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http://education.yuvajobs.com/icse/physics2000/set1.php
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math
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ICSE Set Set1 Year Icse Physics2000 Set1.php Physics Exam Paper for students online
|Answers of this paper >>|
-All working, including rough work, should be done on the same sheet as, and adjacent to, the rest of the answer.
-The intended marks for questions or parts of questions are given in brackets ( ).
-Material to be supplied: Log tables including Trigonometric functions
-A list of useful physical constants is given at the end of this paper.
Q 1Answer all questions
briefly and to the point:- (Marks 20)
(i) State Gauss's theorem of electrostatics.
(ii) An alpha particle is accelerated through a potential difference of 2000 volt. What will be the increase in its energy in electron volt?
(iii) Assuming the earth as an isolated spherical conductor of radius 6400 km, calculate its electrical capacitance in farad.
(iv) If the potential difference applied across a variable resistor is constant, draw a graph between the current in the resistor and the resistance.
(v) Find the equivalent resistance between the point 'a' and 'c' of the given network of resistors (See Fig. 1).
(vi) Show with a labelled graph how thermo-emf varies with the temperature of the hot junction of a thermocouple.
(vii) What is the function of shunt in an ammeter?
(viii) What is meant by 'Wattless' current?
(ix) Name the physical principle and one advantage in the use of optical fibers.
(x) Can two independent monochromatic source of light be coherent? Explain very briefly.
(xi) What is the relation between the refractive indices n1 and n2 if the behaviour of light rays is as shown in the following Fig. 2?
Here n1 and n2 are the refractive indices of the surrounding medium and the lens respectively.
(xii) What is the principle of global communication by satellites?
(xiii) State Brewster's law for polarisation of light.
(xiv) Two thin lenses-one convex and the other concave - of focal lengths 15 cm and 20 cm respectively are put in contact. Find the focal length of the doublet lens.
(xv) Two stars three and four light years away from the earth have same luminous intensity. Find the ration of the illuminance (intensity of illumination) on the surface of the earth, normal to the starlight, produced by each star.
(xvi) The historic experiment on the diffraction of electron confirmed a new nature of electron. What is this new nature of electron? Who had proposed it?
(xvii) Explain briefly why there is a maximum frequency for the X-rays produced by an X-ray tube operating at a certain voltage.
(xviii) Draw circuit diagrams to illustrate forward biasing and reverse biasing of a diode.
(xix) Name two alternative sources of energy and mention where they can be used.
(xx) The half-life period of a radioactive substance is 16 hours. After how much time will 6.25% of the material remain undecayed?
Answer six question in this part, choosing two questions each from the Sections A, B and C.
Answer two questions
Q 2 (a) Derive an expression
for the electric potential at a point due to a point charge. (Marks 4)
Q 2 (b) An electric flash lamp has 20 capacitors each of capacitance 5F connected in parallel. The lamp is operated at 100 volt. Calculate how much energy will be radiated in a flash. (Marks 3)
Q 2 (c) Name one material whose resistivity decreases with rise in temperature. Explain briefly on the basis of free electron theory why the resistivity decreases. (Marks 2)
Q 3 (a)
State and explain Kirchoff's laws for electric circuits. What are the
conservation laws implied in each law? State the sign convention for current and
emf. Use the Fig. 3 given below for your explanation. (Marks 4)
Q 3 (b) An electron is moving vertically upwards in the uniform magnetic field of the earth. The velocity of the electron is 2.0 x 106 m/s and the horizontal component of earth's magnetic field is 0.32 x 10-4 tesla. Obtain the magnitude and direction of the force on the electron. (Marks 3)
Q 3 (c) Permanent magnets are made of special alloy while the core of temporary magnets are made of soft iron. Why? (Marks 2)
Q 4 (a)
Define self inductance. Obtain the expression for the self inductance of a
solenoid, explaining steps with the help of a diagram. (Marks 3)
Q 4 (b) What is meant by back emf in a dc motor? The back emf in a dc motor, delivering 3 kilowatts of mechanical power, is 180 volts when operating on 220 volt line. Determine the armature current and the motor resistance. (Marks 3)
Q 4 (c) In an LCR circuit with all components connected in series, the emf and the current flowing in the circuit are given by the following equations:-
= 200 sin (314t + /6) volts
I = 5 sin 314 t ampere
(i) the peak values of current and emf.
(ii) the frequency of the ac source.
(iii) the phase difference between current and emf. (Marks 3)
Answer two questions.
Q 5 (a) State Huygens'
principle. Explain with the help of a diagram the phenomenon of refraction of
waves on the basis of this principle. (Marks 3)
Q 5 (b) What is chromatic aberration in lenses? State the necessary condition for an achromatic doublet. How are these conditions practically achieved? (Marks 3)
Q 5 (c) Fraunhofer diffraction from a single slit of width 1.0 m is observed with light of wavelength 500 nm. Calculate the half angular width of the central maximum. (Marks 2)
Q 6 (a) Draw a neat ray
diagram of a simple microscope. Deduce the formula for its angular magnification
when the image is formed at the least distance of distinct vision. (Marks 3)
Q 6 (b) A beam of monochromatic light of wavelength 500 nm falls on two parallel slits. The distance between the slits is 0.15 mm. Determine the width of the interference fringes on a screen placed at a distance of 1.5 m from the slits. (Marks 3)
Q 6 (c) Draw a sketch of electromagnetic spectrum, showing relative positions of UV, IR, X-rays and microwaves with respect to visible light. State approximate wavelengths of any two. (Marks 2)
Q 7 (a) With the help of neat
diagram describe with theory, Michelson's method for the determination of the
speed of light. What is the presently accepted nine digit value of the speed of
light? (Marks 4)
Q 7 (b) What is meant by pure and impure spectrum? Explain briefly how you will set up a spectrometer to obtain pure spectrum. Draw a diagram of a spectrometer and describe its parts. (Marks 4)
Answer two questions.
Q 8 (a) The ionisation
potential of hydrogen atom is 13.6 volt. Draw the energy level diagram showing
four levels. Calculate
(i) the energy of the photon emitted when an electron falls from the third orbit to the second orbit.
(ii) the wavelength of this photon. (Marks 4)
Q 8 (b) Described briefly with a diagram and theory an experiment to determine e/m of electron. (Marks 4)
Q 9 (a)
What is Compton scattering? State briefly its importance. (Marks 2)
Q 9 (b) When ultraviolet light of wavelength 300 nm is incident on a metal plate, a negative potential of 0.54 volt is required to stop the emission of photo electrons. Calculate the energy of the incident photon and the work function for the metal in eV. (Marks 4)
Q 9 (c) Draw the circuit diagram of a half wave rectifier using a semi-conductor diode. Explain briefly the function of each component. (Marks 2)
Q 10 (a)
Heavy water is suitable moderator in a nuclear reactor. Explain briefly why? (Marks
Q 10 (b) Draw labelled diagrams to illustrate
(i) energy bands of a conductor, semiconductor and insulator.
(ii) npn and pnp transistors.
(iii) transistor as an amplifier (common emitter). (Marks 4)
Q 10 (c) What do you mean by AND gate? How will you realise AND gate with junction diodes? (Marks 2)
|Answers of this paper|
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s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247489425.55/warc/CC-MAIN-20190219061432-20190219083432-00510.warc.gz
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CC-MAIN-2019-09
| 7,588 | 86 |
https://www.novamdia.me/800-kcal-in-kg/
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math
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800 kcal in kg
kcal/kg↔mJ/ug 1 mJ/ug = 238.845897 kcal/kg, with an individual range of about 11.1 to 12.8 MJ (2 650 to 3 050 kcal)/day or 185 to 200 kJ (44 to 48 kcal)/kg…
[PDF]Day 1-2: Goal kcal – 20 kcal/kg, Developed by: J, the recommended mean energy intake for a male population of this age group with a mean height of 1.70 m and a lifestyle with a mean PAL of 1.75,
This calculator is intended to provide clinicians with a ‘snapshot’ of a patient’s existing enteral and parenteral nutrition status, RD, Rather than providing a recommended nutritional regimen, Large calorie (Cal) is the energy needed to increase 1 kg of water by 1°C at a pressure of 1 atmosphere., kcal kg hard coal , Conversion base : 1 kg SKE = 7000 kcal, The SI base unit for mass is the kilogram, kcal/kg↔mJ/ng 1 mJ/ng = 238845.897 kcal/kg, Subscribe as 1 kg
1 Kilocalories per hour = 1.163 Joules per second: 10 Kilocalories per hour = 11.63 Joules per second: 2500 Kilocalories per hour = 2907.5 Joules per second: 2 Kilocalories per hour = 2.326 Joules per second: 20 Kilocalories per hour = 23.26 Joules per second: 5000 Kilocalories per hour = 5815 Joules per second: 3 Kilocalories per hour = 3.489 Joules per second: 30 Kilocalories per hour = 34
Energy in food can also be measured in kilojoules (kJ),000, Boiling point ,184 joules (J), You can view more details on each measurement unit: calorie or kg, so when you see 800 calories on a food label it actually means 800 kilocalories, Calories: Differences and How to Convert
Calories are a unit of energy, macronutrients, Small calorie (cal) is the energy needed to increase 1 gram of water by 1°C at a pressure of 1 atmosphere., air through the kiln would be 0.9 kcal/kg clinker,The answer is 7716.1791764707, Chart — EndMemo
kcal/kg↔mJ/g 1 kcal/kg = 4186.799993 mJ/g, protein per kg, Greenwood, Learn the difference and what these terms mean, and the same applies when you calculate an activity that burns 800 calories, sometimes call kilogram calorie is a measurement unit of energy, Absolute pressure, but in diet and exercise, Day 2-3: Goal kcal – 25 kcal/kg, 3) Provide thiamine (100 mg) daily x 5 – 7 days, Critical Care Program – Vancouver Coastal Health Authority, is about 11.7 MJ (2 800 kcal)/day or 195 kJ (47 kcal)/kg/day to maintain an optimum population median BMI of 21.0 (WHO/FAO, kilocalorie , We assume you are converting between calorie [burned] and kilogram, 1 calorie is equal to 0.00012959782 kilogram.
Convert calorie to kilograms
The answer is 7716.1791764707, 2002), Day 3-5: Goal kcal – final goal rate, 1 calorie is equal to 0.00012959782 kilogram.
[PDF]From kcal/kg to MJ/kg multiply kcal/kg by 0.004187 From kcal/kg to Btu/lb multiply kcal/kg by 1.8 From MJ/kg to kcal/kg multiply MJ/kg by 238.8 From MJ/kg to Btu/lb multiply MJ/kg by 429.9 From Btu/lb to kcal/kg multiply Btu/lb by 0.5556 From Btu/lb to MJ/kg multiply Btu/lb by 0.002326 Conversions – Gross/Net (per ISO, You can view more details on each measurement unit: calorie or kilograms, Thus 1, for As Received figures)
Men BMR = 88.362 + (13.397 x weight in kg) + (4.799 x height in cm) – (5.677 x age in years) Women BMR = 447.593 + (9.247 x weight in kg) + (3.098 x height in cm) –
Convert kilocalories to kilograms hard coal
You are currently converting energy units from kilocalorie to kg hard coal 1 kcal = 0.00014285714285714 kg SKE, The SI base unit for mass is the kilogram,000 calories = 1 kilocalorie or kcal.
800 kcal to Calorie Conversion
You are currently converting Energy units from Kilocalorie (international) to Calorie, Large calorie is also called food calorie and is used as a unit of food energy.
Energy values in kcal/kg are given on a basis of 4.1868 J, (See Calorie Calculator), kcal/kg↔uJ/ug 1 kcal/kg = 4.1868 uJ/ug.
For example, kcal/kg↔uJ/g 1 kcal/kg = 4186799.99347 uJ/g, If you’re looking to convert calories to kJ, multiply the number of calories by 4.18.
kcal/kg to cal/g Converter, kcal/kg↔mJ/mg 1 kcal/kg = 4.1868 mJ/mg, Switch units Starting unit, the term is used to mean kilocalories (kcal), Specific volume (steam) Density (steam) Specific enthalpy of liquid water (sensible heat) Specific enthalpy of steam
1 kcal/h = 69 766 666 800 000 pJ/min: 69 766 666 800 000 × P kcal/h = P pJ/min: kilocalorie ⁄ hour picojoule ⁄ second: Jelly Beans weighs 1 585.83 kg/m³ (99.00013 lb/ft³) with specific gravity of 1.58583 relative to pure water, 800 Kilocalorie (international) (kcal) =, Refeeding Syndrome
kcal to calories conversion calculator
Calories to kcal conversion calculator Small & large calories, Calculate how much of this gravel is
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s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030334579.46/warc/CC-MAIN-20220925132046-20220925162046-00117.warc.gz
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CC-MAIN-2022-40
| 4,650 | 22 |
https://deepai.org/publication/an-mcmc-approach-to-empirical-bayes-inference-and-bayesian-sensitivity-analysis-via-empirical-processes
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math
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This paper is concerned with two related problems. In the first, there is a function , where is a subset of some Euclidean space, and we wish to obtain confidence sets for . For each , the expression for is analytically intractable; however, we have at our disposal a family of functions
and a sequence of random variables(these are iid or the initial segment of an ergodic Markov chain) such that the random function satisfies for each . We are interested in how we can use to form both a point estimate and a confidence set for .
This problem appears in empirical Bayes analysis and under many forms in likelihood inference. In empirical Bayes analysis, the application that is the focus of this paper, it arises as follows. Suppose we are in a standard Bayesian situation in which we observe a data vectorwhose distribution is (with density with respect to some dominating measure) for some . We have a family of potential prior densities , and because the hyperparameter can have a great impact on subsequent inference, we wish to choose it carefully. Selection of is often guided by the marginal likelihood of the data under the prior , given by
By definition, the empirical Bayes choice of is . Unfortunately, analytic calculation of is not feasible except for a few textbook examples, and estimation of
via Monte Carlo is notoriously difficult—for example, the “harmonic mean estimator” introduced byNewton and Raftery (1994) typically converges at a rate which is much slower than (Wolpert and Schmidler, 2012).
It is very interesting to note that if is a constant, then the information regarding given by the two functions and is the same: the same value of maximizes both functions, and the second derivative matrices of the logarithm of these two functions are identical. In particular, the Hessians of the logarithm of these two functions at the maximum (i.e. the observed Fisher information) are the same and, therefore, the standard point estimates and confidence regions based on and are identical. This is a very useful observation because it turns out that it is usually easy to estimate the entire family for a suitable choice of . Indeed, for any , let denote the posterior corresponding to , let be fixed but arbitrary, and suppose that are either independent and identically distributed according to the posterior , or are the initial segment an ergodic Markov chain with invariant distribution . Let be the likelihood function. Note that given by (1.1) is the normalizing constant in the statement “the posterior is proportional to likelihood times the prior,” i.e.
in which the first equality follows from (1.2) and cancellation of the likelihood. Let . Since is a fixed constant, as noted above, the two functions and give exactly the same information about . If we let , then —this quantity is computable, since it involves only the priors and not the posteriors—so we have precisely the situation discussed in the first paragraph of this paper. Other examples of this situation arising in frequentist inference, and in particular in missing data models, are given in Sung and Geyer (2007) and Doss and Tan (2014).
In Bayesian applications it is rare that Monte Carlo estimates of posterior quantities can be based on iid samples; in the vast majority of cases they are based on Markov chain samples, and that is the case that is the focus of this paper. We show that under suitable regularity conditions,
where can be estimated consistently. Now, in general, almost sure convergence of to pointwise is not enough to imply that converges to under any mode of convergence, and in fact it is trivial to construct a counterexample in which and are deterministic functions defined on , for every , but does not converge to . To obtain results (1.4) and (1.5) above, some uniformity in the convergence is needed. We establish the necessary uniform convergence and show that (1.4) and (1.5) are true under certain regularity conditions on the sequence , the functions , and the function . Result (1.5) enables us to obtain confidence sets for .
The second problem we are interested in pertains to the Bayesian framework discussed earlier and is described as follows. Suppose that is a real-valued function of , and consider , the posterior expectation of given , when the prior is . Suppose that is fixed but arbitrary, and that is an ergodic Markov chain with invariant distribution . A very interesting and well-known fact, which we review in Section 2.3, is that for any , if we define
is a consistent estimate of . Clearly is a weighted average of the ’s. Under additional regularity conditions on the Markov chain and the function , we even have a central limit theorem (CLT):
, and we can consistently estimate the limiting variance. Thus, with a single Markov chain run, using knowledge of only the priors and not the posteriors, we can estimate and form confidence intervals forfor any particular value of . Now in Bayesian sensitivity analysis applications, we will be interested in viewing for many values of . For example, in prior elicitation settings, we may wish to find those aspects of the prior that have the biggest impact on the posterior, so that the focus of the effort is spent on those important aspects. We may also want to determine whether differences in the prior opinions of many experts have a significant impact on the conclusions. (For a discussion of Bayesian sensitivity analysis see Berger (1994) and Kadane and Wolfson (1998).) In these cases we will be interested in forming confidence bands for that are valid globally, as opposed to pointwise.
A common feature of the two problems we study in this paper is the need for uniformity in the convergence: to obtain confidence intervals for we need some uniformity in the convergence of to , and to obtain confidence bands for we need functional CLT’s for the stochastic process . Empirical process theory is a body of results that can be used to establish uniform almost sure convergence and functional CLT’s in very general settings. However, the results hold only under strong regularity conditions; and these conditions are often hard to check in practical settings—indeed the results can easily be false if the conditions are not met. Empirical process theory is fundamentally based on an iid assumption, whereas in our setting, the sequence is a Markov chain. In this paper we show how empirical process methods can be applied to our two problems when the sequence is a Markov chain, and we also show how the needed regularity conditions can be established.
The rest of the paper is organized as follows. In Section 2 we state our theoretical results, the main ones—those that pertain to the Markov chain case—being as follows. Theorem 3 asserts uniform convergence of to when the sequence is a Harris ergodic Markov chain, under certain regularity conditions on the family (the precise details are spelled out in the statement of the theorem), and we show how these regularity conditions can be checked with relative ease in standard settings. We then give a simple result which says that under a mild regularity assumption on , the condition entails . Theorem 4 establishes that under certain regularity conditions, we have asymptotic normality of . Theorem 6 establishes almost sure uniform convergence of to , and also functional weak convergence: the process converges weakly to a mean Gaussian process indexed by . We also show how this result can be used to construct confidence bands for that are valid globally. A by-product is functional weak convergence of to a mean Gaussian process indexed by , and construction of corresponding globally valid confidence bands for . In Section 3 we give two illustrations on Bayesian models in which serious consideration needs to be given to the effect of the hyperparameter and its choice. The first is to the Latent Dirichlet Allocation topic model, where we show how our methodology can be used to do sensitivity analysis, and the second is to a model for Bayesian variable selection in linear regression, where we show how our methodology can be used to select the hyperparameter. In the Appendix we provide the proofs of all the theorems except for Theorem 3; additionally, we show how the regularity conditions in Theorem 1 and Theorem 3 would typically be checked, and we verify these conditions in a simple setting.
2 Convergence of as a Process and Convergence of the Empirical Argmax
This section consists of three parts. Section 2.1 deals with uniform convergence of for the iid case, and introduces the framework that will enable us to obtain results for the Markov chain case; this framework will be used in Section 2.1 and in the rest of the paper. Section 2.2 deals with point estimates and confidence sets for , and Section 2.3 deals with uniform convergence and functional CLT’s for estimates of posterior expectations. Throughout, uniformity refers to a class of functions indexed by .
2.1 Uniform Convergence of
Let be a measurable subset of for some , and let
be a probability measure on, where is the Borel sigma-field on . We assume that are independent and identically distributed according to , and we let be the empirical measure that they induce. We assume that is a convex compact subset of for some , and that for each ,
is measurable. The strong law of large numbers (SLLN) states that
Since we will be interested in versions of (2.1) that are uniform in , there will exist measurability difficulties, so we have to be careful in dealing with measurability issues. Before proceeding, we review some terminology and standard facts from the theory of empirical processes. We will use the following standard empirical process notation: for a signed measure on and a -integrable function , denotes . Let be an arbitrary probability measure on , suppose that are independent and identically distributed according to , and let be the empirical measure induced by . If is a class of functions mapping to , and is a signed measure on , we use the notation . We say that is Glivenko-Cantelli if converges to almost surely; sometimes we will say is -Glivenko-Cantelli, to emphasize the dependence on . Let . Our goal is to establish that is -Glivenko-Cantelli, which is exactly equivalent to the statement that the convergence in (2.1) holds uniformly in .
The IID Case
Theorem 1 (Theorem 6.1 and Lemma 6.1 in Wellner (2005))
Suppose that are independent and identically distributed according to . Suppose that is continuous in for -almost all . If is measurable and satisfies , then the class is -Glivenko-Cantelli.
Let and (the subscript to the expectation indicates that ). Then the conclusion of the theorem is the statement .
The integrability condition seems strong, and an even stronger integrability condition is imposed in Theorem 3. We discuss this issue in Remark 1 following the statement of Theorem 3, where we explain that in fact the two conditions are fairly easy to check in practice.
The next theorem also establishes that the class is Glivenko-Cantelli. In the theorem, the integrability condition on is replaced by an integrability condition on (here, is the gradient vector of with respect to , and is Euclidean norm). The condition on the gradient is sometimes easier to check. We include the theorem in part because a component of its proof is a key element in the proofs of Theorems 5 and 6 of this paper.
Suppose that are independent and identically distributed according to , and that for each , . Assume also that for -almost all , exists and is continuous on . If is measurable and satisfies , then the class is -Glivenko-Cantelli.
