gives a block diagram\nof the system used for this procedure. The feedback gains are computed using the\n\niterative Lyapunov design method described in Chapter 2. The gains depend on\nthe linear model which in turn depends on seven flight parameters. Six of these\nparameters are held constant while the seventh one changes. The resulting gains are\nchecked to see if they depend on this changing parameter. The process is repeated\nfor each flight parameter. It is found that the gains depend on angle of attack, mach\nnumber, and dynamic pressure. This information will be used in the gain scheduling\n\nprocess. The next section will discuss the atmospheric tables used to generate p and\nV from M and Q. The flight parameter generator will then be presented. The third\nsection describes the initializer. The details of the iterative Lyapunov design method\nare given in the fourth section. Finally, the results of the comparison between the\ngains and flight parameters will be given in the fifth section.\n\n5.1 Generating p and V from M and Q\n\nBy inspection, the linear models from Chapter 4 clearly depend on a, 0, p, q, r.\np, V, and the aerodynamic coefficients. The coefficients, however, can be eliminated\nfrom the list because they depend on mach number and angle of attack. It is desirable\n\nto replace p and V with M and Q since the later two can be easily measured on the\nmissile. We know the following.\n\n2\nQ = ^P2(5.1)\n\nM = (5.2)\nVs0\n60\n\nDependent\nFlight\nParameters\n\nFigure 5.1. A block diagram of the system used to determine the dependence of\ngains on flight parameters.\n\nwhere Vsos is the speed of sound. Both p and Vos are functions of altitude.\n\np = fp(h)\n\nV8s0 = f(h)\n\nwhere h is altitude in feet above sea level. Here,\natmospheric tables and are implemented by linear\nand substituting the result into (5.1) gives\n\nQ =\n\nWe generate a third table in the following way.\n\nfp and fs are functions based on\ninterpolation. Solving (5.2) for V\n\n(5.5)\n\nf3(h) := f(h)f(h) = pV\n\n= 2\n\n(5.6)\n\n(5.7)\n\nThe function, f3, is a one-to-one function so that its inverse can be found by reading\nthe table backwards. With this in mind we can solve (5.7) for h.\n\nh f (2Q)\nh= f31M\n\n(5.8)\n\nFrom (5.2) and (5.4)\n\nV = MV,,, = Mf,(h)\n\n(5.9)\n\nSubstituting (5.8) into (5.9) to eliminate h gives\n\nS= Mf, (f- (2Q))\n\nf (2Q))\n-fP ( M2^\n\n(5.10)\n\nAlso from (5.3)\n\n(5.11)\n\n(5.3)\n\n(5.4)\n\nEquations (5.10) and (5.11) are used to eliminate p and V in the linear model. As\na result, the linearized model can now be generated from the following seven flight\n\nparameters.\n\n[M Q ca 3 p q r] (5.12)\n\n5.2 The Flight Parameter Generator\n\nA series of flight conditions are made and used to generate many linear models.\nFeedback gains are generated for each condition. The first of the series is called\nthe nominal flight condition. The values of the parameters for the nominal flight\n\ncondition are\n\nM = 2, Q = 1250 psf, c = 8,/3 = 0,\n\np = 00/s,q = 100/s,r = 00/s.\n\nNext, one of the parameters is allowed to vary while the other six are held constant.\n\nThis process is repeated six times so that each parameter can be tested. Table 5.1\n\nshows the starting point, and the minimum and maximum values of each changing\n\nparameter. Parameters with nonzero starting points begin at the starting point,\n\nincrement to the maximum value, return to the starting point, and then decrement\n\nto the minimum value. Due to symmetry, the remaining variables have starting points\n\nat zero and increment to their maximum values. M and Q sweep through a narrow\n\nrange because of restrictions of the atmospheric tables.\n\n5.3 Initializing the Iterative Lyapunov Design Method\n\nThe iterative Lyapunov design method requires an initial guess, Kqo and Kpro and\ntwo positive definite matrices, Pq and Pp,,. The initializer supplies these values and\n\nwill be discussed in this section.\n\n64\n\nTable 5.1. The initial point and range of changing flight parameters.\nInitial point Minimum value Maximum value\nM 2 1 2.6\nQ 1250 psf 700 psf 5000 psf\na 8 _-80 200\n/3 00 00 100\np 00/s 00/s 500/S\nq 10/s -101/s 200/s\nr 00/s 0/s 200/s\n\nThe initial feedback gains are found by using a pole placement algorithm. At the\nnominal flight condition, the linear models are given by\n\nA [ -1.1345 0.9996 [Bq -0.14631 (5.13)\nS -261.4732 0.6209 ,Bq -123.1091 '\n\nApr -25. [ 24 1 .1350.9996 -017.5143 5. (5.13)\n-0.459 0.140 -.100 0.018 0.117\nAp, = -2255.5 -2.41 .066 ,Bp = -1173.5 -1335.6 (5.14)\n73.0 -0.181 -.648 L 2.01 -114.4\nThe desired eigenvalues for the closed loop pitch dynamics have been chosen to be\n-40jlO0. For the closed loop roll-yaw model the desired pole locations are -20j5,\n-80. The resulting feedback gains are\n\nS\\ 11.1633 -.6233 1 = -1.4166 -.0758 .3758 (5.15)\nL 2.9337 .0086 -.3300 -\n\nFor both closed loop systems, P must be found so that xTpx is a Lyapunov\nfunction. The following problem is stated.\n\nGiven a stable linear system : = Ax, find a positive definite function,\nV = xTpx so that V1 = xT(PA + ATP)x is a negative definite function.\n\nLet A be put into Jordan canonical form.\n\nA = SJS-1 (5.16)\n\nwhere S is an invertible matrix. The diagonals of J are the real parts of the eigen-\n\nvalues of A, and imaginary parts of the eigenvalues lie in skew symmetric locations\n\noff of the diagonals. For example, if the eigenvalues of A are a jb, c, then\n\n'a -b 0\nJ= b a 0 (5.17)\n0 0 c\nLet\n\nJ = Jdiag + Jskew (5.18)\n\nwhere Jdiag is the symmetric part of J and Jakew is the skew symmetric part of J. In\n\nthe above example\n\na 0 0 0 -b 0\nJdiag 0 a 0 Jskew = b 0 0 (5.19)\n0 0 c 0 0\nMaking the following transformation on the system k = Ax, let\n\nx = Sz (5.20)\n\nThen\n\nSi = ASz (5.21)\n\nand\n\nz = S-'ASz = Jz. (5.22)\n\nLet Pz = -Jdiag. We wish to check the velocities of the system z = Jz on the ellipsoid\n\nzTPzz = -zTJd\"gz = 1 (5.23)\n\nThe normal to the ellipsoid at z is -Jdiagz. The projection of z = Jz onto the normal\n\nis JdiagZ. Here the normal velocity component has the same magnitude but opposite\n\ndirection to the normal of the ellipsoid. If the system has no complex eigenvalues\n\nthen the velocities are orthogonal to the ellipsoid everywhere. For this reason, the\n\nchoice of P- = -Jdiag is the best choice for a positive definite function for the system\n\nS= Jz.\n\nV(z) = -zT JdiagZ\n\n(5.24)\n\nMaking the following transformation into the x coordinate system gives\n\nz = S-ix\n\n(5.25)\n\nwhich implies\n\nV(x) = -xT[S-1]TJdiagS-X\n\nOur choice of P is\n\nP = -S-1]TJdiagS-.\n\nFor the nominal flight condition, the Jordan canonical form of Ag -\nApr- BprKpr is found and from (5.27)\n\nS261772 60 1 ] 6792 10.7\nPq 609 14.9 Pp= 10.7 4.02\nP 60 14[9 -316.4 -.509\n\nThe eigenvalues of (PqAq + ATpq PqBqKq KTBTpq) at\n\n-316.4\n-.509 (5.28)\n15.7 j\n\nthe nominal point are\n\n-1.0715 x 106, -40.002\n\n(5.29)\n\nLikewise, for the closed loop roll-yaw system the eigenvalues are\n\n-1.3614 x 105, -320.00,-20.003\n\n(5.30)\n\n5.4 The Iterative LvaDunov Design Method\n\nThe iterative Lyapunov design method generates feedback gains so that xT PqX\nand xTPp'x are Lyapunov functions for each closed loop system. The algorithm\nrequires the initial guesses Kgo and KprO, for the first point, and the positive definite\nmatrices Pq and Ppr. As a given flight condition changes, the feedback gains from\n\n(5.26)\n\n(5.27)\n\nBqKq and\n\nthe previous point become the initial guess for the present point. The result of\n\nthis algorithm is a series of gains; one set for each flight condition generated. The\n\nnext portion of this section discusses some modifications made to the algorithm from\n\nsection 2.4. The material which follows describes how the design constraints are\n\nselected.\n\nThe design method of section 2.4 finds a K so that the eigenvalues of (PA+ATp)\n\nfall between cf and Cf where A is the closed loop system. The design algorithm\n\nused in Figure 5.1 has been modified so that the resulting K satisfies the following\n\nrequirements.\n\ncl < Ai ([P(A-BK) + (A-BK)TP]] < Z,\nc2 < A2 [P(A- BK) + (A -BK)TP]] < 2 (5.31)\n... ... ... ... ,...\n\nc, < A,[I[P(A BK)+ (A BK)TP]] < n\nwhere A1 is the smallest eigenvalue, A2 is the second smallest eigenvalue, and so on.\n\nThe values of the c's are supplied by the designer. This modification has been made\n\nwith the belief if more design constraints are made, then the performance will vary\n\nless with changes in the flight conditions. The modified algorithm is as follows.\n\n1) Choose P, e1f, cif, ... c,, and Z,,f. The selection of clf and cnf must\n\nobey Theorem 2. However, Theorem 2 does not guarantee the nonempti-\n\nness of kC.\n\n2) Compute cl0Cio, ... ,cn,,no so that the initial guess, K0 E Co0 and\n\nJCf C /Co.\n\n3) Let i = 0\n\n4) Let i = i + 1\n\n5) Compute Ki so that Ki E AC-1 and K, 9iC,-I.\n\n6) Compute cli. ci, ... cni, Cni so that Ki E 9C, and KACf C Ci.\n\n68\n\n7) Repeat steps 4) 5) and 6) until one of two events occur.\n\n1. cli= =Cl, Zli = Cll, ... cini = Cnf and Zni = Enf.\n2. (Cli cli-1) *.. (ni- -,i-1) become very small.\n\nEvent 1. implies that the final answer has been found. Event 2. implies that /f\nis empty and Ki satisfies the constraints corresponding to cli, ,i, ... cn, Z,. Steps\n\n2), 5), and 6) are accomplished in the following way.\n\nSteps 2) and 6)\n\nLet\ne = A1 [!(PA + ATp PBKi KTBTP)]\n\ne = An [(PA + ATP- PBKi KTBTP) (5.32)\nThen,\nCli = min[clf, el]\nEli = max[-cl, el]\n... ...... (5.33)\nfc = mzn[f _, e-]\n-,i = max[nf, en]\n\nStep 5)\n\nLet K = Ki\nFor j = 1 to m\nFor k = 1 to n\n\nP1 = -(PBQj,k + QBTP)\nPO = P(A BKO,k) + (A BKo,k)TP 2cI\n\nKoj,k = K kj,kQj,k\n\nSolve det[kPi + P0] = 0 for in terms of k for c = cli, ... Ci.\n\nReject all values of k which do not meet the constraints from\nsteps 2) or 6). The result is 2n intervals whose lower bounds are\n\ndesignated by k1, ... k2,n and whose upper bounds is named k, ..., k2n.\n\nFind the intersection of these intervals by evaluating\n\nk = max[k1, ...,1k2,]\n\nk = mn [kl,...,k2n]\nFind the midpoint of the interval by computing\n\nkj,k = (k+k).\nLet K = kj,kQj,k + Koj,k\n\nNext k\n\nNext j\n\nKA+i = K\n\nRemarks:\n\nLet the feedback sets ),C1, k 2i &2, 2... ,, and C n be defined respectively as the\n\nset of all K so the constraints (5.31) are met. Theorem 2 provides conditions for the\n\nnonemptiness of K, and kAn. But conditions for the nonemptiness of the remaining\n\nsets are still unknown. Also, )CI, ... ,_ are not convex in general. These are the\n\nlimitations of using the modified design algorithm.\n\nWe now turn to selecting constraining values for the eigenvalues of [P(A BK)+\n\n(A BK)TP]. It is necessary to evaluate the uncontrollable normal velocity compo-\n\nnents for each flight condition that will be generated in Figure 5.1. Tables 5.2 and 5.3\n\nshow for both models the minimum and maximum uncontrollable normal velocities\n\nfor each changing flight parameter.\n\nTable 5.2. Uncontrollable normal velocity components for the pitch model\nchanging parameter minimum uncontrollable velocity maximum uncontrollable velocity\nM -43.1456 -41.7253\nQ -45.0144 -42.2289....\n........ -43.3856 -42.2362....\n__________ -42.6047 -42.6047\np -42.6047 -42.6045\nq ..... -42.6047 -42.6047\nr -42.6047 -42.6044\n\nTable 5.3. Uncontrollable normal velocity components for the roll-yaw model\nchanging parameter minimumuncontrollable velocity maximum uncontrollable velocity\nM -21.3937 -20.9751\nQ -22.3016 -21.1574\n_a_______ -21.7136 -21.1140\n________ -21.3180 -21.3180\np -21.3181 -21.3180\nq -21.3184 -21.3180\nr -21.3180 \"-21.3180..\n\nFor the pitch model, the uncontrollable normal velocity components range from\n\n-45.0144 to -41.7253. Theorem 2 requires that\n\n-41.7253 < q2\n!ql < -45.0144\n\n(5.34)\n\nIn addition, from (5.29), we want\n\ncq1 <\n_q2 <\n\n-1.0715 x 106\n-40.002\n\n< Cql\n< Cq2\n\n(5.35)\n\nFrom this the constraining values for the pitch model have been chosen to be\n\n!21 = -1.2 X 106, qC- = -1 x 106,\nSq2 = -45, zq2 = -35\n\n(5.36)\n\nLikewise for the roll-yaw model, when looking at Table 5.3, Theorem 2 requires that\n\n-20.9751\ngprl\n\nCpr3\n-22.3016\n\n(5.37)\n\nThe eigenvalues in (5.30) suggest the following.\n\nCp l < -1.3614 x 105 < cprl\ncpr2 < -320.00 < Cpr2 (5.38)\nCp,3 < -20.003 < Cp,3\nThe following constraining values have been chosen for the roll-yaw model.\n\ncpi = -150000, Cprl = -110000\nCp,2 = -340, Zp, = -300 (5.39)\nCpr3 = -22, Cpr3 = -17\n5.5 Formulation of a State Tracker\n\nThe autopilot of the EMRAAT missile will be a state tracker. That is, we want to\nbe able to change the location of the equilibrium point in order to control the values\n\nof some of the states. The following shows how this will be accomplished.\n\nGiven the linear system\n\nx= Ax + Bu (5.40)\n\ny =Cx, (5.41)\n\nwe would like to find a control law\n\nu = -Kx + KrefV (5.42)\n\nso that y, the output, tracks v ,the reference input, asymptotically. We require y = v\n\nwhen 5 = 0. When 5 = 0, then\n\n0 = Ax BKx + BKfv. (5.43)\n\nSince K is chosen so that the system is stable, then A BK is invertible and\n\nx = -(A BK)-1BKrefV (5.44)\n\nAlso,\n\nv = y = Cx = -C(A BK)-BKref V\n\n(5.45)\n\nBecause (5.45) is true for any v, then\n\nI = -C(A BK)-BKref, (5.46)\n\nControllability of the system implies that C(A-BK)-'B is invertible. Solving (5.46)\nfor Kref gives\n\nKref = -[C(A BK)-'B]-1 (5.47)\n\nThe EMRAAT missile has three inputs and therefore only three states can be\ntracked. Controlling a in the pitch model and /3 and p in the roll yaw model is\ndesirable. For the pitch model y = a, implying that\n\nCq =[1 0] (5.48)\n\nand, therefore,\n\nKef, = -[Cq(Aq BqKq)-'Bq]- (5.49)\n\nFor the roll-yaw model\n\nPi 0 o1 J (5.50)\nY = p 0 1 0\nand, thus,\n\nKrefr = -[Cpr(Apr BpKp1Bpr]'1 (5.51)\n\nKrfq and Krefpr are computed for each flight condition and then compared along\nwith Kq and Kpr to the flight conditions.\n\n5.6 Comparing Gains with Flight Parameters\n\nA series of gains have been generated as a function of different flight conditions.\nEach flight parameter has been swept through a range of points while the remaining\nsix have been held constant. In order to compare different gains on the same input\nit has been decided to use the products of gains and their corresponding terms at\ntypical values. For example a typical value of a is 8. So we set ao equal to 8\n\nTable 5.4. Extreme values and range of the pitch channel control terms as individual\nflight parameters vary. Gains depend mostly on M, Q, and a._ __\n_____M_. M] Q ... p q r\ncontrol min min min min min min min\nterm max max max max max max max\ndiff diff diff diff diff diff diff\nkqlIao -2.15 -3.1 -2.1 -1.60 -1.61 -1.61 -1.60\n-.72 -.16 -1.4 -1.59 -1.59 -1.59 -1.59\n1.43 2.95 .69 .017 .017 .017 .017\nkqi2qo -.1509 -.21 -.16 -.118 -.118 -.118 -.118\n-.0529 -.031 -.10 -.117 -.117 -.117 -.117\n.099 .179 .054 .0005 .0005 .0005 .0005\nkref qllcO -2.53 -3.5 -2.7 -1.98 -1.98 -1.98 -1.98\n-.916 -.53 -1.7 -1.97 -1.96 -1.96 -1.96\nS 1.61 3.0 .94 .017 .017 .017 .017\n\nand look at kqllao. We set qo and aco equal to 10/s and 8 respectively so that\n\nwe can look at kql2qo and kref qllao. The sum of these three terms are fed into to\n\nthe elevator. For the terms which are fed into the remaining inputs, the following\n\n/o = = 2\npo = p = 100/s (5.52)\nro = 25/s\nFigures 5.2a-g show kqi ao plotted against all seven flight parameters. These seven\nfigures show that kqnao changes with M, Q, and a but remains nearly constant when\n\n/3, p, q, and r change. Similar figures exist for the remaining eleven gains and are\n\nsummarized in Tables 5.4 and 5.5. The minimum and maximum values of each term\n\nis listed for each changing flight parameter along with the difference between the\n\nminimum and maximum values. Angles are expressed in radians. From this table it\nwas determined that all gains will be scheduled against M, Q, and a.\n\n-1.59\n\n-I,\n\n.1.61 o v ) ) --------\n0 20 40 6o a6 106 120 140 160 10 206\n\nFigure 5.2. kqliao vs. a)M ; b)Q; c)a; d)3; e) p; f) q; g) r\n\nTable 5.5. Extreme values and range of the roll-yaw channel control terms as indi-\nvidual flight parameters vary. Gains depend mostly on M, Q, and a.\nSM Q 1 p q -r\ncontrol min min min min minm imm mmin\nterm max max max max max max max\ndiff diff diff diff diff diff diff\nkpli/3o -.050 -.11 -.22 -.050 -.050 .050 .050\n.022 .045 -.013 -.038 -.036 -.036 -.036\n.072 .157 .21 .011 .014 .014 .013\nkpTrl2po -.17 -.14 -.29 -.13 -.13 -.13 -.13\n-.011 -.061 -.13 -.13 -.13 -.13 -.13\n.16 .20 .15 .001 .001 .001 .001\nkpr13ro .074 .047 .15 .16 .16 .16 .16\n.18 .28 .25 .16 .17 .17 .17\n.10 .23 .099 .0006 .001 .001 .0009\nkpr2ito .026 .0062 .065 .093 .092 .092 .092\n.11 .167 .15 .10 .101 .101 .101\n.080 .16 .084 .008 .009 .009 .009\nkpr22po -.054 -.18 .0015 .014 .014 .014 .014\n.015 .015 .152 .014 .015 .015 .015\n.068 .20 .15 .0007 .0008 .0005 .0007\nkpr23ro -.151 -.25 -.24 -.15 -.15 -.15 -.15\n-.067 -.034 -.11 -.14 -.14 -.14 .14\n.084 .22 .12 .001 .002 .001 .002\nkref pr11ii/3co -.14 -.21 -.16 -.14 -.14 -.14 -.15\n-.064 -.050 -.13 -.13 -.13 -.13 -.13\n.08 .16 .028 .011 .014 .013 .013\nkref pr12PcO -.071 -.046 -.39 -.040 -.040 -.045 -.040\n.029 .15 .056 -.039 -.040 .013 -.038\n.10 .19 .45 .001 .001 .057 .002\nkref pr21l3co .055 .030 .11 .12 .12 .12 .12\n.126 .19 .17 .13 .13 .13 .13\n.071 .16 .058 .008 .009 .010 .009\nkref pr22PcO -.010 -.33 -.17 -.071 -.071 -.12 -.071\n-.070 -.070 .24 -.070 -.070 -.066 -.070\n.027 .26 .41 .001 .001 .051 .002\n\nCHAPTER 6\nCOMPUTING LOOK-UP TABLES\n\nIn the last chapter we showed that the gains depend mostly on a, M, and Q.\nThis chapter describes the process of generating a look-up table of gains verses M,\n\nQ, and a. First a grid of points is formulated. Design constraints are then formulated\n\nbased on the knowledge of uncontrollable velocity components of the linear models.\n\nFinally, the gains are computed.\n\n6.1 Determining a Grid of Points\n\nA two dimensional grid of points for M and Q has been made and used for each\nentry of a in the table for both the pitch channel and the roll-yaw channel. Q sweeps\nthrough a wide range of values starting with 100 psf and ending at 15,000 psf. The\n\nvalues of M were chosen so that each entry of M and Q lie in the atmospheric tables\n\nused to compute p and V. As a result, the grid points are not rectangular. All\n\nentries are restricted to values between M = .6 and M = 3.5 and must correspond\n\nto altitudes between sea level and 50,000 ft.\n\nTable 6.1 shows the values of a used in the look-up table for both channels. M\nand Q sweep through all values of the grid previously mentioned for each value of a\nin Table 6.1. The spacing between the grid points was determined in a trial and error\n\nprocess. During the iterative procedure for computing feedback gains, the initial\nguess for each point came from the result of an adjacent grid point. The closer the\n\nspacing between adjacent points, the fewer iterations were needed to find the next\n\nfeedback gain. Numerical problems as described in section 2.4 and shown in figure\n2.10 were encountered. When this happened some of the constraints were relaxed\n\nso interior points in the feedback set could be found. Later these constraints were\n\n77\n\nTable 6.1. Values of a in the pitch controller look-up table.\na\nPitch Roll-Yaw\n-3.0 -1.0\n1.0 1.0\n4.0 2.5\n8.0 4.0\n12.0 8.0\n16.0 12.0\n20.0 16.0\n_____ 20.0\n\nmade more restrictive and returned to their original assignments. This became a\n\ntedious process for some parts of the grid. When the number of iterations exceeded\n\n100 it was decided to add more grid points so that the desired feedback gains could\n\nbe found in fewer iterations.\n\n6.2 Formulation of the Design Constraints\n\nBefore using the iterative Lyapunov design algorithm, the uncontrollable normal\n\nvelocity components for the entire M-Q-a grid must be determined. This information\n\nis needed to formulate the design constraints. This process is described by the block\n\ndiagram in Figure 6.1. Table 6.2 presents the minimum and maximum uncontrollable\n\nnormal velocity components of the pitch model throughout the M Q grid for each\n\nvalue of a. Here P = Pq, the matrix computed during the initializing procedure\n\nof the last chapter. Table 6.3 gives the same result for the roll-yaw model where\n\nP = Ppr,. The extreme values of uncontrollable normal velocities for the pitch model\n\nare -49.5985 and -37.3480. Also the eigenvalues of\n\nI[Pq(Aq BqKq) + (Aq BqKq)TPq] (6.1)\n\nP P\nq'^ pr\n\nUncontrollable\nNormal\nVelocities\n\nFormulation\nof Design\nConstraints\n\nFigure 6.1. Formulation of the Design Constraints.\n\nTable 6.2. Uncontrollable normal velocity components for the pitch model.\na(degrees) minimum uncontrollable velocity maximum uncontrollable velocity\n-3 -47.1539 -37.4260\n1 -47.0806 -37.3873\n4 -47.1611 -37.3480\n8 -47.9797 -37.4247\n.. 12 -48.8796 -37.3906\n16 -49.5915 -37.6144\n20 -49.5985 -37.4355\n\nTable 6.3. Uncontrollable normal velocity components for the roll-yaw model.\na(degrees) minimum uncontrollable velocity [maximum uncontrollable velocity\n-1 -23.3399 -19.6307\n1 -23.3774 -19.6409\n2.5 -23.4445 -19.6883\n4 -23.4971 -19.7450\n8 -24.5852 -17.2696\n12 -25.3163 -19.9600\n16 -25.6480 -19.8625\n20 -26.4797 -20.30717\n\nFlight\nParameter\nGenerator\n\nat the nominal flight condition from the previous chapter were found to be -1.0715 x\n\n106 and -40.002. Theorem 2 requires that\n\n-37.3480 < cq2 (6.2)\ncqI < -49.5985\nBut we also want\ncq1 < -1.0715 x 106 < ql (6.3)\n-q2 < -40.002 < (6q.\nsince we desire the eigenvalues of (6.1) to be close to those at the nominal flight\n\ncondition. From this, the constraining values were chosen to be\n\nCqI = -1.2 x 106, q1 = -106,\n\nCq2 =-45, cq2 =-35\n\nLikewise, for the roll-yaw model, the uncontrollable normal velocity components range\n\nfrom -26.4797 to -17.2696. From Theorem 2 we must have\n\n-17.2696 < p3 (6.4)c\ncpr1 < -26.4797 )\nWith the eigenvalues of\n\nS[Pp,(Apr BprKpr) + (Ap. BprKp,)TPp.]\n2\nat the nominal flight condition being -1.3614 x 105, -320.00, and -20.003, the\n\nfollowing is desirable.\n\ncp,1 < -1.3614 x 105 < CplI\ncpr2 < -320.00 < Cp,2 (6.5)\ncpr3 < -20.003 < Cp,3\nWith this in mind, the following selections were made.\n\n.prl = -200000, CprIl =-90000,\n\npr2 = -400, Tp,2 = -250,\n\ncpr3 = -25, Cp,3 = -15\n\n6.3 Generating the Look-Up Table\n\nWith the design constraints set, feedback gains and feedforward gains are gen-\n\nerated for each grid point. Figure 6.2 gives a block diagram of the system used to\n\naccomplish this. For the pitch model, the initial guess comes from one of the gains\n\nthat was generated in the previous chapter when determining the dependence be-\n\ntween gains and flight parameters. The operating point from which this initial guess\n\noriginates is\n\nM = 1, Q = 1250psf, a = 8 (6.6)\n\nand is one of the extreme values listed in Table 5.1. The result of the first point is\n\nused to start adjacent points which, in turn, start new adjacent points until gains\n\nhave been computed for the entire grid. The look-up table for the roll-yaw model is\n\nNumerical problems were encountered in parts of the look-up table for the roll-\n\nyaw model. They were similar to the problems that were predicted in step 5) of the\n\niterative design method given in Chapter 2. To overcome these difficulties, some of the\n\nconstraining values were relaxed for a number of iterations and were later returned to\n\ntheir original assignments in the algorithm. Eventually, the desired feedback matrix\n\nwas found.\n\nFigure 6.3 shows kq ii verses mach number and dynamic pressure when a = 8. As\n\nthis figure would indicate, the gains generated from the iterative Lyapunov method\n\nare smooth with respect to the dependent flight parameters. This fact gives hope\n\nthat the gain scheduling scheme will be easy.\n\nGain\nSchedules\n\nFigure 6.2. Formulation of the look-up table.\n\n10000\n\n5000\n\n0 0.5\n\nFigure 6.3. kql, verses M and Q when a = 80\n\nCHAPTER 7\nGAIN SCHEDULING\n\nThe result of the previous chapter is thirteen tables of gains in terms of mach\n\nnumber, dynamic pressure, and angle of attack. This chapter discusses the process\n\nof curve fitting used to implement the look-up table. The results of a test of this\n\nscheme will follow.\n\n7.1 Curve Fitting\n\nIt was decided to use a combination of polynomial fitting and interpolation to\n\nimplement the autopilot. Third order polynomials were fit to the tables as a function\n\nof mach number. However, low order polynomials could not achieve close fits as a\n\nfunction of angle of attack or dynamic pressure, so linear interpolation was used for\n\nthese two variables. Figure 7.la shows kq11 as a function of M for various constant\n\nvalues of Q at a = 8. An example of polynomial fitting of one of these curves is\n\nshown in Figure 7.1b where Q = 150psf. A polynomial has been made for every\n\na-Q pair and the polynomial coefficients are interpolated as a function of these two\n\nvariables. The tables of the three pitch controller gains each have 7 entries for a\n\nand 22 entries for Q. Since each polynomial has 4 coefficients the total number of\n\ncoefficients for each gain is 7 x 22 x 4 = 616. Similarly the tables for the roll-yaw\n\ncontroller have 8 entries for a and the same number of entries for Q. Each of the ten\n\ngains are then scheduled using 8 x 22 x 4 = 704 polynomial coefficients.\n\n7.2 Testing the Fit\n\nFigure 7.2 gives the system for testing the polynomial and interpolation routines.\nGains were generated from these routines at locations centered between the original\n\ngrid points. These new locations were found by taking the average of the coordinates\n\n83\n\na 0\na\n10\n0_ .......\n\n.20 .- -47\n30,\ni- .. 48.\n\n-60\n-70-50\n.5 1 1.5 2 as 3 3.5 .6 0.65 0.7 0.75 0.8 0.A5 0.9 0.95 1\nM M\n\nFigure 7.1. A plot of the kqii verses M with (a) various constant values of Q and\nQ = 8, and (b) a least squares third order polynomial fit for the plot where Q =\n150psf and ac = 8.\n\nof adjacent grid points. Linear models were also made at each test point where /, p,\n\nq, and r were set to zero. The eigenvalues of\n\nI[Pq(Aq- BqKq) + (Aq BqKq)TPq] (7.1)\n\nand\n\nj[Pp,(Ap,, Bp,Kp,) + (Ap. BpKp)TPP] (7.2)\n\nwere computed to see if they remained within the desired limits. Table 7.1 shows\n\nthe maximum eigenvalue of (7.2) for all of the test points at a = 1.75. These values\n\nare plotted against their indices in Figure 7.3. Most of the eigenvalues of Table\n\n7.1 lie within the desired limits of -25 and -15. The eigenvalue in seventeenth row,\n\nsecond column, however, is -3.96, the worst value found out of all of the test points.\n\nAlthough the deviation is high, this value is still negative indicating stability for the\n\nclosed loop system. For the pitch controller, the actual limiting values of A1 and A2\n\nat all the test points are\n\n-1.37 x 106 < Aq1 < -9.85 x 105 (73)\n-47.7 < Aq2 < -38.1 ( )\n\nFigure 7.2. A test of the curve fitting routines used to implement the autopilot.\n\nTable 7.1. The maximum eigenvalues of (6.1) for all test points at a = 1.75.\nQ/M indices 1 2 3 4 5 6 7[ 8\n1 -20.79 -20.82 -20.84 -20.87 -20.88 -20.89 -20.90 -20.90\n2 -20.68 -20.69 -20.70 -20.72 -20.74 -20.76 -20.76 -20.76\n3 -20.84 -20.85 -20.84 -20.83 -20.81 -20.79 -20.78 -20.79\n4 -20.59 -20.61 -20.49 -20.30 -20.18 -20.33 -20.34 -20.21\n5 -20.30 -20.22 -19.93 -19.71 -19.96 -19.88 -20.01 -20.62\n6 -20.41 -19.97 -19.77 -20.16 -19.88 -20.20 -20.79 -20.97\n7 -19.10 -17.70 -17.86 -18.92 -20.03 -20.42 -19.75 -20.81\n8 -18.62 -19.79 -20.67 -20.37 -16.16 -19.14 -20.21 -19.48\n9 -19.90 -20.97 -19.33 -14.48 -19.58 -21.08 -20.93 -19.64\n10 -20.23 -20.88 -11.67 -16.89 -19.96 -20.40 -19.72 -19.41\n11 -20.57 -20.38 -12.40 -18.26 -20.60 -20.85 -20.28 -20.30\n12 -20.77 -17.82 -17.66 -20.97 -21.85 -21.65 -21.14 -21.32\n13 -19.83 -18.51 -21.91 -20.81 -20.87 -22.17 -22.13 -22.15\n14 -20.45 -22.23 -20.57 -17.42 -20.84 -22.21 -22.42 -22.39\n15 -22.42 -22.13 -18.83 -18.24 -20.65 -21.23 -21.49 -21.90\n16 -21.82 -19.45 -15.17 -20.54 -21.09 -20.92 -20.94 -21.73\n17 -21.85 -3.96 -20.63 -22.26 -22.35 -22.13 -22.03 -22.48\n18 -18.88 -19.74 -21.48 -22.29 -22.58 -22.65 -22.68 -22.87\n19 -20.85 -21.06 -21.71 -22.27 -22.69 -22.98 -23.19 -23.37\n20 -21.65 -21.93 -22.29 -22.63 -22.93 -23.19 -23.39 -23.52\n21 -20.62 -22.15 -22.52 -22.62 -22.61 -22.53 -22.46 -22.42\n\n-10\n\nM~ -20,\nE\nt-30,\n\n-40-\n\n-50,,\n30 -\n\n10 04\nI ,,^ F 0 0 . .\n\nIndices of M\n\nFigure 7.3. A plot of the eigenvalues from Table 6.1.\n\nlll in i c -\n\n87\n\nLikewise, for the roll-yaw controller, the limiting values at all the test points are\n\n-2.03 x 105 < Aprl K -9.28 x 104\n-2392.7 < Apr2 < -182.6 (7.4)\n-26.27 < Apr3 < -3.96\n\nSome of these values differ significantly from the desired constraints; however, since\n\nthese values are still negative, these deviations are acceptable and indicate that the\n\nclosed loop system will be stable. It should also be noted that most eigenvalues\n\nremain well within their desired constraints as shown in Table 7.1.\n\nCHAPTER 8\nNONLINEAR SIMULATIONS\n\nA nonlinear simulation has been used to test the proposed autopilot for the EM-\n\nRAAT missile. First a section follows giving an overview of the nonlinear simulation.\n\nA test module is then made to generate state commands in order to evaluate the\n\nautopilot's tracking ability. Finally, a series of flight scenarios are run to determine\n\nthe ability the missile has to intercept the target.\n\n8.1 The Nonlinear Simulation\n\nFigure 8.1 shows a block diagram of the simulation used to test the EMRAAT\n\nmissile. The program is written in FORTRAN. Initial conditions of the target and\n\nmissile are specified by the user. The simulation is then run and a trajectory of both\n\nthe target and missile results. All target and missile variables can be observed. The\n\ntarget is programmed to fly in a straight line until the range between the target and\n\nmissile falls below 5,000 ft. The target then makes a 9 g turn to the right. The\n\nsimulation terminates when the closing velocity becomes positive.\n\nThe seeker measures the line of sight angles and the range rate of the target.\n\nThe simulation uses exact measurements of these values and does not assume any\n\nnoise. These values are sent to the guidance law which, in this case, implements\n\nproportional navigation. A derivation of this guidance law can be found in Bryson\n\nand Ho . The outputs of the guidance law are two desired accelerations, a, and\n\na,. The BTT logic makes the conversion from the acceleration commands to the\n\nthree state commands ac, 0, and pc. Since the missile can achieve a much higher\n\nacceleration with angle of attack than with sideslip, the BTT logic uses Pc to rotate\n\nthe desired accelerations into the pitch plane. If this roll maneuver is successful then\n\nSeeker\n(RF)\n\nGain\nSchedule\n\nM Q .\n\nK\n\nK~f\n\nn Guidance Law\nR (Pro-Nay)\nOQ\nvc\n\nayc, azc\n\nBTT Logic\n\naI O cl\n\nPC\n\noi p q r\n\nSp\n6q\nSr\n\nNonlinear\nMissile\nDynamics\n\na Iaz\n\nExact Computation of Missile and Target Variables\n\nFigure 8.1. A Block Diagram of the Nonlinear Missile Simulation.\n\nTarget\nPosition\n\nU = -Kx + K,,fv\n\ni[ l, i\n\nay, will become small and a, will become positive. ac and /30, are computed in an\n\nattempt to match a^ and ayo respectively.\n\nThe autopilot implements the control law\n\nu = -Kx + Krefv (8.1)\n\nwhere x is a vector containing the actual states and v contains the state commands\n\nfrom the bank-to-turn (BTT) logic. The states come from exact measurements in the\n\nsimulation. If this autopilot were to be implemented in an actual missile, the states\n\nwould be measured using an inertial platform. The gains K and Krf come from the\n\ngain schedule implemented with a combination of polynomials and interpolation. In\n\nthis simulation there is no delay in the gain schedule and K and h,,f are produced\n\ninstantaneously. The output of the autopilot is the control surface angles 6p, 6q, and\n\n86. Linear and angular accelerations are computed by the missile dynamics module\nof the program. The simulation uses the output of the missile dynamics to compute\n\nall of the flight variables including the position and velocity of the missile.\n\n8.2 A Test of State Tracking\n\nThe model for the EMRAAT missile has five states and three inputs. The autopi-\n\nlot is designed to track three state commands: ac, 3, and pc. Before running missile\n\ntarget scenarios it was decided to test the autopilot's tracking ability. The BTT logic\n\nwas disconnected, and the following commands were applied to the reference inputs\n\nof the autopilot.\n100 for 0 s < t < .5 s\n0 for .5 s < t < 2.75 s (8.2)\nc= 10 for 2.75 s < t <3.75s (.\n0 for 3.75 s < t\n0' for 0 s /= 5 for 2 s 0 for 2.5s < t\n\nTable 8.1. The rise times of each commanded state.\nAltitude(ft) Mach t,(s) t, (s) 4.(s)\n20,000 2.0 .100 .243 .0284\n50,000 3.0 .108 .316 .0181\n\n0/s for 0 s < t < 1 s\n100/s for 1 s < t < 1.5 s\n0/s for 1.5 s < t< 2.75 s .\nPc 100o/s for 2.75 s < t< 3.25 s (8.4)\n-100/s for 3.25 s 0/s for 3.75 s < t\nThe experiment was performed once at an initial altitude of 20,000 ft with an initial\n\nmach number of 2.0 and a second time at an initial altitude of 50,000 ft with an initial\n\nmach number of 3.0. Figure 8.2 shows the results of the first test. In addition to the\n\ncommanded states, figure 8.2 also shows the remaining two states q and r. The rise\n\ntime as defined as the time needed to achieve 90% of the desired value was found\n\nfor each commanded state and is shown in table 8.1. It should be noted that mach\n\nnumber and dynamic pressure do not change significantly during this test. This is a\n\ntest of accuracy in state tracking and is not a valid test of gain scheduling in terms\n\nof M and Q. However, a changes very rapidly and does not appear to hinder the\n\nperformance.\n\nCross coupling is evident from this experiment. The step in the roll rate of figure\n\n8.2c at t = 2 s is due to the sideslip command in figure 8.2b. A lesser degree of cross\n\ncoupling occurs during the roll command for 2.75 s < t < 3.75 s when a which is at\n\n10 is rotated into sideslip. The same effects are present in the second experiment at\n\n50,000 ft.\n\nFigure 8.2d and 8.2e show the two states that have no commands. The spikes in\n\nthe pitch rate occur when ac changes value, because a pitching maneuver is required\n\nto change the angle of attack. The two spikes in the yaw rate occur for the same\n\nb) C--) Co,,m,..c 0ta C-)\n\n0 = a a a r a\n\na) p C -) ~,. Oo,,a~ 0 (-)\n'Sc.\n\n'ool\n\nS.o\n\n-0)\n\nFigure 8.2. A test of state tracking at an initial altitude of 20,000 ft with an initial\n\nmach number of 2.0. The states shown are (a) a, (b) 03, (c) p, (d) q, and (e) r.\n\nS.---\n\n. T\n\n. -a ora.\n\nreason when sideslip changes. The two steps in the yaw rate, however, are due to\n\ncross coupling with the roll rate.\n\nThe rise times in table 8.1 are small and indicate that the missile will perform well\n\nin flight scenarios. While cross coupling effects are noticeable, they are not believed\n\nto be great enough to significantly hinder the performance of the missile; thus it was\n\ndecided to run the missile through a series of flight scenarios to see how well the\n\nmissile can intercept the target.\n\n8.3 Simulation of Flight Scenarios.\n\nA series of flight scenarios has been run to test the autopilot. Figure 8.3 shows\n\nthe trajectory of the missile and target for a flight scenario at an altitude of 20,000\n\nft. The miss distance is .64 ft. A hit is considered to be any miss distance under 10\n\nft. Figure 8.4 shows the commanded and the actual y and z accelerations. The devi-\n\nations from the commanded normal accelerations are mostly due to a simplification\n\nused in implementing the BTT logic. Instead of using two aerodynamic coefficients,\n\nproportionality constants were assumed. The errors are not large enough to prevent\n\nthe interception of the target. The state tracking of the reference inputs appears to\n\nbe working well as seen in figure 8.5 and indicates that the autopilot is performing\n\nwell.\n\nThe trajectories of a 40,000 foot altitude scenario is shown in figure 8.6. The miss\n\ndistance is 0.05 ft. A similar scenario at 10,000 ft is shown in figure 8.7. The miss\n\ndistance is 2.8 ft.\n\nFigure 8.8 gives a case were a miss occurred. The scenario took place at 50,000\n\nft and the target was missed by 452 ft. Figure 8.9a and b shows that the missile was\n\nunable to achieve the desired z acceleration and angle of attack after 1.3 seconds into\n\nthe flight. This is because the elevator reached its -40 limit as figure 8.9c indicates.\n\na) Top View of Missile (-) and Target (-) Trajectory\n\n6000\n\n5000\n\n4000\n\n3000\n\n0 1000 2000 3000 400 5000 6008 7000 800O 9\nPosition in the x direction in ft.\nxl04 b) Side View of Missile (-) and Target (-) Trajectory\n\n-2.05 k\n\n0 1000 2000 3000 4000 5000 6000 7000 8000\nPosition in thex direction in ft.\n\nFigure 8.3. (a) Top view and (b) side view of missile and target\nscenario which occurred at 20,000 ft. The miss distance is 0.64\nnumbers of the target and missile are 2.5 and .92 respectively.\n\ntrajectories of a\nft. Initial mach\n\na) Commanded z Acceleration (-) and Acwal z Acceleration (-)\n\nSeconds\nb) Conmmanded y Acceleration (-) and Actal y Acceleration (-)\n\nSeconds\n\nFigure 8.4. Commanded and actual (a) z accelerations and (b) y accelerations for\nthe scenario in figure 8.3.\n\nFull Text\n68\n7) Repeat steps 4) 5) and 6) until one of two events occur.\n1. Cjj Q-ifi cu C\\}i i Qni Qnf ^nd Cni Cnf.\n2. (clt c1_1) ... (c cm_ 1) become very small.\nEvent 1. implies that the final answer has been found. Event 2. implies that ICf\nis empty and K{ satisfies the constraints corresponding to cu, cu, ... cn, cn. Steps\n2), 5), and 6) are accomplished in the following way.\nSteps 2) and 6)\nLet\nThen,\nei\nAi\n\\{PA + ATP PBKi KjBrP)\n\\{PA + AtP PBKi KjBTP)\\\n(5.32)\ncu = min[clf, ex]\ncu = max[ci/,ei]\nQni ?7iZ7i[cny, en]\nCjii 77]\n(5.33)\nStep 5)\nLet K = Ki\nFor j = 1 to m\nFor k = 1 to n\nPi = ~(PBQi,k + QT,.kBTP)\nPo = P(A BK0j,k) + (A BKaj,t)TP 2cl\nK0j,k kj,kQj,k\n\nusing another autopilot design. The model was linearized, and the fol\nlowing system was taken at 4.00 seconds into the flight.\n43\nX\n= Ax + Bu\n(2.128)\nwhere\n' -.9939\n-.1612\n.0086\n.9997\n0.0 '\n.1613\n-.4100\n.1079\n0.0\n-.9997\nA =\n-64.36\n-1557\n-2.078\n-.1763\n.0442\n(2.129)\n-252.2\n-6.097\n.0173\n-.6403\n.1611\n-.5806\n83.29\n-.0261\n-.1597\n-.5701\n0.0\n-.1296\n0.0 '\n-.0149\n0.0\n.1044\nB =\n-1150\n-31.06\n-1222\n-4.494\n-121.7\n-4.747\n1.234\n-.2802\n-108.6\nand\nx = [a,/3,p,?,r]r\n(2.130)\n(2.131)\nStep 1)\nLet P = I. The minimum and maximum uncontrollable normal velocity\ncomponents were found by computing the limiting values in equations\n(2.47) and (2.48). Since P /, these terms simplify and are evaluated as\nfollows:\nmin -xT [PA + A1 Phi\nX6S 2 J\nand\nmax -xT[PA + AT P\\x\nxes 2\nA rmn[\\Bl(A + A^B,] (2.132)\n-.7263 (2.133)\n\\maX[\\Bl(A + AT)BL] (2.134)\n(2.135)\n-.3017\n\n79\nat the nominal flight condition from the previous chapter were found to be 1.0715 x\n106 and 40.002. Theorem 2 requires that\n-37.3480 < cq 2\ncql < -49.5985\n(6.2)\n(6.3)\nBut we also want\ncql < 1.0715 x 106 < Cqi\ncq 2 < 40.002 < cq2\nsince we desire the eigenvalues of (6.1) to be close to those at the nominal flight\ncondition. From this, the constraining values were chosen to be\nCql = -1.2 X 106, Cql = -106,\ncq2 = -45, Cq2 = -35\nLikewise, for the roll-yaw model, the uncontrollable normal velocity components range\nfrom 26.4797 to 17.2696. From Theorem 2 we must have\n17.2696 < cpr3\ncpn < -26.4797\n(6.4)\nWith the eigenvalues of\nPpT ( Apr\n- B\npr Kpr ) h (-^.pr\nBpr\nKpT)T PpT]\ncondition being 1.3614 x\n105\n, -320.00,\nand 20.003, the\nprl\n<\n-1.3614 x 105\n<\nCpr 1\npr 2\n<\n-320.00\n<\nCpr 2\n(6.5)\nLpr3\n<\n-20.003\n<\nCpr 3\nWith this in mind, the following selections were made.\ncpri = 200000, cpri = 90000,\nCpr2 = -400, Cpr2 = -250,\npr3 CpT 3 = 15\n\n2\nFigure 1.1. The EMRAAT missile .\n\n81\nGain\nSchedules\nFigure 6.2. Formulation of the look-up table.\n\n28\nThen\n~axT[PA + AtP PBK1 K?BTP]x < ac V xTx = 1 (2.65)\nand\n^(l-a)xT[PA+ATP-PBK2-K^BrP]x < (1 -a)c VxTx = 1 (2.66)\nwhere 0 < a < 1. Adding (2.65) to (2.66) gives\nxT[PA + ATP]x \\axT[PBKx + K?BtP]x\n|(1 a)xT[PjBAr2 + K^BTP]yi which becomes\n(2.67)\n^T[PA+ATP-PB(aK1+{l-a)K2)-(aK1+(l-a)K2)TBTP} (2.68)\nSo aKi + (1 ol)K2 G 1C for 0 < a < 1. /C is convex.\nNow that we have conditions on the nonemptiness and convexity of 1C and 1C it\nwould be desirable to find the boundary of these sets. We know that if one of the\neigenvalues of the square matrix Q is c then\ndet[Q cl] = 0. (2.69)\nThis is helpful in understanding the following exhaustive search method for comput\ning the boundary of 1C. A similar procedure exists for finding dlC.\n1) All eigenvalues of\ni[PA + AtP PBK KtBtP] cl (2.70)\nneed to be less than or equal to zero. Fix all but one of the entries of K.\nLet the jth entry be the free entry and be represented by k. Let Kqj\n\n37\nthe line that passes through K{ and is parallel to the kn axis. Point c is found by\ncomputing \\Ki + \\K{2. Since 1C1 is convex, then c 6 AC,. This step is repeated again\nby searching along the kx2 axis. Using similar arguments, point / is also in AC. Kt+\\\nis set equal to point /. For higher order systems, additional boundary points are\nfound and interior points computed by searching along directions which are parallel\nto the axis of the coordinate system.\nUnder normal conditions, this procedure works well. However numerical problems\ndo occur. These problems will now be discussed along with a cure. Figure 2.10b shows\na second order example of when the above method fails. The boundaries are shaped\nin such a way that searching along line lx and l2 yields no new boundary points. A\nsolution to this problem is to relax one of the constraining values by setting, for ex\nample, Ci equal to c + 6. d)Ci will then move so that point b can be found as shown in\nfigure 2.10c. Later, the relaxed constraining value can take on its original assignment.\n2. We need to show that c, > c,+i when c, > Cf. In this case\nCi Ai > Cf (2.114)\nSince Ki+i ICl and Ki+1 £ <9AA, then the following condition holds with strict\ninequality.\n^xt[P(A BKi+1) + (A BKi+1)TP]x < c V x : xTx = 1 (2.115)\nTaking (2.110) for i + 1 then\n^xt[P(A BKi+1) + (A BA'!+1)tF]x < A,-+i V x : xTx = 1 (2.116)\nand there exists an x which satisfies the above condition for equality. Therefore\nCi > At-+i and Ci > max\\c.f,\\i+]] = c,-+1. Using similar arguments, it can be shown\n\nK,¡ was found before 6 < .0001. The program ran 16 iterations. The final\nresult is\n45\n.00046\n2.125\n-.7974\n.0805\n.7535 '\nII\n2.057\n-.0012\n2.339\n-6.577\n.2591\n(2.147)\n-.00029\n-.7259\n.0263\n-.1346\n-.7985 _\nNow to check the result.\nII\n^1\n^min [ ,-y ( ^4\n+ Ar-\nBKf -\n(2.148)\n=\n-818.0\n(2.149)\nXf -\n+ Ar-\nBKj -\nKjBT)]\n(2.150)\n\n-.3001\n(2.151)\nThis meets the desired constraint.\nCf < Xf < \\f < c¡\n\n25\nnp\nc > maxx\nxe5\nP Ax = max\nxes\ntxT[iM + AtP]x\n(2.46)\nConditions (2.45) and (2.46) are respectively equivalent to\nc < Amin[^(y/H)-lBl[AP-x + P~l At]B ^VH)~1] (2.47)\nand\nc > Amax[\\('/HTlBl[AP-1 + P-lAT\\B^VH)-1] (2.48)\nwhere\nH = (P-1BA.)TP~1B (2.49)\nThe matrix H is square, full rank, and has dimension n m. The above is true for\nthe following reasons. Equation (2.46) can be rewritten as\nc > max ^xT[PA + AT P]x (2.50)\n= max )-nTBl[AP_1 + P~lAT]Bxfi (2.51)\nH1 H i=l\nSince H is positive definite then y/H exists, is square, and has an inverse. By making\nthe substitution, // = (v/#)~1z, (2.51) becomes\nc > max T(VH)-1BI[AP~1 + P~lA)B^VH)~1'z (2.52)\nzTz=i2\nwhich is equivalent to\nC > Amax^iVHy'BliAP-1 + P-lA]B^VH)-1} (2.53)\nEquation (2.45) is equivalent to (2.47) for similar reasons. We now give a proof of\nTheorem 2.\n\nTable A.6. Tabular data for Cy0\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-.2226\n-.2052\n-.2175\n-.1717\n-.1758\n-4\n-.2870\n-.2668\n-.2305\n-.2195\n-.1911\n0\n-.3096\n-.3099\n-.2546\n-.2478\n-.2096\n2\n-.3210\n-.3235\n-.2660\n-.2579\n-.2299\n4\n-.3200\n-.3363\n-.2717\n-.2629\n-.2478\n8\n-.3113\n-.3614\n-.2851\n-.2623\n-.2812\n12\n-.2918\n-.3774\n-.3238\n-.2925\n-.3178\n16\n-.2445\n-.3447\n-.3849\n-.3698\n-.3783\n20\n-.2515\n-.2730\n-.4800\n-.4440\n-.4631\nTable A.7. Tabular data for Cn6r\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-1.0842\n-.9703\n-.4520\n-.4236\n-.2831\n-4\n-1.0743\n-.9505\n-.4600\n-.3709\n-.2739\n0\n-1.0446\n-.9683\n-.4900\n-.3818\n-.3015\n2\n-1.0356\n-.9802\n-.4900\n-.4000\n-.3162\n4\n-1.0257\n-1.0050\n-.5080\n-.4091\n-.3272\n8\n-1.0000\n-1.0535\n-.5200\n-.4436\n-.3640\n12\n-.9505\n-1.0891\n-.5300\n-.5200\n-.4412\n16\n-.8851\n-1.0931\n-.5760\n-.6400\n-.5441\n20\n-.7921\n-1.0475\n-.6840\n-.7709\n-.6838\n\n84\nM M\nFigure 7.1. A plot of the kqu verses M with (a) various constant values of Q and\na = 8, and (b) a least squares third order polynomial fit for the plot where Q =\n150psf and a = 8.\nof adjacent grid points. Linear models were also made at each test point where /?, p,\nq, and r were set to zero. The eigenvalues of\n\\[P,(A, ~ B,K,) + (A, B,K,fP,} (7.1)\nand\n^ (Apr Sp'pJ\\.pr'j ~f\" Bpr Kpr ) Ppr ] (7-2)\nwere computed to see if they remained within the desired limits. Table 7.1 shows\nthe maximum eigenvalue of (7.2) for all of the test points at a = 1.75. These values\nare plotted against their indices in Figure 7.3. Most of the eigenvalues of Table\n7.1 lie within the desired limits of 25 and 15. The eigenvalue in seventeenth row,\nsecond column, however, is 3.96, the worst value found out of all of the test points.\nAlthough the deviation is high, this value is still negative indicating stability for the\nclosed loop system. For the pitch controller, the actual limiting values of and A2\nat all the test points are\n-1.37 x 106 < A,i < -9.85 x 105\n-47.7 < A?2 < -38.1\n(7.3)\n\n67\nthe previous point become the initial guess for the present point. The result of\nthis algorithm is a series of gains; one set for each flight condition generated. The\nnext portion of this section discusses some modifications made to the algorithm from\nsection 2.4. The material which follows describes how the design constraints are\nselected.\nThe design method of section 2.4 finds a K so that the eigenvalues of |(PA+ATP)\nfall between c¡ and c/ where A is the closed loop system. The design algorithm\nused in Figure 5.1 has been modified so that the resulting K satisfies the following\nrequirements.\nQi < Ai\n< Ao\n2\n\\[P(A BK) + (A BK)tP]\n\\[P{A BK) + (A BK)tP}\n< cx\n< c2\n(5.31)\ncn < An [\\[P{A-BK) + (A-BK)tP}\\ < cn\nwhere Ai is the smallest eigenvalue, A2 is the second smallest eigenvalue, and so on.\nThe values of the cs are supplied by the designer. This modification has been made\nwith the belief if more design constraints are made, then the performance will vary\nless with changes in the flight conditions. The modified algorithm is as follows.\n1) Choose P, cxj, ci/, ... cnf, and cnf. The selection of and cn¡ must\nobey Theorem 2. However, Theorem 2 does not guarantee the nonempti\nness of K.\n2) Compute c10,Cio, ... ,cn0,cn0 so that the initial guess, Ko £ and\nK,} C /C0.\n3) Let f = 0\n4) Let i i + 1\n5) Compute Kt so that K{ 6 i and ' ^ 5AT,_i.\n6) Compute cu. ci, ... cn, cn so that Kt 6 dICi and C /Ct.\n\nAPPENDIX A\nAERODYNAMIC DATA FOR THE EM R A AT AIRFRAME\nThis appendix gives aerodynamic data for the extended medium range air-to-air\ntechnology (EMRAAT) airframe. The aerodynamic coefficients are given in tabular\nform. All coefficients that correspond to angle rates have no dimensions and must be\nmultiplied by All coefficients that correspond to angle positions are given in per\ndegrees and must be multiplied by ^.\nMissile Reference Diameter\nMissile Reference Area\nd = .625ft\nS = .3067ft2\nTable A.l. Tabular data for C/v\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-4.816\n-5.472\n-5.104\n-4.502\n-3.952\n-4\n-2.408\n-2.736\n-2.552\n-2.251\n-1.976\n0\n0.000\n0.000\n0.000\n0.000\n0.000\n2\n1.204\n1.368\n1.276\n1.126\n.988\n4\n2.408\n2.736\n2.552\n2.251\n1.976\n8\n5.264\n6.072\n5.644\n5.011\n4.456\n12\n8.613\n10.068\n9.456\n8.458\n7.528\n16\n12.467\n14.674\n13.898\n12.490\n11.092\n20\n16.800\n19.730\n18.610\n16.732\n14.860\n106\n\n56\nwhere d is the missile diameter. Again, the aerodynamic coefficients come from\nwind tunnel testing and are tabulated in the first appendix. Eulers three moment\nequations can be written as follows.\nl '\nm\n- J\nP '\nq\nn\nr\nwhere\nJ =\n+ H(p,q,r)\nIxx IxY Ixz\nIxy Iyy Iyz\nIxz Iyz Izz\n(4.29)\n(4.30)\nand\n-(Iyy Izz)qr + Wz(r2 q2) Ixvpq + Ixyrp\nH = ~(Izz Ixx)rp + Ixz(p2 ~ r2) Ixyqr + Iyzpq (4-31)\n-(Ixx lYY)pq + Ixy(q2 p2) Ivzrp + Ixzqr _\nWe will assume that the inertial cross products, Ixy, Ixz, and Iyz are small. Also\ndue to symmetry of the airframe, Iyy and Izz are assumed to be equal. Inertial data\nfor the EMRAAT missile is given in the second appendix. Solving (4.29) for p, q,\nand f gives\nP =\nIxx\nl\nq =\nIyy\nand\n1\nr =\nIzz\n[m + (Izz ~ Ixx) rp]\n[n + (Ixx Iyy) pq]\n(4.32)\n(4.33)\n(4.34)\nSubstituting (4.26), (4.27), and (4.28) into (4.32), (4.33), and (4.34) gives\nQSd\nq =\np =\nQSd\nh\nIxx\n\\plpP + ClpP + Clrr + Cl6pdp + Cl6rST\nYY\n~t~ Cmqq 4\" SJm6c¡8q\n+\nIzz Ixx\nIyy\n-rp\n(4.35)\n(4.36)\nand\nQSd r\nIzz\nCng/3 + CnpP + Cnrr + Cns 8p + Cn¡j.S\n, Ixx Iyy\n+ j pq\nizz\n(4.37;\n\n91\nTable 8.1. The rise times of each commanded state.\nAltitude(ft)\nMach\n%(s)\nir/3(s)\nr-p(s)\n20,000\n2.0\nTOO\n.243\n.0284\n50,000\n3.0\n.108\n.316\n.0181\n0/s\nfor 0 s < t < 1 s\n100%\nfor 1 s < t < 1.5 s\n0%\nfor 1.5 s < t < 2.75 s\n100%\nfor 2.75 s < t < 3.25 s\n-100%\nfor 3.25 s < t < 3.75 s\n0%\nfor 3.75 s < t\n(8.4)\nThe experiment was performed once at an initial altitude of 20,000 ft with an initial\nmach number of 2.0 and a second time at an initial altitude of 50,000 ft with an initial\nmach number of 3.0. Figure 8.2 shows the results of the first test. In addition to the\ncommanded states, figure 8.2 also shows the remaining two states q and r. The rise\ntime as defined as the time needed to achieve 90% of the desired value was found\nt\nfor each commanded state and is shown in table 8.1. It should be noted that mach\nnumber and dynamic pressure do not change significantly during this test. This is a\ntest of accuracy in state tracking and is not a valid test of gain scheduling in terms\nof M and Q. However, a changes very rapidly and does not appear to hinder the\nperformance.\nCross coupling is evident from this experiment. The step in the roll rate of figure\n8.2c at t = 2 s is due to the sideslip command in figure 8.2b. A lesser degree of cross\ncoupling occurs during the roll command for 2.75 s < t < 3.75 s when a which is at\n10 is rotated into sideslip. The same effects are present in the second experiment at\n50,000 ft.\nFigure 8.2d and 8.2e show the two states that have no commands. The spikes in\nthe pitch rate occur when ac changes value, because a pitching maneuver is required\nto change the angle of attack. The two spikes in the yaw rate occur for the same\n\n16\nFigure 2.4. An example of a second order system with negative radial velocity com\nponents\nFigure 2.5. One velocity vector on the ellipse xTPx = 1\n\n63\nEquations (5.10) and (5.11) are used to eliminate p and V in the linear model. As\na result, the linearized model can now be generated from the following seven flight\nparameters.\nM Q a ¡3 p q r\n5.2 The Flight Parameter Generator\n(5.12)\nA series of flight conditions are made and used to generate many linear models.\nFeedback gains are generated for each condition. The first of the series is called\nthe nominal flight condition. The values of the parameters for the nominal flight\ncondition are\nM = 2, Q = 1250 psf, a = 8, (3 = 0,\np = 0/s, q = 10/s, r = 0/s.\nNext, one of the parameters is allowed to vary while the other six are held constant.\nThis process is repeated six times so that each parameter can be tested. Table 5.1\nshows the starting point, and the minimum and maximum values of each changing\nparameter. Parameters with nonzero starting points begin at the starting point,\nincrement to the maximum value, return to the starting point, and then decrement\nto the minimum value. Due to symmetry, the remaining variables have starting points\nat zero and increment to their maximum values. M and Q sweep through a narrow\nrange because of restrictions of the atmospheric tables.\n5.3 Initializing the Iterative Lyapunov Design Method\nThe iterative Lyapunov design method requires an initial guess, Kqo and Kpro and\ntwo positive definite matrices, Pq and Ppr. The initializer supplies these values and\nwill be discussed in this section.\n\n13\ninside one of the positive regions. Since the velocity of the system always has an\nouter radial component, the magnitude of the state vector is always increasing so\nthat the system (2.1) is unstable.\nA direct relationship exists between the positive and negative region boundaries\nand the eigenvectors of the system. In this case, the eigenvectors are rotating clock\nwise at the same angular velocity as the region boundaries. It can be shown that if a\nsystem has constant eigenvectors with time varying eigenvalues which have negative\nreal parts for all time then the system is stable. The stability question, however, is\nnot as easy to answer for the case with moving eigenvectors.\nIf the system has no positive regions, then it will be stable regardless of how\nquickly it changes. However the converse is not true in general. Figure 2.3 shows a\nstate vector and its corresponding velocity for some dynamic system. The angle d\ncan be computed in the following way.\ncos(\\$) =\nT\nXxX\nXX\nFor x to have an inner radial component then\nThis is true when\nwhere\nt 3tt\n- < 9 <\n2 ~ 2\nxTx < 0\nx = Ax\n(2.5)\n(2.6)\n(2.7)\n(2.8)\nSo if xT,4x is negative everywhere for all time, then the system is stable. Since the\nsystem is linear, it is sufficient to check the sign of xTAx on the unit circle. If all\nvelocities on the unit spheroid point inside that boundary, then the same is true for\n\nA CONTROLLER DESIGN METHOD WHICH APPLIES TO TIME VARYING\nLINEAR SYSTEMS\nBy\nKURT WALTER KOENIG\nA DISSERTATION PRESENTED TO THE GRADUATE SCHOOL\nOF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT\nOF THE REQUIREMENTS FOR THE DEGREE OF\nDOCTOR OF PHILOSOPHY\nUNIVERSITY OF FLORIDA\n1994\n\n9\nand A is bounded. If 0 is an asymptotically stable equilibrium point of z = A(t)z for\nall time, then 0 is a locally stable equilibrium point for the system\nX = /(*,x)\nThese ideas can be applied in the following way. Given a nonlinear system\nx = /(x,w,u)\ndefine\nA(x,w) =\ndf\ndx\nB{x,w) =\n(x.wfi\ndf\ndu\n(1.19)\n(1.20)\n(x,w)\nwhere x is a vector containing the states and w is a vector containing additional\nsystem parameters. At regions near the operating point, the system becomes close\nto\nAx = A(x, w)Ax + B(x, w)Au (1.21)\nwhere Ax and Au are small perturbations between the states and inputs and the\noperating point. We would like to find a feedback control law,\nu = K(x, w)x\n(1.22)\nso that\n-P[A(x, w) B(x, w)K(x, w)] + [A(x, w) B(x, w)K(x, w)]rP (1.23)\nis negative definite for all x and w where P is positive definite. If such a control law\nis found, then the perturbations from the nominal trajectory are locally stable for\nthe system\nx = /(x,w,-T(x,w)x) (1.24)\nShahruz and Behtash give one control law which places some of the eigenvalues\nof (1.23) for the case where P = I. However unnecessary limitations on where the\n\n65\nwhere S is an invertible matrix. The diagonals of J are the real parts of the eigen\nvalues of /l, and imaginary parts of the eigenvalues lie in skew symmetric locations\noff of the diagonals. For example, if the eigenvalues of A are a jb, c, then\nJ =\na b 0\nb a 0\n0 0c\nLet\nJ Jdiag ~t\" Jskew\n(5.17)\n(5.18)\nwhere Jdiag is the symmetric part of J and Jskew is the skew symmetric part of J. In\nthe above example\na\n0\n0 '\n' 0\n-b\n0 '\n0\na\n0\nJskew\nb\n0\n0\n0\n0\nc\n0\n0\n0\nJdiag\nMaking the following transformation on the system x = /lx, let\nx = Sz\nThen\nSi = ASz\nand\n(5.19)\n(5.20)\n(5.21)\ni S 1ASz = J z. (5.22)\nLet Pz = Jdiag We wish to check the velocities of the system z = Jz on the ellipsoid\n'Td~- ~Tr- --1 (5.23)\nZTPzZ = -zTJdiagz = 1\nThe normal to the ellipsoid at z is JdiagZ- The projection of z = Jz onto the normal\nis JdiagZ. Here the normal velocity component has the same magnitude but opposite\ndirection to the normal of the ellipsoid. If the system has no complex eigenvalues\nthen the velocities are orthogonal to the ellipsoid everywhere. For this reason, the\n\n95\na) Commanded z Acceleration (-) and Actual z Acceleration ()\nFigure 8.4. Commanded and actual (a) z accelerations and (b) y accelerations for\nthe scenario in figure 8.3.\n\nThe design algorithm finds feedback gains which place the eigenvalues of the\nderivative of a given Lyapunov function. Limitations on placement of these eigenval\nues are stated. This concept is applied to a second order linear time varying system\nwhere ordinary pole placement techniques fail. The design method is then applied\nto a linearized model of the EMRAAT missile which is a function of time varying\nflight parameters. Feedback gains are generated as a function of these flight param\neters. It is discovered that the gains depend on dynamic pressure, mach number,\nand angle-of-attack. A combination of interpolation and polynomial fitting is used\nto create a look-up table for the gains. The resulting controller is programmed into\na nonlinear simulation which runs missile and target scenarios. Small miss distances\nare achieved.\nv\n\n57\nIf gyroscope effects are considered small, then (4.24), (4.25), (4.35), (4.36), and (4.37)\ncan be written as two separate systems: the pitch dynamics, and the roll-yaw dy\nnamics. The pitch dynamics are as follows.\na =\n1 +\nQSCna\n-ll r\nmV\nQSr \\ (y Sr \\\n77CNa ) a + 1 tyCV 9 +\nmV J \\ mV\nQS\nmV\nCNSq ) tig\n9 =\nQSd\niyy\nT Ct -(- Cm<¡q T Crfiiq^q\n(4.38)\n(4.39)\nThe roll-yaw dynamic equations are\n' (W *(+%>) *(-*+3*)^0 *\n(4.40)\n(4.41)\nP =\nr =\nQSd\nIxx\nQSd\n[chp + clpp + Clrr + Chp8p + C,trSr\nizz\n\\Cnpfi + CnpP + CnTr + Cnsp6p + Cner6r\n(4.42)\nThe states of the pitch model are a and q with Sq as its input. For the roll-yaw\nmodel, the states are ¡3, p, and r, and the inputs are 6P and Sr.\n4.2 The Linear Model\nThe previous section gave a model of the pitch dynamics and the roll-yaw dynam\nics in the following form.\n[q q }T = fg(a,q,Sq) (4.43)\n[/5 p r }T = fpr(P,p,r,8p,8r) (4.44)\nWe would like to have two linearized models for use with the proposed design method.\nThe result will be two systems in the following form.\nx = Ax + Bn\n(4.43)\n\nTable A. 16. Tabular data for CUp\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n121.7839\n121.3270\n44.3658\n28.8446\n16.7480\n-4\n98.7909\n94.1548\n41.3766\n27.5697\n16.7480\n0\n76.4321\n66.1137\n37.2075\n26.7729\n16.4228\n2\n64.9356\n51.5798\n34.2183\n25.9761\n15.9450\n4\n53.1219\n37.9147\n30.9931\n25.0996\n15.7317\n8\n30.2081\n9.3997\n24.0708\n22.3904\n13.4959\n12\n7.6115\n-18.1675\n16.9912\n21.0359\n13.0081\n16\n-15.7780\n-46.9984\n10.9341\n20.3487\n12.8455\n20\n-38.5332\n-77.0142\n6.3717\n20.1594\n12.7642\nTable A. 17. Tabular data for Civ,\nM\ncnq\n.8\n-149.068\n1.2\n-134.226\n2.0\n-87.602\n2.7\n-77.176\n3.5\n-75.149\nTable A.18. Tabular data for Cmq\nM\nCmq\n.8\n-1981.268\n1.2\n-1900.576\n2.0\n-1218.300\n2.7\n-1040.346\n3.5\n-1007.925\n\n44\nTheorem 2 guarantees the nonemptyness of 1C and 1C if c¡ > .3017 and\ncj < .7263. With this in mind, we let c¡ = .3 and = 6000. Since\nthe intersection of 1C and 1C is not guaranteed, Cj was chosen to be very\nnegative to increase the probability of getting an answer.\nStep 2)\nSo\nA ^min [^(A + Ar)]\n(2.136)\n= -781.4\n(2.137)\n(2.138)\n^ = + AT)]\n(2.139)\n= 778.9\n(2.140)\n(2.141)\nc0 = mm(cy, A)\n(2.142)\n= -6000\n(2.143)\nCo = max(cf, A)\n(2.144)\n= 778.9\n(2.145)\nSteps 3) 7)\nA program was written for MATLAB to carry out the iterations in steps\n3) 7). The program would terminate if 1C j was found or when\n8 |Ci ct_i| < e = .0001\n(2.146)\n\n40\n1) Choose P, Cf, and c/. This selection must obey Theorem 2. It should\nbe noted that Theorem 2 guarantees the nonemptiness of tCf and ICf, but\nnot their intersections. If the system is time varying, then, in order to\nuse this algorithm, P must be found so that\n\\Bl{t)[A{t)P~l + p-1AT(t)\\BL{t) < 0 Vi (2.119)\nOtherwise this algorithm cannot guarantee stability.\n2) Compute c0 and c0 so that K0 = 0 6 dK-o and JCf C fCo- K0 will be\nthe initial guess in the search for Kf G\n3) Let i = 0\n4) Let i = i T 1\n5) Find Ki so that K{ G and K{\n6) Compute ct- and c so that Kl G dKx and K¡ C or so that G ICf.\n7) Repeat steps 4) 5) and 6) until one of two events occur.\n1. c = Cj- and c = cj.\n2. (c ci_1) and (c2 c_i) become very small.\nRemarks:\nIf event 1 occurs, then )Cf is nonempty and Kt G )Cj. If event 2 occurs, then IC/ is\nempty and Ki yields a closed loop system that meets the constraints corresponding\nto c and c,.\nWe now explain how to perform steps 2), 5), and 6).\nStep 2)\n\n7\nSigma max[E(t)] vs. Time\nFigure 1.4. A plot of crmax[E{t)\\ versus time.\nE(t) is the result of subtracting this matrix from the closed loop system from the\nrest of the trajectory. Solving for P in (1.12) gives\nP =\n17.29 -.0454\n-.0454 .012\n;i.i5)\nComputing the upper bound in (1.13) gives\n1\n= .0578\n(1.16)\nFigure 1.4 shows the plot of crmax[E(t)\\. Obviously since amax[E{t)\\ is greater than\n.0587 for most of the trajectory then (1.13) is not satisfied. This procedure was\nrepeated by taking the constant closed loop matrix from every point in the trajectory,\nand the result was the same. The time varying linear system from this trajectory is\nknown to be stable because a Lyapunov function exists which can show this. The\nupper bound in (1.13) can not be easily used if at all to support the claim of stability.\nDesoer gives a stability limit on the rate of variation of a given closed loop\ntime varying system. Wilson, Cloutier, and Yedavalli give a stability limit on\n\n97\na) Top View of Missile (-) and Target () Trajectory\nFigure 8.6. (a) Top view and (b) side view of missile and target trajectories of\na scenario which occurred at 40,000ft. The miss distance is .05 ft. Initial mach\nnumbers of the target and missile are 3.0 and .92 respectively.\n\n55\nform and are given in the first appendix along with other data pertaining to the\nEMRAAT missile.\nSubstituting (4.18) and (4.19) into (4.16) and (4.17) and solving for and /? gives\n3 pot r + [Cy3/3 + Cypp + Cyrr + CySp6p + Cy6rST\nand\nQSCNa QS ^\n4 77 ot = q-pp ¡7 CNaa + CNqq + CNsq8q\na =\n1 +\nmV\nQSCN\n-il r\nmV\nmV\nQS\nq- p(3 y. (CNaa + CNqq + CNiqSq\n(4.21)\n(4.22)\n(4.23)\nEquations (4.21) and (4.23) are two nonlinear state equations. In order to separate\npitch dynamics from roll-yaw dynamics, /5 is assumed close to zero in (4.23). Thus,\na =\n1 +\nQSCNo\nmV\n-i\nQs \\ f QS\n:Cn0 ) a + ( 1 7777Cjv, ) q +\nmV\nmV\nQS\nmV\na\nNSq q\n(4.24)\nEquation (4.21) can be written as\n*-(&) *(+*(-* + 3*)-O*) *\n(4.25)\nWe now turn to the moment equations of the missile. Since thrust is zero and since\ngravity does not contribute any moment to the missile, then the moments around\nthe x, y, and 2 axis are given by /, m, and n respectively, the moments due to\naerodynamic pressure. They are given by\nl QSd [Cip/3 4- Cipp + Cirr + Ci6pSp 4- Ci6t8t\nm QSd CrriaQ 4 CmtC T Cmqq 4~ Cm6q8q\n(4.26)\n(4.27)\nand\nn QSd Cnpfl + CnpP + Cnrr + CnSp8p 4\" CnSr8-\n(4.28)\n\nCHAPTER 6\nCOMPUTING LOOK-UP TABLES\nIn the last chapter we showed that the gains depend mostly on a, M, and Q.\nThis chapter describes the process of generating a look-up table of gains verses M,\nQ, and a. First a grid of points is formulated. Design constraints are then formulated\nbased on the knowledge of uncontrollable velocity components of the linear models.\nFinally, the gains are computed.\n6.1 Determining a Grid of Points\nA two dimensional grid of points for M and Q has been made and used for each\nentry of a in the table for both the pitch channel and the roll-yaw channel. Q sweeps\nthrough a wide range of values starting with 100 psf and ending at 15,000 psf. The\nvalues of M were chosen so that each entry of M and Q lie in the atmospheric tables\nused to compute p and V. As a result, the grid points are not rectangular. All\nentries are restricted to values between M = .6 and M = 3.5 and must correspond\nto altitudes between sea level and 50,000 ft.\nTable 6.1 shows the values of a used in the look-up table for both channels. M\nand Q sweep through all values of the grid previously mentioned for each value of a\nin Table 6.1. The spacing between the grid points was determined in a trial and error\nprocess. During the iterative procedure for computing feedback gains, the initial\nguess for each point came from the result of an adjacent grid point. The closer the\nspacing between adjacent points, the fewer iterations were needed to find the next\nfeedback gain. Numerical problems as described in section 2.4 and shown in figure\n2.10 were encountered. When this happened some of the constraints were relaxed\nso interior points in the feedback set could be found. Later these constraints were\n76\n\n53\nzero. Also, the weight of the missile is small in comparison to the aerodynamic force\nso gravity is neglected. The aerodynamic force in the x direction is much smaller\nthan the aerodynamic force in the y and 2 directions, and therefore, will be ignored\nfor the rest of the paper. Newtons second law of motion implies the following:\nFy = m[v + ru pw] (4.1)\nFz = m[w + pv qu\\ (4.2)\nFy and Fz are the aerodynamic forces in the y and 2 directions. The quantities\ninside the brackets are the total accelerations in the y and 2 directions. We know\nthat\nV = [u2 + 2 + w2}* (4.3)\nSince v and w are much smaller than u then\nV ~ u\nAngle of attack and sideslip are given by\na arctan\nft = arctan |\nIf we assume that a and ft are small, then\nw\na ~\nu\nand\nu\nEquations (4.1) and (4.2) can be rewritten as\nFy = mu\nV w\nb r-p\n.u u.\n(4.4)\n(4.5)\n(4.6)\n(4.7)\n(4.S)\n(4.9)\n\n96\na) Commanded Alpha (-) and Actual Alpha (--)\nFigure 8.5. Commanded and actual states, (a) a, (b) /3, and (c) p. for the scenario\nin figure 8.3.\n\n105\nvarying system. Under this restriction, the algorithm can be used to achieve local\nstability for a nonlinear system by formulating a linearized model around a time\nvarying operating point. Using this model, feedback gains are then computed as a\nfunction of parameters which characterize the operating point. There is no restriction\non the rate of change of the system parameters.\nIn stabilizing an time varying linear system, it is not known how to choose a\nconstant positive definite P, or even if a P exists so B'[(t)[A(t)P~1 + P~l AT(t)]Bj_(t)\nis positive definite. Further study in this area would be warranted.\n\nTable A. 19. Tabular data for C/v\nM\nCN;\n.8\n163.215\n1.2\n198.109\n2.0\n168.132\n2.7\n172.199\n3.5\n164.634\nTable A.20. Tabular data for Cm\nM\ncm.\n.8\n442.575\n1.2\n532.153\n2.0\n94.406\n2.7\n15.694\n3.5\n-27.525\nTable A.21. Tabular data for Cyr\nM\nCyr\n.8\n219.964\n1.2\n186.303\n2.0\n92.659\n2.7\n53.447\n3.5\n40.197\n\n86\nTable 7.1. The maximum eigenvalues of (6.1) for all test points at a 1.75.\nQ/M indices\n1\n2\n3\n4\n5\n6\n7\n8\n1\n-20.79\n-20.82\n-20.84\n-20.87\n-20.88\n-20.89\n-20.90\n-20.90\n2\n-20.68\n-20.69\n-20.70\n-20.72\n-20.74\n-20.76\n-20.76\n-20.76\n3\n-20.84\n-20.85\n-20.84\n-20.83\n-20.81\n-20.79\n-20.78\n-20.79\n4\n-20.59\n-20.61\n-20.49\n-20.30\n-20.18\n-20.33\n-20.34\n-20.21\n5\n-20.30\n-20.22\n-19.93\n-19.71\n-19.96\n-19.88\n-20.01\n-20.62\n6\n-20.41\n-19.97\n-19.77\n-20.16\n-19.88\n-20.20\n-20.79\n-20.97\n7\n-19.10\n-17.70\n-17.86\n-18.92\n-20.03\n-20.42\n-19.75\n-20.81\n8\n-18.62\n-19.79\n-20.67\n-20.37\n-16.16\n-19.14\n-20.21\n-19.48\n9\n-19.90\n-20.97\n-19.33\n-14.48\n-19.58\n-21.08\n-20.93\n-19.64\n10\n-20.23\n-20.88\n-11.67\n-16.89\n-19.96\n-20.40\n-19.72\n-19.41\n11\n-20.57\n-20.38\n-12.40\n-18.26\n-20.60\n-20.85\n-20.28\n-20.30\n12\n-20.77\n-17.82\n-17.66\n-20.97\n-21.85\n-21.65\n-21.14\n-21.32\n13\n-19.83\n-18.51\n-21.91\n-20.81\n-20.87\n-22.17\n-22.13\n-22.15\n14\n-20.45\n-22.23\n-20.57\n-17.42\n-20.84\n-22.21\n-22.42\n-22.39\n15\n-22.42\n-22.13\n-18.83\n-18.24\n-20.65\n-21.23\n-21.49\n-21.90\n16\n-21.82\n-19.45\n-15.17\n-20.54\n-21.09\n-20.92\n-20.94\n-21.73\n17\n-21.85\n-3.96\n-20.63\n-22.26\n-22.35\n-22.13\n-22.03\n-22.48\n18\n-18.88\n-19.74\n-21.48\n-22.29\n-22.58\n-22.65\n-22.68\n-22.87\n19\n-20.85\n-21.06\n-21.71\n-22.27\n-22.69\n-22.98\n-23.19\n-23.37\n20\n-21.65\n-21.93\n-22.29\n-22.63\n-22.93\n-23.19\n-23.39\n-23.52\n21\n-20.62\n-22.15\n-22.52\n-22.62\n-22.61\n-22.53\n-22.46\n-22.42\nFigure 7.3. A plot of the eigenvalues from Table 6.1.\n\nABSTRACT iii\nCHAPTERS\n1 INTRODUCTION 1\n1.1 Eci.rlier Works 3\n1.2 Purpose 10\n2 THEORY BEHIND THE DESIGN METHOD 11\n2.1 A Geometric Interpretation of Lyapunovs Linear Stability Theorem 11\n2.2 Adding Control to the System 18\n2.3 A Linear Feedback Set to Control xTPx 23\n2.4 One Linear Feedback Matrix to Control xTPx 34\n3 A TIME VARYING SECOND ORDER EXAMPLE 46\n4 DERIVATION OF A MODEL OF THE EMRAAT MISSILE 51\n4.1 The Nonlinear Model 51\n4.2 The Linear Model 57\n5 THE DEPENDENCE OF GAINS ON FLIGHT PARAMETERS 60\n5.1 Generating p and V from M and Q 60\n5.2 The Flight Parameter Generator 63\n5.3 Initializing the Iterative Lyapunov Design Method 63\n5.4 The Iterative Lyapunov Design Method 66\n5.5 Formulation of a State Tracker 71\n5.6 Comparing Gains with Flight Parameters 72\n6 COMPUTING LOOK-UP TABLES 76\n6.1 Determining a Grid of Points 76\n6.2 Formulation of :he Design Constraints 77\n6.3 Generating the Look-Up Table 80\nii\n\n49\nFigure 3.2. The time varying feedback set which satisfies the design constraints.\n\n20\nFigure 2.7. A case in which the normal velocity component cannot be controlled.\nTheorem 1\nConsider the system, x = A(t)x + B(t)u, where dim(x) = n, dim(u) = m\nand A(t) and B(t) have compatible dimensions. Let P be a constant\npositive definite matrix. The normal velocity component, xTPx, can be\narbitrarily set to any value with the right choice of u at a given time t,\nat any point x, on the spheroid, xTx = 1, unless x is contained in the set\nS(t) := P~1span(B(t)) fl {x : xTx = 1} (2.22)\nwhere B(t) is a basis of column vectors orthogonal to span(B(t)). If\nx S(t) then\nxTPx = xTPA(t)x (2.23)\nand this velocity component cannot be controlled at time t.\nTheorem 1 gives the parts of the unit spheroid for which the velocity component\nxTPx is uncontrollable and can therefore be used to determine if a linear time varying\n\n19\nFigure 2.6. A velocity vector (a) without control and (b) with control.\nTo answer question 1), consider a second order single input system frozen at some\ninstant in time.\nx=Ax + F?u (2.21)\nA is a 2 x 2 matrix, and B is a two dimensional column vector, u is a scalar input\nand can take on any real value. If u = 0 then the velocity field of the system is\nx = Ax. Figure 2.6a shows the velocity, x = Ax, of some state vector, x, in the\nsystem. The control vector, B, is also shown. When m / 0, x has the extra term,\nBu. The direction of Bu is constant, but its magnitude is directly proportional to u.\nFigure 2.6b shows many possible values for x by sweeping u through a wide range of\nvalues in small increments. As this diagram shows, the arrow head of each velocity\nvector can be placed on the line drawn parallel to B and intersecting the arrow head\nof Ax. This demonstrates that velocities can be controlled along the space spanned\nby the columns of B. The following theorem answers question 2).\n\nTable A. 10. Tabular data for Ci0\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n.22653\n.35501\n.19609\n.16015\n.10417\n-4\n.20408\n.26612\n.17228\n.12767\n.09510\n0\n.13163\n.12245\n.08447\n.04391\n.02323\n2\n.07857\n.03469\n.01228\n-.00701\n-.01779\n4\n-.02106\n-.06000\n-.05953\n-.06421\n-.05336\n8\n-.17469\n-.27837\n-.18083\n-.17934\n-.10744\n12\n-.24204\n-.33265\n-.22214\n-.22066\n-.12232\n16\n-.24408\n-.30857\n-.22214\n-.22030\n-.11416\n20\n-.20857\n-.27714\n-.19200\n-.19077\n-.10417\nTable A.ll. Tabular data for C¡Sp\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-.13341\n-.16085\n-.07361\n-.06506\n-.05441\n-4\n-.14801\n-.17360\n-.09268\n-.08207\n-.07059\n0\n-.15265\n-.17539\n-.10111\n-.09057\n-.07353\n2\n-.15044\n-.17204\n-.10288\n-.08725\n-.06949\n4\n-.14314\n-.16957\n-.10067\n-.07874\n-.05956\n8\n-.13119\n-.15459\n-.09224\n-.06617\n-.04265\n12\n-.12345\n-.13714\n-.08115\n-.06174\n-.03860\n16\n-.11062\n-.12371\n-.07251\n-.05989\n-.04669\n20\n-.10111\n-.11655\n-.06829\n-.05767\n-.04853\n\nkql 1alphaO\nhq11*alphaO kqirajphaO\n4\n4^\n\n93\nreason when sideslip changes. The two steps in the yaw rate, however, are due to\ncross coupling with the roll rate.\nThe rise times in table 8.1 are small and indicate that the missile will perform well\nin flight scenarios. While cross coupling effects are noticeable, they are not believed\nto be great enough to significantly hinder the performance of the missile; thus it was\ndecided to run the missile through a series of flight scenarios to see how well the\nmissile can intercept the target.\n8.3 Simulation of Flight Scenarios.\nA series of flight scenarios has been run to test the autopilot. Figure 8.3 shows\nthe trajectory of the missile and target for a flight scenario at an altitude of 20,000\nft. The miss distance is .64 ft. A hit is considered to be any miss distance under 10\nft. Figure 8.4 shows the commanded and the actual y and 2 accelerations. The devi\nations from the commanded normal accelerations are mostly due to a simplification\nused in implementing the BTT logic. Instead of using two aerodynamic coefficients,\nproportionality constants were assumed. The errors are not large enough to prevent\nthe interception of the target. The state tracking of the reference inputs appears to\nbe working well as seen in figure 8.5 and indicates that the autopilot is performing\nwell.\nThe trajectories of a 40,000 foot altitude scenario is shown in figure 8.6. The miss\ndistance is 0.05 ft. A similar scenario at 10,000 ft is shown in figure 8.7. The miss\ndistance is 2.8 ft.\nFigure 8.8 gives a case were a miss occurred. The scenario took place at 50,000\nft and the target was missed by 452 ft. Figure 8.9a and b shows that the missile was\nunable to achieve the desired 2 acceleration and angle of attack after 1.3 seconds into\nthe flight. This is because the elevator reached its 40 limit as figure 8.9c indicates.\n\n4\nAlpha vs. Time\nFigure 1.2. Angle of attack from one trajectory of the EMRAAT missile.\nbound on supt>0 ||A(i)|| is given which, if enforced, guarantees asymptotic stability\nof (1.1). As will now be shown, this limit is too restrictive for the EMRAAT missile.\nFigure 1.2 shows the angle of attack of the trajectory of a missile flown using an\nalready existing autopilot. Desoer proposes the Lyapunov function\nV'(x,i) = xT(£l/ + P(f))x (1.2)\nwhere ej > 0 and P(t) is chosen so that\nAT(t)P{t) + P{t)A{t) = -3/\nDesoer gives a bound on V,\nV < xTx[3 + 2eicim +\nwhere the following definitions are made:\nA/ 3m4\n4^0\na\\f := sup ||A(i)|| < oo\nt> o\n(1.3)\n(1.4)\n(1.3)\n\n23\nCase 2: x ^ S(t)\nSuppose we want to set xTPx = c where c is some arbitrarily chosen\nvalue. Then we want\nc = xtPA()x + xtPB()u (2.36)\nSince x S(t) then xTP5(t)u ^ 0 and there exists at least one u such\nthat 2.36 is true.\nOne could solve (2.36) for u to use as a control law, but this would not be practical\nto implement because A and B can not easily be computed. Linear state feedback is\nmore desirable. The next section will develop a procedure for computing the set of\nfeedback gains that will implement a feedback control law keeping the normal velocity\ncomponent, xTPx, within some specified limit.\n2.3 A Linear Feedback Set to Control xTPx\nRecall from Section 2.1 that the stability of a time varying linear system\nx = A(i)x (2.37)\ncould be analyzed by evaluating the bounds of x1 PA(t)x on the spheroid\nxTx = 1 (2.38)\nNow we want to find the set of all linear state feedback gains for the control law,\nu = -K(t)x (2.39)\nfor the system\nx = A(t)x + B(t)u (2.40)\nso that condition (2.19) holds for the closed loop system where c and c are now\nspecified. This section will give conditions for the nonemptiness of such a set followed\n\n41\nThe lower and upper bound of xTPx for the open loop system i\ntively,\nAo ^min[~^{P A + F P)\\\nAo = Amax[(P A + AT P)}\nThen,\nc0 = min[cf, Aq]\nc0 = max[cf, A0]\nStep 5)\nThe algorithm is now given.\nK = Ki\nFor j = 1 to m\nFor k = 1 to n\nPo j,k P kjQj,k\nPi = -(PBQj,k + QlkBTP)\nPo = P(A BKoj'k) + (A BKo,,k)TP 2cl\nSolve detfkPi + P0] = 0 in terms of k for c = c,, and c,-.\nE.eject all values which do not meet the constraints from\nsteps 2) and 6). The result is two intervals whose\nlower bound is ki_ and k2 and whose upper bound is\nk\\ and 2\nFind the intersection of these intervals by evaluating\nrespec-\n(2.120)\n(2.121)\n(2.122)\n(2.123)\n\nCHAPTER 4\nDERIVATION OF A MODEL OF THE EMRAAT MISSILE\nThe next section derives a nonlinear model of the EMRAAT missile. A linearized\nmodel is then generated in the following section and will be used for the design of the\nautopilot. Aerodynamic and inertial cross coupling are assumed negligible in order\nto reduce the order of the model. A linearized pitch model and a linearized roll-yaw\nmodel results. All assumptions are clearly stated.\n4.1 The Nonlinear Model\nBefore deriving the nonlinear model some variables and terms are defined. Figure\n4.1 shows the missile body coordinate frame of the EMRAAT missile . The three\naxes, x, y, and z, are fixed to the missile as shown. The velocity of the missile is\nrepresented by V. Angle of attack, a, is defined as the angle between x and the\nprojection of V onto the x-z plane. Sideslip, is the angle between x and the\nprojection of V onto the x-y plane. The velocity components, u, u, and to, are the\nprojections of V onto the x, y, and z axis respectively. The angle rates, p, q, and r,\nare named roll rate, pitch rate, and yaw rate respectively and are defined as the rate\nof rotation around the x, y, and z axis. Their directions obey the right hand rule.\nThese definitions are similar to those applied to aircraft as given by Etkin .\nThe following derivation of the nonlinear model is based on a similar model found\nin Smith [2 and in Koenig . Assumptions are made here to separate the system\ninto two lower order models: the pitch model and the roll-yaw model. We begin the\nderivation by starting with the force equations. The forces acting on the missile are\nthrust, gravity, and aerodynamic forces. The autopilot design in this paper applies\nto the second phase of the flight when the engine is no longer burning so thrust is\n51\n\n59\npr 21\nb\npr 11\nQSCySp\nmV\nQSdCip\n, CLpr22\nXX\nbprl2\nIxx\nQSCy6r\nmV\n, Clpr23\n\n5pr21\nQSdCn\nQSdCl6p QSdC[(r\nIxx\n\"? bpr22\nIxx\n'np QSdCnp QSdCnr\ndpr31 j i &pr32 j ? &pr33\nIzz\n)pT 31\nl QSdcnsp L\n^nr31 j ) Vpr32\nZZ\nIzz F-J Izz\nQSdCn6r\nIzz\nThe linearized model is given by\n0'\ndprll dprl 2 ^prl3\n/J'\nbprll bpri2\np\n=r\n&pr21 dpr22 dpT 23\nP\n+\n6pr21 bpr22\nr\nCtpr31 dpr32 dpr 33\nr\nbpr31 bpr32\n(4.49)\n\nTable A. 14. Tabular data for C¡.\nol/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n61.3158\n4.0263\n-15.2343\n-9.6414\n-10.8909\n-4\n74.2982\n40.0000\n-1.7188\n-5.3386\n-4.6446\n0\n86.3158\n79.4748\n9.0625\n0.0000\n2.2422\n2\n92.4561\n99.2560\n13.3594\n.1593\n5.2853\n4\n97.6316\n117.4617\n16.3281\n.9562\n7.6877\n8\n91.5789\n151.6849\n16.0156\n3.1076\n3.6036\n12\n86.8421\n179.9562\n15.6250\n4.3825\n0.0000\n16\n88.0702\n198.4245\n21.4063\n6.0558\n-1.2813\n20\n93.9474\n294.0919\n25.9375\n7.8884\n3.0430\nTable A.15. Tabular data for Cyp\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-9.2259\n-7.7656\n-3.9717\n-3.1351\n-2.0118\n-4\n-9.2417\n-7.6484\n-3.9717\n-3.1583\n-2.0197\n0\n-9.2338\n-7.7656\n-3.9152\n-3.1390\n-2.0039\n2\n-9.1945\n-7.9375\n-3.8799\n-3.0579\n-2.0039\n4\n-9.1866\n-8.0937\n-3.8233\n-2.9730\n-2.0039\n8\n-9.1945\n-8.3281\n-3.7385\n-2.8108\n-1.8619\n12\n-9.2259\n-8.3203\n-3.6396\n-2.7027\n-1.7357\n16\n-9.1081\n-8.2969\n-3.5830\n-2.6718\n-1.6331\no\nO\n-9.0688\n-7.9609\n-3.6254\n-2.6873\n-1.6095\n\n5\n(To is positive and\nRe A,[A(t)] < 2ar0 Vz, Vi > 0\nm is a constant and depends only on cr0 and aj^ so that\nAlso,\nexp(T/l(i))|| < mexp( cr0r), Vt > 0 Vi > 0\nhM := sup [|()||.\nz>o\nIf ex is allowed to be very small, then from (1.4), stability would result if\n4cr2\nau <\n77T\n(1.6)\n(1.7)\n(1.8)\n(1.9)\nFor the linear pitch model of the given trajectory, cr0 = 15.75. At .04 seconds into\nthe trajectory the closed loop matrix is\nA =\n-3.2533 .8997\n-1678.2 -83.50\n(1.10)\nAt this instant in time, m = 22.88 and au 2231. ^ = .0036. From this we see\nthat the linear system is changing much faster than the limit shown in (1.9). The\nsystem is not slowly varying. Note that m was only found for i = .04 sec. and not\nfor all time. If a greater m were found for the rest of the trajectory, then the limit in\n1.9 would be even more restrictive and the result would be the same. This test does\nnot show that the system is unstable. In fact, a Lyapunov function is known which\nshows that this linear system is stable but Desoers theory can not support this fact.\nWilson, Cloutier, and Yedavalli give a stability condition for a constant linear\nsystem with time varying uncertainties. Given a system,\nx [A + E(t)]x\n(1.11)\nif a positive definite P can be found which solves the Lyapunov equation.\nPA + AtP + 21 = 0\n(1.12)\n\n52\n\n99\nThe miss occurred due to the reduced elevator effectiveness from the low air density\nat high altitudes.\nAnother miss is shown in figure 8.10. Figure 8.11 shows that the desired z ac\nceleration was not achieved because the commanded angle of attack was at its 18\nlimit. The commanded normal acceleration was too great to be realized. Again, the\nautopilot performed as well as it could, but the miss occurred because of physical\nlimitations of the missile. The results of these simulations indicate that the autopilot\ndoes its job and, in reasonable scenarios, allows the missile to intercept the target.\n\nREFERENCES\n Robert F. Wilson, James R. Cloutier, and R. K. Yedavalli, Control Design for\nRobust Eigenstructure Assignment in Linear Uncertain Systems, Proceedings\nof the 30th IEEE Conference on Decision and Control, Vol. 3, pp. 2982-6, 1991.\n Roger L. Smith, An Autopilot Design Methodology for Bank-to-Turn Missiles,\nAir Force Armament Laboratory, Eglin AFB, August 1989.\n Jeff S. Shamma and Michael Athans, Analysis of Gain Scheduled Control for\nNonlinear Plants, IEEE Transactions on Automatic Control Vol. 35, No.8, pp.\n898-907, 1990.\n Jeff S. Shamma and Michael Athans, Gain Scheduling: Potential Hazards and\nPossible Remedies, IEEE Control Systems Magazine, Vol. 12, No. 3, pp. 101-7,\n1992.\n Jeff S. Shamma and Michael Athans, Stability and Robustness of Slowly Time-\nVarying Linear Systems, Proceedings of the 26th Conference on Decision and\nControl, Los Angeles, December 1987.\n Jeff S. Shamma and James R. Cloutier, Gain-Scheduled Missile Autopilot De\nsign Using Linear Parameter Varying Transformations, Journal of Guidance,\nControl, and Dynamics, Vol. 16, No. 2, pp. 256-63, 1993.\n Kevin A. Wise, Barry C. Mears, and Kameshwar Poolla, Missile Autopilot De\nsign Using Hoo Optimal Control with //-Synthesis, Proceedings of the American\nControl Conference, Vol. 3, San Diego, 1990.\n Mario A. Rotea and Pramod P. Khargonekar, Simultaneous H2/H00 Optimal\nControl with State Feedback, Proceedings of the American Control Conference,\nVol. 3, San Diego, 1990.\n Charles A. Desoer, Slowly Varying System x = A(f)x, IEEE Transactions on\nAutomatic Control, Vol. AC-14, pp. 780-781, December 1969.\n M. Vidyasagar, Nonlinear Systems Analysis, Prentice-Hall, Englewood Cliffs,\nNJ, 1978.\n Wolfgang Hahn, Stability of Motion, Springer-Verlag, New York, 1967.\n S. M. Shahruz and S. Behtash, Design of Controllers for Linear Parameter-\nVarying Systems by the Gain Scheduling Technique, Journal of Mathematical\nAnalysis and Applications, Vol. 168, pp. 195-217, July 1992.\n118\n\n6\nFigure 1.3. Angle of attack from one trajectory of the EMRAAT missile,\nthen the system will be asymptotically stable if\nmax[E{t)\\ < l- Vi (1.13)\n&max\\E )\nwhere crmax is the maximum singular value. This idea is initially interesting because\nit requires the error, E(i), to be bounded, but does not constrain the time variation.\nThe above result can not be easily applied to the EMRAAT missile. This will be\nshown by applying the result to a trajectory of the EMRAAT missile whose angle of\nattack is shown in figure 1.3. In order to check the stability of the EMRAAT missile,\nA must be chosen since it may be any matrix which is stable. It is believed that the\nbest choice of A should come from a closed loop matrix in the trajectory. With this\nin mind, A was taken at t = 2 sec into the trajectory.\n-4.477 .9526\n-1.684 -79.47\nA-BK =\n(1.14)\n\n103\na) Commanded z Acceleration (-) and Actual z Acceleration (--)\nc) Elevator Deflection Angle\nFigure 8.11. Commanded and actual accelerations (a) in the 2 direction, (b) com\nmanded and actual angles of attack, and (c) elevator deflection angle for the scenario\nshown in figure 8.10. Saturation of the a command resulted in the miss.\n\nCHAPTER 7\nGAIN SCHEDULING\nThe result of the previous chapter is thirteen tables of gains in terms of mach\nnumber, dynamic pressure, and angle of attack. This chapter discusses the process\nof curve fitting used to implement the look-up table. The results of a test of this\nscheme will follow.\n7.1 Curve Fitting\nIt was decided to use a combination of polynomial fitting and interpolation to\nimplement the autopilot. Third order polynomials were fit to the tables as a function\nof mach number. However, low order polynomials could not achieve close fits as a\nfunction of angle of attack or dynamic pressure, so linear interpolation was used for\nthese two variables. Figure 7.1a shows kqU as a function of M for various constant\nvalues of Q at a = 8. An example of polynomial fitting of one of these curves is\nshown in Figure 7.1b where Q = 150psf. A polynomial has been made for every\nol-Q pair and the polynomial coefficients are interpolated as a function of these two\nvariables. The tables of the three pitch controller gains each have 7 entries for a\nand 22 entries for Q. Since each polynomial has 4 coefficients the total number of\ncoefficients for each gain is 7 x 22 x 4 = 616. Similarly the tables for the roll-yaw\ncontroller have 8 entries for a and the same number of entries for Q. Each of the ten\ngains are then scheduled using 8 x 22 x 4 = 704 polynomial coefficients.\n7.2 Testing the Fit\nFigure 7.2 gives the system for testing the polynomial and interpolation routines.\nGains were generated from these routines at locations centered between the original\ngrid points. These new locations were found by taking the average of the coordinates\n83\n\n34\n2.4 One Linear Feedback Matrix to Control xT.Px\nThe search procedure given in section 2.3 becomes impractical for high order\nsystems with multiple inputs because the number of grid points for the fixed entries\nof K becomes very large. This section discusses an iterative Lyapunov design method\nwhich saves on computations and finds one element K /C, if it exists, where K, =\n1C fl 1C. By applying this procedure at every instant in time to a time varying linear\nsystem, one can find a control law stabilizing the system if a constant positive definite\nP exists which satisfying the following condition.\n^BTL(t)[A(t)P~l + P~1AT(t)\\BL(t) < 0 Vi (2.107)\nBy Theorem 2, this condition guarantees the existence of a c < 0 such that K{t)\nis nonempty for every instant in time. This procedure applies to all systems which\ncan be stabilized with respect to a Lyapunov function given by a constant positive\ndefinite P. The following is a discussion of the iterative Lyapunov method followed\nby the algorithm itself. An example is then given applying this procedure to one\noperating point of a fifth order linearized model of the EMRAAT missile.\nFigure 2.9 illustrates the iterative procedure. fCf is the feedback set which satisfies\nthe designers predetermined constraints for some specified P. The constraints are\n£f < xt[PA + AtP PBK KTBTP\\x Let ICi be defined as the set of all K which satisfies the following constraints.\nSU < xt[PA + AtP PBK KtBtP]x Given Ki, we would like to find c, and c, so that if Ki ^ 1Cf then Kz 6 d)Ci. We also\nrequire that ICf C K{. If Kz £ JCf, then we want JC{ = K¡. The following definitions\nfor Cj and c, meet these requirements. Let\nA, := max -x.T[P(A BICC + (A BKCTPht\n11*11=1 2\n(2.110)\n\n26\nProof:\nWe will show the nonemptiness of 1C. 1C can be shown to be nonempty in\na similar manner.\nCase 1:\nnr\nmaxx PAx < c\nxeS\nthen\nxTPAx < c V x S.\n(2.54)\n(2.55)\nLet K = kBT P where k is a scalar value. Then\nxTPx = ^xt[PA + AtP PBK KTBTP]x (2.56)\n= \\xT[PA + AtP]x ^k[xTPB}[BTPx\\ (2.57)\nSince xtPB = [BTPx]T then\nxtPBBtPx> OVx^S, VxTx = 1 (2.58)\nIf (2.55) is true and if k is made large enough, then the second term in\n(2.57) will dominate V x S and xTPx < c V x : xTx = 1. Since\nkBTP e X then X is nonempty.\nCase 2:\nmaxxTJDAx > c (2.59)\n-x.es '\nThen\nnr\nx P Ax > c for some x G S.\n(2.60)\nBy Theorem 1, xPAx can not be controlled V x £ <5; so 1C is empty.\n\nBIOGRAPHICAL SKETCH\nKurt Walter Koenig was born on July 30, 1967 in Rolla, Missouri. He graduated\nwith high honors from the University of Florida with a bachelors degree in electrical\nengineering in December of 1989. He received a Florida Graduate Scholarship in\n1990 to pursue a masters degree in electrical engineering at the University of Florida\nwhich he received in December of 1991. He was admitted into the Palace Knight\nprogram in 1992 and received support from the United States Air Force to work on\na Ph.D. in electrical engineering at the University of Florida.\n120\n\n92\nFigure 8.2. A test of state tracking at an initial altitude of 20,000 ft with an initial\nmach number of 2.0. The states shown are (a) a, (b) /3, (c) p, (d) q. and (e) r.\n\n48\nThe uncontrollable velocity for this example is constant for all time. In selecting the\nconstraining values, we must have\nc < -.9548 < c (3.9)\nc -2 (3.10)\nc = .5 (3.11)\nSince the system is second order and has only one input it is possible to plot the set\nof all feedback gains which satisfy the following.\nc < ^xT[P(A(t) B(t)K(t)) + (A(t) B(t)K(t))TP]x < c Vx : xTx = 1, Vi (3.12)\nThis time varying feedback set is shown in figure 3.2. Since the time varying nature\nof the system is periodic, and from inspection of figure 3.2, the following control law\nis chosen.\nu = 3[cos(i) sin(i)]x (3.13)\nThe feedback matrix in (3.13) is shown to be inside the moving feedback set in figure\n3.2. The eigenvalues of the derivative of the resulting Lyapunov function is\nA[A(<) B{t)K{t) + AT{t) KT{t)BT{t)} = -1.1193, -0.8579 Vi (3.14)\nTherefore, the closed loop system is stable. Figure 3.3 shows a trajectory of the\nLyapunov based closed loop system where the initial condition is x0 = [1 0]T.\n\n17\npoint on the ellipse, and its velocity, x, is also shown. The normal to the ellipse at x\nis Px. To ensure stability, it is sufficient to require that the projection of x onto the\nnormal of the ellipse, Px, is negative everywhere. The resulting term, xTPx, is the\nnormal velocity component on the ellipse. The main thrust of this paper is to be able\nto control xTPx and to make this normal velocity component negative everywhere\nfor all time in order to achieve stability. As before, the linearity of the system allows\none to only check xTPx on the unit circle. This generalizes the previous stability\ncondition to\nStability Condition 2: For the system x = A(t)x, if xTPx = xTPA(i)x <\n0 V x : xTx =1, V t > 0 for some constant positive definite matrix P,\nthen the origin is a global asymptotically stable equilibrium point.\nThe above condition applies to linear systems of any order.\nLyapunov stated that if any positive definite function of the system states\nF(x) is always decreasing, i.e. V(x) < 0, then the system is stable (asymptotically\nstable for linear systems) . For the linear case, let\nF(x)=xTPx (2.10)\nV(x) is positive definite if and only if P is a positive definite matrix. Taking the\nderivative gives\nV(x) = xTPx + xTPx\n(2.11)\n= xTArPx + xTPAx\n(2.12)\n= xt[AtP + PA]x\n(2.13)\nLyapunovs stability condition becomes\nxt[AtP + PA]x <0 V x, V t > 0\n(2.14)\n\n15\nFigure 2.3. x and x\nall spheroids centered on the origin. This leads to the following stability condition\nwhich is given without proof.\nStability Condition 1: If xTx = xTAx < 0 V x : xTx = 1, V i > 0 for\nthe system, x = A(i)x, then the origin is a global asymptotically stable\nequilibrium point.\nFigure 2.4 illustrates an example of a second order system which meets Stability\nCondition 1. Note, that while all figures given so far represent second order linear\nsystems, the above stability condition applies to linear systems of any order.\nThe above stability condition can be made less conservative by expanding the\nclass of shapes to ellipsoids defined in the following way.\nxTPx = 1 (2.9)\nwhere P is a symmetric, positive definite matrix. Figure 2.5 shows an ellipse of the\nform (2.9) defined for a second order system. The state vector, x, is drawn to some\n\n35\nFigure 2.9. A geometric view of the iterative Lyapunov design method.\n\nCHAPTER 9\nCONCLUSION\nA controller design algorithm has been made which can stabilize time varying\nlinear systems. The motivation behind this work was the need to achieve local sta\nbility of a nonlinear system by stabilizing a linearized model which is a function of\ntime varying system parameters. This work began by studying second order linear\ntime varying systems that were stable when frozen at any given time but were un\nstable when allowed to vary with time. Geometric techniques were used to rederive\nLyapunovs linear stability theorem. These techniques were carried further to give a\ncondition for the existence of state feedback which places an upper and lower bound\non the eigenvalues of the derivative of a given Lyapunov function. This led to a design\nalgorithm which makes the derivative of the Lyapunov function negative definite. A\ntime varying second order system was then stabilized when ordinary pole placement\ntechniques failed. The algorithm was then applied to a linearized model of the EM-\nRAAT missile and used to formulate an autopilot. Feedback gains were generated\nalong a grid of flight parameters. The gains were scheduled against angle of attack,\nmach number, and dynamic pressure. The resulting autopilot was tested using a non\nlinear simulation. The missile was able to intercept the target in reasonable scenarios\nand flight conditions.\nThe significance of the proposed design algorithm is that no assumption is made\non time invariance of the system. However as stated in chapter 2, this algorithm\ncan only be used to stabilize a system if there exists a constant positive definite P\nsuch that ()[A()P_1 P~lAT(t)]BjL(t) is negative definite for all time. If this\nis true then a K(t) can be found so that xTPx is a Lyapunov function for the time\n104\n\n30\nThen (2.74) becomes\ndet[ kP1 + P0] = 0 (2.78)\nSolve (2.78) for k.\n3) Reject all complex roots. If all the roots are complex then skip the\nnext step.\n4) Test the intervals between the real roots by checking to see if\nAmax[PA + AtP Tel PBK KtBtP] < 0 (2.79)\nThe K's that bound the interval which satisfies (2.79) lie on the boundary\nof 1C. Convexity of 1C implies that no more than two K's bound this\ninterval.\n5) Repeat the process for all possible values for the fixed entries in K.\nThe result is dlC.\ndjC can be computed in a similar way by reversing the inequalities in the above\nprocedure and by replacing \\max with Amn. To find 1C the intersection of 1C and 1C\ncan be found in step 4).