The Markov Chain Case
Suppose now that the sequence is a Markov chain with invariant distribution , and that it is Harris ergodic (that is, it is irreducible, aperiodic, and Harris recurrent; see Meyn and Tweedie (1993, chapter 17) for definitions). Suppose also that for all . The best way to deal with the family of averages , is through the use of “regenerative simulation.” A regeneration is a random time at which a stochastic process probabilistically restarts itself; therefore, the “tours” made by the process in between such random times are iid. For example, if the stochastic process is a Markov chain on a discrete state space , and if is any point to which the chain returns infinitely often with probability one, then the times of return to form a sequence of regenerations. This iid structure will enable us to establish uniform convergence of the family . Before we explain this, we first note that for most of the Markov chains used in MCMC algorithms, the state space is continuous, and there is no point to which the chain returns infinitely often with probability one. Fortunately, Mykland et al. (1995) provided a general technique for identifying a sequence of regeneration times that is based on the construction of a minorization condition. This construction is reviewed at the end of this subsection, and gives rise to regeneration times with the property that
Suppose now that there exists a regeneration sequence which satisfies (2.2). Such a Markov chain will be called regenerative. For any , consider . Let
be the sum of over the tour. Also, let , denote the length of the tour. The ’s do not involve . Note that the pairs are iid. If we run the chain for regenerations, then the total number of cycles is given by
Also, . We have
In (2.4), the convergence statement on the left follows from Harris ergodicity of the chain. The convergence statement on the right follows from two applications of the SLLN: By (2.2), and this, together with the convergence statement on the left, entails convergence of . The SLLN then implies that (if then the SLLN implies that with probability one). We conclude that . Note that continuity in of for almost all sequences follows from continuity in of for almost all , since with probability one, is a finite sum. Suppose in addition that is measurable and satisfies . Then by Theorem 1 we have . Since , we obtain
We summarize this in the following theorem.
We now discuss the integrability condition , and our discussion encompasses the weaker condition assumed in Theorem 1. Suppose that for all . In the Appendix we show that, because is assumed to be compact, it is often possible to prove that for some ,
In this case, since , we obtain
which is finite. Thus, checking that reduces to establishing (2.6). In the Appendix we consider the Bayesian framework discussed in Section 1, in which , where is a family of priors, and , the posterior distribution corresponding to the prior , where is fixed. We show that if is an exponential family, then condition (2.6) holds. Therefore, the integrability condition is satisfied in a large class of examples. Moreover, the method we use for establishing (2.6) can be applied to other examples as well.
The idea to transform results for the iid case to the Markov chain case via regeneration has been around for many decades. Levental (1988) also obtained a Glivenko-Cantelli theorem for the Markov chain setting. In essence, the difference between his approach and ours is that his starting point is a Glivenko-Cantelli theorem for the iid case which requires a condition involving the minimum number of balls of radius in that are needed to cover —he is using metric entropy. This condition is very hard to check. By contrast, our starting point is a Glivenko-Cantelli theorem for the iid case which is based on bracketing entropy—in brief, the main regularity condition is implied by the continuity condition in Theorem 3. This continuity condition is trivial to verify: the parametric families that we are working with in our Bayesian setting satisfy it automatically.
The Minorization Construction
We now describe a minorization condition that can sometimes be used to construct regeneration sequences. Let be the transition function for the Markov chain . The construction described in Mykland et al. (1995) requires the existence of a function , whose expectation with respect to is strictly positive, and a probability measure on , such that satisfies
This is called a minorization condition and, as we describe below, it can be used to introduce regenerations into the Markov chain driven by . Define the Markov transition function by
Note that for fixed , is a probability measure. We may therefore write
which gives a representation of as a mixture of two probability measures, and . This provides an alternative method of simulating from . Suppose that the current state of the chain is . We generate . If , we draw ; otherwise, we draw . Note that if , the next state of the chain is drawn from , which does not depend on the current state. Hence the chain “forgets” the current state and we have a regeneration. To be more specific, suppose we start the Markov chain with and then use the method described above to simulate the chain. Each time , we have and the process stochastically restarts itself; that is, the process regenerates. Mykland et al. (1995) provided a very widely applicable method, the so-called “distinguished point technique”, for constructing a pair that can be used to form a minorization scheme which satisfies (2.2).
For any fixed , consider now the expression
in (2.4). The bivariate CLT gives
. (We have ignored the moment conditions onand that are needed, but we will return to these conditions in Section 2.3, where we give a rigorous development of a functional version of the CLT (2.8), in which the left side of (2.8) is viewed as a process in .) The delta method applied to the function gives the CLT
where (and is evaluated at the vector of means in (2.8)). Moreover, can be estimated in a simple manner using a plug-in estimate. Whether or not this method gives estimates of variance that are useful in the practical sense depends on whether or not the minorization condition we construct yields regenerations which are sufficiently frequent. Successful constructions of minorization conditions have been developed for widely used chains in many papers (we mention in particular Mykland et al. (1995), Roy and Hobert (2007), Tan and Hobert (2009), and Doss et al. (2014)); nevertheless, successful construction of a minorization condition is the exception rather than the norm. In this context, we point out that here regenerative simulation is notable primarily as a device that enables us to prove the theoretical results in the present paper and to arrive at informative expressions for asymptotic variances, but it may be possible to estimate these variances by other methods; this point is discussed further in Section 2.2.
2.2 A Consistent Estimator and Confidence Sets for
This section pertains to as an estimator of . After establishing that (2.5) entails that is consistent, we show that under additional regularity conditions, (i) is asymptotically normal, and (ii) we can consistently estimate the asymptotic variance. Results (i) and (ii) enable us to form asymptotically valid confidence sets for .
Suppose that is a compact subset of Euclidean space, and let and be deterministic real-valued functions defined on . Suppose further that is continuous and has a unique maximizer, and that for each the maximizer of exists and is unique. If converges to uniformly on , then the maximizer of converges to the maximizer of .
The proof of Lemma 1 is routine and is given in the Appendix. Consider now and . By Lemma 1, if is continuous and its maximizer is unique, then implies . Thus, under continuity of and uniqueness of its maximizer, any conditions that imply (2.5)—in particular the conditions of Theorems 1, 2, or 3—are also conditions that imply strong consistency of as an estimator of .
Before stating the next theorem, we need to set some notation and assumptions. We assume that each of and has a unique maximizer, and we denote and . For a function , denotes the gradient vector and denotes the Hessian matrix. We will assume that for every , and exist and are continuous for all . Recall that is defined by (2.3). The Markov chain will be run for regenerations, and in the asymptotic results below, . We will use the notation , , , etc. For almost any realization , the random variable is a finite sum, and therefore . Similarly, . We will assume that the family is such that the interchange of the order of integration and either first or second order differentiation is permissible, i.e.
For , let
Suppose that is a Markov chain on the measurable space and has as an invariant probability measure. Let be the -step Markov transition function. Recall that the chain is called geometrically ergodic if there exist a constant and a function such that for ,
If is a matrix, then a statement of the sort will mean for . We will refer to the following conditions.
The chain is geometrically ergodic.
For every , there exists such that .
The function is twice continuously differentiable and the matrix is nonsingular.
is measurable and .
is measurable and .
is measurable and .
is measurable and has finite expectation.
In the expression for the asymptotic variance given by (2.10), the term is the variance of a certain function of the Markov chain, and the term measures the inverse of the curvature of at its maximum ( is a deterministic function and does not involve the Markov chain): the flatter the surface at its maximum, the higher is the asymptotic variance.
The integrability condition in Assumption A4 was discussed in Remark 1, where we showed that it is satisfied whenever there exist such that for all (cf. (2.6), in which without loss of generality we take the constants to be equal to .) The integrability conditions in A5–A7 are satisfied under (2.13) and (2.14) below, which are very similar to (2.6). To make our explanation notationally less cumbersome and easier to follow, we will assume that , so that , , , and are all scalars. Assume that there exist and constants such that
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https://trinhvietcuong.com/qa/quick-answer-how-do-we-convert-km-to-cm.html
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math
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- Which scale is equivalent to 1 cm in KM?
- Which distance is the same as 5 kilometers in centimeters?
- Which is bigger 6km or 600 cm?
- How many zeros are in a kilometer?
- What is 1000000000000000 called?
- How many feet go into a mile?
- Is 6 kilometers equal to 600 centimeters?
- How do you convert cm to KM?
- What size is a 1 50000 scale map?
- How many zeros are in a gazillion?
- Which is the biggest Centimetre Metre or Kilometre?
- How many cm are in KM?
- How many kilometers does 7 centimeters represent?
- What is the meaning of 1 50000?
- How do you convert cm to km in geography?
- How many kilometer are in a mile?
- What is the highest number?
- What is the greatest measure of length?
- Is 1m 100cm?
- What is ground distance?
- How many meters are in a km?
Which scale is equivalent to 1 cm in KM?
Centimeter to Kilometer Conversion TableCentimetersKilometers1 cm0.00001 km2 cm0.00002 km3 cm0.00003 km4 cm0.00004 km10 more rows.
Which distance is the same as 5 kilometers in centimeters?
Kilometers to Centimeters tableKilometersCentimeters2 km200000.00 cm3 km300000.00 cm4 km400000.00 cm5 km500000.00 cm16 more rows
Which is bigger 6km or 600 cm?
There would be 6000 (600 times 10) mm in 600 cm. There are 100 cm in a meter, and 10 cm in a mm. … The 6,000,000 mm (6 km) is the greatest (largest number).
How many zeros are in a kilometer?
Probably you know that 1 km (kilometer) is equal to 1000 m (meter) . When you are familiar with the imperial system 1 m is between 3ft 3 inches and 3ft 4 inches. Now clearly zero means zero, 0, nothing, but in measuring we have some conventions. It is impossible to measure precise.
What is 1000000000000000 called?
quadrillionThe next named number after trillion is quadrillion, which is a 1 with 15 zeros after it: 1,000,000,000,000,000.
How many feet go into a mile?
5,280 FeetWhy Are There 5,280 Feet in a Mile?
Is 6 kilometers equal to 600 centimeters?
1 Kilometer is equal to 1000 meters. Meters is the SI base unit of length. The prefix kilo, abbreviated “k”, indicates one thousand. 1 km = 1000 m….Please share if you found this tool useful:Conversions Table6 Kilometers to Centimeters = 600000300 Kilometers to Centimeters = 3000000014 more rows
How do you convert cm to KM?
Centimeters to Kilometers Conversion 1 Centimeter (cm) is equal to 0.00001 kilometer (km). To convert centimeters to kilometers, multiply the centimeter value by 0.00001 or divide by 100000.
What size is a 1 50000 scale map?
Understanding Map ScalesScale1 inch represents approximately1 centimeter represents1:50,0004,166 feet500 meters1:62,5001 mile625 meters1:63,3601 mile (exact)633.6 meters1:100,0001.6 miles1 kilometer8 more rows
How many zeros are in a gazillion?
Gazzen, from Latin earthly edge , or end of the earth, abbreviated to gaz (literally 28,819 ancient Greek miles 12, been one full revolution of the globe). Therefore a Gazillion has (28819 x 3) zeros and a Gazillion is…
Which is the biggest Centimetre Metre or Kilometre?
Kilometers (km) are larger than centimeters (cm), so you expect there to be less than one km in a cm. Cm is 10 times smaller than a dm; a dm is 10 times smaller than a m, etc. Since you are going from a smaller unit to a larger unit, divide.
How many cm are in KM?
100000 centimetersKilometers to Centimeters Conversion 1 Kilometer (km) is equal to 100000 centimeters (cm).
How many kilometers does 7 centimeters represent?
7 cm to km conversion. A centimeter, or centimetre, is a unit of length equal to one hundredth of a meter. There are 2.54 centimeters in an inch….Convert 7 Centimeters to Kilometers.cmkm7.007 x 10-57.017.01 x 10-57.027.02 x 10-57.037.03 x 10-596 more rows
What is the meaning of 1 50000?
a representative fraction to describe the ratio between the map and the real world. This can be shown as 1:50,000 or 1/50,000. … This means that 1 inch on the map represents 50,000 inches in the real world, 1 foot on the map represents 50,000 feet on the map, and so forth.
How do you convert cm to km in geography?
working out what 1 cm on the map is equivalent to as a real life distance. A map has a scale of 1:25000. This means that 1 cm on the map is 25000 cm in real life, which is 250 m or 0.25 km. It is also useful to note that if 1 cm is 0.25 km, then 4 cm will represent 1 km in real life.
How many kilometer are in a mile?
1.609344 kilometersThere are 1.609344 kilometers in 1 mile.
What is the highest number?
The biggest named number that we know is googolplex, ten to the googol power, or (10)^(10^100). That’s written as a one followed by googol zeroes.
What is the greatest measure of length?
KilometersAnswer and Explanation: Kilometers are the longest unit of metric measurement. The abbreviation for kilometers is ‘km”.
Is 1m 100cm?
Convert meter to cm, centimeters to meter (1m = 100cm)
What is ground distance?
[′grau̇nd ‚dis·təns] (navigation) The great-circle distance between two ground positions, as contrasted with slant distance or slant range, the straight-line distance between two points. Also known as ground range.
How many meters are in a km?
1000 metersKilometers to Meters Conversion 1 Kilometer (km) is equal to 1000 meters (m).
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https://ttu-ir.tdl.org/items/7ad508bf-38cd-459a-b20e-96fdae727640
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math
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Deformations on tori
Deformations of Riemann surfaces have been studied for some time. The prominent deformation actions along surfaces are conformal weldings and the earthquake maps developed by William Thurston and Curtis McMullen. Previously, these deformations have required that a surface be cut open along a geodesic. We describe actions similar to conformal weldings and earthquakes in which cutting along a geodesic is not required. In particular, we explore the results of cutting a compact, Euclidean Riemann surface along a circle and applying similar actions. Using the discrete conformal approximations of circle packing, we describe a method by which these actions can be numerically estimated.
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http://dipoleantigravity.blogspot.com/2007_04_22_archive.html
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math
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Saturday, April 28, 2007
This means that the matters ejected through the poles by the jet mechanism especially the ones that drift away from the rotation axis can actually come back to the equatorial plane to recycle the whole process again.
Those matters that are on the way of coming back toward the equatorial plane will not be visible because of the lack of the force that can make them stay in the same location for a long enough period of time for the condensation and the gravitation and also because of the significant reduction of the collision cross section between the incoming and outgoing particles off the course from the axis of rotation.
So there is a strong possibility that the dark matters are not only populated around the equatorial plane but also around the whole space surrounding the rotating galactic center like a halo(dark matter halo). They are there but can not be visible. There is an issue of the visibility of the dark matters when they are made of gases and dusts. However, in the jet mechanism, because of its sudden expansion of the volume when they are ejected from the port of the poles, there is a strong chance that the matters will be cooled down quickly and condensed into a much bigger size than gases or dusts.
The force lines in the gravito magnetic concept of dipole gravity makes it a lot clearer on how the matters will behave around the strongly gravitating rotating ultra compact stellar objects.
Due to the point source nature of the active galactic nuclei compared to the sheer size of the volume of the galaxy itself, matters ejected from the ports form a distribution function in such a way that the outgoing flux density is constant, which is the main cause of the flat rotational velocity curves observed in the spiral galaxies.
Of course, this density distribution function is valid only up to the point where the finite boundary of the dark matter halo ends. Naturally, it is expected that the distribution function will be a sharply decreasing function than the typical 1/r as it comes close to the boundary of the dark matter.
Derivation of Lense-Thirring force from Dipole Gravity
Friday, April 27, 2007
At the present level of understanding of gravito magnetism and the available technology, the answer is no.
We do not know exactly why this phenomenon is possible in the first place. It will eventually lead us to the more fundamental understanding of the nature and then we will be able to engineer it.
In all practicality, turning a wheel at the speed of 1,000,000 rpm is barely possible at the moment in the form of the ultra centrifuge. And the gravito magnetic force generated from it is minuscule. However, if one can assume that the proper technology will be available in the future, there is no theoretical limit of speed that people can travel. And also, this mechanism of propulsion does not require continuous supply of energy in principle because of the conservation of the angular momentum of the gravito magnetic dipole rotor.
It is not hard to imagine the process of a star becoming a black hole because of the final crunch of the density and its radius, even without Schwarzschild metric. It's a seamless logical conclusion except that the metric provided the fundamental ground for the mathematical details in its process. Black hole is a logical extension of Newtonian gravity plus the equivalence principle and special relativity.
Now we can perform exactly the same logical extension for the gravito magnetism. This perfectly imaginable mechanical process produces the means to travel the universe with little expenditure of energy. Could the wormhole be the logical macroscopic extension of the gravito magnetism?
In the remote future, when the true cause of the gravity phenomenon is understood in depth in relation to the electricity and magnetism, we will be able to engineer it in a much more efficient way than the brute force turning of the massive wheel.
Thursday, April 26, 2007
After all, it can bee seen that a rotating sphere, donut shaped ring and a rotating football shaped object along the long side of axis etc. creates static gravitational quadrupole moment or two superposed gravito magnets and that it's a very general feature rather than exceptions. The core of the fast rotating galaxy may have the shape of the two superposed saucers due to its strong expansion around the equatorial plane of the rotation axis.
There was a thought about the name of this phenomenon. Basically it's a static gravitational phenomena because it is not radiating. However, in terms of the motion of the source in rotation, it resembles the mass current like in magnetism. In a way, it looks so much like a magnetic counter part of gravity.
However, this entity is originally from one of the terms in the linear expansion of the weakly gravitating object in general relativity, ie. the second order term next to Newtonian(mono pole) gravity. So, it is certainly the dipole gravity. There can hardly be any doubt about that.
The only problem with it is that, if you talk about gravitational dipole moment, it sounds like there are two different types of mass creating one, which is a nonsense, as we are so much used to the nomenclature of the electricity and magnetism.
In terms of the properties, I think there is a lot more similarities with magnetism than electricity since the force lines and the poles behave much like magnetism than electricity.
For example, the negative mass pole doesn't exist in isolation. It exists only with the rotational motion of the source much like the magnetic current. Like in magnetism, it always exists in pairs with the positive mass pole.
So I guess there is a duality of the characteristics on this phenomena. It's magnetic in property but it came out of the womb of the single monopole(electric like) gravity in its origin.
In science, the force of tradition is powerful. A new drastic theory is always looked at with great suspicion no matter how it may sound correct or explain the nature so well. I think it's in human nature. There is nothing strange about it.
Still, it is so remarkable that Lens-Thirring force has never been challenged on its accuracy in regard to the direction of the force. Of course, the reason is because it amalgamated Mach's principle with general relativity so well.
It's like, of course, it should be. Everybody knows there is centrifugal force in nature and general relativity should be able to confirm it and moreover Mach said it is the result of the local rotating object in interaction with the rest of the matter in the universe.
Why would anyone want to challenge such a perfect scenario?
May be the crucial reason for this was because there was no cosmological or experimental data associated to their result at the time of their publication. There was no other guiding principles that can direct their mathematical results. Mach's principle was the only relevant light house, which in a sense had little to do with any empirical data.
Since Lens-Thirring force is basically the force caused by the circular acceleration of the individual mass components within the spherical shell(which is modeled as the universe), which comes from the fundamental postulate of general relativity, ie, an acceleration of a mass creates gravity effect, their result should not be different from the one calculated from the dipole gravity method using the relativistic center of mass shift of the rotating hemispherical shell.
They are basically the same forces.
As such, as the second strongest gravity force in nature, it should be able to explain cosmological phenomena hitherto unexplained. The unexpected obstacle in this effort was the obvious conflict between Lens-Thirring force and the supposedly attractive radial force in the equatorial plane in the rotating galaxy. The dark matter problem demands attractive radial gravity force in the galactic plane but Lens-Thirring force which is the small footprint of the same force at the center of the rotating center has the wrong sign because Lens-Thirring force is the outgoing force from the center as the centrifugal force is like.
What if Lens-Thirring force had come out with the opposite sign that is the correct one? First of all, they would have been disappointed by the result itself. And they would have abandoned the whole idea of publishing it. It's a very interesting speculation.. But I'm sure it could have sparked more debate and deeper investigation into the subject.
The question still remains is then what is the real cause of the centrifugal force? In mechanics, it is caused by the tendency of an object trying to stay straight line in circular motion. I don't know if an object requires the existence of all the other matters in the universe to make it want to go straight line to conserve momentum.
There may be other causes of this. Maybe the inertial ether effect. The reason a balloon tends to stay in the same spot when there is no wind blowing is because the total sum of the air molecules trying to give it a net momentum cancels out. It resists motion when it is pushed because of the canceling forces from all the molecules in the air. So, the cross section, the mass and velocity of the air molecules take part in this resistance effect.
Therefore, it becomes natural to speculate that there may be particles working like these "air molecules" for the gravity effect in the universe.
We can go on into this topic further, but it will take us to too far afield.
Tuesday, April 24, 2007
In the original paper published by Lense-Thirring, the sign of the radial component of the force in the direction perpendicular to the rotation axis was repulsive and harmonic which has been considered as the general relativistic confirmation of the centrifugal force and the Mach's principle. In his paper published in 1999, Jeong also used the result of Lense-Thirring to confirm that his calculation of the superposed dipole gravity force inside a rotating sphere matches with the previoulsy published results of Lense-Thirring.
However, after a careful investigation of the dipole gravity potential and comparing it with the observed cosmological data, it has been found out that there may be sign errors in both Lense-Thirring and Jeong's forces originally published in 1999.
In Lense-Thirring's case, obviously, it was their desire to prove that general relativity conforms to Mach's principle or vice versa. In Jeong's theory of dipole gravity case, it was his desire to make sure his result conforms to Lense-Thirring's in his presentation of the diagram.
It may be that both have been wrong in the signs of their forces.
The major reason for this suspicion is in the direction of the radial component of the dipole gravity (gravito magnetic) force in the equatorial plane. If the lines of the force resembles that of two opposing bar magnets within the sphere, the radial force line will be outgoing (repulsive) around the equatorial plane just like the way Lense-Thirring force behaves near the center of the sphere. And it should be that way if anyone wants to claim it is the evidence of the centrifugal force.
However, the problem with this is in the fact that the direction of the outgoing radial force in the equatorial plane is not consistent with the dark matter problem. Also at the poles, the direction of the dipole gravity force is incoming which makes it hard to explain the jets considering that the Newtonian gravity force line is also incoming. So, the jet phenomena can be explained more logical way if the Z directional potential is like below than its upside down form.
For example, in the above shape of the potential along the Z axis, where the domed side represents the repulsive force, any density overflow from the central region toward the poles will be automatically ejected along the rotating axis of the space. And the longer range Newtonian gravity will bring them back to form two way jet streams. The only problem with this reversion is that the original Lense-Thirring force has to have its sign reversed.
While the direction of the dipole gravity force can be made arbitrary, because the direction of the center of mass shift can be defined either way depending on the convention, it has to be consistent with the observed cosmological data.
To solve the dark matter problem, we need an additional long range radial gravity force in the equatorial plane that is not repulsive but attractive. If this is the case, Lense-Thirring force has its sign reversed.
The derivation of the radial force in the equatorial plane presented in the page has the correct sign, because it was derived directly from the original potential not read off from the diagram.
In the illustration above, the force lines at the center of the two superposed gravito magnets are incoming(attractive) and that of the both poles of the magnet are outgoing(repulsive). At the center of these two magnets facing the same type of poles(attractive) in the middle, the force lines along the equatorial plane will form array of horizontal incoming force lines. The extension of this force lines at the center between the two gravito magnets can not be outgoing, meaning that the following expression of Lense-Thirring force has the wrong sign(remember the centrifugal force is radially outgoing force ).
The fundamental physical reason for this is because the continuity of the force line either in magnetism or gravito magnetism doesn't allow the change of its direction on its path 180 degree.
Therefore, Lense-Thirring force has to be the attractive harmonic force toward the center in the equatorial plane in total contrast to their original claim and also the force along the rotation axis must behave like repulsive harmonic force as shown in the above potential diagram along the Z axis(notice the potential near the center(Z=0) of the diagram) instead of the attractive harmonic force shown in the above expression and in the previous saddle diagram.
This comes to the clear conclusion that Lense-Thirring force has nothing to do with the centrifugal force. It is rather a simple manifestation of the attractive radial dipole gravity force near the center of the rotating spherical shell.
Therefore, it is corrected at this point that the dipole gravity force in the rotating hemisphere is attractive on the flat side and repulsive on the domed side, instead of the other way around. It makes the jet phenomena explained in a much more simple and elegant way and so is the dark matter problem, although the way the dipole moves itself in space makes it not being very well stream lined in the normal common sense of the aerodynamics.
But then there is nothing common sense about cosmology anyways. The alternative method of detecting dipole gravity force experiment mentioned in the following page will prove on this point by showing the direction of its actual movement in space.
Now, it seems that the overall consistency is restored.
Let's blame it on Mach. It was all his fault. :)
Of course, I'm trying to be humorous here if anyone has noticed. It's Lense-Thirring's fault and Jeong's fault as well when he modified the dipole potential diagram to fit the result of Lens-Thirring's. But how about those who didn't check it out all the way through? Let's try to be happy at least we found the error and corrected it.
For those who are not familiar with the origin of dipole gravity, it would be informative to quote from a textbook on gravitation. The following is a clip from the book by A.K. Raychaudhuri, S.Banerji and A. Banerjee titled "General relativity, Astrophysics, and Cosmology". There is no special reason for quoting from this book, because it is the same in "Gravitation" by Kip Thorne et al. as well.
Linear expansion in mathematics is a convenient way to sort out a major complex function into a series of simplified terms in the order of its magnitude or around the region of particular interest. The first major term in the weakly gravitating linearized theory of general relativity which is the strongest is Newtonian monopole and naturally the second one is the gravitational dipole.
The above is quoted from the chapter on "External field of a weakly gravitating source" on page 43.
"The dipole term vanishes, if the origin O is also the center of mass of the source" is what it says and agreed upon. The integral is basically the definition of the dipole gravitational moment that is the length element times the density integrated over the whole volume of the object. It must be noted that this is a very broad general statement regardless of the geometrical shape of the source in consideration.
However, when the rotating hemisphere is introduced as a source, it becomes obvious that the meaning of the statement "if the origin O is also the center of mass of the source" becomes very ambiguous. Although the standard choice of the geometrical shape of the source has been a sphere since it is generally the easiest and the simplest shape to integrate the Newtonian potential over the volume element, this doesn't necessarily have to be the case for the dipole gravitational moment.
Now the question is "which center of mass?" since the rotating hemisphere now has the duality in its center of mass. The one at rest or the one in rotation? If one had set up the origin of the coordinate system at the center of mass of the hemisphere when it was at rest, the new center of mass in rotation would have been shifted due to the relativistic mass increase effect. One may say the origin of the coordinate system has to be the one that has already been shifted. But the problem is not that simple, because no one asks how fast the object is going to rotate before setting up the coordinate system. And furthermore, this question did not exist for the rotating sphere. Why?
This shift is energy dependent and it's not a fluke of accident caused by the misaligned coordinate system. Here we postulate that this relativistic center of mass shift is the real cause of the gravitational dipole moment.
However, it must be noted that even without this postulate, if Lens-Thirring had their calculation using the rotating hemisphere, they would have gotten the same answer. They could have come up with all the details of the dipole gravity field around the rotating hemisphere. The reason for their result which covers the region only close to the center of the sphere was because of the wrong choice of the (gravitational) source(spherical shell). If one chooses the rotating sphere as the source, the integral calculation for the dipole gravity field becomes very difficult other than the region close to the center of the sphere.
It seems like the unique way a finite length element can be related to the rotational motion of an object is by the relativistic center of mass shift from the rotation of a longitudinal axially asymmetric object.
In any case, the rest is to see if the result of this "postulate" conforms with the known results in physics, for example, Lens-Thirring force etc and furthermore, to see if the results can explain cosmological phenomena hitherto unexplained, ie, jet phenomena, dark matter problems and others.
The importance of this force is that it is the strongest long range gravity force right next to Newtonian gravity. And any other higher order gravity effect will be weaker at least by the factor of v/c.
Some time ago, there was a debate regarding the theory with Dr. Choptuik at UT Austin. It must have been almost 9 years ago. He said "dipole gravity is not general relativity". He may be right because Einstein did not say anything about it. In fact, if you have special relativity, and apply it in the same manner to any advanced theory of gravity(for example, Brans-Dicke theory of gravitation), the second order linearized term will show the same result ie, dipole gravity. So, in a way, dipole gravity does not help differentiating general relativity from other similar theories.
However, is it really fair to say dipole gravity is not general relativity? I don't think you can say DPG is not general relativity. But then also, you can't say it is the trade mark of general relativity either. After all, Newtonian gravity can be derived out of both general relativity and Brans-Dicke theory solely because they are designed that way. If your theory doesn't produce Newtonian gravity in its linearized form, your theory of gravity would be simply wrong.
If you say "dipole gravity is not general relativity", isn't it the same as saying "Newtonian gravity is not general relativity" as if Newtonian gravity has nothing to do with general relativity, which is not true. Both Newtonian gravity and dipole gravity are parts of the broader theory like general relativity and/or Brans-Dicke theory. However, neither of them(NG and DPG) may provide the crucial test to differentiate between the two( general relativity and Brans-Dicke theory).
Dipole gravity is a part of general relativity and is also a part of Brans-Dicke theory as well. It is a theory that can not be avoided once special relativity is proven to be correct. Apparently, the rotating hemisphere violates Newton's second law of motion. Because its center of mass changes without the external force in the direction of the shifted center of mass.
We can not take both principles to be absolutely correct when they are contradicting each other in such a glaring fashion. http://dipoleantigravity.blogspot.com/2007/05/what-is-at-stake.html
Monday, April 23, 2007
The GPB probe may have already enough raw data to prove the theory of dipole gravity(gravito magnetism). Some of their anomalous results have been put aside as being caused by some kind of electrostatic disturbance effect present within the device itself.
But with such a high precision device that may have been already tested in the laboratory numerous times, it is hard to believe it can have such a systematic error that has not been expected before hand.
And most importantly, how would they know if their device is performing properly or not without a proper theory that predicts physical anomalies that are measurable within their range of precision. After all, gravito magnetic(dipole gravity) effect is the largest right next to the Newtonian gravity.
As such, any huge unknown anomaly in the raw data must be associated with this effect, ie, the true gravito magnetism based on dipole gravity, not the one based on the modified Maxwell's equation.
Investigation of this possibility by analyzing the raw data can be an excellent PhD dissertation project.
Sunday, April 22, 2007
In the following we will discuss the gradient of the dipole gravity potential. This shows the detailed picture of the actual force associated with dipole gravity. This program is accomplished by taking the gradient in the spherical coordinate system on the isolated dipole potential.
where the function f is replaced by the following expression which is the total gravity, ie, the Newtonian gravity plus the dipole gravity for a rotating hemisphere.
In the case of the radial component(r dependent force) of the dipole gravity force,
where dz is the gravitational dipole moment and the latitude angle theta is measured from positive Z axis of the rotation. Since the center of mass moves toward the flat side of the hemisphere while in rotation, convention has to be made in such a way that the flat side of the hemisphere faces the positive Z axis.
The value of the cosine function changes its sign from positive to negative as it crosses over the equatorial plane of the rotating hemisphere, which is located at half the length of the radius down from the flat side of the hemisphere toward the dome, where the latitude angle is 90 degree.
This is the fundamental cause of the repulsive gravity force at the domed side of the rotating hemisphere and the unidirectional acceleration of the isolated dipole gravitational moment in the matter filled universe and also the justification for the experiments proposed in
It must be noted that this is the same force that produces the Lens-Thirring force near the center of the rotating sphere when the two rotating hemispheres are superposed face to face to form a single rotating sphere.
The latitude angular dependency of the dipole gravity force
is represented by the force in the direction from the flat side of the poles toward the domed side following the latitude circular curvature of the radius r. The strength of it is maximum at the equatorial plane which is at the 90 degree latitude angular position and it is minimum when the latitude angle is either at zero or 180 degree.
At this point, it must be noted that the directions of the force lines resemble exactly that of a dipole bar magnet. In all the similarities and qualifications, this is the gravitational dipole magnet as stated in the introduction, the true "gravito-magnetism" that has been searched for a long time.
After all those effects and calculations, there is one new physical entity stands out, that is this new negative gravitational mass pole on the domed side of the rotating hemisphere which exists temporarily with the angular momentum. Since the angular momentum is conserved, the negative mass pole doesn't disappear as long as the angular momentum stays on. It can also be seen that this negative gravitational mass pole can not exist in the universe in isolation.
The last phi(circumferential angle) directional dipole gravity force is zero because there is no phi dependency in the dipole gravity potential.
We don't know exactly what it means when people say "frame dragging force" in terms of its mathematical form, but at least there is no "dragging" force that follows the rotating surface along the circumferential direction(phi direction) within dipole gravity which is the strongest gravity force next to the Newtonian monopole force.
This term is normally used in association with Lens-Thirring force but its strength is v/c times smaller compared to the radial dipole gravity force. Lens-Thirring force is certainly a part of gravito-magnetism(dipole gravity) but "frame dragging force" is too weak(V/c) to be a part of it. If Gravity Probe B works perfectly, it will detect dipole effect in a much more stronger signal before it detects the "frame dragging" force.
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http://www.algebra.com/algebra/homework/word/mixtures/Mixture_Word_Problems.faq.question.76086.html
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Question 76086: A telephone number consist of seven digits the first three representing the exchange. How many different telephone numbers are possible with in the 537 exchange? Answer by Cintchr(481) (Show Source):
You can put this solution on YOUR website! The last 4 digits determine how many numbers there are in the exchange.
Because this involves INCLUSIVE numbers, we subtract the first number from the last and then ADD 1. The reason we add one is that we have in INCLUDE the fisrt one.
So you can see that this really works, lets look at a simpler problem:
You are supposed to read pages 1 - 5 in your history book. How many pages dod you read?
A total of 5
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http://www.expertsmind.com/questions/regression-to-the-mean-3018745.aspx
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math
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Regression to the mean is the procedure first noted by Sir Francis Galton that 'each peculiarity in man is shared by his kinsmen, but on average to the less degree.' Hence the tendency, for instance, for tall parents to produce tall offspring but then who, on average, is shorter than their own parents. The term is now usually used to label the phenomenon that a variable which is extreme on its first measurement will tend to be closer to the centre of the distribution for the later measurement. For instance, in a screening programme for hypertension, only the persons with high blood pressure are asked to return for the second measure. On an average, the second measure taken will be less than the first measure.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917124371.40/warc/CC-MAIN-20170423031204-00630-ip-10-145-167-34.ec2.internal.warc.gz
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CC-MAIN-2017-17
| 718 | 1 |
https://www.simiode.org/resources/1020
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math
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Bulte, C. H. F. 1992. The differential equation of the deflection curve. International Journal of Mathematical Education in Science and Technology. 23(1): 5-63.
Article Abstract: This paper presents the derivation and the physical meaning of the general fourth-order linear differential equation (with sectionally continuous derivatives) of the deflection curve and its general formulation and solution as a multipoint boundary value problem. An algorithm is presented in which shearing forces, bending moments, deflections, critical load and natural frequency are calculated for (non)-uniform beams and columns, with arbitrary lateral and axial (dis)continuous distributed load functions and concentrated load, on (intermediate) (elastic) supports.
The paper offers a very thorough illustration of modeling building of a physical phenomena with very helpful diagrams and explanations. The diagrams are particualarly well done and very helpful.
Cite this work
Researchers should cite this work as follows:
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s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00457.warc.gz
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CC-MAIN-2022-49
| 1,005 | 5 |
http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/torsion
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math
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Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T equivalent to F × d, which is applied perpendicular to the axis of the bar, as shown in the figure. Such a bar is said to be in torsion.
TORSIONAL SHEARING STRESS, τ
For a solid or hollow circular shaft subject to a twisting moment T, the torsional shearing stress τ at a distance ρ from the center of the shaft is
where J is the polar moment of inertia of the section and r is the outer radius.
For solid cylindrical shaft:
For hollow cylindrical shaft:
ANGLE OF TWIST
The angle θ through which the bar length L will twist is
where T is the torque in N·mm, L is the length of shaft in mm, G is shear modulus in MPa, J is the polar moment of inertia in mm4, D and d are diameter in mm, and r is the radius in mm.
POWER TRANSMITTED BY THE SHAFT
A shaft rotating with a constant angular velocity ω (in radians per second) is being acted by a twisting moment T. The power transmitted by the shaft is
where T is the torque in N·m, f is the number of revolutions per second, and P is the power in watts.
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CC-MAIN-2015-06
| 1,117 | 12 |
https://www.theproblemsite.com/pro-problems/physics/motion-forces/linear/force/tractor-on-a-flatbed
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math
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Tractor on a FlatbedPro Problems > Physics > Motion and Forces > Linear Motion > Force
Tractor on a Flatbed
A 6000 kg tractor rests on a flatbed, held in place by chains. The chains provide a maximum horizontal force of 8000 N. When the flatbed is traveling 20 m/s, what is the minimum stopping distance if the chains are not to break?
SolutionIn order to make it feasible for teachers to use these problems in their classwork, no solutions are publicly visible, so students cannot simply look up the answers. If you would like to view the solutions to these problems, you must have a Virtual Classroom subscription.
Add a Passenger
A 50 slug car can accelerate from 0 to 50 ft/s while traveling 1000 feet. The car then stops, and a passenger gets in. Now the car can accelerate from 0 to 50 ft/s while traveling 1050 feet. What is the weight of the passenger?
Sledding in the Wind
A 50 kg sled starts at rest, and is being pushed across a horizontal frictionless surface with a force of 30 N. The wind is pushing against the sled. The sled travels 100 meters while accelerating to 10 m/s. What is the force of the wind on the sled?
Force and Friction on a Crate
If the coefficient of kinetic friction between a 35 kg crate and the floor is .30, what horizontal force is required to move the crate at a steady speed across the floor? What horizontal force is required if µk is zero?
Force on Two Objects
If a force accelerates 4.5 kg at 40 m/s2, that same force would accelerate 18 kg by how much?
Braking to a Stop
A 1000-kg car moving north at 100km/h brakes to a stop in 50m. What are the magnitude and direction of the force?
A 5.0 kg missle is projected from rest to a speed of 4.5 x 103 m/s. How much time is required to do this by a net force of 5.3 x 105 N?
Cart with Groceries
1.A horizontal force of 10N is applied to a 30 kg cart on a level floor.
- How far will it move in 3.0s, starting from rest? Ignore friction.
- How far will it move in 3.0s if 30N of groceries is added to the cart first?
Out of Control Elevator
An elevator has a mass of 2,000 kg. An upward force of 21,000 Newtons is applied to the elevator. The elevator travels up 50 floors, and each floor is 2.5 meters tall, with a 20 centimeter gap between floors. How long does it take the elevator to arrive, and at what speed will it be traveling?
Moving a Box
Jamal is pulling a 4,900 N box with a force of 500 N. Jakob is pulling in the opposite direction with a force of 350 N. How fast will the box be moving after 5 seconds?
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http://www.ams.org/samplings/mathmoments/mm98-juggling-podcast
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math
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|Publications Meetings The Profession Membership Programs Math Samplings Policy & Advocacy In the News About the AMS|
There’s more mathematics involved in juggling than just trying to make sure that the number of balls (or chainsaws) that hits the ground stays at zero. Subjects such as combinatorics and abstract algebra help jugglers answer important questions, such as whether a particular juggling pattern can actually be juggled. For example, can balls be juggled so that the time period that each ball stays aloft alternates between five counts and one? The answer is "Yes." Math also tells you that the number of balls needed for such a juggling pattern is the average of the counts, in this case three.
Once a pattern is shown to be "juggleable" and the number of balls needed is known, equations of motion determine the speed with which each ball must be thrown and the maximum height it will attain. Obviously the harder a juggler throws, the faster and higher an object will go. Unfortunately hang time increases proportionally to the square root of the height, so the difficulty of keeping many objects in the air increases very quickly. Both math and juggling have been around for millennia yet questions still remain in both subjects. As two juggling mathematicians wrote, "A juggler, like a mathematician, is never finished: there is always another great unsolved problem."1
1. "Fountains, Showers, and Cascades," Joe Buhler and Ron Graham. The Sciences, January-February 1984.
For More Information: The Mathematics of Juggling, Burkard Polster, 2003.
Comments: Email Webmaster
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https://www.podbean.com/media/share/pb-497jb-c7e8fd?utm_campaign=w_share_ep&utm_medium=dlink&utm_source=w_share
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math
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It's been long enough, time to get back on that oar! Craig, Ryan, and Stephanie discuss chapters 1-55 of The Broken Eye, book 3 in Brent Weeks' Lightbringer series. Lots of questions this time around: Are we spending enough time with our villains? Is "fake it 'til you make it" a real thing? What is the most touching of human gestures? Is Ryan a monster for not laughing at Craig's best jokes? Listen and find out!
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Music: "Adventure Time" and "The Seven Seas" courtesy of https://www.philter.no/
#323. The Book of Three (Prydain Chronicles #1)
#322. Heretics of Dune, feat. Jack Butler
#321. Mort (Discworld #4) | Author's Shelf feat. Daniel Greene
#320. The MCU on Disney+ | Wandavision, Falcon & the Winter Soldier, Loki
What Makes Prose Good? | Part 2: Defining Prose
Why (and How) Should Stories End? w/ Brent Weeks
#318. The Stone Sky (Broken Earth #3)
When does a book go from HOMAGE to THEFT?
Is it better to read DEEPLY or WIDELY?
#317. The Wheel of Tangents #9
#316. Jurassic Park (the book, not the movie)
#315. Video Game Movie Redux, starring Mortal Kombat 2021
#314. The Obelisk Gate
#313. Godzilla vs Kong - Overreactions & Overinterpretations
#312. John Carter of Mars | Heroes of Sci-fi
#311. JUSTICE LEAGUE - The Snyder Cut Reaction & DCEU Discussion
#310. Rhythm of War - part 5 part deux
#309. Rhythm of War, part 5
#308. Magic Kingdom For Sale--Sold!
#307. The Golden Compass (His Dark Materials #1)
Write The Book: Conversations on Craft
Myths and Legends
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CC-MAIN-2021-39
| 1,572 | 27 |
http://www.chegg.com/courses/berkeley/IND%20ENG/263A
|
math
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Be the first to write one.
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Parts arrive at a station in a production line according to a Poisson process with rate lambda >0. The amount of time required to machine each part is exponentially distributed with a mean of 1/mu. With probability p, a machined part is found to be defective and is immediately reworked. A part that requires rework is indistinguishable from a newly arriving part, and is reworked until all defects are corrected.
a) What is the probability that a part is worked exactly m times, m = 1,2,3,...?
b) What is the expected amount of time that a part is processed, including all time spent correcting any defects?
c) Solve for the steady state probability that there are n parts at this station, n = 0,1,2,...?
d) Find the expected value of the total time that a part spends in this station.•
We can't find any books for this course.
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CC-MAIN-2013-48
| 933 | 9 |
https://www.hackmath.net/en/math-problem/2651
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math
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Students of all 7, 8, and 9 classes in one school may take up 4,5,6, and 7 abreast, and nobody will be left. How many is the average count of pupils in one class if there are always four classes in each grade?
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In the physical education lesson, students entered the two-step, three-step, four-step, six-step, and eight-step classes, but there was always one student left. How many students practiced if there were more than 40 and less than 50?
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There are more than 20 but less than 40 students in the class. A third of the pupils wrote the mathematics test with a one, a sixth with a two, and a ninth with a three. Nobody got a four. How many students in the class wrote the test on a five?
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There are a total of 56 pupils in three classes in the rural school. In the second grade, there are 20% fewer pupils than in the first grade, and in the third grade, there are 25% more pupils than in the second grade. How many pupils are there in each cla
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In a class are 34 students. The average height of the students is 165 cm. What will be the average height of students in the classroom when two pupils, tall 176 cm and 170 cm, move from this school/class? The average height in meters of all the students i
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There are 30 students in the class. Five students in the class had a mark of three or triple at the end of the math certificate, and the other students had a mark of one or two. The average mark in all students' mathematics in the class at the end of the
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Two pupils painted the class in four hours. How long will it take for six pupils?
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CC-MAIN-2023-50
| 4,630 | 41 |
https://fatspike.com/problems-of-math/
|
math
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Problems Of Math
Above is the Problems Of Math section. Here you will find all we have for Problems Of Math. For instance there are many worksheet that you can print here, and if you want to preview the Problems Of Math simply click the link or image and you will take to save page section.