\nThe fixed entries in step 5) must be assigned to a finite number of grid points if\nthe above procedure is to be executed on a real system. The spacing of these grid\npoints must be smaller than the size of the feedback set. If the system has a high\norder or a multiple number of inputs, the number of grid points will become too large,\nand it will not be practical to implement this method. Understanding this procedure,\nhowever, leads to the formulation of an iterative method that can be used on high\norder, multiple input systems and will be presented in the following section. First,\nan example is given applying the exhaustive search method to a second order, single\ninput system.\n\n64\nTable 5.1.\n'he initial point and range of changing flight parameters\nInitial point\nMinimum value\nMaximum value\nM\n2\n1\n2.6\nQ\n1250 psf\n700 psf\n5000 psf\na\n8\n-8\n20\nP\n0\n0\n10\nP\n0 /s\n0 /s\n500/s\nQ\n10 /s\n1\nH-\no\no\n0)\n200/s\nr\n0 /s\n0 /s\n200/s\nThe initial feedback gains are found by using a pole placement algorithm. At the\nnominal flight condition, the linear models are given by\n4 =\n-1.1345 0.9996\n-261.4732 0.6209\n,Bq =\n-0.1463\n-123.1091\nA\nr-Lrpr\nL\nL\n-0.459\n0.140\n-.100\n0.018\n0.117\n-2255.5\n-2.41\n.066\n, Bpr\n-1173.5\n-1335.6\n73.0\n-0.181\n-.648\n2.01\n-114.4\n(5.13)\n(5.14)\nThe desired eigenvalues for the closed loop pitch dynamics have been chosen to be\n40jl0. For the closed loop roll-yaw model the desired pole locations are 20j5,\n80. The resulting feedback gains are\nKq =\n-11.1633 -.6233\n, KpT\n-1.4166 -.0758 .3758\n2.9337 .0086 -.3300\n(5.15)\nFor both closed loop systems, P must be found so that xTPx is a Lyapunov\nfunction. The following problem is stated.\nGiven a stable linear system x = Ax, find a positive definite function,\nV = xTPx so that V = xt(PA + ATP)x is a negative definite function.\nLet A be put into Jordan canonical form.\nA = SJS\n(5.16)\n\n75\nTable 5.5. Extreme values and range of the roll-yaw channel control terms as indi\nvidual flight parameters vary. Gains depend mostly on M, Q, and a.\nM\nQ\na.\nP\nP\n<7\nr\ncontrol\nmin\nmin\nmin\nmin\nmin\nmin\nmin\nterm\nmax\nmax\nmax\nmax\nmax\nmax\nmax\ndiff\ndiff\ndiff\ndiff\ndiff\ndiff\ndiff\nkprllfio\n-.050\n-.11\n-.22\n-.050\n-.050\n.050\n.050\n.022\n.045\n-.013\n-.038\n-.036\n-.036\n-.036\n.072\n.157\n.21\n.011\n.014\n.014\n.013\nfcprl2PQ\n-.17\n-.14\n-.29\n-.13\n-.13\n-.13\n-.13\n-.011\n-.061\n-.13\n-.13\n-.13\n-.13\n-.13\n.16\n.20\n.15\n.001\n.001\n.001\n.001\n^prl3^0\n.074\n.047\n.15\n.16\n.16\n.16\n.16\n.18\n.28\n.25\n.16\n.17\n.17\n.17\n.10\n.23\n.099\n.0006\n.001\n.001\n.0009\nkpr 2l/^0\n.026\n.0062\n.065\n.093\n.092\n.092\n.092\n.11\n.167\n.15\n.10\n.101\n.101\n.101\n.080\n.16\n.084\n.008\n.009\n.009\n.009\nkpr22Po\n-.054\n-.18\n.0015\n.014\n.014\n.014\n.014\n.015\n.015\n.152\n.014\n.015\n.015\n.015\n.068\n.20\n.15\n.0007\n.0008\n.0005\n.0007\nkpr 23^0\n-.151\n-.25\n-.24\n-.15\n-.15\n-.15\n-.15\n-.067\n-.034\n-.11\n-.14\n-.14\n-.14\n.14\n.084\n.22\n.12\n.001\n.002\n.001\n.002\nkref prll/^cO\n-.14\n-.21\n-.16\n-.14\n-.14\n-.14\n-.15\n-.064\n-.050\n-.13\n-.13\n-.13\n-.13\n-.13\n.08\n.16\n.028\n.011\n.014\n.013\n.013\nkref prl2PcO\n-.071\n1\nO\n-.39\n-.040\n-.040\n-.045\n-.040\n.029\n.15\n.056\n-.039\n-.040\n.013\n1\nO\nCO\n00\n.10\n.19\n.45\n.001\n.001\n.057\n.002\nkref pr2lfic0\n.055\n.030\n.11\n.12\n.12\n.12\n.12\n.126\n.19\n.17\n.13\n.13\n.13\n.13\n.071\n.16\n.058\n.008\n.009\n.010\n.009\nkref pr22PcO\n-.010\n-.33\n-.17\n-.071\n-.071\n-.12\n-.071\n-.070\n-.070\n.24\n-.070\n-.070\n-.066\n-.070\n.027\n.26\n.41\n.001\n.001\n.051\n.002\n\nTable A.8. Tabular data for Ci6r\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-.11703\n-.11700\n-.08247\n-.06413\n-.04209\n-4\n-.12446\n-.12390\n-.08446\n-.05906\n-.04209\n0\n-.13189\n-.12870\n-.08705\n-.06123\n-.04420\n2\n-.13131\n-.13020\n-.08904\n-.06304\n-.04638\n4\n-.13249\n-.13540\n-.09223\n-.06775\n-.04964\n8\n-.13738\n-.14780\n-.09900\n-.07536\n-.05761\n12\n-.14305\n-.16186\n-.11036\n-.08804\n-.06812\n16\n-.14932\n-.16660\n-.11813\n-.10181\n-.08188\no\nO\n(M\n-.14266\n-.15690\n-.13008\n-.11775\n-.09565\nTable A.9. Tabular data for CySr\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n.1635\n.1386\n.06877\n.04874\n.04270\n-4\n.1639\n.1331\n.06996\n.05054\n.03745\n0\n.1649\n.1357\n.07036\n.05269\n.04007\n2\n.1608\n.1408\n.07194\n.05305\n.04232\n4\n.1535\n.1474\n.07292\n.05556\n.04719\n8\n.1452\n.1494\n.07273\n.06201\n.05618\n12\n.1467\n.1553\n.07668\n.07168\n.06780\n16\n.1414\n.1498\n.08360\n.08208\n.08052\no\nO\n.1101\n.1231\n.09368\n.09570\n.08951\n\n42\nk = max[ku k2\\\nk min[ki, 2],\nFind the midpoint of the interval by computing\nk j,k = \\(k+k)-\nLet I\\ kj,kQj,k d- Kqj,k\nNext k\nNext j\nKi+i = K\nStep 6)\nLet\nA = Xmin[\\(PA + ATP-PBKi-KTBTP)\\ (2.124)\nA = Amax[i(PA + ATP-PBKi-KjBTP)] (2.125)\n&\nSo,\nCj = min[cfi A] (2.126)\nCi = max[c/,A] (2.127)\nThe following example illustrates the feasibility in applying this method to the\nEMRAAT missile.\nExample 2\nWe would like to apply the iterative Lyapunov design method to the EM\nRAAT missile. The missile was flown in a simulation through a trajectory\n\nAPPENDIX B\nINERTIAL DATA FOR THE EMRAAT AIRFRAME\nThis appendix gives the weight and moments and products of inertia of the EM\nRAAT missile with no fuel.\nMissile Weight\nW = 227 lb\nMoments of Inertia\nIxx = 1-08 slug-ft2\nIyy = 70.13 slug-ft2\nIzz = 70.66 slug-ft2\nProducts of Inertia\nIxy = 0.274 slug-ft2\nIxz = 0.704 slug-ft2\nIyz 0.017 slug-ft2\n117\n\n7 GAIN SCHEDULING 83\n7.1 Curve Fitting 83\n7.2 Testing the Fit 83\n8 NONLINEAR SIMULATIONS 88\n8.1 The Nonlinear Simulation 88\n8.2 A Test of State Tracking 90\n8.3 Simulation of Flight Scenarios 93\n9 CONCLUSION 104\nAPPENDICES\nA AERODYNAMIC DATA FOR THE EMRAAT AIRFRAME 106\nB INERTIAL DATA FOR THE EMRAAT AIRFRAME 117\nREFERENCES 118\nBIOGRAPHICAL SKETCH 120\n\n38\na b\nkX2 k\\2\nFigure 2.10. Step 5 of the iterative Lyapunov design method for (a) a second order\nexample, (b) how it sometimes fails, and (c) how this problem is corrected.\n\nTable A.2. Tabular data for Cm\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n6.976\n12.336\n11.376\n8.445\n7.472\n-4\n3.488\n6.168\n5.688\n4.222\n3.736\n0\n0.000\n0.000\n0.000\n0.000\n0.000\n2\n-1.746\n-3.084\n-2.844\n-2.111\n-1.868\n4\n-3.488\n-6.168\n-5.688\n-4.222\n-3.736\n8\n-7.762\n-11.238\n-10.662\n-8.174\n-6.908\n12\n-11.802\n-15.530\n-15.074\n-11.439\n-9.972\n16\n-14.880\n-19.330\n-19.282\n-14.603\n-13.624\nO\nCM\n-16.030\n-25.700\n-29.560\n-24.218\n-25.930\nTable A.3. Tabular data for C^6q\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n.153\n.195\n.095\n.077\n.048\n-4\n.181\n.215\n.106\n.073\n.055\n0\n.197\n.219\n.107\n.069\n.056\n2\n.199\n.212\n.104\n.068\n.055\n4\n.198\n.207\n.101\n.067\n.051\n8\n.199\n.195\n.091\n.056\n.046\n12\n.199\n.175\n.076\n.047\n.044\n16\n.188\n.161\n.064\n.044\n.047\n20\n.184\n.150\n.056\n.047\n.055\n\n119\n Hassan K. Khalil, Nonlinear Systems, p. 175, Macmillan Publiching Company,\nNew York, 1992.\n A. M. Lyapunov, Stability of Motion, Academic Press, New York, 1966.\n Shannon Fields, unpublished Ph. D. work, private communication, University of\nFlorida, February 1994.\n H. L. Royden, Real Analysis, The Macmillan Company, New York, 1964.\n Bernard Etkin, Dynamics of Flight Stability and Control, Second Edition, John\nWiley and Sons, New York, 1982.\n Kurt W. Koenig, The Design of a Robust Autopilot for an Air-to-Air Missile,\nM.S. Thesis, University of Florida, December 1991.\n Arthur E. Bryson and Yu-Chi Ho, Applied Optimal Control, p. 155, Hemisphere\nPublishing Corporation, Washington, 1975.\n\n54\nFz mu\n'w v\n+P <7\nLu u\nwhich simplifies to\nFy = mV\nFz = mV\n'v\nf- r pa.\n.u\n'w .\n+ p/3 q\nL u\nWe assume that the forward velocity changes slowly so that\n~ 0.\nThen\nand\nSo (4.11) and (4.12) become\nw\na ~\nu\nP=~\nu\nmV\nFz_\nmV\n= (3 + r pet\nL + p/3 q\nThe aerodynamic forces are given by\nFy = QS [cYl,0 + CYp +CYrr + Cr6r + CyJ,\nFz = QS Cjvaa + Cn0c* + Cjv9 Q is the dynamic pressure and is defined as\nQ = \\pv2\n(4.10)\n(4.11)\n(4.12)\n(4.13)\n(4.14)\n(4.15)\n(4.16)\n(4.17)\n(4.18)\n(4.19)\n(4.20)\nwhere p is the air density. S is the surface area of the wing. The aerodynamic\ncoefficients come from wind tunnel tests. They depend on mach number, and some\ndepend on angle of attack. The values of these coefficients have been put in tabular\n\n20 >\nFigure 6.3. kqU verses M and Q when a = 8\n\n47\nTrajectory of a pole placement time varying system\nFigure 3.1. The trajectory of the above closed loop system based on pole placement.\nThe initial conditions are x0 = [1 0]T. The system is unstable.\nIf the initial conditions of the system are Xo = [1 0]r, the resulting trajectory is\nunstable as figure 3.1 shows.\nWe now turn to the Lyapunov design method. We choose P to be the identity\nmatrix. Before giving the design constraints, we need to check the the value of the\nuncontrollable normal velocity components for all time. At t = 0\nB = [.9848 .1736]t (3.6)\nSince P is the identity matrix, we are interested in the normal velocity component\nwhich is on the part of the unit circle whose tangent is parallel to B. So we let\nx = Bl = [.1736 .98481er (3.7)\nThe uncontrollable normal velocity component at t = 0 is\nxTPx = ^xt[PA(0) + .4r(0)P]x = -.9548\n(3.8)\n\n102\na) Top View of Missile (-) and Target () Trajectory\nFigure 8.10. (a) Top view and (b) side view of missile and target trajectories of a\nscenario which occurred at 35,000ft. The target was missed 767 ft. Initial mach\nnumbers of the target and missile are 1.9 and .92 respectively.\n\n73\nTable 5.4. Extreme values and range of the pitch channel control terms as individual\nflight parameters vary. Gains depend mostly on M, Q, and a.\nM\nQ\na\n0\nP\n<7\nr\ncontrol\nmin\nmin\nmin\nmin\nmin\nmin\nmin\nterm\nmax\nmax\nmax\nmax\nmax\nmax\nmax\ndiff\ndiff\ndiff\ndiff\ndiff\ndiff\ndiff\nkqnao\n-2.15\n-3.1\n-2.1\n-1.60\n-1.61\n-1.61\n-1.60\n-.72\n-.16\n-1.4\n-1.59\n-1.59\n-1.59\n-1.59\n1.43\n2.95\n.69\n.017\n.017\n.017\n.017\nkquQo\n-.1509\n-.21\n-.16\n-.118\n-.118\n-.118\n-.118\n-.0529\n-.031\n-.10\n-.117\n-.117\n-.117\n-.117\n.099\n.179\n.054\n.0005\n.0005\n.0005\n.0005\nkref qll^cO\n-2.53\n-3.5\n-2.7\n-1.98\n-1.98\n-1.98\n-1.98\n-.916\n-.53\n-1.7\n-1.97\n-1.96\n-1.96\n-1.96\n1.61\n3.0\n.94\n.017\n.017\n.017\n.017\nand look at kqncio. We set q0 and equal to 10/s and 8 respectively so that\nwe can look at kqi2q0 and kref 9nac0. The sum of these three terms are fed into to\nthe elevator. For the terms which are fed into the remaining inputs, the following\n/% = &o = 2\nPo = Pco = l00/s (5.52)\nr0 = 25/s\nFigures 5.2a-g show kqUa0 plotted against all seven flight parameters. These seven\nfigures show that kqnflo changes with M, Q, and a but remains nearly constant when\n0, Pi q, and r change. Similar figures exist for the remaining eleven gains and are\nsummarized in Tables 5.4 and 5.5. The minimum and maximum values of each term\nis listed for each changing flight parameter along with the difference between the\nminimum and maximum values. Angles are expressed in radians. From this table it\nwas determined that all gains will be scheduled against M, Q, and a.\n\n22\nalways a control, u, which can place x inside the ellipsoid. This is true because the\nvelocity field can always be controlled in the B direction. This is not true, however,\nfor points on the ellipsoid whose normal is parallel to span{B\\_) as shown in figure\n2.7. These points on the ellipsoid can be found by computing\nP_1 span(B _l) (2.29)\nwhere B_ is a basis of column vectors orthogonal to span(B). We now give the proof\nof Theorem 1.\nProof:\nCase 1: x S(t)\nSubstituting 2.20 into xTPx, we have\nxTPx = xTPA()x + xTPfi(i)u (2.30)\nSince xe5 then\nx = p-'liB^t)) (2.31)\nwhere l(Pj_(£)) is any linear combination of Bj_(t). So\nxTfifi() u = l(P1(i))rp-1PP(i)u (2.32)\n= \\(B(t))TB(t)u (2.33)\n= 0 (2.34)\nand therefore from 2.30 we have,\nxTPi = xTPA(i)x (2.35)\n\n72\nBecause (5.45) is true for any v, then\nI = C(A BK)~lBKref\n(5.46)\nControllability of the system implies that C(A BK) 1B is invertible. Solving (5.46)\nfor Kref gives\nKref = ~[C(A BK)-lB]~l (5.47)\nThe EM R A AT missile has three inputs and therefore only three states can be\ntracked. Controlling a in the pitch model and ¡3 and p in the roll yaw model is\ndesirable. For the pitch model y = a, implying that\nCg = [ 1 0 ] (5.48)\nand, therefore,\nKrefq = ~[Cq(Aq BqKq)-1 Bq]~l (5.49)\nFor the roll-yaw model\nand, thus,\n1 0 0\n0 1 0\nB-cfpr [Cr)rf Apr Bprhpr) B\n1-1\npr J\n(5.50)\n(5.51)\nKrefq and Krefpr are computed for each flight condition and then compared along\nwith Kq and Kpr to the flight conditions.\n5.6 Comparing Gains with Flight Parameters\nA series of gains have been generated as a function of different flight conditions.\nEach flight parameter has been swept through a range of points while the remaining\nsix have been held constant. In order to compare different gains on the same input\nit has been decided to use the products of gains and their corresponding terms at\ntypical values. For example a typical value of a is 8. So we set ao equal to 8\n\nTable A.22. Tabular data for CUr\nM\ncnr\n.8\n-1857.534\n1.2\n-1854.110\n2.0\n-1182.192\n2.7\n-937.671\n3.5\n-849.315\nTable A.23. Tabular data for C\\p\nM\nCip\n.8\n-91.499\n1.2\n-114.634\n2.0\n-67.057\n2.7\n-47.486\n3.5\n-35.671\n\n31\nExample 1\nGiven the system\nwhere\nand\nlet\nx = Ax + B u\n(2.80)\nA =\nB =\n-10 12\n0 -1\ncos(lb)\nsin(75)\nP = I\nWe want to find all feedback gains which satisfy the following constraint.\nc < x1 P < c Vx:x1x = l\n(2.81)\nBefore specifying c and c, we must check the normal velocity components\nfor x E xi = Bl =\nsin 75\ncos 75\nThen\nxf Axi = 12.40\n(2.82)\n(2.83)\nFrom Theorem 2 we must have\nc < -12.40 < c\n(2.84)\nFrom this we choose\nc = -14\nc = 9\n(2.85)\n(2.86)\n\n36\nand\nA := mm ^xT[P(,4 BKt) + (A BI Then let\nc, = max[cj,\\i] (2.112)\nand\nc, = min[cf, A,] (2.113)\nKq is the initial guess in the search for Kf Kf. c0 and Cq are computed using\n(2.112) and (2.113) so that Kq is a member of dJCo and Kf C Kq. Then a new\nfeedback matrix, K\\ is found which lies inside of Kq, but not on the boundary. New\nconstraining values are found in the same way as before so that the boundary of the\nnext feedback set contains K\\. K% is then found so that it lies inside of the present\nfeedback set, but not on its boundary. This process is continued until Kl = Kf Kf.\nThe success of finding Kf depends on the following conditions.\n1. Given that Ki dKi we must be able to find fG+i such that K{+1 G K{.\n2. We must show that c,+i < c when c > c/ and ci+l > c when c{ < Cf.\n3. We must show that Kf C Ki+1 C Ki.\n4. Kf must be nonempty.\nWe now address these four points.\n1. Given Ki dKi, we need to find a second feedback matrix K2 0K{ where\nK2 Ki. Then, due to convexity, \\Ki + \\Ka is a member of Kt. Figure 2.10a shows\na second order example of a procedure for finding Ki+i. The algorithm will be given\nshortly. The horizontal and vertical axis are assigned to kn and 12 respectively. The\nregion Ki is enclosed by 'dK; and dKi- Ki is known. Ki2 is found by searching along\n\n89\nFigure 8.1. A Block Diagram of the Nonlinear Missile Simulation.\n\n39\nthat c, < ci+1 when ct < Cj.\n3. We now show that ICf C JCi+1 C /C. Since c = max[c/, A], then c,- > c/. We have\nalready seen that ct > A+i. So c > max[c/, A+1] = C{+\\. Similarly c < c+1. XA+1 is\nthe set of all K so that\nci+1 < ^xT[P(A BK) + (A M)rP]x < Ci+i V x : xTx = 1 (2.117)\nSince Ci > Ci+i and c < c+1, then for every element of the following holds.\nc,- < ixT[P(^ PA\") + (A PA\")tP]x < c V x : xTx = 1 (2.118)\nTherefore, JCi+1 C /C. Since c+i > Cj and ci+1 < Cf. then using a similar argument\nICf C ICi+1.\n4. In using this design procedure, P is chosen so that the maximum uncontrol\nlable normal velocity component is negative. Then from Theorem 2, c/ can be made\nnegative in an attempt to achieve stability. c¡ must be greater than the maximum\nuncontrollable normal velocity component, and c¡ must be less than the minimum\nuncontrollable normal velocity component. From Theorem 2 this will guarantee the\nnonemptiness of ICf and ICf. The nonemptiness of the intersection of these two sets,\nhowever, is unknown. If ICf is nonempty, then, as i becomes large, A\", £ ICf. If ICf\nis empty then cl and c will converge to values which do not match the desired con\nstraints and K{ will yield a closed loop system that meets the constraints given by c\nand ct. The designer will either have to accept this result or try again with a different\nP or different constraining values or both. Since stability is desired, one approach\nwould be to keep P and c¡ and lower Cf until ICf becomes large enough to intersect 1C/.\nThe outline of the iterative Lyapunov design method is as follows.\n\n32\nIn this example n = 2 and m = 1. It was decided to set i = j 1 so\nthat\nQi,j ~ Qi,i i\n0 '\n(2.87)\nA0t\\j A 01,1 = 0\na2\n(2.88)\nand\nA kQip + Aqi,i\nk K2\n(2.89)\nFrom equations (2.76) and (2.77) come\nAc ~[PQi,i + Qi,iBT]\n(2.90)\nPie = -[BQhi + Qi'1BT]\n(2.91)\nPoc = A + AT 2c/ BKoi'i KqXIBt (2.92)\nPo-c = A + AT 2c/ BKoi'i KqI XBt (2.93)\nThe roots of the following polynomials are computed in terms of k while\nincrementing K2 through a wide range of values.\ndet[hPic + Poe] = 0 (2.94)\ndet[kPic T Toe] = 0 (2.95)\nRejecting complex roots and checking the regions separated by the real\nroots give\nkc(K2), kc(K2) (2.96)\nkc(K2), k-c{K2) (2.97)\nThe intersection of these regions are found.\nk(A2) maxfk^kc]\nk(A'2) = mznfkc, kc]\n(2.98)\n(2.99)\n\nCHAPTER 2\nTHEORY BEHIND THE DESIGN METHOD\nThe material in this section gives the theory leading to the proposed controller\ndesign method. The first section presents a geometric interpretation of Lyapunovs\nlinear stability theorem. The effects of control on the velocity field of a system is\nstudied in the next section. The nonemptiness and convexity of the set of all feedback\ngains which stabilize a system with respect to a given Lyapunov function are then\ndiscussed. Finally, an iterative procedure that finds one element of this set is given.\n2.1 A Geometric Interpretation of Lyapunovs Linear Stability Theorem\nTo understand why time invariance is a necessary assumption for eigenstructure\ndesign methods the following second order linear time varying system was studied.\nThe example given now is from Vidyasagars example 5.3,109 and can also be\nfound in Khalil . Given the following system\nx = A(t)x\n(2.1)\nwhere\nm\nl + acos2(t) 1 a sin(i) cos(i)\n1 a sin() cos(t) l + asin2(i)\nVidyasagar notes that the transition matrix is given by\n*(*,0)\ne( 1) cos(f) e 4 sin(i)\n_eD-i) sin() e-cos()\nand the characteristic equation is\n(2.2)\n(2.3)\nA2 + (2 a)A + (2 a) 0 (2-4)\nThe roots of (2.4) have negative real parts for 1 < a < 2. The exponents in the first\ncolumn of 4>(i, 0) indicate, however, that the system is unstable for these values of a.\n11\n\n66\nchoice of jP, = Jdiag is the best choice for a positive definite function for the system\nz = J z.\nV(z) = ~zTJdiagz (5.24)\nMaking the following transformation into the x coordinate system gives\nz = S x\n(5.25)\nwhich implies\nV(x) = -xT[S~1]T JdiagS'1* (5.26)\nOur choice of P is\nP = -[S'1]7, Jd.za.gS~1. (5.27)\nFor the nominal flight condition, the Jordan canonical form of Aq BqKq and\nApr BpTKpr is found and from (5.27)\nP =\n26772 609\n609 14.9\nP\n1 pr\n6792 10.7\n10.7 4.02\n-316.4 -.509\n-316.4 '\n-.509\n15.7\n(5.28)\nq KqBjPq) at the nominal point are\n-40.002\n(5.29)\n-1.0715 x 106,\nLikewise, for the closed loop roll-yaw system the eigenvalues are\n-1.3614 x 105, -320.00, -20.003 (5.30)\n5.4 The Iterative Lyapunov Design Method\nThe iterative Lyapunov design method generates feedback gains so that xTP,x\nand xTPprx are Lyapunov functions for each closed loop system. The algorithm\nrequires the initial guesses Kqo and Kpr0, for the first point, and the positive definite\nmatrices Pq and PpT. As a given flight condition changes, the feedback gains from\n\n58\nwhere x contains the states and u contains the inputs. A and B are matrices which\nare functions of several time varying flight parameters and are computed as follows.\nA = l\n(x,w )\ndf\n(4.46)\n.ominal\n(4.4?)\n(^fW ) nominal\nwhere w contains additional flight parameters. Note that x and u are now pertur\nbations from the point around which the linearization is taken.\nWe linearize the pitch model first. From inspection of (4.38) we see that\naq\\l 1 1 +\nQSCNa\nmV\n-l\n-QS\nC\nail2=M+^r(1-\nmV\nQS\nNa\nbqll ( 1 +\nmV\nQsc^y1 f-QS\n7}CNq\nmV\nmV\nmV\nCn6,\nTo linearize (4.39), we must first substitute (4.38) in for . Then we differentiate as\nin (4.46) and (4.47).\naq21\naq22\n5921 =\nQSd\nIyy\nQSd\nIyy\nQSd\nIyy\nCma + Cm ( 1 +\nQSCN\n-i\nCm<¡ + Cm* i 1 +\nmV\nQSCN -1\nmV\n-QS\nC\n1 -\nmV\nQS\nNa\nmVNq\nCm6q + Cm f 1 +\nQSCN\nmV\n-i\n-QSr\nlvCN\nThe resulting linearized pitch model is\nCt\nqll\nUql2\na\n+\nbqll\nkq21\n9 .\naq21\naq22\n. q.\n(4.48)\nThe same procedure is applied to the roll-yaw model. Assuming that a is constant\nin (4.40), and from inspection of equations (4.40) to (4.42) the following results.\nttprll\nQSCy\nmV\n}i aprl2 a +\nQSCYp\nmV\ni aprl3 1 +\nQSCYr\nmV '\n\n100\na) Top View of Missile (-) and Target (-) Trajectory\nFigure 8.8. (a) Top view and (b) side view of missile and target trajectories of a\nscenario which occurred at 50,000 ft. The missile missed the target by 452 ft. Initial\nmach numbers of the target and missile are 3.0 and .92 respectively.\n\n71\nThe eigenvalues in (5.30) suggest the following.\nQprl\n<\n-1.3614 x 105\n<\nCprl\npr2\n<\n-320.00\n<\nCpr2\npr 3\n<\n-20.003\n<\nCpr3\n(5.38)\nThe following constraining values have been chosen for the roll-yaw model.\ncprl = -150000, cpTl = -110000\ncpr2 = -340, cpt2 = -300 (5.39)\nQ-ptz 22, Cpr3 17\n5.5 Formulation of a State Tracker\nThe autopilot of the EMRAAT missile will be a state tracker. That is, we want to\nbe able to change the location of the equilibrium point in order to control the values\nof some of the states. The following shows how this will be accomplished.\nGiven the linear system\nx Ax + Bu\n(5.40)\ny = Cx,\n(5.41)\nwe would like to find a control law\nu = Kx + ATe/V\n(5.42)\nso that y, the output, tracks v ,the reference input, asymptotically. We require y = v\nwhen x = 0. When x = 0, then\n0 = Ax BKx + BKTefV.\n(5.43)\nSince K is chosen so that the system is stable, then A BK is invertible and\nx = -(A- BK)-lBKreSv\n(5.44)\nAlso,\nV = y = Cx = C(A BK)~lBKreiv\n(5.45)\n\nCHAPTER 1\nINTRODUCTION\nThe original goal of this dissertation was to formulate an autopilot design method\nwhich stabilizes the flight of a bank-to-turn air-to-air missile. The result was the de\nvelopment of a controller design method which applies to time varying linear systems.\nWhen used on a nonlinear system, local stability is achieved. The remainder of this\nchapter gives a brief description of the missile used in this study, a summary of related\nworks preceding this study, and the purpose and outline of this dissertation.\nAn air-to-air missile is launched from a military aircraft with the intent of in\ntercepting an enemy aircraft or target. Bank-to-turn (BTT) missiles have wings\ngiving them greater maneuverability than the conventional skid-to-turn (STT) mis\nsiles which have fins. Thus, the target must work much harder to evade a BTT\nmissile. The dynamics of BTT airframes are highly unstable, making for a difficult\ncontrols problem.\nThe missile under study in this paper is the extended medium range air-to-air\ntechnology (EMRAAT) missile as shown in Figure 1.1 , The EMRAAT missile\nis a theoretical missile formulated by the Air Force with the intent of studying the\nfeasibility of the BTT concept. The knowledge gained from the study of this missile\nwill be used by the Air Force if it ever decides to make a real BTT missile.\nThe EMRAAT missile is equipped with a seeker. If the seeker is infrared based,\nthen it measures the line of sight angles to the target. In the case that the seeker is\nradar based, then, in addition to the line of sight angles, the range and range rate\nof the target are measured. This information is passed to the guidance law which\ndetermines the desired normal accelerations needed to intercept the target. The\n1\n\n101\na) Commanded z Acceleration (-) and Actual z Acceleration ()\nFigure 8.9. Commanded and actual (a) accelerations in the 2 direction, (b) com\nmanded and actual angles of attack, and (c) the elevator deflection angle, for the\nscenario shown in figure 8.8. Saturation of the elevator angle was the cause of the\nmiss.\n\n24\nby a discussion of its convexity. Finally this section will present an exhaustive search\nmethod for finding the boundary of this set.\nFor the remainder of this chapter, A and B will be frozen at one instant in\ntime. A discussion follows on how to find a controller or set of controllers which\nsatisfies (2.19) at one given instant. If a given system is time varying then the results\nwhich follow must be applied for all time with P being constant and positive definite.\nThese results apply to all systems which can be stabilized with respect to a Lyapunov\nfunction using a constant P. The success of these methods depends on the existence\nof a constant positive definite P so that\n+ p-xAT(t)]BL{t) < 0 Vi (2.41)\nSince we require P to be zero, these results are conservative.\nLet 1C and 1C be the set of all K which satisfy the left and right inequalities\nrespectively of the following expression.\nc < ~xt[PA + AtP PBK KTBTP]x < c V x : xTx = 1 (2.42)\nj\nwhere P is positive definite. The objective is to find\n}C:=lCr\\X (2.43)\nThe following theorem gives conditions for the nonemptiness of 1C and 1C\nTheorem 2 Given the nth order m-input system,\nx = Ax + Bn (2.44)\n1C and K are nonempty if and only if there exists a positive definite P so\nthat the following conditions are true:\nc < minxTPAx = min-^xT[.PA + ATP]x\nxes xgs 2\n(2.45)\n\n29\ncontain all the fixed entries of K and be defined as follows.\n^11 k ij k\\ n\nThen\nwhere\nLOi,j\nkn 0 k{.\nkmi kmj kmn\nK = k Qitj + if.\nQi,j\nm\no i,j\n1 j n\n0 0 0\n0\n0 1 0\n0\n0 0 0\n(2.71)\n(2.72)\n(2.73)\n2) Find all values of the free entry which make\ndet[PA + 1 P 2cl PBK I This problem reduces to finding the roots of a polynomial. After sub\nstituting equation (2.72) into the argument of the above determinant, it\nbecomes\nk(-PBQitj QJjBtP) + (PA + AtP PBKoj K^BTP) (2.75)\nLet\np¡(i,j) := -(PBQ+ Ql,BTP) (2.76)\nP0(Ko¡¡) = PA + AtP PBKoi.i Kju BrP (2.77)\n\n70\nTable 5.2. Uncontrollable normal velocity components for the pitch model\nchanging parameter\nminimum uncontrollable velocity\nmaximum uncontrollable velocity\nM\n-43.1456\n-41.7253\nQ\n-45.0144\n-42.2289\na\n-43.3856\n-42.2362\nP\n-42.6047\n-42.6047\nP\n-42.6047\n-42.6045\n-42.6047\n-42.6047\nr\n-42.6047\n-42.6044\nTable 5.3. Uncontrollable normal velocity components for the roll-yaw model\nchanging parameter\nminimum uncontrollable velocity\nmaximum uncontrollable velocity\nM\n-21.3937\n-20.9751\nQ\n-22.3016\n-21.1574\na\n-21.7136\n-21.1140\n0\n-21.3180\n-21.3180\nP\n-21.3181\n-21.3180\nq\n-21.3184\n-21.3180\nr\n-21.3180\n-21.3180\nFor the pitch model, the uncontrollable normal velocity components range from\n45.0144 to 41.7253. Theorem 2 requires that\n-41.7253 <\nc < -45.0144\n(5.34)\nIn addition, from (5.29), we want\nCql < -1.0715 X 106 < Cqi\nCq2\nC9 < -40.002 < Cq2\n(5.35)\nFrom this the constraining values for the pitch model have been chosen to be\nc?1 = -1.2 x 106, qcq = -1 x 106,\n(5.36)\nCq2 45, Cq2 35\nLikewise for the roll-yaw model, when looking at Table 5.3, Theorem 2 requires that\n-20.9751 <\nCpr3\nS-prl\n-22.3016\n(5.37)\n\nTable A. 12. Tabular data for C,\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n.22168\n.10270\n-.05374\n-.02637\n-.02044\n-4\n.25664\n.16052\n-.02687\n.01319\n.00365\n0\n.31726\n.22611\n.00330\n.05275\n.03285\n2\n.33186\n.24984\n.02379\n.06300\n.04380\n4\n.33496\n.26882\n.03877\n.07692\n.05547\n8\n.32743\n.25070\n.07048\n.10549\n.07372\n12\n.34115\n.19935\n.08899\n.13077-\n.09270\n16\n.36991\n.20453\n.09956\n.12161\n.12190\no\nO\n.35973\n.16915\n.13084\n.15824\n.15547\nTable A.13. Tabular data for CySp\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-.033468\n-.022174\n.001062\n.002825\n-.004925\n-4\n-.040358\n-.030174\n-.002743\n-.002082\n-.000373\n0\n-.049933\n-.035826\n-.006726\n-.008327\n-.009403\n2\n-.051007\n-.036957\n-.007965\n-.009814\n-.010896\n4\n-.050828\n-.037217\n-.009204\n-.010409\n-.011045\n8\n-.049664\n-.032957\n-.010929\n-.011004\n-.010000\n12\n-.050380\n-.025130\n-.011770\n-.011004\n-.007313\n16\n-.055302\n-.023391\n-.010265\n-.009665\n-.009701\n20\n-.037673\n-.014957\n-.014159\n-.015762\n-.015821\n\nb\nT T r *\nT\"^ T j\n7 T t ,\n* A j\n* A i i\nT\n* i i\nT T t\n^ i\nr t t .\nr t [\n7\nr T. t L\n^ / y\n.151 UX-a: L 1 1\n15 -10 -5 0 S 10 15\nXI\nFigure 2.2. The trajectory and velocity field of (2.1)\n\n98\na) Top View of Missile (-) and Target () Trajectory\nFigure 8.7. (a) Top view and (b) side view of missile and target trajectories of\na scenario which occurred at 10,000ft. The miss distance is 2.8 ft. Initial mach\nnumbers of the target and missile are 2.0 and .92 respectively.\n\n94\na) Top View of Missile (-) and Target () Trajectory\nFigure 8.3. (a) Top view and (b) side view of missile and target trajectories of a\nscenario which occurred at 20,000 ft. The miss distance is 0.64 ft. Initial mach\nnumbers of the target and missile are 2.5 and .92 respectively.\n\n85\nFigure 7.2. A test of the curve fitting routines used to implement the autopilot.\n\n61\nDependent\nFlight\nParameters\nFigure 5.1. A block diagram of the systemy used to determine the dependence of\ngains on flight parameters.\n\n18\nSince\nxT Px = xT PAx = ^x.T[AT P + PA\\x (2-15)\nStability Condition 2 is equivalent to Lyapunovs linear stability condition.\nIt would be useful to compute the lower and upper bound of the rate of decay of\na given positive definite function,\nV(x) = xT Px\n(2.16)\nof the states of a time varying linear system by evaluating,\nc := mm\nt\nc := max\nt\nXmin^ \\P A(t) + AT(t)P]\nXmaX\\[PA(t) + AT(t)P]\nSo,\nc < ixT[PA() + AT(t)P]x < c V x : xTx =1, Vi\n(2.17)\n(2.18)\n(2.19)\nThen c and c are the lower and upper bounds respectively of the normal velocity\ncomponents, xTPx, on the unit spheroid. If c < 0, then the system is stable.\n2.2 Adding Control to the System\nThe result in section 2.1 is an analytical tool only. A design procedure is needed\nto control stability for the system\nx = A(t)x + B(t)u (2.20)\nwhere u is the control vector and x is the state vector. This section focuses on two\nquestions:\n1) What effect does u have on the velocity field of (2.20)?\n2) Can the normal velocity component xTPx on the spheroid xTx = 1\nbe controlled?\n\nAbstract of Dissertation Presented to the Graduate School\nof the University of Florida in Partial Fulfillment of the\nRequirements for the Degree of Doctor of Philosophy\nA CONTROLLER DESIGN METHOD WHICH APPLIES TO TIME VARYING\nLINEAR SYSTEMS\nBy\nKurt Walter Koenig\nAugust 1994\nChairman: Dr. Thomas E. Bullock\nMajor Department: Electrical Engineering\nA feedback controller design method has been formulated which applies to linear\ntime varying systems. The motivation behind this technique is the design of an\nautopilot for an air-to-air missile. The missile under study is the extended medium\n' range air-to-air technology (EMRAAT) missile, a theoretical bank-to-turn missile\nwhich is under study by the United States Air Force. Knowledge gained by this\nmissile will be applied to future bank-to-turn missiles.\nConventional autopilot design techniques use pole placement, linear quadratic\nregulators, linear quadratic Gaussian/loop transfer recovery techniques, or eigen-\nstructure assignment methods. These techniques are applied to a linear model which\ndepends on time varying flight parameters. In applying these methods, the false as\nsumption is made that the linear model is slowly varying. Since the system is changing\nrapidly, no theoretical basis exists to support the success of the resulting controller.\nThe proposed design method finds state feedback gains which cause the closed loop\nlinearized system to be stable with respect to a specified Lyapunov function. Even\nthough flight parameters change rapidly, local stability around the operating point is\nachieved.\nIV\n\n80\n6.3 Generating the Look-Up Table\nWith the design constraints set, feedback gains and feedforward gains are gen\nerated for each grid point. Figure 6.2 gives a block diagram of the system used to\naccomplish this. For the pitch model, the initial guess comes from one of the gains\nthat was generated in the previous chapter when determining the dependence be\ntween gains and flight parameters. The operating point from which this initial guess\noriginates is\nM = 1, Q = 1250psf, a = 8 (6.6)\nand is one of the extreme values listed in Table 5.1. The result of the first point is\nused to start adjacent points which, in turn, start new adjacent points until gains\nhave been computed for the entire grid. The look-up table for the roll-yaw model is\nNumerical problems were encountered in parts of the look-up table for the roll-\nyaw model. They were similar to the problems that were predicted in step 5) of the\niterative design method given in Chapter 2. To overcome these difficulties, some of the\nconstraining values were relaxed for a number of iterations and were later returned to\ntheir original assignments in the algorithm. Eventually, the desired feedback matrix\nwas found.\nFigure 6.3 shows kqu verses mach number and dynamic pressure when a = 8. As\nthis figure would indicate, the gains generated from the iterative Lyapunov method\nare smooth with respect to the dependent flight parameters. This fact gives hope\nthat the gain scheduling scheme will be easy.\n\nCHAPTER 8\nNONLINEAR SIMULATIONS\nA nonlinear simulation has been used to test the proposed autopilot for the EM-\nRAAT missile. First a section follows giving an overview of the nonlinear simulation.\nA test module is then made to generate state commands in order to evaluate the\nautopilots tracking ability. Finally, a series of flight scenarios are run to determine\nthe ability the missile has to intercept the target.\n8.1 The Nonlinear Simulation\nFigure 8.1 shows a block diagram of the simulation used to test the EMRAAT\nmissile. The program is written in FORTRAN. Initial conditions of the target and\nmissile are specified by the user. The simulation is then run and a trajectory of both\nthe target and missile results. All target and missile variables can be observed. The\ntarget is programmed to fly in a straight line until the range between the target and\nmissile falls below 5,000 ft. The target then makes a 9 g turn to the right. The\nsimulation terminates when the closing velocity becomes positive.\nThe seeker measures the line of sight angles and the range rate of the target.\nThe simulation uses exact measurements of these values and does not assume any\nnoise. These values are sent to the guidance law which, in this case, implements\nproportional navigation. A derivation of this guidance law can be found in Bryson\nand Ho . The outputs of the guidance law are two desired accelerations, aVc and\naZc. The BTT logic makes the conversion from the acceleration commands to the\nthree state commands ctc, /3C, and pc. Since the missile can achieve a much higher\nacceleration with angle of attack than with sideslip, the BTT logic uses pc to rotate\nthe desired accelerations into the pitch plane. If this roll maneuver is successful then\n88\n\nTable A.4. Tabular data for C,\n'<7\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n-1.035\n-1.278\n-.5951\n-.5269\n-.3970\n-4\n-1.200\n-1.44\n-.6862\n-.5490\n-.4377\n0\n-1.320\n-1.4719\n-.7089\n-.5297\n-.4151\n2\n-1.330\n-1.4875\n-.7008\n-.5269\n-.3925\n4\n-1.330\n-1.456\n-.6862\n-.5297\n-.3849\n8\n-1.330\n-1.325\n-.6276\n-.4772\n-.3396\n12\n-1.330\n-1.216\n-.5626\n-.4200\n-.3245\n16\n-1.286\n-1.114\n-.5100\n-.3641\n-.3547\no\nO\n-1.232\n-1.0594\n-.4715\n-.3724\n-.4226\nTable A.5. Tabular data for Cm/3\na/M\n.8\n1.2\n2.0\n2.7\n3.5\n-8\n1.0702\n.8632\n.2175\n-.0923\n-.3292\n-4\n1.2877\n1.2182\n.5090\n.1754\n-.0234\n0\n1.3544\n1.3825\n.6372\n.3138\n.1846\n2\n1.3544\n1.4414\n.6500\n.3446\n.2338\n4\n1.3509\n1.4772\n.6372\n.3723\n.2677\n8\n1.2561\n1.5172\n.491\n.3200\n.2862\n12\n.9502\n1.4133\n.2385\n.2708\n.3169\n16\n.6989\n1.1228\n.3808\n.4338\n.5046\no\nO\n.5123\n.5179\n.6179\n.6554\n.7446\n\n62\nwhere Vsoa is the speed of sound. Both p and Vsos are functions of altitude.\nP = fAh)\n(5.3)\nVsos = fs(h) (5.4)\nwhere h is altitude in feet above sea level. Here, fp and fs are functions based on\natmospheric tables and are implemented by linear interpolation. Solving (5.2) for V\nand substituting the result into (5.1) gives\nQ = (5.5)\nWe generate a third table in the following way.\nMh) := fP(h)f*(h) = pVs2os (5.6)\nSo\nQ = (5.7)\nThe function, /3, is a one-to-one function so that its inverse can be found by reading\nthe table backwards. With this in mind we can solve (5.7) for h.\nFrom (5.2) and (5.4)\nV = MVS0S = Mfs(h)\nSubstituting (5.8) into (5.9) to eliminate h gives\nAlso from (5.3)\n(5.8)\n(5.9)\n(5.10)\n(5.11)\n\nI certify that I have read this study and that in my opinion it conforms to\nacceptable standards of scholarly presentation and is fully adequate, in scope and\nquality, as a dissertation for the degree of Doctor of Philosophy.\nScott L. Miller\nAssociate Professor of\nElectrical Engineering\nI certify that I have read this study and that in my opinion it conforms to\nacceptable standards of scholarly presentation and is fully adequate, in scope and\nquality, as a dissertation for the degree of Doctor of Philosophy.\nDavid C. Wilson\nProfessor of Mathematics\nThis dissertation was submitted to the Graduate Faculty of the College of\nEngineering and to the Graduate School and was accepted as partial fulfillment\nof the requirements for the degree of Doctor of Philosophy.\nAugust 1994\n^ ^>\nWinfred M. Phillips\nDean, College of Engineering\nKaren A. Holbrook\n\n10\neigenvalues can be placed exist. Also, while many feedback control laws make (1.23)\nnegative definite, Shahruz and Behtash give only a small subset of these control laws.\n1.2 Purpose\nThis paper gives an algorithm which can stabilize a linear time varying system by\nplacing the eigenvalues of (1.23) between desired bounds and has fewer limitations\nthan the control law in Shahruz and Behtash . Limitations on the placement of\nthese bounds are stated. This algorithm applies to all systems for which a constant\npositive definite P exists so that (1-23) can be made negative definite for all time.\nFirst, theory is presented leading to the formulation of this algorithm. Then, to\ndemonstrate the usefulness of this algorithm, it is applied to a linear time varying\nsystem where normal pole placement techniques fail. This algorithm is then applied\nto the EMRAAT missile. A nonlinear model is made, and from this, a time vary\ning linearized model is generated which is a function of several flight parameters.\nGains are computed, and their dependence on flight parameters determined. A gain\nscheduling scheme is then implemented yielding local stability around the operating\npoint. The resulting design is tested in a nonlinear simulation. Finally, the results of\nthis test are given, and the usefulness of this design technique is evaluated.\n\n21\nsystem is stabilizable with an appropriate choice of constant positive definite P. If\nthere exists a constant positive definite P so that\nmax max xTfid(i)x < 0\nxes(t)\n(2.24)\nthe system is stabilizable. The above condition is equivalent to requiring the following\nmatrix to be negative definite for all time.\n1\n(2.25)\nThis fact was established by Fields and is true for the following reasons. Condi\ntion (2.24) is true if and only if\n<0V/i^0, Vi\n(2.26)\nSince the denominator of the above fraction is positive, with some simplification the\nabove statement is equivalent to\nBl{t)[A{t)P~l + p-1Ar(i)]JB1(i)/x <0V^0, Vi (2.27)\nwhich is equivalent to\n\\Bl{t)[A{t)P~r + p-1AT(t)]BL(t) < 0 Vi (2.28)\nIf a constant positive definite P exists satisfying condition (2.28), then the given time\nvarying system can be stabilized. This result is more general than a similar one given\nby Shahruz , and the two are equivalent when P = I.\nBefore giving a proof of Theorem 1, we give an intuitive explanation of Theorem\n1 for the second order case with a single input. Figure 2.7 shows the B vector for\nsuch a system along with some ellipsoid xTPx =1. If a state vector, x, is drawn\nto some point on this ellipse whose normal is not perpendicular to B. then there is\n\n33\nFigure 2.8. A plot of the boundary of 1C which guarantees satisfaction of the design\nconstraints of Example 1.\nFinally,\ndlC = {[k{K2), K2] V K2] U {[k(K2), I<2] V I<2} (2.100)\nA plot of dK, is shown in Figure 2.8. Let K = [7 11]. We can check to see\nif K C 1C by evaluating\nAi = \\mtn(A + AT BK KtBt) (2.101)\n= -12.91 (2.102)\nA2 = XmaX(A + At-BK- KtBt) (2.103)\n= -10.52 (2.104)\nAs the following shows,\nc < Ai (2.105)\nA2 < c (2.106)\n\n78\npr\nFigure 6.1. Formulation of the Design Constraints.\nTable 6.2. Uncontrollable normal velocity components for the pitch model.\na(degrees)\nminimum uncontrollable velocity\nmaximum uncontrollable velocity\n-3\n-47.1539\n-37.4260\n1\n-47.0806\n-37.3873\n4\n-47.1611\n-37.3480\n8\n-47.9797\n-37.4247\n12\n-48.8796\n-37.3906\n16\n-49.5915\n-37.6144\n20\n-49.5985\n-37.4355\nTable 6.3. Uncontrollable normal velocity components for the roll-yaw model.\na(degrees)\nminimum uncontrollable velocity\nmaximum uncontrollable velocity\n-1\n-23.3399\n-19.6307\n1\n-23.3774\n-19.6409\n2.5\n-23.4445\n-19.6883\n4\n-23.4971\n-19.7450\n8\n-24.5852\n-17.2696\n12\n-25.3163\n-19.9600\n16\n-25.6480\n-19.8625\n20\n-26.4797\n-20.3071\n\n3\nautopilot attempts to achieve these accelerations by applying the proper control to\nthe control surfaces.\n1.1 Earlier Works\nEarly autopilot designs separated the missile dynamics into two or three lower\norder models. These models were linearized, and classical techniques were used to\nstabilize them. Other designs applied more advanced techniques to higher order lin\nearized models. Some of these techniques are pole placement, linear quadratic regula\ntors, linear quadratic Gaussian/loop transfer recovery techniques and eigenstructure\nassignment methods Feedback controllers would be designed as a function of the\nlinearized models. Gains would then be scheduled against the flight parameters on\nwhich the models depend. Gain scheduling is a popular technique used in control\ndesigns and has been studied in depth by Shamma, Athans, and Cloutier [3,4,5,6].\nThe problem with these design methods is that they assume that the system is\ntime invariant or slowly varying. The linearized models depend on rapidly changing\nflight parameters. Although these methods often work, no theoretical basis exists\nto support the success of these designs. More recently, H<*, and fi synthesis design\ntechniques have been used to formulate one dynamic controller which would stabilize\nthe system for all modeled uncertainties [7,8]. A disadvantage of Hoo controllers is\nthat they have very high orders. Also, the existence of a robust H^ controller which\nsatisfies the performance requirements is not guaranteed. The only way to determine\nthe stability of a given design is to test it using a nonlinear simulation.\nDesoer states if a given time varying system\nx = A(t)x (1.1)\nis stable for each t when t is frozen, then it is not necessarily stable when t is allowed\nto change. However, (1.1) is stable if A(t) changes slowly. 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DQG ZDV DFFHSWHG DV SDUWLDO IXOILOOPHQW RI WKH UHTXLUHPHQWV IRU WKH GHJUHH RI 'RFWRU RI 3KLORVRSK\\ \\$XJXVW A A! :LQIUHG 0 3KLOOLSV 'HDQ &ROOHJH RI (QJLQHHULQJ .DUHQ \\$ +ROEURRN 'HDQ *UDGXDWH 6FKRRO\n\n27\nRemarks:\nThe nonemptiness of the intersection 1C = 1C fl 1C is not guaranteed. If the designer\ndiscovers that no intersection exists, then the upper and lower velocity bounds will\nAnother useful property of 1C is convexity. This property is valuable in formulating\niterative search techniques to be described in the next section. 1C is convex if when\nx and y are elements of 1C, then ax + (1 a)y is also an element of 1C for 0 < a < 1\n-\nTheorem 3 Let K- be the set of all K so that\nAma*[\\[P{A BK) + (A BK)tP}} < c (2.61)\nLet JC be the set of all K so that\nAmin[^[P(A BK) + (A BKfP]] > c (2.62)\nThen /C, 1C, and 1C = 1C fl 1C are convex.\nProof:\nIf 1C and 1C are convex, then the intersection 1C is convex. Convexity of\n1C will be proven here. In a similar manner the proof for the convexity of\n1C can be written.\nLet Ki and K2 be elements of 1C. We must show that aK\\ + (1 a)K2\nis contained in 1C when 0 < a < 1. We know that\n^xt[PA + ATP PBKi K*BtP]x < c V xTx = 1 (2.63)\n^xt[PA + ATP PBI<2 /i2rPrP]x < c V xTx = 1 (2.64)\nand\n\nCHAPTER 3\nA TIME VARYING SECOND ORDER EXAMPLE\nThe following problem gives a case when pole placement succeeds in giving eigen\nvalues with negative real parts, but fails to stabilize the system. The Lyapunov\ndesign method is then employed, and the resulting closed loop system is shown to be\nasymptotically stable for all time.\nWe would like to find a feedback control law that stabilizes the system\nx = A(i)x + B(t) u\n(3.1)\nwhere\nA(t) =\n1 + 1.5cos()[cos() + cos( + 7T /18)] 1 1.5 sin()[cos() + cos (t + tt/18)]\n1 1.5cos()[sin() + sin( + 7r/18)j 1 + 1.5sin(t)[sin(i) + sin(t + tt/18)]\n(3.2)\nand\nB(t) =\nCOS it + 7r/l8)\nsin(t + 7t/18)\n(3.3)\nThe eigenvalues of A(t) are -.4889 and 1.4661 for all time. The following control law\nis proposed.\nu = [ 1.5cos(i) 1.5sin(t) ]x\n(3.4)\nThe resulting closed loop system can be found in example 5.3,109 by Vidyasagar\n and also in Khalil . The eigenvalues for the resulting closed loop system are\nA = .25 j.6614. Since the eigenvalues have negative real parts, one would expect\nthe closed loop system to be stable. However, Vidyasagar shows that the transition\nmatrix is\n\\$(*, 0)\ne'5 cos() e sin(t)\ne-osin() e_cos()\n(3.5)\n46\n\n50\nTrajectory of the Lyapunov based closed loop time varying system\nFigure 3.3. A trajectory of the closed loop system using the Lyapunov design method.\nXq = [1 0]T. The system is stable.\n\n69\nSolve det[k.Pi + P0] = 0 for in terms of k for c = c1, ... cn.\nReject all values of k which do not meet the constraints from\nsteps 2) or 6). The result is 2n intervals whose lower bounds are\ndesignated by kl5 ... k2n and whose upper bounds is named ki, ..., k2n.\nFind the intersection of these intervals by evaluating\nk = max[ka,...,k2n]\nk = mn[ki,k2].\nFind the midpoint of the interval by computing\nkjlfc = i(k + k).\nLet K kj,kQj,k F Aoj,k\nNext k\nNext j\nKi+1 = K\nRemarks:\nLet the feedback sets /Cl5 1C i, JC2, AF2, jCn, and Kn be defined respectively as the\nset of all K so the constraints (5.31) are met. Theorem 2 provides conditions for the\nnonemptiness of and lCn. But conditions for the nonemptiness of the remaining\nsets are still unknown. Also, K, 1, ... ,)Cn are not convex in general. These are the\nlimitations of using the modified design algorithm.\nWe now turn to selecting constraining values for the eigenvalues of ^[P(A BK)^-\n(A BK)TP]. It is necessary to evaluate the uncontrollable normal velocity compo\nnents for each flight condition that will be generated in Figure 5.1. Tables 5.2 and 5.3\nshow for both models the minimum and maximum uncontrollable normal velocities\nfor each changing flight parameter.\n\nCHAPTER 5\nTHE DEPENDENCE OF GAINS ON FLIGHT PARAMETERS\nThis chapter describes the procedure that was used to determine which flight\nparameters would be scheduled against the gains. Figure 5.1 gives a block diagram\nof the system used for this procedure. The feedback gains are computed using the\niterative Lyapunov design method described in Chapter 2. The gains depend on\nthe linear model which in turn depends on seven flight parameters. Six of these\nparameters are held constant while the seventh one changes. The resulting gains are\nchecked to see if they depend on this changing parameter. The process is repeated\nfor each flight parameter. It is found that the gains depend on angle of attack, mach\nnumber, and dynamic pressure. This information will be used in the gain scheduling\nprocess. The next section will discuss the atmospheric tables used to generate p and\nV from M and Q. The flight parameter generator will then be presented. The third\nsection describes the initializer. The details of the iterative Lyapunov design method\nare given in the fourth section. Finally, the results of the comparison between the\ngains and flight parameters will be given in the fifth section.\n5.1 Generating o and V from M and 0\nBy inspection, the linear models from Chapter 4 clearly depend on a, j3, p, q, r,\np, V, and the aerodynamic coefficients. The coefficients, however, can be eliminated\nfrom the list because they depend on mach number and angle of attack. It is desirable\nto replace p and V with M and Q since the later two can be easily measured on the\nmissile. We know the following.\nQ = \\pV2 (5.1)\n1/\nM = (5.2)\nV SOS\n60\n\nthe size of a time varying uncertainty. Both of these tests were found to be too\nconservative when applied to an already existing stable autopilot, and the results\nwere inconclusive. In addition, these tests are analytic tools. As yet, no design\nprocedure exists which can give a closed loop linear system that will pass either\ntest. It would be desirable to formulate a design method that can yield a closed\nloop time varying linear system which satisfies a less conservative stability condition.\nVidyasagar gives two important results. The first can also be found in Hahn \nand says that given a time varying system\nx = A(t)x\nif a positive definite matrix P can be found so that the matrix\n(1.17)\nPA(t) + AT(t)P\n(1.18)\nis negative definite for all time, then the system is asymptotically stable.\nThe second result allows the first result to be applied to a nonlinear system. Given\na nonlinear system of the form\nx = /(i,x)\nwhere\n/(GO) = o\nand / is continuously differentiable, then let\ndf(t,x)\nm =\ndx\nx=0\nand assume that\n||/(i,x) A(i)x||\n\n12\nFigure 2.1. The trajectory of (a) the system (2.1) and (b) a frozen system.\nThis system was simulated using MATLAB with a = 1.5 and xo = [1,0]T. The\neigenvalues of A(t) are \\ + ^j and Figure 2.1a shows the state trajectory\nof the system 2.1 where Xi is assigned to the horizontal axis, and X2 is assigned to\nthe vertical axis. This plot demonstrates the instability of the system. If the system\nwere frozen, i.e. A(t) = A(0), then the stable trajectory of Figure 2.1b would result.\nThis trajectory is shown with the velocity vector field of the frozen system.\nTo explain the instability in Figure 2.1a, we will now discuss the time varying\nvelocity field of equation (2.1). The plots in Figure 2.2 show the trajectory of the\ntime varying system in the state plane for eight instances in time. For each plot, the\nvelocity vector field for that instant in time is superimposed on to the trajectory. Each\nplot in Figure 2.2 shows the boundaries of four pie-shaped regions which this paper\nrefers to as positive and negative regions. Two positive regions exist that are defined\nas the set of all points whose velocities have an outer radial component, i.e. each of\nthese velocities have a component pointing directly away from the origin. Likewise,\nthere are two negative regions containing velocities with inner radial components.\nThe plots show that the boundaries separating the positive and negative regions\nrotate in a clockwise direction. Also, the current position in the trajectory remains\n\n87\nLikewise, for the roll-yaw controller, the limiting values at all the test points are\n-2.03 x 10s\n<\nXprl\n<\n-9.28 x 10'\n-2392.7\n<\nXpr2\n<\n-182.6\n-26.27\n<\nXpr3\n<\n-3.96\n(7.4)\nSome of these values differ significantly from the desired constraints; however, since\nthese values are still negative, these deviations are acceptable and indicate that the\nclosed loop system will be stable. It should also be noted that most eigenvalues\nremain well within their desired constraints as shown in Table 7.1.\n\n90\naVc will become small and aZc will become positive. ac and (3C are computed in an\nattempt to match aZc and aVc respectively.\nThe autopilot implements the control law\nU = KX + KrefV (8.1)\nwhere x is a vector containing the actual states and v contains the state commands\nfrom the bank-to-turn (BTT) logic. The states come from exact measurements in the\nsimulation. If this autopilot were to be implemented in an actual missile, the states\nwould be measured using an inertial platform. The gains K and Kref come from the\ngain schedule implemented with a combination of polynomials and interpolation. In\nthis simulation there is no delay in the gain schedule and K and Kre¡ are produced\ninstantaneously. The output of the autopilot is the control surface angles Sp, 6q, and\n6r. Linear and angular accelerations are computed by the missile dynamics module\nof the program. The simulation uses the output of the missile dynamics to compute\nall of the flight variables including the position and velocity of the missile.\n8.2 A Test of State Tracking\nThe model for the EMRAAT missile has five states and three inputs. The autopi\nlot is designed to track three state commands: ac, /3C, and pc. Before running missile\ntarget scenarios it was decided to test the autopilots tracking ability. The BTT logic\nwas disconnected, and the following commands were applied to the reference inputs\nof the autopilot.\n10 for 0 s < t < .5 s\n_ 0 for .5 s < t < 2.75 s\nac ~ 10 for 2.75 s < t < 3.75 s\n, 0 for 3.75 s < t\n( 0 for 0 s < t < 2 s\n3C = < 5 for 2 s < t < 2.5 s\n( 0 for 2.5 s < t\n(8.3)\n\nI certify that I have read this study and that in my opinion it conforms to\nacceptable standards of scholarly presentation and is fully adequate, in scope and\nquality, as a dissertation for the degree of Doctor of Philosophy.\nThomas E. Bullock, Chairman\nProfessor of Electrical\nEngineering\nI certify that I have read this study and that in my opinion it conforms to\nacceptable standards of scholarly presentation and is fully adequate, in scope and\nquality, as a dissertation for the degree of Doctor of Philosophy,\nJacob Hamfner\nProfessor of Electrical\nEngineering\nI certify that I have read this study and that in my opinion it conforms to\nacceptable standards of scholarly presentation and is fully adequate, in scope and\nquality, as a dissertation for the degree of Doctor of Philosophy.\nMichel A. Lynch\nLecturer of Electrical\nEngineering\nI certify that I have read this study and that in my opinion it conforms to\nacceptable standards of scholarly presentation and is fully adequate, in scope and\nquality, as a dissertation for the degree of Doctor of Philosophy.\nJohn Staudhammer\nProfessor of Electrical\nEngineering\n\n77\nTable 6.1. Values of a in the pitch control\na\nPitch\nRoll-Yaw\n-3.0\n-1.0\n1.0\n1.0\n4.0\n2.5\n8.0\n4.0\n12.0\n8.0\n16.0\n12.0\n20.0\n16.0\n20.0\ner look-up table.\nmade more restrictive and returned to their original assignments. This became a\ntedious process for some parts of the grid. When the number of iterations exceeded\n100 it was decided to add more grid points so that the desired feedback gains could\nbe found in fewer iterations.\n6.2 Formulation of the Design Constraints\nBefore using the iterative Lyapunov design algorithm, the uncontrollable normal\nvelocity components for the entire M-Q-a grid must be determined. This information\nis needed to formulate the design constraints. This process is described by the block\ndiagram in Figure 6.1. Table 6.2 presents the minimum and maximum uncontrollable\nnormal velocity components of the pitch model throughout the M Q grid for each\nvalue of a. Here P = Pq, the matrix computed during the initializing procedure\nof the last chapter. Table 6.3 gives the same result for the roll-yaw model where\nP = Ppr- The extreme values of uncontrollable normal velocities for the pitch model\nare 49.5985 and 37.3480. Also the eigenvalues of\n5IP,{A, B,K,) + (A, B,K,)tP,}\n(6.1)" ]
[ null ]
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https://www.aimsciences.org/journal/1547-5816/2014/10/2
[ "", null, "", null, "", null, "", null, "", null, "ISSN:\n1547-5816\n\neISSN:\n1553-166X\n\n## Journal Home\n\n• Open Access Articles", null, "All Issues\n\n## Journal of Industrial & Management Optimization\n\nApril 2014 , Volume 10 , Issue 2\n\nSelect all articles\n\nExport/Reference:\n\n2014, 10(2): 363-381 doi: 10.3934/jimo.2014.10.363 +[Abstract](4386) +[PDF](255.3KB)\nAbstract:\nWe consider fractional order optimal control problems in which the dynamic control system involves integer and fractional order derivatives and the terminal time is free. Necessary conditions for a state/control/terminal-time triplet to be optimal are obtained. Situations with constraints present at the end time are also considered. Under appropriate assumptions, it is shown that the obtained necessary optimality conditions become sufficient. Numerical methods to solve the problems are presented, and some computational simulations are discussed in detail.\n2014, 10(2): 383-396 doi: 10.3934/jimo.2014.10.383 +[Abstract](2603) +[PDF](571.4KB)\nAbstract:\nVerification of semiconductor chip designs is commonly driven by single goal orientated measures. With increasing design complexities, this approach is no longer effective. We enhance the effectiveness of coverage driven design verifications by applying multi-objective optimization techniques. The technique is based on genetic evolutionary algorithms. Difficulties with conflicting test objectives and selection of tests to achieve multiple verification goals in the genetic evolutionary framework are also addressed.\n2014, 10(2): 397-412 doi: 10.3934/jimo.2014.10.397 +[Abstract](2991) +[PDF](550.9KB)\nAbstract:\nThis paper presents a new method for linear semi-infinite programming. With the introduction of the so-called generalized ladder point, a ladder method for linear semi-infinite programming is developed. This work includes the generalization of the inclusive cone version of the fundamental theorem of linear programming and the extension of a linear programming ladder algorithm. The extended ladder algorithm finds a generalized ladder point optimal solution of the linear semi-infinite programming problem, which is approximated by a sequence of ladder points. Simple convergence properties are provided. The algorithm is tested by solving a number of linear semi-infinite programming examples. These numerical results indicate that the algorithm is very efficient when compared with other methods.\n2014, 10(2): 413-441 doi: 10.3934/jimo.2014.10.413 +[Abstract](4757) +[PDF](730.5KB)\nAbstract:\nIn this paper we study optimal control problems with multiple time delays in control and state and mixed type control-state constraints. We derive necessary optimality conditions in the form of a Pontryagin type Minimum Principle. A discretization method is presented by which the delayed control problem is transformed into a nonlinear programming problem. It is shown that the associated Lagrange multipliers provide a consistent numerical approximation for the adjoint variables of the delayed optimal control problem. We illustrate the theory and numerical approach on an analytical example and an optimal control model from immunology.\n2014, 10(2): 443-460 doi: 10.3934/jimo.2014.10.443 +[Abstract](2486) +[PDF](598.3KB)\nAbstract:\nWe study optimal cooperative collision avoidance strategies for two participants in a planar close proximity encounter. Previous research focused on special cases of this problem and showed that bang-bang strategies without switching are optimal in most situations, while singular controls only appear for the case of participants with unequal linear speeds under certain conditions. This paper extends the earlier analyses to a general case of a coplanar close proximity encounter, for which both parameters of the problem may take arbitrary admissible values. For such a case, we present a theoretical and numerical study of the structure of optimal controls. We prove that both controls can not be singular simultaneously and that the only possible singular control is a zero control. We derive formulas for the singular surfaces and verify that sufficient conditions hold for the computed extremal solutions. We identify different types of structural changes of the control strategies and show how the control structure changes with the change in the model parameters and initial conditions.\n2014, 10(2): 461-476 doi: 10.3934/jimo.2014.10.461 +[Abstract](2855) +[PDF](253.0KB)\nAbstract:\nIn this paper, we propose a new parallel splitting descent method for solving a class of variational inequalities with separable structure. The new method can be applied to solve convex optimization problems in which the objective function is separable with three operators and the constraint is linear. In the framework of the new algorithm, we adopt a new descent strategy by combining two descent directions and resolve the descent direction which is different from the methods in He (Comput. Optim. Appl., 2009, 42: 195-212.) and Wang et al. (submitted to J. Optimiz. Theory App.). Theoretically, we establish the global convergence of the new method under mild assumptions. In addition, we apply the new method to solve problems in management science and traffic equilibrium problem. Numerical results indicate that the new method is efficient and reliable.\n2014, 10(2): 477-501 doi: 10.3934/jimo.2014.10.477 +[Abstract](3135) +[PDF](4452.7KB)\nAbstract:\nTemporarily-captured Natural Earth Satellites (NES) are very appealing targets for space missions for many reasons. Indeed, NES get captured by the Earth's gravity for some period of time, making for a more cost-effective and time-effective mission compared to a deep-space mission, such as the 7-year Hayabusa mission. Moreover, their small size introduces the possibility of returning with the entire temporarily-captured orbiter (TCO) to Earth. Additionally, NES can be seen as interesting targets when examining figures of their orbits. It requires to expand the current state-of-art of the techniques in geometric optimal control applied to low-thrust orbital transfers. Based on a catalogue of over sixteen-thousand NES, and assuming ionic propulsion for the spacecraft, we compute time minimal rendezvous missions for more than $96%$ of the NES. The time optimal control transfers are calculated using classical indirect methods of optimal control based on the Pontryagin Maximum Principle. Additionally we verify the local optimality of the transfers using second order conditions.\n2014, 10(2): 503-519 doi: 10.3934/jimo.2014.10.503 +[Abstract](3323) +[PDF](1340.4KB)\nAbstract:\nA mathematical model for laser cutting with time-dependent cutting velocity is presented. The model involves two coupled nonlinear partial differential equations for the interacting dynamical behaviors of the free melt boundaries during the process. We define a measurement for the roughness of a cutting surface and introduce an optimal control problem for minimizing the roughness with respect to the cutting velocity and the laser beam intensity along the free melt surface. The optimal control problem involves an additional finite-dimensional averaging constraint. Necessary optimality conditions will be deduced and illustrated by means of numerical examples with data from industrial applications.\n2014, 10(2): 521-542 doi: 10.3934/jimo.2014.10.521 +[Abstract](2543) +[PDF](511.7KB)\nAbstract:\nA new adaptive mesh refinement algorithm is proposed for solving Euler discretization of state- and control-constrained optimal control problems. Our approach is designed to reduce the computational effort by applying the inexact restoration (IR) method, a numerical method for nonlinear programming problems, in an innovative way. The initial iterations of our algorithm start with a coarse mesh, which typically involves far fewer discretization points than the fine mesh over which we aim to obtain a solution. The coarse mesh is then refined adaptively, by using the sufficient conditions of convergence of the IR method. The resulting adaptive mesh refinement algorithm is convergent to a fine mesh solution, by virtue of convergence of the IR method. We illustrate the algorithm on a computationally challenging constrained optimal control problem involving a container crane. Numerical experiments demonstrate that significant computational savings can be achieved by the new adaptive mesh refinement algorithm over the fixed-mesh algorithm. Conceivably owing to the small number of variables at start, the adaptive mesh refinement algorithm appears to be more robust as well, i.e., it can find solutions with a much wider range of initial guesses, compared to the fixed-mesh algorithm.\n2014, 10(2): 543-556 doi: 10.3934/jimo.2014.10.543 +[Abstract](2571) +[PDF](411.8KB)\nAbstract:\nMulticriterion optimization and Pareto optimality are fundamental tools in economics. In this paper we propose a new relaxation method for solving multiple objective quadratic programming problems. Exploiting the technique of the linear weighted sum method, we reformulate the original multiple objective quadratic programming problems into a single objective one. Since such single objective quadratic programming problem is still nonconvex and NP-hard in general. By using the techniques of lifting and doubly nonnegative relaxation, respectively, this single objective quadratic programming problem is transformed to a computable convex doubly nonnegative programming problem. The optimal solutions of this computable convex problem are (weakly) Pareto optimal solutions of the original problem under some mild conditions. Moreover, the proposed method is tested with two examples and a practical portfolio selection example. The test examples are solved by CVX package which is a solver for convex optimization. The numerical results show that the proposed method is effective and promising.\n2014, 10(2): 557-566 doi: 10.3934/jimo.2014.10.557 +[Abstract](3451) +[PDF](345.4KB)\nAbstract:\nIn this paper, the integer programming problem is studied. We introduce the notion of integer basis and show that a given integer set can be converted into a fixed number of linear combinations of the basis elements. By employing the Stiefel manifold and optimal control theory, the combinatorial optimization problem can be converted into a continuous optimization problem over the continuous Stiefel manifold. As a result, gradient descent methods can be applied to find the optimal integer solution. We demonstrate by numerical examples that this approach can obtain good solutions. Furthermore, this method gives new insights into continuous relaxation for solving integer programming problems.\n2014, 10(2): 567-590 doi: 10.3934/jimo.2014.10.567 +[Abstract](3124) +[PDF](3758.3KB)\nAbstract:\nFor the problem of multimodal image registration, an optimal control approach is presented. The geometrical information of the images will be transformed into weighted edge sketches, for which a linear-elastic or hyperelastic registration will be performed. For the numerical solution of this problem, we provide a direct method based on discretization methods and large-scale optimization techniques. A comparison of a separated and a joint access for the generation of the edge sketches and the determination of the matching deformation is made. The quality of the results obtained with the optimal control method competes well with those generated by a standard variational method.\n2014, 10(2): 591-611 doi: 10.3934/jimo.2014.10.591 +[Abstract](2842) +[PDF](640.6KB)\nAbstract:\nThe problem of scheduling with time-dependent processing times has been studied for more than two decades and significant advances have been made over the years. However, most work has paid more attention to the single-criterion models. Furthermore, most heuristics are constructed for the time-dependent scheduling problems in a step-by-step way. Motivated by the observations, this paper studies a two-agent scheduling model with increasing linear deterioration jobs in which the processing time of a job is modeled as an increasing linear function of its starting time. The objective function is to minimize the sum of the maximum weighted tardiness of the jobs of the first agent and the total weighted tardiness of the jobs of the second agent. This problem is known to be strongly NP-hard. Thus, as an alternative, the branch-and-bound, ant colony algorithm and simulated annealing algorithms are developed for the problem. Computational results are also presented to determine the performance of the proposed algorithms.\n2014, 10(2): 613-620 doi: 10.3934/jimo.2014.10.613 +[Abstract](2921) +[PDF](353.5KB)\nAbstract:\nIn inverse scheduling problems, a job sequence is given and the objective is to determine the minimal perturbation to parameters, such as processing times or weights of jobs so that the given schedule becomes optimal with respect to a pre-selected objective function. In this paper we study the necessary and sufficient conditions for optimality of the total completion time problem on parallel machines and inverse scheduling problem of the total completion time objective on parallel machines in which the processing times are minimally adjusted, so that a given target job sequence becomes an optimal schedule.\n2014, 10(2): 621-636 doi: 10.3934/jimo.2014.10.621 +[Abstract](2124) +[PDF](395.0KB)\nAbstract:\nBy using the Least Squares Support Vector Machines (LS-SVMs), we develop a numerical approach to find an approximate solution for affine nonlinear systems with partially unknown functions. This approach can obtain continuous and differential approximate solutions of the nonlinear differential equations, and can also identify the unknown nonlinear part through a set of measured data points. Technically, we first map the known part of the affine nonlinear systems into high dimensional feature spaces and derive the form of approximate solution. Then the original problem is formulated as an approximation problem via kernel trick with LS-SVMs. Furthermore, the approximation of the known part can be expressed via some linear equations with coefficient matrices as coupling square matrices, and the unknown part can be identified by its relationship to the known part and the approximate solution of affine nonlinear systems. Finally, several examples for different systems are presented to illustrate the validity of the proposed approach.\n2014, 10(2): 637-663 doi: 10.3934/jimo.2014.10.637 +[Abstract](3092) +[PDF](536.0KB)\nAbstract:\nWe present a substitution secant/finite difference (SSFD) method to solve the finite minimax optimization problems with a number of functions whose Hessians are often sparse, i.e., these matrices are populated primarily with zeros. By combining of a substitution method, a secant method and a finite difference method, the gradient evaluations can be employed as efficiently as possible in forming quadratic approximations to the functions, which is more effective than that for large sparse unconstrained differentiable optimization. Without strict complementarity and linear independence, local and global convergence is proven and $q$-superlinear convergence result and $r$-convergence rate estimate show that the method has a good convergence property. A handling method of a nonpositive definitive Hessian is given to solve nonconvex problems. Our numerical tests show that the algorithm is robust and quite effective, and that its performance is comparable to or better than that of other algorithms available.\n\n2020 Impact Factor: 1.801\n5 Year Impact Factor: 1.688\n2020 CiteScore: 1.8" ]
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http://swmath.org/software/29239
[ "# Rank Nullity\n\nRank-Nullity Theorem in Linear Algebra. In this contribution, we present some formalizations based on the HOL-Multivariate-Analysis session of Isabelle. Firstly, a generalization of several theorems of such library are presented. Secondly, some definitions and proofs involving Linear Algebra and the four fundamental subspaces of a matrix are shown. Finally, we present a proof of the result known in Linear Algebra as the “Rank-Nullity Theorem”, which states that, given any linear map f from a finite dimensional vector space V to a vector space W, then the dimension of V is equal to the dimension of the kernel of f (which is a subspace of V) and the dimension of the range of f (which is a subspace of W). The proof presented here is based on the one given by Sheldon Axler in his book Linear Algebra Done Right. As a corollary of the previous theorem, and taking advantage of the relationship between linear maps and matrices, we prove that, for every matrix A (which has associated a linear map between finite dimensional vector spaces), the sum of its null space and its column space (which is equal to the range of the linear map) is equal to the number of columns of A.\n\n##", null, "Keywords for this software\n\nAnything in here will be replaced on browsers that support the canvas element\n\n## References in zbMATH (referenced in 2 articles )\n\nShowing results 1 to 2 of 2.\nSorted by year (citations)" ]
[ null, "http://swmath.org/media/img/minus.png", null ]
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https://answers.everydaycalculation.com/lcm/6720-3087
[ "Solutions by everydaycalculation.com\n\n## What is the LCM of 6720 and 3087?\n\nThe lcm of 6720 and 3087 is 987840.\n\n#### Steps to find LCM\n\n1. Find the prime factorization of 6720\n6720 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 5 × 7\n2. Find the prime factorization of 3087\n3087 = 3 × 3 × 7 × 7 × 7\n3. Multiply each factor the greater number of times it occurs in steps i) or ii) above to find the LCM:\n\nLCM = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5 × 7 × 7 × 7\n4. LCM = 987840\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to find LCM of upto four numbers in your own time:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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https://socratic.org/questions/what-is-the-domain-and-range-of-y-x-x-2-1-1
[ "# What is the domain and range of y= -x/( x^2-1)?\n\nDomain is any real value except $x \\ne \\pm 1$\n$y = - \\frac{x}{{x}^{2} - 1}$\nDomain (values of x) : ${x}^{2} - 1 \\ne 0 \\mathmr{and} {x}^{2} \\ne 1 \\mathmr{and} x \\ne \\pm 1$\nDomain is any real value except $x \\ne \\pm 1$" ]
[ null ]
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https://visualfractions.com/calculator/factors/factors-of-516/
[ "# Factors of 516\n\nSo you need to find the factors of 516 do you? In this quick guide we'll describe what the factors of 516 are, how you find them and list out the factor pairs of 516 for you to prove the calculation works. Let's dive in!\n\n## Factors of 516 Definition\n\nWhen we talk about the factors of 516, what we really mean is all of the positive and negative integers (whole numbers) that can be evenly divided into 516. If you were to take 516 and divide it by one of its factors, the answer would be another factor of 516.\n\nLet's look at how to find all of the factors of 516 and list them out.\n\n## How to Find the Factors of 516\n\nWe just said that a factor is a number that can be divided equally into 516. So the way you find and list all of the factors of 516 is to go through every number up to and including 516 and check which numbers result in an even quotient (which means no decimal place).\n\nDoing this by hand for large numbers can be time consuming, but it's relatively easy for a computer program to do it. Our calculator has worked this out for you. Here are all of the factors of 516:\n\n• 516 ÷ 1 = 516\n• 516 ÷ 2 = 258\n• 516 ÷ 3 = 172\n• 516 ÷ 4 = 129\n• 516 ÷ 6 = 86\n• 516 ÷ 12 = 43\n• 516 ÷ 43 = 12\n• 516 ÷ 86 = 6\n• 516 ÷ 129 = 4\n• 516 ÷ 172 = 3\n• 516 ÷ 258 = 2\n• 516 ÷ 516 = 1\n\nAll of these factors can be used to divide 516 by and get a whole number. The full list of positive factors for 516 are:\n\n1, 2, 3, 4, 6, 12, 43, 86, 129, 172, 258, and 516\n\n## Negative Factors of 516\n\nTechnically, in math you can also have negative factors of 516. If you are looking to calculate the factors of a number for homework or a test, most often the teacher or exam will be looking for specifically positive numbers.\n\nHowever, we can just flip the positive numbers into negatives and those negative numbers would also be factors of 516:\n\n-1, -2, -3, -4, -6, -12, -43, -86, -129, -172, -258, and -516\n\n## How Many Factors of 516 Are There?\n\nAs we can see from the calculations above there are a total of 12 positive factors for 516 and 12 negative factors for 516 for a total of 24 factors for the number 516.\n\nThere are 12 positive factors of 516 and 12 negative factors of 516. Wht are there negative numbers that can be a factor of 516?\n\n## Factor Pairs of 516\n\nA factor pair is a combination of two factors which can be multiplied together to equal 516. For 516, all of the possible factor pairs are listed below:\n\n• 1 x 516 = 516\n• 2 x 258 = 516\n• 3 x 172 = 516\n• 4 x 129 = 516\n• 6 x 86 = 516\n• 12 x 43 = 516\n\nJust like before, we can also list out all of the negative factor pairs for 516:\n\n• -1 x -516 = 516\n• -2 x -258 = 516\n• -3 x -172 = 516\n• -4 x -129 = 516\n• -6 x -86 = 516\n• -12 x -43 = 516\n\nNotice in the negative factor pairs that because we are multiplying a minus with a minus, the result is a positive number.\n\nSo there you have it. A complete guide to the factors of 516. You should now have the knowledge and skills to go out and calculate your own factors and factor pairs for any number you like.\n\nFeel free to try the calculator below to check another number or, if you're feeling fancy, grab a pencil and paper and try and do it by hand. Just make sure to pick small numbers!\n\n## Factors Calculator\n\nWant to find the factor for another number? Enter your number below and click calculate.\n\nFactors of 517" ]
[ null ]
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http://fistf.info/10-2-problem-solving-volume-of-prisms-and-cylinders-75/
[ "# 10-2 PROBLEM SOLVING VOLUME OF PRISMS AND CYLINDERS\n\nA right cone has a height of 15 meters and a slant height of 17 meters. Prisms and Cylinders Lesson p. Solving Volume Problems You can use the formulas for the volume of a rectangular prism and the volume of a rectangular pyramid to solve problems. Volume of prism contains rectangular prism, L-blocks, solid blocks, counting cubes, triangular prism and other mixed prisms. Therefore, if the height of the pyramid were tripled, its volume would be tripled.", null, "The base of the sculpture is 2 feet long and 2 feet wide. Round your answers to the nearest tenth, if necessary. Student must know the formula or technique to solve the problems before taking these surface area worksheets. Choose the letter for the best answer. Using Cubes to Find the Volume of a Rectangular Prism You can find the volume of this prism by counting how many cubes tall, long, and wide the prism is and then multiplying.", null, "Volume of Pyramid worksheet Grade A Find the height of each pyramid. All measurements are in centimeters.\n\n## 3D Shapes Volume Problems\n\nSolving problems involving volume and surface area. Volume of prisms, Volume of vollume, cylinder and sphere, Volume of pyramids and Volume of mixed and combined shapes. What is the volume of this cylinder? Monday the 29th Matthew.\n\nGE HEALTHYMAGINATION CASE STUDY ANALYSIS\n\nThe solvinb area of a pyramid is 88 square feet.\n\nSubstitute for r and h. Giza in Egypt is the site of the three great Egyptian pyramids. For example, the solid in Exercise 23 is made up of a pyramid on top of a cube. What was the original price?\n\nIf you wish to download it, please recommend it to your friends in any social system. What is the area of its base? Find the volume of a rectangular pyramid with a length of 14 feet, a width of 12 feet For guided practice, interactive modeling is used to answer the first 3 questions.\n\nVolume The volume of a xolving figure is the amount that fills the figure. Round to the nearest tenth, if necessary. The sides of the 3-in.\n\n## Lesson 5 problem solving practice volume of pyramids\n\nA square pyramid has a base edge of 5. Along with your textbook, daily homework, and class notes, the completed Word Problem Practice Workbookcan help you review for quizzes and tests. Round your answers to two decimal places. Choose the letter solvint the best answer.\n\n# Lesson 5 problem solving practice volume of pyramids\n\nFor sure you need to master these basic equations solvjng the GED. Find the volume of the cone. Share buttons are a little bit lower.\n\nBHAM UNI THESIS\n\nFront answer Example of classical argument essay and industry to echo the endpoints of lesson problem solving volume of pyramids and cones answers arc. Rectangular pyramid;OP 6 2. In this volume of a pyramid worksheet, students solve 5 multiple choice problems. Parts of this lesson 4 principles of prisms and pyramids, surface area.", null, "The volume of the new pyramid is cm3. A simple demonstration explains the relationship in terms of volume between prisms and pyramids. The volume of the prism is ft 3. A rectangular prism has a solvig of cubic centimeters and a height of 4 centimeters.\n\nShow students the volume formula for square pyramids on the poster: The surface area of a pyramid is equal to the sum of its area of the base and the lateral area." ]
[ null, "http://fistf.info/essay.png", null, "https://www.havefunteaching.com/wp-content/uploads/2014/11/volume-worksheet-2.jpg", null, "https://images.slideplayer.com/31/9705096/slides/slide_12.jpg", null ]
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https://brilliant.org/practice/number-theory-warmups-level-4-challenges/
[ "", null, "Number Theory\n\n# Number Theory Warmups: Level 4 Challenges\n\nWhat is the remainder when $1^{2013}+2^{2013}+\\cdots +2012^{2013}+2013^{2013}$ is divided by $2014$?\n\n$\\large \\frac{1}{a}+\\frac{1}{b}=\\frac{1}{100000}$\n\nHow many distinct ordered pairs of positive integers $(a,b)$ are there which satisfy the above equation?\n\n$\\large \\displaystyle \\dfrac{a}{\\frac{b}{c}} = \\dfrac{\\frac{a}{b}}{c}$\n\nThe above equation is a common mistake made when interpreting fractions.\n\nHow many ordered triplets of integers ${(a,b,c)}$ with $-10\\leq a,b,c \\leq 10$ are there, such that the above equation is a true statement?\n\nGiven a positive integer $n$, let $p(n)$ be the product of the non-zero digits of $n$. (If $n$ has one digit, then $p(n)$ is equal to that digit.) Let\n\n$S = p(1) + p(2) + \\cdots + p(999).$\n\nWhat is the largest prime factor of $S$?\n\n$a, b$ and $c$ are distinct positive integers strictly greater than 1. If $abc$ divides $(ab-1)(bc-1)(ca-1)$, what is the value of $abc$?\n\n×" ]
[ null, "https://ds055uzetaobb.cloudfront.net/brioche/chapter/Number%20Theory%20Warmups-VjDjoa.png", null ]
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https://numpy.org/devdocs/reference/generated/numpy.core.records.array.html
[ "# numpy.core.records.array¶\n\ncore.records.array(obj, dtype=None, shape=None, offset=0, strides=None, formats=None, names=None, titles=None, aligned=False, byteorder=None, copy=True)[source]\n\nConstruct a record array from a wide-variety of objects.\n\nA general-purpose record array constructor that dispatches to the appropriate `recarray` creation function based on the inputs (see Notes).\n\nParameters\nobjany\n\nInput object. See Notes for details on how various input types are treated.\n\ndtypedata-type, optional\n\nValid dtype for array.\n\nshapeint or tuple of ints, optional\n\nShape of each array.\n\noffsetint, optional\n\nPosition in the file or buffer to start reading from.\n\nstridestuple of ints, optional\n\nBuffer (buf) is interpreted according to these strides (strides define how many bytes each array element, row, column, etc. occupy in memory).\n\nformats, names, titles, aligned, byteorder :\n\nIf `dtype` is `None`, these arguments are passed to `numpy.format_parser` to construct a dtype. See that function for detailed documentation.\n\ncopybool, optional\n\nWhether to copy the input object (True), or to use a reference instead. This option only applies when the input is an ndarray or recarray. Defaults to True.\n\nReturns\nnp.recarray\n\nRecord array created from the specified object.\n\nNotes\n\nIf obj is `None`, then call the `recarray` constructor. If obj is a string, then call the `fromstring` constructor. If obj is a list or a tuple, then if the first object is an `ndarray`, call `fromarrays`, otherwise call `fromrecords`. If obj is a `recarray`, then make a copy of the data in the recarray (if `copy=True`) and use the new formats, names, and titles. If obj is a file, then call `fromfile`. Finally, if obj is an `ndarray`, then return `obj.view(recarray)`, making a copy of the data if `copy=True`.\n\nExamples\n\n```>>> a = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])\narray([[1, 2, 3],\n[4, 5, 6],\n[7, 8, 9]])\n```\n```>>> np.core.records.array(a)\nrec.array([[1, 2, 3],\n[4, 5, 6],\n[7, 8, 9]],\ndtype=int32)\n```\n```>>> b = [(1, 1), (2, 4), (3, 9)]\n>>> c = np.core.records.array(b, formats = ['i2', 'f2'], names = ('x', 'y'))\n>>> c\nrec.array([(1, 1.0), (2, 4.0), (3, 9.0)],\ndtype=[('x', '<i2'), ('y', '<f2')])\n```\n```>>> c.x\nrec.array([1, 2, 3], dtype=int16)\n```\n```>>> c.y\nrec.array([ 1.0, 4.0, 9.0], dtype=float16)\n```\n```>>> r = np.rec.array(['abc','def'], names=['col1','col2'])\n>>> print(r.col1)\nabc\n```\n```>>> r.col1\narray('abc', dtype='<U3')\n```\n```>>> r.col2\narray('def', dtype='<U3')\n```" ]
[ null ]
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https://hal-u-picardie.archives-ouvertes.fr/hal-03621265v1/datacite
[ "https://hal-u-picardie.archives-ouvertes.fr/hal-03621265Guedda, MohammedMohammedGueddaLAMFA - Laboratoire Amiénois de Mathématique Fondamentale et Appliquée - UMR CNRS 7352 - UPJV - Université de Picardie Jules Verne - CNRS - Centre National de la Recherche ScientifiqueBenlahsen, M.M.BenlahsenUPJV - Université de Picardie Jules VerneLPMC - Laboratoire de Physique de la Matière Condensée - UR UPJV 2081 - UPJV - Université de Picardie Jules VerneSriti, M.M.SritiAchemlal, D.D.AchemlalExact similarity solutions for forced convection flow over horizontal plate in saturated porous medium with temperature-dependent viscosityHAL CCSD2017[MATH] Mathematics [math]DESSAIVRE, Louise2022-03-28 10:01:402023-03-24 14:53:262022-03-28 10:01:40enJournal articles10.1140/epjp/i2017-11652-01In this paper, we revisit a mathematical model representing a two-dimensional forced convection boundary-layer flow over a horizontal impermeable plate with a variable heat flux and viscosity. It is assumed that the fluid viscosity varies as an inverse linear function of temperature, the free stream velocity varies as an inverse linear of x and the wall heat flux varies with x as x.; where lambda > -1 and x measures the distance along the surface. Analytical local similarity solutions are presented which reveal that there are two competing effects: lambda and theta(e); where theta(e) is the variable viscosity parameter. It has been shown that for.e > 0 dual solutions exist and boundary separation occurs, while a unique local similarity solution exists for any theta(e) < 0." ]
[ null ]
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https://socratic.org/questions/58156fe77c0149413cc7cb5e
[ "# Question #7cb5e\n\nOct 31, 2016\n\nA\n\n#### Explanation:\n\nThe cell reaction when it is working is:\n\n$\\textsf{Z n + C {u}^{2 +} r i g h t \\le f t h a r p \\infty n s Z {n}^{2 +} + C u}$\n\nAt $\\textsf{{25}^{\\circ} C}$ a working form of the Nernst Equation is:\n\n$\\textsf{{E}_{c e l l} = {E}^{\\circ} - \\frac{0.0592}{n} \\log Q}$\n\n$\\textsf{Q}$ is the reaction quotient and is given by:\n\n$\\textsf{Q = \\frac{\\left[Z {n}^{2 +}\\right]}{\\left[C {u}^{2 +}\\right]}}$\n\n$\\textsf{n}$ is the number of moles of electrons transferred which, in this case, is 2.\n\nAs current is drawn from the cell as it is working, the potential difference between the two 1/2 cells gradually falls and eventually becomes zero.\n\nThe cell is now flat and $\\textsf{{E}_{c e l l} = 0}$.\n\nPutting this in to the Nernst Equation we get:\n\n$\\textsf{0 = 1.1 - \\frac{0.0592}{2} \\log Q}$\n\n$\\therefore$$\\textsf{\\log Q = \\frac{1.1 \\times 2}{0.0592} = 37.16}$\n\nThis gives $\\textsf{A}$ to be the correct response.\n\nThe system has now reached equilibrium at which point $\\textsf{K = Q}$.\n\nThis means that $\\textsf{K \\cong {10}^{37}}$ which is a fantastically high number and effectively tells us that the reaction has gone to completion as you would expect from our knowledge of displacement reactions." ]
[ null ]
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https://socratic.org/questions/how-do-you-evaluate-7-5-5-8-4
[ "# How do you evaluate [(7-5)^5/8]-4?\n\nFirst, note that the parentheses mean that $7 - 5$ should be computed first to get $7 - 5 = 2$. Then raise that to the fifth power to get ${\\left(7 - 5\\right)}^{5} = {2}^{5} = 32$ (powers come before divisions in order of operations). Then divide by 8 to get $\\frac{{\\left(7 - 5\\right)}^{2}}{8} = \\frac{32}{8} = 4$. Finally, subtract 4 to get $\\frac{{\\left(7 - 5\\right)}^{2}}{8} - 4 = 4 - 4 = 0$.\nThe outer brackets $\\left[\\right]$ are not technically needed in the notation." ]
[ null ]
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https://justaaa.com/finance/253924-what-is-the-simple-rate-of-return-of-a-micro
[ "Question\n\n# What is the simple rate of return of a micro-sprinkler irrigation system that costs \\$120,000 and...\n\n1. What is the simple rate of return of a micro-sprinkler irrigation system that costs \\$120,000 and has returns of \\$50,000 in year one, \\$40,000 in year two, \\$30,000 in year three, and \\$20,000 in year four? All cash outflows occur at the end of the year.\n1. 1%\n2. 7%\n3. 10%\n4. 17%\n5. 70%\n\nAns.\n\nSimple rate of Return = Net Income / Initial Investment\n\nNet Income = Total Cash Flows - Depreciation\n\nTotal Cash Flows = \\$ 50,000 + \\$ 40,000 + \\$ 30,000 + \\$ 20,000 = \\$ 140,000\n\nAs no salvage value is given Total Depreciation in four years = \\$ 120,000 (\\$ 30,000 per Year)\n\nNet Income = \\$ 140,000 - \\$ 120,000 = \\$ 20,000\n\nInitial Investment = \\$ 120,000\n\nSimple rate of Return = \\$ 20,000 / \\$ 120,000 = 0.1666 or 16.67%\n\nSimple rate of Return = 17% (rounded off)\n\nSo the correct option is D.\n\n#### Earn Coins\n\nCoins can be redeemed for fabulous gifts." ]
[ null ]
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https://www.stumblingrobot.com/2016/04/22/find-first-four-nonzero-terms-power-series-solution-y-x2-y2/
[ "Home » Blog » Find the first four nonzero terms of the power series solution of y′ = x2 + y2\n\n# Find the first four nonzero terms of the power series solution of y′ = x2 + y2\n\nConsider the differential equation", null, "with initial conditions", null, "when", null, ". Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.\n\nLet", null, "be the power-series solution of the differential equation. Then we must have", null, "From the initial condition", null, "when", null, "we have", null, ". Therefore, equating like powers of", null, "we have the following equations", null, "Therefore, we have", null, "" ]
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https://us.metamath.org/nfeuni/rabswap.html
[ "", null, "New Foundations Explorer < Previous   Next > Nearby theorems Mirrors  >  Home  >  NFE Home  >  Th. List  >  rabswap GIF version\n\nTheorem rabswap 2790\n Description: Swap with a membership relation in a restricted class abstraction. (Contributed by NM, 4-Jul-2005.)\nAssertion\nRef Expression\nrabswap {x A x B} = {x B x A}\n\nProof of Theorem rabswap\nStepHypRef Expression\n1 ancom 437 . . 3 ((x A x B) ↔ (x B x A))\n21abbii 2465 . 2 {x (x A x B)} = {x (x B x A)}\n3 df-rab 2623 . 2 {x A x B} = {x (x A x B)}\n4 df-rab 2623 . 2 {x B x A} = {x (x B x A)}\n52, 3, 43eqtr4i 2383 1 {x A x B} = {x B x A}\n Colors of variables: wff setvar class Syntax hints:   ∧ wa 358   = wceq 1642   ∈ wcel 1710  {cab 2339  {crab 2618 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-rab 2623 This theorem is referenced by: (None)\n Copyright terms: Public domain W3C validator" ]
[ null, "https://us.metamath.org/nfeuni/nf.gif", null ]
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https://hal.science/hal-02530253v3
[ "", null, "", null, "On the Search Efficiency of Parallel Lévy Walks on ${\\mathbb Z}^2$ - Archive ouverte HAL Access content directly\nReports (Research Report) Year : 2020\n\n## On the Search Efficiency of Parallel Lévy Walks on ${\\mathbb Z}^2$\n\nFrancesco d'Amore\nGeorge Giakkoupis\n• Function : Author\n• PersonId : 928662\nEmanuele Natale\n\n#### Abstract\n\nMotivated by the \\emph{Lévy flight foraging hypothesis} -- the premise that the movement of various animal species searching for food resembles a \\emph{Lévy walk} -- we study the search efficiency of parallel Lévy walks on the infinite 2-dimensional grid. We assume that $k$ independent identical discrete-time Lévy walks, with exponent parameter $\\alpha \\in(1,+\\infty)$, start simultaneously at the origin, and we are interested in the time $h_{\\alpha,k,\\ell}$ until some walk visits a given target node at distance $\\ell$ from the origin. First, we observe that the total work, i.e., the product $k\\cdot h_{\\alpha,k,\\ell}$, is at least $\\Omega(\\ell^2)$, for any combination of the parameters $\\alpha,k,\\ell$. Then we provide a comprehensive analysis of the time and work, for the complete range of these parameters. Our main finding is that for any $\\alpha$, there is a specific choice of $k$ that achieves optimal work, $\\tilde{\\mathcal{O}}\\left(\\ell^2\\right)$, whereas all other choices of $k$ result in sub-optimal work. In particular, in the interesting super-diffusive regime of $2 < \\alpha < 3$, the optimal value for $k$ is $\\tilde \\Theta\\left(\\ell^{1-(\\alpha-2)}\\right)$. Our results should be contrasted with several previous works showing that the exponent $\\alpha = 2$ is optimal for a wide range of related search problems on the plane. On the contrary, in our setting of multiple walks which measures efficiency in terms of the natural notion of work, no single exponent is optimal: for each $\\alpha$ (and $\\ell$) there is a specific choice of $k$ that yields optimal efficiency.\n\n### Dates and versions\n\nhal-02530253 , version 1 (03-04-2020)\nhal-02530253 , version 2 (16-04-2020)\nhal-02530253 , version 3 (03-08-2020)\nhal-02530253 , version 4 (19-02-2021)\nhal-02530253 , version 5 (09-12-2021)\n\n### Identifiers\n\n• HAL Id : hal-02530253 , version 3\n• ARXIV :\n\n### Cite\n\nAndrea Clementi, Francesco d'Amore, George Giakkoupis, Emanuele Natale. On the Search Efficiency of Parallel Lévy Walks on ${\\mathbb Z}^2$. [Research Report] Inria & Université Cote d'Azur, CNRS, I3S, Sophia Antipolis, France; Università degli Studi di Roma \"Tor Vergata\"; Univ Rennes, Inria, CNRS, IRISA, France. 2020. ⟨hal-02530253v3⟩\n\n### Export\n\nBibTeX XML-TEI Dublin Core DC Terms EndNote DataCite\n551 View" ]
[ null, "https://piwik-hal.ccsd.cnrs.fr//matomo.php", null, "https://piwik-hal.ccsd.cnrs.fr//matomo.php", null ]
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https://www.ferdinand-malcher.de/blog/convert-wavelength-in-nm-to-rgb.html
[ "", null, "## Convert wavelength in nm to RGB\n\nMaybe some day you'll need to convert wavelengthes in nanometers to RGB values.\nI just needed it for visualizing emission values of flourescent dyes in a lab information system.\nI didn't want to make that by hand, picking e.g. 20 values from the color range to let them stand for everything between them, but wanted to have a proper and (at least approximately) exact solution.\n\nThe function below is based on an algorithm developed by Dan Bruton I found in the web.\nI just transcribed it to PHP, and it works pretty good.\n\n```\n<?php\nfunction wavtorgb(\\$w){\n\\$w = intval(\\$w);\nif(!\\$w) return false;\n\n//color\nif(\\$w >= 100 AND \\$w < 380){\n\\$r = 0.91;\n\\$g = 0.77;\n\\$b = 1.0;\n\n}elseif(\\$w >= 380 AND \\$w < 440){\n\\$r = abs(\\$w - 440) / (440 - 350);\n\\$g = 0.0;\n\\$b = 1.0;\n\n}elseif(\\$w >= 440 AND \\$w < 490){\n\\$r = 0.0;\n\\$g = abs(\\$w - 440) / (490 - 440);\n\\$b = 1.0;\n\n}elseif(\\$w >= 490 AND \\$w < 510){\n\\$r = 0.0;\n\\$g = 1.0;\n\\$b = abs(\\$w - 510) / (510 - 490);\n\n}elseif(\\$w >= 510 AND \\$w < 580){\n\\$r = abs(\\$w - 510) / (580 - 510);\n\\$g = 1.0;\n\\$b = 0.0;\n\n}elseif(\\$w >= 580 AND \\$w < 645){\n\\$r = 1.0;\n\\$g = abs(\\$w - 645) / (645 - 580);\n\\$b = 0.0;\n\n}elseif(\\$w >= 645 AND \\$w < 780){\n\\$r = 1.0;\n\\$g = 0.0;\n\\$g = 0.0;\n\n}elseif(\\$w >= 780 AND \\$w < 1000000){\n\\$r = 1.0;\n\\$g = 0.31;\n\\$b = 0.31;\n\n}else{\n\\$r = 0.0;\n\\$g = 0.0;\n\\$b = 0.0;\n}\n\n//let the intensity sss fall off near the vision limits\nif(\\$w >= 380 AND \\$w < 420){\n\\$sss = 0.3 + 0.7*(\\$w - 350) / (420 - 350);\n\n}elseif(\\$w >= 420 AND \\$w <= 700){\n\\$sss = 1.0;\n\n}elseif(\\$w > 700 AND \\$w <= 780){\n\\$sss = 0.3 + 0.7*(780 - \\$w) / (780 - 700);\n\n}else{\n\\$sss = 1.0;\n}\n\\$sss *= 255;\n\n\\$r = intval(\\$sss * \\$r);\n\\$g = intval(\\$sss * \\$g);\n\\$b = intval(\\$sss * \\$b);\n\n//label\nif(\\$w >= 100 AND \\$w < 380){\n\\$label = \\$w.\"nm (UV)\";\n\n}elseif(\\$w >= 380 AND \\$w < 780){\n\\$label = \\$w.\"nm\";\n\n}elseif(\\$w >= 780 AND \\$w < 1000000){\n\\$label = \\$w.\"nm (IR)\";\n\n}else{\nreturn false;\n}\n\n\\$return = array(\\$r, \\$g, \\$b, \\$label);\nreturn \\$return;\n}\n?>\n```" ]
[ null, "https://www.ferdinand-malcher.de/media/layout/headline.png", null ]
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https://support.minitab.com/en-us/minitab/19/help-and-how-to/statistical-modeling/anova/how-to/general-manova/interpret-the-results/all-statistics-and-graphs/sscp-matrix/
[ "# SSCP matrix for General MANOVA\n\nMinitab can display three sums of squares and cross products (SSCP) matrices to examine variability, error, and partial correlations. To display the SSCP matrix, go to Stat > ANOVA > General MANOVA > Results and select Matrices (hypothesis, error, partial correlations) under Display of Results.\n\nUse the SSCP matrix to assess the partitioning of variability in the same way you do for univariate sums of squares. The matrix labeled as SSCP Matrix for model term is the hypothesis SSCP matrix for the responses with the specified model term. The diagonal elements are the univariate ANOVA sums of squares for the model term for each response. The off-diagonal elements of this matrix are the cross products.\n\nThe matrix labeled as SSCP Matrix for Error is the error sums of squares and cross products matrix. The diagonal elements of this matrix are the univariate ANOVA error sums of squares for the response variables. The off-diagonal elements of this matrix are the cross products. This matrix is displayed one time, after the SSCP matrix for the first model term.\n\nUse the matrix of partial correlations, labeled as Partial Correlations for the Error SSCP Matrix, to assess how the response variables are related. The off-diagonal elements are the correlations among the residuals or, equivalently, the correlations among the responses conditioned on the model. If the correlation between the responses is low, you may want to perform univariate ANOVA tests for the individual responses. This matrix is displayed one time, after the SSCP matrix for error.\n\n## Interpretation\n\nFor these results, the univariate sums of squares for usability and quality for method are 29.07 and 9,220. The univariate error sums of squares for usability and quality are 49.75 and 68.59. The partial correlation between usability and quality is 0.50413, indicating that the response variables are moderately correlated.\n\nSSCP Matrix (adjusted) for Method Usability Quality Rating Rating Usability Rating 29.07 16.372 Quality Rating 16.37 9.220\nSSCP Matrix (adjusted) for Error Usability Quality Rating Rating Usability Rating 49.75 29.45 Quality Rating 29.45 68.59\nPartial Correlations for the Error SSCP Matrix Usability Quality Rating Rating Usability Rating 1.00000 0.50413 Quality Rating 0.50413 1.00000\nBy using this site you agree to the use of cookies for analytics and personalized content.  Read our policy" ]
[ null ]
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https://www.colorhexa.com/d6cadd
[ "# #d6cadd Color Information\n\nIn a RGB color space, hex #d6cadd (also known as Languid lavender) is composed of 83.9% red, 79.2% green and 86.7% blue. Whereas in a CMYK color space, it is composed of 3.2% cyan, 8.6% magenta, 0% yellow and 13.3% black. It has a hue angle of 277.9 degrees, a saturation of 21.8% and a lightness of 82.9%. #d6cadd color hex could be obtained by blending #ffffff with #ad95bb. Closest websafe color is: #cccccc.\n\n• R 84\n• G 79\n• B 87\nRGB color chart\n• C 3\n• M 9\n• Y 0\n• K 13\nCMYK color chart\n\n#d6cadd color description : Light grayish violet.\n\n# #d6cadd Color Conversion\n\nThe hexadecimal color #d6cadd has RGB values of R:214, G:202, B:221 and CMYK values of C:0.03, M:0.09, Y:0, K:0.13. Its decimal value is 14076637.\n\nHex triplet RGB Decimal d6cadd `#d6cadd` 214, 202, 221 `rgb(214,202,221)` 83.9, 79.2, 86.7 `rgb(83.9%,79.2%,86.7%)` 3, 9, 0, 13 277.9°, 21.8, 82.9 `hsl(277.9,21.8%,82.9%)` 277.9°, 8.6, 86.7 cccccc `#cccccc`\nCIE-LAB 82.785, 7.606, -7.916 61.901, 61.758, 77.062 0.308, 0.308, 61.758 82.785, 10.977, 313.855 82.785, 5.601, -13.498 78.586, 3.076, -3.13 11010110, 11001010, 11011101\n\n# Color Schemes with #d6cadd\n\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #d1ddca\n``#d1ddca` `rgb(209,221,202)``\nComplementary Color\n• #cdcadd\n``#cdcadd` `rgb(205,202,221)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #ddcadb\n``#ddcadb` `rgb(221,202,219)``\nAnalogous Color\n• #caddcd\n``#caddcd` `rgb(202,221,205)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #dbddca\n``#dbddca` `rgb(219,221,202)``\nSplit Complementary Color\n• #caddd6\n``#caddd6` `rgb(202,221,214)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #ddd6ca\n``#ddd6ca` `rgb(221,214,202)``\nTriadic Color\n• #cad1dd\n``#cad1dd` `rgb(202,209,221)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #ddd6ca\n``#ddd6ca` `rgb(221,214,202)``\n• #d1ddca\n``#d1ddca` `rgb(209,221,202)``\nTetradic Color\n• #b29bbf\n``#b29bbf` `rgb(178,155,191)``\n• #beabc9\n``#beabc9` `rgb(190,171,201)``\n• #cabad3\n``#cabad3` `rgb(202,186,211)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #e2dae7\n``#e2dae7` `rgb(226,218,231)``\n• #eee9f1\n``#eee9f1` `rgb(238,233,241)``\n• #faf9fb\n``#faf9fb` `rgb(250,249,251)``\nMonochromatic Color\n\n# Alternatives to #d6cadd\n\nBelow, you can see some colors close to #d6cadd. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #d1cadd\n``#d1cadd` `rgb(209,202,221)``\n• #d3cadd\n``#d3cadd` `rgb(211,202,221)``\n• #d4cadd\n``#d4cadd` `rgb(212,202,221)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #d8cadd\n``#d8cadd` `rgb(216,202,221)``\n• #d9cadd\n``#d9cadd` `rgb(217,202,221)``\n• #dbcadd\n``#dbcadd` `rgb(219,202,221)``\nSimilar Colors\n\n# #d6cadd Preview\n\nText with hexadecimal color #d6cadd\n\nThis text has a font color of #d6cadd.\n\n``<span style=\"color:#d6cadd;\">Text here</span>``\n#d6cadd background color\n\nThis paragraph has a background color of #d6cadd.\n\n``<p style=\"background-color:#d6cadd;\">Content here</p>``\n#d6cadd border color\n\nThis element has a border color of #d6cadd.\n\n``<div style=\"border:1px solid #d6cadd;\">Content here</div>``\nCSS codes\n``.text {color:#d6cadd;}``\n``.background {background-color:#d6cadd;}``\n``.border {border:1px solid #d6cadd;}``\n\n# Shades and Tints of #d6cadd\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #060407 is the darkest color, while #fbfafc is the lightest one.\n\n• #060407\n``#060407` `rgb(6,4,7)``\n• #100c13\n``#100c13` `rgb(16,12,19)``\n• #1b141f\n``#1b141f` `rgb(27,20,31)``\n• #251b2b\n``#251b2b` `rgb(37,27,43)``\n• #2f2337\n``#2f2337` `rgb(47,35,55)``\n• #3a2b42\n``#3a2b42` `rgb(58,43,66)``\n• #44324e\n``#44324e` `rgb(68,50,78)``\n• #4e3a5a\n``#4e3a5a` `rgb(78,58,90)``\n• #594266\n``#594266` `rgb(89,66,102)``\n• #634972\n``#634972` `rgb(99,73,114)``\n• #6e517e\n``#6e517e` `rgb(110,81,126)``\n• #78598a\n``#78598a` `rgb(120,89,138)``\n• #826096\n``#826096` `rgb(130,96,150)``\nShade Color Variation\n• #8c6aa0\n``#8c6aa0` `rgb(140,106,160)``\n• #9576a7\n``#9576a7` `rgb(149,118,167)``\n• #9f82af\n``#9f82af` `rgb(159,130,175)``\n• #a88eb7\n``#a88eb7` `rgb(168,142,183)``\n• #b19abe\n``#b19abe` `rgb(177,154,190)``\n• #baa6c6\n``#baa6c6` `rgb(186,166,198)``\n• #c4b2ce\n``#c4b2ce` `rgb(196,178,206)``\n• #cdbed5\n``#cdbed5` `rgb(205,190,213)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #dfd6e5\n``#dfd6e5` `rgb(223,214,229)``\n• #e8e2ec\n``#e8e2ec` `rgb(232,226,236)``\n• #f2eef4\n``#f2eef4` `rgb(242,238,244)``\n• #fbfafc\n``#fbfafc` `rgb(251,250,252)``\nTint Color Variation\n\n# Tones of #d6cadd\n\nA tone is produced by adding gray to any pure hue. In this case, #d4d1d6 is the less saturated color, while #dfa9fe is the most saturated one.\n\n• #d4d1d6\n``#d4d1d6` `rgb(212,209,214)``\n• #d5cdda\n``#d5cdda` `rgb(213,205,218)``\n• #d6cadd\n``#d6cadd` `rgb(214,202,221)``\n• #d7c7e0\n``#d7c7e0` `rgb(215,199,224)``\n• #d8c3e4\n``#d8c3e4` `rgb(216,195,228)``\n• #d9c0e7\n``#d9c0e7` `rgb(217,192,231)``\n• #dabdea\n``#dabdea` `rgb(218,189,234)``\n• #dab9ee\n``#dab9ee` `rgb(218,185,238)``\n• #dbb6f1\n``#dbb6f1` `rgb(219,182,241)``\n• #dcb3f4\n``#dcb3f4` `rgb(220,179,244)``\n• #ddaff8\n``#ddaff8` `rgb(221,175,248)``\n• #deacfb\n``#deacfb` `rgb(222,172,251)``\n• #dfa9fe\n``#dfa9fe` `rgb(223,169,254)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #d6cadd is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
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https://discuss.pytorch.org/t/cross-entropy-for-multivariate-classification/67471
[ "# Cross entropy for multivariate classification\n\nI have a classification problem where each sample output consists of four values each of which is one of the 6 classes, i.e `y` has shape `(batchsize, 4)`. So far, I have created a toy model which outputs a logits tensor `out` having the shape `(batchsize, 6(=num_classes), 4)`. I am calculating the loss for each of the target value and summing up, followed by backprop:\n\n``````out = model(in)\nout.shape #(batchsize, 6, 4)\n\nout1 = out[:,:,0].squeeze(-1)\nout1.shape #(batchsize, 6)\nout2 = out[:,:,1].squeeze(-1)\nout3 = out[:,:,2].squeeze(-1)\nout4 = out[:,:,3].squeeze(-1)\ny1 = y[:,0].squeeze(-1)\ny1.shape #(batchsize)\ny2 = y[:,1].squeeze(-1)\ny3 = y[:,2].squeeze(-1)\ny4 = y[:,3].squeeze(-1)\n\ntotal_loss = criterion(out1, y1) + criterion(out2, y2) + criterion(out3, y3) + criterion(out4, y4)\n\ntotal_loss.backward()\noptimizer.step()\n``````\n\nI want to know if this is the correct workaround for multivariate classification\n\nHi Hashir!\n\nIf I understand what you are trying to do here (and I am not at all\nsure that I do), then torch.nn.CrossEntropyLoss should do what\nyou want without further manipulation. (Look at the parts where\nthe documentation talks about the “K-dimensional case.”)\n\nLet the output of your model be a tensor of shape `(batchsize, 6, 4)`.\nI will call this tensor `prediction`. I will call the last dimension (the\n“4” dimension) the “channel.” (So in your use case you have four\n“channels.”) For a given sample in the batch and a given channel,\nyour `prediction` consists of six logits, one for each of your six classes.\n\nYou use your loss function to compare your `prediction` with your\n`target`. These are the known class labels that you use for training.\n`target` will be a tensor of shape `(batchsize, 4)`, and for a given\nsample within the batch and for a given channel (the “4” dimension)\nyou will have a single value that is an integer class label ranging\nfrom 0 to 5.\n\n(Note that `prediction` and `target` are not of the same shape:\n`prediction` carries all of the batchsize, class, and channel dimensions,\nwhile `target`, because it uses a single integer class label per item,\ncarries only the batchsize and channel dimensions.)\n\nBy default, `CrossEntropyLoss` will average your loss over both the\nbatchsize and channel dimensions. If you prefer the sum, then pass\nin the optional `reduction = 'sum'` argument when you construct\nyour `CrossEntropyLoss` loss-function object.\n\nBest.\n\nK. Frank\n\n2 Likes\n\n@KFrank I had a look over the “K-dimensional case” in torch.nn.CrossEntropyLoss and it really helped solve my problem. What I was essentially doing can be done with `criterion = torch.nn.CrossEntropyLoss(reduction='sum')`.\n\nThanks for the help!" ]
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{"ft_lang_label":"__label__en","ft_lang_prob":0.8978357,"math_prob":0.951418,"size":2020,"snap":"2022-05-2022-21","text_gpt3_token_len":463,"char_repetition_ratio":0.12400793,"word_repetition_ratio":0.011869436,"special_character_ratio":0.22079208,"punctuation_ratio":0.11557789,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956835,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-21T04:13:02Z\",\"WARC-Record-ID\":\"<urn:uuid:5a13169f-af08-4461-9dae-99e8bc77bd74>\",\"Content-Length\":\"21169\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:2a76d3aa-88aa-48b3-b51c-043f131c25a9>\",\"WARC-Concurrent-To\":\"<urn:uuid:da397b3e-2823-421b-80e2-7f9820e74837>\",\"WARC-IP-Address\":\"159.203.145.104\",\"WARC-Target-URI\":\"https://discuss.pytorch.org/t/cross-entropy-for-multivariate-classification/67471\",\"WARC-Payload-Digest\":\"sha1:WGLXBJYPMCB4DAOX7WLRCLSD7P3XLZKX\",\"WARC-Block-Digest\":\"sha1:ZO4UKEW2LD4XGHRMFHYVIKS7ZGS5X7XY\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662534773.36_warc_CC-MAIN-20220521014358-20220521044358-00750.warc.gz\"}"}
https://bikeloanemi.com/hero-electric-nyx-emi-calculator/
[ "# Hero Electric NYX EMI Calculator\n\nCalculate Your Hero Electric NYX Loan EMI Instantly\n\nHero Electric NYX\n\n* Ex-showroom Price\n\nThe calculated EMI for Hero Electric NYX starts at ₹ 1,937 per month at a bank interest rate of 9.7% for a tenure of 36 months. The below EMI calculator tool gives you a comprehensive information about the total payable amount and guides you in finding the best finance option for your Electric NYX.\n\nDown Payment\n\nInterest Rate (Per/Annum)\n\nLoan Term (in months)\n\nMonths\n• Loan Amount\n• Interest Payable\n• Total loan payment\nEMI\nPer/Month\nCalculated on Ex-showroom Price\n\n#### ₹ document.write(formatNumber('60000'));\n\n* Ex-showroom Price\n\nThe calculated EMI for Hero Electric NYX starts at ₹ 1,937 per month at a bank interest rate of 9.7% for a tenure of 36 months. The below EMI calculator tool gives you a comprehensive information about the total payable amount and guides you in finding the best finance option for your Electric NYX.\n\nDown Payment\nInterest Rate (Per/Annum)\nLoan Term (in months)\nEMI\nPer/Month\nCalculated on Ex-showroom Price" ]
[ null ]
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https://studylib.net/doc/10605481/18.303-problem-set-7-solutions-problem-1--10-10-points--
[ "18.303 Problem Set 7 Solutions Problem 1 (10+10 points):", null, "18.303 Problem Set 7 Solutions\nProblem 1 (10+10 points):\ng(x0 )d3 x0 , the second term is:\nZ\nZ\n&middot; (x0 − x1 )d3 x0 = ∇0 g|x1 &middot;\nx0 d3 x0 − x1 volume(V ) ,\n(a) If we plug the Taylor expansion of g into\n∇0 g|x1\nR\nV\nV\nV\n0\nwhere we have pulled the constant (x -independent) terms out of the integral. Hence, this is zero (for all x0 and\nx) if\nZ\n1\nx 0 d3 x 0 ,\nx1 =\nvolume(V ) V\nwhich is just the centroid of the volume V .\n(b) From the notes, we can write g(x0 ) more explicitly as\ng(x0 ) =\nHence, if we compute the Hessian Hij =\nH=\n16π 2 |x1\n(x0 − x0 )\n(x0 − x)\n&middot;\n.\n4π|x0 − x|3 4π|x0 − x0 |3\n∂2g\n∂x0i ∂x0j ,\nevaluated at x0 = x1 , we get (after some algebra):\n2I\n9(x1 − x)(x1 − x)T (x1 − x0 )(x1 − x0 )T\n+\n3\n3\n− x| |x1 − x0 |\n16π 2 |x1 − x|5 |x1 − x0 |5\n3(x1 − x)(x1 − x)T\n3(x1 − x0 )(x1 − x0 )T\n−\n−\n,\n2\n5\n3\n16π |x1 − x| |x1 − x0 |\n16π 2 |x1 − x|3 |x1 − x0 |5\nwhere I is the 3 &times; 3 identity matrix and T is transposition of the column vectors. The first time comes when\nboth derivatives fall on the numerator, the second term when both fall on the denominator, and the third\nand fourth terms are when one derivative acts on the numerator and one on the denominator. [Note that the\n(x0 − x0 )(x0 − x0 )T is a column vector times a row vector, an “outer product”, which by the rules of matrix\nmultiplication produces a 3 &times; 3 matrix.] You may not have been able to write H in such a compact form.\nThe resulting “quadrupole” term in B̂u0 is then\nZ\nZ\nX\n0\nT\n0\n3 0\nln(c2 /c1 ) (x − x1 ) H(x − x1 )d x = ln(c2 /c1 )\nHij\n(x0i − x1i )(x0j − x1j )d3 x0 .\nV\ni,j\nV\nR\nNote that the integrals V (x0i − x1i )(x0j − x1j )d3 x0 are properties of the geometric shape V only, independent of\nthe source location x0 or the observer location x.\nFrom the formula for H, we can see by inspection that this term scales as ∼ 1/|x − x1 |3 and ∼ 1/|x0 − x1 |3 ,\nwhich means that at large separations it is negligible compared to the dipole term from class.\nProblem 2 (5+5+(5+5) points):\n(a) We first need to compute u0.5 from the initial u0 . We can do that with second-order accuracy using, e.g. a\nCrank-Nicolson scheme:\n−1 A∆t\nA∆t\n0.5\nu = 1−\n1−\nu0 .\n4\n4\nThen, given u0 and u0.5 , or (by induction) from uk and uk+0.5 for k = 0, 1, 2, . . ., we can get uk+1 via the first\nleapfrog equation\nuk+1 = uk + Auk+0.5 ∆t\nand then we can get uk+1.5 from the second leapfrog equation\nuk+1.5 = uk+0.5 + Auk ∆t.\nIn this way we can compute un and un+0.5 for as many n &gt; 0 as desired.\n1\n(b) Applying the first leapfrog equation and then the second equation, we have\nun+1 − un\nun − un−1\n−\n= A un+0.5 − un−0.5 = ∆tA2 un ,\n∆t\n∆t\nor\nun+1 − 2un + un−1\n= A2 un .\n∆t2\n2\nBut the left-hand side is just our familiar finite-difference approximation for ∂∂t2u evaluated at t = n∆t, and so\n2\n∂ ∂u\n∂\n∂u\n2\nthis is an approximation for (deriving from the original PDE) ∂∂t2u = ∂t\n∂t = ∂t Âu = Â ∂t = Â u.\n(c) If we plug in unm = eikm Gn and un+0.5\n= eikm Gn+0.5 a into the first leapfrog equation, evaluated at the point m\nm\nin space, we obtain\n2\nik\n−ik\nun+0.5 − 2un+0.5\n+ un+0.5\nun+1\n− unm\nG−1\nm\nm−1\nm\nikm n+0.5 e − 2 + e\nikm n+0.5 −4 sin (k/2)\n=\nae\nG\n=\nae\nG\n= eikm Gn\n= m+1\n∆t\n∆t\n∆x2\n∆x2\n∆x2\nwhich simplifies to\nG0.5 − G−0.5\n4∆t sin2 (k/2)\n.\n=−\na\n∆x2\nIf we plug into the second leapfrog equation, we similarly get\n2\nun\n− 2unm + unm−1\n− un−0.5\nun+0.5\n1 − G−1\nm\nm\nikm n −4 sin (k/2)\n= aeikm Gn+0.5\n= m+1\n=\ne\nG\n,\n∆t\n∆t\n∆x2\n∆x2\nwhich simplifies to\na(G0.5 − G−0.5 ) = −\n4∆t sin2 (k/2)\n.\n∆x2\n(i) Combining the two equations, we have a = 1/a, or a2 = 1 =⇒ a = &plusmn;1.\n(ii) For a = +1, we get\nG0.5 − G−0.5 = −\n√\n4∆t sin2 (k/2)\n1\n≤ 0 =⇒ G ≤ √ =⇒ G ≤ 1.\n2\n∆x\nG\n2" ]
[ null, "https://s2.studylib.net/store/data/010605481_1-e9cc42e425f2fd2a3a6c1b18e6ea84cd-768x994.png", null ]
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https://www.cemmap.ac.uk/publication/optimal-uniform-convergence-rates-and-asymptotic-normality-for-series-estimators-under-weak-dependence-and-weak-conditions/
[ "# Optimal uniform convergence rates and asymptotic normality for series estimators under weak dependence and weak conditions\n\n22 December 2014\n\n### Type\n\nWorking Paper (CWP46/14)\n\n### DOI\n\nWe show that spline and wavelet series regression estimators for weakly dependent regressors attain the optimal uniform (i.e. sup-norm) convergence rate (n= log n)–p=(2p+d) of Stone (1982), where d is the number of regressors and p is the smoothness of the regression function. The optimal rate is achieved even for heavy-tailed martingale difference errors with finite (2 + (d=p))th absolute moment for d=p < 2.We also establish the asymptotic normality of t statistics for possibly nonlinear, irregular functionals of the conditional mean function under weak conditions. The results are proved by deriving a new exponential inequality for sums of weakly dependent random matrices, which is of independent interest." ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.87214863,"math_prob":0.9736731,"size":909,"snap":"2022-27-2022-33","text_gpt3_token_len":197,"char_repetition_ratio":0.107182324,"word_repetition_ratio":0.0,"special_character_ratio":0.20242023,"punctuation_ratio":0.059210528,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99536324,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-06-30T15:47:25Z\",\"WARC-Record-ID\":\"<urn:uuid:b2b4621a-fc24-47cd-a1b8-360ceb05dd06>\",\"Content-Length\":\"50606\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1a8600fe-1ad1-486e-9eab-6f673ada1697>\",\"WARC-Concurrent-To\":\"<urn:uuid:beafbef4-356a-49c7-b64a-1ffc9319480c>\",\"WARC-IP-Address\":\"35.205.160.190\",\"WARC-Target-URI\":\"https://www.cemmap.ac.uk/publication/optimal-uniform-convergence-rates-and-asymptotic-normality-for-series-estimators-under-weak-dependence-and-weak-conditions/\",\"WARC-Payload-Digest\":\"sha1:M2BOVVYLEXU44NLQ4EGZFAX6ZYVBR3OT\",\"WARC-Block-Digest\":\"sha1:TCPDRRGY5HKDSWZ4IR752B3MPSPRMJNT\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-27/CC-MAIN-2022-27_segments_1656103850139.45_warc_CC-MAIN-20220630153307-20220630183307-00746.warc.gz\"}"}
https://www.cs.bham.ac.uk/~mhe/propositional-truncation-via-rewriting/index.html
[ "-- Martin Escardo, September 2017, based on earlier ideas an Agda files.\n\nmodule index where\n\n-- The large type of propositions:\n\nopen import prop\n\n-- The type of propositions is a set, assuming functional and\n-- propositional extensionality:\n\nopen import propisset\n\n-- We can define large propositional truncation by universal\n-- quantification over propositions. Then by resizing it we get the\n-- usual propositional truncation. But we only apply resizing to large\n-- propositions which arise as truncations.\n\nopen import proptrunc\n\n-- We then develop some amount of logic in the type Prop of\n-- propositions, where we define the logical connectives and their\n-- introduction and elimination rules following the ideas of the HoTT\n-- book. We then prove that\n--\n-- false = ∀ r. r\n-- p ∧ q = ∀ r. (p ⇒ q ⇒ r) ⇒ r\n-- p ∨ q = ∀ r. (p ⇒ r) ⇒ (q ⇒ r) ⇒ r\n-- ∃ p = ∀ r. (∀ x. p(x) ⇒ r) ⇒ r\n\nopen import logic\n\n-- We then prove the axiom of description: for any set X and any\n-- p:X→Prop,\n--\n-- (∃!(x:X).p(x))=true → Σ(x:X).p(x)=true.\n\nopen import description" ]
[ null ]
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https://boklm.eu/prolog/page_11.html
[ "A   Meta logical Constructs\n\nvar(X)\nTrue if X is a variable.var(X)\nnonvar\nTrue if X is not a variable.nonvar(X)\natom(X)\nTrue if X is an atom.atom(X)\ninteger(X)\nTrue if X is an integer.integer(X)\natomic(X)\nTrue if X is an atom or a number.atomic(X)\nfloat(X)\nTrue if X is a floiting point number.float(X)\nfunctor(Term,Name,Arity)\nTrue if the arity of Term is Arity and his name Name.functor()\n\n <-- SWI-Prolog - Input and Output -->" ]
[ null ]
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http://www.html5gamedevs.com/topic/43076-how-do-you-offset-specific-frames-on-a-tilemap/
[ "Recommended Posts", null, "I'm trying to create an isometric game.\n\nThe only trouble is that there's no magical \"Hexagonal Grid\" with Tilemaps.\n\nAs such, I need to offset each tile, but so far none of the properties seem to be doing anything.\n\nHere's what i've tried so far:\n\n``````map = game.add.tilemap('map', 219, 253);\n\nlayer = map.createLayer(0);\n\nlayer.resizeWorld();\n\nfor(var y = 0; y < map.height; ++y){\n\nfor(var x = 0; x < map.width; ++x){\n\nvar tile = map.getTile(x, y);\n\nif (tile){\ntile.Y += Math.random(0,30000);\ntile.worldY += Math.random(0,30000);\ntile.y += Math.random(0,30000);\ntile.centerY += Math.random(0,30000);\n}\n\nconsole.log(tile);\n\n}\n}``````\n\nJoin the conversation\n\nYou can post now and register later. If you have an account, sign in now to post with your account.\nNote: Your post will require moderator approval before it will be visible.", null, "×   Pasted as rich text.   Paste as plain text instead\n\nOnly 75 emoji are allowed." ]
[ null, "https://i.imgur.com/aIjKvro.png", null, "http://www.html5gamedevs.com/uploads/set_resources_1/84c1e40ea0e759e3f1505eb1788ddf3c_default_photo.png", null ]
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https://www.slideserve.com/keona/momentum
[ "", null, "Download", null, "Download Presentation", null, "Momentum\n\n# Momentum\n\nDownload Presentation", null, "## Momentum\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Momentum • What is momentum? • p = mv • How does momentum change? (impulse)‏ • I = mΔv = fΔt • Quiz! • Conservation of momentum • pinitial=pfinal • Elastic and inelastic collisions • Quiz! • Momentum conservation in 2D • Problem Solving • Integrating our knowledge\n\n2. What is Momentum?\n\n3. Momentum in the Vernacular • In everyday experience, momentum is the amount “unf” an object has So what factors affect the momentum of an object?\n\n4. What affects Momentum? Which has more “unf”? • A biker going at 20 mph • A car going at 20 mph\n\n5. What affects Momentum? Which has more “unf”? • A biker going at 20 mph • A car going at 20 mph A car will certainly hurt more, why? Because it is more massive (more mass)‏\n\n6. What affects Momentum? Which has more “unf”? • A car going at 10 mph • A car going at 2 mph\n\n7. What affects Momentum? Which has more “unf”? • A car going at 10 mph • A car going at 2 mph The faster car will have more “unf”, why? Because faster things are harder to stop\n\n8. Momentum Defined Momentum is the product of mass and velocity This is normally written p = m x v Bolded letters denote vectors What are the units of momentum? p = m x v m: kg v: m/s p: kg • m/s :kilogram meters per second\n\n9. p=mv What is the momentum of a bee that weighs 10 grams and flies at 2 m/s? How does that compare to a tortoise that weighs 1kg and moves at .05m/s?\n\n10. p=mv What is the momentum of a bee that weighs 10 grams and flies at 2 m/s? 10g=.01kg p=mv=.01x2= .02kgm/s How does that compare to a tortoise that weighs 1kg and moves at .05m/s? p=mv=1x.05= .05kgm/s The tortoise has more momentum.\n\n11. p=mv Which has more momentum, my car or me?\n\n12. p=mv Which has more momentum, my car or me? vcar=0 vme= something like 1m/s pcar = mcar * vcar = mcar *0 = 0 pme = mme* vme = mme * something like 1m/s = more than 0\n\n13. • How does momentum change? (impulse)‏ • I = mΔv = fΔt\n\n14. How Does Momentum of an Object Change? p=mv Consider Δp=Δmv What does this mean? Why is this not a change in momentum of the object?\n\n15. How Does Momentum of an Object Change? p=mv Consider Δp=mΔv What does this mean?\n\n16. How Does Momentum of an Object Change? p=mv Consider Δp=mΔv This means that velocity is changing. Unlike Δm, Δv does not imply that the object is falling apart or clumping together\n\n17. Introducing Impulse • Δp is know as impulse (I)‏ • Think of impulse as a change from the default path. (Momentum would keep carrying me this direction but I changed it)‏ • I had a sudden impulse to “” (I was suddenly did something that was not planned)‏\n\n18. Δp = I = mΔv A bowling ball (5kg) moving at 5m/s bowls through a set of bowling pins. Right before the ball falls down the shoot it is going 3m/s What is the impulse the bowling pins provide to a bowling ball?\n\n19. Δp = I = mΔv A bowling ball (5kg) moving at 5m/s bowls through a set of bowling pins. Right before the ball falls down the shoot it is going 3m/s What is the impulse the bowling pins provide to a bowling ball? I = mΔv I = m*(vfinal – vinitial)‏ I = 5*(3-5)=-10kgm/s\n\n20. Other Ways to Find Impulse I = mΔv Remember: a = Δv/Δt therefore: Δv = aΔt Substituting in we get: I = maΔt Remember: F=ma Substituting in we get: I=FΔt I = Δp = mΔv = FΔt\n\n21. I = Δp = mΔv = FΔt In football, a field goal is kicked with the football initially at rest. The football (300g) is kicked at 25m/s.  What was the impulse? 300g=.3kg I = mΔv = m*(Vfinal-Vinitial) = .3*(25-0) = 7.5kgm/s The player's foot was in contact with the ball for .1 seconds.  What is the average force during the time of contact? I = FΔt 7.5 = F * .1 F= 75 N\n\n22. I = Δp = mΔv = FΔt I want to open a cracked door by throwing a ball at it. If I have two balls of equal mass, one bouncy, and the other clay, which of the two should I throw to achieve my goal? The bouncy one The clay one It doesn’t matter, they are the same Z) I have no clue\n\n23. Clay ball Bouncy ball\n\n24. Two-Minute Problem A loaded truck collides with a car causing a huge damage to the car. Which of the following is true about the collision? A. The force on the truck is greater than the force on the car B. The force on the car is greater than the force on the truck C. The force on the truck is the same in magnitude as the force on the car D. The car and truck accelerate in the same direction E. During the collision the truck has greater acceleration than the car Z. I have no idea\n\n25. Conservation of Momentum Recall Newton's Third law: Every action has an equal and opposite reaction: F1=-F2 When I push on the desk it pushes back on me with equal force in the opposite direction.\n\n26. Steel ball demo I = Δp = mΔv = FΔt Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1 Compare the time of contact for Ball 1 with Ball 2 to the time of contact for Ball 2 with Ball 1\n\n27. Steel ball demo Compare the force of Ball1 on Ball 2 and Ball 2 on Ball 1 F12 = -F21 Compare the time of contact for Ball 1 with Ball 2 to the time of contact for Ball 2 with Ball 1 Δt12 = Δt21\n\n28. Derivation of Conservation of Momentum Recall from last class that I = Δp = mΔv = FΔt F1 = -F2 Δt1 = Δt2= Δt I1= F1 Δt I2 = F2 Δt = - F1 Δt I1=-I2 In any interaction, momentum gain of one object is equal to the loss of momentum from another\n\n29. Quick check An astronaut (80kg) in space kicks off of his space shuttle at 1.5m/s. What is the impulse that he provides to the space shuttle? (Assume that away from the space shuttle is the positive direction)‏ What is the impulse that the space shuttle provides to him? What is the total change in momentum of the space shuttle and the astronaut together? 120 kgm/s 240 kgm/s 0 kgm/s -120 kgm/s Z) I have no idea\n\n30. Total Momentum of a System is conserved If there are no outside forces acting on an system, the momentum of that system remains constant; it is conserved This is the property momentum conservation. ptotal = Σp = p1 + p2 + p3 + … ptotalinitial = ptotalfinal\n\n31. Momentum Conservation ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal A 10 kg object moves at a constant velocity 2 m/s to the right and collides with a 4 kg object moving at a velocity 5 m/s to the left. Which of the following statements is correct? A. The total momentum before and after the collision is 20 kg·m/s B. The total momentum before and after the collision is 40 kg·m/s C. The total momentum before and after the collision is 10 kg·m/s D. The total momentum before and after the collision is 30 kg·m/s E. The total momentum before and after the collision is zero\n\n32. Momentum Conservation ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal A freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 24,000 kg. After the collision the car A is moving at 3 m/s in the same direction. What is the velocity of car B after the collision? A. 1 m/s B. 3 m/s C. 5 m/s D. 7 m/s D. 11 m/s\n\n33. Momentum Conservation ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal The same situation as before: A freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 24,000 kg. However, after the collision the car A is moving at 3 m/s in the opposite direction. What is the velocity of car B after the collision? A. 1 m/s B. 3 m/s C. 5 m/s D. 7 m/s D. 11 m/s\n\n34. ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal Time to get tricky, you may need a calculator: A loaded freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 8,000 kg. After the collision the cars stick to each other and moves like one object. What is the velocity of two cars after the collision? A. 2 m/s B. 4 m/s C. 6 m/s D. 8 m/s D. 12 m/s\n\n35. Homework: • Periods 2 and 3: • We do not have a quiz tomorrow, however I expect you to be able to do the practice quiz on the website. If you don’t feel comfortable with it take extra care reviewing the physics classroom • Review the Physics Classroom • “Momentum Conservation Principle” • (link on website)‏\n\n36. Today’s Schedule (Jan. 5)‏ • Summary of Equations • Schedule for the next week and a half • Test on Jan. 13th (Friday next week)‏ • Today’s material: Review momentum equation uses\n\n37. Momentum in a Nutshell • p = mv • Δp = I = mΔv = FΔt • ptotal = Σp = p1 + p2 + p3 + … • ptotailnitial = ptotalfinal\n\n38. p=mv What is the momentum of a bee that weighs 10 grams and flies at 2 m/s? 10g=.01kg p=mv=.01x2= .02kgm/s How does that compare to a tortoise that weighs 1kg and moves at .05m/s? p=mv=1x.05= .05kgm/s The tortoise has more momentum.\n\n39. I = Δp = mΔv = FΔt In football, a field goal is kicked with the football initially at rest. The football (300g) is kicked at 25m/s.  What was the impulse? 300g=.3kg I = mΔv = m*(Vfinal-Vinitial) = .3*(25-0) = 7.5kgm/s The player's foot was in contact with the ball for .1 seconds.  What is the average force during the time of contact? I = FΔt 7.5 = F * .1 F= 75 N\n\n40. ptotal = Σp = p1 + p2 + p3 + … • An astronaut (80kg) in space kicks off of his space shuttle at 1.5m/s. What is the impulse that he provides to the space shuttle? (Assume that away from the space shuttle is the positive direction)‏ • What is the impulse that the space shuttle provides to him? • What is the total change in momentum of the space shuttle and the astronaut together? • 120 kgm/s • 240 kgm/s • 0 kgm/s • -120 kgm/s • Z) I have no idea\n\n41. ptotailnitial = ptotalfinal ptotal final = m₁f x v ₁f + m ₂f x v₂f ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotailnitial = ptotalfinal\n\n42. No Official Homework We will be working on more difficult problems from here on out If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems Review kinematics if the projectile motion was difficult\n\n43. Egg Throw! (bring a jacket)‏\n\n44. Other Similar Impulse Applications • Climbing ropes • Airbags • Circus Nets • Diving Golf/Baseball: following through on your swings\n\n45. Sample Test Question: Ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of negligible friction shown above. It makes a head-on collision with a target ball B of mass 0.50 kg at rest at the edge of the table. As a result of the collision, the Ball A rebounds, sliding backwards at 0.70 m/s immediately after the collision. (a) Calculate the speed of Ball B immediately after the collision. (b) Calculate the horizontal displacement d.\n\n46. (a) Calculate the speed of the 0.50 kg target ball immediately after the collision. We use conservation of momentum to solve this one. The tabletop is 1.20 m above a level, horizontal floor. The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision.\n\n47. (b) Calculate the horizontal displacement d. This is a projectile motion problem. We know the ball’s horizontal velocity and the height of the table, so we can easily find the horizontal distance it travels as it falls. Time to fall: The ball has a horizontal velocity of 0.42 m/s (which we just figured out), so the distance d is simply:\n\n48. 2 Dimensional Momentum Which of the following are possible final momenta for the case below? A B C D E F\n\n49. Full 2D Problem on Board Ball 1 (1kg) travelling at 5m/s collides with Ball 2 (4kg) travelling at 3m/s After the collision Ball 1 is travelling 1m/s at 60 degrees above the x-axis. What is the velocity (direction and speed) of Ball 2?\n\n50. Tricky Extension In another experiment on the same table, the target ball B is replaced by target ball C of mass 0.10 kg. The incident ball A again slides at 1.4 m/s, as shown above left, but this time makes a glancing collision with the target ball C that is at rest at the edge of the table. The target ball C strikes the floor at point P, which is at a horizontal displacement of 0.15 m from the point of the collision, and at a horizontal angle of 30 from the + x-axis, as shown above right. (c) Calculate the speed v of the target ball C immediately after the collision. (d) Calculate the y-component of incident ball A's momentum immediately after the collision." ]
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https://howkgtolbs.com/convert/62.14-kg-to-lbs
[ "# 62.14 kg to lbs - 62.14 kilograms to pounds\n\nDo you want to learn how much is 62.14 kg equal to lbs and how to convert 62.14 kg to lbs? You couldn’t have chosen better. This whole article is dedicated to kilogram to pound conversion - both theoretical and practical. It is also needed/We also want to emphasize that all this article is dedicated to only one amount of kilograms - this is one kilogram. So if you need to know more about 62.14 kg to pound conversion - keep reading.\n\nBefore we go to the practice - this is 62.14 kg how much lbs conversion - we will tell you a little bit of theoretical information about these two units - kilograms and pounds. So we are starting.\n\nHow to convert 62.14 kg to lbs? 62.14 kilograms it is equal 136.9952496068 pounds, so 62.14 kg is equal 136.9952496068 lbs.\n\n## 62.14 kgs in pounds\n\nWe are going to start with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, formally known as International System of Units (in short form SI).\n\nSometimes the kilogram can be written as kilogramme. The symbol of the kilogram is kg.\n\nFirst definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. This definition was not complicated but impractical to use.\n\nThen, in 1889 the kilogram was described by the International Prototype of the Kilogram (in short form IPK). The IPK was made of 90% platinum and 10 % iridium. The IPK was used until 2019, when it was replaced by a new definition.\n\nNowadays the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is exactly 0.001 tonne. It can be also divided to 100 decagrams and 1000 grams.\n\n## 62.14 kilogram to pounds\n\nYou know some facts about kilogram, so now we can move on to the pound. The pound is also a unit of mass. It is needed to underline that there are more than one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to concentrate only on pound-mass.\n\nThe pound is in use in the Imperial and United States customary systems of measurements. Of course, this unit is used also in other systems. The symbol of this unit is lb or “.\n\nThe international avoirdupois pound has no descriptive definition. It is just equal 0.45359237 kilograms. One avoirdupois pound could be divided into 16 avoirdupois ounces and 7000 grains.\n\nThe avoirdupois pound was implemented in the Weights and Measures Act 1963. The definition of this unit was given in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### How many lbs is 62.14 kg?\n\n62.14 kilogram is equal to 136.9952496068 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.\n\n### 62.14 kg in lbs\n\nThe most theoretical part is already behind us. In this part we are going to tell you how much is 62.14 kg to lbs. Now you know that 62.14 kg = x lbs. So it is time to get the answer. Just see:\n\n62.14 kilogram = 136.9952496068 pounds.\n\nThat is a correct outcome of how much 62.14 kg to pound. You can also round off the result. After rounding off your outcome is exactly: 62.14 kg = 136.708 lbs.\n\nYou know 62.14 kg is how many lbs, so look how many kg 62.14 lbs: 62.14 pound = 0.45359237 kilograms.\n\nOf course, in this case it is possible to also round off this result. After rounding off your result will be as following: 62.14 lb = 0.45 kgs.\n\nWe also want to show you 62.14 kg to how many pounds and 62.14 pound how many kg outcomes in tables. Let’s see:\n\nWe will start with a table for how much is 62.14 kg equal to pound.\n\n### 62.14 Kilograms to Pounds conversion table\n\nKilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)\n62.14 136.9952496068 136.7080\nNow see a chart for how many kilograms 62.14 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n62.14 0.45359237 0.45\n\nNow you learned how many 62.14 kg to lbs and how many kilograms 62.14 pound, so it is time to move on to the 62.14 kg to lbs formula.\n\n### 62.14 kg to pounds\n\nTo convert 62.14 kg to us lbs a formula is needed. We will show you two formulas. Let’s start with the first one:\n\nNumber of kilograms * 2.20462262 = the 136.9952496068 outcome in pounds\n\nThe first formula give you the most correct result. Sometimes even the smallest difference can be significant. So if you need an accurate result - this formula will be the best for you/option to convert how many pounds are equivalent to 62.14 kilogram.\n\nSo go to the shorer formula, which also enables calculations to learn how much 62.14 kilogram in pounds.\n\nThe second formula is down below, see:\n\nAmount of kilograms * 2.2 = the result in pounds\n\nAs you can see, the second formula is simpler. It could be the best solution if you want to make a conversion of 62.14 kilogram to pounds in fast way, for instance, during shopping. Just remember that your outcome will be not so exact.\n\nNow we want to show you these two formulas in practice. But before we are going to make a conversion of 62.14 kg to lbs we want to show you another way to know 62.14 kg to how many lbs totally effortless.\n\n### 62.14 kg to lbs converter\n\nAnother way to learn what is 62.14 kilogram equal to in pounds is to use 62.14 kg lbs calculator. What is a kg to lb converter?\n\nCalculator is an application. Calculator is based on first version of a formula which we gave you above. Due to 62.14 kg pound calculator you can effortless convert 62.14 kg to lbs. You only need to enter amount of kilograms which you want to convert and click ‘convert’ button. You will get the result in a second.\n\nSo try to calculate 62.14 kg into lbs using 62.14 kg vs pound calculator. We entered 62.14 as a number of kilograms. Here is the result: 62.14 kilogram = 136.9952496068 pounds.\n\nAs you see, this 62.14 kg vs lbs calculator is intuitive.\n\nNow let’s move on to our chief topic - how to convert 62.14 kilograms to pounds on your own.\n\n#### 62.14 kg to lbs conversion\n\nWe will start 62.14 kilogram equals to how many pounds conversion with the first formula to get the most exact result. A quick reminder of a formula:\n\nNumber of kilograms * 2.20462262 = 136.9952496068 the outcome in pounds\n\nSo what have you do to know how many pounds equal to 62.14 kilogram? Just multiply amount of kilograms, this time 62.14, by 2.20462262. It gives 136.9952496068. So 62.14 kilogram is exactly 136.9952496068.\n\nYou can also round it off, for example, to two decimal places. It is equal 2.20. So 62.14 kilogram = 136.7080 pounds.\n\nIt is high time for an example from everyday life. Let’s calculate 62.14 kg gold in pounds. So 62.14 kg equal to how many lbs? As in the previous example - multiply 62.14 by 2.20462262. It gives 136.9952496068. So equivalent of 62.14 kilograms to pounds, when it comes to gold, is 136.9952496068.\n\nIn this case it is also possible to round off the result. It is the outcome after rounding off, this time to one decimal place - 62.14 kilogram 136.708 pounds.\n\nNow we can go to examples calculated with a short version of a formula.\n\n#### How many 62.14 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nAmount of kilograms * 2.2 = 136.708 the outcome in pounds\n\nSo 62.14 kg equal to how much lbs? And again, you need to multiply amount of kilogram, this time 62.14, by 2.2. Let’s see: 62.14 * 2.2 = 136.708. So 62.14 kilogram is equal 2.2 pounds.\n\nMake another calculation using this version of a formula. Now calculate something from everyday life, for example, 62.14 kg to lbs weight of strawberries.\n\nSo let’s convert - 62.14 kilogram of strawberries * 2.2 = 136.708 pounds of strawberries. So 62.14 kg to pound mass is 136.708.\n\nIf you learned how much is 62.14 kilogram weight in pounds and are able to convert it with use of two different versions of a formula, let’s move on. Now we want to show you these results in charts.\n\n#### Convert 62.14 kilogram to pounds\n\nWe are aware that results presented in tables are so much clearer for most of you. We understand it, so we gathered all these outcomes in charts for your convenience. Due to this you can quickly make a comparison 62.14 kg equivalent to lbs outcomes.\n\nBegin with a 62.14 kg equals lbs chart for the first version of a formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n62.14 136.9952496068 136.7080\n\nAnd now look 62.14 kg equal pound chart for the second formula:\n\nKilograms Pounds\n62.14 136.708\n\nAs you can see, after rounding off, if it comes to how much 62.14 kilogram equals pounds, the outcomes are not different. The bigger number the more considerable difference. Please note it when you need to make bigger number than 62.14 kilograms pounds conversion.\n\n#### How many kilograms 62.14 pound\n\nNow you learned how to convert 62.14 kilograms how much pounds but we are going to show you something more. Are you interested what it is? What do you say about 62.14 kilogram to pounds and ounces calculation?\n\nWe are going to show you how you can convert it step by step. Let’s begin. How much is 62.14 kg in lbs and oz?\n\nFirst things first - you need to multiply number of kilograms, in this case 62.14, by 2.20462262. So 62.14 * 2.20462262 = 136.9952496068. One kilogram is exactly 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this case there are 2 pounds.\n\nTo convert how much 62.14 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.\n\nSo final outcome is 2 pounds and 327396192 ounces. It is also possible to round off ounces, for instance, to two places. Then your result is 2 pounds and 33 ounces.\n\nAs you see, calculation 62.14 kilogram in pounds and ounces quite simply.\n\nThe last calculation which we are going to show you is conversion of 62.14 foot pounds to kilograms meters. Both foot pounds and kilograms meters are units of work.\n\nTo calculate it it is needed another formula. Before we show you it, see:\n\n• 62.14 kilograms meters = 7.23301385 foot pounds,\n• 62.14 foot pounds = 0.13825495 kilograms meters.\n\nNow look at a formula:\n\nAmount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters\n\nSo to calculate 62.14 foot pounds to kilograms meters you need to multiply 62.14 by 0.13825495. It gives 0.13825495. So 62.14 foot pounds is exactly 0.13825495 kilogram meters.\n\nIt is also possible to round off this result, for instance, to two decimal places. Then 62.14 foot pounds will be exactly 0.14 kilogram meters.\n\nWe hope that this calculation was as easy as 62.14 kilogram into pounds conversions.\n\nThis article was a big compendium about kilogram, pound and 62.14 kg to lbs in calculation. Due to this conversion you learned 62.14 kilogram is equivalent to how many pounds.\n\nWe showed you not only how to make a conversion 62.14 kilogram to metric pounds but also two another conversions - to check how many 62.14 kg in pounds and ounces and how many 62.14 foot pounds to kilograms meters.\n\nWe showed you also other way to make 62.14 kilogram how many pounds conversions, that is using 62.14 kg en pound converter. This is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.\n\nWe hope that now all of you can do 62.14 kilogram equal to how many pounds calculation - on your own or using our 62.14 kgs to pounds converter.\n\nIt is time to make your move! Let’s convert 62.14 kilogram mass to pounds in the way you like.\n\nDo you want to do other than 62.14 kilogram as pounds calculation? For example, for 15 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so easy as for 62.14 kilogram equal many pounds.\n\n### How much is 62.14 kg in pounds\n\nTo quickly sum up this topic, that is how much is 62.14 kg in pounds , we prepared one more section. Here we have for you all you need to know about how much is 62.14 kg equal to lbs and how to convert 62.14 kg to lbs . It is down below.\n\nWhat is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. Let’s see 62.14 kg to pound conversion formula . Have a look:\n\nThe number of kilograms * 2.20462262 = the result in pounds\n\nSee the result of the conversion of 62.14 kilogram to pounds. The exact answer is 136.9952496068 pounds.\n\nThere is also another way to calculate how much 62.14 kilogram is equal to pounds with second, easier type of the equation. Check it down below.\n\nThe number of kilograms * 2.2 = the result in pounds\n\nSo this time, 62.14 kg equal to how much lbs ? The answer is 136.9952496068 lbs.\n\nHow to convert 62.14 kg to lbs in just a moment? You can also use the 62.14 kg to lbs converter , which will do the rest for you and you will get a correct answer .\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms." ]
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https://math.stackexchange.com/questions/1525511/prove-that-a-n-leq-b-n-forall-n-geq-m-rightarrow-infa-n-n-m-infty-leq
[ "# Prove that $a_n \\leq b_n \\forall n \\geq m \\Rightarrow inf(a_n)_{n=m}^\\infty \\leq inf(b_n)_{n=m}^\\infty$\n\nI'm trying to prove the following:\n\n\"Suppose that $(a_n)_{n=m}^\\infty$ and $(b_n)_{n=m}^\\infty$ are two sequences of real numbers such that $a_n \\leq b_n \\forall n \\geq m$. Then we have the inequalities:\n\n(a) sup$(a_n)_{n=m}^\\infty \\leq$ sup$(b_n)_{n=m}^\\infty$ (where sup$(a_n)_{n=m}^\\infty:=$sup$\\{a_n:n\\geq m\\}$);\n\n(b) inf$(a_n)_{n=m}^\\infty \\leq$ inf$(b_n)_{n=m}^\\infty$ (where inf$(a_n)_{n=m}^\\infty :=$inf $\\{a_n:n \\geq m\\}$)\n\nI've managed to prove (a) by contradiction, and I've tried to do the same for (b) (i.e. show that if we suppose inf$(a_n)_{n=m}^\\infty >$ inf$(b_n)_{n=m}^\\infty$ there exists $n \\geq m$ s.t. $a_n > b_n$), but I haven't been able to do so, so far.\n\nSo, I would appreciate any hint about how to deal with inequality (b).\n\nBest regards,\n\nlorenzo.\n\n• You may try to consider the sequences $(-a_n)_{n=m}^{\\infty}$ and $(-b_n)_{n=m}^{\\infty}$ and try to apply the first exercise. – Iulia Nov 12 '15 at 9:11\n\n## 1 Answer\n\nSuppose $inf(a_n)_{n=m}^\\infty >inf(b_n)_{n=m}^\\infty$.\n\nThen there exist an $n_0$ such that $inf(a_n)_{n=m}^\\infty >b_{n0} \\geq inf(b_n)_{n=m}^\\infty$ and hence $a_{n0}\\geq inf(a_n)_{n=m}^\\infty >b_{n0}$.\n\nDidn't you use the same way to prove (1)?" ]
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https://proxies-free.com/ag-algebraic-geometry-the-compactified-jacobian-is-birational-to-a-mathbbp1-fibration-over-the-jacobian-of-normalization/
[ "# ag.algebraic geometry – The compactified Jacobian is birational to a \\$mathbb{P}^1\\$-fibration over the Jacobian of normalization", null, "Let $$Y$$ be an integral curve whose only singularity is one simple node at a point $$y$$, and\n$$pi:Xrightarrow Y$$ be the normalization with $$pi^{-1}(y)={x,z}$$. $$J(X)$$ is the Jacobian of $$X$$, and $$overline{J}(Y)$$ is the compactified Jacobian, parametrizing torsion free sheaves of rank $$1$$ on $$Y$$.\nI want to show $$overline{J}(Y)$$ is birational to a $$mathbb{P}^1$$-fibration over $$J(X)$$.\n\nTake Poincare bundle $$mathcal{P}rightarrow J(X)times X$$, and set $$mathcal{P}_{+}=mathcal{P}{mid}_{J(X)times{x}}$$, $$mathcal{P}_{-}=mathcal{P}{mid}_{J(X)times{z}}$$.\nThen, I get $$mathbb{P}^1$$-bundle $$mathbb{P}(mathcal{P}_+ oplus mathcal{P}_-)rightarrow J(X)$$.\nThere are two sections $$S_+$$ and $$S_-$$ corresponding to $$mathcal{P}_+$$ and $$mathcal{P}_-$$.\nIt suffices to show that there is a following commutative diagram,\n$$require{AMScd}$$\n$$begin{CD} overline{J}(Y) @>displaystyle cong >> mathbb{P}(mathcal{P}_+ oplus mathcal{P}_-)/S_+ sim S_-\\ @V displaystyle pi^* V V @VV displaystyle{R} V\\ J(X) @>> id displaystyle > J(X) end{CD}$$\n\nGiven $$(L,F(L))in mathbb{P}(mathcal{P}_+ oplus mathcal{P}_-)$$($$L$$ is a line bundle on $$X$$ and $$F(L)in L_xoplus L_y$$ is a $$1$$-dimensional subspace), take the kernel of the surjective map $$pi_*Lrightarrow pi_*Lotimes k(y)(cong pi_*(L_xoplus L_y))rightarrow pi_*(L_xoplus L_y/F(L))$$.Then, I get $$h:mathbb{P}(mathcal{P}_+ oplus mathcal{P}_-)rightarrow overline{J}(Y)$$.\n\nBut I don’t understand how to show it is an isomorphism and how to construct $$R$$.", null, "" ]
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http://downloads.hindawi.com/journals/mpe/2013/540687.xml
[ "MPE Mathematical Problems in Engineering 1563-5147 1024-123X Hindawi Publishing Corporation 540687 10.1155/2013/540687 540687 Research Article Optimal Bandwidth Scheduling of Networked Learning Control System Based on Nash Theory and Auction Mechanism Yan Xiang 1 0000-0003-4410-586X Li Hongbo 2 Wang Lide 1 Shen Ping 1 Wen Guanghui 1 School of Electrical Engineering Beijing Jiaotong University Beijing 100044 China njtu.edu.cn 2 Department of Computer Science and Technology State Key Laboratory of Intelligent Technology and Systems Tsinghua University Beijing 100084 China tsinghua.edu.cn 2013 12 6 2013 2013 29 03 2013 15 05 2013 2013 Copyright © 2013 Xiang Yan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.\n\nThis paper addresses the optimal bandwidth scheduling problem for a double-layer networked learning control system (NLCS). To deal with this issue, auction mechanism is employed, and a dynamic bandwidth scheduling methodology is proposed to allocate the bandwidth for each subsystem. A noncooperative game fairness model is formulated, and the utility function of subsystems is designed. Under this framework, estimation of distribution algorithm (EDA) is used to obtain Nash equilibrium for NLCS. Finally, simulation and experimental results are given to demonstrate the effectiveness of the proposed approach.\n\n1. Introduction\n\nNetworked control systems (NCSs) are the multiple feedback control loops closed via a serial communication channel. Compared with the traditional point-to-point control system, the advantages of NCSs are sharing of information resources, powerful system diagnosis, distributed remote control, modular design, configuration flexibility, and low cost . As a result, NCSs have been widely used in national defense, aircraft, industrial automation, intelligent transportation, process control, and financial management . And some useful results were reported on network protocol , network-induced delay , packet dropout , NCS structure , security [14, 15], scheduling , and network constraints .\n\nIt is notable that most of the aforementioned researches are focused on single-layer network structure; few results have been reported on NCSs with double-layer structure. As pointed out by [21, 22], a networked learning control system (NLCS) with double-layer structure can obtain better control performance and stronger robustness. Nonetheless, in real-time NLCS with limited network resources, random network-induced delay may have a significant impact on the performance and stability of the system . The bandwidth availability is the major concern in many networking problems. A good schedule gives an appropriate resource allocation to network nodes and reduces packet collision. The performance of network applications is directly affected by the amount of available bandwidth and the sampling rate . Therefore, the quality of service (QoS) and the quality of control (QoC) depend not only on the control algorithm and the system structure but also on the allocating and scheduling of the network resources. A more optimal allocation of the network bandwidth is the key to improve QoS and QoC. Due to the above discussion, bandwidth scheduling and optimizing is studied, based on the structure of the NLCS, in this paper.\n\nWe know that if each node is allowed to occupy network resource as much as possible according to its own requirements, the overall system performance will be very poor . The limitations of the network resources cause the competition among many network nodes. These network nodes will maximize their own network utility, which forms a noncooperation game among them . To meet the demands of individual and entirety in the NLCS, a network bandwidth scheduling strategy based on a noncooperation game model and auction mechanism is introduced. The auction mechanism will be used at the high-layer level in centralized control method, and the control system performance will be estimated by the cumulative amount of the output errors of the closed control loops. Considering the difficulty of the precise calculation, EDA is applied to obtain an optimal solution.\n\nThe remainder of this paper is organized as follows: Section 2 describes NLCS and scheduling optimization schemes. Section 3 explains the auction-based methodology of bandwidth allocation and presents a fair noncooperative game model for scheduling in NLCS. Section 4 shows how to use the EDA-based optimization algorithm to find the equilibrium point of the game model. Simulation and experimental results are provided in Section 4. Finally, Section 5 concludes the results of this study.\n\nNotation. Throughout this paper, n denotes the n-dimensional Euclidean space. The superscript “T” denotes matrix transposition. And the notation + stands for the set of nonnegative integers.\n\n2. Problem Formulation 2.1. System Structure\n\nA typical double-layer NLCS structure is shown in Figure 1 [21, 22]. Suppose that the number of closed-loop subsystems in the lower layer communication network (LLN) is M and there are two data transfer nodes (the sensor node and the controller node) in every closed-loop subsystem, the number of the total data transfer points will be N=2M. In this figure, Ci, Ai, and Si represent ith controller, actuator, and sensor, respectively. The actuator is event driven, and the controller and sensor are time driven. The sampling period is longer than the polling period of the system. The data (sampling data, node identities, and others) are packed by sensors and sent to the controller. Then, such data will be used to calculate the control commands in the controller, which will be packed and sent to the actuator, then the actuator will update the control algorithm based on the new control commands. At the same time, the upper layer communication network (ULN) collects data of the control performance of the subsystems from the LLN controllers through shared network (LAN, WAN, or Internet) and then optimizes the sampling period and control parameters based on a self-adaptive and scheduling algorithm.\n\nThe structure of double-layer NLCS.\n\nThe timing diagram of bandwidth allocation is illustrated in Figure 2. The controllers report their bandwidth requirements to the learning and scheduling center through the ULN during TP (the polling period). Then the center collects information and allocates bandwidths during TA (the allocation period). According to the given bandwidth and the scheduling algorithm, all control loops send data during TT (the transmission period). After TO (the total operating period), a new period begins.\n\nTiming diagram of bandwidth allocation.\n\n2.2. The Controlled Plants\n\nThe dynamics of the remote controlled plant is given by the following linear model: (1)x(k+1)=Fx(k)+Gu(k), where k+; x(k)n is the plant state; u(k)m is the control input; F and G are known matrices with appropriate dimensions. Suppose that the full state variables are available in the networked subsystem. The sensor is time driven, that is, at time instant kh, it sends the most recent motor state and its timestamp to the controller via the network. Note that (1) can be considered as discretized from a continuous-time system given by (2)x˙p(t)=Axp(t)+Bu(t), with the sampling period h and (3)F=eAh,G=0heAτdτB.\n\nThe electromechanical dynamics of the networked dc-motor subsystem used in this paper can be described as (4)i˙a=-RLia-KbLω+1Lu,ω˙=KtJia-BJω, where ia is the armature winding current; ω is the rotor angular speed; u is the armature winding input voltage; R is the armature winding resistance; L is the armature winding inductance; Kb is the back-electromotive-force- (EMF-) constant; Kt is the torque constant; B is the system damping coefficient; J is the system moment of inertia. The parameters of the dc-motor subsystem are listed in Table 1.\n\nThe parameters of the dc-motor subsystem.\n\n R Motor winding resistance 4 . 961   Ω L Motor winding inductance 3 . 14 × 1 0 - 3  H K b Back-EMF constant 1 . 276 × 1 0 - 2  Vs/rad K t Electric torque constant 7 . 105 × 1 0 - 2  N·m/A J Motor moment of inertia 1 . 6511 × 1 0 - 4  kg·m2 B Damping coefficient 23.64 × e - 6  N·m·sec/rad\n\nBy letting xp[ia,ω]T and the rotor angular speed y=ω, the dc-motor subsystem as Loop 1 can be expressed by (5)x˙p(t)=[-1579.9-4.1430.3-0.1]xp(t)+[318.4730]u(t),(6)y(t)=xp(t).\n\n2.3. NLCS Performance Analysis\n\nThe performance of the NLCS can be demonstrated jointly by the subsystems, and the performance of the subsystem can be described by a function, such as the integral absolute error (IAE) . The system error will be greater if some control loop is disturbed, indicating that the sampling frequency should be raised to a degree that enables the system to return to the equilibrium point rapidly. If the error is smaller, no higher sampling frequency is needed. Figure 3 shows the cumulative error of a rotor speed for a broad range of sampling rates from h=0.050.5 s. Integral absolute error=0te|θ˙(t)|dt, where te is the evaluation time interval. As it can be seen in Figure 3, the relation between control performance and a range of allowed periods can be approximated by a linear relationship as (7). Consider the following (7)ei(ti)=αi+βiti, where αi and βi are specific for each control loop and can be determined prior to system run time. Due to the calculation of the gradient in solving optimization problems, αi could be ignored.\n\nSampling periods versus IAE.\n\nFor each control loop, the relation between sampling period ti and allocation bandwidth xi is given by (8)ti=cixi, where ci is the control time of the control loop (which may contain data exchange from sensor to controller and from controller to actuator, as well as the time spent to execute the controller); hypothetically ci is within a sampling period.\n\nOn one hand, the QoC of each subsystem should be improved, on the other hand, the demands of network bandwidth of each subsystem should be reduced to provide more resources for extra subsystems. Due to the bandwidth limitations, a multiobjective optimization problem can be expressed as follows (9)OP:minδJ1+(1-δ)J2s.t.i=1Mxi1-γ,ximinxiximax, where J1=i=1M(αi+βiti)=i=1M(αi+βici/xi), J2=i=1Mxi, 0<δ<1. To deal with the accidental overload and avoid the sudden deterioration of the system performance, the scale factor γ is introduced to retain part of the bandwidth. ximin and ximax can be calculated based on the resource constraints of the control loop and the maximum allowable delay bound (MADB) .\n\n3. Main Results\n\nThe game model based on auction mechanism is discussed in this section. Auction theory is first proposed by Vickrey in 1961, which mainly includes four basic types: English auction (ascending-bid auction), Dutch Auction (descending-bid auction), and first price auction and second price auction (Vickrey auction). The highest offer will win the bid no matter what type of auction is used, and the optimum outcome is that the price exactly equals what the second highest bidder can afford.\n\nRemark 1.\n\nIt is worth noting that the first two kinds of auction mechanism are bidding open, and the others are sealed. In sealed auction, each bidder submits his price without any information of others, then the auctioneer announces the winner who offers the highest bid. Both of them are suitable for bandwidth allocation in NLCS. The only difference between first and second price auction is the actual price that the winner paid. For second price auction, the winner just needs to pay the second highest bid of others instead of his own. For ease of operation, first price auction is applied in this paper.\n\nThe closed-loop control subsystem is modeled as a player. Every player is not explicitly aware of the existence of other players and their status. Each player puts forward its own bandwidth strategy denoted as xi. Obviously, the bandwidth consumption of each player will affect the bandwidth allocation of other players, but it is impossible to improve the working performance by only increasing the bandwidth requirement. So, a noncooperative game is formed among the closed-loop control subsystems in NLCS. In the game, ULN plays the roll of auctioneer, and every player will pay the bandwidth based on their own bidding strategy and the final bandwidth they have got. The bidding and auction process is shown as follow.\n\nStep 1.\n\nEvery player has the same amount of money g before every round of bidding begins.\n\nStep 2.\n\nEvery player submits the price yi (yi[0,g]) based on their own bandwidth requirement.\n\nStep 3.\n\nULN will run the bandwidth allocation programs based on the bidding prices and allocate the bandwidth resources to each player.\n\nStep 4.\n\nEvery player will pay for the bandwidth they have got.\n\nThe price of bandwidth xi defined as yi which player i needs to pay can be calculated from (10)yi=λxi, where λ is the price of unity bandwidth. The revenue function of every player can be described as follow: (11)Si(xi)={Ui(xi)-yi,ifyi>0,Ui(0),ifyi=0.\n\nFurther, the revenue function based on auction theory and IAE evaluation method is shown as (12): (12)Si(xi)=ln(1+1ei)-yi=ln(1+1βiti)-λxi=ln(1+xiβici)-λxi.\n\nObviously, the revenue of the player is determined by the received QoC and the payment under certain QoS. Every player will make the bidding strategy based on their own revenue and will not increase the price aimlessly. If a player overly spends its money for a much larger bandwidth than its actual need, it will only get a low payoff due to the reduced network utility. The money each player has to pay is based on the bandwidth they have got, which may not match the initial offer.\n\nEvery player in LLN wants to maximize their revenue under the framework of NLCS. ULN will make a network resources allocation strategy which makes it impossible to get more bandwidth resources through changing the offer. Generally speaking, the network utility of a single node depends on the scheduling strategy of others. If there is no node chooses other scheduling strategies when the scheduling strategies of other nodes that are decided, this equilibrium will not be broken in the network. This equilibrium is called Nash equilibrium (NE) . Thus, the allocation problem of the network resources can be transformed into the solution of the Nash equilibrium with a noncooperation game model.\n\nTo address the modelling problem, we introduce the following definitions to prove the existence and uniqueness of Nash equilibrium in NLCS.\n\nDefinition 2.\n\nA noncooperative networked learning control game (NNLCG) with M players participated in, NG=(Γ,{Xi},{Si(·)}), where Γ={1,2,,M} is the index set of players, {Xi} is a strategy profile, and {Si(·)} is the revenue function set of players. If xi* is the best strategy of player i when other players choose strategies x-i*=(x1*,,xi-1*,xi+1*,,xM*) which is shown in (13), x*=(x1*,,xi*,,xM*) will be a Nash equilibrium. One has the following: (13)Si(xi*,x-  i*)Si(xi,x-  i*),xiXi,iΓ.\n\nDefinition 3.\n\nThe function Si(·):Xi+1 defined on the convex set Xi is quasi concave in xi if and only if (14)Si(ξxi+(1-ξ)xi,x-i)min(Si(xi,x-i),Si(xi,x-i)) for all xi,xiXi and ξ[0,1].\n\nTheorem 4.\n\nA Nash equilibrium exists in game NG=(,{Pj},{uj(·)}) if for all j:\n\nPj is a nonempty, convex, and compact subset of some Euclidean space N.\n\nuj(p) is continuous in p and quasi concave in pj.\n\nProof.\n\nTheorem 4 can be obtained using the classical Kakutani fixed-point theorem and in some sense generalizes Nash’s setting on the strategy space of the players. For the proof of this theorem, please refer to .\n\nTheorem 5.\n\nA Nash equilibrium exists in the NNLCG, NG=(Γ,{Xi},{Si(·)}).\n\nProof.\n\nBy Theorem 4, we know that there exists a Nash equilibrium in NNLCG when the conditions in Theorem 4 are satisfied. Each player in NNLCG has a strategy space [ximin,ximax], ximin>0, and ximinximax. Thus, the first condition is satisfied. It remains to prove that the revenue function Si(x) is quasi concave in xi for all i in NNCG.\n\nThe first-order differential of the revenue function Si(·) is (15)Si(xi)xi=1βici+xi-λ.\n\nThe second-order differential of the revenue function Si(·) is (16)Si2(xi)xi2=-1(βici+xi)2<0.\n\nKnown from the above equations, the revenue function is continuous and differentiable in x and concave in xi. A concave function Si(x) is quasi concave too. This completes the proof of the theorem.\n\nTheorem 6.\n\nThe NNLCG has a unique equilibrium.\n\nProof.\n\nThe proof of the unique equilibrium can be carried out by following similar lines as in the proof of Theorem  2 in and is thus omitted here to avoid unnecessary repetition.\n\nUp to now, the problem of network resources allocation under a general framework of double-layer NLCS has be changed into the problem of Nash-equilibrium-point solving in the noncooperation game model. It is hard to solve the Nash equilibrium point by the traditional numerical method. So, estimation of distribution algorithm (EDA) is introduced in this paper to find the Nash equilibrium point. The EDA algorithm used in this paper is described as follows.\n\nRule 1.\n\nInitialization: generate gain candidates meeting the ratio of bandwidth (RoB) randomly to form an initial population.\n\nRule 2.\n\nRepeat the following steps until the termination criterion is met.\n\nSelection: select the best gain candidates from the parent generation.\n\nUpdating: update using the selected promising gain candidates.\n\nSampling: generate gain candidates meeting RoB based on the updated; copy the best gain candidate in the current population to the next population.\n\nFor more details about EDA, please refer to [34, 35], and the reference therein.\n\n4. Experiments and Result Analysis\n\nIn this section, an illustrative example is presented to show the effectiveness of the proposed method. To this end, let us consider a double-layer NLCS as shown in Figure 1. The considered NLCS has three different networked dc-motor subsystems, where one takes the form (5) and the models of the rest two subsystems are listed in Table 2. Apparently, γ=0.1, TO=0.5 s, TL=0.005 s, TP=0.03 s, and the rest of TO is TT. The allocation period TA is very short compared to TL so that TA can be assumed to be zero.\n\nParameters of the rest two subsystems.\n\nParameter A Parameter B\nLoop 2 [ 0 1 1 - 217.4 ] [ 0 1669.5 ]\nLoop 3 [ - 579.9 - 4.1 30.3 - 0.01 ] [ 18.4713 0 ]\n\nThe auction-based bandwidth allocation (ABA) has an initial population of 100 solutions and ten generations. When the overall fitness value is stabilized, the Nash equilibrium point is reached. The bandwidth vector allocated by the ULN in this case is x*=[0.22  0.37  0.31]T. To illustrate the improving level of QoC and the bandwidth occupied, we compare ABA with equal bandwidth allocation (EBA), which is a simple static bandwidth allocation method. And the ratio of bandwidth given to player i can be quickly computed by xi=(1-γ)/M. The IAE and the RoB of two different strategies are given in Table 3. Accordingly, the step responses of three networked dc-motor subsystems with given corresponding RoB are shown in Figure 4.\n\nSimulation results of comparison test.\n\nEBA ABA\nIAE RoB IAE RoB\nLoop 1 0.8199 0.30 0.9090 0.22\nLoop 2 1.4720 0.30 1.2629 0.37\nLoop 3 1.2310 0.30 1.1375 0.31\n\nTotal 3.5229 0.90 3.3094 0.90\n\nDifferent step responses of three networked dc-motor subsystems.\n\nLoop 1\n\nLoop 2\n\nLoop 3\n\nAs shown in Table 3 and Figure 4, the control performance of networked subsystem degrades when network conditions become worse (given less RoB). This phenomenon is reasonable, since each dc-motor subsystem performance worsens with longer time delays and more packet losses. It is notable that some loop has higher value of IAE in ABA than EBA; the ABA still gives the better overall performance according to the simulation results. So this equal bandwidth allocation method may be ineffective, since some loops may not receive enough bandwidth according to their demands, while others may receive unnecessary bandwidth compared to them.\n\nFigure 5 shows the box-and-whisker diagram of system revenue over 30 paired simulation. And the statistics of the 30 simulation results are listed in Table 4. The same network condition is used in the simulation of networked system using the ABA and the EBA. Note that the 30 paired runs used 30 different network conditions and they were generated randomly under the given network constraints. Obviously, the overall payoff of ABA is higher than EBA. Based on the mean value of system payoff using ABA and EBA (denoted as J¯ABA and J¯EBA, resp.), as listed in Table 4, we can further conclude that, comparing with EBA on the average, ABA method increases the system payoff by 21.56% [(J¯ABA-J¯EBA)100%/J¯EBA].\n\nStatistics of the 30 simulation results.\n\nEBA ABA\nMean value 1.9048 2.3154\nStandard deviation 0.0603 0.0696\n\nThe box-and-whisker diagram.\n\nMotivated by the results, we found that auction-based bandwidth allocation that optimizes resource scheduling strategy can effectively meet the desired objectives in the resource-constrained NLCS.\n\n5. Conclusion\n\nThis paper presents a noncooperation game model based on Nash theory and auction mechanism for bandwidth allocation in NLCS with limited resources. And the estimation of distribution algorithm is introduced to solve the problem effectively. The proposed method forces all players sharing the same network to have allocated bandwidths at Nash equilibrium point. Network resources are allocated in the optimum way to reduce delays and packet losses, and the overall performance of systems with communication constraints is significantly improved. The simulation and experiment results indicate the effectiveness and availability of the proposed approach.\n\nAcknowledgments\n\nThe authors would like to thank C. Hao, D. W. Yang, and T. Tony for their valuable comments and suggestions to improve the quality of the paper. The work of X. Yan is supported by the Fundamental Research Funds for the Central Universities under Grant E11JB00310. The work of L. D. Wang and P. Shen is supported by the Fundamental Research Funds for the Central Universities under Grant E12JB00140. The work of H. Li is supported by National Basic Research Program of China (973 Program) under Grant 2012CB821206, the National Natural Science Foundation of China under Grant 61004021, and Beijing Natural Science Foundation (4122037).\n\nZhang L. Gao H. Kaynak O. 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http://gpfd.net/read/35.4x-20x-2012-b5-e3-b0-cb-bd-e2-b7-bd-b3-cc.html
[ "# 5.4x x 12点八解方程\n\n5.4x+x=12.8(5.4+1)x=12.86.4x=12.8x=12.8/6.4x=2.\n\n5.4x+x=12.8; 6.4x=12.8; x=12.8除以6.4; x=2\n\nx=2 只要两边同时乘于10就可以得出54x+10x=128\n\n7x-12=8 移项,得 7x=8+12 [移项后,-12到右边要变符号] 7x=20 解得 x=20/7 注 20/7 [七分之二]\n\n4x=12解得x=3\n\n5.4x+x=12.8 解:54X+10X=128 64X=128 X=2\n\n5.4x+x=12.8 6.4x=12.8 x=12.8/6.4 x=2 扩展资料复合应用题解题思路:是由两个或两个以上相互联系的简单应用题组合而成的. 1、理解题意,就是弄清应用题中的已知条件和要求问题. 2、分析数量关系,就是分析已知数量与未知数数量,已知数量与未知数数量间的关系,找到解题途径,确定先算什么,再算什么,最好算什么. 3、列式解答,就是根据分析,列出算式并计算出来. 4、验算并给出答案,就是检验解答过程中是否合理,结果是否正确,与原题的条件是否相符,最后写出答案." ]
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https://search.r-project.org/CRAN/refmans/AgroReg/html/linear.linear.html
[ "linear.linear {AgroReg} R Documentation\n\n## Analysis: Linear-Linear\n\n### Description\n\nThis function performs linear linear regression analysis.\n\n### Usage\n\nlinear.linear(\ntrat,\nresp,\nmiddle = 1,\nCI = FALSE,\nbootstrap.samples = 1000,\nsig.level = 0.05,\nerror = \"SE\",\nylab = \"Dependent\",\nxlab = \"Independent\",\ntheme = theme_classic(),\npoint = \"all\",\nwidth.bar = NA,\nlegend.position = \"top\",\ntextsize = 12,\npointsize = 4.5,\nlinesize = 0.8,\nlinetype = 1,\npointshape = 21,\nfillshape = \"gray\",\ncolorline = \"black\",\nround = NA,\nxname.formula = \"x\",\nyname.formula = \"y\",\ncomment = NA,\nfontfamily = \"sans\"\n)\n\n\n### Arguments\n\n trat Numeric vector with dependent variable. resp Numeric vector with independent variable. middle A scalar in [0,1]. This represents the range that the change-point can occur in. 0 means the change-point must occur at the middle of the range of x-values. 1 means that the change-point can occur anywhere along the range of the x-values. CI Whether or not a bootstrap confidence interval should be calculated. Defaults to FALSE because the interval takes a non-trivial amount of time to calculate bootstrap.samples The number of bootstrap samples to take when calculating the CI. sig.level What significance level to use for the confidence intervals. error Error bar (It can be SE - default, SD or FALSE) ylab Variable response name (Accepts the expression() function) xlab treatments name (Accepts the expression() function) theme ggplot2 theme (default is theme_classic()) point defines whether you want to plot all points (\"all\") or only the mean (\"mean\") width.bar Bar width legend.position legend position (default is \"top\") textsize Font size pointsize shape size linesize line size linetype line type pointshape format point (default is 21) fillshape Fill shape colorline Color lines round round equation xname.formula Name of x in the equation yname.formula Name of y in the equation comment Add text after equation fontfamily Font family\n\n### Details\n\nThe linear-linear model is defined by: First curve:\n\ny = \\beta_0 + \\beta_1 \\times x (x < breakpoint)\n\nSecond curve:\n\ny = \\beta_0 + \\beta_1 \\times breakpoint + w \\times x (x > breakpoint)\n\n### Value\n\nThe function returns a list containing the coefficients and their respective values of p; statistical parameters such as AIC, BIC, pseudo-R2, RMSE (root mean square error); breakpoint and the graph using ggplot2 with the equation automatically.\n\n### Author(s)\n\nModel imported from the SiZer package\n\nGabriel Danilo Shimizu\n\nLeandro Simoes Azeredo Goncalves\n\n### References\n\nChiu, G. S., R. Lockhart, and R. Routledge. 2006. Bent-cable regression theory and applications. Journal of the American Statistical Association 101:542-553.\n\nToms, J. D., and M. L. Lesperance. 2003. Piecewise regression: a tool for identifying ecological thresholds. Ecology 84:2034-2041.\n\nlibrary(AgroReg)" ]
[ null ]
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https://agrat.cat/converting-improper-fractions-to-mixed-numbers-worksheet-answers
[ "", null, "", null, "Converting Improper Fractions To Mixed Numbers Worksheet Answers. Worksheet #1 worksheet #2 worksheet #3 worksheet #4 worksheet #5 worksheet #6. Math, fractions, converting, mixed, improper created date:\n\nMaths worksheet — converting mixed fractions to improper. To convert an improper fraction to a mixed fraction, divide the numerator by the denominator, write down the quotient as the whole number and the remainder as the numerator on top of the same denominator. Live worksheets > english > math > fractions > improper fractions to mixed numbers.\n\n### Worksheet #1 Worksheet #2 Worksheet #3 Worksheet #4 Worksheet #5 Worksheet #6.\n\nImproper fractions and mixed numbers worksheets grade 4. This is a worksheet taken from urbrainy and customised into an interactive worksheet. Below are six versions of our grade 4 fractions worksheet on converting mixed numbers to.\n\n### Looking Forward, Students Can Then Progress To Additional Number Worksheets, For Example A.\n\nImproper fractions to mixed numbers. Convert these improper fractions into mixed numbers grade. Math, fractions, converting, mixed, improper created date:\n\n### Very Often Parents Train Their Kids Lots Of Skills Without Having Actually Realising That The Enjoyable They May Be Getting Is A Good Way For A Child To.\n\nWrite improper fractions as mixed numbers. If you are learning to learn calculus or any other innovative math, then you could most likely know the principle associated with this. Grade 5 math worksheet fractions convert fractions to source:\n\n### These Free Improper Fractions To Mixed Numbers Worksheets Exercises Will Have Your Kids Engaged And Entertained While They Improve Their Skills.\n\nHow to add fractions and mixed numbers — steemit. At 4th and 5th grade, with the more abstract worksheets for 5th. Improper fraction and mixed number fractions this is a set of 26 worksheets designed to teach students about improper and mixed fractions (fractions greater than 1) as well as converting between the two.\n\n### Worksheet Improper Fractions To Mixed Numbers Worksheet Fun From Converting Mixed Numbers To Improper Fractions Worksheet, Sourcesiteraven.com For Example, You Would Do The Math Of, If This Was My First Year, What Percentage Of The Amount That I Made Went Back Into My Business, Then I Would Multiply That Number By And Then Sum The.\n\nConvert improper fractions to mixed numbers. To convert an improper fraction to a mixed fraction, divide the numerator by the denominator, write down the quotient as the whole number and the remainder as the numerator on top of the same denominator. Maths worksheet — converting mixed fractions to improper.\n\nCategories: live worksheet" ]
[ null, "https://yess-online.com/close.png", null, "https://yess-online.com/close.png", null ]
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http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_Limits_LLN&diff=8226&oldid=7269
[ "# AP Statistics Curriculum 2007 Limits LLN\n\n(Difference between revisions)\n Revision as of 04:10, 28 April 2008 (view source)IvoDinov (Talk | contribs)m (→SOCR LLN Activity: typo)← Older edit Revision as of 20:06, 5 November 2008 (view source)IvoDinov (Talk | contribs) (→LLN Application)Newer edit → Line 22: Line 22: One demonstration of the law of large numbers provides practical algorithms for estimation of [http://en.wikipedia.org/wiki/Transcendental_number transcendental numbers]. The two most popular transcendental numbers are [http://en.wikipedia.org/wiki/Pi $\\pi$] and [http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 ''e'']. One demonstration of the law of large numbers provides practical algorithms for estimation of [http://en.wikipedia.org/wiki/Transcendental_number transcendental numbers]. The two most popular transcendental numbers are [http://en.wikipedia.org/wiki/Pi $\\pi$] and [http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 ''e'']. - The [[SOCR_EduMaterials_Activities_Uniform_E_EstimateExperiment | SOCR E-Estimate Experiment]] provides the complete details of this simulation. In a nutshell, we can estimate the value of the [http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 natural number e] using random sampling from Uniform distribution. Suppose $X_1, X_2, \\cdots, X_n$ are drawn from [http://www.socr.ucla.edu/htmls/SOCR_Distributions.html uniform distribution on (0, 1)] and define $U= {\\operatorname{argmin}}_n { \\left (X_1+X_2+...+X_n > 1 \\right )}$, note that all $X_i \\ge 0$. + The [[SOCR_EduMaterials_Activities_Uniform_E_EstimateExperiment | SOCR E-Estimate Experiment]] provides the complete details of this simulation. In a nutshell, we can estimate the value of the [http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 natural number e] using random sampling from [http://socr.ucla.edu/htmls/dist/ContinuousUniform_Distribution.html Uniform distribution]. Suppose $X_1, X_2, \\cdots, X_n$ are drawn from [http://socr.ucla.edu/htmls/dist/ContinuousUniform_Distribution.html Uniform distribution on (0, 1)] and define $U= {\\operatorname{argmin}}_n { \\left (X_1+X_2+...+X_n > 1 \\right )}$, note that all $X_i \\ge 0$. Now, the expected value $E(U) = e \\approx 2.7182$. Therefore, by LLN, taking averages of $\\left \\{ U_1, U_2, U_3, ..., U_k \\right \\}$ values, each computed from random samples $X_1, X_2, ..., X_n \\sim U(0,1)$ as described above, will provide a more accurate estimate (as $k \\rightarrow \\infty$) of the natural number ''e''. Now, the expected value $E(U) = e \\approx 2.7182$. Therefore, by LLN, taking averages of $\\left \\{ U_1, U_2, U_3, ..., U_k \\right \\}$ values, each computed from random samples $X_1, X_2, ..., X_n \\sim U(0,1)$ as described above, will provide a more accurate estimate (as $k \\rightarrow \\infty$) of the natural number ''e''. Line 31: Line 31:\n\n+ ===References=== ===References=== * TBD * TBD\n\n## General Advance-Placement (AP) Statistics Curriculum - The Law of Large Numbers\n\n### Motivation\n\nSuppose we independently conduct one experiment repeatedly. Assume that we are interested in the relative frequency of occurrence of one event whose probability to be observed at each experiment is p. The ratio of the observed frequency of that event to the total number of repetitions converges towards p as the number of (identical and independent) experiments increases. This is an informal statement of the Law of Large Numbers (LLN).\n\nFor a more concrete example, suppose we study the average height of a class of 100 students. Compared to the average height of 3 randomly chosen students from this class, the average height of 10 randomly chosen students is most likely closer to the real average height of all 100 students. Because the sample of 10 is a larger than the sample of 3, it is a better representation of the entire class. At one extreme, a sample of 99 of the 100 students will produce a sample average height almost exactly the same as the average height for all 100 students. On the other extreme, sampling a single student will be an extremely variant estimate of the overall class average weight.\n\n### The Law of Large Numbers (LLN)\n\nIt is generally necessary to draw the parallels between the formal LLN statements (in terms of sample averages) and the frequent interpretations of the LLN (in terms of probabilities of various events).\n\nSuppose we observe the same process independently multiple times. Assume a binarized (dichotomous) function of the outcome of each trial is of interest (e.g., failure may denote the event that the continuous voltage measure < 0.5V, and the complement, success, that voltage ≥ 0.5V – this is the situation in electronic chips which binarize electric currents to 0 or 1). Researchers are often interested in the event of observing a success at a given trial or the number of successes in an experiment consisting of multiple trials. Let’s denote p=P(success) at each trial. Then, the ratio of the total number of successes to the number of trials (n) is the average", null, "$\\overline{X_n}={1\\over n}\\sum_{i=1}^n{X_i}$, where", null, "$X_i = \\begin{cases}0,& \\texttt{failure},\\\\ 1,& \\texttt{success}.\\end{cases}$ represents the outcome of the ith trial. Thus,", null, "$\\overline{X_n}=\\hat{p}$, the ratio of the observed frequency of that event to the total number of repetitions, estimates the true p=P(success). Therefore,", null, "$\\hat{p}$ converges towards p as the number of (identical and independent) trials increases.\n\n### SOCR LLN Activity\n\nGo to SOCR Experiments and select the Coin Toss LLN Experiment from the drop-down list of experiments in the top-left panel. This applet consists of a control toolbar on the top followed by a graph panel in the middle and a results table at the bottom. Use the toolbar to flip coins one at a time, 10, 100, 1,000 at a time or continuously! The toolbar also allows you to stop or reset an experiment and select the probability of Heads (p) using the slider. The graph panel in the middle will dynamically plot the values of the two variables of interest (proportion of heads and difference of Heads and Tails). The outcome table at the bottom presents the summaries of all trials of this experiment.\n\n### LLN Application\n\nOne demonstration of the law of large numbers provides practical algorithms for estimation of transcendental numbers. The two most popular transcendental numbers are π and e.\n\nThe SOCR E-Estimate Experiment provides the complete details of this simulation. In a nutshell, we can estimate the value of the natural number e using random sampling from Uniform distribution. Suppose", null, "$X_1, X_2, \\cdots, X_n$ are drawn from Uniform distribution on (0, 1) and define", null, "$U= {\\operatorname{argmin}}_n { \\left (X_1+X_2+...+X_n > 1 \\right )}$, note that all", null, "$X_i \\ge 0$.\n\nNow, the expected value", null, "$E(U) = e \\approx 2.7182$. Therefore, by LLN, taking averages of", null, "$\\left \\{ U_1, U_2, U_3, ..., U_k \\right \\}$ values, each computed from random samples", null, "$X_1, X_2, ..., X_n \\sim U(0,1)$ as described above, will provide a more accurate estimate (as", null, "$k \\rightarrow \\infty$) of the natural number e.\n\nThe Uniform E-Estimate Experiment, part of SOCR Experiments, provides a hands-on demonstration of how the LLN facilitates stochastic simulation-based estimation of e.", null, "• TBD" ]
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https://www.impan.pl/pl/wydawnictwa/czasopisma-i-serie-wydawnicze/fundamenta-mathematicae/all/207/1/88472/homeomorphism-groups-of-sierpinski-carpets-and-erdos-space
[ "# Wydawnictwa / Czasopisma IMPAN / Fundamenta Mathematicae / Wszystkie zeszyty\n\n## Homeomorphism groups of Sierpiński carpets and Erdős space\n\n### Tom 207 / 2010\n\nFundamenta Mathematicae 207 (2010), 1-19 MSC: Primary 57S05. DOI: 10.4064/fm207-1-1\n\n#### Streszczenie\n\nErdős space $\\mathfrak E$ is the “rational” Hilbert space, that is, the set of vectors in $\\ell^2$ with all coordinates rational. Erdős proved that $\\mathfrak E$ is one-dimensional and homeomorphic to its own square $\\mathfrak E \\times \\mathfrak E$, which makes it an important example in dimension theory. Dijkstra and van Mill found topological characterizations of $\\mathfrak E$. Let $M^{n+1}_n$, $n \\in \\mathbb N$, be the $n$-dimensional Menger continuum in $\\mathbb{R}^{n+1}$, also known as the $n$-dimensional Sierpiński carpet, and let $D$ be a countable dense subset of $M^{n+1}_n$. We consider the topological group $\\mathcal{H}(M^{n+1}_n, D)$ of all autohomeomorphisms of $M^{n+1}_n$ that map $D$ onto itself, equipped with the compact-open topology. We show that under some conditions on $D$ the space $\\mathcal{H}(M^{n+1}_n, D)$ is homeomorphic to $\\mathfrak E$ for $n \\in \\mathbb{N} \\setminus \\{3\\}$.\n\n#### Autorzy\n\n• Jan J. DijkstraFaculteit der Exacte Wetenschappen//Afdeling Wiskunde\nVrije Universiteit Amsterdam\nDe Boelelaan 1081\n1081 HV Amsterdam\nThe Netherlands\ne-mail\n• Dave VisserFaculteit der Exacte Wetenschappen//Afdeling Wiskunde\nVrije Universiteit Amsterdam\nDe Boelelaan 1081\n1081 HV Amsterdam\nThe Netherlands\ne-mail\n\n## Przeszukaj wydawnictwa IMPAN\n\nZbyt krótkie zapytanie. Wpisz co najmniej 4 znaki.\n\n## Przepisz kod z obrazka", null, "Odśwież obrazek" ]
[ null, "https://www.impan.pl/pl/wydawnictwa/czasopisma-i-serie-wydawnicze/fundamenta-mathematicae/all/207/1/88472/homeomorphism-groups-of-sierpinski-carpets-and-erdos-space", null ]
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https://www.data-to-viz.com/caveat/spaghetti.html
[ "A Spaghetti plot is a line plot with many lines displayed together. With more than a few (~5?) groups this kind of graphic gets really hard to read, and thus provides little insight about the data. Let’s make an example with the evolution of baby names in the US from 1880 to 2015.\n\n``````# Libraries\nlibrary(tidyverse)\nlibrary(hrbrthemes)\nlibrary(kableExtra)\noptions(knitr.table.format = \"html\")\nlibrary(babynames)\nlibrary(streamgraph)\nlibrary(viridis)\nlibrary(DT)\nlibrary(plotly)\n\ndata <- babynames %>%\nfilter(name %in% c(\"Mary\",\"Emma\", \"Ida\", \"Ashley\", \"Amanda\", \"Jessica\", \"Patricia\", \"Linda\", \"Deborah\", \"Dorothy\", \"Betty\", \"Helen\")) %>%\nfilter(sex==\"F\")\n\n# Plot\ndata %>%\nggplot( aes(x=year, y=n, group=name, color=name)) +\ngeom_line() +\nscale_color_viridis(discrete = TRUE) +\ntheme(\nlegend.position=\"none\",\nplot.title = element_text(size=14)\n) +\nggtitle(\"A spaghetti chart of baby names popularity\") +\ntheme_ipsum()``````" ]
[ null ]
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https://www.ademcetinkaya.com/2023/04/lonhbr-harbour-energy-plc.html
[ "Outlook: HARBOUR ENERGY PLC is assigned short-term Ba1 & long-term Ba1 estimated rating.\nDominant Strategy : Sell\nTime series to forecast n: 09 Apr 2023 for (n+1 year)\nMethodology : Modular Neural Network (Financial Sentiment Analysis)\n\n## Abstract\n\nHARBOUR ENERGY PLC prediction model is evaluated with Modular Neural Network (Financial Sentiment Analysis) and Logistic Regression1,2,3,4 and it is concluded that the LON:HBR stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell\n\n## Key Points\n\n1. How do you know when a stock will go up or down?\n2. What is prediction in deep learning?\n3. Market Risk\n\n## LON:HBR Target Price Prediction Modeling Methodology\n\nWe consider HARBOUR ENERGY PLC Decision Process with Modular Neural Network (Financial Sentiment Analysis) where A is the set of discrete actions of LON:HBR stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4\n\nF(Logistic Regression)5,6,7= $\\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \\dots & {p}_{1n}\\\\ & ⋮\\\\ {p}_{j1}& {p}_{j2}& \\dots & {p}_{jn}\\\\ & ⋮\\\\ {p}_{k1}& {p}_{k2}& \\dots & {p}_{kn}\\\\ & ⋮\\\\ {p}_{n1}& {p}_{n2}& \\dots & {p}_{nn}\\end{array}$ X R(Modular Neural Network (Financial Sentiment Analysis)) X S(n):→ (n+1 year) $∑ i = 1 n r i$\n\nn:Time series to forecast\n\np:Price signals of LON:HBR stock\n\nj:Nash equilibria (Neural Network)\n\nk:Dominated move\n\na:Best response for target price\n\nFor further technical information as per how our model work we invite you to visit the article below:\n\nHow do AC Investment Research machine learning (predictive) algorithms actually work?\n\n## LON:HBR Stock Forecast (Buy or Sell) for (n+1 year)\n\nSample Set: Neural Network\nStock/Index: LON:HBR HARBOUR ENERGY PLC\nTime series to forecast n: 09 Apr 2023 for (n+1 year)\n\nAccording to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell\n\nX axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)\n\nY axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)\n\nZ axis (Grey to Black): *Technical Analysis%\n\n## IFRS Reconciliation Adjustments for HARBOUR ENERGY PLC\n\n1. For example, an entity hedges an exposure to Foreign Currency A using a currency derivative that references Foreign Currency B and Foreign Currencies A and B are pegged (ie their exchange rate is maintained within a band or at an exchange rate set by a central bank or other authority). If the exchange rate between Foreign Currency A and Foreign Currency B were changed (ie a new band or rate was set), rebalancing the hedging relationship to reflect the new exchange rate would ensure that the hedging relationship would continue to meet the hedge effectiveness requirement for the hedge ratio in the new circumstances. In contrast, if there was a default on the currency derivative, changing the hedge ratio could not ensure that the hedging relationship would continue to meet that hedge effectiveness requirement. Hence, rebalancing does not facilitate the continuation of a hedging relationship in situations in which the relationship between the hedging instrument and the hedged item changes in a way that cannot be compensated for by adjusting the hedge ratio\n2. If items are hedged together as a group in a cash flow hedge, they might affect different line items in the statement of profit or loss and other comprehensive income. The presentation of hedging gains or losses in that statement depends on the group of items\n3. In some cases, the qualitative and non-statistical quantitative information available may be sufficient to determine that a financial instrument has met the criterion for the recognition of a loss allowance at an amount equal to lifetime expected credit losses. That is, the information does not need to flow through a statistical model or credit ratings process in order to determine whether there has been a significant increase in the credit risk of the financial instrument. In other cases, an entity may need to consider other information, including information from its statistical models or credit ratings processes.\n4. If there are changes in circumstances that affect hedge effectiveness, an entity may have to change the method for assessing whether a hedging relationship meets the hedge effectiveness requirements in order to ensure that the relevant characteristics of the hedging relationship, including the sources of hedge ineffectiveness, are still captured.\n\n*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.\n\n## Conclusions\n\nHARBOUR ENERGY PLC is assigned short-term Ba1 & long-term Ba1 estimated rating. HARBOUR ENERGY PLC prediction model is evaluated with Modular Neural Network (Financial Sentiment Analysis) and Logistic Regression1,2,3,4 and it is concluded that the LON:HBR stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Sell\n\n### LON:HBR HARBOUR ENERGY PLC Financial Analysis*\n\nRating Short-Term Long-Term Senior\nOutlook*Ba1Ba1\nIncome StatementCBaa2\nBalance SheetBaa2Caa2\nLeverage RatiosCaa2Caa2\nCash FlowBaa2C\nRates of Return and ProfitabilityB2B3\n\n*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.\nHow does neural network examine financial reports and understand financial state of the company?\n\n### Prediction Confidence Score\n\nTrust metric by Neural Network: 91 out of 100 with 619 signals.", null, "## References\n\n1. Bell RM, Koren Y. 2007. Lessons from the Netflix prize challenge. ACM SIGKDD Explor. Newsl. 9:75–79\n2. T. Shardlow and A. Stuart. A perturbation theory for ergodic Markov chains and application to numerical approximations. SIAM journal on numerical analysis, 37(4):1120–1137, 2000\n3. J. Harb and D. Precup. Investigating recurrence and eligibility traces in deep Q-networks. In Deep Reinforcement Learning Workshop, NIPS 2016, Barcelona, Spain, 2016.\n4. R. Rockafellar and S. Uryasev. Optimization of conditional value-at-risk. Journal of Risk, 2:21–42, 2000.\n5. Chamberlain G. 2000. Econometrics and decision theory. J. Econom. 95:255–83\n6. Efron B, Hastie T. 2016. Computer Age Statistical Inference, Vol. 5. Cambridge, UK: Cambridge Univ. Press\n7. Clements, M. P. D. F. Hendry (1997), \"An empirical study of seasonal unit roots in forecasting,\" International Journal of Forecasting, 13, 341–355.\nFrequently Asked QuestionsQ: What is the prediction methodology for LON:HBR stock?\nA: LON:HBR stock prediction methodology: We evaluate the prediction models Modular Neural Network (Financial Sentiment Analysis) and Logistic Regression\nQ: Is LON:HBR stock a buy or sell?\nA: The dominant strategy among neural network is to Sell LON:HBR Stock.\nQ: Is HARBOUR ENERGY PLC stock a good investment?\nA: The consensus rating for HARBOUR ENERGY PLC is Sell and is assigned short-term Ba1 & long-term Ba1 estimated rating.\nQ: What is the consensus rating of LON:HBR stock?\nA: The consensus rating for LON:HBR is Sell.\nQ: What is the prediction period for LON:HBR stock?\nA: The prediction period for LON:HBR is (n+1 year)" ]