Grade 1 Word Problems Numeracy 8 Problem Solving Strategies For The Math Classroom Worksheets On Sets And Venn Diagrams – Proteussheet 016 Third Grade Story Problems New Math Word Practice 016 Third Grade Story Problems New Math Word Practice Pin On Math Teaching Resources Financial Tatements For Business Plans Examples Tatement Worksheet Ideas 12th Grade Math Worksheets With Grade 3 Vocabulary Worksheets – Jackpotprint Pin On School Coloring Books Math Coloring Worksheets 3rd Grade 016 Third Grade Story Problems New Math Word Practice The Math Dude Quick And Dirty Tips To Make Math Easier On Worksheets 2 3 Math Worksheets For Grade Time Word Problems Math Vs Science Free Printable Language Arts Worksheets.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875148850.96/warc/CC-MAIN-20200229083813-20200229113813-00219.warc.gz
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CC-MAIN-2020-10
| 1,011 | 3 |
https://www.amazon.co.uk/Babylonian-Theorem-Peter-S-Rudman/dp/159102773X
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math
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Babylonian Theorem Hardcover – 15 Jul 2010
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Most Helpful Customer Reviews on Amazon.com (beta)
I have to say that it got off to a fairly good start, with a good description of Egyptian and Babylonian number systems and an explanation for how they might have evolved. Although some of the related equations are not difficult to derive, I think that a quick derivation would have been helpful. I also would not have been able to figure out what a greedy algorithm was from the explanation given if I did not already know it, but these are relatively minor points.
The problem comes when the author starts talking about what he calls the Babylonian Theorem mentioned in the title. He claims that the Babylonians knew how to prove the Pythagorean Theorem and he gives as justification a geometric diagram. Now the diagram does geometrically show that (a-b)^2 + 4ab = (a+b)^2, but I have hard time seeing how the Pythagorean Theorem follows, because the diagram contains no right triangles. There is a related diagram that can be used to prove the Pythagorean Theorem, but the author makes no reference to it, and I am not convinced that the Babylonians could have made use of it, because there is some algebraic manipulation required that they might not have been able to handle.
Okay, so at the very least the author showed how the Babylonians came up with a way of solving a particular type of quadratic equation. The author then claims to show how this was used to solve problems. He gives the following problem from a Babylonian text: A number subtracted from its inverse is equal to 7. I was guessing that in modern terms this would be: x - 1/x = 7, though neither this or any other interpretation is presented. My interpretation must be incorrect because it is stated that the equation has an integer solution and you can tell by inspection that this will not be true for my equation. There is then shown how the Babylonian student solved the problem and I have no idea how the manipulations relate to the original problem.
Later on, it is stated that Euclid proved the Babylonian Theorem using the Pythagorean Theorem. What is shown is a simple way of constructing a right triangle have a hypotenuse of (a+b) and a side of (a-b). Since there is a simple general method of constructing right triangles using straightedge and compass, I am not sure what this particular construction proves.
I would strongly suggest that the author do some serious editing of the book, providing explanations. It may yet prove to be useful, but in its present form it is one big mess.
Look for similar items by category
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- Books > Scientific, Technical & Medical > Mathematics
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| 3,551 | 21 |
https://era.ed.ac.uk/handle/1842/1860
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math
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Cross-Sectional Variation in Investment Trust Discount Volatility
Adams, Andrew T
Discount volatility is generally an important component of total risk for closed-end funds but there is considerable cross-sectional variation in the magnitude of this discount volatility. These are interesting aspects of the closed-end fund discount puzzle which have received little attention in the literature, particularly as regards UK investment trusts. This paper seeks to explain the cross-sectional variation in discount volatility for the UK investment trust sector. The sample consists of 59 UK conventional investment trusts in continuous operation over the five years from 1 January 1992 to 31 December 1996. Discount volatility is calculated using monthly intervals. Four explanatory variables are highly significant - trust share turnover, standard deviation of NAV return, market value and percentage of underlying assets which are unquoted. There is no evidence that either small investor sentiment or UK specific sentiment has any impact on discount volatility.
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CC-MAIN-2023-23
| 1,061 | 3 |
https://mathoverflow.net/questions/212316/non-strictly-singular-operators-and-complemented-subspaces
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math
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If $T$ is a bounded operator which is not strictly singular, acting on a separable Banach space $X$, can one always find an infinite dimensional, closed and complemented, subspace $Y$ such that $T$ restricted to $Y$ is an isomorphism on $Y$?
Thanks for the email, Markus.
Let’s agree that “space” means “infinite dimensional Banach space” so that subspaces are always infinite dimensional.
A Banach space $X$ is decomposable if it is the direct sum of two subspaces; in other words, if there is a (bounded, linear) projection $P$ on $X$ s.t. $PX$ and $(I-P)X$ are both infinite dimensional. The first indecomposable Banach space was constructed by Gowers and Maurey; in fact, their space is hereditarily indecomposable. Now we know that indecomposable spaces are very common; see [AFHORSZ] and references therein. In particular, $\ell_p$, $1<p<\infty$, is a subspace of a separable indecomposable space.
For an example that gives a negative answer to Markus’ problem, take an indecomposable space $X$ that contains a decomposable subspace $Y$ ($Y$ can be a Hilbert space). Take a projection $P$ on $Y$ that has infinite dimensional range and infinite dimensional kernel. Extend $P$ to an operator $T$ from $X$ into some injective space that contains $X$. $T$ is obviously not strictly singular since $T$ is the identity on $PY$. Also, the kernel of $T$, being infinite dimensional, has infinite dimensional intersection with every finite codimensional subspace. But since $X$ is indecomposable, all complemented subspaces of $X$ are finite codimensional.
I could not have answered this natural and basic (though I never thought of it until reading this post) question a few years ago. $$ $$ [AFHORSZ] Argyros, S. A.(GR-ATHN); Freeman, D.(1-TX); Haydon, R.(4-OXBR); Odell, E.(1-TX); Raikoftsalis, Th.(GR-ATHN); Schlumprecht, Th.(1-TXAM); Zisimopoulou, D.(GR-ATHN) Embedding uniformly convex spaces into spaces with very few operators. (English summary) J. Funct. Anal. 262 (2012), no. 3, 825–849. $$ $$
EDIT: After I posted this, user19038 gave a reference in a comment above that shows that the OP's question was raised by Vitali Milman in a 1970 paper and solved in the linked paper. The example involves only classical spaces; it is the inclusion mapping from $L \log^\lambda L$ into $L_1$ with $\lambda < 1/2$.
I think that the correct Markus' question concerns operators in L(X).In this case the answer is positive for separable C(K). I do not know what happens in the case of spaces with an unconditional basis. Moreover the following seems interesting. Let X be a separable reflexive space and T in L(X). Does there exist indecomposable Y containing isomorphically X and S in L(Y) that extends T? I also do not know what is the answer if in the previous question if we replace the indecomposable Y by the space C[o,1].
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| 2,838 | 8 |
http://tankinlian.blogspot.com/2008/12/results-do-we-need-new-property-portal.html
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math
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Property agency service for a flat fee favours the richer ones transacting high value properties.E.g. transact a S$150,000 3-room HDB flat for S$1,000, transact a S$3,000,000 bungalow also S$1,000.The responsibility and work rendered by the agent for different property also varies.I'd think that a fairer way should be a flat fee (e.g. S$500) plus a low commission (e.g. 0.5%).
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CC-MAIN-2013-20
| 393 | 2 |
http://wikamn.se/guestbook/?page=1152
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math
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s3://commoncrawl/crawl-data/CC-MAIN-2021-25/segments/1623488552937.93/warc/CC-MAIN-20210624075940-20210624105940-00035.warc.gz
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CC-MAIN-2021-25
| 683 | 5 |
https://www.scribd.com/doc/13628928/802-11-Protocol-Stack-and-Physical-Layer
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math
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This action might not be possible to undo. Are you sure you want to continue?
"The 802.11 Protocol Stack and Physical
Table of Contents
2.2 Wireless Local Area Network
2.3 The Basic Structure of a Wireless LAN
2.4 Comari!" a Wireless LAN to a Wire# Network
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3.1 'rotocol Structure
3.2 'rotocol Stack Architecture
3.2.1 Statio! (STA) Architecture*
3.2.2 Access+'oi!t (A') Architecture*
3.2.3 Basic Service Set (BSS)*
3.2.4 %!#ee!#e!t Basic Service Set (%BSS)*
3.2., &-te!#e# Service Set (&SS)*
3.2.. Service Set %#e!tifier (SS%/)*
3.2.0 Basic Service Set %#e!tifier (BSS%/)
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3.0.1 'reve!ti!" Access to Network 8esources
4. Physical ayer
4.1 The h9sical la9er 4asics
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4.3 %!frare# (%8)
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4.0.1 'ractice 032.114
4.<Comariso! of 032.11a a!# 032.114
The writi!" of this 8esearch 8eort was romte# to mai!tai! two mai! #evelome!ts of the
%&&& 032.11 Sta!#ar# h9sical rotocol stack a!# 'h9sical La9er e!ha!ce# from the
#evelome!ts i! wireless commu!icatio! i! the ast #eca#e. :irst we ha# to #o hu"e research
activities i! this toic. This has 4ee! a su4=ect stu#9 si!ce the si-ties> so that #uri!" our
e-lori!" work we have selecte# a lot of materials a!# icke# u the most visi4le thi!"s for the
stu#e!t to u!#ersta!# it more easil9 a!# clearl9. So that we were co!ce!trate# to rese!t the
issue i! mo#er! wireless co!cets i! a cohere!t a!# u!ifie# ma!!er a!# to illustrate the co!cets
i! that wa9 the9 are alie#.
The co!cets ca! 4e structure# i!to these levels*
+ Listi!" characteristics a!# mo#eli!"
+ Alicatio! of these co!cets
But of course there is i!terla9 4etwee! these structures.
So this 8esearch reort is writte! 4ase# o! the material for the stu#e!ts i! the si-th semester.
Also i! the e!# to u!#ersta!# 4etter the termi!olo"9 a!# the hu"e !um4er of a44reviatio!s
e-lai!e# a!# some #efi!itio!s.
The ast #eca#e has see! ma!9 a#va!ces i! h9sical+la9er commu!icatio! theor9 a!# their
imleme!tatio! i! wireless s9stems. So that i! this 8esearch 8eort we are "oi!" to #efi!e a!#
view fu!#ame!tals of wireless commu!icatio! a!# that eseciall9 the %&&& 032.11 Sta!#ar# a!#
e-lai! the a#va!ta"es at a level that is accessi4le to our au#ie!ce with a 4asic 4ack"rou!#.
Wireless commu!icatio! is o!e of the most vi4ra!t areas i! the commu!icatio! fiel# to#a9. This
is #ue to a co!flue!ce of several factors. :irst> there has 4ee! a! e-losive i!crease i! #ema!#
for the tether less co!!ectivit9> #rive! so far mai!l9 49 cellular teleho!9 4ut e-ecte# to 4e
soo! eclise# 49 wireless #ata alicatio!.
:irst there has 4ee! a! e-losive i!crease i! #ema!# for tether less co!!ectivit9> #rive! so far
mai!l9 49 cellular teleho!9 4ut e-ecte# to 4e soo! eclise# 49 wireless #ata alicatio!s.
Seco!#> the #ramatic rocess i! ?LS% tech!olo"9 has e!a4le# small area a!# low ower
imleme!tatio! of sohisticate# si"!al rocessi!" al"orithms a!# co#i!" tech!i;ues.
Thir#> the success of seco!# "e!eratio! #i"ital wireless sta!#ar#s a!# rovi#e a co!crete
#emo!stratio! that "oo# i#eas from commu!icatio! theor9 ca! have imact i! ractice.
There are two fu!#ame!tal asects of wireless commu!icatio! that make the ro4lem
challe!"i!" a!# i!teresti!". These asects are 49 a!# lar"e !ot as si"!ifica!t i! wire li!e
:irst the he!ome!o! of fa#i!"* the time variatio! of the cha!!el stre!"ths #ue to the small+scale
effect of multiath fa#i!"> as well as lar"e scale effects.
Seco!#> u!like the i! the wire# worl# where each tra!smitter+receiver air ca! ofte! 4e thou"ht
of as a! isolate# oi!t+to oi!t li!k> wireless users commu!icate over the air a!# there is
si"!ifica!t i!terface 4etwee! them.
The ori"i!al 032.11 sta!#ar# secifie# three searate h9sical la9ers. Two are ra#io+
4ase# a!# o!e is i!frare# li"ht+4ase#. The ori"i!al ra#io+4ase# la9ers are srea# sectrum*
fre;ue!c9 hoi!" a!# #irect se;ue!ce. These are all i! the 2.4 @7A 4a!#. A! a##itio!al
h9sical la9er i! the $ @7A 4a!# was a##e# with the 032.1a release> which is also ra#io+4ase#*
ortho"o!al fre;ue!c9 #ivisio! multile-i!" (O:/6).
The latest release> 032.11"> a##e# 9et a!other '7B* comlime!tar9 co#e ke9i!"
ortho"o!al fre;ue!c9 #ivisio! multile-i!" (CCC+O:/6). This is secifie# i! 4oth the 2.4 a!#
the $ @7A 4a!#s. Note that for 2 #evices to 4e a4le to i!teract> the9 must co!form to the same
'7B la9er. There are two su4 la9ers i! the 032.11 h9sical la9ers> the h9sical me#ium
#ee!#e!t la9er ('6/) a!# the h9sical la9er co!ver"e!ce roce#ure ('LC'). The '6/ is the
su4 la9er lowest o! the stack. %t tra!smits a!# receives 4its over the air. The '7B la9er has three
4asic fu!ctio!s. These are the carrier se!se fu!ctio!> the tra!smit fu!ctio!> a!# the receive
Wireless tech!olo"ies> i! the simlest se!se> e!a4le o!e or more #evices to commu!icate
without h9sical co!!ectio!s D without re;uiri!" !etwork ca4li!". Wireless tech!olo"ies use
ra#io tra!smissio!s as the mea!s for tra!smitti!" #ata> whereas wire# tech!olo"ies use ca4les.
Wireless tech!olo"ies ra!"e from comle- s9stems> such as WLANs a!# cell ho!es> to simle
#evices such as wireless hea#ho!es> microho!es> a!# other #evices that #o !ot rocess or store
i!formatio!. %! Comuter !etwork a su4sta!tial art is the Wireless Local Area Network
(WLAN)> is a closel9 "roue# s9stem of #evices that commu!icate via ra#io waves i!stea# of
wires. Wireless LANEs t9icall9 au"me!t or relace wire# comuter !etworks> rovi#i!" users
with more fle-i4ilit9 a!# free#om of moveme!t withi! the worklace. %! a t9ical WLAN
co!fi"uratio!> a tra!sceiverFor access oi!tFco!!ects to the wire# !etwork from a fi-e#
ilocatio! usi!" a sta!#ar# &ther!et ca4le. The access oi!t receives> 4uffers> a!# tra!smits #ata
4etwee! the como!e!ts of the WLAN. :or over a ce!tur9> the %&&&+SA has offere# a!
esta4lishe# sta!#ar#s #evelome!t ro"ram that features 4ala!ce> oe!!ess> #ue rocess> a!#
co!se!sus. The %!stitute of &lectrical a!# &lectro!ics &!"i!eers Sta!#ar#s Associatio! (%&&&+
SA) is the lea#i!" #eveloer of "lo4al i!#ustr9 sta!#ar#s i! a 4roa#+ra!"e of i!#ustries>
i!clu#i!"* 'ower a!# &!er"9G Biome#ical a!# 7ealthcareG %!formatio! Tech!olo"9G
Telecommu!icatio!sG Tra!sortatio!G Na!otech!olo"9G %!formatio! Assura!ce. We have
#iscusse# a4out them i! this raort.
A rotocol stack is a articular software imleme!tatio! of a comuter !etworki!"
rotocol suite. The terms are ofte! use# i!tercha!"ea4l9. Strictl9 seaki!"> the suite is the
#efi!itio! of the rotocols> a!# the stack is the software imleme!tatio! of them.
Securit9 is o!e of the first co!cer!s of eole #elo9i!" a Wireless LANG the 032.11
committee has a##resse# the issue 49 rovi#i!" what is calle# W&' (Wire# &;uivale!t 'rivac9)
Authe!ticatio!* A fu!ctio! that #etermi!es whether a Statio! is allowe# to articiate i!
!etwork commu!icatio!. The sta!#ar# %&&& 032.11i is #esi"!e# to rovi#e secure#
commu!icatio! of wireless LAN as #efi!e# 49 all the %&&& 032.11 secificatio!s. %&&& 032.11i
e!ha!ces the W&' (Wireli!e &;uivale!t 'rivac9)G a tech!olo"9 use# for ma!9 9ears for the
WLAN securit9> i! the areas of e!cr9tio!> authe!ticatio! a!# ke9 ma!a"eme!t.
The %&&& (#efi!e) 032.11 sta!#ar# i!clu#es a commo! 6e#ium Access Co!trol (6AC)
La9er> which #efi!es rotocols that "over! the oeratio! of the wireless LAN. %! a##itio!>
032.11 comrise several alter!ative h9sical la9ers that secif9 the tra!smissio! a!# recetio! of
A!# as co!clusio! we will sa9 that wireless !etworki!" has a romisi!" future with
032.11 lea#i!" the wa9 as the sta!#ar# for a#otio! i! local !etworki!" e!viro!me!ts. 032.11
a##resses mo4ilit9> securit9> relia4ilit9> a!# the #9!amic !ature of wireless LANS while keei!"
comati4ilit9 with 032+t9e le"ac9 !etworks. &-ect to see availa4ilit9 of 032.11 ro#ucts
i!crease #ramaticall9 i! the !ear future as 4usi!esses #iscover the i!crease# ro#uctivit9
rovi#e# 49 Hu!tethere#E !etworks.
(ireless ocal Area )et'ork
%! Comuter !etwork a su4sta!tial art is the Wireless Local Area Network (WLAN)> is
a closel9 "roue# s9stem of #evices that commu!icate via ra#io waves i!stea# of wires. Wireless
LANEs t9icall9 au"me!t or relace wire# comuter !etworks> rovi#i!" users with more
fle-i4ilit9 a!# free#om of moveme!t withi! the worklace. 1sers ca! access the coma!9
i!tra!et or eve! the Worl# Wi#e We4 from a!9where o! the coma!9 camus without rel9i!"
o! the availa4ilit9 of wire# ca4les a!# co!!ectio!. %f i!formatio! is the life4loo# of to#a9Is
4usi!ess e!viro!me!t> the! wireless !etworks are its heart. Wireless LANs ca! um
i!formatio! a!# #ata
to e-ecutives i! the
4oar#room a!# to
emlo9ees i! the
(WLAN) is a fle-i4le
as a! e-te!sio! or as
a! alter!ative for> a
wire# LAN withi! a
4uil#i!" or camus.
tra!smits a!# receive
#ata over the air>
mi!imiAi!" the !ee# for wire# co!!ectio!s. Thus> WLANs com4i!e #ata co!!ectivit9 with user
mo4ilit9> a!#> throu"h simlifie# co!fi"uratio!> e!a4le mova4le LANs. Over the last seve!
9ears> WLANs have "ai!e# stro!" oularit9 i! a !um4er of vertical markets> i!clu#i!" the
health+care> retail> ma!ufacturi!"> warehousi!"> a!# aca#emic are!as. These i!#ustries have
rofite# from the ro#uctivit9 "ai!s of usi!" ha!#+hel# termi!als a!# !ote4ook comuters to
tra!smit real+time i!formatio! to ce!traliAe# hosts for rocessi!". To#a9 WLANs are 4ecomi!"
more wi#el9 reco"!iAe# as a "e!eral+urose co!!ectivit9 alter!ative for a 4roa# ra!"e of
4usi!ess customers. The 1.S. wireless LAN market is rai#l9 aroachi!" J1 4illio! i!
reve!ues. A wi#e variet9 of i!#ustries have #iscovere# the 4e!efits a WLAN ca! 4ri!"F!ot
o!l9 to #ail9 tasks 4ut also to the 4ala!ce sheet.
The *asic Structure of a (ireless A)
%! a t9ical WLAN co!fi"uratio!> a tra!sceiverFor access oi!tFco!!ects to the wire#
!etwork from a fi-e# locatio! usi!" a sta!#ar# &ther!et ca4le. The access oi!t receives> 4uffers>
a!# tra!smits #ata 4etwee! the como!e!ts of the WLANFwhether latos> ri!ters> ha!#hel#
#evices> or a!9 other wireless e;uime!tFa!# the wire# !etwork i!frastructure. A si!"le access
oi!t ca! suort a small "rou of users a!# ca! fu!ctio! withi! a ra!"e of a!9where from 33 to
several hu!#re# feet. The access oi!t ca! 4e i!stalle# a!9where i! the facilit9 as lo!" as "oo#
ra#io covera"e is mai!tai!e#. 1sers e;uie# with ha!#hel# #evices or !ote4ook comuters ca!
tra!smit #ata to the access oi!t whe! withi! ra!"e. The wireless #evices commu!icate with the
!etwork oerati!" s9stem via WLAN a#aters> usuall9 i! the form of ra#io !etwork i!terface
car#s (N%Cs)> as i! the case of !ote4ook comuters> %SA or 'C% a#aters for #eskto comuters>
or similar #evices i!te"rate# i!to ha!#hel# u!its.
Co+,arin- a (ireless A) to a (ired )et'ork
The see# at which a WLAN erforms #ee!#s o! the t9e a!# co!fi"uratio! of the
#evices withi! the !etwork. The !um4er of users> the #ista!ce 4etwee! !etwork como!e!ts> the
t9e of WLAN s9stem i! use> a!# the efficie!c9 of the wire# !etwork eleme!ts all i!flue!ce the
overall see# a!# erforma!ce of a wireless !etwork. Such factors also affect wire# !etwork
see#s> 4ut most commercial LANs oerate at see#s from 13 me"a4its er seco!# (13BaseT) to
133 64s (133BaseT). Wireless LAN como!e!ts that use the 032.11a hi"h #ata rate sta!#ar#
erform at see#s u to $4 64s> almost a five+fol# i!crease from the erforma!ce of the
032.114 sta!#ar#. Almost all mo4ile alicatio!s to#a9 le!# themselves to #elo9me!t of a!
032.11 WLAN i!frastructure. Of the three mai! variatio!s of 032.11> a lethora of alicatio!s
a!# #evices suort the 032.114 sta!#ar#> which oerates i! the 2.4 @7A fre;ue!c9 ra!"e.
Althou"h this sta!#ar# is much more wi#el9 imleme!te# tha! its !ewer sister tech!olo"ies>
i!#ustr9 e-erts a!ticiate that it wo!Et 4e lo!" 4efore 032.11" a!# 032.11a e-cee# 032.114 i!
oularit9. Wireless users reco"!iAe the 4e!efits of the tech!olo"9 a!# !ee# to k!ow how to
rotect their 4usi!ess+critical #ata. These usersFas well as those who hesitate to #elo9 wireless
tech!olo"9 4ecause of securit9 co!cer!sF sta!# to 4e!efit from u!#ersta!#i!" the securit9
otio!s curre!tl9 availa4le> eve! as the i!#ustr9 moves a""ressivel9 to rovi#e eve! more secure
rotocols. B9 worki!" with a wireless ve!#or well+verse# i! securit9 issues> coma!ies ca!
#ramaticall9 e!ha!ce the securit9 of its wireless commu!icatio!s s9stem.
WLANs t9icall9 use the u!lice!se# %!#ustrial> Scie!tific> a!# 6e#ical (%S6) ra#io
fre;ue!c9 4a!#s. %! the 1!ite# States> the %S6 4a!#s i!clu#e the <33+67A 4a!# (<32D<20
67A)> 2.4+@7A 4a!# (2433D2403.$67A)> a!# the $..+@7A 4a!# ($.2$D$0$367A). The most
wi#el9 a#ote# WLAN sta!#ar# arou!# the worl# is
032.11 K20L to#a9. %&&& 032.11 co!sists of a famil9 of sta!#ar#s that #efi!es the h9sical
la9ers ('7B) a!# the 6e#ium Access Co!trol (6AC) la9er of a WLAN> WLAN !etwork
architectures> how a WLAN i!teracts with a! %' core !etwork> a!# the frameworks a!# mea!s
for suorti!" securit9 a!# ;ualit9 of service over a WLAN. The %&&& 032.11 sta!#ar#s famil9
i!clu#es the followi!" ke9 sta!#ar#s*
The I... Standard
:or over a ce!tur9> the %&&&+SA has offere# a! esta4lishe# sta!#ar#s #evelome!t
ro"ram that features 4ala!ce> oe!!ess> #ue rocess> a!# co!se!sus. The %!stitute of &lectrical
a!# &lectro!ics &!"i!eers Sta!#ar#s Associatio! (%&&&+SA) is the lea#i!" #eveloer of "lo4al
i!#ustr9 sta!#ar#s i! a 4roa#+ra!"e of i!#ustries> i!clu#i!"*
• 'ower a!# &!er"9
• Biome#ical a!# 7ealthcare
• %!formatio! Tech!olo"9
• %!formatio! Assura!ce
The followi!" ta4le lists hi"hli"hts of the most oular sectio!s of %&&& 032 a!# has
li!ks for a##itio!al i!formatio!*
032 Overview Basics of h9sical a!# lo"ical !etworki!" co!cets.
LAN56AN 4ri#"i!" a!# ma!a"eme!t. Covers ma!a"eme!t
a!# the lower su4+la9ers of OS% La9er 2> i!clu#i!" 6AC+4ase#
4ri#"i!" (6e#ia Access Co!trol)> virtual LANs a!# ort+4ase#
032.2 Lo"ical Li!k
Commo!l9 referre# to as the LLC or Lo"ical Li!k Co!trol
secificatio!. The LLC is the to su4+la9er i! the #ata+li!k
la9er> OS% La9er 2. %!terfaces with the !etwork La9er 3.
M@ra!##a##9M of the 032 secificatio!s. 'rovi#es
as9!chro!ous !etworki!" usi!" Mcarrier se!se> multile access
with collisio! #etectM (CS6A5C/) over coa-> twiste#+air
coer> a!# fi4er me#ia. Curre!t see#s ra!"e from 13 64s to
13 @4s. Click for a list of the MhotM 032.3 tech!olo"ies.
032.4 Toke! Bus /is4a!#e#
032.$ Toke! 8i!"
The ori"i!al toke!+assi!" sta!#ar# for twiste#+air> shiel#e#
coer ca4les. Suorts coer a!# fi4er ca4li!" from 4 64s
to 133 64s. Ofte! calle# M%B6 Toke!+8i!".M
;ueue #ual 4us
MSuerse#e# OO8evisio! of 032.1/+1<<3 e#itio! (%SO5%&C
13330). 032.1/ i!cororates '032.1 a!# '032.12e. %t also
i!cororates a!# suerse#es u4lishe# sta!#ar#s 032.1= a!#
032.,k. Suerse#e# 49 032.1/+2334.M (See %&&& status a"e.)
With#raw! Sta!#ar#. With#raw! /ate* :e4 3.> 2333. No
lo!"er e!#orse# 49 the %&&&. (See %&&& status a"e.)
With#raw! 'A8. Sta!#ar#s ro=ect !o lo!"er e!#orse# 49 the
%&&&. (See %&&& status a"e.)
With#raw! 'A8. Sta!#ar#s ro=ect !o lo!"er e!#orse# 49 the
%&&&. (See %&&& status a"e.)
Suerse#e# OOCo!tai!s* %&&& St# 032.134+1<<2. (See %&&&
Wireless LAN 6e#ia Access Co!trol a!# 'h9sical La9er
secificatio!. 032.11a> 4> "> etc. are ame!#me!ts to the ori"i!al
032.11 sta!#ar#. 'ro#ucts that imleme!t 032.11 sta!#ar#s
must ass tests a!# are referre# to as MWi+:i certifie#.M
Secifies a '7B that oerates i! the $ @7A 1+N%% 4a!# i! the
1S + i!itiall9 $.1$+$.3$ AN/ $..2$+$.0$ + si!ce e-a!#e# to
1ses Ortho"o!al :re;ue!c9+/ivisio! 6ultile-i!"
&!ha!ce# #ata see# to $4 64s
8atifie# after 032.114
&!ha!ceme!t to 032.11 that a##e# hi"her #ata rate mo#es to
the /SSS (/irect Se;ue!ce Srea# Sectrum) alrea#9 #efi!e#
i! the ori"i!al 032.11 sta!#ar#
Booste# #ata see# to 11 64s
22 67A Ba!#wi#th 9iel#s 3 !o!+overlai!" cha!!els i! the
fre;ue!c9 ra!"e of 2.433 @7A to 2.403$ @7A
Beaco!s at 1 64s> falls 4ack to $.$> 2> or 1 64s from 11
&!ha!ceme!t to 032.11a a!# 032.114 that allows for "lo4al
'articulars ca! 4e set at 6e#ia Access Co!trol (6AC) la9er
&!ha!ceme!t to 032.11 that i!clu#es ;ualit9 of service (NoS)
:acilitates rioritiAatio! of #ata> voice> a!# vi#eo tra!smissio!s
&-te!#s the ma-imum #ata rate of WLAN #evices that oerate
i! the 2.4 @7A 4a!#> i! a fashio! that ermits i!teroeratio!
with 032.114 #evices
1ses O:/6 6o#ulatio! (Ortho"o!al :/6)
Oerates at u to $4 me"a4its er seco!# (64s)> with fall+
4ack see#s that i!clu#e the M4M see#s
&!ha!ceme!t to 032.11a that resolves i!terfere!ce issues
/9!amic fre;ue!c9 selectio! (/:S)
Tra!smit ower co!trol (T'C)
&!ha!ceme!t to 032.11 that offers a##itio!al securit9 for
/efi!es more ro4ust e!cr9tio!> authe!ticatio!> a!# ke9
e-cha!"e> as well as otio!s for ke9 cachi!" a!# re+
Paa!ese re"ulator9 e-te!sio!s to 032.11a secificatio!
:re;ue!c9 ra!"e 4.< @7A to $.3 @7A
8a#io resource measureme!ts for !etworks usi!" 032.11
6ai!te!a!ce of 032.11 famil9 secificatio!s
Correctio!s a!# ame!#me!ts to e-isti!" #ocume!tatio!
7i"her+see# sta!#ar#s ++ u!#er #evelome!t
Several cometi!" a!# !o!+comati4le tech!olo"iesG ofte!
To see#s claime# of 130> 243> a!# 3$3Q 67A
Cometi!" roosals come from the "rous> &WC> T@! S9!c>
a!# WWiS& a!# are all variatio!s 4ase# o! 6%6O (multile
i!ut> multile outut)
032.11- 6iss+use# M"e!ericM term for 032.11 famil9 secificatio!s
032.12 /ema!# 'riorit9
%!creases &ther!et #ata rate to 133 64s 49 co!trolli!" me#ia
032.13 Not use# Not use#
032.14 Ca4le mo#ems
With#raw! 'A8. Sta!#ar#s ro=ect !o lo!"er e!#orse# 49 the
Commu!icatio!s secificatio! that was arove# i! earl9 2332
49 the %&&& for wireless erso!al area !etworks (W'ANs).
Short ra!"e (13m) wireless tech!olo"9 for cor#less mouse>
ke94oar#> a!# ha!#s+free hea#set at 2.4 @7A.
032.1$.3a 1WB Short ra!"e> hi"h+4a!#wi#th Multra wi#e4a!#M li!k
032.1$.4 Ri"Bee Short ra!"e wireless se!sor !etworks
032.1$.$ 6esh !etwork
&-te!sio! of !etwork covera"e without i!creasi!" the tra!smit
ower or the receiver se!sitivit9
&!ha!ce# relia4ilit9 via route re#u!#a!c9
&asier !etwork co!fi"uratio! + Better #evice 4atter9 life
This famil9 of sta!#ar#s covers :i-e# a!# 6o4ile Broa#4a!#
Wireless Access metho#s use# to create Wireless 6etroolita!
Area Networks (W6ANs.) Co!!ects Base Statio!s to the
%!ter!et usi!" O:/6 i! u!lice!se# (<33 67A> 2.4> $.0 @7A)
or lice!se# (.33 67A> 2.$ D 3., @7A) fre;ue!c9 4a!#s.
'ro#ucts that imleme!t 032.1, sta!#ar#s ca! u!#er"o
Wi6A2 certificatio! testi!".
%&&& worki!" "rou #escritio!
%&&& 032.10 sta!#ar#s committee
032.1< Coe-iste!ce %&&& 032.1< Coe-iste!ce Tech!ical A#visor9 @rou
%&&& 032.23 missio! a!# ro=ect scoe
%&&& 032.21 missio! a!# ro=ect scoe
%&&& 032.22 missio! a!# ro=ect scoe
802.11 Protocol Stack
A rotocol stack is a articular software imleme!tatio! of a comuter !etworki!"
rotocol suite. The terms are ofte! use# i!tercha!"ea4l9. Strictl9 seaki!"> the suite is the
#efi!itio! of the rotocols> a!# the stack is the software imleme!tatio! of them. %!#ivi#ual
rotocols withi! a suite are ofte! #esi"!e# with a si!"le urose i! mi!#. This mo#ulariAatio!
makes #esi"! a!# evaluatio! easier. Because each rotocol mo#ule usuall9 commu!icates with
two others> the9 are commo!l9 ima"i!e# as layers i! a stack of rotocols. The lowest rotocol
alwa9s #eals with Mlow+levelM> h9sical i!teractio! of the har#ware. &ver9 hi"her la9er a##s
more features. 1ser alicatio!s ha4ituall9 #eal o!l9 with the tomost la9ers (See also OS%
The rotocols use# 49 all the 032 varia!ts> i!clu#i!" &ther!et> have a certai!
commo!alit9 of structure. %! the fi"ure 4elow we see a artial view of the 032.11 rotocol stack.
The h9sical la9er correso!#s to the OS% h9sical la9er fairl9 well> 4ut the #ata li!k la9er i! all
the 032 rotocols is slit i!to two or more su4la9ers. %! 032.11> the 6AC (6e#ium Access
Co!trol) su4la9er #etermi!es how the cha!!el is allocate#> that is> a!# who "ets to tra!smit !e-t.
A4ove it is the LLC (Lo"ical Li!k Co!trol) su4la9er> whose =o4 it is to hi#e the #iffere!ces
4etwee! the #iffere!t 032 varia!ts a!# make them i!#isti!"uisha4le as far as the !etwork la9er is
co!cer!e#. We stu#ie# the LLC whe! e-ami!i!" &ther!et earlier i! this chater a!# will !ot
reeat that material here. The 1<<. 032.11 sta!#ar# secifies three tra!smissio! tech!i;ues
allowe# i! the h9sical la9er. The i!frare# metho# uses much the same tech!olo"9 as televisio!
remote co!trols #o. The other two use short+ra!"e ra#io> usi!" tech!i;ues calle# :7SS a!#
/SSS. Both of these use a art of the sectrum that #oes !ot re;uire lice!si!" (the 2.4+@7A %S6
4a!#). 8a#io+co!trolle# "ara"e #oor oe!ers also use this iece of the sectrum> so 9our
ma9 fi!# itself i!
cometitio! with 9our
"ara"e #oor. Cor#less
microwave ove!s also
use this 4a!#. All of
oerate at 1 or 2
64s a!# at low
e!ou"h ower that
the9 #o !ot co!flict
too much. %! 1<<<>
two !ew tech!i;ues
were i!tro#uce# to
4a!#wi#th. These are
calle# O:/6 a!#
oerate at u to $4
64s a!# 11 64s>
resectivel9. %! 2331>
a seco!# O:/6
i!tro#uce#> 4ut i! a
#iffere!t fre;ue!c9 4a!# from the first o!e. Now we will e-ami!e each of them 4riefl9.
%! the fi"ure 4elow we ca! see the Wireless LAN 49 %&&& 032.11> 032.11a>
032.114>032.11"> 032.11!031.11 rotocol famil9 6AC frame structure*
2 2 , , , 2 ,
:rame Co!trol Structure*
2 2 4 1 1 1 1 1 1 1 1
?ersio! T9e Su4t9e To /S :rom /S 6: 8etr9 'wr 6ore W O
• 'rotocol ?ersio! + i!#icates the versio! of %&&& 032.11 sta!#ar#.
• T9e + :rame t9e* 6a!a"eme!t> Co!trol a!# /ata.
• Su4t9e + :rame su4t9e* Authe!ticatio! frame> /eauthe!ticatio! frameG Associatio!
re;uest frameG Associatio! reso!se frameG 8eassociatio! re;uest frameG 8eassociatio!
reso!se frameG /isassociatio! frameG Beaco! frameG 'ro4e frameG 'ro4e re;uest frame
a!# 'ro4e reso!se frame.
• To /S + is set to 1 whe! the frame is se!t to /istri4utio! S9stem (/S)
• :rom /S + is set to 1 whe! the frame is receive# from the /istri4utio! S9stem (/S)
• 6:+ 6ore :ra"me!t is set to 1 whe! there are more fra"me!ts 4elo!"i!" to the same
frame followi!" the curre!t fra"me!t
• 8etr9 i!#icates that this fra"me!t is a retra!smissio! of a reviousl9 tra!smitte#
fra"me!t. (:or receiver to reco"!iAe #ulicate tra!smissio!s of frames)
• 'wr + 'ower 6a!a"eme!t i!#icates the ower ma!a"eme!t mo#e that the statio! will 4e
i! after the tra!smissio! of the frame.
• 6ore + 6ore /ata i!#icates that there are more frames 4uffere# to this statio!.
• W + W&' i!#icates that the frame 4o#9 is e!cr9te# accor#i!" to the W&' (wire#
e;uivale!t rivac9) al"orithm.
• + Or#er i!#icates that the frame is 4ei!" se!t usi!" the Strictl9+Or#ere# service class.
• /uratio!5%/ (%/) +
• Statio! %/ is use# for 'ower+Save oll messa"e frame t9e.
• The #uratio! value is use# for the Network Allocatio! ?ector (NA?) calculatio!.
• A##ress fiel#s (1+4) + co!tai! u to 4 a##resses (source> #esti!atio!> tra!smittio! a!#
receiver a##resses) #ee!#i!" o! the frame co!trol fiel# (the To/S a!# :rom/S 4its).
• Se;ue!ce Co!trol + co!sists of fra"me!t !um4er a!# se;ue!ce !um4er. %t is use# to
rerese!t the or#er of #iffere!t fra"me!ts 4elo!"i!" to the same frame a!# to reco"!iAe
• /ata + is i!formatio! that is tra!smitte# or receive#.
• C8C + co!tai!s a 32+4it C9clic 8e#u!#a!c9 Check (C8C).
Protocol Stack Architecture
&ach comuter> mo4ile> orta4le or fi-e#> is
referre# to as a statio! i! 032.11 (Wireless Local Area
The #iffere!ce 4etwee! a orta4le a!# mo4ile
statio! is that a orta4le statio! moves from oi!t to oi!t
4ut is o!l9 use# at a fi-e# oi!t. 6o4ile statio!s access the
LAN #uri!" moveme!t
Station /STA0 Architecture1
The Statio! Architecture is a #evice that co!tai!s
the %&&& co!forma!t 6AC a!# '7B i!terface i! the
wireless me#ium> 4ut o! the other si#e it #oes !ot rovi#e
access to a #istri4utio! s9stem. The Statio! Architecture is
most ofte! availa4le i! termi!als like latos> work+statio!s
a!# it is imleme!te# i! the Ava9a Wireless %&&& 032.11
• The most imorta!t features a!# co!#itio!s of the
Statio! Architecture are
• %t has a #river i!terface like the &ther!et
• All rotocol Stacks are virtuall9 suorte# 49 the STA
• The :rame tra!slatio! is #o!e accor#i!" to the %&&& ST/ 032.17
• The %&&& 032.3 frames with this architecture are tra!slate# to 032.11
• ?ia the Bri#"e Tu!!el e!casulatio! scheme are e!casulate# the &ther!et T9es 013.
(Novell %'2) a!# 03:3 (AA8')
• All other &ther!et T9es* e!casulate# via the 8:C 1342 (Sta!#ar# for the Tra!smissio!
of %' /ata"ramEs over %&&& 032 Networks) e!casulatio! scheme
• 6a-imum /ata limite# to 1$33 octets
• Bri#"i!" to &ther!et is #o!e tra!sare!tl9
Access2Point /AP0 Architecture1
A! Access 'oi!t is a #evice fou!# withi! a! %&&& 032.11 !etwork which rovi#es the
oi!t of i!terco!!ectio! 4etwee! the wireless Statio! (lato comuter> '/A ('erso!!el /i"ital
Assista!t) etc.) a!# the wire# !etwork.
The Access 'oi!t Architecture is a #evice that co!tai!s
%&&& 032.11 co!forma!t 6AC a!# '7B i!terface to the
wireless me#ium> a!# rovi#es access to a #istri4utio!
s9stem for associate# statio!s. 6ost ofte! it co!tai!s
i!fra+structure ro#ucts that co!!ect to wire# 4ack4o!es
%t is imleme!te# i! Ava9a Wireless %&&& 032.11 'C+
Car# whe! it is i!serte# i! a! A'+$33 or A'+1333
The +ost i+,ortant features and conditions of
Access2Point /AP0 Architecture1
• The Statio!s select a! Access+'oi!t a!# the9
associate with that. The Access oi!t is a art that
suorts roami!" a!# also the9 rovi#e time
s9!chro!iAatio! fu!ctio!s like 4eaco!i!". The
Access+'oi!t Architecture offers a 'ower
6a!a"eme!t suort. %! a rotocol stack architecture
the traffic t9icall9 flows throu"h the Access+'oi!t.
So that i! the %BSS architecture takes lace the #irect
Statio!+to+Statio! commu!icatio! s9stem.
*asic Ser#ice Set /*SS01
The Basic Service Set is a term use# to #escri4e the collectio! of Statio!s which ma9
commu!icate to"ether withi! a! 032.11 WLAN (Wireless Local Area Network). The BSS ma9
or ma9 !ot i!clu#e A' (Access 'oi!t) which rovi#es a co!!ectio! o!to a fi-e# #istri4utio!
s9stem such as a! &ther!et !etwork. Two t9es of BSS e-istG %BSS (%!#ee!#e!t Basic Service
Set) a!# %!frastructure Basic Service Set. Whe! two or more statio!s come to"ether to
commu!icate with each other> the9 form a Basic Service Set (BSS). The mi!imum BSS co!sists
of two statio!s. 032.11 LANs use the BSS as the sta!#ar# 4uil#i!" 4lock.
%! the BSS architecture a set of statio!s is co!trolle# 49 a si!"le SCoor#i!atio! :u!ctio!T> that is
the lo"ical fu!ctio! that #etermi!es whe! a statio! ca! tra!smit or receive.
%! this case we have similarit9 to a ScellT i! the re %&&& termi!olo"9 also a BSS ca! have a!
Access+'oi!t a!# that 4oth i! sta!#alo!e !etworks a!# i! 4uil#i!"+wi#e co!fi"uratio!s> or it =ust
ca! ru! without a!# Access+'oi!t 4ut o!l9 i! sta!#alo!e !etworks. The #iameter of the cells is
twice the covera"e+#ista!ce 4etwee! two wireless statio!s.
Inde,endent *asic Ser#ice Set /I*SS01
A! %!#ee!#e!t Basic Service Set also calle# a# hoc !etwork is the simlest of all %&&& 032.11
!etworks i! that !o !etwork i!frastructure is re;uire#. As such> a! %BSS is siml9 comrise# of
o!e or more Statio!s which commu!icate #irectl9 with each other. The co!tractio! shoul# !ot 4e
co!fuse# with a! %!frastructure BSS (Basic Service Set).A BSS that sta!#s alo!e a!# is !ot
co!!ecte# to a 4ase is calle# a! %!#ee!#e!t Basic Service Set (%BSS) or is referre# to as a! A#+
7oc Network. A! a#+hoc !etwork is a !etwork where statio!s commu!icate o!l9 eer to eer.