[ null, "https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEie6zmojlVkFtzhnSDPsL0ofN0Qf8imsRuJLmBsuPhvw7a_V8sO4akz1ZmrC1z138iTEnVBz4WOe7nRaUku8PCVLHIr3blhvleBHEbt1VcY4D4zSFZNC02CJ-SsnJxURqjoZxkZaeFVd_lKWZF2n-hK-At4y-ts0RYd4tsXgAkh6yfcsMyVn7NLarhmsw/s1600/footerlogo.png", null ]
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https://git.blender.org/gitweb/gitweb.cgi/blender.git/blob/04fdec2017461bc9f85b24a74782e6049c0f2df9:/source/blender/blenlib/intern/BLI_kdopbvh.c
[ "1 /**\n2  *\n3  * ***** BEGIN GPL LICENSE BLOCK *****\n4  *\n5  * This program is free software; you can redistribute it and/or\n6  * modify it under the terms of the GNU General Public License\n7  * as published by the Free Software Foundation; either version 2\n8  * of the License, or (at your option) any later version.\n9  *\n10  * This program is distributed in the hope that it will be useful,\n11  * but WITHOUT ANY WARRANTY; without even the implied warranty of\n12  * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the\n13  * GNU General Public License for more details.\n14  *\n15  * You should have received a copy of the GNU General Public License\n16  * along with this program; if not, write to the Free Software Foundation,\n17  * Inc., 59 Temple Place - Suite 330, Boston, MA  02111-1307, USA.\n18  *\n19  * The Original Code is Copyright (C) 2006 by NaN Holding BV.\n20  * All rights reserved.\n21  *\n22  * The Original Code is: all of this file.\n23  *\n24  * Contributor(s): Daniel Genrich, Andre Pinto\n25  *\n26  * ***** END GPL LICENSE BLOCK *****\n27  */\n29 #include \"math.h\"\n30 #include <stdio.h>\n31 #include <stdlib.h>\n32 #include <string.h>\n33 #include <assert.h>\n35 #include \"MEM_guardedalloc.h\"\n37 #include \"BKE_utildefines.h\"\n39 #include \"BLI_kdopbvh.h\"\n40 #include \"BLI_arithb.h\"\n42 #ifdef _OPENMP\n43 #include <omp.h>\n44 #endif\n48 #define MAX_TREETYPE 32\n49 #define DEFAULT_FIND_NEAREST_HEAP_SIZE 1024\n51 typedef struct BVHNode\n52 {\n53         struct BVHNode **children;\n54         struct BVHNode *parent; // some user defined traversed need that\n55         struct BVHNode *skip;\n56         float *bv;              // Bounding volume of all nodes, max 13 axis\n57         int index;              // face, edge, vertex index\n58         char totnode;   // how many nodes are used, used for speedup\n59         char main_axis; // Axis used to split this node\n60 } BVHNode;\n62 struct BVHTree\n63 {\n64         BVHNode **nodes;\n65         BVHNode *nodearray; /* pre-alloc branch nodes */\n66         BVHNode **nodechild;    // pre-alloc childs for nodes\n67         float   *nodebv;                // pre-alloc bounding-volumes for nodes\n68         float   epsilon; /* epslion is used for inflation of the k-dop     */\n69         int     totleaf; // leafs\n70         int     totbranch;\n71         char    tree_type; // type of tree (4 => quadtree)\n72         char    axis; // kdop type (6 => OBB, 7 => AABB, ...)\n73         char    start_axis, stop_axis; // KDOP_AXES array indices according to axis\n74 };\n76 typedef struct BVHOverlapData\n77 {\n78         BVHTree *tree1, *tree2;\n79         BVHTreeOverlap *overlap;\n80         int i, max_overlap; /* i is number of overlaps */\n81         int start_axis, stop_axis;\n82 } BVHOverlapData;\n84 typedef struct BVHNearestData\n85 {\n86         BVHTree *tree;\n87         const float     *co;\n88         BVHTree_NearestPointCallback callback;\n89         void    *userdata;\n90         float proj;                 //coordinates projection over axis\n91         BVHTreeNearest nearest;\n93 } BVHNearestData;\n95 typedef struct BVHRayCastData\n96 {\n97         BVHTree *tree;\n99         BVHTree_RayCastCallback callback;\n100         void    *userdata;\n103         BVHTreeRay    ray;\n104         float ray_dot_axis;\n105         float idot_axis;\n106         int index;\n108         BVHTreeRayHit hit;\n109 } BVHRayCastData;\n110 ////////////////////////////////////////m\n113 ////////////////////////////////////////////////////////////////////////\n114 // Bounding Volume Hierarchy Definition\n115 //\n116 // Notes: From OBB until 26-DOP --> all bounding volumes possible, just choose type below\n117 // Notes: You have to choose the type at compile time ITM\n118 // Notes: You can choose the tree type --> binary, quad, octree, choose below\n119 ////////////////////////////////////////////////////////////////////////\n121 static float KDOP_AXES =\n122 { {1.0, 0, 0}, {0, 1.0, 0}, {0, 0, 1.0}, {1.0, 1.0, 1.0}, {1.0, -1.0, 1.0}, {1.0, 1.0, -1.0},\n123 {1.0, -1.0, -1.0}, {1.0, 1.0, 0}, {1.0, 0, 1.0}, {0, 1.0, 1.0}, {1.0, -1.0, 0}, {1.0, 0, -1.0},\n124 {0, 1.0, -1.0}\n125 };\n127 /*\n128  * Generic push and pop heap\n129  */\n130 #define PUSH_HEAP_BODY(HEAP_TYPE,PRIORITY,heap,heap_size)       \\\n131 {                                                                                                       \\\n132         HEAP_TYPE element = heap[heap_size-1];                  \\\n133         int child = heap_size-1;                                                \\\n134         while(child != 0)                                                               \\\n135         {                                                                                               \\\n136                 int parent = (child-1) / 2;                                     \\\n137                 if(PRIORITY(element, heap[parent]))                     \\\n138                 {                                                                                       \\\n139                         heap[child] = heap[parent];                             \\\n140                         child = parent;                                                 \\\n141                 }                                                                                       \\\n142                 else break;                                                                     \\\n143         }                                                                                               \\\n144         heap[child] = element;                                                  \\\n145 }\n147 #define POP_HEAP_BODY(HEAP_TYPE, PRIORITY,heap,heap_size)       \\\n148 {                                                                                                       \\\n149         HEAP_TYPE element = heap[heap_size-1];                  \\\n150         int parent = 0;                                                                 \\\n151         while(parent < (heap_size-1)/2 )                                \\\n152         {                                                                                               \\\n153                 int child2 = (parent+1)*2;                                      \\\n154                 if(PRIORITY(heap[child2-1], heap[child2]))      \\\n155                         --child2;                                                               \\\n156                                                                                                         \\\n157                 if(PRIORITY(element, heap[child2]))                     \\\n158                         break;                                                                  \\\n159                                                                                                         \\\n160                 heap[parent] = heap[child2];                            \\\n161                 parent = child2;                                                        \\\n162         }                                                                                               \\\n163         heap[parent] = element;                                                 \\\n164 }\n166 int ADJUST_MEMORY(void *local_memblock, void **memblock, int new_size, int *max_size, int size_per_item)\n167 {\n168         int   new_max_size = *max_size * 2;\n169         void *new_memblock = NULL;\n171         if(new_size <= *max_size)\n172                 return TRUE;\n174         if(*memblock == local_memblock)\n175         {\n176                 new_memblock = malloc( size_per_item * new_max_size );\n177                 memcpy( new_memblock, *memblock, size_per_item * *max_size );\n178         }\n179         else\n180                 new_memblock = realloc(*memblock, size_per_item * new_max_size );\n182         if(new_memblock)\n183         {\n184                 *memblock = new_memblock;\n185                 *max_size = new_max_size;\n186                 return TRUE;\n187         }\n188         else\n189                 return FALSE;\n190 }\n193 //////////////////////////////////////////////////////////////////////////////////////////////////////\n194 // Introsort\n195 // with permission deriven from the following Java code:\n196 // http://ralphunden.net/content/tutorials/a-guide-to-introsort/\n197 // and he derived it from the SUN STL\n198 //////////////////////////////////////////////////////////////////////////////////////////////////////\n199 static int size_threshold = 16;\n200 /*\n201 * Common methods for all algorithms\n202 */\n203 static int floor_lg(int a)\n204 {\n205         return (int)(floor(log(a)/log(2)));\n206 }\n208 /*\n209 * Insertion sort algorithm\n210 */\n211 static void bvh_insertionsort(BVHNode **a, int lo, int hi, int axis)\n212 {\n213         int i,j;\n214         BVHNode *t;\n215         for (i=lo; i < hi; i++)\n216         {\n217                 j=i;\n218                 t = a[i];\n219                 while((j!=lo) && (t->bv[axis] < (a[j-1])->bv[axis]))\n220                 {\n221                         a[j] = a[j-1];\n222                         j--;\n223                 }\n224                 a[j] = t;\n225         }\n226 }\n228 static int bvh_partition(BVHNode **a, int lo, int hi, BVHNode * x, int axis)\n229 {\n230         int i=lo, j=hi;\n231         while (1)\n232         {\n233                 while ((a[i])->bv[axis] < x->bv[axis]) i++;\n234                 j--;\n235                 while (x->bv[axis] < (a[j])->bv[axis]) j--;\n236                 if(!(i < j))\n237                         return i;\n238                 SWAP( BVHNode* , a[i], a[j]);\n239                 i++;\n240         }\n241 }\n243 /*\n244 * Heapsort algorithm\n245 */\n246 static void bvh_downheap(BVHNode **a, int i, int n, int lo, int axis)\n247 {\n248         BVHNode * d = a[lo+i-1];\n249         int child;\n250         while (i<=n/2)\n251         {\n252                 child = 2*i;\n253                 if ((child < n) && ((a[lo+child-1])->bv[axis] < (a[lo+child])->bv[axis]))\n254                 {\n255                         child++;\n256                 }\n257                 if (!(d->bv[axis] < (a[lo+child-1])->bv[axis])) break;\n258                 a[lo+i-1] = a[lo+child-1];\n259                 i = child;\n260         }\n261         a[lo+i-1] = d;\n262 }\n264 static void bvh_heapsort(BVHNode **a, int lo, int hi, int axis)\n265 {\n266         int n = hi-lo, i;\n267         for (i=n/2; i>=1; i=i-1)\n268         {\n269                 bvh_downheap(a, i,n,lo, axis);\n270         }\n271         for (i=n; i>1; i=i-1)\n272         {\n273                 SWAP(BVHNode*, a[lo],a[lo+i-1]);\n274                 bvh_downheap(a, 1,i-1,lo, axis);\n275         }\n276 }\n278 static BVHNode *bvh_medianof3(BVHNode **a, int lo, int mid, int hi, int axis) // returns Sortable\n279 {\n280         if ((a[mid])->bv[axis] < (a[lo])->bv[axis])\n281         {\n282                 if ((a[hi])->bv[axis] < (a[mid])->bv[axis])\n283                         return a[mid];\n284                 else\n285                 {\n286                         if ((a[hi])->bv[axis] < (a[lo])->bv[axis])\n287                                 return a[hi];\n288                         else\n289                                 return a[lo];\n290                 }\n291         }\n292         else\n293         {\n294                 if ((a[hi])->bv[axis] < (a[mid])->bv[axis])\n295                 {\n296                         if ((a[hi])->bv[axis] < (a[lo])->bv[axis])\n297                                 return a[lo];\n298                         else\n299                                 return a[hi];\n300                 }\n301                 else\n302                         return a[mid];\n303         }\n304 }\n305 /*\n306 * Quicksort algorithm modified for Introsort\n307 */\n308 static void bvh_introsort_loop (BVHNode **a, int lo, int hi, int depth_limit, int axis)\n309 {\n310         int p;\n312         while (hi-lo > size_threshold)\n313         {\n314                 if (depth_limit == 0)\n315                 {\n316                         bvh_heapsort(a, lo, hi, axis);\n317                         return;\n318                 }\n319                 depth_limit=depth_limit-1;\n320                 p=bvh_partition(a, lo, hi, bvh_medianof3(a, lo, lo+((hi-lo)/2)+1, hi-1, axis), axis);\n321                 bvh_introsort_loop(a, p, hi, depth_limit, axis);\n322                 hi=p;\n323         }\n324 }\n326 static void sort(BVHNode **a0, int begin, int end, int axis)\n327 {\n328         if (begin < end)\n329         {\n330                 BVHNode **a=a0;\n331                 bvh_introsort_loop(a, begin, end, 2*floor_lg(end-begin), axis);\n332                 bvh_insertionsort(a, begin, end, axis);\n333         }\n334 }\n335 static void sort_along_axis(BVHTree *tree, int start, int end, int axis)\n336 {\n337         sort(tree->nodes, start, end, axis);\n338 }\n340 //after a call to this function you can expect one of:\n341 //      every node to left of a[n] are smaller or equal to it\n342 //      every node to the right of a[n] are greater or equal to it\n343 static int partition_nth_element(BVHNode **a, int _begin, int _end, int n, int axis){\n344         int begin = _begin, end = _end, cut;\n345         while(end-begin > 3)\n346         {\n347                 cut = bvh_partition(a, begin, end, bvh_medianof3(a, begin, (begin+end)/2, end-1, axis), axis );\n348                 if(cut <= n)\n349                         begin = cut;\n350                 else\n351                         end = cut;\n352         }\n353         bvh_insertionsort(a, begin, end, axis);\n355         return n;\n356 }\n358 //////////////////////////////////////////////////////////////////////////////////////////////////////\n359 static void build_skip_links(BVHTree *tree, BVHNode *node, BVHNode *left, BVHNode *right)\n360 {\n361         int i;\n363         node->skip = left;\n364         node->skip = right;\n366         for (i = 0; i < node->totnode; i++)\n367         {\n368                 if(i+1 < node->totnode)\n369                         build_skip_links(tree, node->children[i], left, node->children[i+1] );\n370                 else\n371                         build_skip_links(tree, node->children[i], left, right );\n373                 left = node->children[i];\n374         }\n375 }\n377 /*\n378  * BVHTree bounding volumes functions\n379  */\n380 static void create_kdop_hull(BVHTree *tree, BVHNode *node, float *co, int numpoints, int moving)\n381 {\n382         float newminmax;\n383         float *bv = node->bv;\n384         int i, k;\n386         // don't init boudings for the moving case\n387         if(!moving)\n388         {\n389                 for (i = tree->start_axis; i < tree->stop_axis; i++)\n390                 {\n391                         bv[2*i] = FLT_MAX;\n392                         bv[2*i + 1] = -FLT_MAX;\n393                 }\n394         }\n396         for(k = 0; k < numpoints; k++)\n397         {\n398                 // for all Axes.\n399                 for (i = tree->start_axis; i < tree->stop_axis; i++)\n400                 {\n401                         newminmax = INPR(&co[k * 3], KDOP_AXES[i]);\n402                         if (newminmax < bv[2 * i])\n403                                 bv[2 * i] = newminmax;\n404                         if (newminmax > bv[(2 * i) + 1])\n405                                 bv[(2 * i) + 1] = newminmax;\n406                 }\n407         }\n408 }\n410 // depends on the fact that the BVH's for each face is already build\n411 static void refit_kdop_hull(BVHTree *tree, BVHNode *node, int start, int end)\n412 {\n413         float newmin,newmax;\n414         int i, j;\n415         float *bv = node->bv;\n418         for (i = tree->start_axis; i < tree->stop_axis; i++)\n419         {\n420                 bv[2*i] = FLT_MAX;\n421                 bv[2*i + 1] = -FLT_MAX;\n422         }\n424         for (j = start; j < end; j++)\n425         {\n426 // for all Axes.\n427                 for (i = tree->start_axis; i < tree->stop_axis; i++)\n428                 {\n429                         newmin = tree->nodes[j]->bv[(2 * i)];\n430                         if ((newmin < bv[(2 * i)]))\n431                                 bv[(2 * i)] = newmin;\n433                         newmax = tree->nodes[j]->bv[(2 * i) + 1];\n434                         if ((newmax > bv[(2 * i) + 1]))\n435                                 bv[(2 * i) + 1] = newmax;\n436                 }\n437         }\n439 }\n441 // only supports x,y,z axis in the moment\n442 // but we should use a plain and simple function here for speed sake\n443 static char get_largest_axis(float *bv)\n444 {\n445         float middle_point;\n447         middle_point = (bv) - (bv); // x axis\n448         middle_point = (bv) - (bv); // y axis\n449         middle_point = (bv) - (bv); // z axis\n450         if (middle_point > middle_point)\n451         {\n452                 if (middle_point > middle_point)\n453                         return 1; // max x axis\n454                 else\n455                         return 5; // max z axis\n456         }\n457         else\n458         {\n459                 if (middle_point > middle_point)\n460                         return 3; // max y axis\n461                 else\n462                         return 5; // max z axis\n463         }\n464 }\n466 // bottom-up update of bvh node BV\n467 // join the children on the parent BV\n468 static void node_join(BVHTree *tree, BVHNode *node)\n469 {\n470         int i, j;\n472         for (i = tree->start_axis; i < tree->stop_axis; i++)\n473         {\n474                 node->bv[2*i] = FLT_MAX;\n475                 node->bv[2*i + 1] = -FLT_MAX;\n476         }\n478         for (i = 0; i < tree->tree_type; i++)\n479         {\n480                 if (node->children[i])\n481                 {\n482                         for (j = tree->start_axis; j < tree->stop_axis; j++)\n483                         {\n484                                 // update minimum\n485                                 if (node->children[i]->bv[(2 * j)] < node->bv[(2 * j)])\n486                                         node->bv[(2 * j)] = node->children[i]->bv[(2 * j)];\n488                                 // update maximum\n489                                 if (node->children[i]->bv[(2 * j) + 1] > node->bv[(2 * j) + 1])\n490                                         node->bv[(2 * j) + 1] = node->children[i]->bv[(2 * j) + 1];\n491                         }\n492                 }\n493                 else\n494                         break;\n495         }\n496 }\n498 /*\n499  * Debug and information functions\n500  */\n501 #if 0\n502 static void bvhtree_print_tree(BVHTree *tree, BVHNode *node, int depth)\n503 {\n504         int i;\n505         for(i=0; i<depth; i++) printf(\" \");\n506         printf(\" - %d (%ld): \", node->index, node - tree->nodearray);\n507         for(i=2*tree->start_axis; i<2*tree->stop_axis; i++)\n508                 printf(\"%.3f \", node->bv[i]);\n509         printf(\"\\n\");\n511         for(i=0; i<tree->tree_type; i++)\n512                 if(node->children[i])\n513                         bvhtree_print_tree(tree, node->children[i], depth+1);\n514 }\n516 static void bvhtree_info(BVHTree *tree)\n517 {\n518         printf(\"BVHTree info\\n\");\n519         printf(\"tree_type = %d, axis = %d, epsilon = %f\\n\", tree->tree_type, tree->axis, tree->epsilon);\n520         printf(\"nodes = %d, branches = %d, leafs = %d\\n\", tree->totbranch + tree->totleaf,  tree->totbranch, tree->totleaf);\n521         printf(\"Memory per node = %ldbytes\\n\", sizeof(BVHNode) + sizeof(BVHNode*)*tree->tree_type + sizeof(float)*tree->axis);\n522         printf(\"BV memory = %dbytes\\n\", MEM_allocN_len(tree->nodebv));\n524         printf(\"Total memory = %ldbytes\\n\", sizeof(BVHTree)\n525                 + MEM_allocN_len(tree->nodes)\n526                 + MEM_allocN_len(tree->nodearray)\n527                 + MEM_allocN_len(tree->nodechild)\n528                 + MEM_allocN_len(tree->nodebv)\n529                 );\n531 //      bvhtree_print_tree(tree, tree->nodes[tree->totleaf], 0);\n532 }\n533 #endif\n535 #if 0\n538 static void verify_tree(BVHTree *tree)\n539 {\n540         int i, j, check = 0;\n542         // check the pointer list\n543         for(i = 0; i < tree->totleaf; i++)\n544         {\n545                 if(tree->nodes[i]->parent == NULL)\n546                         printf(\"Leaf has no parent: %d\\n\", i);\n547                 else\n548                 {\n549                         for(j = 0; j < tree->tree_type; j++)\n550                         {\n551                                 if(tree->nodes[i]->parent->children[j] == tree->nodes[i])\n552                                         check = 1;\n553                         }\n554                         if(!check)\n555                         {\n556                                 printf(\"Parent child relationship doesn't match: %d\\n\", i);\n557                         }\n558                         check = 0;\n559                 }\n560         }\n562         // check the leaf list\n563         for(i = 0; i < tree->totleaf; i++)\n564         {\n565                 if(tree->nodearray[i].parent == NULL)\n566                         printf(\"Leaf has no parent: %d\\n\", i);\n567                 else\n568                 {\n569                         for(j = 0; j < tree->tree_type; j++)\n570                         {\n571                                 if(tree->nodearray[i].parent->children[j] == &tree->nodearray[i])\n572                                         check = 1;\n573                         }\n574                         if(!check)\n575                         {\n576                                 printf(\"Parent child relationship doesn't match: %d\\n\", i);\n577                         }\n578                         check = 0;\n579                 }\n580         }\n582         printf(\"branches: %d, leafs: %d, total: %d\\n\", tree->totbranch, tree->totleaf, tree->totbranch + tree->totleaf);\n583 }\n584 #endif\n586 //Helper data and structures to build a min-leaf generalized implicit tree\n587 //This code can be easily reduced (basicly this is only method to calculate pow(k, n) in O(1).. and stuff like that)\n588 typedef struct BVHBuildHelper\n589 {\n590         int tree_type;                          //\n591         int totleafs;                           //\n593         int leafs_per_child  ;      //Min number of leafs that are archievable from a node at depth N\n594         int branches_on_level;      //Number of nodes at depth N (tree_type^N)\n596         int remain_leafs;                       //Number of leafs that are placed on the level that is not 100% filled\n598 } BVHBuildHelper;\n600 static void build_implicit_tree_helper(BVHTree *tree, BVHBuildHelper *data)\n601 {\n602         int depth = 0;\n603         int remain;\n604         int nnodes;\n606         data->totleafs = tree->totleaf;\n607         data->tree_type= tree->tree_type;\n609         //Calculate the smallest tree_type^n such that tree_type^n >= num_leafs\n610         for(\n611                 data->leafs_per_child = 1;\n612                 data->leafs_per_child <  data->totleafs;\n613                 data->leafs_per_child *= data->tree_type\n614         );\n616         data->branches_on_level = 1;\n618         //We could stop the loop first (but I am lazy to find out when)\n619         for(depth = 1; depth < 32; depth++)\n620         {\n621                 data->branches_on_level[depth] = data->branches_on_level[depth-1] * data->tree_type;\n622                 data->leafs_per_child  [depth] = data->leafs_per_child  [depth-1] / data->tree_type;\n623         }\n625         remain = data->totleafs - data->leafs_per_child;\n626         nnodes = (remain + data->tree_type - 2) / (data->tree_type - 1);\n627         data->remain_leafs = remain + nnodes;\n628 }\n630 // return the min index of all the leafs archivable with the given branch\n631 static int implicit_leafs_index(BVHBuildHelper *data, int depth, int child_index)\n632 {\n633         int min_leaf_index = child_index * data->leafs_per_child[depth-1];\n634         if(min_leaf_index <= data->remain_leafs)\n635                 return min_leaf_index;\n636         else if(data->leafs_per_child[depth])\n637                 return data->totleafs - (data->branches_on_level[depth-1] - child_index) * data->leafs_per_child[depth];\n638         else\n639                 return data->remain_leafs;\n640 }\n642 /**\n643  * Generalized implicit tree build\n644  *\n645  * An implicit tree is a tree where its structure is implied, thus there is no need to store child pointers or indexs.\n646  * Its possible to find the position of the child or the parent with simple maths (multiplication and adittion). This type\n647  * of tree is for example used on heaps.. where node N has its childs at indexs N*2 and N*2+1.\n648  *\n649  * Altought in this case the tree type is general.. and not know until runtime.\n650  * tree_type stands for the maximum number of childs that a tree node can have.\n651  * All tree types >= 2 are supported.\n652  *\n653  * Advantages of the used trees include:\n654  *  - No need to store child/parent relations (they are implicit);\n655  *  - Any node child always has an index greater than the parent;\n656  *  - Brother nodes are sequencial in memory;\n657  *\n658  *\n659  * Some math relations derived for general implicit trees:\n660  *\n661  *   K = tree_type, ( 2 <= K )\n662  *   ROOT = 1\n663  *   N child of node A = A * K + (2 - K) + N, (0 <= N < K)\n664  *\n665  * Util methods:\n666  *   TODO...\n667  *    (looping elements, knowing if its a leaf or not.. etc...)\n668  */\n670 // This functions returns the number of branches needed to have the requested number of leafs.\n671 static int implicit_needed_branches(int tree_type, int leafs)\n672 {\n673         return MAX2(1, (leafs + tree_type - 3) / (tree_type-1) );\n674 }\n676 /*\n677  * This function handles the problem of \"sorting\" the leafs (along the split_axis).\n678  *\n679  * It arranges the elements in the given partitions such that:\n680  *  - any element in partition N is less or equal to any element in partition N+1.\n681  *  - if all elements are diferent all partition will get the same subset of elements\n682  *    as if the array was sorted.\n683  *\n684  * partition P is described as the elements in the range ( nth[P] , nth[P+1] ]\n685  *\n686  * TODO: This can be optimized a bit by doing a specialized nth_element instead of K nth_elements\n687  */\n688 static void split_leafs(BVHNode **leafs_array, int *nth, int partitions, int split_axis)\n689 {\n690         int i;\n691         for(i=0; i < partitions-1; i++)\n692         {\n693                 if(nth[i] >= nth[partitions])\n694                         break;\n696                 partition_nth_element(leafs_array, nth[i], nth[partitions], nth[i+1], split_axis);\n697         }\n698 }\n700 /*\n701  * This functions builds an optimal implicit tree from the given leafs.\n702  * Where optimal stands for:\n703  *  - The resulting tree will have the smallest number of branches;\n704  *  - At most only one branch will have NULL childs;\n705  *  - All leafs will be stored at level N or N+1.\n706  *\n707  * This function creates an implicit tree on branches_array, the leafs are given on the leafs_array.\n708  *\n709  * The tree is built per depth levels. First branchs at depth 1.. then branches at depth 2.. etc..\n710  * The reason is that we can build level N+1 from level N witouth any data dependencies.. thus it allows\n711  * to use multithread building.\n712  *\n713  * To archieve this is necessary to find how much leafs are accessible from a certain branch, BVHBuildHelper\n714  * implicit_needed_branches and implicit_leafs_index are auxiliar functions to solve that \"optimal-split\".\n715  */\n716 static void non_recursive_bvh_div_nodes(BVHTree *tree, BVHNode *branches_array, BVHNode **leafs_array, int num_leafs)\n717 {\n718         int i;\n720         const int tree_type   = tree->tree_type;\n721         const int tree_offset = 2 - tree->tree_type; //this value is 0 (on binary trees) and negative on the others\n722         const int num_branches= implicit_needed_branches(tree_type, num_leafs);\n724         BVHBuildHelper data;\n725         int depth;\n727         // set parent from root node to NULL\n728         BVHNode *tmp = branches_array+0;\n729         tmp->parent = NULL;\n731         //Most of bvhtree code relies on 1-leaf trees having at least one branch\n732         //We handle that special case here\n733         if(num_leafs == 1)\n734         {\n735                 BVHNode *root = branches_array+0;\n736                 refit_kdop_hull(tree, root, 0, num_leafs);\n737                 root->main_axis = get_largest_axis(root->bv) / 2;\n738                 root->totnode = 1;\n739                 root->children = leafs_array;\n740                 root->children->parent = root;\n741                 return;\n742         }\n744         branches_array--;       //Implicit trees use 1-based indexs\n746         build_implicit_tree_helper(tree, &data);\n748         //Loop tree levels (log N) loops\n749         for(i=1, depth = 1; i <= num_branches; i = i*tree_type + tree_offset, depth++)\n750         {\n751                 const int first_of_next_level = i*tree_type + tree_offset;\n752                 const int  end_j = MIN2(first_of_next_level, num_branches + 1); //index of last branch on this level\n753                 int j;\n755                 //Loop all branches on this level\n756 #pragma omp parallel for private(j) schedule(static)\n757                 for(j = i; j < end_j; j++)\n758                 {\n759                         int k;\n760                         const int parent_level_index= j-i;\n761                         BVHNode* parent = branches_array + j;\n762                         int nth_positions[ MAX_TREETYPE + 1];\n763                         char split_axis;\n765                         int parent_leafs_begin = implicit_leafs_index(&data, depth, parent_level_index);\n766                         int parent_leafs_end   = implicit_leafs_index(&data, depth, parent_level_index+1);\n768                         //This calculates the bounding box of this branch\n769                         //and chooses the largest axis as the axis to divide leafs\n770                         refit_kdop_hull(tree, parent, parent_leafs_begin, parent_leafs_end);\n771                         split_axis = get_largest_axis(parent->bv);\n773                         //Save split axis (this can be used on raytracing to speedup the query time)\n774                         parent->main_axis = split_axis / 2;\n776                         //Split the childs along the split_axis, note: its not needed to sort the whole leafs array\n777                         //Only to assure that the elements are partioned on a way that each child takes the elements\n778                         //it would take in case the whole array was sorted.\n779                         //Split_leafs takes care of that \"sort\" problem.\n780                         nth_positions[        0] = parent_leafs_begin;\n781                         nth_positions[tree_type] = parent_leafs_end;\n782                         for(k = 1; k < tree_type; k++)\n783                         {\n784                                 int child_index = j * tree_type + tree_offset + k;\n785                                 int child_level_index = child_index - first_of_next_level; //child level index\n786                                 nth_positions[k] = implicit_leafs_index(&data, depth+1, child_level_index);\n787                         }\n789                         split_leafs(leafs_array, nth_positions, tree_type, split_axis);\n792                         //Setup children and totnode counters\n793                         //Not really needed but currently most of BVH code relies on having an explicit children structure\n794                         for(k = 0; k < tree_type; k++)\n795                         {\n796                                 int child_index = j * tree_type + tree_offset + k;\n797                                 int child_level_index = child_index - first_of_next_level; //child level index\n799                                 int child_leafs_begin = implicit_leafs_index(&data, depth+1, child_level_index);\n800                                 int child_leafs_end   = implicit_leafs_index(&data, depth+1, child_level_index+1);\n802                                 if(child_leafs_end - child_leafs_begin > 1)\n803                                 {\n804                                         parent->children[k] = branches_array + child_index;\n805                                         parent->children[k]->parent = parent;\n806                                 }\n807                                 else if(child_leafs_end - child_leafs_begin == 1)\n808                                 {\n809                                         parent->children[k] = leafs_array[ child_leafs_begin ];\n810                                         parent->children[k]->parent = parent;\n811                                 }\n812                                 else\n813                                         break;\n815                                 parent->totnode = k+1;\n816                         }\n817                 }\n818         }\n819 }\n822 /*\n823  * BLI_bvhtree api\n824  */\n825 BVHTree *BLI_bvhtree_new(int maxsize, float epsilon, char tree_type, char axis)\n826 {\n827         BVHTree *tree;\n828         int numnodes, i;\n830         // theres not support for trees below binary-trees :P\n831         if(tree_type < 2)\n832                 return NULL;\n834         if(tree_type > MAX_TREETYPE)\n835                 return NULL;\n837         tree = (BVHTree *)MEM_callocN(sizeof(BVHTree), \"BVHTree\");\n839         //tree epsilon must be >= FLT_EPSILON\n840         //so that tangent rays can still hit a bounding volume..\n841         //this bug would show up when casting a ray aligned with a kdop-axis and with an edge of 2 faces\n842         epsilon = MAX2(FLT_EPSILON, epsilon);\n844         if(tree)\n845         {\n846                 tree->epsilon = epsilon;\n847                 tree->tree_type = tree_type;\n848                 tree->axis = axis;\n850                 if(axis == 26)\n851                 {\n852                         tree->start_axis = 0;\n853                         tree->stop_axis = 13;\n854                 }\n855                 else if(axis == 18)\n856                 {\n857                         tree->start_axis = 7;\n858                         tree->stop_axis = 13;\n859                 }\n860                 else if(axis == 14)\n861                 {\n862                         tree->start_axis = 0;\n863                         tree->stop_axis = 7;\n864                 }\n865                 else if(axis == 8) // AABB\n866                 {\n867                         tree->start_axis = 0;\n868                         tree->stop_axis = 4;\n869                 }\n870                 else if(axis == 6) // OBB\n871                 {\n872                         tree->start_axis = 0;\n873                         tree->stop_axis = 3;\n874                 }\n875                 else\n876                 {\n877                         MEM_freeN(tree);\n878                         return NULL;\n879                 }\n882                 //Allocate arrays\n883                 numnodes = maxsize + implicit_needed_branches(tree_type, maxsize) + tree_type;\n885                 tree->nodes = (BVHNode **)MEM_callocN(sizeof(BVHNode *)*numnodes, \"BVHNodes\");\n887                 if(!tree->nodes)\n888                 {\n889                         MEM_freeN(tree);\n890                         return NULL;\n891                 }\n893                 tree->nodebv = (float*)MEM_callocN(sizeof(float)* axis * numnodes, \"BVHNodeBV\");\n894                 if(!tree->nodebv)\n895                 {\n896                         MEM_freeN(tree->nodes);\n897                         MEM_freeN(tree);\n898                 }\n900                 tree->nodechild = (BVHNode**)MEM_callocN(sizeof(BVHNode*) * tree_type * numnodes, \"BVHNodeBV\");\n901                 if(!tree->nodechild)\n902                 {\n903                         MEM_freeN(tree->nodebv);\n904                         MEM_freeN(tree->nodes);\n905                         MEM_freeN(tree);\n906                 }\n908                 tree->nodearray = (BVHNode *)MEM_callocN(sizeof(BVHNode)* numnodes, \"BVHNodeArray\");\n910                 if(!tree->nodearray)\n911                 {\n912                         MEM_freeN(tree->nodechild);\n913                         MEM_freeN(tree->nodebv);\n914                         MEM_freeN(tree->nodes);\n915                         MEM_freeN(tree);\n916                         return NULL;\n917                 }\n920                 for(i=0; i< numnodes; i++)\n921                 {\n922                         tree->nodearray[i].bv = tree->nodebv + i * axis;\n923                         tree->nodearray[i].children = tree->nodechild + i * tree_type;\n924                 }\n926         }\n928         return tree;\n929 }\n931 void BLI_bvhtree_free(BVHTree *tree)\n932 {\n933         if(tree)\n934         {\n935                 MEM_freeN(tree->nodes);\n936                 MEM_freeN(tree->nodearray);\n937                 MEM_freeN(tree->nodebv);\n938                 MEM_freeN(tree->nodechild);\n939                 MEM_freeN(tree);\n940         }\n941 }\n943 void BLI_bvhtree_balance(BVHTree *tree)\n944 {\n945         int i;\n947         BVHNode*  branches_array = tree->nodearray + tree->totleaf;\n948         BVHNode** leafs_array    = tree->nodes;\n950         //This function should only be called once (some big bug goes here if its being called more than once per tree)\n951         assert(tree->totbranch == 0);\n953         //Build the implicit tree\n954         non_recursive_bvh_div_nodes(tree, branches_array, leafs_array, tree->totleaf);\n956         //current code expects the branches to be linked to the nodes array\n957         //we perform that linkage here\n958         tree->totbranch = implicit_needed_branches(tree->tree_type, tree->totleaf);\n959         for(i = 0; i < tree->totbranch; i++)\n960                 tree->nodes[tree->totleaf + i] = branches_array + i;\n963         //bvhtree_info(tree);\n964 }\n966 int BLI_bvhtree_insert(BVHTree *tree, int index, float *co, int numpoints)\n967 {\n968         int i;\n969         BVHNode *node = NULL;\n971         // insert should only possible as long as tree->totbranch is 0\n972         if(tree->totbranch > 0)\n973                 return 0;\n975         if(tree->totleaf+1 >= MEM_allocN_len(tree->nodes)/sizeof(*(tree->nodes)))\n976                 return 0;\n978         // TODO check if have enough nodes in array\n980         node = tree->nodes[tree->totleaf] = &(tree->nodearray[tree->totleaf]);\n981         tree->totleaf++;\n983         create_kdop_hull(tree, node, co, numpoints, 0);\n984         node->index= index;\n986         // inflate the bv with some epsilon\n987         for (i = tree->start_axis; i < tree->stop_axis; i++)\n988         {\n989                 node->bv[(2 * i)] -= tree->epsilon; // minimum\n990                 node->bv[(2 * i) + 1] += tree->epsilon; // maximum\n991         }\n993         return 1;\n994 }\n997 // call before BLI_bvhtree_update_tree()\n998 int BLI_bvhtree_update_node(BVHTree *tree, int index, float *co, float *co_moving, int numpoints)\n999 {\n1000         int i;\n1001         BVHNode *node= NULL;\n1003         // check if index exists\n1004         if(index > tree->totleaf)\n1005                 return 0;\n1007         node = tree->nodearray + index;\n1009         create_kdop_hull(tree, node, co, numpoints, 0);\n1011         if(co_moving)\n1012                 create_kdop_hull(tree, node, co_moving, numpoints, 1);\n1014         // inflate the bv with some epsilon\n1015         for (i = tree->start_axis; i < tree->stop_axis; i++)\n1016         {\n1017                 node->bv[(2 * i)] -= tree->epsilon; // minimum\n1018                 node->bv[(2 * i) + 1] += tree->epsilon; // maximum\n1019         }\n1021         return 1;\n1022 }\n1024 // call BLI_bvhtree_update_node() first for every node/point/triangle\n1025 void BLI_bvhtree_update_tree(BVHTree *tree)\n1026 {\n1027         //Update bottom=>top\n1028         //TRICKY: the way we build the tree all the childs have an index greater than the parent\n1029         //This allows us todo a bottom up update by starting on the biger numbered branch\n1031         BVHNode** root  = tree->nodes + tree->totleaf;\n1032         BVHNode** index = tree->nodes + tree->totleaf + tree->totbranch-1;\n1034         for (; index >= root; index--)\n1035                 node_join(tree, *index);\n1036 }\n1038 float BLI_bvhtree_getepsilon(BVHTree *tree)\n1039 {\n1040         return tree->epsilon;\n1041 }\n1044 /*\n1045  * BLI_bvhtree_overlap\n1046  */\n1047 // overlap - is it possbile for 2 bv's to collide ?\n1048 static int tree_overlap(BVHNode *node1, BVHNode *node2, int start_axis, int stop_axis)\n1049 {\n1050         float *bv1 = node1->bv;\n1051         float *bv2 = node2->bv;\n1053         float *bv1_end = bv1 + (stop_axis<<1);\n1055         bv1 += start_axis<<1;\n1056         bv2 += start_axis<<1;\n1058         // test all axis if min + max overlap\n1059         for (; bv1 != bv1_end; bv1+=2, bv2+=2)\n1060         {\n1061                 if ((*(bv1) > *(bv2 + 1)) || (*(bv2) > *(bv1 + 1)))\n1062                         return 0;\n1063         }\n1065         return 1;\n1066 }\n1068 static void traverse(BVHOverlapData *data, BVHNode *node1, BVHNode *node2)\n1069 {\n1070         int j;\n1072         if(tree_overlap(node1, node2, data->start_axis, data->stop_axis))\n1073         {\n1074                 // check if node1 is a leaf\n1075                 if(!node1->totnode)\n1076                 {\n1077                         // check if node2 is a leaf\n1078                         if(!node2->totnode)\n1079                         {\n1081                                 if(node1 == node2)\n1082                                 {\n1083                                         return;\n1084                                 }\n1086                                 if(data->i >= data->max_overlap)\n1087                                 {\n1088                                         // try to make alloc'ed memory bigger\n1089                                         data->overlap = realloc(data->overlap, sizeof(BVHTreeOverlap)*data->max_overlap*2);\n1091                                         if(!data->overlap)\n1092                                         {\n1093                                                 printf(\"Out of Memory in traverse\\n\");\n1094                                                 return;\n1095                                         }\n1096                                         data->max_overlap *= 2;\n1097                                 }\n1099                                 // both leafs, insert overlap!\n1100                                 data->overlap[data->i].indexA = node1->index;\n1101                                 data->overlap[data->i].indexB = node2->index;\n1103                                 data->i++;\n1104                         }\n1105                         else\n1106                         {\n1107                                 for(j = 0; j < data->tree2->tree_type; j++)\n1108                                 {\n1109                                         if(node2->children[j])\n1110                                                 traverse(data, node1, node2->children[j]);\n1111                                 }\n1112                         }\n1113                 }\n1114                 else\n1115                 {\n1117                         for(j = 0; j < data->tree2->tree_type; j++)\n1118                         {\n1119                                 if(node1->children[j])\n1120                                         traverse(data, node1->children[j], node2);\n1121                         }\n1122                 }\n1123         }\n1124         return;\n1125 }\n1127 BVHTreeOverlap *BLI_bvhtree_overlap(BVHTree *tree1, BVHTree *tree2, int *result)\n1128 {\n1129         int j, total = 0;\n1130         BVHTreeOverlap *overlap = NULL, *to = NULL;\n1131         BVHOverlapData **data;\n1133         // check for compatibility of both trees (can't compare 14-DOP with 18-DOP)\n1134         if((tree1->axis != tree2->axis) && (tree1->axis == 14 || tree2->axis == 14) && (tree1->axis == 18 || tree2->axis == 18))\n1135                 return 0;\n1137         // fast check root nodes for collision before doing big splitting + traversal\n1138         if(!tree_overlap(tree1->nodes[tree1->totleaf], tree2->nodes[tree2->totleaf], MIN2(tree1->start_axis, tree2->start_axis), MIN2(tree1->stop_axis, tree2->stop_axis)))\n1139                 return 0;\n1141         data = MEM_callocN(sizeof(BVHOverlapData *)* tree1->tree_type, \"BVHOverlapData_star\");\n1143         for(j = 0; j < tree1->tree_type; j++)\n1144         {\n1145                 data[j] = (BVHOverlapData *)MEM_callocN(sizeof(BVHOverlapData), \"BVHOverlapData\");\n1147                 // init BVHOverlapData\n1148                 data[j]->overlap = (BVHTreeOverlap *)malloc(sizeof(BVHTreeOverlap)*MAX2(tree1->totleaf, tree2->totleaf));\n1149                 data[j]->tree1 = tree1;\n1150                 data[j]->tree2 = tree2;\n1151                 data[j]->max_overlap = MAX2(tree1->totleaf, tree2->totleaf);\n1152                 data[j]->i = 0;\n1153                 data[j]->start_axis = MIN2(tree1->start_axis, tree2->start_axis);\n1154                 data[j]->stop_axis  = MIN2(tree1->stop_axis,  tree2->stop_axis );\n1155         }\n1157 #pragma omp parallel for private(j) schedule(static)\n1158         for(j = 0; j < MIN2(tree1->tree_type, tree1->nodes[tree1->totleaf]->totnode); j++)\n1159         {\n1160                 traverse(data[j], tree1->nodes[tree1->totleaf]->children[j], tree2->nodes[tree2->totleaf]);\n1161         }\n1163         for(j = 0; j < tree1->tree_type; j++)\n1164                 total += data[j]->i;\n1166         to = overlap = (BVHTreeOverlap *)MEM_callocN(sizeof(BVHTreeOverlap)*total, \"BVHTreeOverlap\");\n1168         for(j = 0; j < tree1->tree_type; j++)\n1169         {\n1170                 memcpy(to, data[j]->overlap, data[j]->i*sizeof(BVHTreeOverlap));\n1171                 to+=data[j]->i;\n1172         }\n1174         for(j = 0; j < tree1->tree_type; j++)\n1175         {\n1176                 free(data[j]->overlap);\n1177                 MEM_freeN(data[j]);\n1178         }\n1179         MEM_freeN(data);\n1181         (*result) = total;\n1182         return overlap;\n1183 }\n1186 /*\n1187  * Nearest neighbour - BLI_bvhtree_find_nearest\n1188  */\n1189 static float squared_dist(const float *a, const float *b)\n1190 {\n1191         float tmp;\n1192         VECSUB(tmp, a, b);\n1193         return INPR(tmp, tmp);\n1194 }\n1196 //Determines the nearest point of the given node BV. Returns the squared distance to that point.\n1197 static float calc_nearest_point(const float *proj, BVHNode *node, float *nearest)\n1198 {\n1199         int i;\n1200         const float *bv = node->bv;\n1202         //nearest on AABB hull\n1203         for(i=0; i != 3; i++, bv += 2)\n1204         {\n1205                 if(bv > proj[i])\n1206                         nearest[i] = bv;\n1207                 else if(bv < proj[i])\n1208                         nearest[i] = bv;\n1209                 else\n1210                         nearest[i] = proj[i];\n1211         }\n1213 /*\n1214         //nearest on a general hull\n1215         VECCOPY(nearest, data->co);\n1216         for(i = data->tree->start_axis; i != data->tree->stop_axis; i++, bv+=2)\n1217         {\n1218                 float proj = INPR( nearest, KDOP_AXES[i]);\n1219                 float dl = bv - proj;\n1220                 float du = bv - proj;\n1222                 if(dl > 0)\n1223                 {\n1225                 }\n1226                 else if(du < 0)\n1227                 {\n1229                 }\n1230         }\n1231 */\n1232         return squared_dist(proj, nearest);\n1233 }\n1236 typedef struct NodeDistance\n1237 {\n1238         BVHNode *node;\n1239         float dist;\n1241 } NodeDistance;\n1243 #define NodeDistance_priority(a,b)      ( (a).dist < (b).dist )\n1245 // TODO: use a priority queue to reduce the number of nodes looked on\n1246 static void dfs_find_nearest_dfs(BVHNearestData *data, BVHNode *node)\n1247 {\n1248         if(node->totnode == 0)\n1249         {\n1250                 if(data->callback)\n1251                         data->callback(data->userdata , node->index, data->co, &data->nearest);\n1252                 else\n1253                 {\n1254                         data->nearest.index     = node->index;\n1255                         data->nearest.dist      = calc_nearest_point(data->proj, node, data->nearest.co);\n1256                 }\n1257         }\n1258         else\n1259         {\n1260                 //Better heuristic to pick the closest node to dive on\n1261                 int i;\n1262                 float nearest;\n1264                 if(data->proj[ node->main_axis ] <= node->children->bv[node->main_axis*2+1])\n1265                 {\n1267                         for(i=0; i != node->totnode; i++)\n1268                         {\n1269                                 if( calc_nearest_point(data->proj, node->children[i], nearest) >= data->nearest.dist) continue;\n1270                                 dfs_find_nearest_dfs(data, node->children[i]);\n1271                         }\n1272                 }\n1273                 else\n1274                 {\n1275                         for(i=node->totnode-1; i >= 0 ; i--)\n1276                         {\n1277                                 if( calc_nearest_point(data->proj, node->children[i], nearest) >= data->nearest.dist) continue;\n1278                                 dfs_find_nearest_dfs(data, node->children[i]);\n1279                         }\n1280                 }\n1281         }\n1282 }\n1284 static void dfs_find_nearest_begin(BVHNearestData *data, BVHNode *node)\n1285 {\n1286         float nearest, sdist;\n1287         sdist = calc_nearest_point(data->proj, node, nearest);\n1288         if(sdist >= data->nearest.dist) return;\n1289         dfs_find_nearest_dfs(data, node);\n1290 }\n1293 #if 0\n1294 static void NodeDistance_push_heap(NodeDistance *heap, int heap_size)\n1295 PUSH_HEAP_BODY(NodeDistance, NodeDistance_priority, heap, heap_size)\n1297 static void NodeDistance_pop_heap(NodeDistance *heap, int heap_size)\n1298 POP_HEAP_BODY(NodeDistance, NodeDistance_priority, heap, heap_size)\n1300 //NN function that uses an heap.. this functions leads to an optimal number of min-distance\n1301 //but for normal tri-faces and BV 6-dop.. a simple dfs with local heuristics (as implemented\n1302 //in source/blender/blenkernel/intern/shrinkwrap.c) works faster.\n1303 //\n1304 //It may make sense to use this function if the callback queries are very slow.. or if its impossible\n1305 //to get a nice heuristic\n1306 //\n1307 //this function uses \"malloc/free\" instead of the MEM_* because it intends to be openmp safe\n1308 static void bfs_find_nearest(BVHNearestData *data, BVHNode *node)\n1309 {\n1310         int i;\n1311         NodeDistance default_heap[DEFAULT_FIND_NEAREST_HEAP_SIZE];\n1312         NodeDistance *heap=default_heap, current;\n1313         int heap_size = 0, max_heap_size = sizeof(default_heap)/sizeof(default_heap);\n1314         float nearest;\n1316         int callbacks = 0, push_heaps = 0;\n1318         if(node->totnode == 0)\n1319         {\n1320                 dfs_find_nearest_dfs(data, node);\n1321                 return;\n1322         }\n1324         current.node = node;\n1325         current.dist = calc_nearest_point(data->proj, node, nearest);\n1327         while(current.dist < data->nearest.dist)\n1328         {\n1329 //              printf(\"%f : %f\\n\", current.dist, data->nearest.dist);\n1330                 for(i=0; i< current.node->totnode; i++)\n1331                 {\n1332                         BVHNode *child = current.node->children[i];\n1333                         if(child->totnode == 0)\n1334                         {\n1335                                 callbacks++;\n1336                                 dfs_find_nearest_dfs(data, child);\n1337                         }\n1338                         else\n1339                         {\n1341                                 if(heap_size >= max_heap_size\n1342                                 && ADJUST_MEMORY(default_heap, (void**)&heap, heap_size+1, &max_heap_size, sizeof(heap)) == FALSE)\n1343                                 {\n1344                                         printf(\"WARNING: bvh_find_nearest got out of memory\\n\");\n1346                                         if(heap != default_heap)\n1347                                                 free(heap);\n1349                                         return;\n1350                                 }\n1352                                 heap[heap_size].node = current.node->children[i];\n1353                                 heap[heap_size].dist = calc_nearest_point(data->proj, current.node->children[i], nearest);\n1355                                 if(heap[heap_size].dist >= data->nearest.dist) continue;\n1356                                 heap_size++;\n1358                                 NodeDistance_push_heap(heap, heap_size);\n1359         //                      PUSH_HEAP_BODY(NodeDistance, NodeDistance_priority, heap, heap_size);\n1360                                 push_heaps++;\n1361                         }\n1362                 }\n1364                 if(heap_size == 0) break;\n1366                 current = heap;\n1367                 NodeDistance_pop_heap(heap, heap_size);\n1368 //              POP_HEAP_BODY(NodeDistance, NodeDistance_priority, heap, heap_size);\n1369                 heap_size--;\n1370         }\n1372 //      printf(\"hsize=%d, callbacks=%d, pushs=%d\\n\", heap_size, callbacks, push_heaps);\n1374         if(heap != default_heap)\n1375                 free(heap);\n1376 }\n1377 #endif\n1380 int BLI_bvhtree_find_nearest(BVHTree *tree, const float *co, BVHTreeNearest *nearest, BVHTree_NearestPointCallback callback, void *userdata)\n1381 {\n1382         int i;\n1384         BVHNearestData data;\n1385         BVHNode* root = tree->nodes[tree->totleaf];\n1387         //init data to search\n1388         data.tree = tree;\n1389         data.co = co;\n1391         data.callback = callback;\n1392         data.userdata = userdata;\n1394         for(i = data.tree->start_axis; i != data.tree->stop_axis; i++)\n1395         {\n1396                 data.proj[i] = INPR(data.co, KDOP_AXES[i]);\n1397         }\n1399         if(nearest)\n1400         {\n1401                 memcpy( &data.nearest , nearest, sizeof(*nearest) );\n1402         }\n1403         else\n1404         {\n1405                 data.nearest.index = -1;\n1406                 data.nearest.dist = FLT_MAX;\n1407         }\n1409         //dfs search\n1410         if(root)\n1411                 dfs_find_nearest_begin(&data, root);\n1413         //copy back results\n1414         if(nearest)\n1415         {\n1416                 memcpy(nearest, &data.nearest, sizeof(*nearest));\n1417         }\n1419         return data.nearest.index;\n1420 }\n1423 /*\n1424  * Raycast - BLI_bvhtree_ray_cast\n1425  *\n1426  * raycast is done by performing a DFS on the BVHTree and saving the closest hit\n1427  */\n1430 //Determines the distance that the ray must travel to hit the bounding volume of the given node\n1431 static float ray_nearest_hit(BVHRayCastData *data, float *bv)\n1432 {\n1433         int i;\n1435         float low = 0, upper = data->hit.dist;\n1437         for(i=0; i != 3; i++, bv += 2)\n1438         {\n1439                 if(data->ray_dot_axis[i] == 0.0f)\n1440                 {\n1441                         //axis aligned ray\n1442                         if(data->ray.origin[i] < bv - data->ray.radius\n1443                         || data->ray.origin[i] > bv + data->ray.radius)\n1444                                 return FLT_MAX;\n1445                 }\n1446                 else\n1447                 {\n1448                         float ll = (bv - data->ray.radius - data->ray.origin[i]) / data->ray_dot_axis[i];\n1449                         float lu = (bv + data->ray.radius - data->ray.origin[i]) / data->ray_dot_axis[i];\n1451                         if(data->ray_dot_axis[i] > 0.0f)\n1452                         {\n1453                                 if(ll > low)   low = ll;\n1454                                 if(lu < upper) upper = lu;\n1455                         }\n1456                         else\n1457                         {\n1458                                 if(lu > low)   low = lu;\n1459                                 if(ll < upper) upper = ll;\n1460                         }\n1462                         if(low > upper) return FLT_MAX;\n1463                 }\n1464         }\n1465         return low;\n1466 }\n1468 //Determines the distance that the ray must travel to hit the bounding volume of the given node\n1469 //Based on Tactical Optimization of Ray/Box Intersection, by Graham Fyffe\n1470 //[http://tog.acm.org/resources/RTNews/html/rtnv21n1.html#art9]\n1471 //\n1472 //TODO this doens't has data->ray.radius in consideration\n1473 static float fast_ray_nearest_hit(const BVHRayCastData *data, const BVHNode *node)\n1474 {\n1475         const float *bv = node->bv;\n1476         float dist;\n1478         float t1x = (bv[data->index] - data->ray.origin) * data->idot_axis;\n1479         float t2x = (bv[data->index] - data->ray.origin) * data->idot_axis;\n1480         float t1y = (bv[data->index] - data->ray.origin) * data->idot_axis;\n1481         float t2y = (bv[data->index] - data->ray.origin) * data->idot_axis;\n1482         float t1z = (bv[data->index] - data->ray.origin) * data->idot_axis;\n1483         float t2z = (bv[data->index] - data->ray.origin) * data->idot_axis;\n1485         if(t1x > t2y || t2x < t1y || t1x > t2z || t2x < t1z || t1y > t2z || t2y < t1z) return FLT_MAX;\n1486         if(t2x < 0.0 || t2y < 0.0 || t2z < 0.0) return FLT_MAX;\n1487         if(t1x > data->hit.dist || t1y > data->hit.dist || t1z > data->hit.dist) return FLT_MAX;\n1489         dist = t1x;\n1490         if (t1y > dist) dist = t1y;\n1491     if (t1z > dist) dist = t1z;\n1492         return dist;\n1493 }\n1495 static void dfs_raycast(BVHRayCastData *data, BVHNode *node)\n1496 {\n1497         int i;\n1499         //ray-bv is really fast.. and simple tests revealed its worth to test it\n1500         //before calling the ray-primitive functions\n1501         float dist = fast_ray_nearest_hit(data, node);\n1502         if(dist >= data->hit.dist) return;\n1504         if(node->totnode == 0)\n1505         {\n1506                 if(data->callback)\n1507                         data->callback(data->userdata, node->index, &data->ray, &data->hit);\n1508                 else\n1509                 {\n1510                         data->hit.index = node->index;\n1511                         data->hit.dist  = dist;\n1513                 }\n1514         }\n1515         else\n1516         {\n1517                 //pick loop direction to dive into the tree (based on ray direction and split axis)\n1518                 if(data->ray_dot_axis[ (int)node->main_axis ] > 0.0f)\n1519                 {\n1520                         for(i=0; i != node->totnode; i++)\n1521                         {\n1522                                 dfs_raycast(data, node->children[i]);\n1523                         }\n1524                 }\n1525                 else\n1526                 {\n1527                         for(i=node->totnode-1; i >= 0; i--)\n1528                         {\n1529                                 dfs_raycast(data, node->children[i]);\n1530                         }\n1531                 }\n1532         }\n1533 }\n1535 static void iterative_raycast(BVHRayCastData *data, BVHNode *node)\n1536 {\n1537         while(node)\n1538         {\n1539                 float dist = fast_ray_nearest_hit(data, node);\n1540                 if(dist >= data->hit.dist)\n1541                 {\n1542                         node = node->skip;\n1543                         continue;\n1544                 }\n1546                 if(node->totnode == 0)\n1547                 {\n1548                         if(data->callback)\n1549                                 data->callback(data->userdata, node->index, &data->ray, &data->hit);\n1550                         else\n1551                         {\n1552                                 data->hit.index = node->index;\n1553                                 data->hit.dist  = dist;\n1555                         }\n1557                         node = node->skip;\n1558                 }\n1559                 else\n1560                 {\n1561                         node = node->children;\n1562                 }\n1563         }\n1564 }\n1566 int BLI_bvhtree_ray_cast(BVHTree *tree, const float *co, const float *dir, float radius, BVHTreeRayHit *hit, BVHTree_RayCastCallback callback, void *userdata)\n1567 {\n1568         int i;\n1569         BVHRayCastData data;\n1570         BVHNode * root = tree->nodes[tree->totleaf];\n1572         data.tree = tree;\n1574         data.callback = callback;\n1575         data.userdata = userdata;\n1577         VECCOPY(data.ray.origin,    co);\n1578         VECCOPY(data.ray.direction, dir);\n1581         Normalize(data.ray.direction);\n1583         for(i=0; i<3; i++)\n1584         {\n1585                 data.ray_dot_axis[i] = INPR( data.ray.direction, KDOP_AXES[i]);\n1586                 data.idot_axis[i] = 1.0f / data.ray_dot_axis[i];\n1588                 if(fabs(data.ray_dot_axis[i]) < FLT_EPSILON)\n1589                 {\n1590                         data.ray_dot_axis[i] = 0.0;\n1591                 }\n1592                 data.index[2*i] = data.idot_axis[i] < 0.0 ? 1 : 0;\n1593                 data.index[2*i+1] = 1 - data.index[2*i];\n1594                 data.index[2*i]   += 2*i;\n1595                 data.index[2*i+1] += 2*i;\n1596         }\n1599         if(hit)\n1600                 memcpy( &data.hit, hit, sizeof(*hit) );\n1601         else\n1602         {\n1603                 data.hit.index = -1;\n1604                 data.hit.dist = FLT_MAX;\n1605         }\n1607         if(root)\n1608         {\n1609                 dfs_raycast(&data, root);\n1610 //              iterative_raycast(&data, root);\n1611         }\n1614         if(hit)\n1615                 memcpy( hit, &data.hit, sizeof(*hit) );\n1617         return data.hit.index;\n1618 }\n1620 float BLI_bvhtree_bb_raycast(float *bv, float *light_start, float *light_end, float *pos)\n1621 {\n1622         BVHRayCastData data;\n1623         float dist = 0.0;\n1625         data.hit.dist = FLT_MAX;\n1627         // get light direction\n1628         data.ray.direction = light_end - light_start;\n1629         data.ray.direction = light_end - light_start;\n1630         data.ray.direction = light_end - light_start;\n1634         data.ray.origin = light_start;\n1635         data.ray.origin = light_start;\n1636         data.ray.origin = light_start;\n1638         Normalize(data.ray.direction);\n1639         VECCOPY(data.ray_dot_axis, data.ray.direction);\n1641         dist = ray_nearest_hit(&data, bv);\n1643         if(dist > 0.0)\n1644         {\n1646         }\n1647         return dist;\n1649 }\n1651 /*\n1652  * Range Query - as request by broken :P\n1653  *\n1654  * Allocs and fills an array with the indexs of node that are on the given spherical range (center, radius)\n1655  * Returns the size of the array.\n1656  */\n1657 typedef struct RangeQueryData\n1658 {\n1659         BVHTree *tree;\n1660         const float *center;\n1663         int hits;\n1665         BVHTree_RangeQuery callback;\n1666         void *userdata;\n1669 } RangeQueryData;\n1672 static void dfs_range_query(RangeQueryData *data, BVHNode *node)\n1673 {\n1674         if(node->totnode == 0)\n1675         {\n1677                 //Calculate the node min-coords (if the node was a point then this is the point coordinates)\n1678                 float co;\n1679                 co = node->bv;\n1680                 co = node->bv;\n1681                 co = node->bv;\n1683         }\n1684         else\n1685         {\n1686                 int i;\n1687                 for(i=0; i != node->totnode; i++)\n1688                 {\n1689                         float nearest;\n1690                         float dist = calc_nearest_point(data->center, node->children[i], nearest);\n1692                         {\n1693                                 //Its a leaf.. call the callback\n1694                                 if(node->children[i]->totnode == 0)\n1695                                 {\n1696                                         data->hits++;\n1697                                         data->callback( data->userdata, node->children[i]->index, dist );\n1698                                 }\n1699                                 else\n1700                                         dfs_range_query( data, node->children[i] );\n1701                         }\n1702                 }\n1703         }\n1704 }\n1706 int BLI_bvhtree_range_query(BVHTree *tree, const float *co, float radius, BVHTree_RangeQuery callback, void *userdata)\n1707 {\n1708         BVHNode * root = tree->nodes[tree->totleaf];\n1710         RangeQueryData data;\n1711         data.tree = tree;\n1712         data.center = co;\n1714         data.hits = 0;\n1716         data.callback = callback;\n1717         data.userdata = userdata;\n1719         if(root != NULL)\n1720         {\n1721                 float nearest;\n1722                 float dist = calc_nearest_point(data.center, root, nearest);" ]
[ null ]
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https://howkgtolbs.com/convert/8.25-kg-to-lbs
[ "# 8.25 kg to lbs - 8.25 kilograms to pounds\n\nDo you want to know how much is 8.25 kg equal to lbs and how to convert 8.25 kg to lbs? You couldn’t have chosen better. In this article you will find everything about kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to point out that whole this article is devoted to one number of kilograms - that is one kilogram. So if you need to learn more about 8.25 kg to pound conversion - read on.\n\nBefore we move on to the more practical part - that is 8.25 kg how much lbs conversion - we are going to tell you a little bit of theoretical information about these two units - kilograms and pounds. So let’s start.\n\nHow to convert 8.25 kg to lbs? 8.25 kilograms it is equal 18.1881366150 pounds, so 8.25 kg is equal 18.1881366150 lbs.\n\n## 8.25 kgs in pounds\n\nWe are going to begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, that is International System of Units (in abbreviated form SI).\n\nFrom time to time the kilogram can be written as kilogramme. The symbol of the kilogram is kg.\n\nFirstly, the definition of a kilogram was formulated in 1795. The kilogram was described as the mass of one liter of water. This definition was simply but hard to use.\n\nLater, in 1889 the kilogram was described by the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was replaced by a new definition.\n\nToday the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”\n\nOne kilogram is 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.\n\n## 8.25 kilogram to pounds\n\nYou learned some information about kilogram, so now we can go to the pound. The pound is also a unit of mass. It is needed to highlight that there are more than one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we are going to to concentrate only on pound-mass.\n\nThe pound is used in the British and United States customary systems of measurements. Of course, this unit is in use also in other systems. The symbol of this unit is lb or “.\n\nThe international avoirdupois pound has no descriptive definition. It is equal 0.45359237 kilograms. One avoirdupois pound can be divided into 16 avoirdupois ounces or 7000 grains.\n\nThe avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”\n\n### How many lbs is 8.25 kg?\n\n8.25 kilogram is equal to 18.1881366150 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.\n\n### 8.25 kg in lbs\n\nTheoretical section is already behind us. In next part we want to tell you how much is 8.25 kg to lbs. Now you learned that 8.25 kg = x lbs. So it is time to get the answer. Have a look:\n\n8.25 kilogram = 18.1881366150 pounds.\n\nIt is a correct outcome of how much 8.25 kg to pound. You may also round it off. After rounding off your result will be as following: 8.25 kg = 18.150 lbs.\n\nYou know 8.25 kg is how many lbs, so have a look how many kg 8.25 lbs: 8.25 pound = 0.45359237 kilograms.\n\nOf course, in this case you may also round off the result. After rounding off your outcome is as following: 8.25 lb = 0.45 kgs.\n\nWe are also going to show you 8.25 kg to how many pounds and 8.25 pound how many kg outcomes in tables. Let’s see:\n\nWe want to begin with a chart for how much is 8.25 kg equal to pound.\n\n### 8.25 Kilograms to Pounds conversion table\n\nKilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)\n8.25 18.1881366150 18.1500\nNow look at a table for how many kilograms 8.25 pounds.\n\nPounds Kilograms Kilograms (rounded off to two decimal places\n8.25 0.45359237 0.45\n\nNow you learned how many 8.25 kg to lbs and how many kilograms 8.25 pound, so it is time to go to the 8.25 kg to lbs formula.\n\n### 8.25 kg to pounds\n\nTo convert 8.25 kg to us lbs a formula is needed. We will show you two formulas. Let’s start with the first one:\n\nNumber of kilograms * 2.20462262 = the 18.1881366150 outcome in pounds\n\nThe first version of a formula will give you the most exact outcome. In some situations even the smallest difference could be considerable. So if you want to get a correct outcome - this version of a formula will be the best for you/option to convert how many pounds are equivalent to 8.25 kilogram.\n\nSo move on to the second version of a formula, which also enables conversions to know how much 8.25 kilogram in pounds.\n\nThe second formula is as following, let’s see:\n\nAmount of kilograms * 2.2 = the result in pounds\n\nAs you can see, the second version is simpler. It can be the best choice if you need to make a conversion of 8.25 kilogram to pounds in fast way, for instance, during shopping. You only need to remember that final outcome will be not so accurate.\n\nNow we want to learn you how to use these two versions of a formula in practice. But before we will make a conversion of 8.25 kg to lbs we are going to show you easier way to know 8.25 kg to how many lbs without any effort.\n\n### 8.25 kg to lbs converter\n\nAn easier way to know what is 8.25 kilogram equal to in pounds is to use 8.25 kg lbs calculator. What is a kg to lb converter?\n\nConverter is an application. Converter is based on first version of a formula which we gave you above. Thanks to 8.25 kg pound calculator you can easily convert 8.25 kg to lbs. You only need to enter amount of kilograms which you need to convert and click ‘calculate’ button. The result will be shown in a flash.\n\nSo try to convert 8.25 kg into lbs using 8.25 kg vs pound converter. We entered 8.25 as an amount of kilograms. Here is the result: 8.25 kilogram = 18.1881366150 pounds.\n\nAs you can see, our 8.25 kg vs lbs converter is easy to use.\n\nNow we can go to our primary issue - how to convert 8.25 kilograms to pounds on your own.\n\n#### 8.25 kg to lbs conversion\n\nWe are going to start 8.25 kilogram equals to how many pounds calculation with the first version of a formula to get the most correct outcome. A quick reminder of a formula:\n\nNumber of kilograms * 2.20462262 = 18.1881366150 the result in pounds\n\nSo what need you do to know how many pounds equal to 8.25 kilogram? Just multiply amount of kilograms, in this case 8.25, by 2.20462262. It is exactly 18.1881366150. So 8.25 kilogram is equal 18.1881366150.\n\nIt is also possible to round it off, for instance, to two decimal places. It is equal 2.20. So 8.25 kilogram = 18.1500 pounds.\n\nIt is high time for an example from everyday life. Let’s convert 8.25 kg gold in pounds. So 8.25 kg equal to how many lbs? And again - multiply 8.25 by 2.20462262. It is exactly 18.1881366150. So equivalent of 8.25 kilograms to pounds, if it comes to gold, is equal 18.1881366150.\n\nIn this case you can also round off the result. Here is the outcome after rounding off, in this case to one decimal place - 8.25 kilogram 18.150 pounds.\n\nNow let’s move on to examples calculated with short formula.\n\n#### How many 8.25 kg to lbs\n\nBefore we show you an example - a quick reminder of shorter formula:\n\nAmount of kilograms * 2.2 = 18.150 the result in pounds\n\nSo 8.25 kg equal to how much lbs? And again, you need to multiply number of kilogram, in this case 8.25, by 2.2. Look: 8.25 * 2.2 = 18.150. So 8.25 kilogram is exactly 2.2 pounds.\n\nLet’s make another calculation using shorer formula. Now convert something from everyday life, for instance, 8.25 kg to lbs weight of strawberries.\n\nSo let’s convert - 8.25 kilogram of strawberries * 2.2 = 18.150 pounds of strawberries. So 8.25 kg to pound mass is exactly 18.150.\n\nIf you learned how much is 8.25 kilogram weight in pounds and can convert it with use of two different versions of a formula, let’s move on. Now we want to show you all outcomes in charts.\n\n#### Convert 8.25 kilogram to pounds\n\nWe realize that results shown in charts are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in tables for your convenience. Thanks to this you can quickly make a comparison 8.25 kg equivalent to lbs results.\n\nBegin with a 8.25 kg equals lbs chart for the first formula:\n\nKilograms Pounds Pounds (after rounding off to two decimal places)\n8.25 18.1881366150 18.1500\n\nAnd now see 8.25 kg equal pound table for the second formula:\n\nKilograms Pounds\n8.25 18.150\n\nAs you see, after rounding off, when it comes to how much 8.25 kilogram equals pounds, the outcomes are the same. The bigger amount the more significant difference. Remember it when you need to make bigger amount than 8.25 kilograms pounds conversion.\n\n#### How many kilograms 8.25 pound\n\nNow you learned how to calculate 8.25 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What about 8.25 kilogram to pounds and ounces calculation?\n\nWe want to show you how you can convert it step by step. Let’s start. How much is 8.25 kg in lbs and oz?\n\nFirst thing you need to do is multiply amount of kilograms, this time 8.25, by 2.20462262. So 8.25 * 2.20462262 = 18.1881366150. One kilogram is exactly 2.20462262 pounds.\n\nThe integer part is number of pounds. So in this case there are 2 pounds.\n\nTo convert how much 8.25 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It gives 327396192 ounces.\n\nSo your result is equal 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your outcome is exactly 2 pounds and 33 ounces.\n\nAs you see, conversion 8.25 kilogram in pounds and ounces easy.\n\nThe last calculation which we want to show you is conversion of 8.25 foot pounds to kilograms meters. Both of them are units of work.\n\nTo convert it it is needed another formula. Before we show you it, look:\n\n• 8.25 kilograms meters = 7.23301385 foot pounds,\n• 8.25 foot pounds = 0.13825495 kilograms meters.\n\nNow look at a formula:\n\nAmount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters\n\nSo to calculate 8.25 foot pounds to kilograms meters you have to multiply 8.25 by 0.13825495. It is equal 0.13825495. So 8.25 foot pounds is 0.13825495 kilogram meters.\n\nYou can also round off this result, for instance, to two decimal places. Then 8.25 foot pounds is equal 0.14 kilogram meters.\n\nWe hope that this conversion was as easy as 8.25 kilogram into pounds conversions.\n\nThis article is a huge compendium about kilogram, pound and 8.25 kg to lbs in calculation. Thanks to this calculation you know 8.25 kilogram is equivalent to how many pounds.\n\nWe showed you not only how to make a calculation 8.25 kilogram to metric pounds but also two other conversions - to check how many 8.25 kg in pounds and ounces and how many 8.25 foot pounds to kilograms meters.\n\nWe showed you also other solution to make 8.25 kilogram how many pounds calculations, this is with use of 8.25 kg en pound converter. This will be the best choice for those of you who do not like converting on your own at all or this time do not want to make @baseAmountStr kg how lbs conversions on your own.\n\nWe hope that now all of you are able to do 8.25 kilogram equal to how many pounds conversion - on your own or using our 8.25 kgs to pounds calculator.\n\nIt is time to make your move! Let’s calculate 8.25 kilogram mass to pounds in the way you like.\n\nDo you need to do other than 8.25 kilogram as pounds calculation? For example, for 10 kilograms? Check our other articles! We guarantee that conversions for other numbers of kilograms are so easy as for 8.25 kilogram equal many pounds.\n\n### How much is 8.25 kg in pounds\n\nAt the end, we are going to summarize the topic of this article, that is how much is 8.25 kg in pounds , we prepared for you an additional section. Here we have for you all you need to remember about how much is 8.25 kg equal to lbs and how to convert 8.25 kg to lbs . Have a look.\n\nHow does the kilogram to pound conversion look? The conversion kg to lb is just multiplying 2 numbers. How does 8.25 kg to pound conversion formula look? . See it down below:\n\nThe number of kilograms * 2.20462262 = the result in pounds\n\nSo what is the result of the conversion of 8.25 kilogram to pounds? The correct result is 18.1881366150 lbs.\n\nIt is also possible to calculate how much 8.25 kilogram is equal to pounds with second, easier version of the equation. Let’s see.\n\nThe number of kilograms * 2.2 = the result in pounds\n\nSo in this case, 8.25 kg equal to how much lbs ? The result is 18.1881366150 lbs.\n\nHow to convert 8.25 kg to lbs in a few seconds? It is possible to use the 8.25 kg to lbs converter , which will make all calculations for you and you will get an accurate result .\n\n#### Kilograms [kg]\n\nThe kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.\n\n#### Pounds [lbs]\n\nA pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.\nRead more related articles:\n 8.01 kg to lbs = 17.659 8.02 kg to lbs = 17.6811 8.03 kg to lbs = 17.7031 8.04 kg to lbs = 17.7252 8.05 kg to lbs = 17.7472 8.06 kg to lbs = 17.7693 8.07 kg to lbs = 17.7913 8.08 kg to lbs = 17.8133 8.09 kg to lbs = 17.8354 8.1 kg to lbs = 17.8574 8.11 kg to lbs = 17.8795 8.12 kg to lbs = 17.9015 8.13 kg to lbs = 17.9236 8.14 kg to lbs = 17.9456 8.15 kg to lbs = 17.9677 8.16 kg to lbs = 17.9897 8.17 kg to lbs = 18.0118 8.18 kg to lbs = 18.0338 8.19 kg to lbs = 18.0559 8.2 kg to lbs = 18.0779 8.21 kg to lbs = 18.0999 8.22 kg to lbs = 18.122 8.23 kg to lbs = 18.144 8.24 kg to lbs = 18.1661 8.25 kg to lbs = 18.1881\n 8.26 kg to lbs = 18.2102 8.27 kg to lbs = 18.2322 8.28 kg to lbs = 18.2543 8.29 kg to lbs = 18.2763 8.3 kg to lbs = 18.2984 8.31 kg to lbs = 18.3204 8.32 kg to lbs = 18.3425 8.33 kg to lbs = 18.3645 8.34 kg to lbs = 18.3865 8.35 kg to lbs = 18.4086 8.36 kg to lbs = 18.4306 8.37 kg to lbs = 18.4527 8.38 kg to lbs = 18.4747 8.39 kg to lbs = 18.4968 8.4 kg to lbs = 18.5188 8.41 kg to lbs = 18.5409 8.42 kg to lbs = 18.5629 8.43 kg to lbs = 18.585 8.44 kg to lbs = 18.607 8.45 kg to lbs = 18.6291 8.46 kg to lbs = 18.6511 8.47 kg to lbs = 18.6731 8.48 kg to lbs = 18.6952 8.49 kg to lbs = 18.7172 8.5 kg to lbs = 18.7393\n 8.51 kg to lbs = 18.7613 8.52 kg to lbs = 18.7834 8.53 kg to lbs = 18.8054 8.54 kg to lbs = 18.8275 8.55 kg to lbs = 18.8495 8.56 kg to lbs = 18.8716 8.57 kg to lbs = 18.8936 8.58 kg to lbs = 18.9157 8.59 kg to lbs = 18.9377 8.6 kg to lbs = 18.9597 8.61 kg to lbs = 18.9818 8.62 kg to lbs = 19.0038 8.63 kg to lbs = 19.0259 8.64 kg to lbs = 19.0479 8.65 kg to lbs = 19.07 8.66 kg to lbs = 19.092 8.67 kg to lbs = 19.1141 8.68 kg to lbs = 19.1361 8.69 kg to lbs = 19.1582 8.7 kg to lbs = 19.1802 8.71 kg to lbs = 19.2023 8.72 kg to lbs = 19.2243 8.73 kg to lbs = 19.2464 8.74 kg to lbs = 19.2684 8.75 kg to lbs = 19.2904\n 8.76 kg to lbs = 19.3125 8.77 kg to lbs = 19.3345 8.78 kg to lbs = 19.3566 8.79 kg to lbs = 19.3786 8.8 kg to lbs = 19.4007 8.81 kg to lbs = 19.4227 8.82 kg to lbs = 19.4448 8.83 kg to lbs = 19.4668 8.84 kg to lbs = 19.4889 8.85 kg to lbs = 19.5109 8.86 kg to lbs = 19.533 8.87 kg to lbs = 19.555 8.88 kg to lbs = 19.577 8.89 kg to lbs = 19.5991 8.9 kg to lbs = 19.6211 8.91 kg to lbs = 19.6432 8.92 kg to lbs = 19.6652 8.93 kg to lbs = 19.6873 8.94 kg to lbs = 19.7093 8.95 kg to lbs = 19.7314 8.96 kg to lbs = 19.7534 8.97 kg to lbs = 19.7755 8.98 kg to lbs = 19.7975 8.99 kg to lbs = 19.8196 9 kg to lbs = 19.8416" ]
[ null ]
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https://blender.stackexchange.com/questions/187380/exponential-distance-array/187410
[ "# Exponential distance array\n\nHow can I create an array with the distances between the objects increasing exponentially? I want something like this:\n\n[].[]..[]...[]....[].....[]\n\nThe distance between the objects should increase.\n\nThx reply ok I made this but only scale object change not distance", null, "• I can reproduce the effect you got if i add a cube to the Vertex in Edit Mode, and scale the Empty in the x-axis alone. If that is indeed what you did, you should add a cube in Object Mode. Then Shift select the Vertex (so both the cube and vertex are selected) and press Ctrl+P > Object(Keep Transform). Now the cube should be a child of the vertex. The Verts Instancing will create a child(cube) for all the vertices of the parent. Then if you scale the Empty (along any or all axes) it should work. – Fjoersteller Jul 22 '20 at 17:57\n\n## 2 Answers\n\nLike the previous answer use a single vertex array and an offset object.\n\nAdd a cube, enter edit mode and merge the vertices to the center. Press M then select \"Merge at center\". Add an array modifier to the single vertex and set it to constant offset with your required distance.", null, "Add an empty to control the offset. In the array modifier on the single vertex object activate \"Object offset\" and select the empty.\n\nSelect the empty then scale it slightly to get an exponential increase in the distances between the vertices.\n\nCreate a new cube and parent it to the single vertex array.", null, "Select the single vertex array and go to its Object Properties (the orange square on the properties tab) and open the Instancing panel and choose \"Verts\".", null, "The parented cube should be duplicated with increasing distances between each cube.\n\n• Thanks you so much Guys it worked, unbelivble how you can figure out this ;) – Planet yo Jul 22 '20 at 17:29\n• @Planetyo Sigh. Just so you know this is exactly the same method as mine. – batFINGER Jul 22 '20 at 17:35\n• How this is different to the other answer? – brockmann Jul 23 '20 at 7:09\n• @planetyo asked for a simpler explanation and i provided one – Samir Rahamtalla Jul 26 '20 at 19:21\n• for anyone else looking, my solution was Make Instances Real – Aubrie Feb 26 at 4:38\n\nSingle vert, array modifier, object offset, duplivert.", null, "Animating the scale of the offset object\n\nWith the \"Add Mesh Extra Objects\" addon enabled can add a single vert object. Or by other means.\n\nSince the single vert has no dimensions can array it and only effect the arrayed vert locations.\n\nAdd an array modifier to the single vert. Give it object offset add an empty.\n\nUnder Properties > Object > Instancing select Verts.\n\nMake the display object you \"really\" want to array a child of the single vert. Will now have this object instanced at location of each vert created by array modifier.", null, "In the example image the empty has been given x location 2, and x scale 2. The duplivert object is default cube with some scale in z.\n\nSome theory. Assume object is at (0, 0, 0)\n\n>>> loc = Vector() # zero\n\n\nGet the matrix of the object offset object (the empty)\n\n>>> C.object\nbpy.data.objects['Empty']\n\n>>> M = C.object.matrix_world\n\n\nOk original is at x = 0. For the next 4 arrayed objects, apply the offset object transform\n\n>>> for i in range(4):\n... loc = M @ loc\n... print(loc.x)\n...\n2.0\n6.0\n14.0\n30.0\n\n• Thx reply but its too advance for me, I was sure it some simple option I can change to make this effect – Planet yo Jul 22 '20 at 12:57\n• Using arrayed vertices as an instancer is a very neat idea :). – Jachym Michal Jul 22 '20 at 13:45\n• @JachymMichal Appreciated.. – batFINGER Jul 22 '20 at 13:52" ]
[ null, "https://i.stack.imgur.com/qD9qo.png", null, "https://i.stack.imgur.com/fNlsk.png", null, "https://i.stack.imgur.com/YOyMb.png", null, "https://i.stack.imgur.com/6Ag82.png", null, "https://i.stack.imgur.com/2YtZs.gif", null, "https://i.stack.imgur.com/WfFLr.png", null ]
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https://www.catalyzex.com/author/Rajen%20D.%20Shah
[ "", null, "# Rajen D. Shah\n\n## The Projected Covariance Measure for assumption-lean variable significance testing\n\nNov 03, 2022", null, "", null, "", null, "", null, "Testing the significance of a variable or group of variables $X$ for predicting a response $Y$, given additional covariates $Z$, is a ubiquitous task in statistics. A simple but common approach is to specify a linear model, and then test whether the regression coefficient for $X$ is non-zero. However, when the model is misspecified, the test may have poor power, for example when $X$ is involved in complex interactions, or lead to many false rejections. In this work we study the problem of testing the model-free null of conditional mean independence, i.e. that the conditional mean of $Y$ given $X$ and $Z$ does not depend on $X$. We propose a simple and general framework that can leverage flexible nonparametric or machine learning methods, such as additive models or random forests, to yield both robust error control and high power. The procedure involves using these methods to perform regressions, first to estimate a form of projection of $Y$ on $X$ and $Z$ using one half of the data, and then to estimate the expected conditional covariance between this projection and $Y$ on the remaining half of the data. While the approach is general, we show that a version of our procedure using spline regression achieves what we show is the minimax optimal rate in this nonparametric testing problem. Numerical experiments demonstrate the effectiveness of our approach both in terms of maintaining Type I error control, and power, compared to several existing approaches.\n\n* 89 pages, 5 figures\nVia", null, "## Cross-validation for change-point regression: pitfalls and solutions\n\nDec 06, 2021", null, "", null, "", null, "", null, "Cross-validation is the standard approach for tuning parameter selection in many non-parametric regression problems. However its use is less common in change-point regression, perhaps as its prediction error-based criterion may appear to permit small spurious changes and hence be less well-suited to estimation of the number and location of change-points. We show that in fact the problems of cross-validation with squared error loss are more severe and can lead to systematic under- or over-estimation of the number of change-points, and highly suboptimal estimation of the mean function in simple settings where changes are easily detectable. We propose two simple approaches to remedy these issues, the first involving the use of absolute error rather than squared error loss, and the second involving modifying the holdout sets used. For the latter, we provide conditions that permit consistent estimation of the number of change-points for a general change-point estimation procedure. We show these conditions are satisfied for optimal partitioning using new results on its performance when supplied with the incorrect number of change-points. Numerical experiments show that the absolute error approach in particular is competitive with common change-point methods using classical tuning parameter choices when error distributions are well-specified, but can substantially outperform these in misspecified models. An implementation of our methodology is available in the R package crossvalidationCP on CRAN.\n\nVia", null, "## High-dimensional regression with potential prior information on variable importance\n\nSep 23, 2021", null, "", null, "", null, "", null, "There are a variety of settings where vague prior information may be available on the importance of predictors in high-dimensional regression settings. Examples include ordering on the variables offered by their empirical variances (which is typically discarded through standardisation), the lag of predictors when fitting autoregressive models in time series settings, or the level of missingness of the variables. Whilst such orderings may not match the true importance of variables, we argue that there is little to be lost, and potentially much to be gained, by using them. We propose a simple scheme involving fitting a sequence of models indicated by the ordering. We show that the computational cost for fitting all models when ridge regression is used is no more than for a single fit of ridge regression, and describe a strategy for Lasso regression that makes use of previous fits to greatly speed up fitting the entire sequence of models. We propose to select a final estimator by cross-validation and provide a general result on the quality of the best performing estimator on a test set selected from among a number $M$ of competing estimators in a high-dimensional linear regression setting. Our result requires no sparsity assumptions and shows that only a $\\log M$ price is incurred compared to the unknown best estimator. We demonstrate the effectiveness of our approach when applied to missing or corrupted data, and time series settings. An R package is available on github.\n\n* 16 pages, 7 figures\nVia", null, "## Structure Learning for Directed Trees\n\nAug 19, 2021", null, "", null, "", null, "", null, "Knowing the causal structure of a system is of fundamental interest in many areas of science and can aid the design of prediction algorithms that work well under manipulations to the system. The causal structure becomes identifiable from the observational distribution under certain restrictions. To learn the structure from data, score-based methods evaluate different graphs according to the quality of their fits. However, for large nonlinear models, these rely on heuristic optimization approaches with no general guarantees of recovering the true causal structure. In this paper, we consider structure learning of directed trees. We propose a fast and scalable method based on Chu-Liu-Edmonds' algorithm we call causal additive trees (CAT). For the case of Gaussian errors, we prove consistency in an asymptotic regime with a vanishing identifiability gap. We also introduce a method for testing substructure hypotheses with asymptotic family-wise error rate control that is valid post-selection and in unidentified settings. Furthermore, we study the identifiability gap, which quantifies how much better the true causal model fits the observational distribution, and prove that it is lower bounded by local properties of the causal model. Simulation studies demonstrate the favorable performance of CAT compared to competing structure learning methods.\n\n* 84 pages, 17 figures\nVia", null, "## Modelling High-Dimensional Categorical Data Using Nonconvex Fusion Penalties\n\nFeb 28, 2020", null, "", null, "", null, "", null, "We propose a method for estimation in high-dimensional linear models with nominal categorical data. Our estimator, called SCOPE, fuses levels together by making their corresponding coefficients exactly equal. This is achieved using the minimax concave penalty on differences between the order statistics of the coefficients for a categorical variable, thereby clustering the coefficients. We provide an algorithm for exact and efficient computation of the global minimum of the resulting nonconvex objective in the case with a single variable with potentially many levels, and use this within a block coordinate descent procedure in the multivariate case. We show that an oracle least squares solution that exploits the unknown level fusions is a limit point of the coordinate descent with high probability, provided the true levels have a certain minimum separation; these conditions are known to be minimal in the univariate case. We demonstrate the favourable performance of SCOPE across a range of real and simulated datasets. An R package CatReg implementing SCOPE for linear models and also a version for logistic regression is available on CRAN.\n\n* 41 pages, 10 figures\nVia", null, "## The xyz algorithm for fast interaction search in high-dimensional data\n\nSep 17, 2018", null, "", null, "", null, "", null, "When performing regression on a dataset with $p$ variables, it is often of interest to go beyond using main linear effects and include interactions as products between individual variables. For small-scale problems, these interactions can be computed explicitly but this leads to a computational complexity of at least $\\mathcal{O}(p^2)$ if done naively. This cost can be prohibitive if $p$ is very large. We introduce a new randomised algorithm that is able to discover interactions with high probability and under mild conditions has a runtime that is subquadratic in $p$. We show that strong interactions can be discovered in almost linear time, whilst finding weaker interactions requires $\\mathcal{O}(p^\\alpha)$ operations for $1 < \\alpha < 2$ depending on their strength. The underlying idea is to transform interaction search into a closestpair problem which can be solved efficiently in subquadratic time. The algorithm is called $\\mathit{xyz}$ and is implemented in the language R. We demonstrate its efficiency for application to genome-wide association studies, where more than $10^{11}$ interactions can be screened in under $280$ seconds with a single-core $1.2$ GHz CPU.\n\n* JMLR. Journal of Machine Learning Research. The xyz algorithm for fast interaction search in high-dimensional data. Gian-Andrea Thanei, Nicolai Meinshausen, Rajen D. Shah. 19.37.1.42. 2018\nVia", null, "## On b-bit min-wise hashing for large-scale regression and classification with sparse data\n\nFeb 26, 2018", null, "", null, "", null, "", null, "Large-scale regression problems where both the number of variables, $p$, and the number of observations, $n$, may be large and in the order of millions or more, are becoming increasingly more common. Typically the data are sparse: only a fraction of a percent of the entries in the design matrix are non-zero. Nevertheless, often the only computationally feasible approach is to perform dimension reduction to obtain a new design matrix with far fewer columns and then work with this compressed data. $b$-bit min-wise hashing (Li and Konig, 2011) is a promising dimension reduction scheme for sparse matrices which produces a set of random features such that regression on the resulting design matrix approximates a kernel regression with the resemblance kernel. In this work, we derive bounds on the prediction error of such regressions. For both linear and logistic models we show that the average prediction error vanishes asymptotically as long as $q \\|\\beta^*\\|_2^2 /n \\rightarrow 0$, where $q$ is the average number of non-zero entries in each row of the design matrix and $\\beta^*$ is the coefficient of the linear predictor. We also show that ordinary least squares or ridge regression applied to the reduced data can in fact allow us fit more flexible models. We obtain non-asymptotic prediction error bounds for interaction models and for models where an unknown row normalisation must be applied in order for the signal to be linear in the predictors.\n\n* 40 pages\nVia", null, "" ]
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https://matplotlib.org/stable/api/projections/polar.html
[ "# matplotlib.projections.polar#\n\nclass matplotlib.projections.polar.InvertedPolarTransform(axis=None, use_rmin=True, _apply_theta_transforms=True)[source]#\n\nBases: Transform\n\nThe inverse of the polar transform, mapping Cartesian coordinate space x and y back to theta and r.\n\nParameters:\naxisAxis, optional\n\nAxis associated with this transform. This is used to get the minimum radial limit.\n\nuse_rminbool, optional\n\nIf True add the minimum radial axis limit after transforming from Cartesian coordinates. axis must also be specified for this to take effect.\n\nhas_inverse = True#\n\nTrue if this transform has a corresponding inverse transform.\n\ninput_dims = 2#\n\nThe number of input dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ninverted()[source]#\n\nReturn the corresponding inverse transformation.\n\nIt holds x == self.inverted().transform(self.transform(x)).\n\nThe return value of this method should be treated as temporary. An update to self does not cause a corresponding update to its inverted copy.\n\noutput_dims = 2#\n\nThe number of output dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ntransform_non_affine(values)[source]#\n\nApply only the non-affine part of this transformation.\n\ntransform(values) is always equivalent to transform_affine(transform_non_affine(values)).\n\nIn non-affine transformations, this is generally equivalent to transform(values). In affine transformations, this is always a no-op.\n\nParameters:\nvaluesarray\n\nThe input values as an array of length input_dims or shape (N, input_dims).\n\nReturns:\narray\n\nThe output values as an array of length output_dims or shape (N, output_dims), depending on the input.\n\nclass matplotlib.projections.polar.PolarAffine(scale_transform, limits)[source]#\n\nBases: Affine2DBase\n\nThe affine part of the polar projection.\n\nScales the output so that maximum radius rests on the edge of the axes circle and the origin is mapped to (0.5, 0.5). The transform applied is the same to x and y components and given by:\n\n$x_{1} = 0.5 \\left [ \\frac{x_{0}}{(r_{\\max} - r_{\\min})} + 1 \\right ]$\n\n$$r_{\\min}, r_{\\max}$$ are the minimum and maximum radial limits after any scaling (e.g. log scaling) has been removed.\n\nParameters:\nscale_transformTransform\n\nScaling transform for the data. This is used to remove any scaling from the radial view limits.\n\nlimitsBboxBase\n\nView limits of the data. The only part of its bounds that is used is the y limits (for the radius limits).\n\nget_matrix()[source]#\n\nGet the matrix for the affine part of this transform.\n\nclass matplotlib.projections.polar.PolarAxes(*args, theta_offset=0, theta_direction=1, rlabel_position=22.5, **kwargs)[source]#\n\nBases: Axes\n\nA polar graph projection, where the input dimensions are theta, r.\n\nTheta starts pointing east and goes anti-clockwise.\n\nBuild an Axes in a figure.\n\nParameters:\nfigFigure\n\nThe Axes is built in the Figure fig.\n\n*args\n\n*args can be a single (left, bottom, width, height) rectangle or a single Bbox. This specifies the rectangle (in figure coordinates) where the Axes is positioned.\n\n*args can also consist of three numbers or a single three-digit number; in the latter case, the digits are considered as independent numbers. The numbers are interpreted as (nrows, ncols, index): (nrows, ncols) specifies the size of an array of subplots, and index is the 1-based index of the subplot being created. Finally, *args can also directly be a SubplotSpec instance.\n\nsharex, shareyAxes, optional\n\nThe x- or y-axis is shared with the x- or y-axis in the input Axes.\n\nframeonbool, default: True\n\nWhether the Axes frame is visible.\n\nbox_aspectfloat, optional\n\nSet a fixed aspect for the Axes box, i.e. the ratio of height to width. See set_box_aspect for details.\n\n**kwargs\n\nOther optional keyword arguments:\n\nProperty\n\nDescription\n\nadjustable\n\n{'box', 'datalim'}\n\nagg_filter\n\na filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array and two offsets from the bottom left corner of the image\n\nalpha\n\nscalar or None\n\nanchor\n\n(float, float) or {'C', 'SW', 'S', 'SE', 'E', 'NE', ...}\n\nanimated\n\nbool\n\naspect\n\n{'auto', 'equal'} or float\n\nautoscale_on\n\nbool\n\nautoscalex_on\n\nunknown\n\nautoscaley_on\n\nunknown\n\naxes_locator\n\nCallable[[Axes, Renderer], Bbox]\n\naxisbelow\n\nbool or 'line'\n\nbox_aspect\n\nfloat or None\n\nclip_box\n\nBboxBase or None\n\nclip_on\n\nbool\n\nclip_path\n\nPatch or (Path, Transform) or None\n\nfacecolor or fc\n\ncolor\n\nfigure\n\nFigure\n\nframe_on\n\nbool\n\ngid\n\nstr\n\nin_layout\n\nbool\n\nlabel\n\nobject\n\nmouseover\n\nbool\n\nnavigate\n\nbool\n\nnavigate_mode\n\nunknown\n\npath_effects\n\nlist of AbstractPathEffect\n\npicker\n\nNone or bool or float or callable\n\nposition\n\n[left, bottom, width, height] or Bbox\n\nprop_cycle\n\nCycler\n\nrasterization_zorder\n\nfloat or None\n\nrasterized\n\nbool\n\nsketch_params\n\n(scale: float, length: float, randomness: float)\n\nsnap\n\nbool or None\n\nsubplotspec\n\nunknown\n\ntitle\n\nstr\n\ntransform\n\nTransform\n\nurl\n\nstr\n\nvisible\n\nbool\n\nxbound\n\n(lower: float, upper: float)\n\nxlabel\n\nstr\n\nxlim\n\n(left: float, right: float)\n\nxmargin\n\nfloat greater than -0.5\n\nxscale\n\nunknown\n\nxticklabels\n\nunknown\n\nxticks\n\nunknown\n\nybound\n\n(lower: float, upper: float)\n\nylabel\n\nstr\n\nylim\n\n(bottom: float, top: float)\n\nymargin\n\nfloat greater than -0.5\n\nyscale\n\nunknown\n\nyticklabels\n\nunknown\n\nyticks\n\nunknown\n\nzorder\n\nfloat\n\nReturns:\nAxes\n\nThe new Axes object.\n\nclass InvertedPolarTransform(axis=None, use_rmin=True, _apply_theta_transforms=True)[source]#\n\nBases: Transform\n\nThe inverse of the polar transform, mapping Cartesian coordinate space x and y back to theta and r.\n\nParameters:\naxisAxis, optional\n\nAxis associated with this transform. This is used to get the minimum radial limit.\n\nuse_rminbool, optional\n\nIf True add the minimum radial axis limit after transforming from Cartesian coordinates. axis must also be specified for this to take effect.\n\nhas_inverse = True#\n\nTrue if this transform has a corresponding inverse transform.\n\ninput_dims = 2#\n\nThe number of input dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ninverted()[source]#\n\nReturn the corresponding inverse transformation.\n\nIt holds x == self.inverted().transform(self.transform(x)).\n\nThe return value of this method should be treated as temporary. An update to self does not cause a corresponding update to its inverted copy.\n\noutput_dims = 2#\n\nThe number of output dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ntransform_non_affine(values)[source]#\n\nApply only the non-affine part of this transformation.\n\ntransform(values) is always equivalent to transform_affine(transform_non_affine(values)).\n\nIn non-affine transformations, this is generally equivalent to transform(values). In affine transformations, this is always a no-op.\n\nParameters:\nvaluesarray\n\nThe input values as an array of length input_dims or shape (N, input_dims).\n\nReturns:\narray\n\nThe output values as an array of length output_dims or shape (N, output_dims), depending on the input.\n\nclass PolarAffine(scale_transform, limits)[source]#\n\nBases: Affine2DBase\n\nThe affine part of the polar projection.\n\nScales the output so that maximum radius rests on the edge of the axes circle and the origin is mapped to (0.5, 0.5). The transform applied is the same to x and y components and given by:\n\n$x_{1} = 0.5 \\left [ \\frac{x_{0}}{(r_{\\max} - r_{\\min})} + 1 \\right ]$\n\n$$r_{\\min}, r_{\\max}$$ are the minimum and maximum radial limits after any scaling (e.g. log scaling) has been removed.\n\nParameters:\nscale_transformTransform\n\nScaling transform for the data. This is used to remove any scaling from the radial view limits.\n\nlimitsBboxBase\n\nView limits of the data. The only part of its bounds that is used is the y limits (for the radius limits).\n\nget_matrix()[source]#\n\nGet the matrix for the affine part of this transform.\n\nclass PolarTransform(axis=None, use_rmin=True, _apply_theta_transforms=True, *, scale_transform=None)[source]#\n\nBases: Transform\n\nThe base polar transform.\n\nThis transform maps polar coordinates $$\\theta, r$$ into Cartesian coordinates $$x, y = r \\cos(\\theta), r \\sin(\\theta)$$ (but does not fully transform into Axes coordinates or handle positioning in screen space).\n\nThis transformation is designed to be applied to data after any scaling along the radial axis (e.g. log-scaling) has been applied to the input data.\n\nPath segments at a fixed radius are automatically transformed to circular arcs as long as path._interpolation_steps > 1.\n\nParameters:\naxisAxis, optional\n\nAxis associated with this transform. This is used to get the minimum radial limit.\n\nuse_rminbool, optional\n\nIf True, subtract the minimum radial axis limit before transforming to Cartesian coordinates. axis must also be specified for this to take effect.\n\nhas_inverse = True#\n\nTrue if this transform has a corresponding inverse transform.\n\ninput_dims = 2#\n\nThe number of input dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ninverted()[source]#\n\nReturn the corresponding inverse transformation.\n\nIt holds x == self.inverted().transform(self.transform(x)).\n\nThe return value of this method should be treated as temporary. An update to self does not cause a corresponding update to its inverted copy.\n\noutput_dims = 2#\n\nThe number of output dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ntransform_non_affine(values)[source]#\n\nApply only the non-affine part of this transformation.\n\ntransform(values) is always equivalent to transform_affine(transform_non_affine(values)).\n\nIn non-affine transformations, this is generally equivalent to transform(values). In affine transformations, this is always a no-op.\n\nParameters:\nvaluesarray\n\nThe input values as an array of length input_dims or shape (N, input_dims).\n\nReturns:\narray\n\nThe output values as an array of length output_dims or shape (N, output_dims), depending on the input.\n\ntransform_path_non_affine(path)[source]#\n\nApply the non-affine part of this transform to Path path, returning a new Path.\n\ntransform_path(path) is equivalent to transform_path_affine(transform_path_non_affine(values)).\n\nBases: Locator\n\nEnsures that all ticks are strictly positive. For all other tasks, it delegates to the base Locator (which may be different depending on the scale of the r-axis).\n\nnonsingular(vmin, vmax)[source]#\n\nAdjust a range as needed to avoid singularities.\n\nThis method gets called during autoscaling, with (v0, v1) set to the data limits on the axes if the axes contains any data, or (-inf, +inf) if not.\n\n• If v0 == v1 (possibly up to some floating point slop), this method returns an expanded interval around this value.\n\n• If (v0, v1) == (-inf, +inf), this method returns appropriate default view limits.\n\n• Otherwise, (v0, v1) is returned without modification.\n\nset_axis(axis)[source]#\nview_limits(vmin, vmax)[source]#\n\nSelect a scale for the range from vmin to vmax.\n\nSubclasses should override this method to change locator behaviour.\n\nclass ThetaFormatter[source]#\n\nBases: Formatter\n\nUsed to format the theta tick labels. Converts the native unit of radians into degrees and adds a degree symbol.\n\nclass ThetaLocator(base)[source]#\n\nBases: Locator\n\nUsed to locate theta ticks.\n\nThis will work the same as the base locator except in the case that the view spans the entire circle. In such cases, the previously used default locations of every 45 degrees are returned.\n\nset_axis(axis)[source]#\nview_limits(vmin, vmax)[source]#\n\nSelect a scale for the range from vmin to vmax.\n\nSubclasses should override this method to change locator behaviour.\n\ncan_pan()[source]#\n\nReturn whether this Axes supports the pan/zoom button functionality.\n\nFor a polar Axes, this is slightly misleading. Both panning and zooming are performed by the same button. Panning is performed in azimuth while zooming is done along the radial.\n\ncan_zoom()[source]#\n\nReturn whether this Axes supports the zoom box button functionality.\n\nA polar Axes does not support zoom boxes.\n\nclear()[source]#\n\nClear the Axes.\n\ndrag_pan(button, key, x, y)[source]#\n\nCalled when the mouse moves during a pan operation.\n\nParameters:\nbuttonMouseButton\n\nThe pressed mouse button.\n\nkeystr or None\n\nThe pressed key, if any.\n\nx, yfloat\n\nThe mouse coordinates in display coords.\n\nNotes\n\nThis is intended to be overridden by new projection types.\n\ndraw(renderer)[source]#\n\nDraw the Artist (and its children) using the given renderer.\n\nThis has no effect if the artist is not visible (Artist.get_visible returns False).\n\nParameters:\nrendererRendererBase subclass.\n\nNotes\n\nThis method is overridden in the Artist subclasses.\n\nend_pan()[source]#\n\nCalled when a pan operation completes (when the mouse button is up.)\n\nNotes\n\nThis is intended to be overridden by new projection types.\n\nformat_coord(theta, r)[source]#\n\nReturn a format string formatting the x, y coordinates.\n\nget_data_ratio()[source]#\n\nReturn the aspect ratio of the data itself. For a polar plot, this should always be 1.0\n\nget_rlabel_position()[source]#\nReturns:\nfloat\n\nThe theta position of the radius labels in degrees.\n\nget_rmax()[source]#\nReturns:\nfloat\n\nget_rmin()[source]#\nReturns:\nfloat\n\nget_rorigin()[source]#\nReturns:\nfloat\nget_rsign()[source]#\nget_theta_direction()[source]#\n\nGet the direction in which theta increases.\n\n-1:\n\nTheta increases in the clockwise direction\n\n1:\n\nTheta increases in the counterclockwise direction\n\nget_theta_offset()[source]#\n\nGet the offset for the location of 0 in radians.\n\nget_thetamax()[source]#\n\nReturn the maximum theta limit in degrees.\n\nget_thetamin()[source]#\n\nGet the minimum theta limit in degrees.\n\nReturns:\ntransformTransform\n\nThe transform used for drawing x-axis labels, which will add pad_points of padding (in points) between the axis and the label. The x-direction is in data coordinates and the y-direction is in axis coordinates\n\nvalign{'center', 'top', 'bottom', 'baseline', 'center_baseline'}\n\nThe text vertical alignment.\n\nhalign{'center', 'left', 'right'}\n\nThe text horizontal alignment.\n\nNotes\n\nThis transformation is primarily used by the Axis class, and is meant to be overridden by new kinds of projections that may need to place axis elements in different locations.\n\nReturns:\ntransformTransform\n\nThe transform used for drawing secondary x-axis labels, which will add pad_points of padding (in points) between the axis and the label. The x-direction is in data coordinates and the y-direction is in axis coordinates\n\nvalign{'center', 'top', 'bottom', 'baseline', 'center_baseline'}\n\nThe text vertical alignment.\n\nhalign{'center', 'left', 'right'}\n\nThe text horizontal alignment.\n\nNotes\n\nThis transformation is primarily used by the Axis class, and is meant to be overridden by new kinds of projections that may need to place axis elements in different locations.\n\nget_xaxis_transform(which='grid')[source]#\n\nGet the transformation used for drawing x-axis labels, ticks and gridlines. The x-direction is in data coordinates and the y-direction is in axis coordinates.\n\nNote\n\nThis transformation is primarily used by the Axis class, and is meant to be overridden by new kinds of projections that may need to place axis elements in different locations.\n\nParameters:\nwhich{'grid', 'tick1', 'tick2'}\nReturns:\ntransformTransform\n\nThe transform used for drawing y-axis labels, which will add pad_points of padding (in points) between the axis and the label. The x-direction is in axis coordinates and the y-direction is in data coordinates\n\nvalign{'center', 'top', 'bottom', 'baseline', 'center_baseline'}\n\nThe text vertical alignment.\n\nhalign{'center', 'left', 'right'}\n\nThe text horizontal alignment.\n\nNotes\n\nThis transformation is primarily used by the Axis class, and is meant to be overridden by new kinds of projections that may need to place axis elements in different locations.\n\nReturns:\ntransformTransform\n\nThe transform used for drawing secondart y-axis labels, which will add pad_points of padding (in points) between the axis and the label. The x-direction is in axis coordinates and the y-direction is in data coordinates\n\nvalign{'center', 'top', 'bottom', 'baseline', 'center_baseline'}\n\nThe text vertical alignment.\n\nhalign{'center', 'left', 'right'}\n\nThe text horizontal alignment.\n\nNotes\n\nThis transformation is primarily used by the Axis class, and is meant to be overridden by new kinds of projections that may need to place axis elements in different locations.\n\nget_yaxis_transform(which='grid')[source]#\n\nGet the transformation used for drawing y-axis labels, ticks and gridlines. The x-direction is in axis coordinates and the y-direction is in data coordinates.\n\nNote\n\nThis transformation is primarily used by the Axis class, and is meant to be overridden by new kinds of projections that may need to place axis elements in different locations.\n\nParameters:\nwhich{'grid', 'tick1', 'tick2'}\nname = 'polar'#\nset(*, adjustable=<UNSET>, agg_filter=<UNSET>, alpha=<UNSET>, anchor=<UNSET>, animated=<UNSET>, aspect=<UNSET>, autoscale_on=<UNSET>, autoscalex_on=<UNSET>, autoscaley_on=<UNSET>, axes_locator=<UNSET>, axisbelow=<UNSET>, box_aspect=<UNSET>, clip_box=<UNSET>, clip_on=<UNSET>, clip_path=<UNSET>, facecolor=<UNSET>, frame_on=<UNSET>, gid=<UNSET>, in_layout=<UNSET>, label=<UNSET>, mouseover=<UNSET>, navigate=<UNSET>, path_effects=<UNSET>, picker=<UNSET>, position=<UNSET>, prop_cycle=<UNSET>, rasterization_zorder=<UNSET>, rasterized=<UNSET>, rgrids=<UNSET>, rlabel_position=<UNSET>, rlim=<UNSET>, rmax=<UNSET>, rmin=<UNSET>, rorigin=<UNSET>, rscale=<UNSET>, rticks=<UNSET>, sketch_params=<UNSET>, snap=<UNSET>, subplotspec=<UNSET>, theta_direction=<UNSET>, theta_offset=<UNSET>, theta_zero_location=<UNSET>, thetagrids=<UNSET>, thetalim=<UNSET>, thetamax=<UNSET>, thetamin=<UNSET>, title=<UNSET>, transform=<UNSET>, url=<UNSET>, visible=<UNSET>, xbound=<UNSET>, xlabel=<UNSET>, xlim=<UNSET>, xmargin=<UNSET>, xscale=<UNSET>, xticklabels=<UNSET>, xticks=<UNSET>, ybound=<UNSET>, ylabel=<UNSET>, ylim=<UNSET>, ymargin=<UNSET>, yscale=<UNSET>, yticklabels=<UNSET>, yticks=<UNSET>, zorder=<UNSET>)[source]#\n\nSet multiple properties at once.\n\nSupported properties are\n\nProperty\n\nDescription\n\nadjustable\n\n{'box', 'datalim'}\n\nagg_filter\n\na filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array and two offsets from the bottom left corner of the image\n\nalpha\n\nscalar or None\n\nanchor\n\n(float, float) or {'C', 'SW', 'S', 'SE', 'E', 'NE', ...}\n\nanimated\n\nbool\n\naspect\n\n{'auto', 'equal'} or float\n\nautoscale_on\n\nbool\n\nautoscalex_on\n\nunknown\n\nautoscaley_on\n\nunknown\n\naxes_locator\n\nCallable[[Axes, Renderer], Bbox]\n\naxisbelow\n\nbool or 'line'\n\nbox_aspect\n\nfloat or None\n\nclip_box\n\nBboxBase or None\n\nclip_on\n\nbool\n\nclip_path\n\nPatch or (Path, Transform) or None\n\nfacecolor or fc\n\ncolor\n\nfigure\n\nFigure\n\nframe_on\n\nbool\n\ngid\n\nstr\n\nin_layout\n\nbool\n\nlabel\n\nobject\n\nmouseover\n\nbool\n\nnavigate\n\nbool\n\nnavigate_mode\n\nunknown\n\npath_effects\n\nlist of AbstractPathEffect\n\npicker\n\nNone or bool or float or callable\n\nposition\n\n[left, bottom, width, height] or Bbox\n\nprop_cycle\n\nCycler\n\nrasterization_zorder\n\nfloat or None\n\nrasterized\n\nbool\n\nrgrids\n\ntuple with floats\n\nrlabel_position\n\nnumber\n\nrlim\n\nunknown\n\nrmax\n\nfloat\n\nrmin\n\nfloat\n\nrorigin\n\nfloat\n\nrscale\n\nunknown\n\nrticks\n\nunknown\n\nsketch_params\n\n(scale: float, length: float, randomness: float)\n\nsnap\n\nbool or None\n\nsubplotspec\n\nunknown\n\ntheta_direction\n\nunknown\n\ntheta_offset\n\nunknown\n\ntheta_zero_location\n\nstr\n\nthetagrids\n\ntuple with floats, degrees\n\nthetalim\n\nunknown\n\nthetamax\n\nunknown\n\nthetamin\n\nunknown\n\ntitle\n\nstr\n\ntransform\n\nTransform\n\nurl\n\nstr\n\nvisible\n\nbool\n\nxbound\n\n(lower: float, upper: float)\n\nxlabel\n\nstr\n\nxlim\n\n(left: float, right: float)\n\nxmargin\n\nfloat greater than -0.5\n\nxscale\n\nunknown\n\nxticklabels\n\nunknown\n\nxticks\n\nunknown\n\nybound\n\n(lower: float, upper: float)\n\nylabel\n\nstr\n\nylim\n\n(bottom: float, top: float)\n\nymargin\n\nfloat greater than -0.5\n\nyscale\n\nunknown\n\nyticklabels\n\nunknown\n\nyticks\n\nunknown\n\nzorder\n\nfloat\n\nSet the radial gridlines on a polar plot.\n\nParameters:\n\nlabelstuple with strings or None\n\nThe labels to use at each radial gridline. The matplotlib.ticker.ScalarFormatter will be used if None.\n\nanglefloat\n\nThe angular position of the radius labels in degrees.\n\nfmtstr or None\n\nFormat string used in matplotlib.ticker.FormatStrFormatter. For example '%f'.\n\nReturns:\nlineslist of lines.Line2D\n\nlabelslist of text.Text\n\nThe tick labels.\n\nOther Parameters:\n**kwargs\n\nkwargs are optional Text properties for the labels.\n\nWarning\n\nThis only sets the properties of the current ticks. Ticks are not guaranteed to be persistent. Various operations can create, delete and modify the Tick instances. There is an imminent risk that these settings can get lost if you work on the figure further (including also panning/zooming on a displayed figure).\n\nUse set_tick_params instead if possible.\n\nset_rlabel_position(value)[source]#\n\nUpdate the theta position of the radius labels.\n\nParameters:\nvaluenumber\n\nThe angular position of the radius labels in degrees.\n\nset_rlim(bottom=None, top=None, *, emit=True, auto=False, **kwargs)[source]#\n\nSet the radial axis view limits.\n\nThis function behaves like Axes.set_ylim, but additionally supports rmin and rmax as aliases for bottom and top.\n\nset_rmax(rmax)[source]#\n\nParameters:\nrmaxfloat\nset_rmin(rmin)[source]#\n\nParameters:\nrminfloat\nset_rorigin(rorigin)[source]#\n\nParameters:\nroriginfloat\nset_rscale(*args, **kwargs)[source]#\nset_rticks(*args, **kwargs)[source]#\nset_theta_direction(direction)[source]#\n\nSet the direction in which theta increases.\n\nclockwise, -1:\n\nTheta increases in the clockwise direction\n\ncounterclockwise, anticlockwise, 1:\n\nTheta increases in the counterclockwise direction\n\nset_theta_offset(offset)[source]#\n\nSet the offset for the location of 0 in radians.\n\nset_theta_zero_location(loc, offset=0.0)[source]#\n\nSet the location of theta's zero.\n\nThis simply calls set_theta_offset with the correct value in radians.\n\nParameters:\nlocstr\n\nMay be one of \"N\", \"NW\", \"W\", \"SW\", \"S\", \"SE\", \"E\", or \"NE\".\n\noffsetfloat, default: 0\n\nAn offset in degrees to apply from the specified loc. Note: this offset is always applied counter-clockwise regardless of the direction setting.\n\nset_thetagrids(angles, labels=None, fmt=None, **kwargs)[source]#\n\nSet the theta gridlines in a polar plot.\n\nParameters:\nanglestuple with floats, degrees\n\nThe angles of the theta gridlines.\n\nlabelstuple with strings or None\n\nThe labels to use at each theta gridline. The projections.polar.ThetaFormatter will be used if None.\n\nfmtstr or None\n\nFormat string used in matplotlib.ticker.FormatStrFormatter. For example '%f'. Note that the angle that is used is in radians.\n\nReturns:\nlineslist of lines.Line2D\n\nThe theta gridlines.\n\nlabelslist of text.Text\n\nThe tick labels.\n\nOther Parameters:\n**kwargs\n\nkwargs are optional Text properties for the labels.\n\nWarning\n\nThis only sets the properties of the current ticks. Ticks are not guaranteed to be persistent. Various operations can create, delete and modify the Tick instances. There is an imminent risk that these settings can get lost if you work on the figure further (including also panning/zooming on a displayed figure).\n\nUse set_tick_params instead if possible.\n\nset_thetalim(*args, **kwargs)[source]#\n\nSet the minimum and maximum theta values.\n\nCan take the following signatures:\n\n• set_thetalim(minval, maxval): Set the limits in radians.\n\n• set_thetalim(thetamin=minval, thetamax=maxval): Set the limits in degrees.\n\nwhere minval and maxval are the minimum and maximum limits. Values are wrapped in to the range $$[0, 2\\pi]$$ (in radians), so for example it is possible to do set_thetalim(-np.pi / 2, np.pi / 2) to have an axis symmetric around 0. A ValueError is raised if the absolute angle difference is larger than a full circle.\n\nset_thetamax(thetamax)[source]#\n\nSet the maximum theta limit in degrees.\n\nset_thetamin(thetamin)[source]#\n\nSet the minimum theta limit in degrees.\n\nset_yscale(*args, **kwargs)[source]#\n\nSet the yaxis' scale.\n\nParameters:\nvalue{\"linear\", \"log\", \"symlog\", \"logit\", ...} or ScaleBase\n\nThe axis scale type to apply.\n\n**kwargs\n\nDifferent keyword arguments are accepted, depending on the scale. See the respective class keyword arguments:\n\nNotes\n\nBy default, Matplotlib supports the above-mentioned scales. Additionally, custom scales may be registered using matplotlib.scale.register_scale. These scales can then also be used here.\n\nstart_pan(x, y, button)[source]#\n\nCalled when a pan operation has started.\n\nParameters:\nx, yfloat\n\nThe mouse coordinates in display coords.\n\nbuttonMouseButton\n\nThe pressed mouse button.\n\nNotes\n\nThis is intended to be overridden by new projection types.\n\nclass matplotlib.projections.polar.PolarTransform(axis=None, use_rmin=True, _apply_theta_transforms=True, *, scale_transform=None)[source]#\n\nBases: Transform\n\nThe base polar transform.\n\nThis transform maps polar coordinates $$\\theta, r$$ into Cartesian coordinates $$x, y = r \\cos(\\theta), r \\sin(\\theta)$$ (but does not fully transform into Axes coordinates or handle positioning in screen space).\n\nThis transformation is designed to be applied to data after any scaling along the radial axis (e.g. log-scaling) has been applied to the input data.\n\nPath segments at a fixed radius are automatically transformed to circular arcs as long as path._interpolation_steps > 1.\n\nParameters:\naxisAxis, optional\n\nAxis associated with this transform. This is used to get the minimum radial limit.\n\nuse_rminbool, optional\n\nIf True, subtract the minimum radial axis limit before transforming to Cartesian coordinates. axis must also be specified for this to take effect.\n\nhas_inverse = True#\n\nTrue if this transform has a corresponding inverse transform.\n\ninput_dims = 2#\n\nThe number of input dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ninverted()[source]#\n\nReturn the corresponding inverse transformation.\n\nIt holds x == self.inverted().transform(self.transform(x)).\n\nThe return value of this method should be treated as temporary. An update to self does not cause a corresponding update to its inverted copy.\n\noutput_dims = 2#\n\nThe number of output dimensions of this transform. Must be overridden (with integers) in the subclass.\n\ntransform_non_affine(values)[source]#\n\nApply only the non-affine part of this transformation.\n\ntransform(values) is always equivalent to transform_affine(transform_non_affine(values)).\n\nIn non-affine transformations, this is generally equivalent to transform(values). In affine transformations, this is always a no-op.\n\nParameters:\nvaluesarray\n\nThe input values as an array of length input_dims or shape (N, input_dims).\n\nReturns:\narray\n\nThe output values as an array of length output_dims or shape (N, output_dims), depending on the input.\n\ntransform_path_non_affine(path)[source]#\n\nApply the non-affine part of this transform to Path path, returning a new Path.\n\ntransform_path(path) is equivalent to transform_path_affine(transform_path_non_affine(values)).\n\nBases: YAxis\n\nThis overrides certain properties of a YAxis to provide special-casing for a radial axis.\n\nParameters:\naxesAxes\n\nThe Axes to which the created Axis belongs.\n\nThe acceptance radius for containment tests. See also Axis.contains.\n\nclearbool, default: True\n\nWhether to clear the Axis on creation. This is not required, e.g., when creating an Axis as part of an Axes, as Axes.clear will call Axis.clear. .. versionadded:: 3.8\n\nclear()[source]#\n\nClear the axis.\n\nThis resets axis properties to their default values:\n\n• the label\n\n• the scale\n\n• locators, formatters and ticks\n\n• major and minor grid\n\n• units\n\n• registered callbacks\n\nset(*, agg_filter=<UNSET>, alpha=<UNSET>, animated=<UNSET>, clip_box=<UNSET>, clip_on=<UNSET>, clip_path=<UNSET>, data_interval=<UNSET>, gid=<UNSET>, in_layout=<UNSET>, inverted=<UNSET>, label=<UNSET>, label_coords=<UNSET>, label_position=<UNSET>, label_text=<UNSET>, major_formatter=<UNSET>, major_locator=<UNSET>, minor_formatter=<UNSET>, minor_locator=<UNSET>, mouseover=<UNSET>, offset_position=<UNSET>, path_effects=<UNSET>, picker=<UNSET>, pickradius=<UNSET>, rasterized=<UNSET>, remove_overlapping_locs=<UNSET>, sketch_params=<UNSET>, snap=<UNSET>, tick_params=<UNSET>, ticklabels=<UNSET>, ticks=<UNSET>, ticks_position=<UNSET>, transform=<UNSET>, units=<UNSET>, url=<UNSET>, view_interval=<UNSET>, visible=<UNSET>, zorder=<UNSET>)[source]#\n\nSet multiple properties at once.\n\nSupported properties are\n\nProperty\n\nDescription\n\nagg_filter\n\na filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array and two offsets from the bottom left corner of the image\n\nalpha\n\nscalar or None\n\nanimated\n\nbool\n\nclip_box\n\nBboxBase or None\n\nclip_on\n\nbool\n\nclip_path\n\nPatch or (Path, Transform) or None\n\ndata_interval\n\nunknown\n\nfigure\n\nFigure\n\ngid\n\nstr\n\nin_layout\n\nbool\n\ninverted\n\nunknown\n\nlabel\n\nobject\n\nlabel_coords\n\nunknown\n\nlabel_position\n\n{'left', 'right'}\n\nlabel_text\n\nstr\n\nmajor_formatter\n\nFormatter, str, or function\n\nmajor_locator\n\nLocator\n\nminor_formatter\n\nFormatter, str, or function\n\nminor_locator\n\nLocator\n\nmouseover\n\nbool\n\noffset_position\n\n{'left', 'right'}\n\npath_effects\n\nlist of AbstractPathEffect\n\npicker\n\nNone or bool or float or callable\n\npickradius\n\nfloat\n\nrasterized\n\nbool\n\nremove_overlapping_locs\n\nunknown\n\nsketch_params\n\n(scale: float, length: float, randomness: float)\n\nsnap\n\nbool or None\n\ntick_params\n\nunknown\n\nticklabels\n\nunknown\n\nticks\n\n1D array-like\n\nticks_position\n\n{'left', 'right', 'both', 'default', 'none'}\n\ntransform\n\nTransform\n\nunits\n\nunits tag\n\nurl\n\nstr\n\nview_interval\n\nunknown\n\nvisible\n\nbool\n\nzorder\n\nfloat\n\nBases: Locator\n\nEnsures that all ticks are strictly positive. For all other tasks, it delegates to the base Locator (which may be different depending on the scale of the r-axis).\n\nnonsingular(vmin, vmax)[source]#\n\nAdjust a range as needed to avoid singularities.\n\nThis method gets called during autoscaling, with (v0, v1) set to the data limits on the axes if the axes contains any data, or (-inf, +inf) if not.\n\n• If v0 == v1 (possibly up to some floating point slop), this method returns an expanded interval around this value.\n\n• If (v0, v1) == (-inf, +inf), this method returns appropriate default view limits.\n\n• Otherwise, (v0, v1) is returned without modification.\n\nset_axis(axis)[source]#\nview_limits(vmin, vmax)[source]#\n\nSelect a scale for the range from vmin to vmax.\n\nSubclasses should override this method to change locator behaviour.\n\nBases: YTick\n\nThis subclass of YTick provides radial ticks with some small modification to their re-positioning such that ticks are rotated based on axes limits. This results in ticks that are correctly perpendicular to the spine. Labels are also rotated to be perpendicular to the spine, when 'auto' rotation is enabled.\n\nbbox is the Bound2D bounding box in display coords of the Axes loc is the tick location in data coords size is the tick size in points\n\nset(*, agg_filter=<UNSET>, alpha=<UNSET>, animated=<UNSET>, clip_box=<UNSET>, clip_on=<UNSET>, clip_path=<UNSET>, gid=<UNSET>, in_layout=<UNSET>, label=<UNSET>, label1=<UNSET>, label2=<UNSET>, mouseover=<UNSET>, pad=<UNSET>, path_effects=<UNSET>, picker=<UNSET>, rasterized=<UNSET>, sketch_params=<UNSET>, snap=<UNSET>, transform=<UNSET>, url=<UNSET>, visible=<UNSET>, zorder=<UNSET>)[source]#\n\nSet multiple properties at once.\n\nSupported properties are\n\nProperty\n\nDescription\n\nagg_filter\n\na filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array and two offsets from the bottom left corner of the image\n\nalpha\n\nscalar or None\n\nanimated\n\nbool\n\nclip_box\n\nBboxBase or None\n\nclip_on\n\nbool\n\nclip_path\n\nPatch or (Path, Transform) or None\n\nfigure\n\nFigure\n\ngid\n\nstr\n\nin_layout\n\nbool\n\nlabel\n\nunknown\n\nlabel1\n\nunknown\n\nlabel2\n\nunknown\n\nmouseover\n\nbool\n\npad\n\nfloat\n\npath_effects\n\nlist of AbstractPathEffect\n\npicker\n\nNone or bool or float or callable\n\nrasterized\n\nbool\n\nsketch_params\n\n(scale: float, length: float, randomness: float)\n\nsnap\n\nbool or None\n\ntransform\n\nTransform\n\nurl\n\nstr\n\nvisible\n\nbool\n\nzorder\n\nfloat\n\nupdate_position(loc)[source]#\n\nSet the location of tick in data coords with scalar loc.\n\nclass matplotlib.projections.polar.ThetaAxis(*args, **kwargs)[source]#\n\nBases: XAxis\n\nA theta Axis.\n\nThis overrides certain properties of an XAxis to provide special-casing for an angular axis.\n\nParameters:\naxesAxes\n\nThe Axes to which the created Axis belongs.\n\nThe acceptance radius for containment tests. See also Axis.contains.\n\nclearbool, default: True\n\nWhether to clear the Axis on creation. This is not required, e.g., when creating an Axis as part of an Axes, as Axes.clear will call Axis.clear. .. versionadded:: 3.8\n\naxis_name = 'theta'#\n\nclear()[source]#\n\nClear the axis.\n\nThis resets axis properties to their default values:\n\n• the label\n\n• the scale\n\n• locators, formatters and ticks\n\n• major and minor grid\n\n• units\n\n• registered callbacks\n\nset(*, agg_filter=<UNSET>, alpha=<UNSET>, animated=<UNSET>, clip_box=<UNSET>, clip_on=<UNSET>, clip_path=<UNSET>, data_interval=<UNSET>, gid=<UNSET>, in_layout=<UNSET>, inverted=<UNSET>, label=<UNSET>, label_coords=<UNSET>, label_position=<UNSET>, label_text=<UNSET>, major_formatter=<UNSET>, major_locator=<UNSET>, minor_formatter=<UNSET>, minor_locator=<UNSET>, mouseover=<UNSET>, path_effects=<UNSET>, picker=<UNSET>, pickradius=<UNSET>, rasterized=<UNSET>, remove_overlapping_locs=<UNSET>, sketch_params=<UNSET>, snap=<UNSET>, tick_params=<UNSET>, ticklabels=<UNSET>, ticks=<UNSET>, ticks_position=<UNSET>, transform=<UNSET>, units=<UNSET>, url=<UNSET>, view_interval=<UNSET>, visible=<UNSET>, zorder=<UNSET>)[source]#\n\nSet multiple properties at once.\n\nSupported properties are\n\nProperty\n\nDescription\n\nagg_filter\n\na filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array and two offsets from the bottom left corner of the image\n\nalpha\n\nscalar or None\n\nanimated\n\nbool\n\nclip_box\n\nBboxBase or None\n\nclip_on\n\nbool\n\nclip_path\n\nPatch or (Path, Transform) or None\n\ndata_interval\n\nunknown\n\nfigure\n\nFigure\n\ngid\n\nstr\n\nin_layout\n\nbool\n\ninverted\n\nunknown\n\nlabel\n\nobject\n\nlabel_coords\n\nunknown\n\nlabel_position\n\n{'top', 'bottom'}\n\nlabel_text\n\nstr\n\nmajor_formatter\n\nFormatter, str, or function\n\nmajor_locator\n\nLocator\n\nminor_formatter\n\nFormatter, str, or function\n\nminor_locator\n\nLocator\n\nmouseover\n\nbool\n\npath_effects\n\nlist of AbstractPathEffect\n\npicker\n\nNone or bool or float or callable\n\npickradius\n\nfloat\n\nrasterized\n\nbool\n\nremove_overlapping_locs\n\nunknown\n\nsketch_params\n\n(scale: float, length: float, randomness: float)\n\nsnap\n\nbool or None\n\ntick_params\n\nunknown\n\nticklabels\n\nunknown\n\nticks\n\n1D array-like\n\nticks_position\n\n{'top', 'bottom', 'both', 'default', 'none'}\n\ntransform\n\nTransform\n\nunits\n\nunits tag\n\nurl\n\nstr\n\nview_interval\n\nunknown\n\nvisible\n\nbool\n\nzorder\n\nfloat\n\nclass matplotlib.projections.polar.ThetaFormatter[source]#\n\nBases: Formatter\n\nUsed to format the theta tick labels. Converts the native unit of radians into degrees and adds a degree symbol.\n\nclass matplotlib.projections.polar.ThetaLocator(base)[source]#\n\nBases: Locator\n\nUsed to locate theta ticks.\n\nThis will work the same as the base locator except in the case that the view spans the entire circle. In such cases, the previously used default locations of every 45 degrees are returned.\n\nset_axis(axis)[source]#\nview_limits(vmin, vmax)[source]#\n\nSelect a scale for the range from vmin to vmax.\n\nSubclasses should override this method to change locator behaviour.\n\nclass matplotlib.projections.polar.ThetaTick(axes, *args, **kwargs)[source]#\n\nBases: XTick\n\nA theta-axis tick.\n\nThis subclass of XTick provides angular ticks with some small modification to their re-positioning such that ticks are rotated based on tick location. This results in ticks that are correctly perpendicular to the arc spine.\n\nWhen 'auto' rotation is enabled, labels are also rotated to be parallel to the spine. The label padding is also applied here since it's not possible to use a generic axes transform to produce tick-specific padding.\n\nbbox is the Bound2D bounding box in display coords of the Axes loc is the tick location in data coords size is the tick size in points\n\nset(*, agg_filter=<UNSET>, alpha=<UNSET>, animated=<UNSET>, clip_box=<UNSET>, clip_on=<UNSET>, clip_path=<UNSET>, gid=<UNSET>, in_layout=<UNSET>, label=<UNSET>, label1=<UNSET>, label2=<UNSET>, mouseover=<UNSET>, pad=<UNSET>, path_effects=<UNSET>, picker=<UNSET>, rasterized=<UNSET>, sketch_params=<UNSET>, snap=<UNSET>, transform=<UNSET>, url=<UNSET>, visible=<UNSET>, zorder=<UNSET>)[source]#\n\nSet multiple properties at once.\n\nSupported properties are\n\nProperty\n\nDescription\n\nagg_filter\n\na filter function, which takes a (m, n, 3) float array and a dpi value, and returns a (m, n, 3) array and two offsets from the bottom left corner of the image\n\nalpha\n\nscalar or None\n\nanimated\n\nbool\n\nclip_box\n\nBboxBase or None\n\nclip_on\n\nbool\n\nclip_path\n\nPatch or (Path, Transform) or None\n\nfigure\n\nFigure\n\ngid\n\nstr\n\nin_layout\n\nbool\n\nlabel\n\nunknown\n\nlabel1\n\nunknown\n\nlabel2\n\nunknown\n\nmouseover\n\nbool\n\npad\n\nfloat\n\npath_effects\n\nlist of AbstractPathEffect\n\npicker\n\nNone or bool or float or callable\n\nrasterized\n\nbool\n\nsketch_params\n\n(scale: float, length: float, randomness: float)\n\nsnap\n\nbool or None\n\ntransform\n\nTransform\n\nurl\n\nstr\n\nvisible\n\nbool\n\nzorder\n\nfloat\n\nupdate_position(loc)[source]#\n\nSet the location of tick in data coords with scalar loc." ]
[ null ]
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https://art.torvergata.it/handle/2108/303688
[ "A simple model for the localization of the category $\\mathbf{CLoc}_2$ of oriented and time-oriented globally hyperbolic conformal Lorentzian $2$-manifolds at all Cauchy morphisms is constructed. This provides an equivalent description of $2$-dimensional conformal algebraic quantum field theories (AQFTs) satisfying the time-slice axiom in terms of only two algebras, one for the $2$-dimensional Minkowski spacetime and one for the flat cylinder, together with a suitable action of two copies of the orientation preserving embeddings of oriented $1$-manifolds. The latter result is used to construct adjunctions between the categories of $2$-dimensional and chiral conformal AQFTs whose right adjoints formalize and generalize Rehren's chiral observables.\n\nBenini, M., Giorgetti, L., Schenkel, A. (2022). A skeletal model for 2d conformal AQFTs. COMMUNICATIONS IN MATHEMATICAL PHYSICS, 395(1), 269-298 [10.1007/s00220-022-04428-4].\n\n### A skeletal model for 2d conformal AQFTs\n\n#### Abstract\n\nA simple model for the localization of the category $\\mathbf{CLoc}_2$ of oriented and time-oriented globally hyperbolic conformal Lorentzian $2$-manifolds at all Cauchy morphisms is constructed. This provides an equivalent description of $2$-dimensional conformal algebraic quantum field theories (AQFTs) satisfying the time-slice axiom in terms of only two algebras, one for the $2$-dimensional Minkowski spacetime and one for the flat cylinder, together with a suitable action of two copies of the orientation preserving embeddings of oriented $1$-manifolds. The latter result is used to construct adjunctions between the categories of $2$-dimensional and chiral conformal AQFTs whose right adjoints formalize and generalize Rehren's chiral observables.\n##### Scheda breve Scheda completa Scheda completa (DC)\n2022\nPubblicato\nRilevanza internazionale\nArticolo\nEsperti anonimi\nSettore MAT/05 - ANALISI MATEMATICA\nEnglish\nMathematical Physics\nMathematical Physics\nHigh Energy Physics - Theory\nMathematics - Category Theory\nMathematics - Mathematical Physics\n81Txx, 53C50\nhttp://arxiv.org/abs/2111.01837v2\nBenini, M., Giorgetti, L., Schenkel, A. (2022). A skeletal model for 2d conformal AQFTs. COMMUNICATIONS IN MATHEMATICAL PHYSICS, 395(1), 269-298 [10.1007/s00220-022-04428-4].\nBenini, M; Giorgetti, L; Schenkel, A\nArticolo su rivista\nFile in questo prodotto:\nNon ci sono file associati a questo prodotto.\n\nI documenti in IRIS sono protetti da copyright e tutti i diritti sono riservati, salvo diversa indicazione.\n\nUtilizza questo identificativo per citare o creare un link a questo documento: https://hdl.handle.net/2108/303688\n##### Citazioni\n•", null, "ND\n•", null, "4\n•", null, "3" ]
[ null, "https://art.torvergata.it/sr/cineca/images/thirdparty/pmc_small.png", null, "https://art.torvergata.it/sr/cineca/images/thirdparty/scopus_small.png", null, "https://art.torvergata.it/sr/cineca/images/thirdparty/isi_small.png", null ]
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https://answers.everydaycalculation.com/percent-is/2646-210
[ "Solutions by everydaycalculation.com\n\n## 2646 is what percent of 210?\n\n2646 of 210 is 1260%\n\n#### Steps to solve \"what percent is 2646 of 210?\"\n\n1. 2646 of 210 can be written as:\n2646/210\n2. To find percentage, we need to find an equivalent fraction with denominator 100. Multiply both numerator & denominator by 100\n\n2646/210 × 100/100\n3. = (2646 × 100/210) × 1/100 = 1260/100\n4. Therefore, the answer is 1260%\n\nIf you are using a calculator, simply enter 2646÷210×100 which will give you 1260 as the answer.\n\nMathStep (Works offline)", null, "Download our mobile app and learn how to work with percentages in your own time:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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http://curious.astro.cornell.edu/the-universe/extrasolar-planets/57-our-solar-system/planets-and-dwarf-planets/orbits/244-why-do-planets-have-elliptical-orbits-beginner
[ "## Why do planets have elliptical orbits? (Beginner)\n\nWhy do planets have elliptical orbits? And why do some satellites, when launched in lower orbits, go around Earth in elliptical orbits?\n\nAt first glance it may seem odd that a force such as gravity, which pulls the planets straight in toward the center of mass, should result in elliptical orbits! But in fact it is quite straightforward to understand why this should be so.\n\nIt is certainly possible to set up a satellite so that it has a circular orbit (a circle is just an ellipse whose foci coincide). Gravity can only pull in the direction toward the planet. The inertia of the satellite makes it want to travel in a straight line, but if it does so, its velocity is no longer perfectly perpendicular to the pull of gravity, so gravity pulls it in; this will remove part of the velocity, but as the satellite is also falling inward, it gets a new component of velocity due to the acceleration of gravity. In a circular orbit, we know that the ground speed is constant, so these two effects must perfectly cancel one another out to leave the speed of the satellite unchanged. Now imagine that we fire the satellite's boosters so that its ground speed increases. Now the desire of the satellite to go straight is stronger, so the two effects do not cancel perfectly, and the ground speed will vary. You can see how this corresponds to an elliptical orbit, and how a planet orbiting the Sun behaves in the same way. (Of course, planets have no boosters, but think about what effect the initial velocity of the planet due to the process of its formation would have--what happens if a planet is formed with only a small initial velocity, far from the Sun, or if it is formed with a large velocity, very near to the Sun? What happens if the inital velocity of the planet is zero?).\n\n#### Sara Slater\n\nSara is a former Cornell undergraduate and now a physics graduate student at Harvard University, where she works on cosmology and particle physics.\n\n## Our Reddit AMAs\n\nAMA = Ask Me (Us) Anything" ]
[ null ]
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https://www.geeksforgeeks.org/smallest-number-whose-set-bits-maximum-given-range/?ref=rp
[ "# Smallest number whose set bits are maximum in a given range\n\nGiven a positive integer ‘l‘ and ‘r‘. Find the smallest number ‘n‘ such that l <= n <= r and count of number of set bits(number of ‘1’s in binary representation) is maximum as possible.\nExamples :\n\n```Input: 1 4\nOutput: 3\nExplanation:\nBinary representation from '1' to '4':\n110 = 0012\n210 = 0102\n310 = 0112\n110 = 1002\nThus number '3' has maximum set bits = 2\n\nInput: 1 10\nOutput: 7\n```\n\n## Recommended: Please try your approach on {IDE} first, before moving on to the solution.\n\nSimple approach is to traverse from ‘l’ to ‘r’ and count the set bits for each ‘x'(l <= n <= r) and print the number whose count is maximum among them. Time complexity of this approach is O(n*log(r)).\n\n## C++\n\n `// C++ program to find number whose set ` `// bits are maximum among 'l' and 'r' ` `#include ` `using` `namespace` `std; ` ` `  `// Returns smallest number whose set bits ` `// are maximum in given range. ` `int` `countMaxSetBits(``int` `left, ``int` `right) ` `{ ` `    ``// Initialize the maximum count and ` `    ``// final answer as 'num' ` `    ``int` `max_count = -1, num; ` `    ``for` `(``int` `i = left; i <= right; ++i) { ` `        ``int` `temp = i, cnt = 0; ` ` `  `        ``// Traverse for every bit of 'i' ` `        ``// number ` `        ``while` `(temp) { ` `            ``if` `(temp & 1) ` `                ``++cnt; ` `            ``temp >>= 1; ` `        ``} ` ` `  `        ``// If count is greater than previous ` `        ``// calculated max_count, update it ` `        ``if` `(cnt > max_count) { ` `            ``max_count = cnt; ` `            ``num = i; ` `        ``} ` `    ``} ` `    ``return` `num; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 1, r = 5; ` `    ``cout << countMaxSetBits(l, r) << ``\"\\n\"``; ` ` `  `    ``l = 1, r = 10; ` `    ``cout << countMaxSetBits(l, r); ` `    ``return` `0; ` `} `\n\n## Java\n\n `// Java program to find number whose set ` `// bits are maximum among 'l' and 'r' ` `class` `gfg ` `{ ` `     `  ` `  `// Returns smallest number whose set bits ` `// are maximum in given range. ` `static` `int` `countMaxSetBits(``int` `left, ``int` `right) ` `{ ` `    ``// Initialize the maximum count and ` `    ``// final answer as 'num' ` `    ``int` `max_count = -``1``, num = ``0``; ` `    ``for` `(``int` `i = left; i <= right; ++i)  ` `    ``{ ` `        ``int` `temp = i, cnt = ``0``; ` ` `  `        ``// Traverse for every bit of 'i' ` `        ``// number ` `        ``while` `(temp > ``0``) ` `        ``{ ` `            ``if` `(temp % ``2` `== ``1``) ` `                ``++cnt; ` `            ``temp >>= ``1``; ` `        ``} ` ` `  `        ``// If count is greater than previous ` `        ``// calculated max_count, update it ` `        ``if` `(cnt > max_count) ` `        ``{ ` `            ``max_count = cnt; ` `            ``num = i; ` `        ``} ` `    ``} ` `    ``return` `num; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `l = ``1``, r = ``5``; ` `    ``System.out.println(countMaxSetBits(l, r)); ` ` `  `    ``l = ``1``; r = ``10``; ` `    ``System.out.print(countMaxSetBits(l, r)); ` `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `\n\n## Python 3\n\n `# Python code to find number whose set ` `# bits are maximum among 'l' and 'r' ` ` `  `def` `countMaxSetBits( left, right): ` `    ``max_count ``=` `-``1` `    ``for` `i ``in` `range``(left, right``+``1``): ` `        ``temp ``=` `i ` `        ``cnt ``=` `0` ` `  `        ``# Traverse for every bit of 'i' ` `        ``# number ` `        ``while` `temp: ` `            ``if` `temp & ``1``: ` `                ``cnt ``+``=``1` `            ``temp ``=` `temp >> ``1` `         `  `        ``# If count is greater than previous ` `        ``# calculated max_count, update it ` `        ``if` `cnt > max_count: ` `            ``max_count ``=` `cnt ` `            ``num``=``i ` `    ``return` `num ` ` `  `# driver code  ` `l ``=` `1` `r ``=` `5` `print``(countMaxSetBits(l, r)) ` `l ``=` `1` `r ``=` `10` `print``(countMaxSetBits(l, r))  ` ` `  `# This code is contributed by \"Abhishek Sharma 44\" `\n\n## C#\n\n `// C# program to find number whose set ` `// bits are maximum among 'l' and 'r' ` `using` `System; ` `     `  `class` `gfg ` `{ ` `     `  `// Returns smallest number whose set bits ` `// are maximum in given range. ` `static` `int` `countMaxSetBits(``int` `left, ``int` `right) ` `{ ` `    ``// Initialize the maximum count and ` `    ``// final answer as 'num' ` `    ``int` `max_count = -1, num = 0; ` `    ``for` `(``int` `i = left; i <= right; ++i)  ` `    ``{ ` `        ``int` `temp = i, cnt = 0; ` ` `  `        ``// Traverse for every bit of 'i' ` `        ``// number ` `        ``while` `(temp > 0) ` `        ``{ ` `            ``if` `(temp % 2 == 1) ` `                ``++cnt; ` `            ``temp >>= 1; ` `        ``} ` ` `  `        ``// If count is greater than previous ` `        ``// calculated max_count, update it ` `        ``if` `(cnt > max_count) ` `        ``{ ` `            ``max_count = cnt; ` `            ``num = i; ` `        ``} ` `    ``} ` `    ``return` `num; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `l = 1, r = 5; ` `    ``Console.WriteLine(countMaxSetBits(l, r)); ` ` `  `    ``l = 1; r = 10; ` `    ``Console.Write(countMaxSetBits(l, r)); ` `} ` `} ` ` `  `/* This code contributed by PrinciRaj1992 */`\n\n## PHP\n\n `>= 1; ` `        ``} ` ` `  `        ``// If count is greater than ` `        ``// previous calculated  ` `        ``// max_count, update it ` `        ``if` `(``\\$cnt` `> ``\\$max_count``)  ` `        ``{ ` `            ``\\$max_count` `= ``\\$cnt``; ` `            ``\\$num` `= ``\\$i``; ` `        ``} ` `    ``} ` `    ``return` `\\$num``; ` `} ` ` `  `// Driver code ` `\\$l` `= 1; ``\\$r` `= 5; ` `echo` `countMaxSetBits(``\\$l``, ``\\$r``), ``\"\\n\"``; ` ` `  `\\$l` `= 1; ``\\$r` `= 10; ` `echo` `countMaxSetBits(``\\$l``, ``\\$r``); ` `     `  `// This code is contributed by m_kit ` `?> `\n\nOutput :\n\n```3\n7\n```\n\nEfficient approach is to use bit-manipulation. Instead of iterating for every number from ‘l’ to ‘r’, iterate only after updating the desired number(‘num’) i.e., take the bitwise ‘OR’ of number with the consecutive number. For instance,\n\n```Let l = 2, and r = 10\n1. num = 2\n2. x = num OR (num + 1)\n= 2 | 3 = 010 | 011 = 011\nnum = 3(011)\n3. x = 3 | 4 = 011 | 100 = 111\nnum = 7(111)\n4. x = 7 | 8 = 0111 | 1000 = 1111\nSince 15(11112) is greater than\n10, thus stop traversing for next number.\n```\n\n## C++\n\n `// C++ program to find number whose set ` `// bits are maximum among 'l' and 'r' ` `#include ` `using` `namespace` `std; ` ` `  `// Returns smallest number whose set bits ` `// are maximum in given range. ` `int` `countMaxSetBits(``int` `left, ``int` `right) ` `{ ` `    ``while` `((left | (left + 1)) <= right) ` `        ``left |= left + 1; ` ` `  `    ``return` `left; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `l = 1, r = 5; ` `    ``cout << countMaxSetBits(l, r) << ``\"\\n\"``; ` ` `  `    ``l = 1, r = 10; ` `    ``cout << countMaxSetBits(l, r) ; ` `    ``return` `0; ` `} `\n\n## Java\n\n `// Java program to find number  ` `// whose set bits are maximum  ` `// among 'l' and 'r' ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// Returns smallest number  ` `    ``// whose set bits are  ` `    ``// maximum in given range. ` `    ``static` `int` `countMaxSetBits(``int` `left,  ` `                               ``int` `right) ` `    ``{ ` `    ``while` `((left | (left + ``1``)) <= right) ` `        ``left |= left + ``1``; ` ` `  `    ``return` `left; ` `    ``} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `l = ``1``; ` `    ``int` `r = ``5``; ` `    ``System.out.println(countMaxSetBits(l, r)); ` ` `  `    ``l = ``1``; ` `    ``r = ``10``; ` `    ``System.out.println(countMaxSetBits(l, r)); ` `} ` `} ` ` `  `// This code is contributed by @ajit `\n\n## Python3\n\n `# Python code to find number whose set ` `# bits are maximum among 'l' and 'r' ` ` `  `def` `countMaxSetBits( left, right): ` `     `  `    ``while``(left | (left``+``1``)) <``=` `right: ` `        ``left |``=` `left``+``1` `    ``return` `left ` `     `  `# driver code ` `l ``=` `1` `r ``=` `5` `print``(countMaxSetBits(l, r)) ` `l ``=` `1` `r ``=` `10` `print``(countMaxSetBits(l, r))  ` ` `  `# This code is contributed by \"Abhishek Sharma 44\" `\n\n## C#\n\n `// C# program to find number  ` `// whose set bits are maximum  ` `// among 'l' and 'r' ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `    ``// Returns smallest number  ` `    ``// whose set bits are  ` `    ``// maximum in given range. ` `    ``static` `int` `countMaxSetBits(``int` `left,  ` `                               ``int` `right) ` `    ``{ ` `    ``while` `((left | (left + 1)) <= right) ` `        ``left |= left + 1; ` ` `  `    ``return` `left; ` `    ``} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `    ``int` `l = 1; ` `    ``int` `r = 5; ` `    ``Console.WriteLine(countMaxSetBits(l, r)); ` `     `  `    ``l = 1; ` `    ``r = 10; ` `    ``Console.WriteLine(countMaxSetBits(l, r)); ` `} ` `} ` ` `  `// This code is contributed by @ajit `\n\n## PHP\n\n ` `\n\nOutput :\n\n```3\n7\n```\n\nTime complexity: O(log(n))\nAuxiliary space: O(1)\n\nAttention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.\n\nMy Personal Notes arrow_drop_up", null, "Check out this Author's contributed articles.\n\nIf you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to [email protected]. See your article appearing on the GeeksforGeeks main page and help other Geeks.\n\nPlease Improve this article if you find anything incorrect by clicking on the \"Improve Article\" button below.\n\nArticle Tags :\nPractice Tags :\n\n1\n\nPlease write to us at [email protected] to report any issue with the above content." ]
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http://www.sciencepublishinggroup.com/journal/paperinfo?journalid=141&doi=10.11648/j.pamj.s.2017060301.12
[ "Archive\nSpecial Issues\nAn Introduction to Differential Geometry: The Theory of Surfaces\nPure and Applied Mathematics Journal\nVolume 6, Issue 3-1, June 2017, Pages: 6-11\nReceived: Feb. 6, 2017; Accepted: Feb. 14, 2017; Published: May 13, 2017\nAuthors\nKande Dickson Kinyua, Department of Mathematics, Moi University, Eldoret, Kenya; Department of Mathematics, Karatina University, Karatina, Kenya\nKuria Joseph Gikonyo, Department of Mathematics, Karatina University, Karatina, Kenya\nArticle Tools", null, "Abstract", null, "PDF (218KB)\nAbstract\nFrom a mathematical perspective, a surface is a generalization of a plane which does not necessarily require being flat, that is, the curvature is not necessarily zero. Often, a surface is defined by equations that are satisfied by some coordinates of its points. A surface may also be defined as the image, in some space of dimensions at least three, of a continuous function of two variables (some further conditions are required to insure that the image is not a curve). In this case, one says that one has a parametric surface, which is parametrized by these two variables, called parameters. Parametric equations of surfaces are often irregular at some points. This is formalized by the concept of manifold: in the context of manifolds, typically in topology and differential geometry, a surface is a manifold of dimension two; this means that a surface is a topological space such that every point has a neighborhood which is homeomorphic to an open subset of the Euclidean plane. A parametric surface is the image of an open subset of the Euclidean plane by a continuous function, in a topological space, generally a Euclidean space of dimension at least three. The paper aims at giving an introduction to the theory of surfaces from differential geometry perspective.\nKeywords\nCurvature, Differential Geometry, Geodesics, Manifolds, Parametrized, Surface\nKande Dickson Kinyua, Kuria Joseph Gikonyo, An Introduction to Differential Geometry: The Theory of Surfaces, Pure and Applied Mathematics Journal. Special Issue: Advanced Mathematics and Geometry. Vol. 6, No. 3-1, 2017, pp. 6-11. doi: 10.11648/j.pamj.s.2017060301.12\nReferences\n\nK. D. Kinyua. An Introduction to Differentiable Manifolds, Mathematics Letters. Vol. 2, No. 5, 2016, pp. 32-35.\n\nLang, Serge, Introduction to Differentiable Manifolds, 2nd ed. Springer-Verlag New York. ISBN 0-387-95477-5, 2002.\n\nM. Deserno, Notes on Difierential Geometry with special emphasis on surfaces in R3, Los Angeles, USA, 2004.\n\nM. P. do-Carmo, Differential Geometry of curves and surfaces, Prentice-Hall, Inc., Englewood Cliffs, New Zealand, USA, 1976.\n\nM. Raussen, Elementary Differential Geometry: Curves and Surfaces, Aalborg University, Denmark, 2008.\n\nM. Spivak, A Comprehensive Introduction to Differential Geometry, Vol. 1, Third Edition, Publish or Perish Inc., Houston, USA, 1999.\n\nR. Palais, A Modern Course on Curves and Surfaces, 2003.\n\nT. Shifrin, Differential Geometry: A First Course in Curves and Surfaces, Preliminary Version, University of Georgia, 2016.\n\nV. G. Ivancevic and T. T. Ivancevic Applied Differential Geometry: A Modern Introduction, World Scientific Publishing Co. Pte. Ltd., Toh Tuck Link, Singapore, 2007.\n\nW. Zhang, Geometry of Curves and Surfaces, Mathematics Institute, University of Warwick, 2014.\nPUBLICATION SERVICES", null, "" ]
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https://stats.stackexchange.com/questions/57887/bayesian-analysis-with-histogram-prior-why-draw-simulations-from-the-posterior
[ "# Bayesian analysis with histogram prior. Why draw simulations from the posterior?\n\nThis is a beginner’s question on an exercise in Jim Albert’s “Bayesian Computation with R”. Note that while this might be homework, in my case it is not, as I am learning Bayesian methods in R because I think I might use it in my future analyses.\n\nAnyways, while this is a specific question, it probably involves basic understanding of Bayesian methods.\n\nSo, in exercise 2.2, Jim Albert asks us to analyse the experiment of a penny throw. See here. We are to use a histogram prior, that is, divide the space of possible p values in 10 intervals of length .1 and assign a prior probability to these.\n\nSince I know that the true probability will be .5, and I think it is highly unlikely that the universe has changed laws of probability or the penny is rugged, my priors are:\n\nprior <- c(1,5,20,100,5000,5000,100,20,5,1)\nprior <- prior/sum(prior)\n\n\nalong the interval midpoints\n\nmidpt <- seq(0.05, 0.95, by=0.1)\n\n\nSo far so good. Next, we spin the penny 20 times and record the number of successes (heads) and failures (tail). Easily done:\n\ny <- rbinom(n=20,p=.5,size=1)\ns <- sum(y==1)\nf <- sum(y==0)\n\n\nIn my expeniment, s == 7 and f == 13. Next comes the part that I do not understand:\n\nSimulate from the posterior distribution by (1) computing the posterior density of p on a grid of values on (0,1) and (2) taking a simulated sample with replacement from the grid. (The function histprior and sample are helpful in this computation). How have the interval probabilities changed on the basis of your data?\n\nThis is how it is done:\n\np <- seq(0,1, length=500)\npost <- histprior(p,midpt,prior) * dbeta(p,s+1,f+1)\npost <- post/sum(post)\n\nps <- sample(p, replace=TRUE, prob = post)\n\n\nBut why do we do that?\n\nWe can easily obtain the posterior density by multiplying the prior with the appropriate likelihood, as done in line two of the block above. This is a plot of the posterior distribution:", null, "As the posterior distribution is ordered, we can obtain results for the intervals introduced in the histogram prior by summarizing over elements of the posterior density:\n\npost.vector <- vector()\npost.vector <- sum(post[p < 0.1])\npost.vector <- sum(post[p > 0.1 & p <= 0.2])\npost.vector <- sum(post[p > 0.2 & p <= 0.3])\npost.vector <- sum(post[p > 0.3 & p <= 0.4])\npost.vector <- sum(post[p > 0.4 & p <= 0.5])\npost.vector <- sum(post[p > 0.5 & p <= 0.6])\npost.vector <- sum(post[p > 0.6 & p <= 0.7])\npost.vector <- sum(post[p > 0.7 & p <= 0.8])\npost.vector <- sum(post[p > 0.8 & p <= 0.9])\npost.vector <- sum(post[p > 0.9 & p <= 1])\n\n\n(R experts might find a better way to create that vector. I guess it might have something to do with sweep?)\n\nround(cbind(midpt,prior,post.vector),3)\n\nmidpt prior post.vector\n[1,] 0.05 0.000 0.000\n[2,] 0.15 0.000 0.000\n[3,] 0.25 0.002 0.003\n[4,] 0.35 0.010 0.022\n[5,] 0.45 0.488 0.737\n[6,] 0.55 0.488 0.238\n[7,] 0.65 0.010 0.001\n[8,] 0.75 0.002 0.000\n[9,] 0.85 0.000 0.000\n[10,] 0.95 0.000 0.000\n\n\nFurthermore, we have 500 draws from the posterior distribution that tell us nothing different. Here is a plot of the density of the simulated draws:", null, "Now we use the simulated data to obtain probabilities for our intervals by counting what proportion of simulations are within the interval:\n\nsim.vector <- vector()\nsim.vector <- length(ps[ps < 0.1])/length(ps)\nsim.vector <- length(ps[ps > 0.1 & ps <= 0.2])/length(ps)\nsim.vector <- length(ps[ps > 0.2 & ps <= 0.3])/length(ps)\nsim.vector <- length(ps[ps > 0.3 & ps <= 0.4])/length(ps)\nsim.vector <- length(ps[ps > 0.4 & ps <= 0.5])/length(ps)\nsim.vector <- length(ps[ps > 0.5 & ps <= 0.6])/length(ps)\nsim.vector <- length(ps[ps > 0.6 & ps <= 0.7])/length(ps)\nsim.vector <- length(ps[ps > 0.7 & ps <= 0.8])/length(ps)\nsim.vector <- length(ps[ps > 0.8 & ps <= 0.9])/length(ps)\nsim.vector <- length(ps[ps > 0.9 & ps <= 1])/length(ps)\n\n\n(Again: Is there a more efficient way to do this?)\n\nSummarize results:\n\nround(cbind(midpt,prior,post.vector,sim.vector),3)\n\nmidpt prior post.vector sim.vector\n[1,] 0.05 0.000 0.000 0.000\n[2,] 0.15 0.000 0.000 0.000\n[3,] 0.25 0.002 0.003 0.000\n[4,] 0.35 0.010 0.022 0.026\n[5,] 0.45 0.488 0.737 0.738\n[6,] 0.55 0.488 0.238 0.236\n[7,] 0.65 0.010 0.001 0.000\n[8,] 0.75 0.002 0.000 0.000\n[9,] 0.85 0.000 0.000 0.000\n[10,] 0.95 0.000 0.000 0.000\n\n\nIt comes to no surprise that the simultion produces no other results than the posterior, on which it was based. Thus, why did we draw those simulations in the first place?\n\n• I'm not totally sure, since I'm also a Bayes newbie. But my guess would be that simulations of posterior densities are introduced early in Bayesian texts so that more advanced techniques like MCMC are more intuitive. Just a guess, though. – Sycorax May 2 '13 at 12:02\n• Bayesian specialist here. DJE's guess is 100% correct. – Cyan May 2 '13 at 15:58\n• Alright. So if I assume, later on simulations well be used instead of the posterior distributions. But the simulations can only be drawn if the posterior distribution is known, as seen in ps <- sample(p, replace=TRUE, prob = post)! Am I right in assuming that this will change for more advanced simulation techniques? – mzuba May 3 '13 at 8:17\n\npost.vector <- vector()\npost.vector <- sum(post[p < 0.1])\npost.vector <- sum(post[p > 0.1 & p <= 0.2])\npost.vector <- sum(post[p > 0.2 & p <= 0.3])\npost.vector <- sum(post[p > 0.3 & p <= 0.4])\npost.vector <- sum(post[p > 0.4 & p <= 0.5])\npost.vector <- sum(post[p > 0.5 & p <= 0.6])\npost.vector <- sum(post[p > 0.6 & p <= 0.7])\npost.vector <- sum(post[p > 0.7 & p <= 0.8])\npost.vector <- sum(post[p > 0.8 & p <= 0.9])\npost.vector <- sum(post[p > 0.9 & p <= 1])\n\n\nThe easiest way to do it using base R is:\n\ngroup <- cut(p, breaks=seq(0,1,0.1), include.lowest = T)\npost.vector.alt <- aggregate(post, FUN=sum, by=list(group))\n\n\nNote that the breaks go from 0 to 1. This yields:\n\n Group.1 x\n1 [0,0.1] 3.030528e-13\n2 (0.1,0.2] 1.251849e-08\n3 (0.2,0.3] 6.385088e-06\n4 (0.3,0.4] 6.732672e-04\n5 (0.4,0.5] 2.376448e-01\n6 (0.5,0.6] 7.372805e-01\n7 (0.6,0.7] 2.158296e-02\n8 (0.7,0.8] 2.691182e-03\n9 (0.8,0.9] 1.205200e-04\n10 (0.9,1] 3.345072e-07\n\n\nAnd we have:\n\n> all.equal (post.vector.alt\\$x, post.vector)\n TRUE\n\n\nMy understanding is that since the posterior density obtained from the product of prior density and likelihood is just an APPROXIMATE of the posterior density, thus we cannot make any EXACT inference from it directly.\n\nConsequently, we need to take a random sample from it, and conduct inference from the sample, just like the simulation method for the posterior from beta family.\n\n• The posterior density obtained from the product of the prior and likelihood IS the posterior density, not an approximation thereto, except insofar as the prior and likelihood functions are themselves approximations - an issue which simulation from the posterior won't fix. – jbowman Sep 23 '18 at 17:15" ]
[ null, "https://i.stack.imgur.com/Kmz3V.jpg", null, "https://i.stack.imgur.com/tGYqr.jpg", null ]
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http://www.toontricks.com/2019/02/tutorial-generalized-length-check-on.html
[ "# Tutorial :generalized .length check on functions", null, "### Question:\n\nWhen i write a function in JS or jQuery and a parameter is required, i use the `if something.length` trick...\n\n``this.example = function(e) { if (\\$(e).length) { /*blablabla*/ } else { return false; } } ``\n\nBut i dont want to do this in all functions. Is there a way to generalize this?\n\nlike:\n\n``this.ck = function(e) { return function(e){ if (!(e.length)) { return false;} } } this.example = function(e) { ck(e) /*blablabla*/ } ``\n\n### Solution:1\n\nMaybe this, but see below:\n\n``function ifNonEmpty(f) { return function(e) { if (!\\$(e).length) return false; return f(e); }; } ``\n\nYou'd use that like this:\n\n``var myCoolFunction = ifNonEmpty(function myCoolFunction(e) { // your implementation }; ``\n\nI would however suggest that instead of writing functions that take jQuery objects as parameters, you write those functions as your very own jQuery plugins.\n\n### Solution:2\n\nI would suggest the following.\n\nThis example would first test whether the parameter actually contains anything (i.e. is not null), then your .length check could be on whether it\n\n``function isNonEmpty(p){ if ( (p != null) && (p.length > 0) ){ //test } } ``\n\nI would be wary of using .length, anyway. For example, if you pass a string `.length` will have a value!\n\nTo test if something is a jQuery object:\n\n``alert(p instanceof jQuery); //will alert \"true\" if p is a jQuery object. ``\n\nSo, this gives us:\n\n``function isNonEmpty(p){ if ( (p != null) && (p instanceof jQuery) && (p.length > 0) ){ //test } } ``\n\nOr something of that ilk. Either way, there should be sufficient code there to tailor to your intent.\n\nNote:If u also have question or solution just comment us below or mail us on [email protected]\nPrevious\nNext Post »" ]
[ null, "https://4.bp.blogspot.com/-te3nq62xJRQ/WIzbeIW_hiI/AAAAAAAAAaU/T9jgUxascMELRPlYee11mkSEA3KFhImRQCLcB/s1600/lucky%2Brathore%2B%2Buser%2Basked%2BQuestion_logo.jpg", null ]
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https://www.bbc.co.uk/bitesize/topics/z36tyrd/resources/1
[ "Calculating size - scaling up and down (signed)\n\nA model maker describes how he changes the size of a model by scaling it up or down.\n\nDividing by five with remainders (signed)\n\nA demonstration of how a variety of numbers can be divided by five, leaving remainders.\n\nDividing by ten (signed)\n\nAn illustration of dividing by ten is given by dancing numbers in the gym.\n\nDivision with remainders (signed)\n\nAn introduction to division with remainders, illustrated with a real life problem.\n\nHow many chairs are needed?\n\nMultiplication is used to calculate chairs needed to seat 30 guests at six tables.\n\nHow many chairs are needed? (signed)\n\nMultiplication is used to calculate chairs needed to seat 30 guests at six tables.\n\nMultiplication with square turtles\n\nHow a multiplication fact can be written in a number of different ways.\n\nMultiplication with square turtles (signed)\n\nHow a multiplication fact can be written in a number of different ways.\n\nProblem solving using division and remainders\n\nThe Chuckle Brothers work out how many tables are needed at a wedding reception.\n\nSolving division problems (signed)\n\nTwo simple division problems are solved using animated turtles.\n\n5 learner guides\n\nWe have a selection of learner guides" ]
[ null ]
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https://mathradical.com/calculator-with-radical/linear-inequalities/use-for-hyperbola-chart.html
[ "Algebra Tutorials!", null, "Home", null, "Exponents and Radicals", null, "Division of Radicals", null, "Exponents and Radicals", null, "RADICALS & RATIONAL EXPONENTS", null, "Radicals and Rational Exponents", null, "Radical Equations", null, "Solving Radical Equations", null, "Roots and Radicals", null, "RADICAL EQUATION", null, "Simplifying Radical Expressions", null, "Radical Expressions", null, "Solving Radical Equations", null, "Solving Radical Equations", null, "Exponents and Radicals", null, "Exponents and Radicals", null, "Roots;Rational Exponents;Radical Equations", null, "Solving and graphing radical equations", null, "Solving Radical Equations", null, "Radicals and Rational Exponents", null, "exponential_and_radical_properties", null, "Roots, Radicals, and Root Functions", null, "Multiplication of Radicals", null, "Solving Radical Equations", null, "Radical Expressions and Equations", null, "SOLVING RADICAL EQUATIONS", null, "Equations Containing Radicals and Complex Numbers", null, "Square Roots and Radicals", null, "Solving Radical Equations in One Variable Algebraically", null, "Polynomials and Radicals", null, "Roots,Radicals,and Fractional Exponents", null, "Adding, Subtracting, and Multiplying Radical Expressions", null, "Square Formula and Powers with Radicals", null, "Simplifying Radicals", null, "Exponents and Radicals Practice", null, "Solving Radical Equations", null, "Solving Radical Equations", null, "Solving Radical Equations", null, "Lecture-Radical Expressions", null, "Radical Functions\nTry the Free Math Solver or Scroll down to Tutorials!\n\n Depdendent Variable\n\n Number of equations to solve: 23456789\n Equ. #1:\n Equ. #2:\n\n Equ. #3:\n\n Equ. #4:\n\n Equ. #5:\n\n Equ. #6:\n\n Equ. #7:\n\n Equ. #8:\n\n Equ. #9:\n\n Solve for:\n\n Dependent Variable\n\n Number of inequalities to solve: 23456789\n Ineq. #1:\n Ineq. #2:\n\n Ineq. #3:\n\n Ineq. #4:\n\n Ineq. #5:\n\n Ineq. #6:\n\n Ineq. #7:\n\n Ineq. #8:\n\n Ineq. #9:\n\n Solve for:\n\n Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:\n\nuse for hyperbola chart\nRelated topics:\nhow to solve rational inequalities ;ppt | exponent complex quadratic form | factoring 3 exponent polynomials | merrill algebra 1 | graphing non linear inequalities | find least common denominator calculator for rational equations | how to write root numbers as fractions | printable division sheets for 8 year olds | conceptual physics prentice hall | math algebra( simultaneous equation) help\n\nAuthor Message\nfearness_fled", null, "Registered: 02.09.2004\nFrom: Europe", null, "Posted: Sunday 31st of Dec 09:36 Hi folks I would really value some help with use for hyperbola chart on which I’m really stuck. I have this homework due and don’t know where to solve logarithms, absolute values and complex fractions . I would sure appreciate your assistance rather than hiring a math tutor who are very costly .\nkfir", null, "Registered: 07.05.2006\nFrom: egypt", null, "Posted: Sunday 31st of Dec 17:15 I have been in your place some time agowhen I was learning use for hyperbola chart. What part of matrices and roots poses more problems ? Because I think that what you really need is a good software to help you figure out the basic concepts and ways of solving the exercises. Did you ever use a software like that? I have tried a few but I have to say that Algebrator is the best and the easiest to use. It's not like those other programs because it teaches you how to solve, it doesn't just give you the solutions.\nthicxolmed01", null, "Registered: 16.05.2004\nFrom: Welly, NZ", null, "Posted: Monday 01st of Jan 20:03 I agree. Algebrator not only gets your homework done faster, it actually improves your understanding of the subject by providing useful information on how to solve similar problems . It is a very popular software among students so you should try it out.\ndaujk_vv7", null, "Registered: 06.07.2001\nFrom: I dunno, I've lost it.", null, "Posted: Tuesday 02nd of Jan 18:07 Algebrator is a user friendly product and is definitely worth a try. You will also find many interesting stuff there. I use it as reference software for my math problems and can say that it has made learning math much more fun ." ]
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https://www.thephysicsforum.com/homework-help/6903-help-uncertainty-principle-virtual-particles-problem-part-b-hints.html
[ "# Thread: !HELP! The Uncertainty Principle: Virtual Particles PROBLEM (PART B + HINTS)\n\n1. The uncertainty principle can be expressed as a relation between the uncertainty ΔE in the energy state of a system and the time interval Δt during which the system remains in that state. In symbols,\n\nΔEΔt≥ℏ/2,\n\nwhere ℏ=h/2π, h is Planck's constant.\n\nThe energy-time uncertainty principle says that the longer a system remains in the same energy state, the higher the accuracy (or the smaller the uncertainty) a measurement of that energy can be. Another implication is that physical processes can violate the law of energy conservation as long as the violation occurs for only a short time, determined by the uncertainty principle. This idea is at the base of the theory of virtual particles.\n\n---\n\nPART A\n\nConsider two electrons that interact with each other. Classically, their interaction would be described in terms of the electrostatic force. In quantum mechanics, their interaction is interpreted in terms of emission and absorption of photons: One of the two electrons emits a photon with energy ΔE, which is then absorbed by the other electron after a short period of time.\n\nHow long can the photon survive before it is absorbed without violating the uncertainty principle?\n\nℏ/2ΔE\n\n----\n\nPART B - MAIN PROBLEM - HELP!!!!!\n\nThe same concepts used to describe the emission and absorption of virtual photons in electromagnetic interactions can be used to describe nuclear interactions. According to this theory, when two protons of a nucleus interact with each other, a virtual particle is created. In order to effectively mediate the nuclear force, the virtual particle must exist long enough to travel a distance comparable to the size of the nucleus: a distance on the order of 1.5×10−15m.\n\nAssuming that the virtual particle moves at a speed close to c, what is its minimum mass?\n\nTo simplify the calculation, assume that the kinetic energy of the particle is negligible compared to its rest energy. This assumption is not really valid for particles moving at close to the speed of light, but it will lead to an answer of a reasonable order of magnitude.\n\n---\n\nHint 1.\n\nFind the time interval during which the virtual particle survives\nIf the particle has to travel at least 1.5⋅10^−15m, how long does it survive?\n\nΔt = ??\n\n---\n\nHint 2.\n\nFind the minimum necessary uncertainty in energy.\n\nWhat is the minimum uncertainty in energy that is necessary to create such a particle?\n\n(ΔE)min = ??\n\n---\n\nΔm = ?? kg\n\n----\n\nI'm really unsure how to even solve Part B even not knowing how to solve the two hints listed above.\n\nHow do I go about this, knowing the formula for delta time in the first Part, and also that the\n\ndistance travelled by the particle has to be at least 1.5⋅10^−15m during its survival, so what do I do to solve the two hints (uncertainty in time, uncertainty in E(min)) then go for solving the uncertainty in mass for Part B?\n\nThanks!", null, "", null, "2.", null, "Originally Posted by QuantumDynamite99", null, "---\n\nHint 1.\n\nFind the time interval during which the virtual particle survives\nIf the particle has to travel at least 1.5⋅10^−15m, how long does it survive?\n\nΔt = ??\n\nThe particle travels at", null, "", null, "", null, "Posting Permissions\n You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Forum Rules" ]
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https://www.wikicalculator.com/fdbcalculatorlist.php?lang=en&sub=4&cat=401
[ "1 debt service coverage ratio,DSCR,debt coverage ratio,DCR", null, "2 Relative Strength Index,RSI,Wilder's RSI", null, "3 Cutler's RSI", null, "4 relative strength,relative strength factor,RS", null, "5 Equated Monthly Installment,EMI,Amortization of loans", null, "6 Equated Monthly Installment,EMI,Amortization of loans", null, "7 annualized return on tangible equity,annualized ROTE,annualized rate of return on the tangible common equity,annualized return on average tangible common shareholders' equity", null, "8 Return on tangible equity,ROTE,rate of return on the tangible common equity,return on average tangible common shareholders' equity", null, "9 tangible common shareholders' equity", null, "10 net interest spread,net interest rate spread", null, "11 operating ratio", null, "12 envy ratio,ER", null, "13 retention rate,RR,earnings retention ratio,plowback ratio", null, "14 gross margin,GM,gross profit margin,GPM,gross profit rate,GPR", null, "15 gross margin,GM,gross profit margin,GPM,gross profit rate,GPR", null, "16 operating margin,OM,operating income margin,OIM,operating profit margin,OPM,return on sales,ROS", null, "17 net margin,net profit margin,profit margin,net profit ratio,PM,NM,NPM,NPR", null, "18 return on equity,ROE", null, "19 ROE DuPont,DuPont formula,strategic profit model,return on equity", null, "20 ROE DuPont,DuPont formula,strategic profit model,return on equity", null, "21 ROA DuPont,return on assets DuPont,return on assets ratio", null, "22 return on assets,ROA", null, "23 net working capital,NWC", null, "24 return on net assets,RONA", null, "25 return on net assets,RONA", null, "previous1234567next" ]
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https://pbil.univ-lyon1.fr/CRAN/web/packages/kdensity/vignettes/tutorial.html
[ "# Basic usage\n\nkdensity is called using a syntax similar to stats::density, but with some additional arguments. A call to kdensity returns a density function (with class kdensity), which can be used as ordinary R-functions.\n\nlibrary(\"kdensity\")\nkde = kdensity(mtcars$mpg, start = \"gumbel\", kernel = \"gaussian\") kde(20) #> 0.06829002 Hence it can be used to calculate functionals, such as the expectation. integrate(function(x) x*kde(x), lower = -Inf, upper = Inf)$value\n#> 20.15896\n\nPlotting works as with stats::density. If kdensity was called with a parametric start, use plot_start = TRUE inside a plotting function in order to plot the estimated parametric start density instead of the kernel density.\n\nplot(kde, main = \"Miles per Gallon\")\nlines(kde, plot_start = TRUE, col = \"red\")\nrug(mtcars$mpg)", null, "If the R-package magrittr is installed, you can use pipes to plot. library(\"magrittr\") kde %>% plot(main = \"Miles per Gallon\") %>% lines(plot_start = TRUE, col = \"red\") The generics coef, logLik, AIC and BIC are supported, but work only on the parametric start. coef(kde) #> Maximum likelihood estimates for the Gumbel model #> mu sigma #> 17.321 4.865 logLik(kde) #> 'log Lik.' 1.644676 (df=2) AIC(kde) #> 0.710647 To view information about the kernel density, use the summary generic. summary(kde) #> #> Call: #> kdensity(x = mtcars$mpg, kernel = \"gaussian\", start = \"gumbel\")\n#>\n#> Data: mtcars$mpg (32 obs.) #> Bandwidth: 2.742 ('RHE') #> Support: (-Inf, Inf) #> Kernel: gaussian #> Start: gumbel #> Range: (10.4, 33.9) #> NAs in data: FALSE #> Adjustment: 1 #> #> Parameter estimates: #> mu: 17.32 #> sigma: 4.865 Access members of the kernel density with $ or [[.\n\nkde$start_str #> \"gumbel\" kde[[\"x\"]] #> 21.0 21.0 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 17.8 16.4 17.3 15.2 10.4 #> 10.4 14.7 32.4 30.4 33.9 21.5 15.5 15.2 13.3 19.2 27.3 26.0 30.4 15.8 19.7 #> 15.0 21.4 To make changes to the kernel density estimator, you can use the generic update function, replace values with $, or make a new one.\n\nkde$bw = \"RHE\" update(kde, start = \"normal\") kdensity supports all parametric starts implemented by the package univariateML. See its Github page for a complete list, or take a look at univariateML::univariateML_models. # Custom densities and kernels ## Parametric starts You must supply two functions and one vector to kdensity in order to use a custom parametric start. The first is a density function, the second is a function that estimates the named parameters of the density, and the third is the support of the density. normal = list( density = dnorm, estimator = function(data) { c(mean = mean(data), sd = sd(data)) }, support = c(-Inf, Inf) ) The density function must take the data as its first argument, and all its parameters must be named. In addition, the function estimator must return a vector containing named parameters that partially match the parameter names of the density function. For instance, the arguments of dnorm are x, mean, sd, log, where log = TRUE means that the logarithm of the density is returned. Since estimator returns a named vector with names mean and sd, the names are completely matched. The estimator function doesn’t need to be simple, as the next example shows. ### Example: A skew hyperbolic t-distribution start The built-in data set LakeHuron contains annual measurements of the water level of Lake Huron from 1875 to 1972, measured in feet. We will take a look at the distribution of differences in water level across two consecutive years. Since the data are supported on the real line and there is no good reason to assume anything else, we will use a normal start. LH = diff(LakeHuron) kde_normal = kdensity(LH, start = \"normal\") plot(kde_normal, lwd = 2, col = \"black\", main = \"Lake Huron differences\")", null, "The density is clearly non-normal. Still, it looks fairly regular, and it should be possible to model it parametrically with some success. One of many parametric families of distributions more flexible than the normal family is the skew hyperbolic t-distribution, which is implemented in the R package SkewHyperbolic. This package contains the density function dskewhyp and a maximum likelihood estimating function in skewhypFit. Using these functions, we make the following list: skew_hyperbolic = list( density = SkewHyperbolic::dskewhyp, estimator = function(x) { SkewHyperbolic::skewhypFit(x, printOut = FALSE)$param\n},\n\nsupport = c(-Inf, Inf)\n)\n\nNow we are ready to run kdensity and do some plotting. Note the plot option plot_start = TRUE. With this option on, the estimated parametric density is plotted without any non-parametric correction.\n\nkde_skewhyp = kdensity(LH, start = skew_hyperbolic)\nplot(kde_skewhyp, lwd = 2, col = \"blue\",\nmain = \"Lake Huron differences\")\nlines(kde_normal)\nlines(kde_skewhyp, plot_start = TRUE, lty = 2, lwd = 2)\nrug(LH)", null, "Since all the curves are in agreement, kernel density estimation appears to add unnecessary complexity without sufficient compensation in fit. We are justified in using the skew hyperbolic t-distribution if this simplifies our analysis down the line.\n\n## Kernels\n\nIf you want to make custom kernel, you will need to supply the kernel function, with arguments y, x, h. Here x is the random data you put into kdensity, h is the final bandwidth, and y is the point you want to evaluate at. The kernel is called as 1/h*kernel(y, x, h), and should be able to take vector inputs x and y. In addition to the kernel function, you must supply a support argument, which states the domain of definition of the kernel. For instance, the gcopula kernel is defined on c(0, 1). In addition, you can optionally supply the standard deviation of the kernel. This is only used for symmetric kernels, and is useful since it makes them comparable. For example, the implementation of the gaussian kernel is\n\ngaussian = list(\nkernel = function(y, x, h) dnorm((y-x)/h),\nsd = 1,\nsupport = c(-Inf, Inf))\n\nCustom kernels can be complicated, and do not have to be symmetric." ]
[ null, 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", null, 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", null, 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BHAMWubKgMYmjADTibtmxlwkMoQemgrcBPeAmCUQ7ATA14vnB7aCtqEiQM24xxmwNAmvAXAmD00uAlvADDmIbieC4P1B23AAmmeshZ8wpI64Aq5h3YmDNYhpAEL7B46ggkTB1zh9tAVPGEGnFgNYJh2kgashNtDV+DPHDYAOFZNK7SDi7MYMArBOWnygLF7aCswE6YOGP0QbAXmpOkCNg4vDw9dwUXS1AHnYcBuqgTQLSCAJ3x9ViTAmXhoAUcYxoJHdxFeVtwSiRyyHE6imQyHbjCQi/a+SDpEcQuUkwFDmjDUGHx027CEKW62sjLgystnJQt2Mguysoqhq/oztOGdNGnA2XhoK2vCQfuGKmCtfAzYCsBJQwNOuuN7doABCEe24Kg7vufmoSsNOPAwTNlF52bA3lwpXMsJA97lCDi4kwZLdCTd8T2zOZITBGEYwGl3fLdJjvwAQxAGAZx4v+jckhyNHOCAgRbQ06Q6T5lgx/d8DdhPaIUiTNCCrQFnCdgopJOGGoOtCScYg3M2YKeAhIGi6IQ7vudvwFVIwkTnwTnzDfv0nwGjU9hQmibgnPkGJkwS8C5vwEEJEwNMIsJqE15bVPATkxRXF2sTBZkDru9jPWFagN07P3e+ViEIkwJcTzCIADaE1w3DxADT4huCMCXAkbbYjiVRP3lYc0O0AEf7wrgYEjXh3QrClABH+zaxSPIJLy8k+IlJitMK/5HExApBmBBgiHUBqeUSN8sJUwG8M+ZLyEH7WkGYBmB1/zuActFoeaBFALAJM0XoYhHJfkRl0c1lD3jX4ksSsLAZukV3lzngnU31UOa7jnDWgHcucKbsn5XMcj1FePZyuowBu7jZbZhEl29V5y3nE84WcI3XFUiar9N8wpkC7uCNs94YgxThWYizBLxZvObD5rMIZwh4s3gd4FmE8wPs8420FwQiOcKTEecG2MMbba8PXFL3PINwXoAbvPG2ckGp6UacFWD3wMiQ3TBf+wafgjgjwO55vuO6Zb7TB6hsALuk86ap+hLTjBgIcOhdduwjI3cVU1aaFE/DAA68y06b7rYHX18G8GXPBgI47B4d3vNefSXzbWQ+KXxpe0gQwCF32WnTZbxdGcS7uICDWXCHbrX10HlQFrGXAWq9OrmYOXUG2WVn50eJbLkXJBrGvQ+GA0XRa3fZ2YkuXQZ8ScJn3H5OPrmIwC0ae8GNJ17czHCnyD2FwAlY+GriwrB10tdAr0EDnralv8e2+XtZhawYQdZ4KU7/ZkVSXMDzi1tTr3A/6odP+j+hI03hXhaVEO5fIVyeTLgXzDX2hCbeMycK4QWtrU931gGELUF0QwlRCf+XMbMbmdyG7z8GzIA9hXvYwIAHbywx4IAPGxjw4I2lBRzyYQMDHryxtIBDPmxgwIM3xhZcX8uAA5+oFHBLfwY8eGOJAQfc0p8BD95YasA46mXACQGzIikR4LkKUn2IQmg1BLC8FNWj6Vc0DQEsL0X1aPoVTUMAy0tRPZp+RdMQwPJSVI+mX9E0BLC8FNWj6Vc0DQEsL0X1aPoVTUMAy0tRPZp+RdMQwPJSVI+mX9E0BLA8FjIxYOJiwMTFgImLARMXAyYuBkxcDJi4GDBxMWDiYsDExYCJiwETV1rAejnqyCKJOXp6f2w96yQdhbh6SN6KcN3hKyng873s11Ls15ZzOowuWJ6io2zFcT3hla0I1x0tJQX8dHNXjW/oMlnHCyvSJ8isqSvW9uvKVlTBuqMtBGPwWuM5itux1ayTFKZf17aiKSjEaNEIAeBifb+sA/zew9oiArSiVoDu8JUe8OhS1DllrOkUYzIBDCcI4BDd4Ss54GOIoIIO4CDd4Ss14DBvWDIuOrT9pgNcmhlfueqGSjdtRBBkrW2F1rruGFRaC673hFipVV0baJoUAHCo7vCVeB4c6A27rmsDJTpWAw7WHb6SAi7NYvXk4U0ZJlW5FnCw7vCVOshiAYsBExcDJi4GTFwMmLgYMHExYOJiwMTFgImLARMXAyYuBkxcDJi4GDBxMWDiYsDExYCJiwETFwMmLgZMXAyYuBgwcTFg4mLAxMWAiYsBExcDJi4kgM/3Zl1PMfjNxHbt0NF+8WZ7CV7x3IKe9rqy5nvq9ZqzO/+L63tX+uW6BUzmO5T3fmtKv1Hn+ztZst4RyaxuPd+bhYvq3gp70fWjPOf0JugqpGGhAWxWtg8CLmTHmkW8lerfzjpPc0k5Tnhs4aDq4sMlwK1Vh2r9sP7DrBevvNaouo814XJfacC6mepwC3B9f7rw0S/aDic0gF/qfhsCfDqoVZWlfaW3UNtcYk4a1FLArXXDunyNqrWGULXGAHTri1VxT+//YKmqf4YBq13TpLGPNTqY0ADel7qbBgC39tDor6F1gFVfFcZXng5/Ohinqfzql1ffaDDqQHH9y6E5/tWNdJa/HH5/03X7lamqu/Jfoyr3/im1C3aAFf/Tbx89qsOA5TnNuxZQeABrRLYDnm68gdUCNgz6mwxZF+06WfnL00FtGSf/V2cfZTGqI9UWctIoTTX2uLFge3K33v7eHfqs4tfNfoOmNS0XbUE/a8FVc2ugwgNYv52HLNhaku7BAU9cuLCl0kGLOlufJX8xV8ohXHVk8bvrR8lMu1B33AC+rZoh3lN3952j5no6qCYWuhmuNV4w5g5ZqqUeg+vIqg6ybv1bAxUiwGrMuxBkacADb3lzidt+6Ki8rbZS+cNYnxlrT2++ev8HeUC95h+vT+5V299e6XzvmleXoFsg/3BDhyvIUb2rLlnwhcghlBABVr02PE2S7/sX/3wzEoTZQ6qX5cj64u83DbOyxlDcPn34y5sH6T/Va+UkwAPbK9W0rQ/QVbfGameT+oaebvYVA9aygcrt0BhspPfqHeoQe4nsYN25TzcDFizd/3f7qvjk3rw2asHtMbi/vVLtUjV825qWpTcWrG7o2JsHbxvw03uvR8NKHXAOjVnOgq8fNbej56ItIuXh3//stirVVKwZg8tnLLg1TTKXHO27yLxFbGu8I70xWI0uFwBvawyulC8enAfvL+036MbgO2u80mhqZia+dXH1UUdi6jV3XOEYBdxOdKhqNPLC+BqvNf4Y3ImiVUjGUXTV6o/+iyo5aHpiaDs6L1Wps4k2OLYk9Xz3wYNw8ubBD+od9csoYJedtOOHy0fWv9StKbyN2q0lO6qluB1KVVqw25kH09GF1Paqc5eKAYeWn+l6RhvKRRPS9ATzlp4msaDEgImLARMXAyYuBkxcDJi4GDBxMWDiYsDExYCJiwETFwMmLgZMXAyYuBgwcTFg4mLAxMWAiYsBExcDJi4GTFz/B8M2ugad52NBAAAAAElFTkSuQmCC", null ]
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http://brg.a2hosted.com/?page_id=4249
[ "Sunday Times Teaser 2834 – Degrees of Freedom\n\nby Stephen Hogg\n\nI bought an odd thermometer from an old curiosity shop. On its linear scale the freezing point and boiling point of water were higher than they are on the centigrade scale. In fact the freezing point was a prime number and, higher up the scale, the boiling point was a perfect square. There was only one number on the scale where it actually agreed with the corresponding centigrade temperature. That number was the negative of an odd prime (and not the same prime as the one mentioned earlier).\n\nOn this new scale, what are the freezing and boiling points of water?\n\n1.", null, "Let $$t_f$$ (an odd prime) and $$t_b$$ (a perfect square) be the freezing and boiling points of water on the odd thermometer’s scale. To convert its reading ($$t$$) into centigrade we have to use: $100(t – t_f)/(t_b-t_f)$ So the point at which the two scales give an equal value is given by:$t=100t_f/(100+t_f-t_b)$ Let this be the negative of an odd prime $$p$$, which leads to the equation: $t_b=100+t_f+100t_f/p$ Since $$t_f$$ is an odd prime and $$t_b$$ is an integer perfect square (say $$n^2$$), $$p$$ must be 5 (it is greater than 2 and not equal to $$t_f$$), which gives:$t_f=(n-10)(n+10)/21$ Since $$t_f$$ is an odd prime, $$(n-10)(n+10)$$ must be divisible by 21. But if one of these terms was divisible by 3 and the other by 7, then $$t_f$$ would be the product of two integers and hence not a prime (if one of these integers is 1, it can be shown that the other cannot be prime) . So one of these terms must be divisible by 21.\n\nIf $$(n+10)$$ is divisible by 21, then $$t_f=k(21k-20)$$ for some integer $$k$$, which cannot be prime. So $$(n-10)$$ must be divisible by 21, giving $$t_f=41$$ and $$t_b=961$$.\n\n2.", null, "A bit of analysis gives us the value at which the temperature scales coincide (v), and a formula for the boiling point (b) in terms of the freezing point (f), so we can just consider primes for the freezing point (f) until we find a corresponding boiling point (b) that is a square:\n\nThe enigma.py library is available at [ https://www.magwag.plus.com/jim/enigma.html ].\n\n3.", null, "Did this curious thermometer lead to the expression “It ain’t half hot, Mum”?" ]
[ null, "http://1.gravatar.com/avatar/422387c6c3f420a0f8466c6f6e5cc17c", null, "http://1.gravatar.com/avatar/75cb243345fac925a2622aa00b628d69", null, "http://0.gravatar.com/avatar/9a43d1be8fb231c2efb28b6b793965dc", null ]
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https://www.theorie.physik.uni-muenchen.de/activities/research_seminars/strings_and_fields/archive_sose10/2010_06_17/index.html
[ "", null, "", null, "# Generalized Kaehler Geometry and Supersymmetry\n\nRikard von Unge (Masaryk U)\n\n17.06.2010 at 16:15\n\nI will give a basic introduction to a recently developed mathematical field called Generalized Complex Geometry. In particular I will show how it can be used to describe the target space geometry of two dimensional sigma models with N=(2,2) supersymmetry leading to a definition of Generalized Calabi-Yau manifolds.\n\nArnold Sommerfeld Center\nTheresienstrasse 37\nRoom 348/349" ]
[ null, "http://piwik.physik.uni-muenchen.de/piwik.php", null, "https://cms-static.uni-muenchen.de/default/lmu/img/header-print.gif", null ]
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https://patents.justia.com/patent/5691631