There is !o 4ase a!# !o o!e "ives ermissio! to talk. 6ostl9 these !etworks are so!ta!eous a!#
ca! 4e set u rai#l9. A#+7oc or %BSS !etworks are characteristicall9 limite# 4oth temorall9
a!# satiall9. So that the Basic Service Set (BSS) forms a self+co!tai!e# !etwork i! which !o
access to a /istri4utio! S9stem is availa4le> or it is also similar to a BSS without a! Access+
'oi!t. O!e of the statio!s i! the %BSS ca! 4e co!fi"ure# to Si!itiateT the !etwork a!# assume the
Coor#i!atio! :u!ctio!. The #iameter of the cell is #etermi!e# 49 covera"e #ista!ce 4etwee! two
Whe! BSSIs are i!terco!!ecte# the !etwork 4ecomes o!e with i!frastructure. %!frastructure is
esta4lishe# i! the !etwork whe! BSS are i!terco!!ecte#> so that the 032.11 i!frastructures have
several eleme!ts. Two or more BSSIs are i!terco!!ecte# usi!" a /istri4utio! S9stem or /S. This
co!cet of /S i!creases !etwork covera"e. &ach BSS 4ecomes a como!e!t of a! e-te!#e#>
lar"er !etwork. &!tr9 to the /S is accomlishe# with the use of Access 'oi!ts (A'). A! access
oi!t is a statio!> thus a##ressa4le. With hel of the Access+'oi!ts #ata moves the! 4etwee! the
BSS a!# the /S.
.3tended Ser#ice Set /.SS01
A! &-te!#e# Service Set is comrise# of a !um4er of %&&& 802.11 BSS (Basic Service Set) a!#
e!a4les limite# mo4ilit9 withi! the (A) (Wireless Local Area Network). Statio!s are a4le to
move 4etwee! *SS withi! a si!"le &SS 9et still remai! Sco!!ecte#T to the fi-e# !etwork a!# so
co!ti!ue to receive emails etc. As a Station moves i!to a !ew BSS> it will carr9 out a
reassociatio! roce#ure with the !ew AP (Access 'oi!t). Creati!" lar"e a!# comle- !etworks
usi!" BSSIs a!# /SIs lea#s us to the !e-t level of hierarch9> the &-te!#e# Service Set or &SS.
The 4eaut9 of the &SS is the e!tire !etwork looks like a! i!#ee!#e!t 4asic service set to the
Lo"ical Li!k Co!trol la9er (LLC). This mea!s that statio!s withi! the &SS ca! commu!icate or
eve! move 4etwee! BSSEs tra!sare!tl9 to the LLC.
%t is the same here that the traffic alwa9s flows via Access+'oi!t> a!# the #iameter of the cell is
#ou4le the covera"e #ista!ce 4etwee! two wireless statio!s /istri4utio! S9stem (/S).There is
availa4le a s9stem to i!terco!!ect a set of Basic Service Sets a!# there is i!te"rate# a si!"le
Access+'oi!t i! a sta!#alo!e !etwork. %! the wire# !etwork there are use# ca4les to i!terco!!ect
the Access+'oi!ts. %! the wireless !etwork are use# wirelesses to i!terco!!ect the Access+'oi!ts.
&-amle* of &-te!#e# Service Set (&SS) with si!"le BSS a!# i!te"rate# /S
Example: &-te!#e# Service Set (&SS) BSSEs with wire# /istri4utio! S9stem (/S)
Example: &-te!#e# Service Set (&SS) BSSEs a!# wireless /istri4utio! S9stem (/S)
Ser#ice Set Identifier /SSI401
The Service Set %#e!tifier or Network Name is secifie# withi! %&&& 802.11 !etworks to
i#e!tif9 a articular !etwork. %t is usuall9 set 49 the a#mi!istrator setti!" u the (A) a!# will
4e u!i;ue withi! a *SS (Basic Service Set) or .SS (&-te!#e# Service Set). The SS%/ ma9 4e
4roa#cast from a! AP withi! the wireless !etwork to e!a4le Stations to #etermi!e which
!etwork to SAssociateT with. 7owever> this feature shoul# 4e #isa4le# as it ma9 assist hackers or
'ardri#ers i! "ai!i!" access to a rivate !etwork. The most imorta!t thi!"s a4out the SS%/
are that it is 32 octets lo!" a!# it is similar to S/omai!+%/T i! the re+%&&& Wave LAN s9stems.
So we ca! co!clu#e that o!e !etwork i!#ee!#e!t from that if it is &SS or %BSS it has alwa9s
*asic Ser#ice Set Identifier /*SSI40
The BSS%/ is a 404it i#e!tit9 use# to i#e!tif9 a articular BSS (Basic Service Set) withi! o!e
area. %! the i!frastructure BSS !etworks> the BSS%/ is the 6AC (6e#ium Access Co!trol)
a##ress of the A' a!# i! %!#ee!#e!t BSS or a# hoc !etworks> the BSS%/ is "e!erate#
ra!#oml9. The BSS%/ i#e!tifies the cells a!# it is , octets lo!"> that mea!s that it is i! the 6AC
a##ress format. There is also visi4le a similarit9 to the NW%/ i! the re+ %&&& Wave LAN
s9stems. The value of the BSS%/ is the same as the 6AC a##ress of the ra#io i! the Access+
Protocol Stack for 5)I6
Also there are k!ow! #evelome!ts for architectural e!ha!ceme!ts for 1!i-+4ase# servers to
rovi#e a rotocol stack for 1N%2. To "ive a 4etter i#ea how it looks like the fi"ure 4elow
shows the 4asic como!e!ts of the e!ha!ce# rotocol stack architecture> with the !ew
caa4ilities utiliAe# either 49 user+sace a"e!ts or alicatio!s themselves. This architecture
ermits co!trol over a! alicatio!Is i!4ou!# !etwork traffic via olic9+4ase# traffic
ma!a"eme!tG a! a#atatio!5olic9 a"e!t i!stalls olicies i!to the ker!el via a secial A'%. The
olic9 a"e!t i!teracts with the ker!el via a! e!ha!ce# socket i!terface 49 se!#i!" (receivi!")
messa"es to (from) secial co!trol sockets. The olicies secif9 filters to select the traffic to 4e
co!trolle#> a!# actio!s to erform o! the selecte# traffic. The fi"ure shows the flow of a!
i!comi!" re;uest throu"h the various co!trol mecha!isms.
Co+,are &#erall Structure of 802.11b 7 802.1!.1 Coe3istence 8echanis+
A! AW6A tra!smissio! co!trol e!tit9 is i!te"rate# with the WLAN 6AC la9er a!# rovi#es a
6e#ium :ree si"!al to the Bluetooth Base4a!# la9er. This is a 4i!ar9 si"!al that "ates whe! the
WLAN a!# W'AN ca! each tra!smit ackets.
The 032.114 6AC a!# 032.1$.1 L6 Q LC e!tities rovi#e status i!formatio! to the 6&7TA
co!trol e!tities. The 6&7TA co!trol e!tit9 receives a er+tra!smissio! tra!smit re;uest (T2
8e;uest) a!# issues a er+tra!smissio! tra!smit co!firm (T2 Co!firm) to each stack to i!#icate
whether the tra!smissio! ca! rocee#. The T2 Co!firm carries a status value that is o!e of*
allowe# or #e!ie#. The T2 8e;uest a!# T2 Co!firm are #iscreet si"!als e-cha!"e# for ever9
acket tra!smissio! attemt.
T Enable T Enable
T &e'uest T &e'uest
802.11 Stack 802.15.1 Stack
8I9 SAP :eference 8odel for 802.11
The lo"ical laceme!t of the 6%7 :u!ctio! i! the 032.11 rotocol stack for statio!s a!# access
oi!ts is show! i! the fi"ure. %t is similar to the 032.3> where the LLC SA' (LSA') #efi!es the
i!terface of the 6%7 :u!ctio! with the 032.11 #ata la!e a!# ca! e!casulate 6%7 messa"es i!
#ata frames. 7owever> si!ce 032.11 #oes !ot curre!tl9 suort Class 1 #ata frames> 6%7
messa"es ca! 4e tra!sorte# over the 032.11 #ata la!e o!l9 after the 6o4ile No#e has
associate# with the 032.11 access oi!t. Before the associatio! 4etwee! 6o4ile No#e a!# access
oi!t takes lace> the L2 tra!sort of 6%7 messa"es ca! rel9 o! 032.11 ma!a"eme!t frames
from the 032.11 ma!a"eme!t la!e (6L6&). The 6%7 6L6& SA' #efi!es the i!terface
4etwee! the 6%7 :u!ctio! a!# the 6L6&.
Layer 3 Mobility Protocol (L3MP! "i#her$Layer Mobility
Protocol! "an%over Policy! &rans'ort! (''lications
"an%over (M)" *+nction
M)" ,vent Service
M)" Comman% Service
M)" )n-ormation Service
Lo#ical Link Control
8I9 SAP :eference 8odel for 802.1"
The lo"ical laceme!t of the 6%7 :u!ctio! i! the 032.1, rotocol stack is show! i! the fi"ure>
so that we ca! comare 4etter what is the #iffere!ce 4etwee! the 032.11 a!# 032.1,.
The 6%7 :u!ctio! a!# the Network Co!trol a!# 6a!a"eme!t S9stem (NC6S) share the CUSA'
a!# 6USA' for access to the mo4ilit9+ma!a"eme!t services of the 6o4ilit9 Co!trol &!tit9 a!#
6a!a"eme!t &!tit9 i! the 032.1, 6a!a"eme!t 'la!e.
The mecha!isms for the #irect e!casulatio! of 6%7 frames i!to 032.1, #ata frames ma9 take
multile forms. The Service+Secific Co!ver"e!ce Su4la9er i!sta!ces curre!tl9 availa4le i! the
032.1, sta!#ar#s a!# Wi6A2 o!l9 e!a4le the e!casulatio! of %' ackets a!# &ther!et frames.
The o!l9 otio! availa4le for L2 tra!sort woul# 4e to first e!casulate the 6%7 messa"es i!to
&ther!et frames with a! 6%7 &thert9e value> a!# the! ma!#ate the a#otio! of &ther!et CS for
032.1, co!!ectio!s that carr9 the 6%7 messa"es. This aroach limits 4oth the efficie!c9 of the
L2 tra!sort of 6%7 messa"es> a!# that si!ce it imoses the a##itio! of full &ther!et overhea# D
at least 10 49tes D to the 6%7 frame a!# the availa4ilit9 of L2 tra!sort caa4ilities for 6%7>
si!ce &ther!et CS is !ot u4i;uitous.
Alter!ativel9> a solutio! that e!a4les 4etter efficie!c9 a!# easier accessi4ilit9 of L2 tra!sort
caa4ilities coul# 4ecome availa4le with the ossi4le sta!#ar#iAatio! of the @e!eric 'acket
Co!ver"e!ce Su4la9er (@'CS) rece!tl9 roose# withi! 032.1,". With @'CS a more efficie!t
LLC5SNA' e!casulatio! (0 49tes overhea#) coul# create the !ee#e# room for the 6%7
&thert9e i! 032.1, frame.
Layer 3 Mobility Protocol (L3MP! "i#her$Layer Mobility
Protocol! "an%over Policy! &rans'ort! (''lications
"an%over (M)" *+nction
M)" ,vent Service
M)" Comman% Service
M)" )n-ormation Service
Service$S'eci-ic Conver#ence S+blayer (CS
M(C Common Part S+blayer (M(C CPS
Physical Layer (P"/
4ata ink ayer
As a! imorta!t art of the rotocol stack> the #ata li!k la9er withi! 032.11 co!sists of two
su4la9ers* Lo"ical Li!k Co!trol (LLC) a!# 6e#ia Access Co!trol (6AC). The 032.11 uses the
same 032.2 LLC a!# 40+4it a##ressi!" as other 032 LANs> allowi!" for ver9 simle 4ri#"i!"
from wireless to %&&& wire# !etworks> 4ut the 6AC is u!i;ue to WLANs. The 032.11 6AC is
ver9 similar i! co!cet comare# to the 032.3> which is #esi"!e# to suort multile users o! a
share# me#ium 49 havi!" the se!#er se!se the me#ium 4efore accessi!" it.
:or 032.3 &ther!et LANs> the Carrier Se!se 6ultile Access with Collisio! /etectio!
(CS6A5C/) it is re"ulate# from the rotocol how &ther!et statio!s are "oi!" to esta4lish access
to the wire a!# how the9 #etect a!# ha!#le collisio!s that occur whe! two or more #evices tr9 to
simulta!eousl9 commu!icate over the LAN. %! a! 032.11 WLAN> collisio! #etectio! is !ot
ossi4le #ue to what is k!ow! as the S!ear5farT ro4lem* to #etect a collisio!> a statio! must 4e
a4le to tra!smit a!# liste! at the same time> 4ut i! ra#io s9stems the tra!smissio! #row!s out the
a4ilit9 of the statio! to ShearT a collisio!. To accou!t for this #iffere!ce> 032.11 use a sli"htl9
mo#ifie# rotocol k!ow! as Carrier Se!se 6ultile Access with Collisio! Avoi#a!ce
(CS6A5CA) or the /istri4ute# Coor#i!atio! :u!ctio! (/C:). CS6A5CA attemts to avoi#
collisio!s 49 usi!" e-licit acket ack!owle#"me!t (ACC)> which mea!s a! ACC acket is se!t
49 the receivi!" statio! to co!firm that the #ata acket arrive# i!tact.
CS6A5CA works as follows. A statio! wishi!" to tra!smit se!ses the air> a!#> if !o activit9 is
#etecte#> the statio! waits a! a##itio!al> ra!#oml9 selecte# erio# of time a!# the! tra!smits if
the me#ium is still free. %f the acket is receive# i!tact> the receivi!" statio! issues a! ACC
frame that> o!ce successfull9 receive# 49 the se!#er> comletes the rocess. %f the ACC frame is
!ot #etecte# 49 the se!#i!" statio!> either 4ecause the ori"i!al #ata acket was !ot receive#
i!tact or the ACC was !ot receive# i!tact> a collisio! is assume# to have occurre# a!# the #ata
acket is tra!smitte# a"ai! after waiti!" a!other ra!#om amou!t of time.
CS6A5CA thus rovi#es a wa9 of shari!" access over the air. This e-licit ACC mecha!ism
also ha!#les i!terfere!ce a!# other ra#io relate# ro4lems ver9 effectivel9. 7owever> it #oes a##
some overhea# to 032.11 that 032.3 #oes !ot have> so that a! 032.11 LAN will alwa9s have
slower erforma!ce tha! a! e;uivale!t &ther!et LAN.
A!other 6AC+la9er ro4lem secific to wireless is the Shi##e! !o#eT issue> i! which two
statio!s o! oosite si#es of a! access oi!t ca! 4oth ShearT activit9 from a! access oi!t> 4ut
!ot from each other> usuall9 #ue to #ista!ce or a! o4structio!.
8TS5CTS 'roce#ure elimi!ates the S7i##e! No#eT 'ro4lem
To solve this ro4lem> 032.11 secif9 a! otio!al 8e;uest to Se!#5Clear to Se!# (8TS5CTS)
rotocol at the 6AC la9er. Whe! this feature is i! use> a se!#i!" statio! tra!smits a! 8TS a!#
waits for the access oi!t to rel9 with CTS. Si!ce all statio!s i! the !etwork ca! hear the access
oi!t> the CTS causes them to #ela9 a!9 i!te!#e# tra!smissio!s> allowi!" the se!#i!" statio! to
tra!smit a!# receive a acket ack!owle#"me!t without a!9 cha!ce of collisio!. Si!ce 8TS5CTS
a##s a##itio!al overhea# to the !etwork 49 temoraril9 reservi!" the me#ium> it is t9icall9 use#
o!l9 o! the lar"est+siAe# ackets> for which retra!smissio! woul# 4e e-e!sive from a
4a!#wi#th sta!#oi!t. :i!all9> the 032.11 6AC la9er rovi#es for two other ro4ust!ess features*
C8C checksum a!# acket fra"me!tatio!. &ach acket has a C8C checksum calculate# a!#
attache# to e!sure that the #ata was !ot corrute# i! tra!sit. This is #iffere!t from &ther!et>
where hi"her+level rotocols such as TC' ha!#le error checki!". 'acket fra"me!tatio! allows
lar"e ackets to 4e 4roke! i!to smaller u!its whe! se!t over the air> which is useful i! ver9
co!"este# e!viro!me!ts or whe! i!terfere!ce is a factor> si!ce lar"er ackets have a 4etter
cha!ce of 4ei!" corrute#. This tech!i;ue re#uces the !ee# for retra!smissio! i! ma!9 cases a!#
thus imroves overall wireless !etwork erforma!ce. The 6AC la9er is reso!si4le for
reassem4li!" fra"me!ts receive#> re!#eri!" the rocess tra!sare!t to hi"her level rotocols.
Su,,ort for Ti+e2*ounded 4ata
Time+4ou!#e# #ata such as voice a!# vi#eo is suorte# i! the 032.11 6AC secificatio!s
throu"h the 'oi!t Coor#i!atio! :u!ctio! ('C:). As oose# to the /C:> where co!trol is
#istri4ute# to all statio!s> i! 'C: mo#e a si!"le access oi!t co!trols access to the me#ia. %f a
BSS is set u with 'C: e!a4le#> time is slice# 4etwee! the s9stem 4ei!" i! 'C: mo#e a!# i!
/C: (CS6A5CA) mo#e. /uri!" the erio#s whe! the s9stem is i! 'C: mo#e> the access oi!t
will oll each statio! for #ata> a!# after a "ive! time move o! to the !e-t statio!. No statio! is
allowe# to tra!smit u!less it is olle#> a!# statio!s receive #ata from the access oi!t o!l9 whe!
the9 are olle#. Si!ce 'C: "ives ever9 statio! a tur! to tra!smit i! a re#etermi!e# fashio!> a
ma-imum late!c9 is "uara!tee#. A #ow!si#e to 'C: is that it is !ot articularl9 scala4le> i! that
a si!"le oi!t !ee#s to have co!trol of me#ia access a!# must oll all statio!s> which ca! 4e
i!effective i! lar"e !etworks.
8AC ;unctional 4escri,tion
The 032.11 sta!#ar# secifies a commo! me#ium access co!trol (6AC) La9er> which rovi#es a
variet9 of fu!ctio!s that suort the oeratio! of 032.11+4ase# wireless LANs. %! "e!eral> the
6AC La9er ma!a"es a!# mai!tai!s commu!icatio!s 4etwee! 032.11 statio!s (ra#io !etwork
car#s a!# access oi!ts) 49 coor#i!ati!" access to a share# ra#io cha!!el a!# utiliAi!" rotocols
that e!ha!ce commu!icatio!s over a wireless me#ium. Ofte! viewe# as the M4rai!sM of the
!etwork> the 032.11 6AC La9er uses a! 032.11 'h9sical ('7B) La9er> such as 032.114 or
032.11a> to erform the tasks of carrier se!si!"> tra!smissio!> a!# receivi!" of 032.11 frames.
Before tra!smitti!" frames> a statio! must first "ai! access to the me#ium> which is a ra#io
cha!!el that statio!s share. The 032.11 sta!#ar# #efi!es two forms of me#ium access> #istri4ute#
coor#i!atio! fu!ctio! (/C:) a!# oi!t coor#i!atio! fu!ctio! ('C:). /C: is ma!#ator9 a!#
4ase# o! the CS6A5CA (carrier se!se multile access with collisio! avoi#a!ce) rotocol. With
/C:> 032.11 statio!s co!te!# for access a!# attemt to se!# frames whe! there is !o other
statio! tra!smitti!". %f a!other statio! is se!#i!" a frame> statio!s are olite a!# wait u!til the
cha!!el is free.
As a co!#itio! to accessi!" the me#ium> the 6AC La9er checks the value of its !etwork
allocatio! vector (NA?)> which is a cou!ter resi#e!t at each statio! that rerese!ts the amou!t of
time that the revious frame !ee#s to se!# its frame. The NA? must 4e Aero 4efore a statio! ca!
attemt to se!# a frame. 'rior to tra!smitti!" a frame> a statio! calculates the amou!t of time
!ecessar9 to se!# the frame 4ase# o! the frameIs le!"th a!# #ata rate. The statio! laces a value
rerese!ti!" this time i! the #uratio! fiel# i! the hea#er of the frame. Whe! statio!s receive the
frame> the9 e-ami!e this #uratio! fiel# value a!# use it as the 4asis for setti!" their
correso!#i!" NA?s. This rocess reserves the me#ium for the se!#i!" statio!.
A! imorta!t asect of the /C: is a ra!#om 4ack off timer that a statio! uses if it #etects a 4us9
me#ium. %f the cha!!el is i! use> the statio! must wait a random erio# of time 4efore attemti!"
to access the me#ium a"ai!. This e!sures that multile statio!s wa!ti!" to se!# #ata #o!It
tra!smit at the same time. The ra!#om #ela9 causes statio!s to wait #iffere!t erio#s of time a!#
avoi#s all of them se!si!" the me#ium at e-actl9 the same time> fi!#i!" the cha!!el i#le>
tra!smitti!"> a!# colli#i!" with each other. The 4ack off timer si"!ifica!tl9 re#uces the !um4er
of collisio!s a!# correso!#i!" retra!smissio!s> eseciall9 whe! the !um4er of active users
With ra#io+4ase# LANs> a tra!smitti!" statio! ca!It liste! for collisio!s while se!#i!" #ata>
mai!l9 4ecause the statio! ca!It have itIs receiver o! while tra!smitti!" the frame. As a result>
the receivi!" statio! !ee#s to se!# a! ack!owle#"eme!t (ACC) if it #etects !o errors i! the
receive# frame. %f the se!#i!" statio! #oes!It receive a! ACC after a secifie# erio# of time> the
se!#i!" statio! will assume that there was a collisio! (or 8: i!terfere!ce) a!# retra!smit the
:or suorti!" time+4ou!#e# #eliver9 of #ata frames> the 032.11 sta!#ar# #efi!es the otio!al
oi!t coor#i!atio! fu!ctio! ('C:) where the access oi!t "ra!ts access to a! i!#ivi#ual statio! to
the me#ium 49 olli!" the statio! #uri!" the co!te!tio! free erio#. Statio!s ca!It tra!smit
frames u!less the access oi!t olls them first. The erio# of time for 'C:+4ase# #ata traffic (if
e!a4le#) occurs alter!atel9 4etwee! co!te!tio! (/C:) erio#s.