[ "# Power supply apparatus\n\n- Fujitsu Limited\n\nThe apparatus of the present invention has an interface circuit which returns power in H-level output state, a first power circuit supplying a first supply voltage, and a second power circuit supplying a second supply voltage. The second power circuit receives the entered first supply voltage from the first power circuit, and in a first operation mode (L-level output condition), brings the first supply voltage of 3.3 V down to the second supply voltage of 1.5 V and supplies same to the interface circuit. In a second operation mode (H-level output condition) of the load circuit, a current output from the load circuit is brought up to the first supply voltage of 3.3 V and feeds same back to the first power circuit.\n\n## Latest Fujitsu Limited Patents:\n\nDescription\nBACKGROUND OF THE INVENTION\n\nThe present invention relates to a two-way power supply apparatus which operates with high and low values of voltage and has an operating mode of consuming power from the high-voltage power supply, and another operating mode of sending power back to the low-voltage power supply. More particularly, the present invention relates to a power supply apparatus having a high-speed interface which reduces an excessive power consumption in the power returning mode.\n\nA circuit provided with a comparator is now becoming more popular as a high-speed interface circuit. In FIG. 1, a high-speed interface circuit 12 has an invertor 16 and a comparator 18. The high-speed interface circuit 12 is supplied with 3.3 V supply voltage, for example, from a first power circuit 10. Supply voltage of 1.5 V is supplied from a second power circuit 14. Supply voltage of 3.3 V from the first power circuit 10 is used as a driving power supply for invertor 16 and comparator 18. Supply voltage of 1.5 V from the second power circuit 14 pulls up an output line of the invertor 16 through a resistance R20. A reference voltage for the comparator 18 is set from voltage division by resistances 22 and 24.\n\nWhen an input terminal 34 is a H-level, internal impedance between an output terminal of the invertor 16 and a minus line takes an value of almost zero, and an output of the invertor 16 is an L-level. At this point, current as represented by a solid line 202 flows through a path including a pull-up resistor 20 and the invertor 16, resulting in a decreased output voltage of the invertor 16 of about 0 V. Consequently, output voltage of the invertor 16 becomes lower than the reference voltage, and an output terminal 36 of the comparator 18 becomes an L-level.\n\nWhen the input terminal 34 is an L-level, internal impedance between a supply terminal and the output terminal of the invertor 16 becomes almost zero, and output of the invertor 16 becomes a H-level, i.e., a supply voltage of 3.3 V is applied to the output line. As a result, output voltage of the invertor 16 becomes lower than the reference voltage, and an output terminal of the comparator 18 becomes a H-level.\n\nIn the high-speed interface circuit of FIG. 1, however, when a supply voltage of almost 3.3 V is applied to the output line due to output from the invertor 16 becoming a H-level, current flows through a path as represented by a broken line 204 via the pull-up resistor into a second power circuit 14 having a lower supply voltage of 1.5 V. When current thus flows into the second power circuit 14, output voltage thereof increases and becomes less stable, and along with this, the reference voltage of the comparator 18 increases as well. An operating point of the comparator 18 changes accordingly and becomes less stable.\n\nIn order to solve this problem, it suffices, as shown in FIG. 2, to connect a dummy resistor 200 capable of consuming maximum current flowing into the second power circuit 14. More specifically, the flow of current into the second power circuit 14 can be eliminated, thus keeping output voltage in a stable state, by always causing the dummy resistor 200 to consume the fraction of current flowing out from the high-speed interface circuit 12 on the load side. However, this would result in constant consumption of outflowing current I by the dummy resistor 200 in the H-level state of the invertor 16 as shown in FIG. 3A, and it would accordingly be necessary to increase the power circuit capacity by a value corresponding to the consumption by the dummy resistor 200. Since power is always consumed by the dummy resistor, another problem of decreased power efficiency is encountered.\n\nSUMMARY OF THE INVENTION\n\nAccording to the present invention, there is provided a power supply apparatus which permits returning power from the load side without causing fluctuations of supply voltage, eliminates the necessity of increasing the power supply capacity, and simultaneously improves service efficiency of power.\n\nThe interface circuit serving as a load circuit in the present invention operates by supply of a prescribed first supply voltage (3.3 V) and a prescribed second supply voltage (1.5 V) lower than the first supply voltage. In a first operation mode, the first and second supply voltages are consumed by load, and in a second operation mode, the second supply voltage is returned to the supply side. More specifically, the interface circuit has a gate circuit, for example an invertor, which outputs an L-level signal in the first operation mode, and outputs a H-level signal in the second operation mode.\n\nThe output line of the invertor is pulled up via a pull-up resistor to a supply line of the second supply voltage. An output voltage of the invertor is compared with a prescribed reference voltage in the comparator, outputting an L-level signal when the output voltage is lower than the reference voltage, and a H-level signal when the former is higher than the latter. The invertor and the comparator operate with the first supply voltage of 3.3 V. The reference voltage of the comparator is derived from the second supply voltage of 1.5 V.\n\nIn the present invention, a first power circuit supplying the first supply voltage and a second power circuit supplying the second supply voltage are provided to the interface circuit as such a load circuit. The second power circuit receives the first supply voltage from the first power circuit, and in the first operation mode of the interface circuit (L-level output condition of the inverter), converts the first supply voltage into the second supply voltage which is supplied to the load circuit. In the second operation mode (H-level output condition of the inverter), it converts a current output from the interface circuit into the first supply voltage to feed back to the first power circuit. This second power circuit is provided with a down-converter which supplies the first supply voltage of 3.3 V converted into the second supply voltage of 1.5 V to the interface circuit, an up-converter which converts a current output from the interface circuit into the first supply voltage of 3.3 V to feed back to the first power circuit, and a control circuit which performs switching control of each of them. Each of the down-converter and the up-converter employs a non-insulating type single converter provided with, for example, a switching element. To simplify the circuit configuration, the down-converter and the up-converter are combined into an integral body.\n\nThe second power circuit integrating the down-converter and the up-converter is composed of an inductance connected from connecting points of a pair of switching elements to a plus output terminal of the second supply voltage, a capacitor connected to an output terminal of the second supply voltage, and a control circuit which on-off controls alternately, by a pulse width modulating signal for example, the pair of switching elements in the first mode and the second mode of the interface circuit. A diode is connected in parallel in the reverse polar direction to each of the pair of switching elements. When using MOS-FET, a parasitic diode produced because of the element construction is connected in parallel in the reverse polar direction to the switch section, so that it is not necessary to connect a separate diode.\n\nAs another example of the second circuit, an insulating-type converter and a control circuit thereof may be used as a down-converter and an up-converter. The insulating-type converter converts, in the second operation mode of the interface circuit, current output from the interface circuit by constant-current control into the first supply voltage of 3.3 V to feed back to the first power circuit.\n\nFurthermore, a switching regulator which converts an input voltage into the first supply voltage by constant-voltage control to output, i.e., an insulating-type converter and a control circuit thereof, may be provided. In this case, output of the insulating-type up-converter provided in the second power circuit may be feedback-connected to an output terminal of an insulating-type converter provided between the first power circuit and the interface circuit.\n\nAccording to the power circuit for interface of the present invention, when power is sent back with a supply voltage of 3.3 V from the first power circuit to the second power circuit on the side of a supply voltage of 1.5 V in the switching condition of the interface circuit serving as the lead circuit, the second power circuit acts as an up-converter relative to the returned power, thus permitting return to the first power circuit by increasing voltage to 3.3 V without consuming the returned power. At this point, the second supply voltage of 1.5 V from the second power circuit does not vary and is stable. Even when the reference voltage of the comparator provided in the interface circuit has been worked out from the second supply voltage, it does not vary and stable operation of the comparator can thus be ensured. Because the returned power is not consumed, it is not necessary to increase the power supply capacity by an extent corresponding to that consumed, as is necessary when the returned power is consumed by the dummy resistor, thereby naturally improving the service efficiency of power.\n\nThe above and other objects, features, and advantages of the present invention will become more apparent from the following detailed description with reference to the drawings.\n\nBRIEF DESCRIPTION OF THE DRAWINGS\n\nFIG. 1 is a circuit diagram of a conventional apparatus;\n\nFIG. 2 is a circuit diagram of a conventional apparatus using a dummy resistor;\n\nFIGS. 3A and 3B are timing charts of consumption of return current by the dummy resistor of FIG. 2;\n\nFIG. 4 is a circuit block diagram of a basic embodiment of the present invention;\n\nFIG. 5 is a circuit diagram of an embodiment in which a down-converter and an up-converter are integrated;\n\nFIG. 6 is an equivalent circuit diagram of FIG. 5;\n\nFIGS. 7A and 7B are descriptive views of operations during down time;\n\nFIGS. 8A-8G are timing charts; and during down time;\n\nFIGS. 9A and 9B are descriptive views of operations during up time;\n\nFIGS. 10A-10G are timing charts during up time;\n\nFIG. 11 is a circuit diagram of an embodiment in which a down-converter and an up-converter are separately provided;\n\nFIG. 12 is a circuit diagram of an embodiment in which a switching regulator is used for down and up operations; and\n\nFIG. 13 is a circuit diagram of a variation of the embodiment shown in FIG. 12.\n\nDETAILED DESCRIPTION OF THE PREFERRED EMBODIMENTS\n\nFIG. 4 shows a basic embodiment of the present invention. A first power circuit 10 outputs a first supply voltage of, for example, DC 3.3 V. The first power circuit 10 may be a battery, or an ordinary power circuit which creates a DC power supply by rectification-smoothing an AC power supply. Supply voltage of DC 3.3 V from the first power circuit 10 is supplied from supply terminals 44 and 46 to supply terminals 26 and 28 of an interface circuit 12. The interface circuit 12 serving as a load circuit is provided, for example, with a converter 16 as a gate circuit, and a comparator 18. The inverter 16 and the comparator 18 operate with supply voltage of DC 3.3 V from the first power circuit 10 supplied between the supply terminals 26 and 28. The inverter 16 turns over an input signal from an input terminal 34. The comparator 18 compares an output voltage of the inverter 16 with a prescribed reference voltage: an L-level output is generated at an output terminal 36 if the output voltage is lower than the reference voltage, and a H-level output is generated at the output terminal 36 if the output voltage is higher than the reference voltage. An output line of the inverter 16 is pulled up to a second supply voltage of, for example, DC 1.5 V, supplied from the second power circuit 14 via a pull-up resistor 20. The reference voltage of the comparator 18 is created by dividing the supply voltage of DC 1.5 V from the second power circuit 14 by means of resistors 22 and 24.\n\nThe second power circuit 14 is provided with a down-converter 38 with a control circuit section 39 thereof, and an up-converter 40 with a control circuit section 41 thereof. This circuit provided with the converter 38 and the control circuit 39 is usually known as a switching regulator. In the first operation mode in which input is a H-level at the input terminal 34 of the interface circuit 12, becomes an L-level through turnover by the inverter 16, and output of the comparator 18 is an L-level, the down-converter 38 of the second power circuit 14 operates. The down-converter 38 brings the supply voltage of DC 3.3 V from the first power circuit 10 between power input terminals 48 and 50 down to DC 1.5 V, and supplies it through power output terminals 52 and 54 to the interface circuit 12. In the second operation mode, on the other hand, in which input of the input terminal 34 of the interface circuit 12 is an L-level, and the comparator 18 gives an H-level output as a result of inversion into H-level by the inverter 16, the up-converter 40 of the second power circuit 14 operates. When the comparator 18 is in the H-level state in the interface circuit 12, the supply voltage of DC 3.3 V from the first power circuit 10 to the supply terminal 26 is applied to the power output terminal 52 of the second power circuit 14 from the supply terminal of the inverter 16 through an output terminal and the pull-up resistor 20, to send the power back to the second power circuit 14 side. Upon return of current from the interface circuit 12, the up-converter 40 brings the input voltage of 1.5 V on the side of the power output terminals 52 and 54 up to 3.3 V, and feeds the power back to the first power circuit 10 from the power input terminals 48 and 50. There is also available a method in which the second power circuit 14 always acts to return a constant current to the first power circuit 10.\n\nFIG. 5 shows a concrete embodiment of the down-converter 38 and the up-converter 40 provided in the second power circuit of FIG. 4. This embodiment is characterized in that the down-converter 38, the up-converter 40 and the control circuit sections thereof are integrated to simplify the circuit configuration. In the second power circuit 14, the converter composing the main circuit section comprises MOS-type FETs 56 and 58, an inductance 64 and a capacitor 66. This is a non-insulating type DC--DC converter. Diodes 60 and 62 connected in parallel in the reverse directions to the FETs 56 and 58 can be achieved by parasitic diodes generated by MOS-type FET element structure. It is not therefore necessary to connect separate diodes 60 and 62 to the FETs 56 and 58. The control circuit of the converter is composed of an error amplifier 68, a resistor 70, a Zener diode 72, a pulse width modulating circuit (hereinafter referred to as the \"PWM circuit\") 74 and an inverter 76. The resistor 70 and the Zener diode 72 generate a reference voltage. When there occurs a difference between an output voltage between the supply output terminals 52 and 54 of a pulse input terminal of the error amplifier 68 and the reference voltage, the error amplifier 68 outputs an error voltage corresponding to this difference to the PWM circuit 74. The PWM circuit 74 performs a pulse width control so as to bring an error voltage to zero, i.e., controls duty ratio. An output pulse from the PWM circuit 74 is supplied directly to the gate of the FET 56, and is simultaneously inverted in the inverter 76 for supply to the gate of the FET 58. The FETs 56 and 58 are therefore alternately turned on and off by a pulse signal from the PWM circuit 74.\n\nFIG. 6 shows an equivalent circuit to that shown in FIG. 5. The interface circuit 12 can be deemed as a load circuit 12' as an equivalent circuit. The reversing function of the input level by the inverter 16 is replaced with a switch 78. When an output of the inverter 16 is an H-level, i.e., in the second operation mode in which power is returned to the second power circuit 14, the switch 78 is closed on the side of an equivalent resistance R1 connected to a line from the supply terminal 26. As a result, the supply voltage of DC 3.3 V from the first power circuit 10 is returned through the resistance R1 and the switch 78 to the portion between the power output terminals 52 and 54. At this point, there is present an equivalent resistance R2 on the common terminal side of the switch 78. Accordingly, the first supply voltage of DC 3.3 V between the terminals 26 and 28 is divided by the equivalent resistances R1 and R2. In a stable state, the equivalent resistance R1 has an end voltage of 1.8 V f and the equivalent resistance R2 has an end voltage of 1.5 V.\n\nIn the L-level condition of the inverter 16, on the other hand, the switch 78 is changed over to connection of an equivalent resistance R3 connected on the minus line side. In this state, power is supplied only from the second power circuit 14 to the equivalent resistance R3. The pair of FETs 56 and 58 provided in the second power circuit 14 are changed over by switches 80 and 82. Control circuit sections for the switches 80 and 82 are omitted here.\n\nFIG. 7A shows the operation with the switch 80 being on and the switch 82 being off during down operation of the equivalent circuit shown in FIG. 6. FIG. 7B shows the operation when the switch 82 is turned on and the switch 80 is turned off then. In this case, since the interface circuit 12, in a state in which the switch 78 is changed over to the equivalent resistance R3 side by an L-level output of the inverter 16, is represented by an equivalent resistance R0 which is the parallel resistance value of the equivalent resistances R2 and\n\nWhen the switch 80 is turned on in FIG. 7A, current flows through the switch 80 and an inductance 64 to a capacitor 66 and the additional resistance R0 under the effect of the supply voltage of DC 3.3 V from the first power circuit 10. In FIG. 7A, current flowing inductance 64 is represented by I1, current flowing to the switch 80, by I2, potential at the connecting point of the switches 80 and 82, by V1, and the end voltage of the inductance 64, by V2. Energy is charged to the inductance 64 by the current flowing through the inductance 64. At the same time, the capacitor 66 is charged, and in addition, current is supplied to the equivalent resistance R0 of the interface circuit 12. In the state of this switch 82 being on, the inductance has a reverse polarity as compared with that shown in FIG. 7A, with an end voltage V2 of -1.5 V, and constant voltage control is effected so as to give an end voltage of 1.5 V of the capacitor.\n\nFIGS. 8A to 8G are timing charts during down time of the equivalent circuit shown in FIG. 3. FIG. 8A shows on/off of the switch 80 and FIG. 8B shows on/off of the switch 82: one being on corresponding to the other being off. When the switch 80 shown in FIG. 8A is on, potential V2 at the connecting point of the switches 80 and 82 as shown in FIG. 7A is 3.3 B as shown in FIG. 8C. At this point, the inductance 64 has an end voltage V2 of 1.8 V as shown in FIG. 8D. Current I1 flowing through the inductance 64 increases with a slow gradient as shown in FIG. 8E. Current flowing through the switch 80 is as shown in FIG. 8F and has the same value as current I1 flowing through the inductance 64 of FIG. 8E. Furthermore, current I3 of the switch 82 is zero as shown in FIG. 8G since the switch 82 is off.\n\nThen, when the switch 80 is off in FIG. 8A, and the switch 82 is on in FIG. 8B, voltage V1 in FIG. 8C becomes 0 V because connection with the first power circuit is cut off. The end voltage V2 of the inductance 64 in FIG. 8D is changed over to -1.5 V with a reverse polarity. At this point, current I1 flowing through the inductance 64 slowly decreases at a certain gradient as shown in FIG. 8E. Current I3 flowing through the switch 82 decreases at the same gradient as that of current I1 in FIG. 8E. However, current flows in the minus direction since current direction is reverse from bottom to top. A constant voltage of 1.5 V is thus supplied to the interface circuit 12 in a state in which, during down time, DC 3.3 V from the first power circuit 10 is brought down to 1.5 V and stabilized on this level.\n\nFIG. 9A shows the operation when the switch 80 is on during up time of the second power circuit 14, and FIG. 9B shows the operation when the switch 82 is on. During up time, the inverter 16 of the interface circuit 12 generates an H-level output and is in a state in which the switch 78 is closed to the equivalent resistance R1, as shown in the equivalent circuit of the interface circuit 12 in FIG. 6. For up time, the state of the switch 82 being on in FIG. 9B is described first below. When the switch 82 is turned on, a supply voltage of 3.3 V is feedback-entered from the constant-voltage source 10 of the interface circuit 12 to the second power circuit 14 through the resistance R1. Current flows through a path including the inductance 64 and the switch 82 in parallel with charging of the capacitor 66, and energy is charged into the inductance 64. In this drawing, current of the inductance 64 is represented by I1, current of the switch 82, by I3, end voltage of the inductance 64, by V2, and voltage of the switch 82, by V1.\n\nThen, it is assumed that the switch 82 is turned off from the on-state of the switch 82 in FIG. 9B, and the state is changed over to that shown in FIG. 9A where the switch 80 is on. In the state of FIG. 9A, current flows through the path comprising the switch 80, the first power circuit 10 and the capacitor 66 by the energy charged into the inductance 64. Current flowing through the switch 80 at this moment is represented by I2. The capacitor 66 has an end voltage of 1.5 V, and the inductance 64 has an end voltage V2 of 1.8 V. A feedback power supply of 3.3 V is created by configuring an up-circuit through connection in series. Consequently, the energy charged from the interface circuit 12 into the inductance 64 is fed back to the first power circuit 10 by up operation.\n\nFIGS. 10A to 10G are timing charts in the up operation of FIGS. 9A and 9B. When the switch 80 is off in FIG. 10A, and the switch 82 is on in FIG. 10B, charging is accomplished with the power fed back from the interface circuit 12 so as to give a voltage V1 of 0 V in FIG. 10C and an end voltage V2 of -1.5 V for the inductance 64 in FIG. 10D. At this point, current I1 of the inductance 64 in FIG. 10E is slowly increasing. Current I3 flowing through the switch 82 in FIG. 10G, being a current in the plus direction, shows the same change as that of current I1 flowing through the inductance 64 in FIG. 10E.\n\nWhen the switch 80 is on in FIG. 9A, and the switch 82 is off in FIG. 9B, voltage V1 at the connecting point between the switch 80 and the inductance 64 increases to 3.3 V as shown in FIG. 10C, and end voltage V2 of the inductance 64 in FIG. 10D takes a value of 1.8 V along with constant-voltage control of end voltage of the capacitor 66 to 1.5 V. Current I1 of the inductance 64 of FIG. 10E slowly decreases, and current I2 flowing through the switch 80, represented by the minus direction, shows the same change. The power returned by the interface circuit 12 through these operations during up time is brought up and fed back to the first power circuit 10, so that power is never consumed by, for example, a load resistance.\n\nIn alternate on/off condition of the switches 80 and 82, there is a timing when the switches 80 and 82 are simultaneously turned off upon changing over. At this point, an initial current flows through a path bypassing the switches 80 and 82, passing through each of diodes 60 and 62, and thereafter, the status is switched over to the original path of current by turning on the switch 80 or 82. Therefore, even when there is a timing when the switches 80 and 82 are simultaneously turned off, bypass connection by the diodes 60 and 62 is ensured, thus permitting prevention of occurrence of spike noise by complete separation.\n\nFIG. 11 shows another concrete embodiment of the down-converter 38 and the up converter 40 provided in the second power circuit 14 shown in FIG. 4. In this embodiment, converters and control circuit sections thereof are not common, but separately provided. The down-converter 38 has a main circuit section constituting a non-insulating type DC--DC converter provided with an MOS-type FET166, and connected with a diode 170 in series with the FET166 with a reverse polarity. An end of an inductance 172 is connected to the connecting point of the FET166 and the diode 170, and the other end is connected to a power output terminal 52. A capacitor 174 is connected between the power output terminals 52 and 54. The control circuit section of the down-converter 38 comprises an error amplifier 176 and a PWM circuit 180. The error amplifier 176 compares a reference voltage of 1.5 V based on a reference voltage source 178 and an output voltage and outputs the resultant error voltage to the PWM circuit 180. The PWM circuit 180 controls the duty ratio of a PWM signal to FET166 so as to achieve an error voltage of zero.\n\nThe down-converter 38 using such a non-insulating type single converter charges energy into the inductance 172 upon turn-on of the FET166 by the PWM circuit 180, and constant-voltage-controls an output voltage having charged the capacitor 174 to a prescribed voltage of 1.5 V by altering the time interval between energy charges caused by turning on the FET166. The up-converter 40 has an MOS-type FET185 formed by connecting an inductance 184 in series to a main circuit section composing a non-insulating type DC--DC converter, and further a current detecting resistance 188 is connected in series to the FET185. When the FET185 is turned on, the inductance 184 is charged by being supplied with power from the interface circuit 12. When the FET185 is turned off, energy charged into the inductance 184 is discharged through a path comprising a diode 196, the first power circuit 10, and a capacitor 174, to feed back the energy having been charged into the inductance 184 to the first power circuit 10. The control circuit section of the up-converter 40 is provided with an error amplifier 190 and a PWM circuit 194. A current detecting voltage from the current detecting resistance 188 connected in series to the FET185 and the reference voltage from the reference voltage source are entered into the error amplifier 190. The error amplifier 190 therefore outputs an error signal corresponding to an error in the detected current relative to a certain current determined by the reference voltage source, and the PWM circuit 194 PWM-controls the FET185 so as to maintain output current on a constant level.\n\nFIG. 12 shows another embodiment of the present invention. This embodiment is characterized by the use of a switching regulator provided with an insulating type DC--DC converter which electrically separates the primary side from the secondary side for down and up operations of the second power circuit and a control circuit thereof. Also for power supply from the first power circuit, a switching regulator having an insulating type DC--DC converter for constant-voltage control and a control circuit thereof is provided. In a power supply line from the first power circuit 10 to the interface circuit 12, a switching regulator 86 for constant-voltage control of the first supply voltage of 3.3 V is provided. The main circuit section of the switching regulator 86 is provided with an insulating type DC--DC converter, i.e., a transistor 92 is connected in series to a primary winding 90 of a transformer 88. Following a secondary winding 94 of the transformer 88, a rectifying circuit using diodes 95 and 96 is provided, with furthermore a choke coil 98 and a capacitor 100 for smoothing. A control circuit 102 of the switching regulator 86 compares a received output voltage with a predetermined reference voltage and creates a PWM signal on the basis of an error voltage thereof to on/off-control the transistor 92. In the control circuit 102 also, the primary side and the secondary side are internally isolated. The switching regulator 86 stabilizes an input voltage from the first power circuit 10 to power input terminals 104 and 106 at a value of 3.3 V, and supplies the thus stabilized power from power output terminals 108 and 110 to the interface circuit 102. The switching regulator 86 is effective when supplying always stable first supply voltage of 3.3 V relative to variations in the power supply capacity for the purpose of driving a plurality of interface circuits 12 provided in parallel.\n\nThe second power circuit 14 is provided with a first switching regulator 112 acting as a down-circuit, and a second switching regulator 132 acting as an up-circuit. The switching regulator 124 for down-operations has a transistor 130 connected in series to a primary winding 116 of a transformer 114 as an insulating type DC--DC converter of a main circuit section. Following a secondary winding 118 of the transformer 112, there are provided a rectifying circuit of diodes 120 and 122, with furthermore a choke coil 124 and a capacitor 126 for smoothing. A control circuit 128 receives an output voltage, compares same with a predetermined reference voltage of 1.5 V to create an error voltage, generates a PWM signal having a duty ratio so as to achieve an error of zero, and thus on/off-controls the transistor 130. In the control circuit 128, the primary side and the secondary side are internally isolated. This switching regulator 112 for down-operations is basically the same as the switching regulator provided on the side of the first power circuit 10. A switching regulator 132 for up-operations has a transistor 138 connected in series to a primary winding 136 of a transformer 134 as an insulating type DC--DC converter forming a main circuit section, and further connected to a current detecting resistance 140. The control circuit section is provided with an error amplifier 142 and a PWM circuit 146. The error amplifier 142 outputs a differential voltage as compared with a voltage detected by the current detecting resistance 140 relative to a certain current set by a reference voltage source 144. The PWM circuit 146 outputs a PWM signal controlled so as to give a duty ratio achieving an error voltage of zero to on/off-control the transistor 138. A voltage brought up by a winding ratio is induced in a secondary winding 145 of the transformer 134, rectified at a diode 148, and fed back to the first power circuit 10 under constant-current control.\n\nFIG. 13 shows a variation of the embodiment shown in FIG. 12. This embodiment is characterized in that an output of the switching regulator 132 for up-operations of the second power circuit 14 is fed back to an output end of the switching regulator provided between the first power circuit 10 and the interface circuit 12. Consequently, power feedback lines 160 and 162 are pulled out from output terminals 154 and 156 of the switching regulator 132 for up-operations, and are connected to supply lines from output terminals 108 and 110 of the switching regulator 86 to the interface circuit 12. Output voltage between power output terminal 108 and 110 of the switching regulator 86 is affected by the current feedback from the switching regulator 132 for up-operations. However, since constant-current control is applied to inhibit fluctuation of output voltage caused by feedback in the control circuit 102 of the switching regulator 86, a stable first supply voltage of 3.3 V can be supplied to the interface circuit 12 even in the presence of power feedback. To the extent of power feedback, the power consumption by the first power circuit 10 can be saved.\n\nAccording to the present invention, as described above, in the operation mode of the interface circuit serving as the load circuit, even if power is returned by a supply voltage from a higher power circuit section to a lower power circuit section, the lower power circuit section brings up the returned power while keeping a constant voltage to feedback same to a lower supply voltage, whereby the lower supply voltage to the interface circuit does not vary with the return of power, and stable operations can be ensured while keeping a constant reference voltage of the comparator provided in the interface circuit.\n\nBecause the power returned from the interface circuit serving as the load circuit is brought up and fed back to the original down-side power circuit section, it is possible to utilize the power again as supply power, and improve service efficiency of the power supply through this regeneration of power.\n\nSince the power returned by the interface circuit is never consumed wastefully by a dummy resistor, for example, it is possible to avoid such an uneconomical situation in which the power supply capacity of a power circuit producing a high supply voltage is increased by an amount corresponding to the consumption of feedback power.\n\nIn the embodiments shown in FIGS. 12 and 13, the description has covered a case in which a switching regulator 86 has been provided between the first power circuit 10 and the interface circuit 12. It is however needless to mention that a configuration may be adopted in which a switching regulator 86 is not provided, but a switching regulator 112 for down-operations and a switching regulator 132 for up-operations are provided only for the second power circuit.\n\nAs the power-down circuit used in the second power circuit 14 and the power-up circuit for power feedback in the present invention, any appropriate switching regulator may be used, not limiting to no-insulating and insulating switching regulators of the embodiments. While the above-mentioned embodiments have covered the cases with a first supply voltage of 3.3 V and a second supply voltage of 1.5 V, the voltage used for the interface circuit is not limited to these cases.\n\nAn inverter has been used as an example of the gate circuit for the interface circuit 12, whereas applicable gate circuits include an AND gate, an OR gate, an NAND gate and other appropriate gates only if the output line to the comparator is pulled up under the second supply voltage. The present invention is not of course limited to figures given in the embodiments.\n\n## Claims\n\n1. A power supply apparatus comprising:\n\na load circuit which operates by supply of a prescribed first supply voltage and a second supply voltage different from said first supply voltage, consumes said first supply voltage and said second supply voltage by a load in a first operation mode, and returns said first supply voltage to a supply side of said second supply voltage in a second operation mode;\na first power circuit which supplies said first supply voltage to said load circuit; and\na second power circuit which receives input of said first supply voltage from said first power circuit, converts said first supply voltage into said second supply voltage and supplies the converted voltage to said load circuit in said first operation mode of said load, and converts a current output from said load circuit into said first supply voltage and feeds back the converted voltage to said first power circuit in the second operation mode of said load circuit, said second power circuit comprises:\na down-converter which converts said first supply voltage into said second supply voltage and supplies same to said load circuit in said first operation mode of said load circuit,\nan up-converter which converts a current output from said load circuit into said first supply voltage and feeds same back to said first power circuit in the second operation mode of said load circuit, and\na control circuit which switching-controls each of said down-converter and said up-converter.\n\n2. An apparatus according to claim 1, wherein:\n\na gate circuit which processes an input signal, outputs an L-level signal in said first operation mode, and outputs a H-level signal in said second operation mode;\na pull-up resistor which pulls up an output line of said gate circuit to a supply line of said second supply voltage; and\na comparator which compares an output voltage of said gate circuit with a prescribed reference voltage, outputs an L-level signal when the output voltage is lower than said reference voltage, and outputs an H-level signal when the output voltage is higher than said reference voltage; and\nsaid gate circuit and said comparator operating under said first supply voltage, and said reference voltage being created from said second supply voltage.\n\n3. An apparatus according to claim 1, wherein said down-converter and said up-converter are connected in parallel.\n\n4. An apparatus according to claim 1, wherein:\n\neach of said down-converter and said up-converter is a non-isolating type single converter provided with a switching element.\n\n5. An apparatus according to claim 1, wherein:\n\nsaid down-converter, said up-converter and said control circuit are integrated into a body.\n\n6. An apparatus according to claim 5, wherein:\n\nthe second power circuit further comprises:\na pair of switching elements connected in series between input terminals of said first supply voltage;\nan inductance connected from a connecting point of said pair of switching elements to between plus output terminal of said second supply voltage;\na capacitor connected between output terminals of said second supply voltage; and\nsaid control circuit which alternately on/off controls said pair of switching elements by means of a pulse width modulating signal so as to keep an end voltage of the output terminal of the second supply voltage to said load circuit at a prescribed value in the first and second operation modes of said load circuit.\n\n7. An apparatus according to claim 6, wherein:\n\ndiodes are connected in the direction of the reverse polarity in parallel with each of said pair of switching elements.\n\n8. An apparatus according to claim 7, wherein:\n\nsaid pair of switching elements are an MOS-FET in which said diodes are parasitic diodes which are generated because of the element structure and are connected in the direction of the reverse polarity in parallel with a switch section.\n\n9. An apparatus according to claim 1, wherein:\n\neach of said down-converter and said up-converter comprises an isolating type converter and a control section.\n\n10. An apparatus according to claim 9, wherein:\n\nsaid isolating type up-converter converts a current output from said load circuit into said first supply voltage through constant-current control and feeds same back to said first power circuit in the second operation mode of said load circuit.\n\n11. An apparatus according to claim 9, wherein:\n\nsaid isolating type converter and said control section thereof which convert an input voltage into said first supply voltage through constant-voltage control and output same are provided between said first power circuit and said load circuit.\n\n12. An apparatus according to claim 11, wherein:\n\nan output of the isolating type up-converter provided in said second power circuit is feedback-connected to an output terminal of the isolating type converter provided between said first power circuit and said load circuit.\nPatent History\nPatent number: 5691631\nType: Grant\nFiled: Jun 5, 1995\nDate of Patent: Nov 25, 1997\nAssignee: Fujitsu Limited (Kanagawa)\nInventors: Hiroshi Shimamori (Kawasaki), Ryuichi Yanagisawa (Kawasaki), Shinichi Ohtsu (Kawasaki)\nPrimary Examiner: Jeffrey L. Sterrett\nLaw Firm: Nikaido Marmelstein Murray & Oram LLP\nApplication Number: 8/462,289" ]
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https://www.geeksforgeeks.org/converting-epsilon-nfa-to-dfa-using-python-and-graphviz/?ref=leftbar-rightbar
[ "# Converting Epsilon-NFA to DFA using Python and Graphviz\n\n• Last Updated : 13 Feb, 2022\n\nFinite Automata (FA) is a simple machine used to match patterns in an input string. Finite Automata is a quintuple i.e., it has five elements. In this article we are going to see how to convert epsilon-NFA to DFA using Python and Graphviz.\n\nThe quintuple of FA is represented as :", null, "The Five Elements are :\n\n1. A finite set of States (Q)\n2. A finite set of Input Alphabets (", null, ")\n3. Start State (", null, ")\n4. A finite set of Final States (F)\n5. Transition Function (", null, ")\n\nFinite Automata (FA) are of two types :\n\n• Deterministic Finite Automata (DFA)\n• Non-Deterministic Finite Automata (NFA)\n\n### Deterministic Finite Automata (DFA)\n\nDeterministic Finite Automata (DFA) is a FA where there is only one fixed state the machine can go to for an input alphabet as defined in the transition function.\n\nDFA does not allow", null, "(null) alphabet which means that the machine will not change state if no alphabet is detected.\n\nFor a DFA,", null, "Example, DFA accepts all the strings made up of f’s and g’s which contains “gfg” as a substring.", null, "DFA\n\n### Non-Deterministic Finite Automata (NFA)\n\nNon-Deterministic Finite Automata (NFA) is a FA where the machine can go to more than one states for an input alphabet.\n\nFor an NFA,", null, "NFA which allows", null, "moves is called E-NFA or Epsilon-NFA.\n\nNFA allowing", null, "(null) alphabet means that machine can change state even if no input alphabet is detected.\n\nFor an E-NFA,", null, "Example, NFA that accepts all the strings made up of f’s and g’s which contains “gfg” as a substring.", null, "NFA\n\nThe", null, "is because, for every transition, there are 2 possibilities, to transit or not transit. And this is for every Q state. So", null, "will be the total possible configurations for each transition.\n\n### Why the conversion?\n\nComputers can understand FA in its basic form that is in DFA. But because of the features of NFA, we humans can understand NFA and better E-NFA with ease. So we need to convert the E-NFA to DFA.\n\n### Steps For Converting E-NFA to DFA :", null, "– closure : It is the set of states to which we can go without any input i.e., with", null, "moves.\n\nStep 1: Find", null, "– closure of the start state of NFA and that will be the start state of DFA.\n\nStep 2: Starting with this set, for each alphabet, evaluate", null, "– closure of the transition set for this alphabet.\n\nStep 3: For each new set of closure set we come across, we will repeat Step 2 until no new set is left.\n\nStep 4: The set in DFA which contains the final state of NFA will be the final set state.\n\n#### For example,\n\nLet the given E-NFA be :", null, "NFA\n\nNFA :", null, "Q : {A, B, C, D}", null, ": {a, b, c}", null, ": A\n\nF : {D}", null, ":\n\nIt is an NFA for language which accepts string of the type : {", null, ", where", null, "}\n\n#### Steps to convert :\n\nStep 1: Find", null, "– closure of the start state of NFA and that will be the start state of DFA.", null, "– closure of start state of NFA.", null, "– closure (A) : {A,B,C}\n\nSteps 2,3: Starting with this set, for each alphabet, evaluate", null, "– closure of the transition set for this alphabet. and  For each new set of closure set we come across, we will repeat Step 2 until no new set is left.\n\nCurrent set of states of DFA: {ABC}\n\n• {ABC} -> a = A -> a : ABC\n• {ABC} -> b = BD -> b : BD\n• {ABC} -> c = CD -> c : CD\n\nCurrent set of states of DFA: {ABC, BD, CD}\n\n• {BD} -> a =", null, "• {BD} -> b = BD\n• {BD} -> c =", null, "• {CD} -> a =", null, "• {CD} -> b =", null, "• {CD} -> c = CD\n\nStates of DFA, Q : {ABC, BD, CD,", null, "}\n\nTransition Function", null, ":\n\nStep 4: The set in DFA which contains the final state of NFA will be the final set state.\n\nD was the final state in NFA. So all the states having D in its set will be final state in DFA.\n\nSo final states of DFA, F : {BD, CD}\n\nDFA obtained :", null, "DFA", null, "is also known as dead state because there is no outgoing edge from it. So after machine arrives in the dead state, it cannot reach the final state.\n\nTools for implementation with code:\n\nGraphviz: It is a python library for visualizing graph diagrams.\n\nTo install graphviz in python, run this command in the terminal:\n\npip install graphviz\n\nPrerequisite:\n\nInputs\n\n1. Number of States : no_state\n2. Array of States : states\n3. Number of Alphabets : no_alphabet\n4. Array of Alphabets : alphabets\n5. Start State : start\n6. Number of Final States : no_final\n7. Array of Final States : finals\n8. Number of Transitions : no_transition\n9. Array of Transitions : transitions\nTransitions are of type : [From State, Alphabet, To State]\n\nUtility :\n\n1. Dictionary/Map to get Index from state : states_dict\n2. Dictionary/Map to get Index from alphabet : alphabets_dict\n3. Transition Table Dictionary to get array of ‘to’ states from ‘state’ and ‘alphabet’ pair : transition_table\n4. Digraph object to store graph of the NFA : graph\n\nMethods/Functions :\n\n1. Constructor of the class NFA : __init__()\nIt initializes all the input variables and evaluates the utility variables values from input.\n2. Get input from user : fromUser()\nClass Method to get input from user.\n3. Representation of the quintuple of NFA : __repr__()\n4. Find Epsilon Closure of a state : getEpsilonClosure(state)\nTo find epsilon closure of a state, we maintain a stack to get what state to evaluate next and a dictionary to track which states have been evaluated.\nWe start a while loop from the start state, and find its epsilon transition states.\nWe push all this states to the stack (stack.push(stata)) and mark them in the dictionary (dict[state]=0).\nWe also mark this state complete in the dictionary (dict[state]=1).\nFor each next iteration, we pop the top of the stack each time and evaluate it if it is not evaluated by checking from the dictionary.\n5. Find appropriate name of state for diagram of converted DFA from array of states : getStateName(state_list)\nAs we will get a list of states from the evaluation, to display in the DFA diagram, we need a proper name which will be concatenation of all the names in te set.\nFor ex : To state set is ={A,B,D} , then this function will return a string = “ABD”.\n6. Check if the array contains a final state of NFA to find if the array will be final state in DFA : isFinalDFA(state_list)\nThis function checks if a list of states contains a state which is a final state in NFA, which in turn will tell if the set is final or not.\nFor ex : The set we are checking is = {A,B,D}, and D is a final state in NFA. So this function will return True for this input set. So this set will be final state in DFA.\n\nApproach:\n\n1. Make an object of NFA class: nfa, and initialize it with predefined values using a constructor or using user input.\n2. Initialize the nfa.graph with nodes and edges according to the input values.\n3. Display/Render the NFA graph.\n4. Make another Digraph object to store values of the obtained DFA and to render the diagram: dfa.\n5. Evaluate epsilon closure of all the states of NFA to not recalculate each time and store it in the dictionary with key-value pair as [state -> list of states of closure]: epsilon_closure{}\n6. Make a stack to track which DFA state to evaluate next: dfa_stack[]\n7. Add the epsilon closure of the start state of NFA as the start state of DFA.\n8. Make a list to maintain all the states present in the DFA : dfa_states[]\n9. Start a while loop which we continue till there are no new states in dfa_stack.\n10. We pop the top of dfa_stack to evaluate the current set of states: cur_state.\n11. We traverse through all the alphabets for the current set of states.\n12. We make a set to maintain the epsilon closure of all the states in the current set : from_closure{}\n13. If this set is not empty,\n• We make another set to maintain the to_state set.\n• If this set is not present in dfa_states, then we append it in dfa_stack and dfa_states.\n• Then we add this node in dfa graph and add an edge between cur_state and to_state.\n14. Else this set is empty, then\n• This case is for dead state.\n• If dead state not present in dfa, then we add the new state", null, ". And we make all transitions for all alphabets to itself so that the machine can never leave the dead state.\n• We make transition of cur_state to this dead_state.\n15. At last, all the states have been evaluated. So we will render/view this dfa graph.\n\nBelow is the full implementation:\n\n## Python3\n\n # Conversion of epsilon-NFA to DFA and visualization using Graphviz from graphviz import Digraph class NFA:    def __init__(self, no_state, states, no_alphabet, alphabets, start,                 no_final, finals, no_transition, transitions):        self.no_state = no_state        self.states = states        self.no_alphabet = no_alphabet        self.alphabets = alphabets                 # Adding epsilon alphabet to the list        # and incrementing the alphabet count        self.alphabets.append('e')        self.no_alphabet += 1        self.start = start        self.no_final = no_final        self.finals = finals        self.no_transition = no_transition        self.transitions = transitions        self.graph = Digraph()         # Dictionaries to get index of states or alphabets        self.states_dict = dict()        for i in range(self.no_state):            self.states_dict[self.states[i]] = i        self.alphabets_dict = dict()        for i in range(self.no_alphabet):            self.alphabets_dict[self.alphabets[i]] = i                     # transition table is of the form        # [From State + Alphabet pair] -> [Set of To States]        self.transition_table = dict()        for i in range(self.no_state):            for j in range(self.no_alphabet):                self.transition_table[str(i)+str(j)] = []        for i in range(self.no_transition):            self.transition_table[str(self.states_dict[self.transitions[i]])                                  + str(self.alphabets_dict[                                      self.transitions[i]])].append(                                          self.states_dict[self.transitions[i]])     # Method to get input from User    @classmethod    def fromUser(cls):        no_state = int(input(\"Number of States : \"))        states = list(input(\"States : \").split())        no_alphabet = int(input(\"Number of Alphabets : \"))        alphabets = list(input(\"Alphabets : \").split())        start = input(\"Start State : \")        no_final = int(input(\"Number of Final States : \"))        finals = list(input(\"Final States : \").split())        no_transition = int(input(\"Number of Transitions : \"))        transitions = list()        print(\"Enter Transitions (from alphabet to) (e for epsilon): \")        for i in range(no_transition):            transitions.append(input(\"-> \").split())        return cls(no_state, states, no_alphabet, alphabets, start,                   no_final, finals, no_transition, transitions)     # Method to represent quintuple    def __repr__(self):        return \"Q : \" + str(self.states)+\"\\nΣ : \"        + str(self.alphabets)+\"\\nq0 : \"        + str(self.start)+\"\\nF : \"+str(self.finals) + \\            \"\\nδ : \\n\" + str(self.transition_table)     def getEpsilonClosure(self, state):               # Method to get Epsilon Closure of a state of NFA        # Make a dictionary to track if the state has been visited before        # And a array that will act as a stack to get the state to visit next        closure = dict()        closure[self.states_dict[state]] = 0        closure_stack = [self.states_dict[state]]         # While stack is not empty the loop will run        while (len(closure_stack) > 0):                       # Get the top of stack that will be evaluated now            cur = closure_stack.pop(0)                         # For the epsilon transition of that state,            # if not present in closure array then add to dict and push to stack            for x in self.transition_table[                    str(cur)+str(self.alphabets_dict['e'])]:                if x not in closure.keys():                    closure[x] = 0                    closure_stack.append(x)            closure[cur] = 1        return closure.keys()     def getStateName(self, state_list):               # Get name from set of states to display in the final DFA diagram        name = ''        for x in state_list:            name += self.states[x]        return name     def isFinalDFA(self, state_list):               # Method to check if the set of state is final state in DFA        # by checking if any of the set is a final state in NFA        for x in state_list:            for y in self.finals:                if (x == self.states_dict[y]):                    return True        return False  print(\"E-NFA to DFA\") # INPUT# Number of States : no_state# Array of States : states# Number of Alphabets : no_alphabet# Array of Alphabets : alphabets# Start State : start# Number of Final States : no_final# Array of Final States : finals# Number of Transitions : no_transition# Array of Transitions : transitions nfa = NFA(    4,  # number of states    ['A', 'B', 'C', 'D'],  # array of states    3,  # number of alphabets    ['a', 'b', 'c'],  # array of alphabets    'A',  # start state    1,  # number of final states    ['D'],  # array of final states    7,  # number of transitions    [['A', 'a', 'A'], ['A', 'e', 'B'], ['B', 'b', 'B'],     ['A', 'e', 'C'], ['C', 'c', 'C'], ['B', 'b', 'D'],     ['C', 'c', 'D']]       # array of transitions with its element of type :    # [from state, alphabet, to state]) # nfa = NFA.fromUser() # To get input from user# print(repr(nfa)) # To print the quintuple in console # Making an object of Digraph to visualize NFA diagramnfa.graph = Digraph() # Adding states/nodes in NFA diagramfor x in nfa.states:    # If state is not a final state, then border shape is single circle    # Else it is double circle    if (x not in nfa.finals):        nfa.graph.attr('node', shape='circle')        nfa.graph.node(x)    else:        nfa.graph.attr('node', shape='doublecircle')        nfa.graph.node(x) # Adding start state arrow in NFA diagramnfa.graph.attr('node', shape='none')nfa.graph.node('')nfa.graph.edge('', nfa.start) # Adding edge between states in NFA from the transitions arrayfor x in nfa.transitions:    nfa.graph.edge(x, x, label=('ε', x)[x != 'e']) # Makes a pdf with name nfa.graph.pdf and views the pdfnfa.graph.render('nfa', view=True) # Making an object of Digraph to visualize DFA diagramdfa = Digraph() # Finding epsilon closure beforehand so to not recalculate each timeepsilon_closure = dict()for x in nfa.states:    epsilon_closure[x] = list(nfa.getEpsilonClosure(x))  # First state of DFA will be epsilon closure of start state of NFA# This list will act as stack to maintain till when to evaluate the statesdfa_stack = list()dfa_stack.append(epsilon_closure[nfa.start]) # Check if start state is the final state in DFAif (nfa.isFinalDFA(dfa_stack)):    dfa.attr('node', shape='doublecircle')else:    dfa.attr('node', shape='circle')dfa.node(nfa.getStateName(dfa_stack)) # Adding start state arrow to start state in DFAdfa.attr('node', shape='none')dfa.node('')dfa.edge('', nfa.getStateName(dfa_stack)) # List to store the states of DFAdfa_states = list()dfa_states.append(epsilon_closure[nfa.start]) # Loop will run till this stack is not emptywhile (len(dfa_stack) > 0):    # Getting top of the stack for current evaluation    cur_state = dfa_stack.pop(0)     # Traversing through all the alphabets for evaluating transitions in DFA    for al in range((nfa.no_alphabet) - 1):        # Set to see if the epsilon closure of the set is empty or not        from_closure = set()        for x in cur_state:            # Performing Union update and adding all the new states in set            from_closure.update(                set(nfa.transition_table[str(x)+str(al)]))         # Check if epsilon closure of the new set is not empty        if (len(from_closure) > 0):            # Set for the To state set in DFA            to_state = set()            for x in list(from_closure):                to_state.update(set(epsilon_closure[nfa.states[x]]))             # Check if the to state already exists in DFA and if not then add it            if list(to_state) not in dfa_states:                dfa_stack.append(list(to_state))                dfa_states.append(list(to_state))                 # Check if this set contains final state of NFA                # to get if this set will be final state in DFA                if (nfa.isFinalDFA(list(to_state))):                    dfa.attr('node', shape='doublecircle')                else:                    dfa.attr('node', shape='circle')                dfa.node(nfa.getStateName(list(to_state)))             # Adding edge between from state and to state            dfa.edge(nfa.getStateName(cur_state),                     nfa.getStateName(list(to_state)),                     label=nfa.alphabets[al])                     # Else case for empty epsilon closure        # This is a dead state(ϕ) in DFA        else:                       # Check if any dead state was present before this            # if not then make a new dead state ϕ            if (-1) not in dfa_states:                dfa.attr('node', shape='circle')                dfa.node('ϕ')                 # For new dead state, add all transitions to itself,                # so that machine cannot leave the dead state                for alpha in range(nfa.no_alphabet - 1):                    dfa.edge('ϕ', 'ϕ', nfa.alphabets[alpha])                 # Adding -1 to list to mark that dead state is present                dfa_states.append(-1)             # Adding transition to dead state            dfa.edge(nfa.getStateName(cur_state,),                     'ϕ', label = nfa.alphabets[al]) # Makes a pdf with name dfa.pdf and views the pdfdfa.render('dfa', view = True)\n\nOutput:", null, "DFA\n\nThe complexity of algorithm to convert NFA to DFA :\n\nTime Complexity:", null, "Space Complexity:", null, "Where, Q = No. of States of NFA , N = No. of Alphabets of NFA.\n\nMy Personal Notes arrow_drop_up" ]
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https://jb.asm.org/highwire/markup/188084/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed
[ "TABLE 2.\n\nDiscrepancies between experimental identification and model prediction for essential and nonessential genesa\n\nRationaleSubsystemGene(s) (Blattner no.)\nFalse negatives\nMolecules not included in biomassUbiquinone biosynthesisubiG (b2232); ubiH (b2907)\nPyridoxine biosynthesispdxA (b0052); pdxB (b2320); pdxH (b1638); pdxJ (b2564)\nThiamine biosynthesisiscC (b2530)\nModel includes alternative pathways/     isozymesAmino acid biosynthesiscarA (b0032); carB (b0033); glpD (b3426); glyA (b2551); proA (b0243); proB (b0242); thrB (b0003); thrC (b0004)\nModel predicts impaired but not lethal     phenotypeATP synthaseatpA (b3734); atpB (b3738); atpC (b3731); atpF (b3736); atpG (b3733); atpH (b3735)\nRegulatory effect on glpKPTS/PEP metabolismcrr (b2417); glpK (b3926); ppc (b3956); ptsI (b2416); fruR (b0080)\nFalse positives\nModel biomass components which     might not be essential componentsFatty acid and lipid biosynthesiscls (b1249); fabF (b1095)\nGlycogenglgA (b3429); glgC (b3430)\nLPS synthesisdgkA (b4042); gmhA (b0222); gmhB (b0200); lpxL (b1054); msbB (b1855); rfaC (b3621); rfaD (b3619); rfaE (b3052); rfaF (b3620); rfaG (b3631); rfaI (b3627); rfaJ (b3626); rfaL (b3622)\nSpermidine synthesispfs (b0159); speD (b0120); speE (b0121)\nUnaccounted-for transport mechanismsAmmonium transportamtB (b0451)\nGlycerol transportglpF (b3927)\nSulfate transportcysW (b2423)\nUnaccounted-for metabolic enzymesArginine biosynthesisargB (b3959); argC (b3958); argD (b3359); argG (b3172)\nAspartate biosynthesisaspC (b0928)\nBranced amino acid biosynthesisilvY (b3773); ilvE (b3770); lrp (b0889)\nCentral metabolicaldA (b1415)\nCofactor biosynthesiscoaA (b3974); coaE (b0103); pabC (b1096)\nGlycolyticpgi (b4025)\nLysine biosynthesisdapF (b3809); ushA (b0480); lysR (b2839)\nNucleotide biosynthesis and salvagepyrI (b4244); trxB (b0888); ndk (b2518)\n• a Twenty-six false-negative cases in which the model incorrectly predicted growth of the gene deletion strain were identified, in addition to 42 false-positive cases in which the model incorrectly predicted that genes were essential. Each case is grouped based on the likely rationale for the discrepancy and the gene functional annotation." ]
[ null ]
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https://bmcsystbiol.biomedcentral.com/articles/10.1186/1752-0509-6-42/tables/1
[ "PFO: pyruvate ferredoxin oxidoreductase $C o A + p y r u v a t e + F d Ox → C O 2 + a c e t y l − C o A + H + + F d Red$ 0 1000\nHYDA: hydrogenase $2 H + + F d Red → H 2 + F d Ox$ 0 1000\nFNO: ferredoxin NAD+ oxidoreducatase $N A D + + H + + F d Red → N A D H + F d Ox$ 0 1000\nFNPO: ferredoxin NADP+ oxidoreductase $N A D P + + H + + F d Red → N A D P H + F d Ox$ 0 1000" ]
[ null ]
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http://www.numbersaplenty.com/1112210001
[ "Search a number\nBaseRepresentation\nbin100001001001010…\n…1111101001010001\n32212111211010022000\n41002102233221101\n54234211210001\n6302210304213\n736363425520\noct10222575121\n92774733260\n101112210001\n115208a4921\n12270588069\n13149567436\n14a79da4b7\n1567998a86\nhex424afa51\n\n1112210001 has 32 divisors (see below), whose sum is σ = 1948051200. Its totient is φ = 613630080.\n\nThe previous prime is 1112209963. The next prime is 1112210027. The reversal of 1112210001 is 1000122111.\n\nAdding to 1112210001 its reverse (1000122111), we get a palindrome (2112332112).\n\nIt is a happy number.\n\nIt is not a de Polignac number, because 1112210001 - 29 = 1112209489 is a prime.\n\nIt is a super-2 number, since 2×11122100012 = 2474022172648840002, which contains 22 as substring.\n\nIt is a Harshad number since it is a multiple of its sum of digits (9).\n\nIt is not an unprimeable number, because it can be changed into a prime (1112210041) by changing a digit.\n\nIt is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 95980 + ... + 106941.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (60876600).\n\nAlmost surely, 21112210001 is an apocalyptic number.\n\nIt is an amenable number.\n\n1112210001 is a deficient number, since it is larger than the sum of its proper divisors (835841199).\n\n1112210001 is a wasteful number, since it uses less digits than its factorization.\n\n1112210001 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 202966 (or 202960 counting only the distinct ones).\n\nThe product of its (nonzero) digits is 4, while the sum is 9.\n\nThe square root of 1112210001 is about 33349.8126081692. The cubic root of 1112210001 is about 1036.0855067864.\n\nThe spelling of 1112210001 in words is \"one billion, one hundred twelve million, two hundred ten thousand, one\"." ]
[ null ]
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https://byjus.com/ncert-solutions-for-class-10-maths-chapter-15-probability-ex-15-1/
[ "", null, "# NCERT Solutions for class 10 Maths Chapter 15- Exercise 15.1\n\n## NCERT Solutions For Class 10 Maths Ex 15.1 PDF Free Download\n\nExercise 15.1 of NCERT Solutions for class 10 Maths Chapter 15- Probability is available here in the downloadable PDF format. These exercise questions are considered to be very important since it covers the basic concepts of probability, including an event, possible outcome of an event, etc. There are 25 questions in this exercise which covers all the concepts present in the chapter.\n\nThe subject experts at BYJU’S solve these questions keeping the NCERT Syllabus and guidelines in mind. These NCERT solutions for class 10 Maths are the best study material for the Class 10 students for the board exam preparation.\n\n### Access Answers of Maths NCERT class 10 Chapter 15 – Probability Exercise 15.1", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "### Access other exercise solutions of class 10 Maths Chapter 15- Probability\n\nExercise 15.2 Solutions 8 Question ( 7 long, 1 short)\n\n## NCERT Solutions for class 10 Maths Chapter 15- Probability Exercise 15.1\n\nSome of the topics that Exercise 15.1 of 10th standard Maths cover are as follows:\n\n• A Theoretical Approach to probability: In theoretical approach, as the name suggests, the occurrence of an event is predicted without actually performing the experiment. In such cases, we assume that the outcomes of an event are equally likely. Here, we infer that the experimental probability of an event approaches its theoretical probability if the number of trials of an experiment is very large.\n• Random Experiment: An experiment is said to be random when the result varies as an experiment is repeated several times. These repetitions are called trials. The theoretical probability (classical probability) of an event ‘E’, P(E), is defined as the ratio of Number of outcomes favourable to E to that of the number of all possible outcomes of the experiment.\n• Compound Event: An event, related to a random experiment is said to be a compound event if it is obtained by combining two or more elementary events related to the random experiment.\n\nHere, solving the problems from Exercise 15.1 of chapter 15, Probability, will give the students an idea about the points mentioned above. Similarly, Solving all the problems in the NCERT solutions will give an idea about the concepts covered in all the chapters along with helping the students improve their problem-solving skills." ]
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https://www2.mdpi.com/1099-4300/25/4/588
[ "", null, "Next Article in Journal\nSynchrony-Division Neural Multiplexing: An Encoding Model\nPrevious Article in Journal\nGL-YOLO-Lite: A Novel Lightweight Fallen Person Detection Model\nPrevious Article in Special Issue\nJoint Design of Polar Coding and Physical Network Coding for Two−User Downlink Non−Orthogonal Multiple Access\n\nFont Type:\nArial Georgia Verdana\nFont Size:\nAa Aa Aa\nLine Spacing:\nColumn Width:\nBackground:\nArticle\n\n# Lossy State Communication over Fading Multiple Access Channels\n\nKTH Royal Institute of Technology, 114 28 Stockholm, Sweden\nEntropy 2023, 25(4), 588; https://doi.org/10.3390/e25040588\nReceived: 26 February 2023 / Revised: 25 March 2023 / Accepted: 28 March 2023 / Published: 29 March 2023\n\n## Abstract\n\n:\nJoint communications and sensing functionalities integrated into the same communication network have become increasingly relevant due to the large bandwidth requirements of next-generation wireless communication systems and the impending spectral shortage. While there exist system-level guidelines and waveform design specifications for such systems, an information-theoretic analysis of the absolute performance capabilities of joint sensing and communication systems that take into account practical limitations such as fading has not been addressed in the literature. Motivated by this, we undertake a network information-theoretic analysis of a typical joint communications and sensing system in this paper. Towards this end, we consider a state-dependent fading Gaussian multiple access channel (GMAC) setup with an additive state. The state process is assumed to be independent and identically distributed (i.i.d.) Gaussian, and non-causally available to all the transmitting nodes. The fading gains on the respective links are assumed to be stationary and ergodic and available only at the receiver. In this setting, with no knowledge of fading gains at the transmitters, we are interested in joint message communication and estimation of the state at the receiver to meet a target distortion in the mean-squared error sense. Our main contribution here is a complete characterization of the distortion-rate trade-off region between the communication rates and the state estimation distortion for a two-sender GMAC. Our results show that the optimal strategy is based on static power allocation and involves uncoded transmissions to amplify the state, along with the superposition of the digital message streams using appropriate Gaussian codebooks and dirty paper coding (DPC). This acts as a design directive for realistic systems using joint sensing and transmission in next-generation wireless standards and points to the relative benefits of uncoded communications and joint source-channel coding in such systems.\n\n## 1. Introduction\n\nThe scarcity of spectrum, as well as the bandwidth requirements of key emerging applications such as 6G, necessitate a rethinking of resource consumption. In such systems, it appears prudent to co-design sensing and communication functionalities. This method enables significant gains in spectral, energy, hardware, and cost efficiency. This is known as joint sensing and communication, and it represents a paradigm shift in which sensing and communication operations can be jointly optimized by utilizing a single hardware platform and a joint signal processing framework. These ideas have already been used in a number of novel applications, including vehicular networks, indoor positioning, and covert communications. Joint sensing and communication scenarios have recently received a lot of attention from the signal processing community (see for instance [1,2,3,4]), the communications community (see [5,6,7,8,9,10,11]), and the information theory community (see [12,13,14,15,16]). This work belongs to the final category, where we take an information-theoretic view of joint sensing and communication in a multi-terminal setting.\nJoint sensing and communication also arises in multi-user networks with several sensor nodes observing a common analog source phenomenon, and communicating to a base station (destination) over a wireless fading medium, see Figure 1.\nIn this setting, the sensor nodes must convey a description of the source process to the base station, which then tries to estimate the source process subject to a fidelity criterion. Some of the sensor nodes might also have additional digital data to convey to the base station, which must reliably recover them as well. Since the source process, as well as the data from each node, are of interest to the base station, a tension naturally arises between the rates of data communication and source estimation fidelity. The trade-off between these objectives is of particular interest in such systems, which is among the primary motivations for this work. The analog phenomenon in this example can be thought of as a channel state that affects the digital communication of messages, with the receiver being required to reliably estimate this channel state while also recovering the transmitted messages. As far as the fading process is concerned, it is reasonable in practice to assume that the receiver can track the channel variations, for example, via the use of pilot transmission sequences.\nIn this work, we consider an information-theoretic abstraction of the communication setting in Figure 1. In particular, we focus on joint communication and state estimation over a state-dependent fading Gaussian multiple access channel with no fading knowledge at the transmitters. At each encoder, the state process is assumed to be known non-causally. The fading processes encountered on the respective links are assumed to be stationary and ergodic and to be known only at the receiver. The dual goals of message communication and state estimation at the receiver must be met with a distortion tolerance with respect to a squared-error metric. The trade-off between the average message communication rates and the average distortion in receiver state estimation is of interest. We completely characterize the optimal trade-off region between the communication rates of the different transmitters and the state estimation distortion at the receiver. The details of the setting as well as the motivation for investigating it, will be elucidated in the section that follows.\nHaving introduced the general problem framework, we now discuss the other relevant contributions in the literature, emphasizing the differences from our setting under consideration in the following section.\n\n## 2. Literature Review\n\nIn this section, we discuss the related literature and place our contributions in the context of the state of the art. In particular, we enlist prior works on joint communication and channel state estimation in both point-to-point as well as network information theoretic settings and identify several knowledge gaps.\nSystems such as the one in Figure 1 can be modeled as state-dependent channels, where the channel state typically refers to a variable used to model unknown parameters of the channel statistics. A canonical form of such state-dependent channel models consisting of an additive state over an additive white Gaussian noise (AWGN) channel was investigated in , popularly known as the dirty paper coding (DPC) setting. Surprisingly, ref. demonstrated that regardless of the presence of the state, the channel capacity of this setting remained the same as that of an AWGN channel, independent of the variance of the channel state. This phenomenon later found widespread applications in settings such as digital watermarking and multiple-input-multiple-output wireless broadcast channels .\nIn certain state-dependent channels, in addition to communicating messages, the transmitter may wish to assist the receiver in estimating the channel state (as in the sensor network scenario described in Figure 1). Splitting the average available power between the dual tasks of uncoded transmission of the state and DPC for the message was found to be optimal for the mean squared error distortion measure in a point-to-point (single-user) AWGN channel. In , joint communication and state estimation were considered in a different scenario where the transmitters were unaware of the channel state. In an interesting variation of , Tian et al. characterized the distortion-transmit power trade-off in a point-to-point Gaussian model with noisy state observations at the transmitter in the absence of messages. However, in the presence of messages , a complete characterization of the rate-distortion trade-off region remains unknown for the case of noisy state observations at the transmitter.\nAs far as channel state estimation at the transmitter is concerned, this line of work originated in , where a point-to-point channel with generalized feedback signals to aid the state estimation was investigated. Such models are motivated by joint radar and communications systems where the radar, as well as data communication, share the same frequency band. Following this, ref. considered a multiple access channel extension of the same, where both the senders obtained generalized feedback and obtained an achievable trade-off region. An improved achievable scheme for the multiple access setting was derived recently in . Most recently, a broadcast channel variant was investigated in (see also ), where inner and outer bounds were given for general broadcast channels while a complete characterization was obtained for the special case of physically degraded broadcast channels.\nFading multiple access channels without state (or state estimation requirements) and different degrees of channel state information have been explored in the literature. For instance, the ergodic capacity region for fast-fading Gaussian multiple access channels (GMAC) with perfect channel state information at the transmitters and the receiver was characterized in . More general configurations for transmitter channel state availability were analyzed in , where the capacity region for time-varying models was determined via optimization over appropriate power control laws. Slow fading multiple access channels with distributed channel state information at the transmitters were studied in .\nState-dependent point-to-point fading channels with no state estimation have received some attention in the literature. Vaze and Varanasi , for example, investigated a model with full state knowledge and partial knowledge of the fading process at the transmitter, and the high-SNR achievable rate was characterized. Rini and Shamai examined the impact of phase fading in the DPC setting when the receiver was informed of the fading process. We also note that has addressed point-to-point fading channels (without any state process or state estimation requirements) with channel gains known at both the sender and the receiver.\nHaving reviewed the related literature, we now identify the key knowledge gaps in prior work and the necessity of our work.\n• Analysis of the State of the art and Research gaps: We identify the following crucial aspects.\n• We note that none of the works above consider joint state estimation along with message communication over state-dependent multi-terminal settings with noncausal transmitter state information, which is highly relevant in applications like the joint sensing and communication setting in Figure 1. This is addressed in this paper.\n• While there exist system-level guidelines and waveform design specifications for such systems, a network information-theoretic analysis of the absolute performance capabilities of joint sensing and communication systems that take into account practical limitations has not been addressed in the literature, which we undertake here.\n• Moreover, none of the works on joint communication and estimation mentioned above take fading links into account. Fading is an impairment that must be accounted for in practical wireless communication channel models, such as the joint sensing and communication application shown in Figure 1. This is another gap in the literature that this paper seeks to fill by investigating joint communication and estimation over state-dependent multi-user fading channels, the point-to-point counterpart of which was addressed by the author in .\nNovelty and relevance: In this paper, we address the problem of joint communication and state estimation over a state-dependent fading GMAC with no fading knowledge at the transmitters. The key scientific question we address here is: what is the best possible trade-off between the competing goals of message communication from multiple senders and the fidelity in state estimation at the receiver?\n• The key novelty of our work is that it is the first instance where joint communication and estimation have been considered in a multiple-user setting that also accounts for fading links, as opposed to previous works, which focused only on non-fading links.\n• Moreover, it is the first work that considers non-causal state information (as opposed to causal or strictly causal) at the transmitter in a fading multi-user scenario which is practically relevant as described in the sensor network example from Figure 1.\n• Furthermore, we undertake a comprehensive network information-theoretic study of the fundamental performance limits of such joint communication and estimation settings, which is lacking in the literature. Please refer to Table 1, which highlights our contributions in this paper with respect to the existing works.\nThe key relevance of our study is that it serves as a design guideline for practical systems employing joint sensing and communication envisioned in future 6G wireless standards and broadly applies to systems that involve joint compression and communication/rate-distortion trade-offs. It also points to the relative benefits of uncoded transmission versus joint source-channel coding in such systems. The progress embodied herein builds up towards a better understanding of joint state estimation and communication problems in multi-terminal settings (such as multiple access channels), which is relatively less explored in the literature (with or without the fading aspect).\n• Summary of contributions: We list them below. See also the contribution summary Table 1, which emphasizes the novelty of our work with respect to the existing works.\n• One of our main contributions in the paper is a complete characterization (Theorem 1) of the rate-distortion trade-off region for joint state estimation and communication over a two-user fading GMAC with the state. The key non-trivial part is the proof of converse, which is given in Section 5.\n• We prove that the optimal strategy for the setting under consideration involves uncoded transmissions to amplify the state, along with the superposition of the digital message streams using appropriate Gaussian codebooks and DPC.\n• We prove the optimality of uncoded state amplification in the special case where there are no messages to communicate—please refer to the section on implications of our result given after the statement of Theorem 1 for the details.\n• Our framework naturally generalizes the results of to multiple users, ref. to fading links and the work of to fading links with multiple users, thereby providing a unified framework that encompasses all these prior works on joint communication and estimation.\n• Our study gives a network information-theoretic analysis of the fundamental performance limits of joint sensing and communication systems that take into account practical limitations such as fading. This acts as a design directive for realistic systems using joint sensing and transmission anticipated in upcoming wireless standards and points to the relative merits of uncoded communications and joint source-channel coding in such systems.\nNotations: Random variables are denoted by capital letters, while their realizations are denoted by the corresponding lower-case letters. We use $P ( · )$ to denote the probability of an event. The joint probability distribution of two random variables $( X , Y )$ is denoted by $p X , Y ( x , y )$. Let $E [ · ]$ denote the expected value of a random variable. At times, we will use an explicit subscript in the expectation operation, $E X [ · ]$, to denote that the expectation is taken with respect to the probability distribution of the random variable X. All logarithms are in base 2, unless mentioned otherwise. We denote random sequences of length n with a superscript notation, i.e., $U n : = U 1 , U 2 , ⋯ , U n$. An indexed set of random sequences each of length n is denoted with a subscript for the random variable and a superscript for the length, i.e., $U j n : = U j 1 , U j 2 , ⋯ , U j n$, where $U j i$ stands for the $i$-th component of $U j n$. The covariance of a random vector $X n$ is denoted by $Cov ( X n )$. Calligraphic letters represent alphabets of random variables. $∥ . ∥$ denotes the Euclidean norm of a vector, while $Conv ( · )$ denotes the convex closure of a set. The absolute value of a number is denoted $| · |$, while the transpose of a matrix A is denoted as $A T$. The notation $A ⫫ B$ is used to denote independent random variables $( A , B )$. The Gaussian (normal) distribution with mean $μ$ and variance $σ 2$ is denoted by $N ( μ , σ 2 )$. The set of real numbers is denoted by $R$, while the set of $n −$tuples of positive real numbers is denoted by $R + n$. The Shannon entropy of a discrete-valued random variable X is denoted by $H ( X )$, while the differential entropy of a continuous-valued random variable Y is denoted by $h ( Y )$. The mutual information between any two random variables V and W is denoted by $I ( V ; W )$. The corresponding conditional quantities given a random variable Z are conditional entropy $H ( X | Z )$, conditional differential entropy $h ( Y | Z )$, and conditional mutual information $I ( V ; W | Z )$. If n is an integer variable, $ϕ ( n )$ is a positive function and $f ( n )$ is an arbitrary function, we say that $f = o ( ϕ )$ provided that $lim n → ∞ f ( n ) / ϕ ( n ) = 0$. For any three random variables $( A , B , C )$, we say that $A → B → C$ is a Markov chain if A and C are conditionally independent given B.\nThe rest of the paper is organized as follows: in Section 3, we introduce the system model and state our main results. Section 4 and Section 5 contain the achievable coding scheme and the converse to the rate-distortion trade-off region, respectively. Section 6 contains concluding remarks. The Appendix A.1, Appendix A.2, and Appendix A.3 contain all the details of the proofs that are skipped in the main discussion, to maintain the readability of the paper.\n\n## 3. System Model and Results\n\nConsider the fading multiple access channel shown in Figure 2. The observed samples at the receiver at time instant $i ∈ { 1 , 2 , … , n }$ are given by\n$Y i = ∑ j = 1 2 Θ j , i X j , i + S i + Z i .$\nHere, the samples of the additive noise process are independent and identically distributed (i.i.d.) according to $Z i ∼ N ( 0 , σ Z 2 )$, while the samples of the state process are i.i.d. according to $S i ∼ N ( 0 , σ S 2 )$. The state process is assumed to be known non-causally at each encoder. The fading processes encountered on the respective links are represented by $Θ j n , j ∈ { 1 , 2 }$, with these fading gains being known only at the receiver. The fading processes encountered on both links are assumed to be stationary and ergodic. The codeword lengths can be chosen arbitrarily long to average over the fading of the channel. The given model represents a fast-fading multiple access channel with no knowledge of the fading processes at the transmitters. The state, fading, and additive noise processes are assumed to be independent of each other. In our model, the power constraint on the inputs is assumed to be across blocks, averaged over the random state and the codebook. The dual goals of message ($( W 1 , W 2 )$ in Figure 2) communication, and estimation of the state $S n$ at the receiver to meet a distortion tolerance with respect to a squared error metric must be met. The trade-off between the average message communication rates ($R 1 , R 2 )$ and the average distortion incurred in state estimation (D) at the receiver is sought.\nWe take $W j$ to be uniformly drawn from the set $W j ≜ { 1 , 2 , ⋯ , 2 n R j }$ for $j = { 1 , 2 }$, and independent of each other. The messages $( W 1 , W 2 )$ are assumed to be independent of the state process $S n$. The power constraint on the transmissions is:\n$1 n E W j , S n ∑ i = 1 n X j i 2 ( W j , S n ) ≤ P j , j = { 1 , 2 } .$\nAfter n observations, the decoder estimates $S ^ n = ϕ ( Y n , Θ 1 n , Θ 2 n )$ using a (state) reconstruction map $ϕ ( · ) : Y n × ∏ j = 1 2 Θ j n → R n$, and also decodes the messages $( W 1 , W 2 )$. The (message) decoding map is given by $ψ : Y n × ∏ j = 1 2 Θ j n → W 1 × W 2$. Our aim here is to maintain the average squared-error distortion incurred in state estimation below a given threshold, while also ensuring that the average error probability of decoding the messages is small enough.\nDefinition 1.\nA scheme using the encoder mappings $E j : { 1 , ⋯ , 2 n R j } × S n → X j$ for $j = 1 , 2$ satisfying the power constraints in (2), along with two mappings $ϕ ( · )$ and $ψ ( · )$ at the receiver is called an $( n , R 1 , R 2 , P 1 , P 2 )$ communication-estimation scheme.\nA triple $( R 1 , R 2 , D )$ is said to be achievable if there exists a sequence of $( n , R 1 , R 2 , P 1 , P 2 )$ communication-estimation schemes such that\n$lim sup n → ∞ 1 n E Θ 1 n , Θ 2 n E [ ∥ S n − ϕ ( Y n , Θ 1 n , Θ 2 n ) ∥ 2 ] ≤ D ,$\nand\n$lim sup n → ∞ E Θ 1 n , Θ 2 n P ( ψ ( Y n , Θ 1 n , Θ 2 n ) ≠ ( W 1 , W 2 ) ) = 0 .$\nLet $C est fad - mac ( P 1 , P 2 )$ be the closure of the set of all achievable $( R 1 , R 2 , D )$ triples, with $0 ≤ D ≤ σ S 2$. The main result of the paper is stated below.\nTheorem 1.\nFor the fading Gaussian MAC with state, the trade-off region $C est fad - mac ( P 1 , P 2 )$ is characterized by the convex closure of all $( R 1 , R 2 , D ) ∈ R + 3$ such that\n$R 1 ≤ E Θ 1 1 2 log 1 + γ Θ 1 2 P 1 σ Z 2 ,$\n$R 2 ≤ E Θ 2 1 2 log 1 + β Θ 2 2 P 2 σ Z 2 ,$\n$R 1 + R 2 ≤ E Θ 1 , Θ 2 1 2 log 1 + γ Θ 1 2 P 1 + β Θ 2 2 P 2 σ Z 2 ,$\n$D ≥ E Θ 1 , Θ 2 σ S 2 ( σ Z 2 + γ Θ 1 2 P 1 + β Θ 2 2 P 2 ) Θ 1 2 P 1 + Θ 2 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 γ ¯ P 1 σ S 2 + 2 Θ 2 β ¯ P 2 σ S 2 + 2 Θ 1 Θ 2 γ ¯ β ¯ P 1 P 2 ,$\nfor some fractions $γ ∈ [ 0 , 1 ]$ and $β ∈ [ 0 , 1 ]$, with $γ ¯ ≜ 1 − γ$ and $β ¯ ≜ 1 − β$.\nProof.\nThe proof is given in the following two sections, wherein Section 4 contains the achievability proof while the converse is proved in Section 5. □\nImplications of our result: We now discuss the main consequences of our Theorem 1 for the sensor network scenario outlined earlier in Figure 1. If a given transmitter (sensor node) has a message to communicate to the receiver (base station), then the optimal strategy involves splitting its available power budget into two parts: one part is used to send a scaled version of the state (uncoded state amplification), while the other part is used to communicate the message using dirty paper coding. The parameters $γ$ and $β$ in Theorem 1 precisely perform this role of power-sharing between the dual goals of communication and estimation. This will be elaborated upon in the proof of achievability in Section 4.\nOn the other hand, if a given transmitter (sensor node) has no messages to communicate to the receiver (base station), then the optimal strategy simply involves utilizing its entire power budget for uncoded state amplification, i.e., sending the scaled version\n$X j = P j σ S 2 S .$\nIn other words, uncoded transmission is optimal for such nodes. This phenomenon resembles that of , albeit the latter is in the context of non-fading links with no message communication and no state-dependence. We close this section with a series of remarks that shed further light on the implications of our Theorem 1 and its connection with existing results in the literature.\nRemark 1.\nWhen the second sender is absent, i.e., $P 2 = 0$, and with constant fading gains $Θ 1 ( = Θ 2 ) = 1$ almost surely, our Theorem 1 recovers the characterization of for the point-to-point non-fading scenario as a special case.\nRemark 2.\nWhen the fading gains are constant, i.e., $Θ 1 = Θ 2 = 1$ almost surely, our Theorem 1 recovers the characterization of for the multiple-access non-fading scenario as a special case.\nRemark 3.\nWhen $γ = β = 0$, we obtain an extreme point of the region with zero rates, i.e, $R 1 = R 2 = 0$, and the best state estimate, i.e., minimum possible distortion\n$D min = E Θ 1 , Θ 2 σ S 2 σ Z 2 Θ 1 2 P 1 + Θ 2 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 P 1 σ S 2 + 2 Θ 2 P 2 σ S 2 + 2 Θ 1 Θ 2 P 1 P 2 .$\nThis corresponds to each encoder utilizing its entire power budget for uncoded state amplification, and therefore no message communication is possible.\nRemark 4.\nOn the other hand, when $γ = β = 1$, we obtain the other extreme point of the region with the maximum possible rates for a fading Gaussian MAC, and the worst state estimate, i.e., maximum possible distortion\n$R 1 ≤ R 1 , max = E Θ 1 1 2 log 1 + Θ 1 2 P 1 σ Z 2 ,$\n$R 2 ≤ R 2 , max = E Θ 2 1 2 log 1 + Θ 2 2 P 2 σ Z 2 ,$\n$R 1 + R 2 ≤ E Θ 1 , Θ 2 1 2 log 1 + Θ 1 2 P 1 + Θ 2 2 P 2 σ Z 2 ,$\n$D ≥ D max = E Θ 1 , Θ 2 σ S 2 ( σ Z 2 + Θ 1 2 P 1 + Θ 2 2 P 2 ) Θ 1 2 P 1 + Θ 2 2 P 2 + σ S 2 + σ Z 2 ,$\nThis corresponds to each encoder utilizing its entire power budget for message communication, and therefore no state amplification is possible, and maximum distortion is incurred in state estimation.\n\n## 4. Achievability\n\nThe achievability builds upon well-known techniques like dirty paper coding and successive cancellation, along with appropriate power splitting. The power $P 1$ available at encoder 1 is split into two parts: $γ P 1$ for message transmission and $γ ¯ P 1$ for state amplification, for some $γ ∈ [ 0 , 1 ]$. Similarly, the power $P 2$ available at the second encoder is split into $β P 2$ (message transmission) and $β ¯ P 2$ (state amplification) for some $β ∈ [ 0 , 1 ]$. Then, the following state amplification signals are generated\n$X 1 s j = γ ¯ P 1 σ S 2 S j and X 2 s j = β ¯ P 2 σ S 2 S j , 1 ≤ j ≤ n$\nat the respective encoders. In other words, the power fractions $γ ¯ P 1$ and $β ¯ P 2$ at encoders 1 and 2 respectively are used for uncoded state amplification. Hence, (1) can be rewritten as\n$Y j = Θ 1 j X 1 m j + Θ 1 j X 1 s j + Θ 2 j X 2 m j w w w + Θ 2 j X 2 s j + S j + Z j = Θ 1 j X 1 m j + Θ 2 j X 2 m j w w w + 1 + Θ 1 j γ ¯ P 1 σ S 2 + Θ 2 j β ¯ P 2 σ S 2 S j + Z j ,$\nwhere $E [ | | X 1 m n | | 2 ] ≤ n γ P 1$ and $E [ | | X 2 m n | | 2 ] ≤ n β P 2$, with both $X 1 m n$ and $X 2 m n$ being independent of $S n$. The subscript $m$ in (16) indicates that the corresponding signals are intended for message transmission, whereas the subscript $s$ indicates state amplification signals. To communicate the messages across to the receiver, we invoke the writing on dirty paper result for a Gaussian MAC .\nFrom the DPC result , we recall that a known state process over an AWGN channel can be completely canceled. In particular, a rate R that satisfies\n$R ≤ I ( U ; Y ) − I ( U ; S ) ,$\nwhen evaluated for some feasible joint probability distribution $p U , S , X ( u , s , x ) p Y | X , S ( y | x , s )$, can be achieved by Gelfand-Pinsker coding for a point-to-point channel with a non-causally known state. In order to prove the achievability of the rates (5)–(7), we first consider a dirty paper channel with input $Θ 1 j X 1 m j$, known state\n$S j ′ = 1 + Θ 1 j γ ¯ P 1 / σ S 2 + Θ 2 j β ¯ P 2 / σ S 2 S j ,$\nand unknown noise $Θ 2 j X 2 m j + Z j$. We choose $U 1 j = Θ 1 j X 1 m j + α 1 j S j ′$, $X 1 m j ⫫ S j$ with $X 1 m j ∼ N ( 0 , γ P 1 )$ and\n$α 1 j = γ Θ 1 j 2 P 1 γ Θ 1 j 2 P 1 + β Θ 2 j 2 P 2 + σ Z 2 .$\nThis yields the following rate for user-1 at time instant j with the error probability approaching zero\n$1 2 log 1 + γ Θ 1 j 2 P 1 β Θ 2 j 2 P 2 + σ Z 2 .$\nThe achievable rate for user-1 averaged over a time interval ${ 1 , 2 , … , n }$ is\n$1 n ∑ j = 1 n 1 2 log 1 + γ Θ 1 j 2 P 1 β Θ 2 j 2 P 2 + σ Z 2 ,$\nwhich converges as $n → ∞$ to\n$E 1 2 log 1 + γ Θ 1 2 P 1 β Θ 2 2 P 2 + σ Z 2$\ndue to the stationarity and ergodicity of the fading processes. The decoded codeword $U 1 j$ is then subtracted from the channel output to obtain another dirty paper channel\n$Y ˜ j = Y j − U 1 j = Θ 2 j X 2 m j + ( 1 − α 1 j ) S j ′ + Z j ,$\nwith input $Θ 2 j X 2 m j$, known state $S j ′ ′ = ( 1 − α 1 j ) S j ′$ and unknown noise $Z j$. We pick $U 2 j = Θ 2 j X 2 m j + α 2 j S j ′ ′$, $X 2 m j ⫫ S j$ with $X 2 m j ∼ N ( 0 , β P 2 )$ and\n$α 2 j = β Θ 2 j 2 P 2 β Θ 2 j 2 P 2 + σ Z 2 .$\nThis yields the following rate for user-2 at time instant j with the error probability approaching zero\n$1 2 log 1 + β Θ 2 j 2 P 2 σ Z 2 .$\nThe achievable rate for user-2 averaged over a time interval ${ 1 , 2 , … , n }$ is\n$1 n ∑ j = 1 n 1 2 log 1 + β Θ 2 j 2 P 2 σ Z 2 ,$\nwhich converges as $n → ∞$ to\n$E 1 2 log 1 + β Θ 2 2 P 2 σ Z 2$\ndue to the stationarity and ergodicity of the fading processes. By reversing the decoding order and using time-sharing, the region in expressions (5) through (7) can be achieved. Note that the right-hand sides of expressions (20) and (23) add up to the right-hand side of the sum rate expression in (7). For the state estimate, the receiver forms the linear minimum mean-squared error (MMSE) estimate $S ^ j ( Y j )$ of $S j$ based on $Y j$:\n$S ^ j ( Y j ) = ( σ S 2 + Θ 1 j γ ¯ P 1 σ S 2 + Θ 2 j β ¯ P 2 σ S 2 ) Y j Θ 1 j 2 P 1 + Θ 2 j 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 j γ ¯ P 1 σ S 2 + 2 Θ 2 j β ¯ P 2 σ S 2 + 2 Θ 1 j Θ 2 j γ ¯ β ¯ P 1 P 2 ≜ c 1 c 2 Y j ,$\nwhere\n$c 1 = σ S 2 + Θ 1 j γ ¯ P 1 σ S 2 + Θ 2 j β ¯ P 2 σ S 2 , c 2 = Θ 1 j 2 P 1 + Θ 2 j 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 j γ ¯ P 1 σ S 2 + 2 Θ 2 j β ¯ P 2 σ S 2 + 2 Θ 1 j Θ 2 j γ ¯ β ¯ P 1 P 2 .$\nFor the achievable distortion at time instant j, we evaluate the expected squared error between $S j$ and $S ^ j ( Y j )$, above. The resulting MMSE can be easily verified to be\n$E [ | S j − S ^ j | 2 ] = σ S 2 − c 1 2 c 2 = σ S 2 ( σ Z 2 + γ Θ 1 j 2 P 1 + β Θ 2 j 2 P 2 ) Θ 1 j 2 P 1 + Θ 2 j 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 j γ ¯ P 1 σ S 2 + 2 Θ 2 j β ¯ P 2 σ S 2 + 2 Θ 1 j Θ 2 j γ ¯ β ¯ P 1 P 2 .$\nThe achievable distortion averaged over a time interval ${ 1 , 2 , … , n }$ is\n$1 n ∑ j = 1 n σ S 2 ( σ Z 2 + γ Θ 1 j 2 P 1 + β Θ 2 j 2 P 2 ) Θ 1 j 2 P 1 + Θ 2 j 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 j γ ¯ P 1 σ S 2 + 2 Θ 2 j β ¯ P 2 σ S 2 + 2 Θ 1 j Θ 2 j γ ¯ β ¯ P 1 P 2 ,$\nwhich converges as $n → ∞$ to\n$E σ S 2 ( σ Z 2 + γ Θ 1 2 P 1 + β Θ 2 2 P 2 ) Θ 1 2 P 1 + Θ 2 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 γ ¯ P 1 σ S 2 + 2 Θ 2 β ¯ P 2 σ S 2 + 2 Θ 1 Θ 2 γ ¯ β ¯ P 1 P 2$\ndue to the stationarity and ergodicity of the fading processes. This concludes the proof of achievability.\n\n## 5. Converse\n\nIn this section, we prove that any successful scheme (that has a vanishing probability of error and is within the distortion tolerance) must satisfy the rate-distortion constraints of Theorem 1. This is proved in two subsections: in Section 5.1, we construct an outer bound on the rate-distortion trade-off region. Subsequently, we shall prove in the next Section 5.2 that this outer bound is achievable, thereby proving Theorem 1.\n\n#### 5.1. Outer Bound\n\nThe proof of our outer bound is aided by the following lemma, adapted from (Equation (2), ).\nLemma 1.\nAny communication estimation scheme achieving a distortion $D n ≜ 1 n E | | S n − S ^ n | | 2$ over blocklength n satisfies\n$n 2 log σ S 2 D n ≤ I ( S n ; Y n , Θ 1 n , Θ 2 n ) .$\nProof.\nThe proof is given in Appendix A.1. □\nAnother useful property is the differential entropy maximizing property of the Gaussian distribution , i.e., for $X g n ∼ N ( 0 , K )$,\n$h ( X n ) ≤ h ( X g n ) whenever Cov ( X n ) ⪯ K .$\nThe above facts will be extensively used in our proofs.\nFor $( η 1 , η 2 , λ ) ∈ R + 3$, we define\n$L ( η 1 , η 2 , λ ) = max η 1 R 1 + η 2 R 2 + λ 2 log σ S 2 D ,$\nwhere the maximum is over all $( R 1 , R 2 , D )$ obeying (5)–(8). We note that it suffices to restrict attention to $η i ≥ 0$, since $η i < 0$ will trivially correspond to $R i = 0$ in the maximization, a case already accounted for by $η i = 0$. Likewise, since $D ≤ σ S 2$, only $λ ≥ 0$ need be considered. Therefore, we only consider non-negative weighting coefficients in the sequel. The converse is established by showing that if $( R 1 , R 2 , D n )$ is achievable using block length n, then, for all $η 1 , η 2 , λ ≥ 0$,\n$η 1 R 1 + η 2 R 2 + λ 2 log σ S 2 D n ≤ L ( η 1 , η 2 , λ ) + o ( 1 ) ,$\nwhere $o ( 1 )$ has the usual meaning in standard Landau notation. We note that since the tuple $( W 1 , W 2 , S n , Θ 1 n , Θ 2 n )$ is independent, the Markov chain $X 1 n → S n → X 2 n$ holds. Denoting\n$σ X | Y n 2 ≜ min α ∈ R n × 1 E [ X − α T Y n ] 2 ,$\nwe have for the i-th entry in a block of transmissions,\n$σ X 1 i + X 2 i | S n 2 = σ X 1 i | S n 2 + σ X 2 i | S n 2 .$\nWe define the empirical covariance matrix $K i$ of the vector $( X 1 i , X 2 i , S i )$ with $K i ( p , l )$ denoting its entries. Let us denote\n$K i ( j , j ) = E [ X j i 2 ] = P j i , j = 1 , 2$\nwhere $P j i , j = 1 , 2$ satisfy the power constraints\n$P 1 ≥ 1 n ∑ i = 1 n P 1 i , P 2 ≥ 1 n ∑ i = 1 n P 2 i .$\nNext, we introduce two parameters $γ i ∈ [ 0 , 1 ]$ and $β i ∈ [ 0 , 1 ]$ such that\n$σ X 1 i | S n 2 ≜ γ i P 1 i ,$\n$σ X 2 i | S n 2 ≜ β i P 2 i .$\nWe now define two parameters $γ ∈ [ 0 , 1 ]$ and $β ∈ [ 0 , 1 ]$ such that\n$γ = 1 n P 1 ∑ i = 1 n γ i P 1 i , β = 1 n P 2 ∑ i = 1 n β i P 2 i .$\nWith this, we are ready to prove (27). Firstly, considering the case where $η 1 ≥ η 2$ is sufficient, as a simple renaming of the indices will give us the other case. For $η 2 > 0$, since $λ$ is an arbitrary positive number, maximizing the left-hand side of (27) is equivalent to maximizing $η 1 R 1 + η 2 R 2 + η 2 λ 1 2 log σ S 2 D n$. Dividing by $η 2$, and then renaming $η 1 η 2$ as $η$, the maximization becomes $∀ η ≥ 1 , λ ≥ 0$,\n$max η R 1 + R 2 + λ 2 log σ S 2 D n .$\nFor a given $η > 1$, three regimes of $λ$ arise, as shown in Figure 3. Let $R 1 ( γ ) , R 2 ( β ) , R s u m ( γ , β )$ and $D ( γ , β )$, respectively, denote the right-hand side of Equations (5)–(8). The following two lemmas are crucial to our proofs.\nLemma 2.\nFor $λ ≤ 1$, and $γ , β$ defined in (30), we have\n$η R 1 + R 2 + λ 2 log σ S 2 D n w ≤ ( η − 1 ) R 1 ( γ ) + R s u m ( γ , β ) + λ 2 log σ S 2 D ( γ , β ) + o ( 1 ) .$\nProof.\nThe proof is given in Appendix A.2. □\nLemma 3.\nThe function $g ( γ , β ) : = R s u m ( γ , β ) + 1 2 log σ S 2 D ( γ , β )$ is a non-increasing function in each of the arguments when the other argument is held fixed, for $γ ∈ [ 0 , 1 ]$ and $β ∈ [ 0 , 1 ]$.\nProof.\nWe first note that $D ( γ , β )$ increases with $γ$ (or $β$), see (8). Furthermore, a straightforward inspection reveals that $g ( γ , β )$ is non-increasing in each of the arguments. □\nWe now consider the different regimes for $λ$ (see Figure 3).\nCase 1\n($λ ≤ 1 and η ≥ 1$): In this regime, Lemma 2 directly gives a bound on the weighted sum rate.\nCase 2\n($λ ≥ η and η ≥ 1$): Since $η ≥ 1$, we have\n$η R 1 + R 2 + λ 2 log σ S 2 D n ≤ η R 1 + η R 2 + λ 2 log σ S 2 D n = η ( R 1 + R 2 ) + λ 2 log σ S 2 D n = η R 1 + R 2 + 1 2 log σ S 2 D n + λ − η 2 log σ S 2 D n ≤ ( a ) η R s u m ( 0 , 0 ) + 1 2 log σ S 2 D ( 0 , 0 ) + λ − η 2 log σ S 2 D n ≤ ( b ) 0 + η 2 log σ S 2 D ( 0 , 0 ) + λ − η 2 log σ S 2 D ( 0 , 0 ) ,$\nwhere $( a )$ follows from an application of Lemma 2 followed by Lemma 3, and $( b )$ follows from the fact that uncoded transmission of the state by the two users acting as a super-user with power $( P 1 + P 2 ) 2$ results in the minimal distortion possible .\nCase 3\n($1 ≤ λ ≤ η and η ≥ 1$): Since $λ ≥ 1$, we have\nSince $λ ≥ 1$, we have\n$η R 1 + R 2 + λ 2 log σ S 2 D n ≤ η R 1 + λ R 2 + λ 2 log σ S 2 D n = ( η − λ ) R 1 + λ R 1 + R 2 + 1 2 log σ S 2 D n ≤ ( η − λ ) R 1 + λ R s u m ( γ , 0 ) + 1 2 log σ S 2 D ( γ , 0 ) ,$\nwhere the last step follows from Lemmas 2 and 3. From (A11) in Appendix A.2, it follows that the inequality $R 1 ≤ E Θ 1 1 2 log ( 1 + γ Θ 1 2 P 1 / σ Z 2 )$ holds. Thus,\n$η R 1 + R 2 + λ 2 log σ S 2 D n ≤ η R s u m ( γ , 0 ) + λ 2 log σ S 2 D ( γ , 0 ) .$\nWe next prove that (32)–(34) define the region in Theorem 1.\n\n#### 5.2. Equivalence of Inner and Outer Bounds\n\nWe now show that the regions defined by the inner and outer bounds in Section 4 and Section 5.1 coincide, thereby establishing the capacity region. We will consider three regimes for $λ ≥ 0$ for each $η ≥ 1$, and prove that the maximal value of $η R 1 + R 2 + λ 2 log σ S 2 D$ in the outer bound specified by (32)–(34) can be achieved.\nWhile maximizing $η R 1 + R 2 + λ 2 log σ S 2 D n$, we already proved that $λ ≥ η$ corresponds to an extreme point with zero sum-rate (Case 2 in Section 5.1). Clearly, the corresponding distortion lower bound $D ( 0 , 0 )$ for this case specified by (33) can be achieved by uncoded state transmission by both transmitters using all the available powers. As a result, the condition $λ = η$ encompasses all $λ ≥ η$. Moreover, the regime $1 ≤ λ ≤ η$ (Case 3 of Section 5.1) corresponds to the case where $R 2 = 0$. This implies that we only need to consider $λ = 1$ rather than $λ ∈ [ 1 , η )$. Clearly, the region with $R 2 = 0$ follows from the single-user results of , but for a state process with variance $( P 2 + σ S 2 ) 2$. This proves the achievability of the bound specified by (34). This leaves us with proving the achievability for those cases in which $0 < λ < 1$ holds, corresponding to (32). In this regime, the following lemma is crucial.\nLemma 4.\nFor $( 0 < λ < 1 , η ≥ 1 )$, the function $k ( γ , β ) : = ( η − 1 ) R 1 ( γ ) + R s u m ( γ , β ) + λ 2 log σ S 2 D ( γ , β )$ is jointly strictly concave in $( γ , β )$ for $0 ≤ γ ≤ 1$ and $0 ≤ β ≤ 1$.\nProof.\nThe proof is given in Appendix A.3. □\nSince we know that $η R 1 + R 2 + λ 2 log σ S 2 D n ≤ k ( γ , β )$ for some value of $( γ , β ) ∈ [ 0 , 1 ] × [ 0 , 1 ]$, the strict concavity of $k ( · )$ suggests the existence of a unique $( γ * , β * )$ for which $η R 1 + R 2 + λ 2 log σ S 2 D n$ is maximized for the given $η > 1$ and $0 ≤ λ ≤ 1$. Evidently, choosing these maximizing parameters $( γ * , β * )$ in our achievable theorem will give us the same operating point. Reversing the roles of $R 1$ and $R 2$, we have covered the whole region. Thus, we have established the achievability of the outer bound specified with (32)–(34). This completes the proof of Theorem 1.\n\n## 6. Conclusions\n\nWe investigated joint message transmission and state estimation in a state-dependent fading Gaussian multiple access channel in this paper and characterized the trade-off region between message rates and state estimation distortion. It was shown that the optimal strategy involved static power allocation and uncoded state amplification combined with Gaussian signaling and dirty paper coding. While the role of uncoded communications has been examined before for non-fading settings without state dependence, ours is the first result that brings out its significance in the context of state-dependent fading systems.\nOur framework naturally generalizes previous results concerning state estimation on point-to-point fading channels to multiple users as well as point-to-point non-fading settings to fading links with multiple users. Our results contribute to a better understanding of joint state estimation and communication problems in multi-terminal settings. They can be used as design guidelines for practical systems employing joint sensing and communication envisioned in future 6G wireless standards and broadly applies to systems that involve joint compression and communication/rate-distortion trade-offs.\nHowever, we assumed perfect state observation at the transmitters in this work. A long-standing open problem is that of communicating state and message streams in a fading GMAC with noisy state observations at the transmitters, which is left for future work. Moreover, there could be settings when the receiver cannot track the channel fading gains either, unlike this work. Thus, another interesting avenue for further research is an investigation of the current setup when the encoders, as well as the decoder, are totally uninformed about the fading coefficients on the links.\n\n## Funding\n\nThis research received no external funding.\n\nNot applicable.\n\nNot applicable.\n\nNot applicable.\n\n## Acknowledgments\n\nThe author gratefully acknowledges the insightful comments and suggestions of the three anonymous reviewers, which have greatly helped to improve the presentation of the results in this manuscript.\n\n## Conflicts of Interest\n\nThe author declares no conflict of interest.\n\n## Appendix A\n\n#### Appendix A.1. Proof of Lemma 1\n\nThe proof is along the lines of (Equation (2), ), with appropriate modifications to incorporate the fading coefficients $( Θ 1 n , Θ 2 n )$. We provide a full proof for completeness. Consider the following chain of inequalities, starting with the right-hand side in Lemma 1:\n$1 n I ( S n ; Y n , Θ 1 n , Θ 2 n ) = 1 n ( h ( S n ) − h ( S n | Y n , Θ 1 n , Θ 2 n ) ) = ( a ) 1 n ( h ( S n ) − h ( S n − S ^ n ( Y n , Θ 1 n , Θ 2 n ) | Y n , Θ 1 n , Θ 2 n ) ) ≥ 1 n ( h ( S n ) − h ( S n − S ^ n ) ) = ( b ) 1 n ( ∑ i = 1 n h ( S i ) − h ( S n − S ^ n ) ) ≥ 1 n ∑ i = 1 n ( h ( S i ) − h ( S i − S ^ i ) ) = 1 n ∑ i = 1 n 1 2 log ( 2 π e σ S 2 ) − h ( S i − S ^ i ) ≥ ( c ) 1 n ∑ i = 1 n 1 2 log ( 2 π e σ S 2 ) − 1 2 log ( 2 π e E ( S i − S ^ i ) 2 ) ≥ ( d ) 1 2 log ( 2 π e σ S 2 ) − 1 2 log 2 π e 1 n ∑ i = 1 n E ( S i − S ^ i ) 2 = 1 2 log ( 2 π e σ S 2 ) − 1 2 log 2 π e 1 n E ∥ S n − S ^ n ∥ 2 = 1 2 log σ S 2 1 n E ∥ S n − S ^ n ∥ 2 = 1 2 log ( σ S 2 D n ) ,$\nwhere (a) follows since $S ^ n ( Y n , Θ 1 n , Θ 2 n )$ is a function of $( Y n , Θ 1 n , Θ 2 n )$, (b) follows since $S n$ is i.i.d., (c) follows since the Gaussian distribution maximizes differential entropy for a given variance, while (d) follows from Jensen’s inequality.\n\n#### Appendix A.2. Proof of Lemma 2\n\nBy Fano’s inequality , we can write for any $ϵ > 0$, for large enough n, $H ( W 1 , W 2 | Y n ) ≤ n ϵ$. Then, we have\n$n η R 1 + n R 2 + n λ 2 log σ S 2 D n = n ( η − 1 ) R 1 + n ∑ j = 1 2 R j + n λ 2 log σ S 2 D n ≤ ( a ) ( η − 1 ) H ( W 1 ) + H ( W 1 , W 2 ) + λ I ( S n ; Y n , Θ 1 n , Θ 2 n ) = ( b ) ( η − 1 ) H ( W 1 | X 2 n , S n , Θ 1 n ) + H ( W 1 , W 2 | S n , Θ 1 n , Θ 2 n ) w w w + λ I ( S n ; Y n , Θ 1 n , Θ 2 n ) = ( c ) ( η − 1 ) H ( W 1 | X 2 n , S n , Θ 1 n ) + H ( W 1 , W 2 | S n , Θ 1 n , Θ 2 n ) w w w + λ I ( S n ; Y n | Θ 1 n , Θ 2 n ) ≤ ( d ) ( η − 1 ) I ( W 1 ; Y n | X 2 n , S n , Θ 1 n ) + n ϵ w w + I ( W 1 , W 2 ; Y n | S n , Θ 1 n , Θ 2 n ) + n ϵ + λ I ( S n ; Y n | Θ 1 n , Θ 2 n ) = ( η − 1 ) I ( W 1 ; Y n | X 2 n , S n , Θ 1 n ) w w w + λ I ( W 1 , W 2 , S n ; Y n | Θ 1 n , Θ 2 n ) w w w + ( 1 − λ ) I ( W 1 , W 2 ; Y n | S n , Θ 1 n , Θ 2 n ) + 2 n ϵ = ( η − 1 ) ( h ( Y n | X 2 n , S n , Θ 1 n ) − h ( Y n | W 1 , X 2 n , S n , Θ 1 n ) ) w + λ ( h ( Y n | Θ 1 n , Θ 2 n ) − h ( Y n | W 1 , W 2 , S n , Θ 1 n , Θ 2 n ) ) w + ( 1 − λ ) ( h ( Y n | S n , Θ 1 n , Θ 2 n ) w w w w w w w w − h ( Y n | W 1 , W 2 , S n , Θ 1 n , Θ 2 n ) ) + 2 n ϵ ≤ ∑ i = 1 n ( η − 1 ) ( h ( Y i | X 2 i , S i , Θ 1 i ) − h ( Z i ) ) w + ∑ i = 1 n { λ h ( Y i | Θ 1 i , Θ 2 i ) + ( 1 − λ ) h ( Y i | S i , Θ 1 i , Θ 2 i ) w w w w w w w w w w w w w w w w w w w w w w w − h ( Z i ) } + 2 n ϵ = E ∑ i = 1 n ( η − 1 ) ( h ( Y i | X 2 i , S i , Θ 1 i = Θ 1 ) − h ( Z i ) ) w + E [ ∑ i = 1 n ( λ h ( Y i | Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) w w w w w + ( 1 − λ ) h ( Y i | S i , Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) w w w w w w w w w w w w w w w w w w w w w w w − h ( Z i ) ) ] + 2 n ϵ ,$\nwhere $( a )$ uses Lemma 1 and the fact that the messages are uniformly distributed on their respective alphabets, $( b )$ follows since $( W 1 , W 2 ) ⫫ ( S n , Θ 1 n , Θ 2 n )$ and $W 1 ⫫ ( X 2 n , S n , Θ 1 n )$, (c) follows since $( Θ 1 n , Θ 2 n ) ⫫ S n$ and $( d )$ follows from Fano’s inequality. We now upper bound the term $λ h ( Y i | Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) + ( 1 − λ ) h ( Y i | S i , Θ 1 i = Θ 1 , Θ 2 i = Θ 2 )$ in (A2). Notice that for a given covariance matrix $K i$, $λ h ( Y i | Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) + ( 1 − λ ) h ( Y i | S i , Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) , λ ∈ [ 0 , 1 ]$, is simultaneously maximized when $( Θ 1 X 1 i + Θ 2 X 2 i )$ is jointly Gaussian with $S i$.\nWithout loss of generality, for the purposes of finding an upper bound on $η R 1 + R 2 + λ 2 log σ S 2 D n$, we can express $X j i$ in terms of its linear least squares estimate given $S n$ and an error term. Therefore,\n$X j i = V j i + W j i ,$\nwhere $V j i$ is uncorrelated with $S n$, and $W j i = ∑ k = 1 n β j i k S k$ for appropriate coefficients ${ β j i k , k ∈ { 1 , 2 , … , n } }$. But it is readily verified from expressions (28), (29) and (A3) that $E [ W 1 i 2 ] = ( 1 − γ i ) P 1 i$ and $E [ W 2 i 2 ] = ( 1 − β i ) P 2 i$. Now, for a fixed $E [ W 1 i 2 ] = ( 1 − γ i ) P 1 i$, it follows that the variance of $Θ 1 X 1 i + S i = Θ 1 V 1 i + ( Θ 1 W 1 i + S i )$ would be maximized when $Θ 1 W 1 i$ is a scaled version of $S i$, i.e.,\n$Θ 1 X 1 i = Θ 1 V 1 i + μ 1 i Θ 1 ( 1 − γ i ) P 1 i σ S 2 S i ,$\nwhere $μ 1 i ∈ { − 1 , + 1 }$ is the sign of the correlation between $X 1 i$ and $S i$. Likewise, we can express\n$Θ 2 X 2 i = Θ 2 V 2 i + μ 2 i Θ 2 ( 1 − β i ) P 2 i σ S 2 S i ,$\nwhere $μ 2 i ∈ { − 1 , + 1 }$ is the sign of the correlation between $X 2 i$ and $S i$. Adding expressions (A4) and (A5), we obtain\n$∑ j = 1 2 Θ j X j i = V i + μ 1 i Θ 1 ( 1 − γ i ) P 1 i σ S 2 w w w w w w + μ 2 i Θ 2 ( 1 − β i ) P 2 i σ S 2 S i ,$\nwhere $V i = Θ 1 V 1 i + Θ 2 V 2 i$ is zero mean Gaussian, and independent of $S n$. The second term on the right-hand side of (A6) can be understood as the linear estimate of $( Θ 1 X 1 i + Θ 2 X 2 i )$ given $S i$. Since $V i = Θ 1 V 1 i + Θ 2 V 2 i$ and $S n$ are independent, it follows from (28) and (29) that\n$Var [ V i ] = σ Θ 1 X 1 i + Θ 2 X 2 i | S n 2 = γ i Θ 1 2 P 1 i + β i Θ 2 2 P 2 i .$\nUsing this, it follows that\n$E [ ( Θ 1 X 1 i + Θ 2 X 2 i ) 2 ] ≤ Θ 1 2 P 1 i + Θ 2 2 P 2 i + 2 Θ 1 Θ 2 ( 1 − γ i ) ( 1 − β i ) P 1 i P 2 i ,$\nwhere we have taken $μ 1 i = μ 2 i = 1$ as the sign of correlation in (A6), since negative correlation can only be detrimental for the right-hand side. Denoting $κ ( x ) = ( 1 / 2 ) log ( 2 π e x )$ and using the differential entropy maximizing property of Gaussian random variables for a given variance, we have\n$h ( Y i | X 2 i , S i , Θ 1 i = Θ 1 ) ≤ κ ( γ i Θ 1 2 P 1 i + σ Z 2 ) ,$\n$h ( Y i | S i , Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) w w w w ≤ κ ( γ i Θ 1 2 P 1 i + β i Θ 2 2 P 2 i + σ Z 2 ) ,$\n$h ( Y i | Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) w w w ≤ κ ( Θ 1 2 P 1 i + Θ 2 2 P 2 i + σ S 2 + σ Z 2 w w w w w w w + 2 Θ 1 γ ¯ i P 1 i σ S 2 + 2 Θ 2 β ¯ i P 2 i σ S 2 w w w w w w w + 2 Θ 1 Θ 2 γ ¯ i β ¯ i P 1 i P 2 i ) ,$\nwhere (A8) is under the choice of $( Θ 1 X 1 i + Θ 2 X 2 i )$ which maximizes $λ h ( Y i | Θ 1 i = Θ 1 , Θ 2 i = Θ 2 ) + ( 1 − λ ) h ( Y i | S i , Θ 1 i = Θ 1 , Θ 2 i = Θ 2 )$, for all $λ ∈ [ 0 , 1 ]$. Continuing the sequence of inequalities from (A2):\n$n η R 1 + n R 2 + n λ 2 log σ S 2 D n − 2 n ϵ ≤ ( a ) E ∑ i = 1 n ( η − 1 ) 1 2 log γ i Θ 1 2 P 1 i + σ Z 2 σ Z 2 − n 2 log ( σ Z 2 ) w + E ∑ i = 1 n λ 2 log Θ 1 2 P 1 i + Θ 2 2 P 2 i + σ S 2 + σ Z 2 + 2 Θ 1 γ ¯ i P 1 i σ S 2 + 2 Θ 2 β ¯ i P 2 i σ S 2 + 2 Θ 1 Θ 2 γ ¯ i β ¯ i P 1 i P 2 i w + E ∑ i = 1 n ( 1 − λ ) 2 log γ i Θ 1 2 P 1 i + β i Θ 2 2 P 2 i + σ Z 2 ≤ ( b ) E ( η − 1 ) n 2 log γ Θ 1 2 P 1 + σ Z 2 σ Z 2 − n 2 log ( σ Z 2 ) w + E λ n 2 log Θ 1 2 P 1 + Θ 2 2 P 2 + σ S 2 + σ Z 2 + 2 Θ 1 γ ¯ P 1 σ S 2 + 2 Θ 2 β ¯ P 2 σ S 2 + 2 Θ 1 Θ 2 γ ¯ β ¯ P 1 P 2 w + E n ( 1 − λ ) 2 log γ Θ 1 2 P 1 + β Θ 2 2 P 2 + σ Z 2 = ( c ) E ( η − 1 ) n 2 log γ Θ 1 2 P 1 + σ Z 2 σ Z 2 w w + n R s u m ( γ , β ) + λ n 2 log σ S 2 D ( γ , β ) .$\nwhere (a) follows from the fact that both $λ$ and $( 1 − λ )$ are non-negative for $λ ∈ [ 0 , 1 ]$ and expressions (A7)–(A9), (b) follows from Jensen’s Inequality, and (c) follows from the definitions of $R s u m ( γ , β )$, $D ( γ , β )$ from (7) and (8). Thus, the lemma is proved for all $λ ∈ [ 0 , 1 ]$. Following similar lines, the individual rate $R 1$ can be bounded using Fano’s inequality and (A7), to obtain\n$R 1 ≤ E 1 2 log γ Θ 1 2 P 1 + σ Z 2 σ Z 2 .$\n\n#### Appendix A.3. Proof of Lemma 4\n\nConsider a concave function $T ( ω ̲ )$, where $ω ̲ = [ ω 1 ω 2 … ω N ] ∈ [ 0 , 1 ] N$, and define\n$f ( ω ̲ ) : = ∑ i = 1 N δ i log 1 + ∑ j = 1 i ω j P j + λ 2 log T ( ω ̲ ) ,$\nwhere $δ i , 1 ≤ i ≤ N$ and $λ$ are non-negative constants.\nLemma A1.\nFor $0 < λ ≤ 1$, $f ( · )$ is strictly concave in $ω ̲ ∈ [ 0 , 1 ] N$, whenever $δ i , 1 ≤ i ≤ N$ are not identically zero.\nProof.\nThe first term, which is a linear combination of logarithms, is strictly concave. We next consider the second term. Let $y 1 ̲$ and $y 2 ̲$ be two $N −$ dimensional vectors in $R N$. Notice that for $ι ∈ [ 0 , 1 ]$,\n$ι log T ( y 1 ̲ ) + ( 1 − ι ) log T ( y 2 ̲ ) w w w w w w w w w w w w w w w ≤ log ( ι T ( y 1 ̲ ) + ( 1 − ι ) T ( y 2 ̲ ) ) w w w w w w w w w w w w w w w ≤ log T ( ι y 1 ̲ + ( 1 − ι ) y 2 ̲ ) ,$\nsince $T ( · )$ itself is concave by assumption. □\nLet us now proceed to prove Lemma 4. We denote\n$T ( γ , β ) = Θ 1 2 P 1 + Θ 2 2 P 2 + σ S 2 + σ Z 2 w + 2 Θ 1 γ ¯ P 1 σ S 2 + 2 Θ 2 β ¯ P 2 σ S 2 w + 2 Θ 1 Θ 2 γ ¯ β ¯ P 1 P 2 ,$\nfor convenience. Note that the function\n$k ′ ( γ , β ) ≜ ( η − 1 ) 2 log 1 + γ Θ 1 2 P 1 σ Z 2 w w w + λ 2 log T ( γ , β ) σ Z 2 w w w + ( 1 − λ ) 2 log 1 + γ Θ 1 2 P 1 + β Θ 2 2 P 2 σ Z 2$\nis a sum similar to (A12). We first show that $k ′ ( γ , β )$ is strictly concave by proving $T ( · )$ in (A14) to be a concave function. For $d 0 > 0$, and non-negative constants $d 1 , ⋯ , d N$, the function\n$T ( ω ̲ ) = d 0 + ∑ i = 1 N ∑ j = 1 N d i d j ( 1 − ω i ) ( 1 − ω j )$\nis concave. To see this, we note that $x$ is strictly concave in $x ≥ 0$. Moreover, $x z$ is jointly concave in $( x , z ) ∈ [ 0 , 1 ] 2$, implying that $T ( ω )$ is a concave function. Notice that concavity in the range of interest is unaffected if every variable $x ∈ [ 0 , 1 ]$ is replaced by $1 − x$. Finally, it follows that the function in Lemma 4, $k ( γ , β ) = E [ k ′ ( γ , β ) ]$, is strictly concave as well as a result of Jensen’s inequality. This concludes the proof of the lemma.\n\n## References\n\n1. Chiriyath, A.R.; Paul, B.; Jacyna, G.M.; Bliss, D.W. Inner bounds on performance of radar and communications co-existence. IEEE Trans. Signal Process. 2015, 64, 464–474. [Google Scholar] [CrossRef]\n2. Liu, F.; Liu, Y.F.; Li, A.; Masouros, C.; Eldar, Y.C. Cramér-Rao Bound Optimization for Joint Radar-Communication Beamforming. IEEE Trans. Signal Process. 2021, 70, 240–253. [Google Scholar] [CrossRef]\n3. Shi, C.; Wang, F.; Salous, S.; Zhou, J. 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Cover, T.M.; Thomas, J.A. Elements of Information Theory; John Wiley & Sons: Hoboken, NJ, USA, 2012. [Google Scholar]\nFigure 1. Sensor network example.\nFigure 1. Sensor network example.\nFigure 2. State estimation over a fading Gaussian MAC with state, without fading knowledge at the transmitters.\nFigure 2. State estimation over a fading Gaussian MAC with state, without fading knowledge at the transmitters.\nFigure 3. Range of $λ$ for a given $η$.\nFigure 3. Range of $λ$ for a given $η$.\nTable 1. Summary of paper contributions. Note that single-user (noncausal) refers to a point-to-point state-dependent channel with noncausal transmitter state information, BC (causal) refers to a state-dependent broadcast channel with causal transmitter state information, while MAC (noncausal) refers to a state-dependent multiple access channel with noncausal state information at all the transmitters.\nTable 1. Summary of paper contributions. Note that single-user (noncausal) refers to a point-to-point state-dependent channel with noncausal transmitter state information, BC (causal) refers to a state-dependent broadcast channel with causal transmitter state information, while MAC (noncausal) refers to a state-dependent multiple access channel with noncausal state information at all the transmitters.\nSingle-User (Noncausal)BC (Causal)MAC (Noncausal)\nNo State EstimationThis work\nState EstimationThis workThis work\n Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.\n\n## Share and Cite\n\nMDPI and ACS Style\n\nRamachandran, V. Lossy State Communication over Fading Multiple Access Channels. Entropy 2023, 25, 588. https://doi.org/10.3390/e25040588\n\nAMA Style\n\nRamachandran V. Lossy State Communication over Fading Multiple Access Channels. Entropy. 2023; 25(4):588. https://doi.org/10.3390/e25040588\n\nChicago/Turabian Style\n\nRamachandran, Viswanathan. 2023. \"Lossy State Communication over Fading Multiple Access Channels\" Entropy 25, no. 4: 588. https://doi.org/10.3390/e25040588\n\nNote that from the first issue of 2016, this journal uses article numbers instead of page numbers. See further details here." ]
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[ "", null, "Clustering in r tutorial\n\n### Clustering in r tutorial", null, "Cluster analysis Lecture / Tutorial outline • Cluster analysis • Example of cluster analysis • Work on the assignment. In the retail sector, it can be used to categorize both products and customers. . R supports various functions and packages to perform cluster analysis. In this tutorial I want to show you how to use K means in R ## K-means clustering with 3 clusters of Introduction to K-means Clustering: A Tutorial. 6 \\$\\begingroup\\$ Is it possible to do 2-stage cluster analysis in R? Can anybody provide me resource on it? r clustering. R comes with an easy interface to run hierarchical clustering. R script used to simulate iterative http likes 180. I am a newbie to R and I am trying to do some clustering on a data table where rows represent individual objects and columns represent the features that have been measured for these objects. I also recommend RStudio, which provides a simple interface for writing and executing R code: download it here. We have a workaround this. Tutorial at Melbourne Data Science Week. Detailed tutorial on Practical Guide to Clustering Algorithms & Evaluation in R to improve your understanding of Machine Learning. Clustering categorical data with R. It is used in many elds, such as machine learning, data mining, pattern recognition, image analysis, genomics, systems biology, etc. 28 Dec 2015 Hello everyone, hope you had a wonderful Christmas! In this post I will show you how to do k means clustering in R. 246 ## ## Clustering R comes with an easy interface to run hierarchical clustering. I found the process extremely easy!! Getting StartedR basics: Learn how to perform a k-means with R. vq import kmeans cluster_centers, distortion = kmeans(df[['scaled_red', 'scaled_green', 'scaled_blue']], 2) The final step in k-means clustering is generating cluster labels. max=10) x A numeric matrix of data, or an Euclidean Cluster Extraction. 428 1. complete linkage cluster analysis, because a cluster is formed when all the dissimilarities (‘links’) between pairs of objects in the cluster are less then a particular level. Before we go into how you can use R to perform this type of customer grouping using clustering in SQL Server 2016, we will look at the scenario in R. 5. This is my first try at using R and I have spent a LOT of time pouring over the manual/help pages and internet tutorials on how to do this. The first argument which is passed to this function, is the dataset from Columns 1 to 4 (dataset [,1:4]). csv\" ## c) MDS coordinate file: \"cmd1. If you recall from the post about k means clustering, it requires us to specify the number of clusters, and finding the optimal number of clusters can often be hard. We will use the iris dataset again, like we did for K means clustering. The pattern fit is analogous to factor analysis and is based upon the model = P x Structure where Structure is Pattern * Phi. In this tutorial, you will learn . 38, 72076 Tubingen, Germany ulrike. Clustering ¶ Clustering of (r(i, k)\\), which is the accumulated evidence that sample \\(k\\) “A Tutorial on Spectral Clustering You have customers. Implement the k-means algorithm There is a built-in R function kmeans for the implementation of the k-means clustering algorithm. Clustering using the pam algorithm in R. Points to Remember. Note that there are four dimensions in the data and that only the first two dimensions are used to draw the plot below. K-means Clustering (from \"R in Action\") In R’s partitioning approach, observations are divided into K groups and reshuffled to form the most cohesive clusters possible according to a given criterion. You will need to know how to read in data, subset data and plot items in order to use this video. Kmeans function in R helps us to do k-mean clustering in R. strategies for the EM algorithm, and bootstrap-based inference, making it a full-featured R package for data analysis via finite mixture modelling. Cluster Analysis is a statistical technique for unsupervised learning, which works only with X variables (independent variables) and no Y variable (dependent variable). Density-based Clustering I Group objects into one cluster if they are connected to one another by densely populated area I The DBSCAN algorithm from package fpc provides a density-based clustering for numeric data. In this method, we have known that cluster has a higher density than the rest of the dataset. Download slides in PDF ©2011-2019 Yanchang Zhao. A loose definition of clustering could be “the process of organizing objects into groups whose members are similar in some way”. Use spectral clustering and its variant for community detection in a network. A cluster analysis allows you summarise a dataset by grouping similar observations together into clusters. Clustering can be explained as organizing data into groups where members of a group are similar in some way. Statistics and computing 17:395<U+2013>416. Now data variables are ready for clustering. You only need to specify the data to be clustered and the number of clusters, which we set to 4. xpr, centers=20) I am quite aware that there are a few other questions on the subject, but the answers are very broad and none permits to do what I would like to accomplish. The second argument is the number of cluster or centroid, which I specify number 5. From supervised to unsupervised clustering, comprehensive and detailed version of the following tutorial is available as a R notebook on this Github Gist. K-means clustering is a method of vector quantization, originally from signal processing, that is popular for cluster analysis in data mining. What is Cluster analysis? K-means algorithm ; Optimal k ; What is Cluster analysis? Cluster analysis is part of the unsupervised learning. K-mean clustering In R, writing R codes inside Power BI: Part 6. Although cluster analysis can be run in the R-mode when seeking relationships among variables, this discussion will assume that a Q-mode analysis is being run. method:K means Clustering in R example Iris Data. So data item (1) belongs to cluster 2, data item (2) belongs to cluster 1, and so, through data item (8), which belongs to cluster 2. It does not require us to pre-specify the number of clusters to be generated as is required by the k-means approach. With the distance matrix found in previous tutorial, we can use various techniques of cluster analysis for relationship discovery. A cluster of data objects can be treated as one group. ## 5) Open the R software by double clicking its icon. This video tutorial shows you how to use the means function in R to do K-Means clustering. Learn Data Mining - Clustering Segmentation Using R,Tableau is designed to cover majority of the capabilities of R from Analytics & Data Science perspective, which includes the following: Course is structured to start with introduction to the tool & the principles behind Clustering Using R tool. html K Means clustering may be biased on initial centroids - called cluster seeds Maximum clusters is typically inputs and may also impacts the clusters getting created In the next blog, we focus on creating clusters using R. A comparison on performing hierarchical cluster analysis using the hclust method in core R vs rpuHclust in rpudplus. from scipy. The following example demonstrates how to run the k-means clustering algorithm in R. This tutorial is on how to apply agglomerative hierarchical clustering in R with hclust. R is similar to the award-winning 1 S system, which was Hierarchical Clustering. xpr = read. The pre-validation steps of cluster analysis are already explained in the previous tutorial - Cluster Analysis with R. [R] - k-means clustering tutorial. After plotting a subset of below data, how many clusters will be appropriate? Clustering can be used to identify and group similar objects within your data. 29/9/2013 · In this video I go over how to perform k-means clustering using r statistical computing. Compute cluster centroids : The centroid of data points in the red cluster is shown using red cross and those in grey cluster using grey cross. It can be shown that spectral clustering methods boil down to graph partitioning. ). This R tutorial provides a condensed introduction into the usage of the R environment and its utilities for general data analysis and clustering. 2 ClustOfVar: An R Package for the Clustering of Variables Clustering of variables is an alternative since it makes possible to arrange variables intoData Mining Algorithms In R/Clustering/Density-Based Spatial clustering techniques are a subset of Algorithms_In_R/Clustering/Density-Based_Clustering&oldid Enterotyping : the A summary of the R code that reproduces the results of this tutorial can Here is an example function to perform PAM clustering in R The Segmentation and Clustering course provides students with the foundational knowledge to build and apply clustering models to develop more sophisticated This example illustrates the use of k-means clustering with WEKA The sample data set used for this example is based on the \"bank data\" available in comma-separated I try SOM maps and clustering in R. Clustering: An Introduction. R script tricks. Two-stage clustering in R. Also try practice problems to test ConsensusClusterPlus (Tutorial) Matthew D. The name of the package refers to the Multi-SOM method which represents the multi-map extension of the Kohonen Self-Organizing Map model. You only need to specify the data to Single Cell Genomics Day. ©2011-2019 Yanchang Zhao. I need to take about 200k sentences and cluster them to groups based on text similarity. Introduction mclust (Fraley et al. Clustering of variables is an alternative since it makes possible to arrange variables into. There are two main types of techniques: a bottom-up and a top-down approach. But I hope this tutorial gave you a more accurate view of R’s potential and an interesting introduction to applied text clustering on real data. There are two methods—K-means and partitioning around mediods (PAM). In this post, only base R function This example and sample code-packed example will teach you the essential techniques you need to do text mining in R. Where k is the cluster,x ij is the value of the j th variable for the i th observation, and x kj-bar is the mean of the j th variable for the k th cluster. For each cluster in hierarchical clustering, quantities called p -values are calculated via multiscale bootstrap resampling. You will also learn how to assess the quality of clustering analysis. All we have to define is the clustering criterion and the pointwise distance matrix. This section describes three of the many approaches: hierarchical agglomerative, partitioning, and model based. Next, you'll study the types of clustering methods, such as connectivity-, centroid-, distribution- and density-based clustring. This article is an introduction to clustering and its An Introduction to Clustering and different methods of clustering. Mark each of the linkage types in the connecting line. In R’s partitioning approach, observations are divided into K groups and reshuffled to form the most cohesive clusters possible according to a given criterion. Hello everyone! In this post, I will show you how to do hierarchical clustering in R. Clustering and Feature Extraction in MLlib This tutorial goes over the background knowledge, API interfaces and sample code for clustering, feature extraction and data transformation algorithm in MLlib. Mar 14, 2018 Learn all about clustering and, more specifically, k-means in this R Tutorial, where you'll focus on a case study with Uber data. Oct 13, 2015: Mixture Models, R If you’ve been exposed to machine learning in your work or studies, chances are you’ve heard of the term mixture model. You'll first take a look at the different types of clustering: hard and soft clustering. 462 0. A Tutorial on Spectral Clustering Chris Ding Computational Research Division Lawrence Berkeley National Laboratory University of CaliforniaA Wikibookian suggests that Data Mining Algorithms In R/Clustering/Expectation Maximization be merged into this book or chapter. In this article, we will look at the R clustering visual power BI. Clustering and Data Mining in R Introduction Slide 4/40. Short Course at University of Canberra. What is hierarchical clustering? In k means clustering, it This video tutorial shows you how to use the means function in R to do K-Means clustering. Density-based R Clustering: In regards to the density measurement it creates clusters. Compute distance between every pairs of Clustering for Utility Cluster analysis provides an abstraction from in-dividual data objects to the clusters in which those data objects reside. Sometimes it can be used to predict which group a new object will fall in. In this post, I will show you how to do hierarchical clustering in R. I have RNA-seq data (FPKMs) from Cufflinks and would like to cluster it by gene and produce a heatmap. txt\") # Rows = 250 genes, cols = 32 individuals clusters = kmeans (x = data. We will use the iris dataset fromIn this tutorial, you will learn What is Cluster analysis? K-means algorithm Optimal k What is Cluster analysis? Cluster analysis is part of the unsupervised learning. Time series clustering is to partition time series data into groups based on similarity or distance, so that time series in the same cluster are similar. There are several alternatives to complete linkage as a clustering criterion, and we only discuss two of these: minimum and average clustering. This information provides greater insights about the customer’s needs when used with customer demographics. We will not be using distortion in this tutorial. R has multiple packages for performing K means clustering. Data Mining Cluster Analysis - Learn Data Mining in simple and easy steps starting from basic to advanced concepts with examples Overview, Tasks, Data Mining, Issues Chapter 15 CLUSTERING METHODS Abstract This chapter presents a tutorial overview of the main clustering methods used r +s q +r +s+tThis is a short tutorial for producing heatmaps in R using a modified data set provided by Leanne Wickens. an R object of class \"kmeans\", typically the result ob of ob <- kmeans(. k -means clustering is a technique used to uncover categories. 3 Clustering and Ordinationk-means clustering with R Apply kmeans to newiris, and store the clustering result in kc. As far as I know, the pamk() function serves as a wrapper to pam(), and evaluates the optimal number of clusters. com Jan 30 '16 at 17:31. Data Clustering with R. Density in data space is the measure. Centroid models. Roe ConsensusClusterPlus Bioconductor version: Release (3. 246 ## ## Clustering K-means Clustering (from \"R in Action\") In R’s partitioning approach, observations are divided into K groups and reshuffled to form the most cohesive clusters possible according to a given criterion. Here's a link to a download page: Inside R Download Page. Clustering Tutorial What is Clustering? Clustering is the use of multiple computers, typically PCs or UNIX workstations, multiple storage devices, and redundant interconnections, to form what appears to users as a single highly available Spectral clustering and its variant. A cluster is a group of data that share similar features. As a simple illustration of a k-means algorithm, consider the following data set Clustering Tutorial What is Clustering? Clustering is the use of multiple computers, typically PCs or UNIX workstations, multiple storage devices, and redundant Learn to use the R Clustering with Outliers Power BI. Clustering can be explained as organizing data into groups where members of a group are How to cluster your customer data — with R code examples Clustering customer data helps find hidden patterns in your data by grouping similar things for you. You will learn the implementation of k-means clustering on movie dataset in R. In this article, we include some of the common problems encountered while executing clustering in R. Determine the number of clusters in data. Brock, Charrad). Apply kmeans to newiris, and store the clustering result in kc. With the distance matrix found in previous tutorial, we can use various techniques of cluster analysis for relationship discovery. The first one starts with small clusters composed by a single object and, at each step, merge the current clusters into greater ones, successively, SPSS Tutorial AEB 37 / AE 802 Week 7. This docu-ment provides a tutorial of how to use ConsensusClusterPlus. table (\"my_data. In order to not complicate the tutorial, certain elements of it such as the plane segmentation algorithm, will not be explained here. One of the oldest methods of cluster analysis is known as k-means cluster analysis, and is available in R through the function. How They Work Given a set of N items to be clustered, and an N*N distance (or similarity) matrix, the basic process of Hierarchical Clustering Description. Usage hclust(d, method = \"complete Step 3: K Means Clustering. Intro. Hello everyone, hope you had a wonderful Christmas! In this post I will show you how to do k means clustering in R. • Clustering is a process of partitioning a set of data (or objects) into a set of meaningful sub-classes, called clusters. Where can one find a simple example utilizing the data mining clustering capabilities in SQL Server Analysis Services? Tutorials DBA Data Mining Clustering cluster structure is pronounced in the matrix! This is not a coincidence, and MCL uses this, modifying the random walk process to further emphasize the divide between clusters in the matrix. In this article, we learn to use the clustering with outliers visual. R clustering tutorial for cluster analysis with R by K-mean clustering,hierarchical clustering, similarity aggregation,amap package & clustering applicationWe provide an overview of clustering methods and quick start R codes. In this article, based on chapter 16 of R in Action, Second Edition, author Rob Kabacoff discusses K-means clustering. The main techniques for data mining include classi cation and prediction, clustering, outlier detection, association rules, sequence analysis, time series analysis and text mining, and also some new techniques such as social network analysis and sentiment analysis. Ensure that you are logged in and have the required permissions to access the test. The Matlab function kMeansCluster above call function DistMatrix as shown in the code below. The graph must be partitioned such that edges connecting different clusters should have low weigths, and edges within the same cluster must have high values. K-means clustering can handle larger datasets than hierarchical cluster approaches. before diving into more advanced methods to examine areas where k-means clustering falls short. In order to follow this tutorial, you will need to have R set up on your computer. PDF file at the link. Von Luxburg, U (2007) A tutorial on spectral clustering. We see the cluster centers (means) for the two groups across the four variables ( Murder, Assault, UrbanPop, Rape ). This is a first attempt at a tutorial, and is based around using the Mac version. Hierarchical clustering is an alternative approach to k-means clustering for identifying groups in the dataset. Perform k-means clustering on a data matrix. 8) algorithm for determining cluster count and membership by stability evidence in unsupervised analysis 1 Clustering Techniques. This tutorial demonstrates k-means clustering with R. stackexchange. In this tutorial, we are going to get ourselves familiar with clustering. We have a set of visuals that we can use in Power BI, not write any R code and still leverage the power of BI. Kardi Teknomo – K Mean Clustering Tutorial 9. It is the task of grouping together a set of objects in a way that objects in the same cluster are more similar to each other than to objects in other clusters. Detailed tutorial on Practical Guide to Clustering Algorithms & Evaluation in R to improve your understanding of Machine Learning. Clustering customer data using adendogram (tree diagram) The values on the left refer to the row numbers of the original data set (the values on the bottom refer to a measurement of distance). What is hierarchical clustering? If you recall from the post about k means clustering, it requires us to specify the number of clusters, and finding the optimal number of clusters can often be hard. a code >function which accepts as first argument a (data) matrix like code >x, second argument, say k, k >= 2, the number of clusters desired, and returns a code >list with a component named (or shortened to) code >cluster which is a vector of length code >n = nrow(x) of integers in code >1:k determining the clustering or grouping of the code >n observations. How to cluster text sentences unsupervised? document_clustering. K Means Clustering. A Tutorial on Spectral Clustering Ulrike von Luxburg Max Planck Institute for Biological Cybernetics Spemannstr. ). We can feed in our data into R from many different data file formats, including ASCII formatted text files, Excel spreadsheets and so on. k-Means: Step-By-Step Example. e. Call Detail Record (CDR) is the information captured by the telecom companies during Call, SMS, and Internet activity of a customer. K-means Clustering – Example 1: A pizza chain wants to open its delivery centres across a city. CONTRIBUTED RESEARCH ARTICLES 289 mclust 5: Clustering, Classification and Density Estimation Using Gaussian Finite Mixture Models by Luca Scrucca, Michael Fop, T Tutorial at AusDM 2018. Clustering can be considered the most important unsupervised learning problem; so, as every other problem of this kind, it deals with finding a structure in a collection of unlabeled data. Now you just need to run the script \"Install R Bindings\" and ArcGIS will take care of the rest. It's fairly common to have a lot of dimensions (columns, variables) in your data. If you're already somewhat advanced in R and interested in machine learning, try this: Kaggle Tutorial on who survived the Titanic. In this tutorial, everything you need to know on k-means and clustering in R programming is covered!K-Means Clustering Description. Cluster Analysis sing u R. Refs: Spectral Clustering: A quick overview. The tables contain purchasing and return data based on orders. This tutorial covers various clustering techniques in R. PDF Tutorials. in order to use this code. Online Tutorials. MachineLearning) submitted 5 years ago * by JST_79 Hi All - I'm fairly new to R and am playing around with some of the UCI datasets. Categories K Means, R for Data Science, Segmentation Tags Clustering, K Means, K Means Clustering, k means clustering algorithm r, k means clustering example, k means clustering in r example, k means clustering in r tutorial, r code for k-means, Subjective Segmentation, visualize kmeans Post navigation Validate Cluster Analysis. 16 Mar 201710 Jul 2017 R clustering tutorial for cluster analysis with R by K-mean clustering,hierarchical clustering, similarity aggregation,amap package & clustering 4 Mar 2019 In this tutorial, you will learn What is Cluster analysis? K-means algorithm Optimal k What is Cluster analysis? Cluster analysis is part of the To replicate this tutorial's analysis you will need to load the following packages: library(tidyverse) # data manipulation library(cluster) # clustering algorithms In the k-means cluster analysis tutorial I provided a solid introduction to one of the most popular clustering methods. Step 2. We covered the topic in length and breadth in a series of SAS based articles (including video tutorials), let's now explore the same on R platform. Tags: Clustering, Data Analysis, K-means, Telecom. In this approach, it compares all pairs of data points and merge the one with the closest distance. These are iterative clustering algorithms. How to cluster your customer data — with R code examples towardsdatascience. Clustering validation process can be done with 4 methods (Theodoridis and Koutroubas, G. Euclidean Cluster In this tutorial we will learn how to extract A simple data clustering approach in an Euclidean sense can be implemented by A TUTORIAL ON SUBSPACE CLUSTERING Ren´e Vidal Johns Hopkins University The past few years have witnessed an explosion in the availability of data from multiple Statistical Clustering. Wilkerson October 30, 2018 1 Summary ConsensusClusterPlus is a tool for unsupervised class discovery. This tutorial will cover how to perform combined clustering analysis with CPPTRAJ, which is a way of comparing structure populations between two or more independent trajectories or between different parts of a single trajectory. Clustering is the process of making a group of abstract objects into classes of similar objects. I use this tutorial : https://www. Outline Introduction The k-Means Clustering The k-Medoids Clustering Hierarchical Clustering Density-based Clustering Online Resources 19 / 30 22. r-bloggers. training/blogs/r-clusteringLearn what is R Clustering, R cluster analysis types-K means clustering, DBSCAN clustering and hierarchical clustering,applications of R cluster analysis“Algorithm AS 136: A k-means clustering algorithm You now have all of the bare bones for using kmeans clustering in R. In the k-means cluster analysis tutorial I provided a solid introduction to one of the most popular clustering methods. 2 ClustOfVar: An R Package for the Clustering of Variables. Clustering also helps in classifying documents on the web for information discovery. Andrea Trevino's step-by-step tutorial on the K-means clustering unsupervised machine learning algorithm. luxburg@tuebingen A comparison of computing the distance matrix in CPU with dist function in core R, and in GPU with rpuDist in rpud. A short tutorial to visualize high dimensional data (vector) using t-SNE, Barnes-Hut-SNE, and Clusplot in R. I´m using the pam() R function to perform clustering. Ask Question 6. About Install Get Started Frequently Asked Questions Frequently Requested Vignettes ContactThis tutorial explains tree based modeling which includes decision trees, random forest, bagging, boosting, ensemble methods in R and pythonAfter getting SQL Server with Machine Learning Services installed and your R IDE configured on your machine, you can now proceed and perform clustering using R. . Relative Clustering Validation. Jul 24, 2018 In this tutorial, you will learn to perform hierarchical clustering on a dataset in R. Then R* = R - model and fit is the ratio of sum (r*^2)/sum (r^2) for the off diagonal elements. View Java code. Clustering is a broad set of techniques To perform a cluster analysis in R, A future tutorial will illustrate the PAM clustering The density-based clustering In this chapter, we’ll describe the DBSCAN algorithm and demonstrate how to compute DBSCAN using the fpc R package. TUTORAL C4: Combined Clustering Analysis with CPPTRAJ This tutorial will cover how to perform combined clustering analysis with CPPTRAJ, which is a way of comparing structure populations between two or more independent trajectories or between different parts of a single trajectory. Cluster analysis or clustering is the task of assigning a set of objects into groups (called. pvclust is an R package for assessing the uncertainty in hierarchical cluster analysis. homogeneous clusters and thus to obtain meaningful structures. With this affinity matrix, clustering is replaced by a graph-partition problem, where connected graph components are interpreted as clusters. Dr. A Tutorial on Clustering Algorithms. Implement the k-means algorithm. However, using the same data and parameters I get different results. Each of the following tutorials are in PDF format. Clustering explained using Iris Data. The methods are as follows -. What is Cluster Analysis? • Cluster: a collection of data objects K-Means Clustering in R kmeans(x, centers, iter. Hierarchical clustering is an alternative approach which builds a hierarchy from the bottom-up, and doesn’t require us to specify the number of clusters beforehand. It is structured as follows: Generate clustered data. • Help users understand the natural grouping or structure in a Question: Self-Learning Gene-Expression K-Means Clustering In R. ## R functions needed for RF clustering and results assessment ## b) A test data file: \"testData. Single Cell Genomics Day. You can implement spectral clustering in R using the kernlab package. - The Elements of Statistical Learning 2ed (2009), chapter 14 This is an introduction to R (“GNU S”), a language and environment for statistical computing and graphics. Clustering is an unsupervised learning technique. This document is a gentle introduction to Redis Cluster, that does not use complex to understand distributed systems concepts. For this tutorial, we assume that our data is formatted as Comma-Separated Values (CSV); probably one of the most common data file formats. clusters) so that the objects in the same cluster are more similar (in some sense or another) to each other than to those in other cluste rs. A Tutorial on Spectral Clustering Ulrike von Luxburg Abstract. More examples on data clustering with R and other data mining techniques can be found in my book \"R and Data Mining: Examples and Case Studies\", which is downloadable as a . Next K-means clustering is the popular unsupervised clustering algorithm used to find the PCA, 3D Visualization, and Clustering in R. This page demonstrates k-means clustering with R. *Redis cluster tutorial. Login | Register;This R tutorial provides a condensed introduction into the usage of the R environment and its utilities for general data analysis and clustering. Learn to use the R Clustering with Outliers Power BI. 1. Here’s the full code for this tutorial. In this tutorial, we are going to get ourselves familiar with clustering. GitHub Gist: instantly share code, notes, and snippets. There are several alternative ways of de ning the average and de ning the closeness, and hence a huge number of average linkage methods. Roe. Clustering analysis is performed and the results are interpreted Author: InfluxityViews: 197KR Clustering Tutorial - R Cluster Analysis - DataFlairhttps://data-flair. This page shows R code examples on time series clustering and classification with R. Clustering is the classi cation of data objects into similarity groups (clusters) according to a de ned distance measure. 2 Brief description of Consensus Clustering Consensus Clustering is a method that provides quantitative evidence for de- Confused by clusters? We're not talking grapes. Self Organizing Maps (SOM): Example using RNAseq reads Written by: Ciera Martinez Tutorial is for learning about how to run clustering analysis using Self Organizing Maps using the kohonen package in R. com/how-to-cluster-your-customer-data-with-r-code-examples-6c7e4aa6c5b1Jun 13, 2017 In another post, we talked about how to use the traits you know about for your customers in order to build personas by manually labeling Machine Learning Tutorial for K-means Clustering Algorithm using language R. Join Barton Poulson for an in-depth discussion in this video, Clustering in R, part of Data Science Foundations: Data Mining. Exercise 1. Finding Clusters . Conceptually, when clusters are created, you are interested in distinctive groups of data points, such that the distance between them within clusters ( or compactness) is minimal while the distance between groups ( separation) is as large as possible. Objective First of all we will see what is R Clustering, then we will see the Applications of Clustering, Clustering by Similarity Aggregation, use of R amap…Cluster Analysis: Tutorial with R 5 Fuzzy Clustering 12 1 Introduction In this tutorial We can prune the top level fusions to highlight the clustering: R This first example is to learn to make cluster analysis with R. Clustering wines. ,2016) is a popular R package for model-based clustering, classification, and density estimation based on finite Gaussian mixture modelling. iv. Introductory tutorial to text clustering with R. More specifically you will learn about: What clustering is, when Jul 10, 2017 R clustering tutorial for cluster analysis with R by K-mean clustering,hierarchical clustering, similarity aggregation,amap package & clustering To replicate this tutorial's analysis you will need to load the following packages: library(tidyverse) # data manipulation library(cluster) # clustering algorithms Mar 16, 2017 Make sure to like & comment if you liked this video! This is the second video for our course Unsupervised Learning in R by Hank Roark. In data mining and statistics, hierarchical clustering (also called hierarchical cluster analysis or HCA) is a method of cluster analysis which seeks to build a hierarchy of clusters. K-Means clustering is one of the most common clustering approaches. k-means clustering with R. Previous post. Time Series Clustering. r-bloggers. Cluster analysis or clustering is the task of assigning a set of objects into groups (called clusters) so that the objects in the same Factoextra R Package: Easy Multivariate Data Analyses and Elegant Visualization; Factoextra R It contains also many functions facilitating clustering analysis and Analysis Tutorials. Here is the code: data. up vote 5 down vote favorite. mat, . After writing a few responses, I realized that it would probably benefit not only the Illinois R community but also the larger R community if this information was more widely available. Cluster computing can be used for load balancing as well as for high availability. This tutorial will give you a good idea of how to make text clustering in R and satisfy our needs of data acquisition, data processing and data science. I've worked through some clustering tutorials and I do get some output, however, the heatmap that I get after clustering does not correspond at all to the Clustering categorical data with R. We will be using the Kmeans algorithm to perform the clustering of customers. Installing R and RStudio. Hastie et al. RDataMining Slides Series: Data Clustering with R We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. Using Mixture Models for Clustering. We will be using the Ward's method as the clustering criterion. Here's a sweet tutorial -- now updated -- on clustering, high availability, redundancy, and replication. Read more » R news and tutorials contributed by (750) R bloggers Hierarchical Cluster Analysis. I assume the reader is reasonably au fait with R Studio and able to install packages, load libraries etc…. 1. The cluster number is set to 3. Andrea Trevino presents a beginner introduction to the widely-used K-means clustering algorithm in this tutorial This tutorial covers various clustering techniques in R. Until then, the diagonal was included in the cluster fit statistics. txt\" ## 4) Unzip all the files into the same directory. In addition, numerous clustering validity indices are available in the package to estimate the number R Tutorial 16. Clustering is the grouping of objects together so that objects belonging in the same group (cluster) are more similar to each other than those in other groups (clusters). This is a short tutorial for producing heatmaps in R using a modified data set provided by Leanne Wickens. How to perform a cluster analysis and plot a dendrogram in R. Given a set of observations , where each observation is a -dimensional real vector, -means clustering aims to partition the n observations into () so as to minimize the within-cluster sum of squares (WCSS). A A A A A A A A A A B B B B B B B B B B B B B B B + Figure 1: Distance between two clusters A and B de ned by single, complete and average linkage. Also try practice problems to test & improve your skill level. Discuss whether or not this merger Since I never worked with R-Bridge before, Today I started doing some testing and I decided that the best way to learn it was to create a simple Toolbox to do K-Means clustering on point shapefiles, which I think is a function not available in ArcGIS. js-Shiny-App is provided. In this intro cluster analysis tutorial, we'll check out a few algorithms in Python so you can get a basic understanding of the fundamentals of clustering on a real dataset. Introduction to K-means Clustering: A Tutorial. Applications of Cluster Analysis. We will use R to implement the k-means algorithm for cluster analysis or the DavisThin data set. k represents the number of categories identified, with each category’s average (mean) characteristics being appreciably different from that of other categories. In this article, we include some of Cluster Analysis: Tutorial with R Hierarchic clustering (function hclust) is in standard R and available with- 2. Ad-Cluster analysis is sensitive to both the distance metric selected and the criterion for deter-mining the order of clustering. By Rathnadevi Manivannan, Treselle Systems. By Daniel R. What is Clustering? Clustering is the use of multiple computers, typically PCs or UNIX workstations, multiple storage devices, and redundant interconnections, to form what appears to users as a single highly available system. Hierarchical Clustering in R. Ng A, Jordan M, Weiss Y (2002) On spectral clustering iclust: Item Cluster Analysis – Hierarchical cluster analysis using psychometric principles #long output shows clustering history #ICLUST(r. One way to see and understand patterns from data is by means of visualization. December 2013. K means Clustering in R example In this tutorial I want to show you how to use K means in R with Iris Data example. I will introduce you to the 5 most common clustering – k-means clustering tutorial. There are a huge number of different clustering algorithms available in R. ConsensusClusterPlus (Tutorial) Matthew D. cluster. This question came from our site for people interested in statistics, machine learning, data analysis, data References. In other words, its objective is to find:: where is the mean of points in . In Wikipedia ‘s current words, it is: the task of grouping a set of objects in such a way that objects in the same group (called a cluster) are more similar (in some sense or another) to each other than to those in other groups Most “advanced analytics” tools have some ability to cluster in them. TUTORIAL C1: This tutorial will cover how to perform combined clustering analysis with CPPTRAJ, Documentation. The first step (and certainly not a trivial one) when using k-means cluster analysis is to specify the number of clusters (k) that will be formed in the final solution. It also helps in the identification of groups of houses in a city according to house type, value, and geographic location. Hierarchical Cluster Analysis. This Edureka k-means clustering algorithm tutorial video will take you through the machine learning introduction, cluster analysis, types of clustering algorithms, k-means clustering, how it works along with a demo in R. Two versions of the method are provided : Stochastic and Batch. K-Means Clustering Tutorial. Find the patterns in your data sets using these Clustering. R tutorial for Spatial Statistics Since we are clustering events based on their geographical location we are working with two variables, A crashcourse on the 5 most common clustering methods – with code in R. Different approaches may yield 2. We only use one of these methods commonly known as upgma. 4 ClustOfVar: An R Package for the Clustering of Variables (a) X~ k is the standardized version of the quantitative matrix X k, (b) Z~ k = JGD 1=2 is the standardized version of the indicator matrix G of the quali- K-means Algorithm using R. 2 Brief description of Consensus Clustering Consensus Clustering is a method that provides quantitative evidence for de- Where W_k (I use the underscore to indicate the subscripts) is the within-cluster variation for the cluster k, n_k is the total number of elements in the cluster k, p is the total number of variables we are considering for clustering and x_ij is one variable of one event contained in cluster k. PCA, 3D Visualization, and Clustering in R. Note: For those who prefer Python, I also have a short tutorial for Heatmaps, Hierarchical Clustering, If you want to change the default clustering method K-means Clustering with R: Call Detail Record Analysis. Clustering is also used in outlier detection applications such as detection of credit card fraud. Luxburg - A Tutorial on Spectral Clustering. It's fairly common to have a lot of dimensions (columns, variables) in your data. Tutorial: An app in R shiny visualizing biopsy data — in a pharmaceutical company Learn how to build a shiny app for the visualization of clustering results. By Eiko Fried 2016-10-19, 4:48 pm 2018-05-02 clustering, community detection, fruchterman-reingold, R, tutorial, visualization A problem we see in psychological network papers is that authors sometimes over-interpret the visualization of their data. For example, millions of cameras have been installed in build- ings, streets, airports and cities around the world. Introduction. 0. A Tutorial for Clustering with XCluster Many people have requested additional documentation for using XCluster (not surprising since there wasn't any). In average linkage, the distance between the clusters 2. As you can see the data (fitbit data) is in variable “dataset”. The past few years have witnessed an explosion in the availability of data from multiple sources and modalities. mclust is a contributed R package for model-based clustering, classification, and density estimation based on finite normal mixture modelling. You will then learn about the k-means clustering algorithm, an Learn R functions for cluster analysis. We will use the iris dataset We provide an overview of clustering methods and quick start R codes. Definition. R clustering and decision tree examples (self. Computing k-means clustering in R. A hierarchical clustering method consists of grouping data objects into a tree of clusters. Plot the clusters and their centres. 3. Saurav assets in this tutorial, Take the first step into image analysis in Python by using k-means clustering to analyze the dominant colors in an image in this free data science tutorial. I. Hierarchical clustering. Also try practice problems to test Introduction. In this blog, you will learn the concepts of Machine Learning and clustering. The first argument which is passed to this function, is the dataset from Columns 1 to 4 (dataset[,1:4]). However, this still requires R coding. Hello everyone, hope you had a wonderful Christmas! In this post I will show you how to do k means clustering in R. In this tutorial we will learn how to extract Euclidean clusters with the pcl::EuclideanClusterExtraction class. clustering in r tutorialLearn R functions for cluster analysis. Sunday February 3, 2013. K-means clustering Use the k-means algorithm and Euclidean distance to cluster the following 8 examples into 3 clusters: Hierarchical Clustering on Categorical Data in R. R has many facilities for clustering analysis. K Means Clustering using R Details. We also get the cluster assignment for each observation (i. This can for example be used to target a specific group of customers for marketing efforts. But how should you categorize them to target sales? How many of such categories exist? To answer these questions, we can use cluster analysis. com/self-organising-maps-for-customer-segmentation-using-r/SOM maps work fine, but A Wikibookian suggests that Data Mining Algorithms In R/Clustering/Expectation Maximization be merged into this book or chapter. for i=1:k c(i,:)=mean(m(find(g==i),:)); end end y=[m,g]; end. A new R package for Multi-SOM clustering. Data Clustering Using R. As such, clustering Distortion is the sum of squared distances between each point and its nearest cluster center. For educational purposes a D3. Implement the k-means algorithm There is a built-in R function kmeans for the implementation of the k-means clustering algorithm. Randomly assign each data point to a cluster : Let’s assign three points in cluster 1 shown using red color and two points in cluster 2 shown using grey color. Sometimes we will refer to a bicluster of patients as a submatrix of the original gene- expression array: • mD = the number of patients within the bicluster • n = the number of genes involved in the bicluster In this tutorial we’ll slowly walk through a biclustering analysis of a particular gene- expression data set. As you read from left to right, you can see the order in which clusters were merged together to create larger clusters. csv\" ## d) The tutorial file: \"RFclusteringTutorial. In this tutorial we propose four of the most used clustering K-means is an exclusive clustering algorithm, K-means Cluster Analysis. This algorithm finds the groups that exist organically in the data and the results allow the user to label new data quickly. ## K-means clustering with 3 clusters of sizes 8, 12, 5 ## ## Cluster means: In R’s partitioning approach, observations are divided into K groups and reshuffled to form the most cohesive clusters possible according to a given criterion. Hierarchical cluster analysis on a set of dissimilarities and methods for analyzing it. Both R and RStudio are totally free and easy to install. Not to mention failover, load balancing, CSM, and resource sharing. Learn data science with data scientist Dr. The number of clusters is set to 3. Clustering Tutorial. About Install Get Started Frequently Asked Questions Frequently Requested Vignettes Contact R Tutorial and Exercise Solution eBook. For example, in the data set 14 Mar 2018 K-Means Clustering in R Tutorial. is the distance between cluster centroids. Cluster analysis. For example, in the data set mtcars, we can run the distance matrix with hclust, and plot a dendrogram that displays a hierarchical relationship among the vehicles. clustering in r tutorial This text provides R tutorials on statistics including hypothesis testing, ANOVA and linear regressions. A TUTORIAL ON SUBSPACE CLUSTERING Ren´e Vidal Johns Hopkins University The past few years have witnessed an explosion in the availability of data from multiple sources and modalities. Discuss whether or not this merger Join Barton Poulson for an in-depth discussion in this video Clustering in Python, part of Data Science Foundations: Data MiningCluster Analysis sing u R . k-Means. Hierarchical clustering is an alternative Machine Learning Tutorial for K-means Clustering Algorithm using language R. I've worked through some clustering tutorials and I do get some output, however, the heatmap that I get after clustering does not correspond at all to the How to perform a cluster analysis and plot a dendrogram in R. Cluster analysis is a classification of objects from the data, where by classification we mean a labeling of objects with class (group) labels. The intuition of the formula is that we would like to find groups that are different with each other and each member of each group should be similar with the other members of each cluster. We have the capability to integrate R code in power BI and then leverage R visuals. 006 3. k-means clustering in Excel tutorial 2017-12-19. com/self-organising-maps-for-customer-segmentation-using-r/ Detailed tutorial on Practical Guide to Text Mining and Feature Engineering in R to improve your understanding of Machine Learning. While doing cluster analysis, we first partition the set of data into groups based on data similarity and then assign the labels to the groups. Hierarchical agglomera- tive cluster analysis begins by calculating a matrix of distances among items in this data ma- trix. Alabama was assigned to cluster 2, Arkansas was assigned to cluster 1, etc. Happy coding! This blog post was written by This Video, will show you how to do hierarchical clustering in R. I try SOM maps and clustering in R. This tutorial will help you set up and interpret a k-means Clustering in Excel using the XLSTAT software. 4 Tutorial on Spectral Clustering, ICML 2004, Chris Ding © University of California 7 Properties of Graph Laplacian Laplacian matrix of the Graph: L =D −W • L is Tutorial exercises Clustering – K-means, Nearest Neighbor and Hierarchical. Cluster Analysis in Data Mining from University of Illinois at Urbana-Champaign. Take the first step into image analysis in Python by using k-means clustering to analyze the dominant colors in an image in this free data science tutorial. May 27, 2014. The first step is to find the clusters within a set of points and which cluster each point belongs to. This implies running standard clustering methods such as k-means clustering on a reduced dataset. Wilkerson implements the Consensus Clustering method in R For this tutorial, migrated from stats. This free R tutorial by DataCamp is a great way to get started. Often times I receive inquiries on how to deploy R packages or conduct simulation studies on the Illinois Campus Cluster (ICC). Association Rule Mining with R. In the space of AI, Data Mining, or Machine Learning, often knowledge is captured and represented in the form of high dimensional vector or matrix. The indices of the vector (not displayed) indicate the indices of the source data items and the vector values are cluster IDs. Clustering in R. Discover the basic concepts of cluster analysis, and then study a set of typical clustering methodologies, algorithms, and applications. K-means clustering is a type of unsupervised learning, which is used when the resulting categories or groups in the data are unknown. The primary packages to get started are cluster and fpc. Big Data Analytics - K-Means Clustering. We will use the iris dataset from the datasets library. Being a newbie in R, I'm not very sure how to choose the best number of clusters to do a k-means analysis. Hierarchical Clustering Algorithms. 1 Load the sample data Restore the sample DB The dataset used in this tutorial is hosted in several SQL Server tables. A TUTORIAL ON SUBSPACE CLUSTERING Ren´e Vidal Johns Hopkins University. In recent years, spectral clustering has become one of the most popular modern clustering algorithms", null, "", null, "" ]
[ null, "http://germanclassesinpune.in/ns0mvuz/elements/", null, "http://germanclassesinpune.in/ns0mvuz/elements/techarticles/", null, "http://germanclassesinpune.in/ns0mvuz/elements/techarticles/", null, "http://germanclassesinpune.in/ns0mvuz/bvdacu5.php", null ]
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https://jelinekstat.readthedocs.io/en/latest/
[ "# jelinekstat: Jelinek’s Statistics for Second-Order Tensors¶\n\nApplication software in Python 3 to apply the statistical model for a sample of $$n$$ $$2^\\mathrm{nd}$$-order tensors Jelínek (1978) in order to obtain the mean tensor $$\\boldsymbol{k}$$ of the sample, the $$\\boldsymbol{k}$$‘s principal values $$k_1, k_2, k_3$$, with their confidence intervals, and the $$\\boldsymbol{k}$$‘s principal directions $$\\boldsymbol{p}_1, \\boldsymbol{p}_2, \\boldsymbol{p}_3$$ with their confidence regions.\n\nThis application program is able to plot the summary of the statistical model described above in a stereographic projection for a better understanding of the outcomes." ]
[ null ]
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https://math.stackexchange.com/questions/1033932/solve-system-of-equations
[ "# Solve system of equations\n\n$$\\sin(x+y)+1.6x=0$$\n\n$$x^2+y^2=-1$$\n\nCan this system be solved? Please help me with it. I managed to make graphs of it but can't get it solved without graph.\n\nGraph:", null, "• in which domain is that to be solve? – Dr. Sonnhard Graubner Nov 22 '14 at 17:52\n• I think no imaginary numbers only. I will need to use Newton method after solution to find intersection. – Neone Nov 22 '14 at 17:54\n\nHint: For what real $x$ and $y$ will we have $$x^2+y^2=-1,$$ if any?\n\nIf the domain is real number no square numbers sum will be negative.\n\nSo no x and y satisfy the equation\n\nsince we have for all real numbers $x^2+y^2\\geq 0>-1$ we get no real solutions for this system. you can only plug $y=\\sqrt{-x^2-1}$ or $y=-\\sqrt{-x^2-1}$ in your equation.\n\n• So it is not possible to use Newton method with this system? Or can we choose any x and y to start with? Because I have a graph. – Neone Nov 22 '14 at 17:59\n• it is also possible to use the newton method for complex numbers, or is a typo in your system? – Dr. Sonnhard Graubner Nov 22 '14 at 18:01\n• And what would be solution if we will use complex? – Neone Nov 22 '14 at 18:02\n\nSolved:\n$$sin(x+y)+1.6x=0$$ $$x^2+y^2=-1$$\n$$y=√(|-x^2-1|)$$\n\n$$sin(x+√(|-x^2-1|)+1.6*x=0$$\n\nx=-0.393536\n\n$$y=√(|0.393536^2-1|)$$\n\ny=0.91931\n\nLooking at the graph it seems these are correct values. Thank you, especially Dr. Sonnhard Graubner.\n\n• Your graph is incorrect. The values of $x$ and $y$ that you found are approximate solutions to $x^2+y^2=1,$ not to $x^2+y^2=-1.$ From $y^2=-x^2-1,$ we can conclude that $|y|=\\sqrt{-x^2-1},$ not that $y=\\sqrt{|-x^2-1|}.$ – Cameron Buie Nov 22 '14 at 19:07\n• Approximate values is what I need because later I will find them as accurate as needed according to error value with Newton method. If there is no x and y values which correspondent to $$x^2+y^2=−1$$ then I will calculate with 1 instead because work has to be done anyway.. Maybe this was a mistake made when filling equations. – Neone Nov 22 '14 at 19:33" ]
[ null, "https://i.stack.imgur.com/TESHH.png", null ]
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https://paperzz.com/doc/2061157/thermochemistry-%E2%80%93-hw-psi-chemistry-name-energy-1---njctl
[ "", null, "# Thermochemistry – HW PSI Chemistry Name Energy 1 - NJCTL\n\n```Thermochemistry – HW\nPSI Chemistry\nName _____________________________\nEnergy\n1) Objects can possess energy as:\n(a) endothermic energy (b) potential energy\nA) a only\nB) b only\nC) c only\nD) a and c\nE) b and c\n(c) kinetic energy\n2) The internal energy of a system\nA) is the sum of the kinetic energy of all of its components\nB) is the sum of the rotational, vibrational, and translational energies of all of its\ncomponents\nC) refers only to the energies of the nuclei of the atoms of the component molecules\nD) is the sum of the potential and kinetic energies of the components\nE) none of the above\n3) Which one of the following conditions would always result in an increase in the internal\nenergy of a system?\nA) The system loses heat and does work on the surroundings.\nB) The system gains heat and does work on the surroundings.\nC) The system loses heat and has work done on it by the surroundings.\nD) The system gains heat and has work done on it by the surroundings.\nE) None of the above is correct.\n4) The value of О”E for a system that performs 111 kJ of work on its surroundings and gains\n89 kJ of heat is __________ kJ.\nA) -111\nB) -200\nC) 200\nD) -22\nE) 22\n5) The value of О”E for a system that performs 13 kJ of work on its surroundings and loses 9\nkJ of heat is __________ kJ.\nA) 22\nB) -22\nC) -4\nD) 4\nE) -13\n6) Calculate the value of О”E in joules for a system that loses 50 J of heat and has 150 J of\nwork performed on it by the surroundings.\nA) 50\nB) 100\nC) -100\nD) -200\nE) +200\n7) The change in the internal energy of a system that absorbs 2,500 J of heat and that does\n7,655 J of work on the surroundings is __________ J.\nA) 10,155\nB) 5,155\nC) -5,155\nD) -10,155\nE) 1.91 x107\n8) The change in the internal energy of a system that releases 2,500 J of heat and that does\n7,655 J of work on the surroundings is __________ J.\nA) -10,155\nB) -5,155\nC) -1.91x107\nD) 10,155\nE) 5,155\n9) When a system __________, О”E is always negative.\nA) absorbs heat and does work\nB) gives off heat and does work\nC) absorbs heat and has work done on it\nD) gives off heat and has work done on it\nE) none of the above is always negative.\n10) Of the following, which one is a state function?\nA) H\nB) q\nC) w\nD) heat\nE) none of the above\nEnergy Changes\n11) Which one of the following is an endothermic process?\nA) ice melting\nB) water freezing\nC) boiling soup\nD) Hydrochloric acid and barium hydroxide are mixed at 25 В°C: the temperature\nincreases. E) Both A and C\n12) Which one of the following is an exothermic process?\nA) ice melting\nB) water evaporating\nC) boiling soup\nD) condensation of water vapor\nE) Ammonium thiocyanate and barium hydroxide are mixed at 25 В°C: the temperature\ndrops.\n13) A __________ О”H corresponds to an __________ process.\nA) negative, endothermic\nB) negative, exothermic\nC) positive, exothermic\nD) zero, exothermic\nE) zero, endothermic\n14) A __________ О”H corresponds to an __________ process.\nA) negative, endothermic\nB) positive, exothermic\nC) positive, endothermic\nD) zero, exothermic\nE) zero, endothermic zero, endothermic\n15) О”H for an endothermic process is _____ while О”H for an exothermic process is _______.\nA) zero, positive\nB) zero, negative\nC) positive, zero\nD) negative, positive\nE) positive, negative\n16) A chemical reaction that absorbs heat from the surroundings is said to be __________\nand has a __________ О”H.\nA) endothermic, positive\nB) endothermic, negative\nC) exothermic, negative\nD) exothermic, positive\nE) exothermic, neutral\n17) The reaction 4Al(s) + 3O2 (g) пѓ 2Al2O3 (s)\nО”HВ° = -3351 kJ\nis __________, and therefore heat is __________ by the reaction.\nA) endothermic, released\nB) endothermic, absorbed\nC) exothermic, released\nD) exothermic, absorbed\nE) thermoneutral, neither released nor absorbed\nCalorimetry\n*18) The molar heat capacity of a compound with the formula C H SO is 88.0 J/mol-K. The\n2\n6\nspecific heat of this substance is __________ J/g-K.\nA) 88.0\nB) 1.13\nC) 4.89\nD) 6.88x103\nE) -88.0\n19) A sample of aluminum metal absorbs 9.86 J of heat, and the temperature of the sample\nincreases from 23.2 В°C to 30.5 В°C. Since the specific heat capacity of aluminum is 0.90 J/gK, the mass of the sample is __________ g.\nA) 72\nB) 1.5\nC) 65\nD) 8.1\nE) 6.6\n20) The specific heat capacity of lead is 0.13 J/g-K. How much heat (in J) is required to raise\nthe temperature of 15g of lead from 22 В°C to 37 В°C?\nA) 2.0\nB) -0.13\nC) 5.8 x10-4\nD) 29\nE) 0.13\n21) The temperature of a 15-g sample of lead metal increases from 22 В°C to 37 В°C upon the\naddition of 29.0 J of heat. The specific heat capacity of the lead is __________ J/g-K.\nA) 7.8\nB) 1.9\nC) 29\nD) 0.13\nE) -29\n*22) The О”H for the solution process when solid sodium hydroxide dissolves in water is 44.4\nkJ/mol. When a 13.9-g sample of NaOH dissolves in 250.0 g of water in a coffee-cup\ncalorimeter, the temperature increases from 23.0 В°C to __________ В°C. Assume that the\nsolution has the same specific heat as liquid water, i.e., 4.18 J/g-K.\nA) 35.2 В°C\nB) 24.0 В°C\nC) 37.8 В°C\nD) 37.0 В°C\nE) 40.2 В°C\n23) An 8.29 g sample of calcium carbonate [CaCO3(s)] absorbs 50.3 J of heat, upon which\nthe temperature of the sample increases from 21.1 В°C to 28.5 В°C. What is the specific heat of\ncalcium carbonate?\nA) .63\nB) .82\nC) 1.1\nD) 2.2\nE) 4.2\n24) A sample of iron absorbs 67.5 J of heat, upon which the temperature of the sample\nincreases from 21.5 В°C to 28.5 В°C. If the specific heat of iron is 0.450 J/g-K, what is the mass\n(in grams) of the sample?\nA) 4.3\nB) 11\nC) 21\nD) 1100\nE) 1.1x 103\n25) A 22.44 g sample of iron absorbs 180.8 J of heat, upon which the temperature of the\nsample increases from 21.1 В°C to 39.0 В°C. What is the specific heat of iron?\nA) 0.140\nB) 0.450\nC) 0.820\nD) 0.840\nE) 0.900\n26) When 45 g of an alloy at 1000C is dropped into 100.0 g of water at 25В°C, the final\ntemperature is 37В°C. What is the specific heat of the alloy (J/goC)? 31) ______\nA)9.88\nB) 48.8\nC) 0.423\nD) 1.77\n*27) A 50.0-g sample of liquid water at 25.0 C is mixed with 29.0 g of water at 45.0 В°C. The\nfinal temperature of the water is __________.\nA) 102\nB) 27.6\nC) 35.0\nD) 142\nE) 32.3\n28) A 6.50-g sample of copper metal at 25.0 В°C is heated by the addition of 84.0 J of energy.\nThe final temperature of the copper is _____ В°C. The specific heat of copper is 0.38 J/g-K\nA) 29.9\nB) 25.0\nC) 9.0\nD) 59.0\nE) 34.0\nHeating curves\n29) How much heat is required to melt 1.5 moles of NaCl at its melting point (пЃ„Hfus = 30\nkJ/mol)\nA) 1.5 kJ\nB) 15 kJ\nC) 20 kJ\nD) 30 kJ\nE) 45 kJ\n30) A substance releases 500 kJ of heat as 25 mol of it condenses from a gas to a liquid .\nWhat is the heat of vaporization (пЃ„Hvap) of this substance?\nA) 20 kJ/mol\nB) 25 kJ/mol\nC) 475 kJ/mol\nD) 525 kJ/mol\n**31) If you supply 36 kJ of heat, how many moles of ice a 0 C can be melted, heated to its\no\nboiling point and completely boiled away?\nпЃ„Hvap = 40.5 kJ/mol; пЃ„Hfus = 6.0 kJ/mol; specific heat for water, C = 0.0753 kJ/mol-oC\n**32) Given the equation:\nH2O (l) пѓ H2O(g)\nпЃ„Hrxn = 40.7 kJ at 100 п‚°C\nCalculate the mass of liquid water (in grams) at 100 п‚°C that can converted to vapor by\nabsorbing 2.40 kJ of heat.\n33) Given the equation: H2O (l) пѓ H2O(g)\nпЃ„Hrxn = 40.7 kJ at 100 п‚°C\nCalculate the heat required to convert 3.00 grams of liquid water at 100 В°C to vapor.\n*34) Based on the following information, which compound has the strongest intermolecular\nforces?\nSubstance\nArgon (Ar)\nBenzene ( C 6 H 6 )\nEthanol ( C 2 H 5OH )\nWater ( H 2 O )\nО”Hvap (kJ/mol)\n6.3\n31.0\n39.3\n40.8\nMethane ( CH 4 )\nA) Argon\nB) Benzene\nC) Ethanol\nD) Water\nE) Methane\n9.2\nUse the information in this table to answer the next four questions.\nSpecific heat of ice\nSpecific heat of water\nSpecific heat of steam\nHeat of fusion of water ( пЃ„ Hfus)\n2.09 J/g-oC\n4.18 J/g-oC\n1.84 J/g-oC\n6.01 kJ/mol\nHeat of vaporization of water ( пЃ„ Hvap)\n40.7 kJ/mol\n35) What amount of heat (in kJ) is required to melt 35.0 g of ice at 0 oC?\nA) 3.09 kJ\nB) 11.7 kJ\nC) 20.9 kJ\nD) 79.1 kJ\n36) What amount of heat (in kJ) is required to completely convert 190.0 g of liquid water at\n18oC to steam at 100.0oC?\nA) 6.51 kJ\nB) 6.94 kJ\nC) 430 kJ\nD) 559 kJ\n37) The enthalpy change for converting 1.00 mol of ice at -50.0 В°C to water at 70.0 В°C is\n________kJ.\nA) 12.28\nB) 6.41\nC) 13.16\nD) 7154\nE) 9.40\n38) The enthalpy change for converting 10.0 g of ice at -25.0 В°C to water at 80.0 В°C is\n__________ kJ.\nA) 12.28\nB) 6.16\nC) 3870\nD) 7.21\nE) 9.88\n**39) The heat of fusion of water is 6.01 kJ/mol. The heat capacity of liquid water is 75.3\nJ/mol-K The conversion of 50.0 g of ice at 0.00 В°C to liquid water at 22.0 В°C requires\n__________ kJ of heat.\nA) 3.8 x102\nB) 21.3\nC) 17.2\nD) 0.469\nE) Insufficient data are given.\nEnthalpy\n40) Which of the following statements is false?\nA) Internal energy is a state function.\nB) Enthalpy is an intensive property.\nC) The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to\nthe enthalpy change for the reverse reaction.\nD) The enthalpy change for a reaction depends on the state of the reactants and\nproducts.\nE) The enthalpy of a reaction is equal to the heat of the reaction.\n41) The value of пЃ„H0 for the reaction below is -72 kJ. __________ kJ of heat are released\nwhen 1.0 mol of HBr is formed in this reaction: H2(g) + Br2 (g) пѓ 2HBr(g)\nA) 144\nB) 72\nC) 0.44\nD) 36\nE) -72\n42) The value of пЃ„H0 for the reaction below is -126 kJ. __________ kj are released when\n2.00 mol of NaOH is formed in the reaction? 2Na2O2(s) + 2H2O(l) пѓ 4NaOH(s) + O2(g)\nA) 252\nB) 63\nC) 3.9\nD) 7.8\nE) -126\n43) The value of пЃ„H0 for the reaction below is -790 kJ. The enthalpy change accompanying\nthe reaction of 0.95 g of S is __________ kJ. 2S(s) + 3O2 (g) пѓ 2SO3(g)\nA) 23\nB) -23\nC) -12\nD) 12\nE) -790\n44) The value of пЃ„H0 for the reaction below is -6535 kJ. __________ kJ of heat are released\nin the combustion of 16.0 g of C6H6(l)? 2C6H6(l) + 15O2 (g) пѓ 12CO2 (g) + 6H2O(l)\nA) 1.34 x 103\nB) 5.23 x014\nC) 669\nD) 2.68 x103\nE) -6535\n45) The value of пЃ„H0 for the reaction below is -482 kJ. Calculate the heat (kJ) released to\nthe surroundings when 12.0 g of CO (g) reacts completely. 2CO(g) + O2(g) пѓ 2CO2(g)\nA) 2.89x103\nB) 207\nC) 103\nD) 65.7\nE) -482\n46) The value of пЃ„H0 for the reaction below is -336 kJ. Calculate the heat (kJ) released to\nthe surroundings when 23.0 g of HCl is formed. CH4(g) +3Cl2 (g) пѓ CHCl3(l) + 3HCl(g)\nA) 177\nB) 2.57x103\nC) 70.7\nD) 211\nE) -336\n47) The value of пЃ„H0 for the reaction below is -186 kJ. Calculate the heat (kJ) released from\nthe reaction of 25 g of Cl2. H2 (g) + Cl2 (g) пѓ 2HCl (g)\nA) 66\nB) 5.3x102\nC) 33\nD) 47\nE) -186\n48) The enthalpy change for the following reaction is -483.6 kJ: 2H2 (g) + O2 (g) пѓ 2H2O(g)\nTherefore, the enthalpy change for the following reaction is __________ kJ:\n4H2 (g) + 2O2 (g) пѓ 4H2O(g)\nA) -483.6\nB) -967.2\nC) 2.34x105\nD) 483.6\nE) 967.2\n49) The value of пЃ„HВ° for the reaction below is +128.1 kJ: CH3OH (l) пѓ CO(g) + 2H2(g)\nHow many kJ of heat are consumed when 15.5 g of CH3OH(l) decomposes as shown in the\nequation?\nA) 0.48\nB) 62.0\nC) 1.3x102\nD) 32\nE) 8.3\n50) The value of пЃ„H0 for the reaction below is +128.1 kJ: CH3OH (l) пѓ CO(g) + 2H2(g)\nHow many kJ of heat are consumed when 5.10 g of H2(g) is formed as shown in the\nequation?\nA) 162\nB) 62.0\nC) 128\nD) 653\nE) 326\n51) The value of пЃ„H0 for the reaction below is +128.1 kJ: CH3OH (l) пѓ CO(g) + 2H2(g)\nHow many kJ of heat are consumed when 5.10 g of CO (g) are formed?\nA) 0.182\nB) 162\nC) 8.31\nD) 23.3\nE) 62.0\n52) The value of пЃ„H0 for the reaction below is +128.1 kJ: CH3OH (l) пѓ CO(g) + 2H2(g)\nHow many kJ of heat are consumed when 5.75 g of CO(g) is formed?\nA) 23.3\nB) 62.0\nC) 26.3\nD) 162\nE) 8.3\n53) The value of пЃ„H0 for the reaction below is -1107 kJ: 2Ba(s) + O2 (g) пѓ 2BaO (s)\nHow many kJ of heat are released when 5.75 g of Ba(s) reacts completely with oxygen to\nform BaO(s)?\nA) 96.3\nB) 26.3\nC) 46.4\nD) 23.2\nE) 193\n54) The value of пЃ„H0 for this reaction is -1107 kJ: 2Ba(s) + O2 (g) пѓ 2BaO (s)\nHow many kJ of heat are released when 5.75 g of BaO(s) is produced?\nA) 56.9\nB) 23.2\nC) 20.8\nD) 193\nE) 96.3\n55) The value of пЃ„H0 for the reaction below is -1107 kJ: 2Ba(s) + O2 (g) пѓ 2BaO (s)\nHow many kJ of heat are released when 15.75 g of Ba(s) reacts completely with oxygen to\nform BaO(s)?\nA) 20.8\nB) 63.5\nC) 114\nD) 70.3\nE) 35.1\nHess’ Law of Heat Summation\n56) Which of the following is a statement of Hess's law?\nA) If a reaction is carried out in a series of steps, the О”H for the reaction will equal the\nsum of the enthalpy changes for the individual steps.\nB) If a reaction is carried out in a series of steps, the О”H for the reaction will equal the\nproduct of the enthalpy changes for the individual steps.\nC) The О”H for a process in the forward direction is equal in magnitude and opposite in\nsign to the О”H for the process in the reverse direction.\nD) The О”H for a process in the forward direction is equal to the О”H for the process in\nthe reverse direction.\nE) The О”H of a reaction depends on the physical states of the reactants and products.\n*57) О”H for the reaction IF (g) пѓ IF (g) + F (g) is __________ kJ, give the data below.\n5\n3\n2\nIF (g) + F2(g) пѓ IF3 (g) )\nIF (g) + 2F2(g) пѓ IF5 (g)\nО”H = -390 kJ\nО”H = -745 kJ\nA) +355\nB) -1135\nC) +1135\nD) +35\nE) -35\n*58) Given the following reactions the enthalpy of the reaction of the nitrogen to produce\nnitric oxide N2 (g) + O2(g) пѓ 2NO (g) is __________ kJ.\nN2 (g) + 2O2(g) пѓ 2NO2 (g)\nО”H = 66.4 kJ\n2NO (g) + O2(g) пѓ 2NO2 (g)\nО”H = -114.2 kJ\nA) 180.6\nB) -47.8\nC) 47.8\nD) 90.3\nE) -180.6\n*59) Given the following reactions the enthalpy of the reaction of nitrogen with oxygen to\nproduce nitrogen dioxide N2 (g) + 2O2(g) пѓ 2NO2 (g) is __________ kJ.\n(1) 2NO пѓ N2 + O2\n(2) 2NO + O2 пѓ 2NO2\nО”H = -180 kJ\nО”H = -112 kJ\nA) 68\nB) -68\nC) -292\nD) 292\nE) -146\n*60) Given the following reactions the enthalpy of the reaction\nCaCO3 (s) пѓ CaO(s) + C (s) + O2 (g) is __________ kJ.\nCaCO3 (s) пѓ CaO(s) + CO2(g)\nC(s) + O2 (g) пѓ CO2(g)\nA) 215.4\nB) 571.6\nC) -215.4\nD) -571.6\nE) 7.01x104\nО”H = 178.1 kJ\nО”H = -393.5 kJ\n*61) Given the following reactions the enthalpy for the decomposition of liquid water into\ngaseous hydrogen and oxygen 2H2O (l) пѓ 2H2 (g) + O2(g) is __________ kJ.\nH2O (l) пѓ H2O (g)\nО”H = 44.01 kJ\n2H2 (g) O2(g) пѓ 2H2O (g)\nО”H = -483.64 kJ\nA) -395.62\nB) -527.65\nC) 439.63\nD) 571.66\nE) 527.65\n*62) Given the following reactions the enthalpy for the decomposition of nitrogen dioxide into\nmolecular nitrogen and oxygen 2NO2 (g) пѓ N2(g) + 2O2 (g) is __________ kJ.\nN2 (g) O2(g) пѓ 2NO(g)\nО”H = +180.7 kJ\n2NO (g) + O2 (g) пѓ 2NO2 (g)\nО”H = -113.1 kJ\nA) 67.6\nB) -67.6\nC) 293.8\nD) -293.8\nE) 45.5\n*63) Given the following reactions the enthalpy of reaction for 4NO(g) пѓ 2NO (g) + N (g)\n2\nis ____ kJ.\nN2 (g) + O2(g) пѓ 2NO (g)\n2NO(g) +O2 (g) пѓ 2NO2 (g)\nA) 67.6\nB) 45.5\nC) -293.8\nD) -45.5\nО”H = +180.7 kJ\nО”H = -113.1 kJ\n2\nE) 293.8\n*64) Given the following reactions the enthalpy of reaction for 2N O(g) пѓ 2NO(g) +N (g)\n2\n2\nis __________ kJ.\nО”H = +180.7 kJ\nО”H = -163.2 kJ\nN2 (g) + O2(g) пѓ 2NO(g)\n2N2O(g) пѓ O2 (g) + 2N2(g)\nA) 145.7\nB) 343.9\nC) -343.9\nD) 17.5\nE) -145.7\nHeats of Formation\n65) Of the following, О”Hf0 is not zero for __________.\nA) O2(g)\nB) C (graphite)\nC) N2(g)\nD) F2 (s)\nE) Cl2(g)\n66) In the reaction below, О”Hf0 is zero for __________.\nNi(s) + 2CO (g) +2PF3 (g) пѓ Ni(CO)2 (PF3)2(l)\nA) Ni(s)\nB) CO(g)\nC) PF3(g)\nD) Ni(CO)2 (PF3)2(l)\nE) both CO and PF3\n67) For the species in the reaction below, О”Hf0is zero for __________.\n2Co(s) + H2(g) + 8PF3 (g) пѓ 2HCo(PF3)4 (I)\nA) Co(s)\nB) H2(g)\nC) PF3 (g)\nD) HCo(PF3)4\nE) both Co(s) and H2(g)\n**68) For which one of the following equations is\nО”H0 rxn equal to О”Hf0 for the product?\nA) Xe (g) + 2F2 (g) пѓ XeF4 (g)\nB) CH4 (g) +2Cl2(g) пѓ CH2Cl2 (l) + 2HCl(g)\nC) N2(g) + O3 (g) пѓ N2O3 (g)\nD) 2CO(g) +O2(g) пѓ 2CO2\nE) C(diamond) + O2(g) пѓ CO2 (g)\n**69) For which of the following reactions is the value of О”H rxn\n0\nfor the product?\nA) 2Ca(s) + O2 (g) пѓ 2CaO(s)\nB) C2H2(g) + H2(g) пѓ C2H4 (g)\nequal to О”Hf0\nC) 2C(graphite) +O2(g) пѓ 2CO (g)\nD) 3Mg(s) + N2(g) пѓ Mg3N2(s)\nE) C(diamond) + O2(g) пѓ CO2 (g)\n**70) For which of the following reactions is the value of О”H rxn\n0\nequal to О”Hf0 for the\nproduct?\nA) 2C(graphite) + 2H2(g) пѓ C2H4(g)\nB) N2(g) +O2(g) пѓ 2NO(g)\nC) 2H2(g) +O2(g) пѓ 2H2O(l)\nD) 2H2(g) + O2(g) пѓ 2H2O(g)\nE) H2O(l) +1/2 O2(g) пѓ 2H2O(l)\n71) For which of the following reactions is the value of О”H0 rxn equal to О”Hf0 for the product?\nA) H2O(l) +1/2 O2(g) пѓ H2O2(l)\nB) N2(g) +O2(g) пѓ 2NO(g)\nC) 2H2(g) + O2(g) пѓ 2H2O(l)\nD) 2H2(g) + O2(g) пѓ 2H2O(g)\nE) none of the above\n72) For which of the following reactions is the value of О”H0 rxn equal to О”Hf0 for the\nproduct?\nA) H2(g) + 1/2O2(g) пѓ H2O(l)\nB) H2(g) + O2(g) пѓ H2O2 (l)\nC) 2C(graphite) + 2H2(g) пѓ C2H4(g)\nD) 1/2N2(g) + O2(g) пѓ NO2(g)\nE) all of the above\n73) The value of пЃ„Hп‚° for the reaction below is -186 kJ.\nThe value of О”Hf0 for HCl (g) is __________ kJ/mol.\nA) -3.72x102\nB) -1.27x102\nC) -93.0\nD) -186\nE) +186\nH2(g) + Cl2(g) пѓ 2HCl(g)\n74) The value of пЃ„Hп‚° for the following reaction is -3351 kJ: 2Al(s) +3O2(g) пѓ 2Al2O3\nThe value of О”Hf0 for Al2O3 is __________ kJ.\nA) -3351\nB) -1676\nC) -32.86\nD) -16.43\nE) +3351\n75) Given the data in the table below, О”H0 rxn for this reaction is __________ kJ.\nCa(OH)2 + 2H3AsO4 пѓ Ca(H2AsO4)2 + 2H2O\nSubstance\nО”Hf0 (kJ/mol)\nCa(OH)2\nH3AsO4\nCa(H2AsO4)2\nH2O\n-986.6\n-900.4\n-2346.0\n-285.9\nA) -744.9\nB) -4519\nC) -4219\nD) -130.4\nE) -76.4\n76) Given the data in the table below, О”H0 rxn for this reaction is __________ kJ.\n4NH3 (g) + 5O2)g) пѓ 4NO(g) + 6H2O(l)\nSubstance\nО”Hf0 (kJ/mol)\nH2O(l)\nNO (g)\nNO2 (g)\nHNO3 (aq)\nNH3 (g)\n-286\n90\n34\n-207\n-46\nA) -1172\nB) -150\nC) -1540\nD) -1892\nE) The О”Hf0 of O 2 (g) is needed.\n77) Given the data in the table below, О”H0 rxn for this reaction is __________ kJ.\nC2H5OH(l) +O2(g) пѓ CH3CO2H(l) + H2O(l)\nSubstance\nО”Hf0 (kJ /mol)\nC2H4(g)\nC2H5OH(l)\nCH3CO2H(l)\nH2O(l)\n523\n-277.7\n-484.5\n-285.8\nA) -79.0\nB) -1048.0\nC) -476.4\nD) -492.6\nE) The value of О”Hf0 of O 2 (g) is required\n78) Given the data in the table below, О”H0 rxn for this reaction is __________ kJ.\n3NO2 +H2O (l) пѓ 2HNO3 (aq) + NO(g)\nSubstance\nО”Hf0 (kJ/mol)\nH2O(l)\nNO (g)\nNO2 (g)\nHNO3 (aq)\nNH3 (g)\nA) 64\nB) 140\nC) -140\nD) -508\nE) -64\n-286\n90\n34\n-207\n-46\n79) Given the data in the table below, О”H0 rxn for this reaction is __________ kJ.\nIF5(g) + F2(g) пѓ IF7(g)\nSubstance\nIF(g)\nIF5(g)\nIF7(g)\nО”Hf0 (kJ /mol)\n-95\n-840\n-941\nA) 1801\nB) -1801\nC) 121\nD) -121\nE) -101\n80) Given the data in the table below, пЃ„Hп‚° for the reaction is __________ kJ.\n2CO (g) + O2 пѓ 2CO2(g)\nSubstance\nCO(g)\nCO2(g)\nCaCO3(s)\nО”Hf0 (kJ /mol)\n-110.5\n-393.7\n-1207.0\nA) -566.4\nB) -283.2\nC) 283.2\nD) -677.0\nE) The О”Hf0 of O 2 (g) is needed\n*81) The value of О”HВ° for the following reaction is 177.8 kJ. The value of О”H\nf\n0\nfor\nCaO(s) is __________ kJ/mol.\nCaCO3(s) пѓ CaO(s) + CO2(g)\nSubstance\nCO(g)\nCO2(g)\nCaCO3(s)\nО”Hf0 (kJ /mol)\n-110.5\n-393.7\n-1207.0\nA) -1600\nB) -813.4\nC) -635.5\nD) 813.4\nE) 177.8\n82) Given the data in the table below, О”H0 rxn for the reaction is __________ kJ.\n2Ag2S(s) +O2(g) пѓ 2Ag2O(s) +2S(s)\nSubstance\nAg2O(s)\nAg2S(s)\nH2S(g)\nH2O(l)\nО”Hf0 (kJ /mol)\n-31.0\n-326\n-20.6\n-286\nA) -1.6\nB) +1.6\nC) -3.2\nD) +3.2\nE) The О”Hf0 of S(s) and of O2(g) are needed.\n83) Given the data in the table below, О”H0 rxn for the reaction is __________ kJ.\nPCl3(g) +3HCl(g) пѓ 3Cl2(g) +PH3(g)\nSubstance\nPCl3(g)\nHCl(g)\nPH3(g)\nО”Hf0 (kJ /mol)\n-288.07\n-92.30\n5.40\nA) -570.37\nB) -385.77\nC) 570.37\nD) 385.77\nE) The О”Hf0 of Cl2(g) is needed for the calculation.\n.\n*84) Given the data in the table below and О”H rxn\n0\nfor the reaction, О”Hf0 of HCl(g) is\n__________ kJ/mol.\nSO2Cl2(g) +2H2O(l) пѓ H2SO4(l) + 2HCl(g)\nSubstance\nО”Hf0 (kJ /mol)\nSO2(g)\n-297\nSO3(g)\n-396\nSO2Cl2(g) -364\nH2SO4(l)\n-814\nH2O(l)\n-286\nA) -184\nB) 60\nC) -92\nD) 30\nE) Insufficient data are given.\nО”Hf0 = -62 kJ\nEnergy\n1) E\n2) D\n3) D\n4) D\n5) B\n6) B\n7) C\n8) A\n9) B\n10) A\n11) E\n12) D\n13) B\n14) C\n15) E\n16) A\n17) C\nCalorimetry\n18) B\n19) B\n20) D\n21) D\n22) D\n23) B\n24) C\n25) B\n26) D\n27) E\n28) D\nHeating Curves\n29) E\n30) A\n31) 0.67 mol water\n32) 1.06 g\n33) 6.78 kJ\n34) D\n35) B\n36) D\n37) C\n38) D\n39) B\nEnthalpy\n40) B\n41) D\n42) B\n43) A\n44) C\n45) C\n46) C\n47) A\n48) B\n49) B\n50) A\n51) D\n52) C\n53) D\n54) C\n55) B\nHess’ Law\n56) A\n57) A\n58) A\n59) A\n60) B\n61) D\n62) B\n63) C\n64) D\nHeats of Formation\n65) D\n66) A\n67) E\n68) A\n69) D\n70) A\n71) E\n72) E\n73) C\n74) B\n75) D\n76) A\n77) D\n78) C\n79) E\n80) A\n81) C\n82) D\n83) C\n84) C\n```\n###### Document\nCategory\nEducation\nViews\n457\nFile Size\n352 KB\nTags\n1/--pages\nReport inappropriate content" ]
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https://www.edwinwenink.xyz/posts/65-stemming_and_lemmatizing_with_sklearn_vectorizers/