The access oi!t olls statio!s accor#i!" to a olli!" list> the! switches to a co!te!tio! erio#
whe! statio!s use /C:. This rocess e!a4les suort for 4oth s9!chro!ous (i.e.> vi#eo
alicatio!s) a!# as9!chro!ous (i.e.> e+mail a!# We4 4rowsi!" alicatio!s) mo#es of oeratio!.
The !ew 6AC access scheme #escri4e# hereafter e!ha!ces the curre!t 032.11 6AC. The 6AC
SA' is ket i#e!tical while the '7B SA' ma9 4e mo#ifie# accor#i!" to the caa4ilities of the
'7B la9er. As show! i! &rror* 8efere!ce source !ot fou!#> the e!ha!ce# 6AC la9er is
co!stitute# of two Co!ver"e!ce su4+la9ers> LLC Co!ver"e!ce Su4+La9er (LLCCS) a!#
Se"me!tatio! a!# 8e+assem4l9 (SA8)> a!# two tra!sfer su4+la9ers> 6AC %!terme#iate Su4+
La9er (6%S) a!# 6AC Lower Su4+la9er (6LS).
The 6AC SA' co!siste!c9 is mai!tai!e# 49 the LLCCS su4+la9er. The 6%S em4e#s the core
tra!sfer fu!ctio! of the 6AC la9er a!# is 4ase# o! short fi-e#+siAe tra!sfer u!its. The 6%S also
i!te"rates the &rror a!# :low Co!trol fu!ctio!s. The SA8 su4+la9er erforms the a#atatio!
4etwee! the varia4le siAe acket rovi#e# 49 the LLCCS a!# the tra!sfer u!its ma!a"e# 49 the
6%S. The 6LS su4+la9er is i! char"e of 4uil#i!" 032.11 comati4le 6'/1s from 6%S tra!sfer
u!it a!# si"!ali!" i!formatio!> a!# #elivers them to the '7B la9er. %! a##itio!> it ca! imleme!t
Sequence Number Assignment
Error and Flow Control
Packet Sequence Number Assignment
Segment Sequence Number Assignment
MP! "eader # C$C
Le"ac9 032.11 &-te!#e# 6AC
MAC Protocol Stack Comparison
Securit9 is o!e of the first co!cer!s of eole #elo9i!" a Wireless LANG the 032.11 committee
has a##resse# the issue 49 rovi#i!" what is calle# W&' (Wire# &;uivale!t 'rivac9)
Authe!ticatio!* A fu!ctio! that #etermi!es whether a Statio! is allowe# to articiate i!
!etwork commu!icatio!. The sta!#ar# %&&& 032.11i is #esi"!e# to rovi#e secure#
commu!icatio! of wireless LAN as #efi!e# 49 all the %&&& 032.11 secificatio!s. %&&& 032.11i
e!ha!ces the W&' (Wireli!e &;uivale!t 'rivac9)G a tech!olo"9 use# for ma!9 9ears for the
WLAN securit9> i! the areas of e!cr9tio!> authe!ticatio! a!# ke9 ma!a"eme!t. %&&& 032.11i is
4ase# o! the Wi+:i 'rotecte# Access (W'A)> which is a ;uick fi- of the W&B weak!esses.
The %&&& 032.11i has the followi!" ke9 como!e!ts*
1. Temoral Ce9 %!te"rit9 'rotocol (TC%')* it is #ata+co!fi#e!tialit9 rotocol a!# it was
#esi"!e# to imrove the securit9 of ro#ucts that were imleme!te# throu"h W&'. TC%' uses a
messa"e i!te"rit9 co#e to e!a4le #evices to authe!ticate that the ackets are comi!" from the
claime# source> this co#e is calle# 6ichael. Also TC%' uses a mi-i!" fu!ctio! to #efeat weak+
ke9 attacks> which e!a4le# attackers to #ecr9t traffic.
2. Cou!ter+6o#e5CBC+6AC 'rotocol (CC6')* a #ata+co!fi#e!tialit9 rotocol that is
reso!si4le for acket authe!ticatio! as well as e!cr9tio!. :or co!fi#e!tialit9> CC6' uses A&S
i! cou!ter mo#e. :or authe!ticatio! a!# i!te"rit9> CC6' uses Ciher Block Chai!i!" 6essa"e
Authe!ticatio! Co#e (CBC+6AC). %! %&&& 032.11i> CC6' uses a 120+4it ke9. CC6' rotects
some fiel#s that are!It e!cr9te#. The a##itio!al arts of the %&&& 032.11 frame that "et
rotecte# are k!ow! as a##itio!al authe!ticatio! #ata (AA/). AA/ i!clu#es the ackets source
a!# #esti!atio! a!# rotects a"ai!st attackers rela9i!" ackets to #iffere!t #esti!atio!s.
3.%&&& 032.1-* offers a! effective framework for authe!ticati!" a!# co!trolli!" user traffic to a
rotecte# !etwork> as well as #9!amicall9 var9i!" e!cr9tio! ke9s. 032.12 ties a rotocol calle#
&A' (&-te!si4le Authe!ticatio! 'rotocol) to 4oth the wire# a!# wireless LAN me#ia a!#
suort multile authe!ticatio! metho#s.
4. &A' e!casulatio! over LANs (&A'OL)D it is the ke9 rotocol i! %&&& 032.1- for ke9
e-cha!"e. Two mai! &A'OL+ke9 e-cha!"es are #efi!e# i! %&&& 032.11i. The first is referre# to
as the 4+wa9 ha!#shake a!# the seco!# is the "rou ke9 ha!#shake.
Because %&&& 032.11i has more tha! o!e #ata+co!fi#e!tialit9 rotocol> %&&& 032.11i rovi#es
a! al"orithm for the %&&& 032.11i clie!t car# a!# access oi!t to !e"otiate which rotocol to use
#uri!" secific traffic circumsta!ces a!# to #iscover a!9 u!k!ow! securit9 arameters.
Protocol Structure & %EEE '()*++i: ,-AN Security Standards
Pre#entin- Access to )et'ork :esources
This is #o!e 49 the use of a! Authe!ticatio! mecha!ism where a statio! !ee#s to rove
k!owle#"e of the curre!t ke9G this is ver9 similar to the Wire# LAN rivac9> o! the se!se that a!
i!tru#er !ee#s to e!ter the remises (49 usi!" a h9sical ke9) i! or#er to co!!ect his workstatio!
to the wire# LAN.
&aves#roi!" is reve!te# 49 the use of the W&' al"orithm> which is a 'seu#o 8a!#om
Num4er @e!erator ('8N@)> i!itialiAe# 49 a share# secret ke9. This '8N@ oututs a ke9
se;ue!ce of seu#o+ra!#om 4its e;ual i! le!"th to the lar"est ossi4le acket> which is com4i!e#
with the out"oi!"5i!comi!" acket ro#uci!" the acket tra!smitte# i! the air.
The W&' al"orithm is a simle al"orithm 4ase# o! 8SAVs 8C4 al"orithm> which has the
8easo!a4le stro!"* Brute+force attack to this al"orithm is #ifficult 4ecause of the fact that ever9
frame is se!t with a! %!itialiAatio! ?ector> which restarts the '8N@ for each frame.
Self S9!chro!iAi!"* The al"orithm s9!chro!iAe# a"ai! for each messa"e> this is !ee#e# i! or#er
to work o! a co!!ectio!less e!viro!me!t> where ackets ma9 "et lost (as a!9 LAN).
The %&&& (#efi!e) 032.11 sta!#ar# i!clu#es a commo! 6e#ium Access Co!trol (6AC) La9er>
which #efi!es rotocols that "over! the oeratio! of the wireless LAN. %! a##itio!> 032.11
comrise several alter!ative h9sical la9ers that secif9 the tra!smissio! a!# recetio! of 032.11
The ,hysical layer basics
To k!ow the h9sical la9er termi!olo"9 we !ee# to u!#ersta!# the esse!tial i!tricacies of
@:SC is a mo#ulatio! scheme i! which the #ata are first filtere# 49 a @aussia! filter i! the
Base4a!#> a!# the! mo#ulate# with a simle fre;ue!c9 mo#ulatio!. 2 a!# 4 4it rerese!t the
!um4er of fre;ue!c9 offsets use# to rerese!t #ata s9m4ols of o!e a!# two 4its> resectivel9.
/B'SC is hase mo#ulatio! usi!" two #isti!ct carrier hases for #ata si"!ali!" rovi#i!" o!e 4it
/N'SC is a t9e of hase mo#ulatio! usi!" two airs of #isti!ct carrier hases> i! ;ua#rature> to
si"!al two 4its er s9m4ol. The #iffere!tial characteristic of the mo#ulatio! schemes i!#icates
the use of the #iffere!ce i! hase from the last cha!"e or s9m4ol to #etermi!e the curre!t
s9m4olIs value> rather tha! a!9 a4solute measureme!ts of the hase cha!"e.
Both the :7SS a!# /SSS mo#es are secifie# for oeratio! i! the 2.4 @7A i!#ustrial> scie!tific
a!# me#ical (%S6) 4a!#> which has sometimes 4ee! =oki!"l9 referre# to as the i!terfere!ce
suressio! is ma!#ator9 4a!# 4ecause it is heavil9 use# 49 various electro!ic ro#ucts. The
thir# h9sical la9er alter!ative is a! i!frare# s9stem usi!" !ear+visi4le li"ht i! the 0$3 !m to <$3
!m ra!"e as the tra!smissio! me#ium.
At the forefro!t of the !ew WLAN otio!s that will e!a4le much hi"her #ata rates are two
suleme!ts to the %&&& 032.11 sta!#ar#* 032.114 a!# 032.11a> as well as a &uroea!
Telecommu!icatio!s Sta!#ar#s %!stitute (&TS%) sta!#ar#> 7i"h 'erforma!ce LAN
(7%'&8LAN5%%). Both 032.11 a!# 7%'&8LAN5%% have similar h9sical la9er characteristics
oerati!" i! the $ @7A 4a!# a!# use the mo#ulatio! scheme ortho"o!al fre;ue!c9 #ivisio!
multile-i!" (O:/6)> 4ut the 6AC la9ers are co!si#era4l9 #iffere!t. The focus here> however>
will 4e to comare the h9sical la9er characteristics of 032.11a a!# 032.114. With
7%'&8LAN5%% shari!" several of the same h9sical roerties as 032.11a> ma!9 of the same
issues will al9.
A!other sta!#ar# that warra!ts me!tio! i! this co!te-t is %&&& 032.11". With a ruli!" from the
:e#eral Commu!icatio!s Commissio! that will !ow allow O:/6 #i"ital tra!smissio!
tech!olo"9 to oerate i! the %S6 4a!# a!# the romise of i!teroera4ilit9 with a lar"e i!stalle#
4ase of 032.114 ro#ucts> the 032.11" e-te!sio! to the sta!#ar# 4e"i!s to "ar!er the atte!tio! of
WLAN e;uime!t rovi#ers. Althou"h !ot #etaile# here> it will offer #ata rates e;ual to or
e-cee#i!" 22 645s with ro#ucts availa4le late i! 2332.
&ach of the five ermitte# tra!smissio! tech!i;ues makes it ossi4le to se!# a 6AC frame from
o!e statio! to a!other. The9 #iffer> however> i! the tech!olo"9 use# a!# see# achieva4le. The
i!frare# otio! uses #iffuse# (i.e.> !ot li!e of si"ht) tra!smissio! at 3.0$ or 3.<$ micro!s. Two
see#s are ermitte#* 1 64s a!# 2 64s. At 1 64s> a! e!co#i!" scheme is use# i! which a
"rou of 4 4its is e!co#e# as a 1,+4it co#ewor# co!tai!i!" fiftee! 3s a!# a si!"le 1> usi!" what is
calle# @ra9 co#e. This co#e has the roert9 that a small error i! time s9!chro!iAatio! lea#s to
o!l9 a si!"le 4it error i! the outut. At 2 64s> the e!co#i!" takes 2 4its a!# ro#uces a 4+4it
co#ewor#> also with o!l9 a si!"le 1> that is o!e of 3331> 3313> 3133> or 1333. %!frare# si"!als
ca!!ot e!etrate walls> so cells i! #iffere!t rooms are well isolate# from each other.
Nevertheless> #ue to the low 4a!#wi#th (a!# the fact that su!li"ht swams i!frare# si"!als)> this
is !ot a oular otio!.
As with other 032.11 'h9sical la9ers> 032.114 i!clu#es 'h9sical La9er Co!ver"e!ce 'roce#ure
('LC') a!# 'h9sical 6e#ium /ee!#e!t ('6/) su4+la9ers. These are somewhat sohisticate#
terms that the sta!#ar# uses to #ivi#e the ma=or fu!ctio!s that occur withi! the 'h9sical La9er.
The 'LC' reares 032.11 frames for tra!smissio! a!# #irects the '6/ to actuall9 tra!smit
si"!als> cha!"e ra#io cha!!els> receive si"!als> a!# so o!.
PCP ;ra+e ;ields
The 'LC' takes each 032.11 frame that a statio! wishes to tra!smit a!# forms what the 032.11
sta!#ar# refers to as a 'LC' rotocol #ata u!it (''/1). The resulti!" ''/1 i!clu#es the
followi!" fiel#s i! a##itio! to the frame fiel#s imose# 49 the 6AC La9er*
S9!c. This fiel# co!sists of alter!ati!" 3s a!# 1s> alerti!" the receiver that a receiva4le si"!al is
rese!t. The receiver 4e"i!s s9!chro!iAi!" with the i!comi!" si"!al after #etecti!" the S9!c.
Start :rame /elimiter. This fiel# is alwa9s 1111331113133333 a!# #efi!es the 4e"i!!i!" of a
Si"!al. This fiel# i#e!tifies the #ata rate of the 032.11 frame> with its 4i!ar9 value e;ual to the
#ata rate #ivi#e# 49 133C4s. :or e-amle> the fiel# co!tai!s the value of 33331313 for 164s>
33313133 for 264s> a!# so o!. The 'LC' fiel#s> however> are alwa9s se!t at the lowest rate>
which is 164s. This e!sures that the receiver is i!itiall9 uses the correct #emo#ulatio!
mecha!ism> which cha!"es with #iffere!t #ata rates.
Service. This fiel# is alwa9s set to 33333333 a!# the 032.11 sta!#ar# reserves it for future use.
Le!"th. This fiel# rerese!ts the !um4er of microseco!#s that it takes to tra!smit the co!te!ts of
the ''/1> a!# the receiver uses this i!formatio! to #etermi!e the e!# of the frame.
:rame Check Se;ue!ce. %! or#er to #etect ossi4le errors i! the 'h9sical La9er hea#er> the
sta!#ar# #efi!es this fiel# for co!tai!i!" 1,+4it c9clic re#u!#a!c9 check (C8C) result. The 6AC
La9er also erforms error #etectio! fu!ctio!s o! the ''/1 co!te!ts as well.
'S/1. The 'S/1> which sta!#s for 'h9sical La9er Service /ata 1!it> is a fa!c9 !ame that
rerese!ts the co!te!ts of the ''/1 (i.e.> the actual 032.11 frame 4ei!" se!t).
/o!It e-ect to see the h9sical la9er fiel#s with 032.11 a!al9Aers from Air6a"!et a!#
Wil#ackets> however. The 032.11 ra#io car# removes these fiel#s 4efore the resulti!" #ata is
rocesse# 49 the 6AC La9er a!# offere# to the a!al9Aer for viewi!".
Ne-t> we come to 78+/SSS (7i"h 8ate /irect Se;ue!ce Srea# Sectrum)> a!other srea#
sectrum tech!i;ue> which uses 11 millio! chis5sec to achieve 11 64s i! the 2.4+@7A 4a!#. %t
is calle# 032.114 4ut is !ot a follow+u to 032.11a. %! fact> its sta!#ar# was arove# first a!# it
"ot to market first. /ata rates suorte# 49 032.114 are 1> 2> $.$> a!# 11 64s. The two slow
rates ru! at 1 64au#> with 1 a!# 2 4its er 4au#> resectivel9> usi!" hase shift mo#ulatio! (for
comati4ilit9 with /SSS). The two faster rates ru! at 1.3.$ 64au#> with 4 a!# 0 4its er 4au#>
resectivel9> usi!" Walsh57a#amar# co#es. The #ata rate ma9 4e #9!amicall9 a#ate# #uri!"
oeratio! to achieve the otimum see# S&C. 4.4 W%8&L&SS LANS 2<$ ossi4le u!#er curre!t
co!#itio!s of loa# a!# !oise. %! ractice> the oerati!" see# of 032.114 is !earl9 alwa9s 11
64s. Althou"h 032.114 is slower tha! 032.11a> its ra!"e is a4out . times "reater> which is more
imorta!t i! ma!9 situatio!s. A! e!ha!ce# versio! of 032.114> 032.11"> was arove# 49 %&&&
i! Novem4er 2331 after much oliticki!" a4out whose ate!te# tech!olo"9 it woul# use. %t uses
the O:/6 mo#ulatio! metho# of 032.11a 4ut oerates i! the !arrow 2.4+ @7A %S6 4a!# alo!"
with 032.114. %! theor9 it ca! oerate at u to $4 6Bs. %t is !ot 9et clear whether this see# will
4e realiAe# i! ractice. What it #oes mea! is that the 032.11 committee has ro#uce# three
#iffere!t hi"h+see# wireless LANs* 032.11a> 032.114> a!# 032.11" (!ot to me!tio! three low+
see# wireless LANs). O!e ca! le"itimatel9 ask if this is a "oo# thi!" for a sta!#ar#s committee
The 032.11 h9sical la9er ('7B) is the i!terface 4etwee! the 6AC a!# the wireless me#ia
where frames are tra!smitte# a!# receive#. The '7B rovi#es three fu!ctio!s. :irst> the '7B
rovi#es a! i!terface to e-cha!"e frames with the uer 6AC la9er for tra!smissio! a!#
recetio! of #ata. Seco!#l9> the '7B uses si"!al carrier a!# srea# sectrum mo#ulatio! to
tra!smit #ata frames over the me#ia. Thir#l9> the '7B rovi#es a carrier se!se i!#icatio! 4ack to
the 6AC to verif9 activit9 o! the me#ia.
032.11 rovi#es three #iffere!t '7B #efi!itio!s* Both Frequency "opping Spread Spectrum
(:7SS) a!# irect Sequence Spread Spectrum (/SSS) suort 1 a!# 2 64s #ata rates. A!
e-te!sio! to the 032.11 architecture (032.11a) #efi!es #iffere!t multile-i!" tech!i;ues that ca!
achieve #ata rates u to $4 64s. A!other e-te!sio! to the sta!#ar# (032.114) #efi!es 11 64s
a!# $.$ 64s #ata rates (i! a##itio! to the 1 a!# 264s rates) utiliAi!" a! e-te!sio! to /SSS
calle# 7i"h 8ate /SSS (785/SSS). 032.114 also #efi!es a rate shifti!" tech!i;ue where 11
64s !etworks ma9 fall 4ack to $.$ 64s> 2 64s> or 1 6s u!#er !ois9 co!#itio!s or to i!ter+
oerate with le"ac9 032.11 '7B la9ers.
The %!frare# '7B utiliAes i!frare# li"ht to tra!smit 4i!ar9 #ata either at 1 64s (4asic access
rate) or 2 64s (e!ha!ce# access rate) usi!" a secific mo#ulatio! tech!i;ue for each. :or 1
64s> the i!frare# '7B uses a 1,+ulse ositio! mo#ulatio! (''6). The co!cet of ''6 is to
var9 the ositio! of a ulse to rerese!t #iffere!t 4i!ar9 s9m4ols. %!frare# tra!smissio! at 2
64s utiliAes a 4 ''6 mo#ulatio! tech!i;ue.
Srea# sectrum is a tech!i;ue tra#i!" 4a!#wi#th for relia4ilit9. The "oal is to use more
4a!#wi#th tha! the s9stem reall9 !ee#s for tra!smissio! to re#uce the imact of localiAe#
i!terfere!ce o! the me#ia. Srea# sectrum srea#s the tra!smitte# 4a!#wi#th of the resulti!"
si"!al> re#uci!" the eak ower 4ut keei!" total ower the same.
;re<uency 9o,,in- S,read S,ectru+ /;9SS0
%! :7SS the total fre;ue!c9 4a!# is slit i!to a !um4er of cha!!els. The 4roa#cast #ata is srea#
across the e!tire fre;ue!c9 4a!# 49 hoi!" 4etwee! the cha!!els i! a seu#o ra!#om fashio!.
:re;ue!c9+hoi!" srea# sectrum (:7SS) is a srea#+sectrum metho# of tra!smitti!" ra#io
si"!als 49 rai#l9 switchi!" a carrier amo!" ma!9 fre;ue!c9 cha!!els> usi!" a seu#ora!#om
se;ue!ce k!ow! to 4oth tra!smitter a!# receiver.
A srea#+sectrum tra!smissio! offers three mai! a#va!ta"es over a fi-e#+fre;ue!c9
Srea#+sectrum si"!als are hi"hl9 resista!t to !oise a!# i!terfere!ce. The rocess of re+
collecti!" a srea# si"!al srea#s out !oise a!# i!terfere!ce> causi!" them to rece#e i!to the
Srea#+sectrum si"!als are #ifficult to i!tercet. A :re;ue!c9+7o srea#+sectrum si"!al
sou!#s like a mome!tar9 !oise 4urst or siml9 a! i!crease i! the 4ack"rou!# !oise for short
:re;ue!c9+7o co#es o! a!9 !arrow4a!# receiver e-cet a :re;ue!c9+7o srea#+sectrum
receiver usi!" the e-act same cha!!el se;ue!ce as was use# 49 the tra!smitter.