[ "# Stemming and lemmatizing with sklearn vectorizers { programmingsklearnNLPIR } · 705 words · 4 minute read\n\nThis post is part of the programming series.\n\nOne of the most basic techniques in Natural Language Processing (NLP) is the creation of feature vectors based on word counts.\n\nscikit-learn provides efficient classes for this:\n\nfrom sklearn.feature_extraction.text import CountVectorizer, TfidfVectorizer\n\n\nIf we want to build feature vectors over a vocabulary of stemmed or lemmatized words, how can we do this and still benefit from the ease and efficiency of using these sklearn classes?\n\n## Vectorizers: the basic use case ¶\n\nConceptually, these vectorizers first build up the vocabulary of your whole text corpus. The size of the vocabulary determines the length of the feature vectors, unless you specify a maximum amount of features (which you probably should, cf. Zipfs law ). The vectorizers check for each document how often a certain word (or n-gram, technically) occurs in that document. The CountVectorizer only takes the term frequency (TF) into account. However, words that occur in almost all the documents (like stop words) are not very useful for characterizing individual documents and distinguishing them from others.\n\nWe should treat matches on non-frequent terms as more valuable than ones on frequent terms, without disregarding the latter altogether. The natural solution is to correlate a term’s matching value with its collection frequency. (Karen Spärk Jones, 1972)\n\nThe TfidfVectorizer therefore additionally weighs the word frequency with how common the word is in the whole corpus. If a word occurs a lot in document $t$ but is quite rare throughout the whole corpus, then this is a useful word to characterize the current document. Conversely, if a term is frequent in document $t$ but it occurs a lot in literally every other document as well, then it is a poor descriptor. In its most basic form (without smoothing etc.) TF*IDF scoring looks like this:\n\n$$TF(t,d) * log (\\frac{N}{DF(t)})$$\n\nWhere $TF(t,d)$ is the frequency of term $t$ in document $d$, $N$ is the total amount of documents in the corpus, and $DF(t)$ is the amount of documents in which term $t$ occurs. The logarithm is called the Inverse Document Frequency (IDF), hence we get TF*IDF. The logarithm prevents that very rare words completely dominate the score. Additionally, it punishes the most frequent words relatively heavy.\n\n## Composing a new tokenizer ¶\n\nIt is very convenient and efficient to use the sklearn vectorizers, but how can we use them when we want to do additional natural language processing during the building of the corpus vocabulary?\n\nVectorizers can be customised with three arguments: 1) preprocessor, 2) tokenizer, and 3) analyzer:\n\n1. The preprocessor is a callable that operates on a whole string and returns a whole string.\n2. The tokenizer takes the preprocessor output and returns a list of tokens.\n3. The analyzer is a callable that replaces the whole pipeline, including preprocessing and tokenization, and I think also including N-gram extraction and stop word filtering.\n\nSo in order to add stemming or lemmatization to the sklearn vectorizers, a good approach is to include this in a custom tokenize function. This does assume our stemming and lemmatization functions only need access to tokens, instead of the whole input strings (may be documents, sections, paragraphs, sentences etc.).\n\nThis is a very nice snippet to compose functions using functools\n\nimport functools\n\ndef compose(*functions):\n'''\nCompose an arbitary amount of functions into a single function\nSource: https://mathieularose.com/function-composition-in-python\n'''\ndef comp(f, g):\nreturn lambda x: f(g(x))\nreturn functools.reduce(comp, functions, lambda x: x)\n\n\nAssuming we have some class where we can assign a stemmer, a lemmatizer, or neither, we can override the tokenizer as follows:\n\n# If a stemmer or lemmatizer is provided in the configuration\n# compose a new tokenization function that includes stemming/lemmatization after tokenization.\n# This allows stemming or lemmatization to be integrated e.g. with CountVectorizer\nif stemmer:\nself._tokenize = compose(self._stemmer.stem, self._tokenizer.tokenize)\nelif lemmatizer:\nself._tokenize = compose(self._lemmatizer.lemmatize, self._tokenizer.tokenize)\nelse:\nself._tokenize = self._tokenizer.tokenize\n\n\nNote that the order of the composition matters, because the function signatures differ. A tokenization function takes a string as an input and outputs a list of tokens, and our stemming or lemmatization function then operates on this list of tokens. We can now define a TfidfVectorizer with our custom callable!\n\nngram_range = (1,1)\nmax_features = 1000\nuse_idf = True\n\ntfidf = TfidfVectorizer(tokenizer=self._tokenize,\nmax_features=max_features,\nngram_range=ngram_range,\nmin_df=1,\nmax_df=1.0,\nuse_idf=use_idf)\n\n\nThe vocabulary will now consist of stems and lemmas." ]
[ null ]
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https://answers.everydaycalculation.com/compare-fractions/7-9-and-9-50
[ "Solutions by everydaycalculation.com\n\n## Compare 7/9 and 9/50\n\n7/9 is greater than 9/50\n\n#### Steps for comparing fractions\n\n1. Find the least common denominator or LCM of the two denominators:\nLCM of 9 and 50 is 450\n\nNext, find the equivalent fraction of both fractional numbers with denominator 450\n2. For the 1st fraction, since 9 × 50 = 450,\n7/9 = 7 × 50/9 × 50 = 350/450\n3. Likewise, for the 2nd fraction, since 50 × 9 = 450,\n9/50 = 9 × 9/50 × 9 = 81/450\n4. Since the denominators are now the same, the fraction with the bigger numerator is the greater fraction\n5. 350/450 > 81/450 or 7/9 > 9/50\n\nMathStep (Works offline)", null, "Download our mobile app and learn to work with fractions in your own time:" ]
[ null, "https://answers.everydaycalculation.com/mathstep-app-icon.png", null ]
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http://www.hextobinary.com/unit/byte/from/Mb/to/PiB/49
[ "### Megabit Pebibyte\n\n##### How many Pebibytes are in 49 Megabits?\n\nPeople do ask questions such as 49 Megabits are equal to how many Pebibytes? The answer is 49 Megabits are equal to 5.440092820663267e-9 Pebibytes. You can also Convert 50 Megabits to Pebibyte\n\n##### How to Convert 49 Megabits to Pebibytes (mb to pib)\n\nWe are going to use very simple Megabit to Pebibyte conversion formula for that. By using our Megabit to Pebibyte conversion tool, you know that one Megabit is equivalent to 1.1102230246251565e-10 Pebibyte. Hence, to convert Megabit to Pebibyte, we just need to multiply the number by 1.1102230246251565e-10. Pleas see the calculation example given below.\n\nConvert 49 Megabit to Pebibyte 49 Megabit = 49 × 1.1102230246251565e-10 = 5.440092820663267e-9 Pebibyte\n\n##### What is Megabit Unit of Measure?\n\nMegabit is a unit of digital information about data. One megabit is equal to 1000000 bits.\n\n##### What is Pebibyte Unit of Measure?\n\nPebibyte is a unit of digital information about data. One pebibyte is equal to 1125899906842624 bytes.\n\n##### Megabit to Pebibyte Conversion Chart\n Megabit [Mb] Pebibyte [PiB] 1 1.1102230246251565e-10 2 2.220446049250313e-10 3 3.3306690738754696e-10 4 4.440892098500626e-10 5 5.551115123125783e-10 6 6.661338147750939e-10 7 7.771561172376096e-10 8 8.881784197001252e-10 9 9.992007221626409e-10 10 1.1102230246251565e-9 100 1.1102230246251565e-8 1000 1.1102230246251565e-7\n##### Megabit to Other Units Conversion Chart\n Megabit [Mb] Output 49 Megabit in Bit equals to 49000000 49 Megabit in Byte equals to 6125000 49 Megabit in Kilobit equals to 49000 49 Megabit in Kibibit equals to 47851.56 49 Megabit in Kilobyte equals to 6125 49 Megabit in Kibibyte equals to 5981.45 49 Megabit in Mebibit equals to 46.73 49 Megabit in Megabyte equals to 6.13 49 Megabit in Mebibyte equals to 5.84 49 Megabit in Gigabit equals to 0.049 49 Megabit in Gibibit equals to 0.04563480615615845 49 Megabit in Gigabyte equals to 0.006125 49 Megabit in Gibibyte equals to 0.005704350769519806 49 Megabit in Terabit equals to 0.000049 49 Megabit in Tebibit equals to 0.000044565240386873484 49 Megabit in Terabyte equals to 0.000006125 49 Megabit in Tebibyte equals to 0.0000055706550483591855 49 Megabit in Petabit equals to 4.9e-8 49 Megabit in Pebibit equals to 4.3520742565306136e-8 49 Megabit in Petabyte equals to 6.125e-9 49 Megabit in Pebibyte equals to 5.440092820663267e-9 49 Megabit in Exabit equals to 4.9e-11 49 Megabit in Exbibit equals to 4.2500725161431774e-11 49 Megabit in Exabyte equals to 6.125e-12 49 Megabit in Exbibyte equals to 5.312590645178972e-12 49 Megabit in Zettabit equals to 4.9e-14 49 Megabit in Zebibit equals to 4.1504614415460717e-14 49 Megabit in Zettabyte equals to 6.125e-15 49 Megabit in Zebibyte equals to 5.1880768019325896e-15 49 Megabit in Yottabit equals to 4.9e-17 49 Megabit in Yobibit equals to 4.0531850015098356e-17 49 Megabit in Yottabyte equals to 6.125e-18 49 Megabit in Yobibyte equals to 5.0664812518872945e-18" ]
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https://www.colorhexa.com/00cd1b
[ "# #00cd1b Color Information\n\nIn a RGB color space, hex #00cd1b is composed of 0% red, 80.4% green and 10.6% blue. Whereas in a CMYK color space, it is composed of 100% cyan, 0% magenta, 86.8% yellow and 19.6% black. It has a hue angle of 127.9 degrees, a saturation of 100% and a lightness of 40.2%. #00cd1b color hex could be obtained by blending #00ff36 with #009b00. Closest websafe color is: #00cc33.\n\n• R 0\n• G 80\n• B 11\nRGB color chart\n• C 100\n• M 0\n• Y 87\n• K 20\nCMYK color chart\n\n#00cd1b color description : Strong lime green.\n\n# #00cd1b Color Conversion\n\nThe hexadecimal color #00cd1b has RGB values of R:0, G:205, B:27 and CMYK values of C:1, M:0, Y:0.87, K:0.2. Its decimal value is 52507.\n\nHex triplet RGB Decimal 00cd1b `#00cd1b` 0, 205, 27 `rgb(0,205,27)` 0, 80.4, 10.6 `rgb(0%,80.4%,10.6%)` 100, 0, 87, 20 127.9°, 100, 40.2 `hsl(127.9,100%,40.2%)` 127.9°, 100, 80.4 00cc33 `#00cc33`\nCIE-LAB 72.054, -72.418, 66.953 22.028, 43.739, 8.318 0.297, 0.59, 43.739 72.054, 98.626, 137.245 72.054, -67.926, 85.772 66.136, -56.284, 38.838 00000000, 11001101, 00011011\n\n# Color Schemes with #00cd1b\n\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #cd00b2\n``#cd00b2` `rgb(205,0,178)``\nComplementary Color\n• #4ccd00\n``#4ccd00` `rgb(76,205,0)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #00cd82\n``#00cd82` `rgb(0,205,130)``\nAnalogous Color\n• #cd004c\n``#cd004c` `rgb(205,0,76)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #8200cd\n``#8200cd` `rgb(130,0,205)``\nSplit Complementary Color\n• #cd1b00\n``#cd1b00` `rgb(205,27,0)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #1b00cd\n``#1b00cd` `rgb(27,0,205)``\n• #b2cd00\n``#b2cd00` `rgb(178,205,0)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #1b00cd\n``#1b00cd` `rgb(27,0,205)``\n• #cd00b2\n``#cd00b2` `rgb(205,0,178)``\n• #008111\n``#008111` `rgb(0,129,17)``\n• #009a14\n``#009a14` `rgb(0,154,20)``\n• #00b418\n``#00b418` `rgb(0,180,24)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #00e71e\n``#00e71e` `rgb(0,231,30)``\n• #01ff22\n``#01ff22` `rgb(1,255,34)``\n• #1bff39\n``#1bff39` `rgb(27,255,57)``\nMonochromatic Color\n\n# Alternatives to #00cd1b\n\nBelow, you can see some colors close to #00cd1b. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #18cd00\n``#18cd00` `rgb(24,205,0)``\n• #07cd00\n``#07cd00` `rgb(7,205,0)``\n• #00cd0a\n``#00cd0a` `rgb(0,205,10)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #00cd2c\n``#00cd2c` `rgb(0,205,44)``\n• #00cd3d\n``#00cd3d` `rgb(0,205,61)``\n• #00cd4e\n``#00cd4e` `rgb(0,205,78)``\nSimilar Colors\n\n# #00cd1b Preview\n\nThis text has a font color of #00cd1b.\n\n``<span style=\"color:#00cd1b;\">Text here</span>``\n#00cd1b background color\n\nThis paragraph has a background color of #00cd1b.\n\n``<p style=\"background-color:#00cd1b;\">Content here</p>``\n#00cd1b border color\n\nThis element has a border color of #00cd1b.\n\n``<div style=\"border:1px solid #00cd1b;\">Content here</div>``\nCSS codes\n``.text {color:#00cd1b;}``\n``.background {background-color:#00cd1b;}``\n``.border {border:1px solid #00cd1b;}``\n\n# Shades and Tints of #00cd1b\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #000901 is the darkest color, while #f4fff6 is the lightest one.\n\n• #000901\n``#000901` `rgb(0,9,1)``\n• #001c04\n``#001c04` `rgb(0,28,4)``\n• #003006\n``#003006` `rgb(0,48,6)``\n• #004409\n``#004409` `rgb(0,68,9)``\n• #00570b\n``#00570b` `rgb(0,87,11)``\n• #006b0e\n``#006b0e` `rgb(0,107,14)``\n• #007f11\n``#007f11` `rgb(0,127,17)``\n• #009213\n``#009213` `rgb(0,146,19)``\n• #00a616\n``#00a616` `rgb(0,166,22)``\n• #00b918\n``#00b918` `rgb(0,185,24)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\n• #00e11e\n``#00e11e` `rgb(0,225,30)``\n• #00f420\n``#00f420` `rgb(0,244,32)``\n• #09ff29\n``#09ff29` `rgb(9,255,41)``\n• #1cff3a\n``#1cff3a` `rgb(28,255,58)``\n• #30ff4b\n``#30ff4b` `rgb(48,255,75)``\n• #44ff5c\n``#44ff5c` `rgb(68,255,92)``\n• #57ff6d\n``#57ff6d` `rgb(87,255,109)``\n• #6bff7e\n``#6bff7e` `rgb(107,255,126)``\n• #7fff8f\n``#7fff8f` `rgb(127,255,143)``\n• #92ffa0\n``#92ffa0` `rgb(146,255,160)``\n• #a6ffb2\n``#a6ffb2` `rgb(166,255,178)``\n• #b9ffc3\n``#b9ffc3` `rgb(185,255,195)``\n• #cdffd4\n``#cdffd4` `rgb(205,255,212)``\n• #e1ffe5\n``#e1ffe5` `rgb(225,255,229)``\n• #f4fff6\n``#f4fff6` `rgb(244,255,246)``\nTint Color Variation\n\n# Tones of #00cd1b\n\nA tone is produced by adding gray to any pure hue. In this case, #5f6e61 is the less saturated color, while #00cd1b is the most saturated one.\n\n• #5f6e61\n``#5f6e61` `rgb(95,110,97)``\n• #57765b\n``#57765b` `rgb(87,118,91)``\n• #4f7e55\n``#4f7e55` `rgb(79,126,85)``\n• #47864f\n``#47864f` `rgb(71,134,79)``\n• #3f8e49\n``#3f8e49` `rgb(63,142,73)``\n• #379644\n``#379644` `rgb(55,150,68)``\n• #2f9e3e\n``#2f9e3e` `rgb(47,158,62)``\n• #27a638\n``#27a638` `rgb(39,166,56)``\n``#20ad32` `rgb(32,173,50)``\n• #18b52c\n``#18b52c` `rgb(24,181,44)``\n• #10bd27\n``#10bd27` `rgb(16,189,39)``\n• #08c521\n``#08c521` `rgb(8,197,33)``\n• #00cd1b\n``#00cd1b` `rgb(0,205,27)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #00cd1b is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.5000817,"math_prob":0.78700846,"size":3654,"snap":"2021-31-2021-39","text_gpt3_token_len":1641,"char_repetition_ratio":0.13452055,"word_repetition_ratio":0.011090573,"special_character_ratio":0.545977,"punctuation_ratio":0.23250565,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98359215,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-24T02:13:54Z\",\"WARC-Record-ID\":\"<urn:uuid:2852df06-95a2-4002-a0e7-aef025a19bd5>\",\"Content-Length\":\"36207\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:10517315-50ca-467b-a977-7f54e24aec11>\",\"WARC-Concurrent-To\":\"<urn:uuid:3edd1378-2c3a-412a-b848-cd009580ab11>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/00cd1b\",\"WARC-Payload-Digest\":\"sha1:JOMNTVX5NEVJS6FE5GJVBAAHRZUWN2SH\",\"WARC-Block-Digest\":\"sha1:I3ZLIQZV45TYH53X674GEFSPCMOZLLDG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046150067.87_warc_CC-MAIN-20210724001211-20210724031211-00316.warc.gz\"}"}
https://www.colorhexa.com/443a3a
[ "# #443a3a Color Information\n\nIn a RGB color space, hex #443a3a is composed of 26.7% red, 22.7% green and 22.7% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 14.7% magenta, 14.7% yellow and 73.3% black. It has a hue angle of 0 degrees, a saturation of 7.9% and a lightness of 24.7%. #443a3a color hex could be obtained by blending #887474 with #000000. Closest websafe color is: #333333.\n\n• R 27\n• G 23\n• B 23\nRGB color chart\n• C 0\n• M 15\n• Y 15\n• K 73\nCMYK color chart\n\n#443a3a color description : Very dark grayish red.\n\n# #443a3a Color Conversion\n\nThe hexadecimal color #443a3a has RGB values of R:68, G:58, B:58 and CMYK values of C:0, M:0.15, Y:0.15, K:0.73. Its decimal value is 4471354.\n\nHex triplet RGB Decimal 443a3a `#443a3a` 68, 58, 58 `rgb(68,58,58)` 26.7, 22.7, 22.7 `rgb(26.7%,22.7%,22.7%)` 0, 15, 15, 73 0°, 7.9, 24.7 `hsl(0,7.9%,24.7%)` 0°, 14.7, 26.7 333333 `#333333`\nCIE-LAB 25.444, 4.368, 1.61 4.661, 4.561, 4.638 0.336, 0.329, 4.561 25.444, 4.656, 20.236 25.444, 5.452, 1.173 21.356, 1.583, 2.074 01000100, 00111010, 00111010\n\n# Color Schemes with #443a3a\n\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #3a4444\n``#3a4444` `rgb(58,68,68)``\nComplementary Color\n• #443a3f\n``#443a3f` `rgb(68,58,63)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #443f3a\n``#443f3a` `rgb(68,63,58)``\nAnalogous Color\n• #3a3f44\n``#3a3f44` `rgb(58,63,68)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #3a443f\n``#3a443f` `rgb(58,68,63)``\nSplit Complementary Color\n• #3a3a44\n``#3a3a44` `rgb(58,58,68)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #3a443a\n``#3a443a` `rgb(58,68,58)``\n• #443a44\n``#443a44` `rgb(68,58,68)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #3a443a\n``#3a443a` `rgb(58,68,58)``\n• #3a4444\n``#3a4444` `rgb(58,68,68)``\n• #1b1717\n``#1b1717` `rgb(27,23,23)``\n• #282323\n``#282323` `rgb(40,35,35)``\n• #362e2e\n``#362e2e` `rgb(54,46,46)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #524646\n``#524646` `rgb(82,70,70)``\n• #605151\n``#605151` `rgb(96,81,81)``\n• #6d5d5d\n``#6d5d5d` `rgb(109,93,93)``\nMonochromatic Color\n\n# Alternatives to #443a3a\n\nBelow, you can see some colors close to #443a3a. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #443a3d\n``#443a3d` `rgb(68,58,61)``\n• #443a3c\n``#443a3c` `rgb(68,58,60)``\n• #443a3b\n``#443a3b` `rgb(68,58,59)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #443b3a\n``#443b3a` `rgb(68,59,58)``\n• #443c3a\n``#443c3a` `rgb(68,60,58)``\n• #443d3a\n``#443d3a` `rgb(68,61,58)``\nSimilar Colors\n\n# #443a3a Preview\n\nThis text has a font color of #443a3a.\n\n``<span style=\"color:#443a3a;\">Text here</span>``\n#443a3a background color\n\nThis paragraph has a background color of #443a3a.\n\n``<p style=\"background-color:#443a3a;\">Content here</p>``\n#443a3a border color\n\nThis element has a border color of #443a3a.\n\n``<div style=\"border:1px solid #443a3a;\">Content here</div>``\nCSS codes\n``.text {color:#443a3a;}``\n``.background {background-color:#443a3a;}``\n``.border {border:1px solid #443a3a;}``\n\n# Shades and Tints of #443a3a\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #040404 is the darkest color, while #faf9f9 is the lightest one.\n\n• #040404\n``#040404` `rgb(4,4,4)``\n• #0f0d0d\n``#0f0d0d` `rgb(15,13,13)``\n• #1a1616\n``#1a1616` `rgb(26,22,22)``\n• #241f1f\n``#241f1f` `rgb(36,31,31)``\n• #2f2828\n``#2f2828` `rgb(47,40,40)``\n• #393131\n``#393131` `rgb(57,49,49)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #4f4343\n``#4f4343` `rgb(79,67,67)``\n• #594c4c\n``#594c4c` `rgb(89,76,76)``\n• #645555\n``#645555` `rgb(100,85,85)``\n• #6e5e5e\n``#6e5e5e` `rgb(110,94,94)``\n• #796767\n``#796767` `rgb(121,103,103)``\n• #847070\n``#847070` `rgb(132,112,112)``\n• #8d7a7a\n``#8d7a7a` `rgb(141,122,122)``\n• #968484\n``#968484` `rgb(150,132,132)``\n• #a08f8f\n``#a08f8f` `rgb(160,143,143)``\n• #a99a9a\n``#a99a9a` `rgb(169,154,154)``\n• #b2a4a4\n``#b2a4a4` `rgb(178,164,164)``\n• #bbafaf\n``#bbafaf` `rgb(187,175,175)``\n• #c4b9b9\n``#c4b9b9` `rgb(196,185,185)``\n• #cdc4c4\n``#cdc4c4` `rgb(205,196,196)``\n• #d6cfcf\n``#d6cfcf` `rgb(214,207,207)``\n• #dfd9d9\n``#dfd9d9` `rgb(223,217,217)``\n• #e8e4e4\n``#e8e4e4` `rgb(232,228,228)``\n• #f1eeee\n``#f1eeee` `rgb(241,238,238)``\n• #faf9f9\n``#faf9f9` `rgb(250,249,249)``\nTint Color Variation\n\n# Tones of #443a3a\n\nA tone is produced by adding gray to any pure hue. In this case, #3f3f3f is the less saturated color, while #790505 is the most saturated one.\n\n• #3f3f3f\n``#3f3f3f` `rgb(63,63,63)``\n• #443a3a\n``#443a3a` `rgb(68,58,58)``\n• #493535\n``#493535` `rgb(73,53,53)``\n• #4e3030\n``#4e3030` `rgb(78,48,48)``\n• #532b2b\n``#532b2b` `rgb(83,43,43)``\n• #572727\n``#572727` `rgb(87,39,39)``\n• #5c2222\n``#5c2222` `rgb(92,34,34)``\n• #611d1d\n``#611d1d` `rgb(97,29,29)``\n• #661818\n``#661818` `rgb(102,24,24)``\n• #6b1313\n``#6b1313` `rgb(107,19,19)``\n• #700e0e\n``#700e0e` `rgb(112,14,14)``\n• #740a0a\n``#740a0a` `rgb(116,10,10)``\n• #790505\n``#790505` `rgb(121,5,5)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #443a3a is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population" ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.5032995,"math_prob":0.6852463,"size":3661,"snap":"2021-43-2021-49","text_gpt3_token_len":1674,"char_repetition_ratio":0.12332513,"word_repetition_ratio":0.011070111,"special_character_ratio":0.5583174,"punctuation_ratio":0.23522854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99082834,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T13:46:58Z\",\"WARC-Record-ID\":\"<urn:uuid:8d9eb2fe-f510-45b1-a3f8-590c06542d96>\",\"Content-Length\":\"36095\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:42393c48-1255-442f-917d-d2ee7c59c0ff>\",\"WARC-Concurrent-To\":\"<urn:uuid:ab109857-f018-4bc9-a7e2-583662dfb34b>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/443a3a\",\"WARC-Payload-Digest\":\"sha1:KQLLGNC7FZFXDDSYSP5KGZMBCDB4YOWK\",\"WARC-Block-Digest\":\"sha1:E75WRKWI2HBIGHQZZ5OOKSSK6JVHP5RF\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363510.40_warc_CC-MAIN-20211208114112-20211208144112-00054.warc.gz\"}"}
https://tex.stackexchange.com/questions/30956/what-does-lettoken-do
[ "# What does \\@let@token do?\n\nIn source2e.pdf I've seen \\@let@token used in several places, for example in the definition of \\@ifnextchar:\n\n\\long\\def\\@ifnextchar#1#2#3{%\n\\let\\reserved@d=#1%\n\\def\\reserved@a{#2}%\n\\def\\reserved@b{#3}%\n\\futurelet\\@let@token\\@ifnch}\n\n\nHowever, I couldn't find a description of \\@let@token, so my question is what exactly does \\@let@token do?\n\n\\@let@token is assigned by \\futurelet to the next token after the \\@ifnextchar, i.e. it is the next character. The \\futurelet\\@let@token\\@ifnch code means \"assign the next token to \\@let@token and then process \\@ifnch\". Inside \\@ifnch the \\@let@token macro is tested if it is equal to the (first token of the) first argument of \\@ifnextchar, i.e. \\reserved@d (see the \\let\\reserved@d=#1).\n\nSo, it does nothing, but it is used as a temporary variable. In theory \\@tempa could have been used as well but wouldn't be as save as a named special macro. The benefit here is that you can check \\@let@token in the false clause of \\@ifnextchar using \\ifx against other tokens without going to the more complex \\@ifnextchar macro again.\n\n\\@ifnextchar{\\relax}{It was a relax!}{%\n\\ifx\\@let@token\\$\nIt was a dollar!\n\\else\n\\ifx\\@let@token&\nIt was an ampersand!\n\\else\nsomething else\n\\fi\n\\fi\n}\n\n\nThis is useful for parser code as in e.g. tikz-timing and ydoc." ]
[ null ]
{"ft_lang_label":"__label__en","ft_lang_prob":0.87612295,"math_prob":0.46854722,"size":943,"snap":"2019-51-2020-05","text_gpt3_token_len":293,"char_repetition_ratio":0.15015975,"word_repetition_ratio":0.0,"special_character_ratio":0.2640509,"punctuation_ratio":0.10471204,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96496874,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-07T11:00:28Z\",\"WARC-Record-ID\":\"<urn:uuid:98c3fc23-0f69-43f9-98ca-0b52c42a1cfc>\",\"Content-Length\":\"132451\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ec36027d-c591-4583-8424-7d4b0ff920f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:91ee27b6-639a-4145-b431-6a0f198916c2>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://tex.stackexchange.com/questions/30956/what-does-lettoken-do\",\"WARC-Payload-Digest\":\"sha1:DCWW25DDQNRG34VAGGY2L7QNTDJ4UVGX\",\"WARC-Block-Digest\":\"sha1:3BGUHP5OXKG3FNHXYHQA44JQDY3ND3CM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540499389.15_warc_CC-MAIN-20191207105754-20191207133754-00451.warc.gz\"}"}
https://www.r-bloggers.com/introduction-to-statistical-methods-in-r/
[ "# Introduction to Statistical Methods in R\n\nJanuary 18, 2016\nBy\n\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't.\n\nData analyses are the product of many different tasks, and statistical methods are one key aspect of any data analysis. There is a common workflow in the related areas of informatics, data mining, data science, machine learning, and statistics. The workflow tasks include data preparation, the development of predictive mathematical models, and the interpretation and preparation of analysis results (including the development of visualizations to communicate findings).\n\nThe presentation provides information on the last two steps of this workflow and reproducible code examples and presents a walk-through of many common statistical methods (including regression, clustering (e.g. K-means and hiearchical), and dimensionality reduction (e.g. prinical component analysis (PCA)) used to explore data with examples in R.\n\nNovice users are shown how to navigate the resuting R object to extract specific elements of interest, such as correlation p-values, regression coefficients, etc. The presentation additionally tries to tackle of some of the key concerns about these introductory methods by providing guidance on the interpretation of analyses results, such as understanding the approximately 10 values returned in a simple linear regression; the importance of and how to deal with missing values through imputation in real world problems; determining the quality of clustering results; and understanding the data transformations that take place in dimension reduction methods. Also provided is information about more sophisticated methodologies, such as regularized regression methods: LASSO, Ridge, and Elastic Net regression, and packages to make use of these more advanced methods in R, such as glmnet for regularized regression.\n\nUsage of these statistical methods for modeling can help users to understand their data sets, and these methodologies can be coupled with other aspects of R and RStudio to develop interactive analyses using the Shiny R package.\n\nThe post Introduction to Statistical Methods in R appeared first on Lunean.\n\nR-bloggers.com offers daily e-mail updates about R news and tutorials about learning R and many other topics. Click here if you're looking to post or find an R/data-science job.\nWant to share your content on R-bloggers? click here if you have a blog, or here if you don't." ]
[ null ]
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https://nike-huarache.com/math-question-237
[ "In this blog post, we discuss how Math answer generator free can help students learn Algebra.", null, "## 8 Free Math Problem Solving Websites and Applications\n\nMicrosoft Math Solver - Math Problem Solver & Calculator Get step-by-step solutions to your math problems Try Math Solver Type a math problem Quadratic equation x2 − 4x − 5 = 0", null, "They use our service\n\n## Wolfram Problem Generator: Online Practice Questions\n\nCreate endless addition, subtraction, multiplication, division, and fact family worksheets with our math worksheet generator. Simply adjust the difficulty to match grade-level expectations for\n\n• Decide mathematic equation\n• Top Teachers" ]
[ null, "https://nike-huarache.com/images/459403f40c03fad2/dbhkenjgomicfaqpl-student-image.jpg", null, "https://nike-huarache.com/images/459403f40c03fad2/lqhnoamkfjgedpbic-home-img-2.png", null ]
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https://spec2000.net/water-resistivity/formation-water-salinity-calculations-and-units-conversions.htm
[ "", null, "WATER SALINITY and Total dissolved solids\nThe salinity of a water sample can be described numerically in several different ways, each with a confusing set of measurement units. This page covers the summation of water chemistry analysis data and conversions between total dissolved solids (TDS), equivalent NaCl content, and chloride content. Conversion of salinity to water resistivity is covered on the next Section in this Chapter.", null, "Equivalent NaCl Water Salinity from Water Analysis\nThe resistivity of a water sample can be calculated from its chemical analysis. To do this, an equivalent NaCl concentration must be determined based on the ionic activity of each ion.", null, "Enter chart with total solids concentration of the sample in ppm (mg/kg) to find weighting factors for each ion present. The concentration of each ion is multiplied by its weighting factor, and the products for all ions are summed to obtain equivalent NaCl concentration.\n\nThe math is pretty simple:\n1: TDS = SUM (IONi)\n2: WSe = SUM (IONi * FACTRi)\n\nWhere:\nTDS = total dissolved solids (ppm)\nIONi = ion concentration of ith component (ppm)\nFACTRi = multiplier factor for ith component (ppm)\nWSe = equivalent NaCl concentration (ppm)", null, "NUMERICAL EXAMPLE:\nAssume formation-water sample analysis\n460 ppm Ca,\n1400 ppm SO4\n19,000 ppm Na plus Cl.\n\nTotal dissolved solids concentration\nTDS = 460 + 1400 + 19,000 = 20,860 ppm\nEntering the chart with this total solids concentration\nCa multiplier = 0.81\nSO4 multiplier = 0.45\nNa+CL multiplier = 1.00\nEquivalent NaCl concentration = 460 ´ 0.81 + 1400 ´ 0.45 + 19,000 ´ 1.0 = 20,000 ppm.", null, "Water Salinity from Chloride Content\nSometimes salinity is reported at the well site in ppm Chlorides instead of ppm NaCl equivalent.\n6: WSa = Ccl * 1.645\n\nWhere:\nCcl = water salinity (ppm Cl)\nWSa = water salinity (ppm NaCl)", null, "COMMENTS:\nUse this relationship when chloride content of the water sample is known. Usually Cl content is derived at the well site from a drill stem test recovery. It is useful as a first approximation until the water sample is analyzed more accurately at a laboratory. The relationship is for pure NaCl solutions and the factor may be higher or lower if other ions are present.", null, "NUMERICAL EXAMPLE:\n1. Chloride concentration to salinity.\nWS = 11,600 ppm Cl * 1.645 = 19,000 ppm NaCl\n\nPage Views ---- Since 01 Jan 2015" ]
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https://www.geeksforgeeks.org/java-zipentry-getcrc-function-with-examples/
[ "Open In App\n\n# Java ZipEntry getCrc() function with examples\n\nThe getCrc() function is a part of java.util.zip package. The function returns the CRC – 32 checksum value of a specific ZipEntry .\nThe CRC is an error detecting code used to detect error in raw data.\nFunction Signature :\n\n`public long getCrc()`\n\nSyntax :\n\n`zip_entry.getCrc();`\n\nParameters :No parameter is required for this function.\nReturn value :The function returns a long value, the Crc of this ZipEntry.\nExceptions :The function does not throw any exception.0\n\nBelow programs illustrates the use of getCrc() function\n\nExample 1: We will create a file named zip_file and get the zip file entry using getEntry() function and then get the CRC-32 of the specified ZipEntry.”file.zip” is a zip file present in f: directory. we will take a “.zip” file as ZipEntry\n\n `// Java program to demonstrate the``// use of getCrc() function`` ` `import` `java.util.zip.*;``import` `java.util.Enumeration;``import` `java.util.*;``import` `java.io.*;`` ` `public` `class` `solution {``    ``public` `static` `void` `main(String args[])``    ``{``        ``try` `{``            ``// Create a Zip File``            ``ZipFile zip_file = ``new` `ZipFile(``\"f:\\\\file1.zip\"``);`` ` `            ``// get the Zip Entry using``            ``// the getEntry() function``            ``ZipEntry entry = zip_file.getEntry(``\"file.zip\"``);`` ` `            ``// Get the Crc``            ``// using the getCrc()``            ``// function``            ``long` `input = entry.getCrc();`` ` `            ``// Display the Crc``            ``System.out.println(``\"Crc : \"` `+ input);``        ``}``        ``catch` `(Exception e) {``            ``System.out.println(e.getMessage());``        ``}``    ``}``}`\n\nOutput:\n\n```Crc : 1798120178\n```\n\nExample 2: We will create a file named zip_file and get the zip file entry using getEntry() function and then get the CRC-32 of the specified ZipEntry.”file.zip” is a zip file present in f: directory. we will take a “.cpp” file as ZipEntry\n\n `// Java program to demonstrate the``// use of getCrc() function`` ` `import` `java.util.zip.*;``import` `java.util.Enumeration;``import` `java.util.*;``import` `java.io.*;`` ` `public` `class` `solution {``    ``public` `static` `void` `main(String args[])``    ``{``        ``try` `{``            ``// Create a Zip File``            ``ZipFile zip_file = ``new` `ZipFile(``\"f:\\\\file1.zip\"``);`` ` `            ``// get the Zip Entry using``            ``// the getEntry() function``            ``ZipEntry entry = zip_file.getEntry(``\"file1.cpp\"``);`` ` `            ``// Get the Crc``            ``// using the getCrc()``            ``// function``            ``long` `input = entry.getCrc();`` ` `            ``// Display the Crc``            ``System.out.println(``\"Crc : \"` `+ input);``        ``}``        ``catch` `(Exception e) {``            ``System.out.println(e.getMessage());``        ``}``    ``}``}`\n\nOutput:\n\n```Crc : 3528251335\n```" ]
[ null ]
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https://stromcv.com/how-to-write-exponential-functions-44
[ "# How to write exponential functions\n\nExponentiation is formed by the combination of base and exponent or power. ( base) (exp) like xn, ak. Power or exponent will always be defined by superscript, for which latex\n\n## Equations of Exponential Functions\n\nThe basic idea is to transform the computation of the standard exponential function into computation of a power of 2: expf (x) = exp2f (x / logf (2.0f)) = exp2f (x * 1.44269504). We split\n\n## Write an Exponential Function\n\nWe know that the general formula for an exponential function is given by: f (x) = abx (or y = abx) Using the first point (1, 10), we substitute x = 1 and y = 10 to get: y = abx 10 = ab1 10 = ab Now\n\n• 808 Math Consultants\n• 4.6/5 Ratings\n• 93542+ Orders completed\n\n## Exponential Functions", null, "• Clarify math equations\n• Scan\n• Get support from expert tutors", null, "", null, "", null, "## Exponential Functions: Writing Equations from Tables\n\nThus, an exponential function can be in one of the following forms. f (x) = b x f (x) = ab x f (x) = ab cx f (x) = e x f (x) = e kx f (x) = p e kx Here, apart from 'x' all other letters are constants, 'x' is a variable, and f (x) is an exponential function in\n`\n\n## The exponential function", null, "Do math problems\nTop Specialists" ]
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http://www.pratiyogi.com/assessment/electrochemistry-test-4/412
[ "# Electrochemistry Test 4\n\nTotal Questions:50 Total Time: 75 Min\n\nRemaining:\n\n## Questions 1 of 50\n\nQuestion:Which colourless gas evolves, when $$N{H_4}Cl$$ reacts with zinc in a dry cell battery\n\n$$N{H_4}$$\n\n$${N_2}$$\n\n$${H_2}$$\n\n$$C{l_2}$$\n\n## Questions 2 of 50\n\nQuestion:Which of the substances $$Na,\\,\\,Hg,\\,\\,S,\\,\\,Pt$$ and graphite can be used as electrodes in electrolytic cells having aqueous solutions\n\n$$Na,\\,\\,Pt$$and graphite\n\n$$Na$$and $$Hg$$\n\n$$Pt$$and graphite only\n\n$$Na$$and $$Al$$only\n\n## Questions 3 of 50\n\nQuestion:In electrolysis of dilute $${H_2}S{O_4}$$using platinum electrodes\n\n$${H_2}$$ is evolved at cathode\n\n$$N{H_3}$$ is produced at anode\n\n$$C{l_2}$$ is obtained at cathode\n\n$${O_2}$$ is produced\n\n## Questions 4 of 50\n\nQuestion:Saturated solution of $$KN{O_3}$$ is used to make ‘salt-bridge’ because\n\nVelocity of $${K^ + }$$ is greater than that of $$NO_3^ -$$\n\nVelocity of $$NO_3^ -$$ is greater than that of $${K^ + }$$\n\nVelocities of both $${K^ + }$$and $$NO_3^ -$$ are nearly the same\n\n$$KN{O_3}$$is highly soluble in water\n\n## Questions 5 of 50\n\nQuestion:In balancing the half reaction $${S_2}O_3^{2 - } \\to {S_{(s)}}$$ the number of electrons that must be added is\n\n4 on the left\n\n3 on the right\n\n2 on the left\n\n2 on the right\n\n## Questions 6 of 50\n\nQuestion:Which one of the following statement is true for a electrochemical cell\n\n$${H_2}$$is cathode and $$Cu$$ is anode\n\n$${H_2}$$ is anode and $$Cu$$ is cathode\n\nReduction occurs at $${H_2}$$electrode\n\nOxidation occurs at $$Cu$$ electrode\n\n## Questions 7 of 50\n\nQuestion:In a electrochemical cell\n\nPotential energy changes into kinetic energy\n\nKinetic energy changes into potential energy\n\nChemical energy changes into electrical energy\n\nElectrical energy changes into chemical energy\n\n## Questions 8 of 50\n\nQuestion:In galvanic cell, the salt bridge is used to\n\nComplete the circuit\n\nReduce the electric resistance in the cell\n\nSeparate cathode from anode\n\nCarry salts for the chemical reaction\n\n## Questions 9 of 50\n\nQuestion:If hydrogen electrode dipped in 2 solution of $$pH = 3$$ and $$pH = 6$$ and salt bridge is connected the e.m.f. of resulting cell is\n\n0.177 V\n\n0.3 V\n\n0.052 V\n\n0.104 V\n\n## Questions 10 of 50\n\nQuestion:The tendency of an electrode to lose electrons is known as\n\nElectrode potential\n\nReduction potential\n\nOxidation potential\n\ne.m.f.\n\n## Questions 11 of 50\n\nQuestion:It has been observed that gaseous hydrogen chloride is a very poor conductor of electricity but a solution of hydrogen chloride gas in water is a good conductor of electricity. This is due to the fact that\n\nWater is good conductor of electricity\n\nHydrogen chloride gas in water solution ionizes\n\nA gas is non-conductor but a liquid conducts electricity\n\nGas does not obey Ohm's law whereas solution does\n\n## Questions 12 of 50\n\nQuestion:Electrolytic conduction differs from metallic conduction in that in the case of electrolytic conduction\n\nThe resistance increases with increasing temperature\n\nThe resistance decreases with increasing temperature\n\nThe flow of current does not generate heat\n\nThe resistance is independent of the length of the conductor\n\n## Questions 13 of 50\n\nQuestion:Which of the following is a highly corrosive salt\n\n$$FeC{l_2}$$\n\n$$PbC{l_2}$$\n\n$$H{g_2}C{l_2}$$\n\n$$HgC{l_2}$$\n\n## Questions 14 of 50\n\nQuestion:Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are\n\n$$Fe$$ is oxidised to $$F{e^{2 + }}$$ and dissolved oxygen in water is reduced to $$\\mathop {OH}\\limits^ \\oplus$$\n\n$$Fe$$ is oxidised to $$F{e^{3 + }}$$ and $${H_2}O$$ is reduced to $$O_2^{2 - }$$\n\n$$Fe$$ is oxidised to $$F{e^{2 + }}$$ and $${H_2}O$$ is reduced to $$O_2^ -$$\n\n$$Fe$$ is oxidised to $$F{e^{2 + }}$$ and $${H_2}O$$ is reduced to $${O_2}$$\n\n## Questions 15 of 50\n\nQuestion:$${E^0} = \\frac{{RT}}{{nF}}$$ In $${K_{eq}}$$. This is called\n\nGibb's equation\n\nGibb's–Helmholtz equation\n\nNernst's equation\n\nVander Waal's equation\n\n## Questions 16 of 50\n\nQuestion:Four alkali metals A, B, C and D are having respectively standard electrode potential as –3.05,–1.66,–0.40 and 0.80. Which one will be the most reactive\n\nA\n\nB\n\nC\n\nD\n\n## Questions 17 of 50\n\nQuestion:If an iron rod is dipped in $$CuS{O_4}$$ solution\n\nBlue colour of the solution turns green\n\nBrown layer is deposited on iron rod\n\nNo change occurs in the colour of the solution\n\nBlue colour of the solution vanishes\n\nNone of the above\n\n## Questions 18 of 50\n\nQuestion:$${E^o}$$ values of $$M{g^{2 + }}/Mg$$ is $$- 2.37V$$, of $$Z{n^{2 + }}/Zn$$ is $$- 0.76V$$ and $$F{e^{2 + }}/Fe$$ is $$- 0.44\\,V$$. Which of the following statements is correct\n\n$$Zn$$ will reduce $$F{e^{2 + }}$$\n\n$$Zn$$ will reduce $$M{g^{2 + }}$$\n\n$$Mg$$ oxidises $$Fe$$\n\n$$Zn$$ oxidises $$Fe$$\n\n## Questions 19 of 50\n\nQuestion:Expression representing the cell potential $$(E\\,\\,cell)$$\n\n$${E_{{\\rm{cathode}}}} + {E_{{\\rm{anode}}}}$$\n\n$${E_{{\\rm{anode}}}} - {E_{{\\rm{cathode}}}}$$\n\n$${E_{{\\rm{cathode}}}}- {E_{{\\rm{anode}}}}$$\n\n$${{\\rm{E}}_{{\\rm{left}}}}- {E_{{\\rm{right}}}}$$\n\n## Questions 20 of 50\n\nQuestion:Iron displaces copper from its salt solution, because\n\nAtomic number of iron is less than that of copper\n\nThe standard reduction potential of iron is less than that of copper\n\nThe standard reduction potential of iron is more than that of copper\n\nThe iron salt is more soluble in water than the copper salt\n\n## Questions 21 of 50\n\nQuestion:Which of the following displaces $$B{r_2}$$ from an aqueous solution containing bromide ions\n\n$$C{l_2}$$\n\n$$C{l^ - }$$\n\n$${I_2}$$\n\n$$I_3^ -$$\n\n## Questions 22 of 50\n\nQuestion:For the cell reaction $$C{u^{2 + }}({C_1}aq) + Zn(s) = Z{n^{2 + }}({C_2}aq) + Cu(s)$$ of an electrochemical cell, the change in free energy at a given temperature is a function of\n\nln $$({C_1})$$\n\nln $$({C_2})$$\n\nln $$({C_1} + {C_2})$$\n\nln $$({C_2}/{C_1})$$\n\n## Questions 23 of 50\n\nQuestion:$$KMn{O_4}$$ acts as an oxidising agent in the neutral medium and gets reduced to $$Mn{O_2}$$. The equivalent weight of $$KMn{O_4}$$ in neutral medium\n\nmol. wt/2\n\nmol.wt/3\n\nmol. wt/4\n\nmol .wt/7\n\n## Questions 24 of 50\n\nQuestion:Which of the following condition will increase the voltage of the cell, represented by the equation $$C{u_{(s)}} + 2A{g^ + }_{(aq)} \\to C{u^{2 + }}_{(aq)} + 2A{g_{(s)}}$$\n\nIncrease in the concentration of $$A{g^ + }$$ion\n\nIncrease in the concentration of $$C{u^ + }$$ ion\n\nIncrease in the dimension of silver electrode\n\nIncrease in the dimension of copper electrode\n\n## Questions 25 of 50\n\nQuestion:$$C{u^ + }$$ ion is not stable in aqueous solution because of disproportionation reaction. $${E^o}$$ value for disproportionation of $$C{u^ + }$$ is (Given $$E_{C{u^{2 + }}/C{u^ + }}^o = 0.15$$, $$E_{C{u^{2 + }}/Cu}^o = 0.34V$$)\n\n0.49 V\n\n0.49 V\n\n0.38 V\n\n0.38 V\n\n## Questions 26 of 50\n\nQuestion:$${E^o}$$ of a cell $$aA + bB \\to cC + dD$$ is\n\n$$- \\frac{{RT}}{{nF}}\\log \\frac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$$\n\n$$- RT\\,\\log \\frac{{{{[a]}^A}{{[b]}^B}}}{{{{[a]}^C}{{[d]}^D}}}$$\n\n$$- \\frac{{RT}}{{nF}}\\log \\frac{{{{[C]}^c}{{[d]}^D}}}{{{{[A]}^a}{{[B]}^b}}}$$\n\n$$- \\frac{{RT}}{{nF}}\\log \\frac{{{{[C]}^c}{{[d]}^D}}}{{{{[a]}^A}{{[B]}^b}}}$$\n\n## Questions 27 of 50\n\nQuestion:A galvanic cell with electrode potential of $$'A' = + 2.23\\;V$$and $$'B' = - 1.43\\;V$$. The value of $$E{^\\circ _{cell}}$$ is\n\n3.66 V\n\n0.80 V\n\n0.80 V\n\n3.66 V\n\n## Questions 28 of 50\n\nQuestion:The e.m.f. of a cell whose half cells are given below is $$M{g^{2 + }} + 2{e^ - } \\to Mg(s)\\;E^\\circ = - 2.37\\;V$$$$C{u^{2 + }} + 2{e^ - } \\to Cu(s)\\;E^\\circ = + 0.34\\;V$$\n\n+ 1.36 V\n\n+ 2.71 V\n\n+ 2.17 V\n\n3.01 V\n\n## Questions 29 of 50\n\nQuestion:Calculate the electrode potential at $${298^o}K$$ for $$Zn|Z{n^{ + + }}$$electrode in which the activity of zinc ions is 0.001 M and $${E^o}_{Zn/Z{n^{ + + }}}$$is –0.74 volts\n\n0. 38 volts\n\n0.83 volts\n\n0.40 volts\n\n0.45 volts\n\n## Questions 30 of 50\n\nQuestion:Which of the following expression is correct\n\n$$\\Delta {G^o} = - nFE_{cell}^o$$\n\n$$\\Delta {G^o} = + nFE_{cell}^o$$\n\n$$\\Delta {G^o} = - 2.303RT\\,nFE_{cell}^o$$\n\n$$\\Delta {G^o} = - nF\\,\\log {K_C}$$\n\n## Questions 31 of 50\n\nQuestion:An electric current is passed through an aqueous solution of the following. Which one shall decompose\n\nUrea\n\nGlucose\n\n$$AgN{O_3}$$\n\nEthyl alcohol\n\n## Questions 32 of 50\n\nQuestion:The electric conduction of a salt solution in water depends on the\n\nShape of its molecules\n\nSize of its molecules\n\nSize of solvent molecules\n\nExtent of its ionization\n\n## Questions 33 of 50\n\nQuestion:Electrolysis is a process in which the cations and anions of the electrolyte are\n\nHydrated\n\nHydrolysed\n\nCharged\n\nDischarged\n\n## Questions 34 of 50\n\nQuestion:Degree of ionisation of a solution depends upon\n\nTemperature\n\nNature of the electrolyte\n\nNature of the solvent\n\nNone of these\n\n## Questions 35 of 50\n\nQuestion:An apparatus used for the measurement of quantity of electricity is known as a\n\nCalorimeter\n\nCathetometer\n\nCoulometer\n\nColorimeter\n\n## Questions 36 of 50\n\nQuestion:The unit of electrochemical equivalent is\n\nGram\n\nGram/ampere\n\nGram/coulomb\n\nCoulomb/gram\n\n## Questions 37 of 50\n\nQuestion:How many Faradays are required to generate one gram atom of magnesium from $$MgC{l_2}$$\n\n1\n\n2\n\n3\n\n4\n\n## Questions 38 of 50\n\nQuestion:To deposit 0.6354 gm of copper by electrolysis of aqueous cupric sulphate solution, the amount of electricity required (in coulombs) is\n\n9650\n\n4825\n\n3860\n\n1930\n\n## Questions 39 of 50\n\nQuestion:Then during electrolysis of a solution of AgNO3, 9650 coulombs of charge pass through the electroplating bath, the mass of silver deposited in the cathode will be\n\n1.08 g\n\n10.8 g\n\n21.6 g\n\n108 g\n\n## Questions 40 of 50\n\nQuestion:Total charge on 1 mole of a monovalent metal ion is equal to\n\n$$9.65 \\times {10^4}\\,\\,\\,Coulomb$$\n\n$$6.28 \\times {10^{18}}\\,\\,\\,Coulomb$$\n\n$$1.6 \\times {10^{ - 19}}\\,\\,\\,Coulomb$$\n\nNone of these\n\n## Questions 41 of 50\n\nQuestion:When an electric current is passed through acidulated water 112 ml of hydrogen gas at N.T.P. collect at the cathode in 965 seconds. The current passed, in amperes is\n\n1\n\n0.5\n\n0.1\n\n2\n\n## Questions 42 of 50\n\nQuestion:A current of strength 2.5 amp was passed through $$CuS{O_4}$$ solution for 6 minutes 26 seconds. The amount of copper deposited is (Atomic weight of $$Cu = 63.5$$)(1 faraday = 96500 coulombs)\n\n0.3175 g\n\n3.175 g\n\n0.635 g\n\n6.35 g\n\n## Questions 43 of 50\n\nQuestion:A certain quantity of electricity is passed through an aqueous solution of $$AgN{O_3}$$ and cupric salt solution connected in series. The amount of $$Ag$$ deposited is $$1.08\\,gm$$, the amount of copper deposited is (atomic weight of Cu = 63.5; Ag = 108)\n\n0.6454 g\n\n6.354 g\n\n0.3177 g\n\n3.177 g\n\n## Questions 44 of 50\n\nQuestion:The number of electrons required to deposit 1gm atom of aluminium (at. wt. = 27) from a solution of aluminium chloride will be (where N is Avogadro's number)\n\n1 N\n\n2 N\n\n3 N\n\n4 N\n\n## Questions 45 of 50\n\nQuestion:2.5 F of electricity are passed through a $$CuS{O_4}$$ solution. The number of gm equivalent of Cu deposited on anode is\n\nZero\n\n1.25\n\n2.5\n\n5\n\n## Questions 46 of 50\n\nQuestion:The equivalent weight of a certain trivalent element is 20. Molecular weight of its oxide is\n\n152\n\n56\n\n168\n\n68\n\n## Questions 47 of 50\n\nQuestion:Silver is removed electrically from 200 ml of a 0.1 N solution of $$AgN{O_3}$$ by a current of 0.1 ampere. How long will it take to remove half of the silver from the solution\n\n16 sec\n\n96.5 sec\n\n100 sec\n\n10 sec\n\n## Questions 48 of 50\n\nQuestion:An electric current is passed through silver nitrate solution using silver electrodes. 10.79 g of silver was found to be deposited on the cathode if the same amount of electricity is passed through copper sulphate solution using copper electrodes, the weight of copper deposited on the cathode is\n\n6.4 g\n\n2.3 g\n\n12.8 g\n\n1.6 g\n\n3.2 g\n\n## Questions 49 of 50\n\nQuestion:The law of electrolysis were proposed by\n\nKohlraush\n\nNernst\n\nBerthelot\n\n## Questions 50 of 50\n\nQuestion:How many atoms of calcium will be deposited from a solution of $$CaC{l_2}$$by a current 0.25 mA following for 60 seconds\n\n$$4.68 \\times {10^{18}}$$\n$$4.68 \\times {10^{15}}$$\n$$4.68 \\times {10^{12}}$$\n$$4.68 \\times {10^9}$$" ]
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https://hannes.emotisys.de/2007/01/the-button-code/
[ "## the button code\n\nJust for the record: This code will drive our arduino boards (two of them and yes! they are connected AND talking!).\n\nIt took Larissa and me quite some time and a lot more nerves. And Tomek, too! He spent hours with us on ICQ altruisticly, even got his two Arduinos out of the box and gave us the right hints finally – by remote!\nSo we learnt a lot. Most important: With two communicating Boards, never forget to connect the grounds. Always remember.\n\nThe code for the geeks:\n\n```//################## //variables that might be changed for interface adjustments```\n\n``` //1 - distance sensor int distancePin = 3; //analog //rgbLED: //red = pin 8; green = 9; blue = pin 10; //choose color for idle status (\"nothing in sight...\") int dist_idleLED = 10; //digital //choose color for active status (\"A FRIEND!\") int dist_activeLED = 8; //digital //if the distance sensor switches on, start a counter //every interruption in the sensor's sight will start the counter again //so the switchback to \"idle\" will only occur //if there is \"nothing in sight\" for a given time (switchbackdelay) //to prevent the light from flickering when the sensor loses \"contact\" int switchbackdelay = 6000; int smoothfactor = 10; //loop x times and than take an average value int smoothdelay = 50; //ms to wait between measurements (of distance_raw) //sensor characteristics for Sharp GP2D12 //distance = 100: nothing in sight, can be as low as 40 (distance > 70cm) //distance = 600; //largest value the sensor put out (around 10cm) int distance_switch = 130; // most probably sth is in the way (around 60cm) //(this threshold represents sensitivity as well) //2 - arduino communication int arduinoOutput = 2; //pin to \"send\" data int arduinoInput = 3; // pin to \"receive\" data int schalterInput = 4; // pin to read, if switch is pressed int led_both = 5; // pin for led that lights up if both arduinos have a button pressed //################## //variables that are needed for internal calculations //1 - distance sensor //a time variable for counters long time; long time_switchon; //starttime of the counter int smoothloops = 0; //number of loops that were already run through for smoothing int smoothcollector = 0; //adds all distance_raw values (will be devided by smoothfactor) int distance_raw = 0; //the value measure directly int distance = 0; //calibrated distance int distance_avg = 0; //the distance smoothened //2 - arduino communication //variable to keep the code blocks separate int both_pressed = 0; // =1 if the button at each arduino is pressed void setup() { //1 - distance sensor pinMode (dist_idleLED, OUTPUT); pinMode (dist_activeLED, OUTPUT); //2 - arduino communication pinMode(led_both, OUTPUT); pinMode(arduinoOutput, OUTPUT); pinMode(arduinoInput, INPUT); pinMode(schalterInput, INPUT); beginSerial(9600); //start serial communciation } void loop(){ //first, check if button is pressed //if not, use distance sensor //if it is pressed, continue to use distance sensor //untill second button (of the other arduino) is pressed as well //########## 1 - read and process distance-sensor data######## //run distance-sensing only, if buttons are not pressed if (both_pressed == 0) { // get raw data distance_raw = analogRead(distancePin); /* output to check single values Serial.print (distance_raw); Serial.print (\" \"); */ //collect data for smoothing smoothcollector += distance_raw; smoothloops++; if (smoothloops >= smoothfactor) { //compute average value distance = smoothcollector / smoothfactor; /* debugging Serial.print (\" //smooth \"); Serial.print (distance); Serial.print (\" // \"); */ //reset smooth-variables for next run smoothcollector = 0; smoothloops = 0; //check if sth is in the way: //start counter time = millis(); //\"distance\" as variable contains sensor-value //which is higher when the object is closer, so: if (distance > distance_switch) { Serial.print (\"A FRIEND!\"); //store time of switching for counter time_switchon = time; //Serial.println (time_switchon); //debugging digitalWrite (dist_idleLED, LOW); digitalWrite (dist_activeLED, HIGH); } else { //only switch back after waiting for some time if (time > time_switchon + switchbackdelay) { Serial.print (\"nobody in sight...\"); digitalWrite (dist_activeLED, LOW); digitalWrite (dist_idleLED, HIGH); } } Serial.print (10, BYTE); //this is a LF linefeed character in ASCII } //end of if-smoothloop... delay(smoothdelay); } // end of if-both_pressed else { //switch the lights of the distance sensor off if both digitalWrite(dist_idleLED, LOW); digitalWrite(dist_activeLED, LOW); } //####### 2 - build a two-switch-AND-connection#### //if the switch is pressed, the Board will set an Output-Pin to HIGH //that Pin is connected to an Input-Pin of another board where it is read as Input //if the switch of that board is pressed, too (>both buttons pressed) //an LED will light up //in place of Serial communication a simpler way via two digitalPins is used /* debugging if (digitalRead(arduinoInput) == HIGH) { Serial.print(\"das andere Arduino sagt: schalter ist an - \"); } else { Serial.print(\"das andere Arduino sagt: nischt - \"); } */ //the switch will connect the Pin to gnd, so it will be low when the switch gets pressed if (digitalRead(schalterInput) == LOW) { Serial.print(\"schalter ist an - \"); digitalWrite(arduinoOutput, HIGH); if (digitalRead(arduinoInput) == HIGH) { //Serial.print(\"das andere Arduino sagt: schalter ist an - \"); digitalWrite(led_both, HIGH); both_pressed = 1; } else { digitalWrite(led_both, LOW); both_pressed = 0; } } else{ //Serial.print(\"nischt is an - \"); digitalWrite(led_both, LOW); both_pressed = 0; digitalWrite(arduinoOutput, LOW); } ```\n\n``` //Serial.print(10, BYTE); //a linefeed-character in ASCII //Serial.print(13, BYTE); } ```\n\nno Responses so far to the button code" ]
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http://eprints.undip.ac.id/24592/
[ "# SOLVING SYSTEM OF NONLINEAR EQUATIONS\n\nSidik Sasongko, Priyo (2010) SOLVING SYSTEM OF NONLINEAR EQUATIONS. PROSIDING SEMINAR NASIONAL ILMU KOMPUTER UNIVERSITAS DIPONEGORO 2010 .\n\nFull text not available from this repository.\n\n## Abstract\n\nFundamental problems from engineering, chemistry, medicine, etc. can be formulated as a system of nonlinear equations. Solving systems of nonlinear equations is of great importance, because these systems frequently arise in many branches of computational mathematics. Finding the solution of n-dimensional nonlinear equations can be a challenging problem, even when a unique solution exists. Finding a solution for such a system requires sometimes high computational efforts. In this paper, we try to solve them using two different approaches. Both approaches transform the problem into an optimization problem. One approach uses Newton method and the other one a genetic algorithm. The solution of a nonlinear equation with computer simulation illustrates the algorithm. Several experiments are performed in order to emphasize the advantages and disadvantages of the two methods. Keywords : Nonlinear equation, optimization problem, Newton technique, Genetic Algorithm.\n\nItem Type: Article Q Science > QA Mathematics > QA75 Electronic computers. Computer science Faculty of Science and Mathematics > Department of Computer Science 24592 Admin Departemen Matematika 10 Dec 2010 17:13 10 Dec 2010 17:13\n\nRepository Staff Only: item control page" ]
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https://www.traditionaloven.com/tutorials/angle/convert-angular-mil-unit-to-second-of-arc-unit.html
[ " Convert mil to '' | angular mil to seconds of arc\n\nangle units conversion\n\nAmount: 1 angular mil (mil) of angle Equals: 202.50 seconds of arc ('') in angle\n\nConverting angular mil to seconds of arc value in the angle units scale.\n\nTOGGLE :   from seconds of arc into angular mils in the other way around.\n\nangle from angular mil to second of arc conversion results\n\nEnter a new angular mil number to convert\n\n* Whole numbers, decimals or fractions (ie: 6, 5.33, 17 3/8)\n* Precision is how many digits after decimal point (1 - 9)\n\nEnter Amount :\nDecimal Precision :\n\nCONVERT :   between other angle measuring units - complete list.\n\nHow many seconds of arc are in 1 angular mil? The answer is: 1 mil equals 202.50 ''\n\n202.50 '' is converted to 1 of what?\n\nThe seconds of arc unit number 202.50 '' converts to 1 mil, one angular mil. It is the EQUAL angle value of 1 angular mil but in the seconds of arc angle unit alternative.\n\n mil/'' angle conversion result From Symbol Equals Result Symbol 1 mil = 202.50 ''\n\nConversion chart - angular mils to seconds of arc\n\n1 angular mil to seconds of arc = 202.50 ''\n\n2 angular mils to seconds of arc = 405.00 ''\n\n3 angular mils to seconds of arc = 607.50 ''\n\n4 angular mils to seconds of arc = 810.00 ''\n\n5 angular mils to seconds of arc = 1,012.50 ''\n\n6 angular mils to seconds of arc = 1,215.00 ''\n\n7 angular mils to seconds of arc = 1,417.50 ''\n\n8 angular mils to seconds of arc = 1,620.00 ''\n\n9 angular mils to seconds of arc = 1,822.50 ''\n\n10 angular mils to seconds of arc = 2,025.00 ''\n\n11 angular mils to seconds of arc = 2,227.50 ''\n\n12 angular mils to seconds of arc = 2,430.00 ''\n\n13 angular mils to seconds of arc = 2,632.50 ''\n\n14 angular mils to seconds of arc = 2,835.00 ''\n\n15 angular mils to seconds of arc = 3,037.50 ''\n\nConvert angle of angular mil (mil) and seconds of arc ('') units in reverse from seconds of arc into angular mils.\n\nAngles\n\nThis calculator is based on conversion of two angle units. An angle consists of two rays (as in sides of an angle sharing a common vertex or else called the endpoint.) Some belong to rotation measurements - spherical angles measured by arcs' lengths, pointing from the center, plus the radius. For a whole set of multiple units of angle on one page, try that Multiunit converter tool which has built in all angle unit-variations. Page with individual angle units.\n\nConverter type: angle units\n\nFirst unit: angular mil (mil) is used for measuring angle.\nSecond: second of arc ('') is unit of angle.\n\nQUESTION:\n15 mil = ? ''\n\n15 mil = 3,037.50 ''\n\nAbbreviation, or prefix, for angular mil is:\nmil\nAbbreviation for second of arc is:\n''\n\nOther applications for this angle calculator ...\n\nWith the above mentioned two-units calculating service it provides, this angle converter proved to be useful also as a teaching tool:\n1. in practicing angular mils and seconds of arc ( mil vs. '' ) measures exchange.\n2. for conversion factors between unit pairs.\n3. work with angle's values and properties.\n\nTo link to this angle angular mil to seconds of arc online converter simply cut and paste the following.\nThe link to this tool will appear as: angle from angular mil (mil) to seconds of arc ('') conversion.\n\nI've done my best to build this site for you- Please send feedback to let me know how you enjoyed visiting." ]
[ null ]
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https://helpcenter.onlyoffice.com/en/onlyoffice-editors/onlyoffice-spreadsheet-editor/Functions/hlookup.aspx
[ "Close\nClose\nClose\n\n# HLOOKUP Function\n\nThe HLOOKUP function is one of the lookup and reference functions. It is used to perform the horizontal search for a value in the top row of a table or an array and return the value in the same column based on a specified row index number.\n\nThe HLOOKUP function syntax is:\n\nHLOOKUP (lookup-value, table-array, row-index-num[, [range-lookup-flag]])\n\nwhere\n\nlookup-value is a value to search for.\n\ntable-array are two or more rows containing data sorted in ascending order.\n\nrow-index-num is a row number in the same column of the table-array, a numeric value greater than or equal to 1 but less than the number of rows in the table-array.\n\nrange-lookup-flag is an optional argument. It is a logical value: TRUE or FALSE. Enter FALSE to find an exact match. Enter TRUE to find an approximate match, in this case if there is not a value that strictly matches the lookup-value, then the function will choose the next largest value less than the lookup-value. If this argument is absent, the function will find an approximate match.\n\nIf the range-lookup-flag is set to FALSE, but no exact match is found, then the function will return the #N/A error.\n\n## How to use HLOOKUP\n\nTo apply the HLOOKUP function,\n\n1. select the cell where you wish to display the result,\n2. click the Insert function", null, "icon situated at the top toolbar,\nor right-click within a selected cell and select the Insert Function option from the menu,\nor click the", null, "icon situated at the formula bar,\n3. select the Lookup and Reference function group from the list,\n4. click the HLOOKUP function,\n5. enter the required arguments separating them by comma,\n6. press the Enter button.\n\nThe result will be displayed in the selected cell.", null, "" ]
[ null, "https://helpcenter.onlyoffice.com/OfficeWeb/apps/spreadsheeteditor/main/resources/help/en/images/insertfunction.png", null, "https://helpcenter.onlyoffice.com/OfficeWeb/apps/spreadsheeteditor/main/resources/help/en/images/function.png", null, "https://helpcenter.onlyoffice.com/OfficeWeb/apps/spreadsheeteditor/main/resources/help/en/images/hlookup_function.gif", null ]
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https://metanumbers.com/31839
[ "# 31839 (number)\n\n31,839 (thirty-one thousand eight hundred thirty-nine) is an odd five-digits composite number following 31838 and preceding 31840. In scientific notation, it is written as 3.1839 × 104. The sum of its digits is 24. It has a total of 2 prime factors and 4 positive divisors. There are 21,224 positive integers (up to 31839) that are relatively prime to 31839.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 24\n• Digital Root 6\n\n## Name\n\nShort name 31 thousand 839 thirty-one thousand eight hundred thirty-nine\n\n## Notation\n\nScientific notation 3.1839 × 104 31.839 × 103\n\n## Prime Factorization of 31839\n\nPrime Factorization 3 × 10613\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 31839 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 31,839 is 3 × 10613. Since it has a total of 2 prime factors, 31,839 is a composite number.\n\n## Divisors of 31839\n\n1, 3, 10613, 31839\n\n4 divisors\n\n Even divisors 0 4 2 2\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 42456 Sum of all the positive divisors of n s(n) 10617 Sum of the proper positive divisors of n A(n) 10614 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 178.435 Returns the nth root of the product of n divisors H(n) 2.99972 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 31,839 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 31,839) is 42,456, the average is 10,614.\n\n## Other Arithmetic Functions (n = 31839)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 21224 Total number of positive integers not greater than n that are coprime to n λ(n) 10612 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3426 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 21,224 positive integers (less than 31,839) that are coprime with 31,839. And there are approximately 3,426 prime numbers less than or equal to 31,839.\n\n## Divisibility of 31839\n\n m n mod m 2 3 4 5 6 7 8 9 1 0 3 4 3 3 7 6\n\nThe number 31,839 is divisible by 3.\n\n## Classification of 31839\n\n• Arithmetic\n• Semiprime\n• Deficient\n\n• Polite\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (31839)\n\nBase System Value\n2 Binary 111110001011111\n3 Ternary 1121200020\n4 Quaternary 13301133\n5 Quinary 2004324\n6 Senary 403223\n8 Octal 76137\n10 Decimal 31839\n12 Duodecimal 16513\n20 Vigesimal 3jbj\n36 Base36 okf\n\n## Basic calculations (n = 31839)\n\n### Multiplication\n\nn×y\n n×2 63678 95517 127356 159195\n\n### Division\n\nn÷y\n n÷2 15919.5 10613 7959.75 6367.8\n\n### Exponentiation\n\nny\n n2 1013721921 32275892242719 1027632133115930241 32718779486278102943199\n\n### Nth Root\n\ny√n\n 2√n 178.435 31.6947 13.358 7.95412\n\n## 31839 as geometric shapes\n\n### Circle\n\n Diameter 63678 200050 3.1847e+09\n\n### Sphere\n\n Volume 1.35197e+14 1.27388e+10 200050\n\n### Square\n\nLength = n\n Perimeter 127356 1.01372e+09 45027.1\n\n### Cube\n\nLength = n\n Surface area 6.08233e+09 3.22759e+13 55146.8\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 95517 4.38954e+08 27573.4\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.75582e+09 3.80375e+12 25996.4\n\n## Cryptographic Hash Functions\n\nmd5 05a6cd7ec4ebca7b256ee86cefd9c488 56aef13cd41629b0064cb851a751cfab940db5db a8f29bfe4c0856a9c86329097ff2c2a32e17c692c9f806af68cbe1d1e06cbff0 a42d6e5733e633fdeeb9fa5f7926cf1412dd84e1c73b0d8fd6166593a62293636543a17f337f90c843648f412252754151e3d7493ccbefaaa927797f8819df47 94109115d9244b20beec96bdb705096dbb126000" ]
[ null ]
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https://www.mathematik.rwth-aachen.de/go/id/ckus/lidx/1/file/698365
[ "# Reconstructing Colourings of Finite Groups\n\nAachen (2017)\nDissertation / PhD Thesis\n\nDissertation, RWTH Aachen University, 2017\n\nAbstract\n\nThis thesis strives to promote a consistently combinatorial approach to study reconstruction problems on finite groups and $G$-sets. At the centre of this endeavour is the notion of the combinatorial $k$-deck coming in its two flavours for colourings and subsets. The combinatorial $k$-deck of a colouring $c$ of a finite $G$-set $X$ is defined as the multiset of equivalence classes of restrictions of $c$ to $k$-element subsets of $X$ (the so-called $k$-patches) that are moved around by the action of $G$, thus providing the notion of equivalence. Put another way, the combinatorial $k$-deck counts, for each $k$-patch, how many equivalent copies of it appear in $c$. The dominating notion of $k$-deck in the literature is a bit different: There, arithmetic calculations with the \"colours\" (that are indeed numbers) are performed resulting in analytic expressions satisfying certain invariance properties. The name analytical $k$-deck (as opposed to the combinatorial $k$-deck) is suggested and consistently used here. The combinatorial $k$-deck always determines the analytical $k$-deck. Conversely, the colour set $F$ and the analytical $\\bigl((|F|-1)k\\bigr)$-deck together determine the combinatorial $k$-deck provided that $F\\subseteq\\mathbb{R}_{\\geq 0}$. The gap between the smallest $k_1$ for which the combinatorial $k_1$-deck is able to distinguish between two given colourings and the smallest $k_2$ for which the analytical $k_2$-deck is able to do so may become arbitrarily large - even after a suitable change of colour values in favour of the analytical deck (whose reconstruction qualities are sensitive to the choice of the colour values). However, even for the combinatorial deck there is no global reconstruction number $k_0\\in\\mathbb{N}$, such that for all finite groups $G$, all finite $G$-sets $X$ and all (finite) colour sets $F$ each colouring $c:X\\longrightarrow F$ is reconstructible from its combinatorial $\\operatorname{min}\\{k_0,|X|\\}$-deck. It is unknown whether there is a global subset reconstruction number for all finite groups. In this respect the 5-transitive Mathieu groups $M_{12}$ and $M_{24}$ show that such a global subset reconstruction number would have to be at least 6. A full combinatorial understanding of all 3-deck failures of subsets in cyclic groups up to (and including) order 24 is obtained, mainly by considering so-called self-complementary subsets $A\\subseteq G$ for which there is some $g\\in G$ with $gA=G\\smallsetminus A$. Self-complementary subsets also explain a great amount of 3-deck failures in dicyclic groups. This thesis contains several positive reconstruction results, too. While the focus in the literature was so far mainly on cyclic or Abelian groups, the techniques suggested here are, in principle, applicable to all finite groups. One of them is the method of anchors, where the merits of sticking consistently to the combinatorial approach become particularly evident. Anchors show, for instance, that a colouring of a finite group with non-empty colour class $A$ is reconstructible from the combinatorial $\\left(\\left\\lfloor\\operatorname{log}_2 |A|\\right\\rfloor+2\\right)$-deck. (This was previously shown for subsets but the proof was more involved.) Another way to obtain positive reconstruction results (for the 3-deck only) is by means of certain matrices derived from the group matrix. The approach is a generalization of an idea which was previously applied to cyclic groups only. Applications include results for certain dihedral groups and for $p$-groups. One if these results is the following one: If $G$ is a finite $p$-group of order $|G|\\geq 3$ and $A\\subseteq G$ with $p\\nmid|A|$ then any colouring of $G$ with one of its colour classes equal to $A$ is reconstructible from the 3-deck.\n\nInstitutions\n\n• Department of Mathematics \n• Chair of Mathematics II (for Engineers)" ]
[ null ]
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https://export.arxiv.org/list/math.PR/2102?skip=200&show=25
[ "# Probability\n\n## Authors and titles for math.PR in Feb 2021, skipping first 200\n\n[ total of 333 entries: 1-25 | ... | 126-150 | 151-175 | 176-200 | 201-225 | 226-250 | 251-275 | 276-300 | ... | 326-333 ]\n[ showing 25 entries per page: fewer | more | all ]\n  arXiv:2102.00782 (cross-list from math.AG) [pdf, ps, other]\nTitle: How many roots of a system of random trigonometric polynomials are real?\nSubjects: Algebraic Geometry (math.AG); Probability (math.PR)\n  arXiv:2102.01049 (cross-list from math.AP) [pdf, ps, other]\nTitle: (Un-)bounded transition fronts for the parabolic Anderson model and the randomized F-KPP equation\nSubjects: Analysis of PDEs (math.AP); Probability (math.PR)\n  arXiv:2102.01162 (cross-list from math.NA) [pdf, ps, other]\nTitle: Strong rates of convergence of space-time discretization schemes for the 2D Navier-Stokes equations with additive noise\nSubjects: Numerical Analysis (math.NA); Probability (math.PR)\n  arXiv:2102.01200 (cross-list from cs.IT) [pdf, ps, other]\nTitle: Group Testing in the High Dilution Regime\nSubjects: Information Theory (cs.IT); Probability (math.PR)\n  arXiv:2102.01214 (cross-list from math.OA) [pdf, ps, other]\nTitle: Tree convolution for probability distributions with unbounded support\nSubjects: Operator Algebras (math.OA); Probability (math.PR)\n  arXiv:2102.01306 (cross-list from math.ST) [pdf, ps, other]\nTitle: An Asymptotic Theory of Joint Sequential Changepoint Detection and Identification for General Stochastic Models\nSubjects: Statistics Theory (math.ST); Probability (math.PR)\n  arXiv:2102.01377 (cross-list from math-ph) [pdf, other]\nTitle: Effective Mori-Zwanzig equation for the reduced-order modeling of stochastic systems\nAuthors: Yuanran Zhu, Huan Lei\nSubjects: Mathematical Physics (math-ph); Statistical Mechanics (cond-mat.stat-mech); Functional Analysis (math.FA); Numerical Analysis (math.NA); Probability (math.PR)\n  arXiv:2102.01419 (cross-list from math.OC) [pdf, ps, other]\nTitle: Community Detection with a Subsampled Semidefinite Program\nSubjects: Optimization and Control (math.OC); Probability (math.PR); Machine Learning (stat.ML)\n  arXiv:2102.01484 (cross-list from math.OC) [pdf, ps, other]\nTitle: A Modified Method of Successive Approximations for Stochastic Recursive Optimal Control Problems\nSubjects: Optimization and Control (math.OC); Probability (math.PR)\n  arXiv:2102.01947 (cross-list from math.RT) [pdf, ps, other]\nTitle: Infinite-dimensional groups over finite fields and Hall-Littlewood symmetric functions\nSubjects: Representation Theory (math.RT); Combinatorics (math.CO); Probability (math.PR)\n  arXiv:2102.02037 (cross-list from math.MG) [pdf, ps, other]\nTitle: The isometry group of Wasserstein spaces: the Hilbertian case\nSubjects: Metric Geometry (math.MG); Mathematical Physics (math-ph); Functional Analysis (math.FA); Probability (math.PR)\n  arXiv:2102.02047 (cross-list from math.DS) [pdf, other]\nTitle: On the convergence rate of the chaos game\nComments: 29 pages, 3 figures, 2 tables\nSubjects: Dynamical Systems (math.DS); Classical Analysis and ODEs (math.CA); Metric Geometry (math.MG); Probability (math.PR)\n  arXiv:2102.02280 (cross-list from math.NT) [pdf, other]\nTitle: Prime zeta function statistics and Riemann zero-difference repulsion\nComments: New introduction and conclusion sections, minor edits throughout\nJournal-ref: J. Stat. Mech. (2021) 073206\nSubjects: Number Theory (math.NT); Statistical Mechanics (cond-mat.stat-mech); Mathematical Physics (math-ph); Probability (math.PR)\n  arXiv:2102.02450 (cross-list from math.ST) [pdf, ps, other]\nTitle: Sharper Sub-Weibull Concentrations: Non-asymptotic Bai-Yin Theorem\nSubjects: Statistics Theory (math.ST); Probability (math.PR); Machine Learning (stat.ML)\n  arXiv:2102.02710 (cross-list from math.OC) [pdf, other]\nTitle: Matching Impatient and Heterogeneous Demand and Supply\nSubjects: Optimization and Control (math.OC); Performance (cs.PF); Systems and Control (eess.SY); Probability (math.PR)\n  arXiv:2102.02954 (cross-list from math-ph) [pdf, ps, other]\nTitle: Superballistic and superdiffusive scaling limits of stochastic harmonic chains with long-range interactions\nAuthors: Hayate Suda\nSubjects: Mathematical Physics (math-ph); Probability (math.PR)\n  arXiv:2102.03172 (cross-list from math.OC) [pdf, ps, other]\nTitle: Noether theorem in stochastic optimal control problems via contact symmetries\nJournal-ref: Mathematics (MDPI) 9, no. 9 (2021)\nSubjects: Optimization and Control (math.OC); Probability (math.PR); Mathematical Finance (q-fin.MF)\n  arXiv:2102.03209 (cross-list from math.NA) [pdf, ps, other]\nTitle: Convergence analysis of explicit stabilized integrators for parabolic semilinear stochastic PDEs\nSubjects: Numerical Analysis (math.NA); Probability (math.PR)\n  arXiv:2102.03225 (cross-list from math.CO) [pdf, ps, other]\nTitle: Expected Value of Statistics on Type-B Permutation Tableaux\nSubjects: Combinatorics (math.CO); Probability (math.PR)\n  arXiv:2102.03589 (cross-list from math.ST) [pdf, ps, other]\nTitle: Edgeworth approximations for distributions of symmetric statistics\nComments: This paper dates back to 2005 including literature up to this year\nSubjects: Statistics Theory (math.ST); Probability (math.PR)\n  arXiv:2102.03680 (cross-list from math.MG) [pdf, ps, other]\nTitle: Reverse isoperimetric inequalities for parallel sets\nAuthors: Piotr Nayar\nSubjects: Metric Geometry (math.MG); Probability (math.PR)\n  arXiv:2102.04083 (cross-list from math.DS) [pdf, other]\nTitle: On Convergence of Random Walks on Moduli Space\nComments: 11 pages; small improvements in presentation in Section 3 on pathwise equidistribution\nSubjects: Dynamical Systems (math.DS); Probability (math.PR)\n  arXiv:2102.04224 (cross-list from math.NA) [pdf, other]\nTitle: Numerical approximation and simulation of the stochastic wave equation on the sphere\nSubjects: Numerical Analysis (math.NA); Probability (math.PR)\n  arXiv:2102.04380 (cross-list from math-ph) [pdf, ps, other]\nTitle: Space-time statistical solutions for an inhomogeneous chain of harmonic oscillators\nAuthors: T.V. Dudnikova" ]
[ null ]
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https://mathematica.stackexchange.com/questions/111467/perform-matrix-tensor-contractions-more-efficiently
[ "# Perform matrix/tensor contractions more efficiently\n\nProblem: explicit index contractions for matrices using the Sum command take too long, and I want to improve the performance for complicated computations.\n\nLet me give you an example, let M be a random $8\\times 8$ matrix:\n\nM = RandomReal[{0, 1}, {8, 8}];\n\n\nIf I want to multiply 8 of those matrices and take a trace I would write something like\n\nTr[M.M.M.M.M.M.M.M]\n\n\nThis gives me the result almost instantaneously.\n\nI can write the same computation in a more complicated way by using the Sum or equivalently ParallelSum command and writing out the index contractions explicitly:\n\nParallelSum[\nM[[i1, i2]] M[[i2, i3]] M[[i3, i4]] M[[i4, i5]]\nM[[i5, i6]] M[[i6, i7]] M[[i7, i8]] M[[i8, i1]],\n{i1, 1, 8}, {i2, 1, 8}, {i3, 1, 8}, {i4, 1, 8},\n{i5, 1, 8}, {i6, 1, 8}, {i7, 1, 8}, {i8, 1, 8}]\n\n\nThis computation takes literally ages, and I actually never waited till the end to see the final result.\n\nQuestion: how can I implement explicit contractions like in the last example efficiently? I need to do this because in more complicated computations with multidimensional tensors contractions cannot be written using matrix multiplication and traces. What is the best practise for this kind of computations?\n\n• Are you absolutely certain that they can't be done using Dot (which is more general than just matrix multiplication), Transpose, Flatten, and Tr? I'm willing to bet that it can be done with those optimized built-ins. Also, at least, please provide an example of one that can't be done with matrix multiplication and Tr for us to play with. – march Mar 30 '16 at 21:12\n• The important fact you have to know is that Dot[] always contracts the last index of one tensor into the first of another. Therefore, you just need to Transpose or Flatten appropriately and then use Dot. As @march said this will always work and you can do the most complicated contractions. – Michael Weyrauch Mar 30 '16 at 22:27\n• You've seen MatrixPower[], right? – J. M. will be back soon Mar 30 '16 at 22:58\n\nI think the simplest way to handle the general case is to use TensorProduct and TensorContract, as follows:\n\nTake a rank 3 array for example, in dimension 100:\n\nIn:= A = RandomReal[{-1, 1}, {100, 100, 100}];\n\n\nConstruct a rank 12 array. Note the use of Inactive, to avoid TensorProduct constructing a large intermediate array:\n\nIn:= A4 = Inactive[TensorProduct][A, A, A, A];\n\nIn:= TensorRank[A4]\nOut= 12\n\n\nNow choose any random contraction of all levels:\n\nIn:= contraction = Partition[RandomSample[Range], 2]\nOut= {{1, 3}, {11, 6}, {8, 7}, {12, 9}, {4, 5}, {2, 10}}\n\n\nThat means a contraction of the 1st and 3rd levels, the 11th and 6th levels, etc. You can easily see the levels as indices.\n\nFinally TensorContract will perform the contraction, and quite fast. Note again the use of Activate to eliminate the Inactive head:\n\nIn:= TensorContract[A4, contraction] // Activate // AbsoluteTiming\nOut= {0.006712, -115212.}\n\n• Thanks for the answer, I actually only now really got to use it! Could you maybe explain why I get an out of memory message if I don't use Inactive for the TensorProduct? – Stan May 19 '16 at 23:37\n• The tensor product of small arrays can be a very large array. The tensor contraction will then reduce that large array to a small result, and the use of Inactive allows getting to that final result without actually constructing the intermediate large array. If you don't use Inactive, TensorProduct will construct a huge array, and that's why you get the memory error. In my example above A has 100^3 = 10^6 entries and A4 would have 100^12 = 10^24 entries if allowed to expand, which is too much for computer memory. – jose May 24 '16 at 3:18\n\nI am illustrating my answer with a very short code. Rather, it is a long comment. The computational complexity of your two examples is cardinally different. In the first case you perform 7 matrix multiplications and perform a trace on the result. The last operation has a quadratic scaling. Therefore leading complexity is $7 N^3$, where $N$ is the matrix dimension.\n\nIn your second example you do all multiplications at once. Computational complexity is $N^8$. Quite clear $N^8\\gg 7N^3$ !\n\nNow, can we do even better? Yes it is indeed possible. You can do the following sequence of operations:\n\n M=M.M\nM=M.M\nM=M.M\nTr[M]\n\n\nThe computational complexity is $3 N^3$. We have a speed up of 7/3.\n\nCan we do even better? Yes it is indeed possible. In the above example $3\\times 8^3=1536$ operations (count only multiplications) is required. We can use the simplest fast matrix multiplication algorithm due to Strassen. It requires 7 multiplications to multiply matrix $2\\times2$. Matrices with dimensions that are powers of 2 are well suitable to apply the algorithm recursively. To multiply 2 matrices $8\\times8$ only $7^3=343$ operations are required. In total we need only 1029 multiplications. Considering that we only need diagonal elements for the trace we can reduce the estimate even further $2\\times 7^3+8^2=750$.\n\nCan we do even better? Quite probably yes, however, the algorithm for fast matrix multiplications of matrices $4\\times4$ with less then 49 multiplications is currently unknown. It was conjectured, however, that asymptotically ($N\\rightarrow \\infty$) the scaling is quadratic.\n\nTo summarise:\n\n$16777216\\gg1536>1029>750\\ge$computational complexity$>192$. Thus, there is still a lot of room for improvement :)" ]
[ null ]
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https://www.gsrikar.com/2017/06/how-to-cast-integer-to-float-and-vice.html
[ "### How to Cast Integer to Float and Vice Versa in TensorFlow\n\nIn this article, we will learn to convert the data types of tensors to integer and float.\n\nFirst, let's create two nodes.\n``````node1 = tf.constant(5) # dtype is int32\nnode2 = tf.constant(6.0) # dtype is float32``````\nLet's multiply the two nodes to create a new node\n``node3 = tf.multiply(node1, node2)``\nThis will throw a\n``````TypeError: Input 'y' of 'Mul' Op has type float32 that does not match type\nint32 of argument 'x'.``````\nThis is because the node1 data type is an integer while the data type of node2 is a float.\n\nMultiply method doesn't convert the types of its variables automatically. For now, we need to change them before performing any type of arithmetic operations on them. I hope this changes in the future releases.\n\nThe correct way is to convert either of the node's type to match the other. Here, we are converting the node1 which is an integer to a float.\n``tf.to_float(node1, name='ToFloat')``\nReplacing the node1 with the above line in the multiply method will resolve the issue. To learn, more about other types of casting you can check the official website.\n\nBelow is a full code that correctly performs multiplication of two nodes when there are of different types.\n\nIn a different scenario, you'll get an error\n``````ValueError: Tensor conversion requested dtype int32 for Tensor with dtype\nfloat64 in Tensorflow.``````\nThe solution is same. Convert the first tensor or the second tensor type to match the other." ]
[ null ]
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https://help.scilab.org/docs/5.5.0/fr_FR/section_021dc814e3abca38503f35dd83ec0ed0.html
[ "Change language to:\nEnglish - 日本語 - Português - Русский\n\nSee the recommended documentation of this function\n\nAide de Scilab >> Optimisation et Simulation > Semidefinite Programming\n\n### Semidefinite Programming\n\n• aplatFlattens a list.\n• list2vecConcatenates list entries in a matrix.\n• lmisolverSolve linear matrix inequations.\n• lmitoolGraphical tool for solving linear matrix inequations.\n• packCompress a list of block-diagonal symmetric matrices.\n• reconsInverse function for aplat.\n• semidefSolve semidefinite problems.\n• unpackUncompress a list of block-diagonal symmetric matrices.\n• vec2listInverse function of list2vec.\n Report an issue << Optimization simplex Optimisation et Simulation fsolve >>" ]
[ null ]
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http://mathcentral.uregina.ca/QandQ/topics/debt
[ "", null, "", null, "Math Central - mathcentral.uregina.ca", null, "", null, "Quandaries & Queries", null, "", null, "", null, "", null, "Q & Q", null, "", null, "", null, "", null, "Topic:", null, "debt", null, "", null, "", null, "start over\n\n2 items are filed under this topic.", null, "", null, "", null, "", null, "Page1/1", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Paying off a debt 2007-04-26", null, "From Ema:How would you enter a formula in Excel for the following: payment is applied to principal balance, then 8% interest compounded daily, then .25% monthly penaltyAnswered by Stephen La Rocque and Penny Nom.", null, "", null, "", null, "", null, "", null, "Three debts 2005-02-03", null, "From Kat:If I have three debts and 49% percent of total debt is loaned at 9% intrest, 34% of the debt is at 21% intrest and 17% of the total debt is at 14% intrest, how do I calculate the average rate of intrest on total debt?Answered by Penny Nom.", null, "", null, "", null, "", null, "", null, "", null, "Page1/1", null, "", null, "", null, "", null, "Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.", null, "", null, "", null, "", null, "about math central :: site map :: links :: notre site français" ]
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https://metanumbers.com/10465
[ "## 10465\n\n10,465 (ten thousand four hundred sixty-five) is an odd five-digits composite number following 10464 and preceding 10466. In scientific notation, it is written as 1.0465 × 104. The sum of its digits is 16. It has a total of 4 prime factors and 16 positive divisors. There are 6,336 positive integers (up to 10465) that are relatively prime to 10465.\n\n## Basic properties\n\n• Is Prime? No\n• Number parity Odd\n• Number length 5\n• Sum of Digits 16\n• Digital Root 7\n\n## Name\n\nShort name 10 thousand 465 ten thousand four hundred sixty-five\n\n## Notation\n\nScientific notation 1.0465 × 104 10.465 × 103\n\n## Prime Factorization of 10465\n\nPrime Factorization 5 × 7 × 13 × 23\n\nComposite number\nDistinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 10465 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 10,465 is 5 × 7 × 13 × 23. Since it has a total of 4 prime factors, 10,465 is a composite number.\n\n## Divisors of 10465\n\n1, 5, 7, 13, 23, 35, 65, 91, 115, 161, 299, 455, 805, 1495, 2093, 10465\n\n16 divisors\n\n Even divisors 0 16 8 8\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 16128 Sum of all the positive divisors of n s(n) 5663 Sum of the proper positive divisors of n A(n) 1008 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 102.299 Returns the nth root of the product of n divisors H(n) 10.3819 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 10,465 can be divided by 16 positive divisors (out of which 0 are even, and 16 are odd). The sum of these divisors (counting 10,465) is 16,128, the average is 1,008.\n\n## Other Arithmetic Functions (n = 10465)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 6336 Total number of positive integers not greater than n that are coprime to n λ(n) 132 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1280 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 6,336 positive integers (less than 10,465) that are coprime with 10,465. And there are approximately 1,280 prime numbers less than or equal to 10,465.\n\n## Divisibility of 10465\n\n m n mod m 2 3 4 5 6 7 8 9 1 1 1 0 1 0 1 7\n\nThe number 10,465 is divisible by 5 and 7.\n\n## Classification of 10465\n\n• Arithmetic\n• Deficient\n\n• Polite\n\n• Heptagonal\n\n• Square Free\n\n### Other numbers\n\n• LucasCarmichael\n\n## Base conversion (10465)\n\nBase System Value\n2 Binary 10100011100001\n3 Ternary 112100121\n4 Quaternary 2203201\n5 Quinary 313330\n6 Senary 120241\n8 Octal 24341\n10 Decimal 10465\n12 Duodecimal 6081\n20 Vigesimal 1635\n36 Base36 82p\n\n## Basic calculations (n = 10465)\n\n### Multiplication\n\nn×i\n n×2 20930 31395 41860 52325\n\n### Division\n\nni\n n⁄2 5232.5 3488.33 2616.25 2093\n\n### Exponentiation\n\nni\n n2 109516225 1146087294625 11993803538250625 125515154027792790625\n\n### Nth Root\n\ni√n\n 2√n 102.299 21.8732 10.1143 6.36719\n\n## 10465 as geometric shapes\n\n### Circle\n\n Diameter 20930 65753.5 3.44055e+08\n\n### Sphere\n\n Volume 4.80072e+12 1.37622e+09 65753.5\n\n### Square\n\nLength = n\n Perimeter 41860 1.09516e+08 14799.7\n\n### Cube\n\nLength = n\n Surface area 6.57097e+08 1.14609e+12 18125.9\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 31395 4.74219e+07 9062.96\n\n### Triangular Pyramid\n\nLength = n\n Surface area 1.89688e+08 1.35068e+11 8544.64\n\n## Cryptographic Hash Functions\n\nmd5 41263b9a46f6f8f22668476661614478 f494954618da5b94a9a1b7b45127df29e5fe7450 0d4112730fb1723be69671d9bd0627119b58fbfa04d0455dcfddd2c7cb7d6342 54e6866bf4b6a064744cfc0c3da12eaea9b65146eee30c81aee70ecf6f8039cc90a4aae3a91af39260dc88fef92c00897b8c68a729fc76299db343ac0730b1a6 7d3355ff8a0e8f8fcaa6c31d585ccd463ec95ed1" ]
[ null ]
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https://sportsbizusa.com/20-worksheet-works-calculating-volume/worksheet-works-calculating-volume-worksheet-works-probability-word-problems-3/
[ "# Worksheet Works Calculating Volume Worksheet Works Probability Word Problems\n\nHome20 Worksheet Works Calculating Volume ➟ Worksheet Works Calculating Volume Worksheet Works Probability Word Problems\n\nWorksheet Works Calculating Volume Worksheet Works Probability Word Problems one of Worksheet From Home - ideas, to explore this Worksheet Works Calculating Volume Worksheet Works Probability Word Problems idea you can browse by and . We hope your happy with this Worksheet Works Calculating Volume Worksheet Works Probability Word Problems idea. You can download and please share this Worksheet Works Calculating Volume Worksheet Works Probability Word Problems ideas to your friends and family via your social media account. Back to 20 Worksheet Works Calculating Volume\n\n## 20 Letter W Worksheet for Preschool\n\nSay and Trace Letter W Beginning Sound Words Worksheet letter c worksheets for preschool pdf, letter k worksheet preschool, letter f tracing worksheets preschool, letter b worksheet preschool free, letter b worksheet preschool, via: pinterest.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind […]\n\n## 20 Shapes Worksheet for Kindergarten\n\nSummer Review with a freebie Kindergarten Smarts identify shapes worksheet kindergarten, 3d shapes worksheet kindergarten, shapes worksheet kindergarten pdf, identifying 3d shapes worksheet kindergarten, solid figures worksheet kindergarten, via: pinterest.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math […]\n\n## 20 Estimating Numbers Worksheets\n\nRounding Whole Numbers Worksheets from The Teacher s Guide estimating mixed numbers worksheets, estimate quotients using patible numbers worksheet answers, estimating quotients using patible numbers worksheets, via: pinterest.com Numbering Worksheets for Kids. Kids are usually introduced to this topic matter during their math education. The main reason behind this is that learning math can be […]" ]
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https://www.igi-global.com/chapter/isomorphism/73148
[ "", null, "# Isomorphism\n\nPayman Biukaghazadeh (Amirkabir University of Technology, Iran)\nDOI: 10.4018/978-1-4666-2661-4.ch003\n\n## Abstract\n\nIn this chapter, the isomorphism application in graph theory is discussed. Various types of the isomorphism such as the automorphism and the homomorphism are introduced. The theorems and hints to reject or accept the isomorphism of graphs are the next section. In the some sections, knowledge about matrices may be necessary; if so, reading chapters one, and especially two, are necessary to understand this chapter.\nChapter Preview\nTop\n\n## Isomorphism\n\nSuppose that ψG (e) = {u, v} is defined in simple graphs. If in graphs G (V, E) and H (V, E), V (G) = V (H), E (G) = E (H), and", null, "then these graphs are called identical. Identical graphs could be shown by the same diagrams. On the other hand, for the graphs which are not identical it is possible to show them by the same diagrams. For instance, it could be seen in figure 1 that graphs G and H have similar structures, as in the other shape of the H. Only the labels of the vertices and the edges of these two graphs are not similar. It is clear that the graphs H and G just have the same structures, drawings, and could not be considered as identical graphs, but they have similar structures.\n\nThis similarity of structure of non-identical graphs is called isomorphism. Two bijections for the vertices and edges of isomorphic graphs, written G ≅ H, must be found in general: θ: is a bijection from the vertices of the graph G to the vertices of the graph H, and φ: is a bijection from the edges of the graph G to the edges of the graph H. The necessary and sufficient conditions for these bijections are:", null, "This bijections hold the isomorphism of the graphs G and H (Bondy & Murty, 2008).\n\nIsomorphic graphs in some of the references such as Diestel (2005) is shown by", null, "The graph property is a group of graphs which is closed for isomorphism. For example, a graph property is the existence of a triangle: the isomorphic graphs to the graph which has a triangle, all must have three adjacent vertices. Isomorphic graphs share common values for some variables, which is called the graph invariants. The most simple graph invariants are the number of the vertices and the number of the edges of a graph; another one is the maximum number of pair wise adjacent vertices (Diestel, 2005).\n\nThe definition of the isomorphism for simple graphs could be described more exactly. The bijection φ could be determined by θ, if the pair of θ and φ holds the same structure of graphs G and H. However, φ (e) = θ (u) θ (v) is true for all the edges of this graph. Hence, isomorphism of two graphs G and H could be defined in terms of the defined permutation θ: V (G) → V (H). This definition holds the adjacency (Godsil & Royle, 2000).\n\n## Complete Chapter List\n\nSearch this Book:\nReset" ]
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https://vmt.in/collections/large-diaphragm
[ "## Large Diaphragm\n\nShowing all 48 Items\n(0)\nRs. 12,150.00\n(0)\nRs. 42,120.00\n(0)\nRs. 22,140.00\n(0)\nRs. 21,540.00\n(0)\nRs. 67,680.00\n(0)\nRs. 4,784.00\n(0)\nRs. 6,384.00\n(0)\nRs. 14,200.00\n(0)\nRs. 75,810.00\n(0)\nRs. 25,500.00\n(0)\nRs. 35,160.00\n(0)\nRs. 11,070.00\n(0)\nRs. 67,680.00\n(0)\nRs. 38,428.00\n(0)\nRs. 48,909.00\n(0)\nRs. 31,138.00\n(0)\nRs. 18,830.00\n(0)\nRs. 33,190.00\n(0)\nRs. 13,920.00\n(0)\nRs. 12,960.00\n(0)\nRs. 8,540.00\n(0)\nRs. 13,990.00\n(0)\nRs. 14,990.00\n(0)\nRs. 9,584.00\n(0)\nRs. 17,850.00\n(0)\nRs. 35,160.00\n(0)\nRs. 20,990.00\n(0)\nRs. 7,260.00\n(0)\nRs. 23,400.00\n(0)\nRs. 36,200.00\n(0)\nRs. 36,800.00\n(0)\nRs. 35,240.00\n(0)\nRs. 22,384.00\n(0)\nRs. 7,984.00\n(0)\nRs. 14,384.00\n(0)\nRs. 6,750.00\n(0)\nRs. 37,480.00\n(0)\nRs. 24,230.00\n(0)\nRs. 18,810.00\n(0)\nRs. 4,230.00\n(0)\nRs. 9,410.00\n(0)\nRs. 31,350.00\n(0)\nRs. 23,670.00\n(0)\nRs. 41,120.00\n(0)\nRs. 32,450.00" ]
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http://math203sp18.wikidot.com/maa-questions-20-22-post-write
[ "Questions 20 & 22 Post Write\n\n20. Statement: The polynomial ring $\\mathbb{Z}[x]$ over the integers is countably infinite.\nDefinitions:\nA countable union of countably infinite sets is countably infinite.\nThe finite product of infinitely countable sets is infinitely countable.\n$\\mathbb{Z}[x] := { \\sum^n _{i=0} \\alpha _i x^i| \\alpha_i \\in \\mathbb{Z} \\forall i \\wedge n \\in \\mathbb{N} }$\nProof: We can prove the polynomial ring is countably infinite by showing the union of all elements of $\\mathbb{Z}[x]$ ($S_oUS_1US_2U...US_nU... \\forall{n}\\in\\mathbb{N}$) is countable and that each set of $\\mathbb{Z}[x]$ is countably infinite.\nWe use induction to show that $S_n$ is countably infinite $\\forall{n}\\in\\mathbb{N}$. The base case $S_o$ is equal to the integers which are countably infinite. We then assume $S_n$ is countably infinte $\\forall n < N$. Next, we want to show that $S_{n+1}$ is countably infinite. $S_{n+1}\\simeq \\mathbb{Z} \\times S_n$. As $\\mathbb{Z}$ and $S_n$ are countably infinite, and thus so is $S_{n+1}$.\nThe union of all elements of $\\mathbb{Z}[x]$ can be shown by proving a bijection $f: S_n \\rightarrow \\mathbb{N}$. $n \\in \\mathbb{N}$ so all elements in $n$ correspond to one element in $\\mathbb{N}$\n\n(Prof. Simmons, 4-2-18: The LaTeX command for union is \\cup $\\cup$. The argument is basically right, but you're done after showing that $S_{n+1}$ is countable if $S_n$ is because there are obviously countably many $S_i$. At the beginning you can simply say that you will show $\\mathbb{Z}[x]$ to be a countable union of countable sets.)\n\n22. Let $(R,0, 1, +, \\cdot,\\leq)$ be an ordered ring with identity.\na. Statement: If $x>0$, then $-x<0$.\n$x + (-x) = 0$ is the definition of an inverse\n$x>0$\n$x + (-x) > 0 + (-x)$ add -x to both sides (And addition respects inequality in ordered rings)\n$0 > -x$ by the definition of an inverse and the additive property of 0.\nb. Statement: $1>0$.\n$1·n=n$ by the identity property\n$1>0$ (This is what you're trying to prove….)\n$1·n>0·n$ multiply by n on both sides\n$n>0$ by the identity property and the multiplicative identity of zero\n\npage revision: 14, last edited: 02 Apr 2018 19:45\nUnless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License" ]
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https://www.dsprelated.com/showthread/matlab/2662-1.php
[ "# Finding a Faster Way\n\nStarted by January 2, 2004\n Here is the code snipet i need help with... for i = 1:end_frame for j = 1:1024 fout(i,j) = ifft(Z(i,j)); end end As you can see, if the end_frame is large, the for loop would take so long. What alternatives can I do? I've tried fout = ifft(Z); but it doesn't seem to give out the right answer. I've also tried for i = 1:end_frame fout(i,:) = ifft(Z(i,:)); end but again it gives a different answer. Which one gives the right answer anyway? And could you help me out becuase this part takes very long in my program. By the way, the dimension of Z is Z(1:end_frame,1:1024)\n Hi, ifft(z) where z is a matrix gives inverse DFT of each column of the matrix. > for i = 1:end_frame > for j = 1:1024 > fout(i,j) = ifft(Z(i,j)); > end > end Here above, you are just taking one element of the matrix for computation of idft. I guess this is not what you wanted. and the other way you tried > for i = 1:end_frame > fout(i,:) = ifft(Z(i,:)); > end Here you are computing idft for each row of the matrix, so you get a different answer in all the three cases. Just check your input matrix and apply the appropriate method. regards, Ravi > Here is the code snipet i need help with... > > for i = 1:end_frame > for j = 1:1024 > fout(i,j) = ifft(Z(i,j)); > end > end > > As you can see, if the end_frame is large, the for loop would take > so long. What alternatives can I do? > > I've tried fout = ifft(Z); but it doesn't seem to give out the right > answer. > > I've also tried > for i = 1:end_frame > fout(i,:) = ifft(Z(i,:)); > end > but again it gives a different answer. > > Which one gives the right answer anyway? And could you help me out > becuase this part takes very long in my program. By the way, the > dimension of Z is Z(1:end_frame,1:1024) > V Ravi Chander 200211014 MTech (ICT) DA-IICT ________________________________________________________________________ \"If everything seems under control, you're just not going fast enough.\" - Mario Andretti ________________________________________________________________________" ]
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