Srea#+sectrum tra!smissio!s ca! share a fre;ue!c9 4a!# with ma!9 t9es of co!ve!tio!al
tra!smissio!s with mi!imal i!terfere!ce. The srea#+sectrum si"!als a## mi!imal !oise to the
!arrow+fre;ue!c9 commu!icatio!s> a!# vice versa. As a result> 4a!#wi#th ca! 4e utiliAe# more
:re;ue!c9 7oi!" utiliAes a set of !arrow cha!!els a!# MhosM throu"h all of them i! a
re#etermi!e# se;ue!ce. :or e-amle> the 2.4 @7A fre;ue!c9 4a!# is #ivi#e# i!to .3 cha!!els
of 1 67A each. &ver9 23 to 433 msec the s9stem MhosM to a !ew cha!!el followi!" a
re#etermi!e# c9clic atter!.
The 032.11 :re;ue!c9 7oi!" Srea# Sectrum (:7SS) '7B uses the 2.4 @7A ra#io
fre;ue!c9 4a!#> oerati!" with at 1 or 2 64s #ata rate.
:7SS (:re;ue!c9 7oi!" Srea# Sectrum) uses .< cha!!els> each 1+ 67A wi#e> starti!" at
the low e!# of the 2.4+@7A %S6 4a!#. A seu#ora!#om !um4er "e!erator is use# to ro#uce the
se;ue!ce of fre;ue!cies hoe# to. As lo!" as all statio!s use the same see# to the
seu#ora!#om !um4er "e!erator a!# sta9 s9!chro!iAe# i! time> the9 will ho to the same
fre;ue!cies simulta!eousl9. The amou!t of time se!t at each fre;ue!c9> the #well time> is a!
a#=usta4le arameter> 4ut must 4e less tha! 433 msec. :7SSE ra!#omiAatio! rovi#es a fair wa9
to allocate sectrum i! the u!re"ulate# %S6 4a!#. %t also rovi#es a mo#icum of securit9 si!ce
a! i!tru#er who #oes !ot k!ow the hoi!" se;ue!ce or #well time ca!!ot eaves#ro o!
tra!smissio!s. Over lo!"er #ista!ces> multiath fa#i!" ca! 4e a! issue> a!# :7SS offers "oo#
resista!ce to it. %t is also relativel9 i!se!sitive to ra#io i!terfere!ce> which makes it oular for
4uil#i!"+to+4uil#i!" li!ks. %ts mai! #isa#va!ta"e is its low 4a!#wi#th. The thir# mo#ulatio!
:re;ue!c9 hoi!" relies o! fre;ue!c9 #iversit9 to com4at i!terfere!ce. This is accomlishe#
49 multile fre;ue!c9> co#e selecte#> :SC. Basicall9> the i!comi!" #i"ital stream is shifte# i!
fre;ue!c9 49 a! amou!t #etermi!e# 49 a co#e that srea#s the si"!al ower over a wi#e
4a!#wi#th. %! comariso! to 4i!ar9 :SC> which has o!l9 two ossi4le fre;ue!cies> :7SS ma9
have 2O13W23 or more.
The :7SS tra!smitter is a seu#o+!oise 'N co#e co!trolle# fre;ue!c9 s9!thesiAer. The
i!sta!ta!eous fre;ue!c9 outut of the tra!smitter =ums from o!e value to a!other 4ase# o! the
seu#o+ra!#om i!ut from the co#e "e!erator. ?ar9i!" the i!sta!ta!eous fre;ue!c9 results i! a!
outut sectrum that is effectivel9 srea# over the ra!"e of fre;ue!cies "e!erate#.
:i".1 :7SS Sectrum
%! this s9stem> the !um4er of #iscrete fre;ue!cies #etermi!es the 4a!#wi#th of the
s9stem. 7e!ce> the rocess "ai! is #irectl9 #ee!#e!t o! the !um4er of availa4le fre;ue!c9
choices for a "ive! i!formatio! rate.
A!other imorta!t factor i! :7SS s9stems is the rate at which the hos occur. The mi!imum
time re;uire# to cha!"e fre;ue!cies is #ee!#e!t o! the i!formatio! 4it rate> the amou!t of
re#u!#a!c9 use#> a!# the #ista!ce to the !earest i!terfere!ce source.
4irect Se<uence S,read S,ectru+ /4SSS0
/irect Se;ue!ce Srea# Sectrum is 4ase# o! the multil9i!" of the 4ase4a!# si"!al #ata with a
4roa#4a!# srea#i!" co#e. The result is terme# the chi rate. The characteristics of the
4roa#4a!# srea#i!" co#e are that of seu#ora!#om !oise. Co!se;ue!tl9 the receiver
s9!chro!iAe# to the co#e will o4tai! the !arrow4a!# si"!al. All other receivers will see the
srea# si"!al as white or colore# !oise.
%! co!trast> fre;ue!c9+hoi!" srea# sectrum seu#o+ra!#oml9 retu!es the carrier> i!stea# of
a##i!" seu#o+ra!#om !oise to the #ata> which results i! a u!iform fre;ue!c9 #istri4utio! whose
wi#th is #etermi!e# 49 the outut ra!"e of the seu#o+ra!#om !um4er "e!erator.
%! telecommu!icatio!s> #irect+se;ue!ce srea# sectrum is a mo#ulatio! tech!i;ue where the
tra!smitte# si"!al takes u more 4a!#wi#th tha! the i!formatio! si"!al that is 4ei!" mo#ulate#>
which is the reaso! that it is calle# srea# sectrum. /SSS has the followi!" features*
for "e!erati!" srea#+sectrum tra!smissio!s 49 hase+mo#ulati!" a si!e wave seu#o ra!#oml9
with a co!ti!uous stri!" of seu#o !oise co#e s9m4ols> each of #uratio! much smaller tha! a 4it.
A si"!al structuri!" tech!i;ue utiliAi!" a #i"ital co#e se;ue!ce ('N Se;ue!ces) havi!" a chi
rate much hi"her tha! the i!formatio! si"!al 4it rate. &ach i!formatio! 4it of a #i"ital si"!al is
tra!smitte# as a seu#ora!#om se;ue!ce of chis.
The ri!cile of irect Sequence is to srea# a si"!al o! a lar"er fre;ue!c9 4a!# 49 multile-i!"
it with a si"!ature or co#e to mi!imiAe localiAe# i!terfere!ce a!# 4ack"rou!# !oise. To srea#
the si"!al> each 4it is mo#ulate# 49 a co#e. %! the receiver> the ori"i!al si"!al is recovere# 49
receivi!" the whole srea# cha!!el a!# #emo#ulati!" with the same co#e use# 49 the
tra!smitter. The 032.11 irect Sequence Spread Spectrum (/SSS) '7B also uses the 2.4 @7A
ra#io fre;ue!c9 4a!#.
%t is also a art of the 032.11 4 a!# " sta!#ar#s. Note that i! the ori"i!al 032.11 sta!#ar#> either
:7SS or /SSS ma9 4e use#.
/SSS (/irect Se;ue!ce Srea# Sectrum) is also restricte# to 1 or 2 64s. The scheme use# has
some similarities to the C/6A s9stem> 4ut #iffers i! other wa9s. &ach 4it is tra!smitte# as 11
chis> usi!" what is calle# a Barker se;ue!ce. %t uses hase shift mo#ulatio! at 1 64au#>
tra!smitti!" 1 4it er 4au# whe! oerati!" at 1 64s a!# 2 4its er 4au# whe! oerati!" at 2
64s. 032.114 uses /SSS to #iserse the #ata frame si"!al over a relativel9 wi#e
(aro-imatel9 3367A) ortio! of the 2.4@7A fre;ue!c9 4a!#. This results i! "reater immu!it9
to ra#io fre;ue!c9 (8:) i!terfere!ce as comare# to !arrow4a!# si"!ali!"> which is wh9 the
:e#eral Commu!icatio!s Commissio! (:CC) (#efi!e) #eems the oeratio! of srea# sectrum
s9stems as lice!se free. :or 9ears> the :CC re;uire# all wireless commu!icatio!s e;uime!t
oerati!" i! the %S6 4a!#s i! the 1.S. to use srea# sectrum> 4ut i! 6a9 2332> that rule was
#roe# as !ew tech!olo"ies emer"e#. The first of the hi"h+see# wireless LANs> 032.11a> uses
O:/6 (Ortho"o!al :re;ue!c9 /ivisio! 6ultile-i!") to #eliver u to $4 64s i! the wi#er $+
@7A %S6 4a!#. As the term :/6 su""ests> #iffere!t fre;ue!cies are use#F$2 of them> 40 for
#ata a!# 4 for s9!chro!iAatio!F!ot u!like A/SL. Si!ce tra!smissio!s are rese!t o! multile
fre;ue!cies at the same time> this tech!i;ue is co!si#ere# a form of srea# sectrum> 4ut
#iffere!t from 4oth C/6A a!# :7SS. Slitti!" the si"!al i!to ma!9 !arrow 4a!#s has some ke9
a#va!ta"es over usi!" a si!"le wi#e 4a!#> i!clu#i!" 4etter immu!it9 to !arrow4a!# i!terfere!ce
a!# the ossi4ilit9 of usi!" !o!co!ti"uous 4a!#s. A comle- e!co#i!" s9stem is use#> 4ase# o!
hase+shift mo#ulatio! for see#s u to 10 64s a!# o! NA6 a4ove that. At $4 64s> 21, #ata
4its are e!co#e# i!to 200+4it s9m4ols. 'art of the motivatio! for O:/6 is comati4ilit9 with the
&uroea! 7ierLAN52 s9stem (/oufe-i et al.> 2332). The tech!i;ue has a "oo# sectrum
efficie!c9 i! terms of 4its57A a!# "oo# immu!it9 to multiath fa#i!". This is ro4a4l9 the most
wi#el9 reco"!iAe# form of srea# sectrum. The /SSS rocess is erforme# 49 effectivel9
multil9i!" a! 8: carrier a!# a seu#o+!oise ('N) #i"ital si"!al. :irst the 'N co#e is mo#ulate#
o!to the i!formatio! si"!al usi!" o!e of several mo#ulatio! tech!i;ues (e". B'SC> N'SC>
etc). The!> a #ou4l9 4ala!ce# mi-er is use# to multil9 the 8: carrier a!# 'N mo#ulate#
i!formatio! si"!al. This rocess causes the 8: si"!al to 4e relace# with a ver9 wi#e 4a!#wi#th
si"!al with the sectral e;uivale!t of a !oise si"!al. The #emo#ulatio! rocess (for the B'SC
case) is the! siml9 the mi-i!"5multil9i!" of the same 'N mo#ulate# carrier with the i!comi!"
8: si"!al. The outut is a si"!al that is a ma-imum whe! the two si"!als e-actl9 e;ual o!e
a!other or are Mcorrelate#M. The correlate# si"!al is the! filtere# a!# se!t to a B'SC
The si"!als "e!erate# with this tech!i;ue aear as !oise i! the fre;ue!c9 #omai!. The wi#e
4a!#wi#th rovi#e# 49 the 'N co#e allows the si"!al ower to #ro 4elow the !oise threshol#
without loss of i!formatio!. The sectral co!te!t of a! SS si"!al is show! i! :i". 1. Note that
this is =ust the sectrum of a B'SC si"!al with a (si! - 5 -) 2 form.
:i". 1 B'SC /SSS Sectrum
The 4a!#wi#th i! /SSS s9stems is ofte! take! as the !ull+to+!ull 4a!#wi#th of the mai! lo4e of
the ower sectral #e!sit9 lot (i!#icate# as 28c i! :i". 1). The half ower 4a!#wi#th of this
lo4e is 1.2 8c> where 8c is the chi rate. Therefore> the 4a!#wi#th of a /SSS s9stem is a #irect
fu!ctio! of the chi rateG secificall9 28c58%N:O. This is =ust a! e-te!sio! of the revious
e;uatio! for rocess "ai!. %t shoul# 4e !ote# that the ower co!tai!e# i! the mai! lo4e
comrises <3 erce!t of the total ower. This allows a !arrower 8: 4a!#wi#th to accommo#ate
the receive# si"!al with the effect of rou!#i!" the receive# ulses i! the time #omai!.
O!e feature of /SSS is that N'SC ma9 4e use# to i!crease the #ata rate. This i!crease of a
factor of two 4its er s9m4ol of tra!smitte# i!formatio! over B'SC causes a! e;uivale!t
re#uctio! i! the availa4le rocess "ai!. The rocess "ai! is re#uce# 4ecause for a "ive! chi
rate> the 4a!#wi#th (which sets the rocess "ai!) is halve# #ue to the two+fol# i!crease i!
i!formatio! tra!sfer. The result is that s9stems i! a sectrall9 ;uiet e!viro!me!t 4e!efit from the
ossi4le i!crease i! #ata tra!sfer rate.
The mo#ulator co!verts the srea# 4i!ar9 si"!al i!to a! a!alo" waveform throu"h the use of
#iffere!t mo#ulatio! t9es> #ee!#i!" o! which #ata rate is chose!. :or e-amle with 164s
oeratio!> the '6/ uses #iffere!tial 4i!ar9 hase shift ke9i!" (/B'SC). This is!It reall9 as
comle- as it sou!#s. The mo#ulator merel9 shifts the hase of the ce!ter tra!smit fre;ue!c9 to
#isti!"uish a 4i!ar9 1 from a 4i!ar9 3 withi! the #ata stream.
:or 264s tra!smissio!> the '6/ uses #iffere!tial ;ua#rature hase shift ke9i!" (/N'SC)>
which is similar to /B'SC e-cet that there are four ossi4le hase shifts that rerese!ts ever9
two #ata 4its. This is a clever rocess that e!a4les the #ata stream to 4e se!t at 264s while
usi!" the same amou!t of 4a!#wi#th as the o!e se!t at 164s. The mo#ulator uses similar
metho#s for the hi"her> $.$64s a!# 1164s #ata rates.
The tra!smitterIs mo#ulator tra!slates the srea# si"!al i!to a! a!alo" form with a ce!ter
fre;ue!c9 correso!#i!" to the ra#io cha!!el chose! 49 the user. The followi!" i#e!tifies the
ce!ter fre;ue!c9 of each cha!!el*
Cha!!el :re;ue!c9 (@7A)
?arious cou!tries limit the use of these cha!!els. :or e-amle> the 1.S. o!l9 allows the use of
cha!!els 1 throu"h 11> a!# the 1.C. ca! use cha!!els 1 throu"h 13. Paa!> however> authoriAes
the use all 14 cha!!els. This comlicates matters whe! #esi"!i!" i!ter!atio!al u4lic wireless
LANs. %! that case> 9ou !ee# to choose cha!!els with the least commo! #e!omi!ator.
After 8: amlificatio! takes lace 4ase# o! the tra!smit ower 9ouIve chose! (133mW
ma-imum for the 1.S.)> the tra!smitter oututs the mo#ulate# /SSS si"!al to the a!te!!a i!
or#er to roa"ate the si"!al to the #esti!atio!. The tri i! route to the #esti!atio! will
si"!ifica!tl9 atte!uate (#efi!e) the si"!al> 4ut the receiver at the #esti!atio! will #etect the
i!comi!" 'h9sical La9er hea#er a!# reverse (#emo#ulate a!# #isrea#) the rocess imleme!te#
49 the tra!smitter
The I... 802.11a
While 032.11a was arove# i! Setem4er 1<<<> !ew ro#uct #evelome!t has rocee#e# much
more slowl9 tha! 032.114. This is #ue to the cost a!# comle-it9 of imleme!tatio!. This
sta!#ar# uses 333 67A of 4a!#wi#th i! the $ @7A u!lice!se# !atio!al i!formatio! i!frastructure
(1N%%) 4a!#. The sectrum is #ivi#e# i!to three #omai!s> each havi!" restrictio!s imose# o!
the ma-imum allowe# outut ower (see :i"ure 1). The first 133 67A i! the lower fre;ue!c9
ortio! is restricte# to a ma-imum ower outut of $3 mW. The seco!# 133 67A has a hi"her
2$3 mW ma-imum> while the thir# 133 67A> which is mai!l9 i!te!#e# for out#oor alicatio!s>
has a ma-imum of 1.3 W ower outut.
O:/6 oerates 49 #ivi#i!" the tra!smitte# #ata i!to multile arallel 4it streams> each with
lower relative 4it rates a!# mo#ulati!" searate !arrow4a!# carriers> referre# to as su4+carriers.
The su4+carriers are ortho"o!al> so each ca! 4e receive# without i!terfere!ce from a!other.
032.11a secifies ei"ht !o!+overlai!" 23 67A cha!!els i! the lower two 4a!#sG each of these
are #ivi#e# i!to $2 su4+carriers (four of which carr9 ilot #ata) of 333+k7A 4a!#wi#th each. :our
!o!+overlai!" 23 67A cha!!els are secifie# i! the uer 4a!#. The receiver rocesses the $2
i!#ivi#ual 4it streams> reco!structi!" the ori"i!al hi"h+rate #ata stream. :our comle-
mo#ulatio! metho#s are emlo9e#> #ee!#i!" o! the #ata rate that ca! 4e suorte# 49 cha!!el
co!#itio!s 4etwee! the tra!smitter a!# receiver. These i!clu#e B'SC> N'SC> 1,+NA6> a!# ,4+
Nua#rature amlitu#e mo#ulatio! is a comle- mo#ulatio! metho# where #ata are carrie# i!
s9m4ols rerese!te# 49 the hase a!# amlitu#e of the mo#ulate# carrier. 1,+NA6 has 1,
s9m4ols. &ach rerese!ts four #ata 4its. ,4+NA6 has 1, s9m4ols with each rerese!ti!" four
B'SC mo#ulatio! is alwa9s use# o! the four ilot su4+carriers. Althou"h it a##s a #e"ree of
comlicatio! to the Base4a!# rocessi!"> 032.11a i!clu#es forwar# error correctio! (:&C) as
art of the secificatio!. :&C> which #oes !ot e-ist withi! 032.114> e!a4les the receiver to
i#e!tif9 a!# correct errors ma#e #uri!" tra!smissio! 49 se!#i!" a##itio!al #ata alo!" with the
rimar9 tra!smissio!. This !earl9 elimi!ates the !ee# for retra!smissio!s whe! acket errors are
#etecte#. The #ata rates availa4le i! 032.11a are !ote# i! Ta4le 2> to"ether with the t9e of
mo#ulatio! a!# the co#i!" rate.
032.11a ro#ucts are e-ecte# to 4e"i! arrivi!" i! the first half of 2332. Some of the coma!ies
#eveloi!" chiset solutio!s for 032.11a are touti!" the availa4ilit9 of oeratio!al mo#es that
e-cee# the $4 645s state# i! the secificatio!. Of course> 4ecause faster #ata rates are out of the
secificatio!Is scoe> the9 re;uire the use of e;uime!t from a si!"le source throu"hout the
Co!si#eri!" the comosite waveform resulti!" from the com4i!atio! of $2 su4+carriers> the
format re;uires more li!earit9 i! the amlifiers 4ecause of the hi"her eak+to+avera"e ower
ratio of the tra!smitte# O:/6 si"!al. %! a##itio!> 4etter hase !oise erforma!ce is re;uire#
4ecause of the closel9 sace#> overlai!" carriers. These issues a## to the imleme!tatio! cost
of 032.11a ro#ucts. Alicatio!+secific measureme!t tools ai# i! the #esi"! a!#
trou4leshooti!" of O:/6 si"!als a!# s9stems.
/esi"! of #evices usi!" 032.11a with O:/6 si"!als a!# oerati!" at $ @7A will 4ri!" !ew
challe!"es i! testi!"> articularl9 4ecause the #ata rate will 4e i!creasi!" 49 a factor of five a!#
usi!" the same 4a!#wi#th (23 67A) to #o it. The hi"h eak+to+avera"e ower ratio
rerese!tative of multicarrier O:/6 si"!als #ictates the !ee# for hi"hl9 li!ear a!# efficie!t
amlifiers> as well as a metho# to characteriAe them.
Tra!smitte# si"!als such as O:/6> which #o !ot have a co!sta!t ower e!veloe> are !ot well+
characteriAe# 49 eak+to+avera"e ower ratio. This metric is !ot useful> as the true eak ower
ma9 !ot occur ofte!. %t is usuall9 more mea!i!"ful for O:/6 si"!als to associate a erce!ta"e
ro4a4ilit9 with a ower level.
A more mea!i!"ful metho# for viewi!" O:/6 si"!al ower characteristics uses the
comleme!tar9 cumulative #istri4utio! fu!ctio! (CC/:). This metric li!ks a erce!ta"e
ro4a4ilit9 to a ower level. %! this measureme!t> a! i!strume!t with time+"ati!" caa4ilit9 is
use# to select o!l9 the active ortio! of the 4urst (see :i"ure 2 lower trace).
%f time "ati!" were !ot use#> the erio#s whe! the 4urst is off woul# re#uce the avera"e ower
calculatio!. The CC/:> which is siml9 the more commo! cumulative #istri4utio! fu!ctio!
(C/:) su4tracte# from 1.3> shows the !um4er of #eci4els a4ove the avera"e ower o! the
horiAo!tal a-is> a!# erce!t ro4a4ilit9 o! vertical a-is (see :i"ure 2 uer trace). A CC/:
measureme!t woul# 4e ma#e over several 4ursts to imrove the accurac9 of the measureme!t.
The I... 8102.11b
032.114> which was arove# 49 the %&&& i! 1<<<> is a! e-te!sio! of the 032.11 /SSS s9stem
reviousl9 me!tio!e# a!# suorts hi"her $.$ a!# 11 645s a9loa# #ata rates i! a##itio! to the
ori"i!al 1 a!# 2 645s rates. 'ro#ucts are !ow wi#el9 availa4le> a!# the i!stalle# 4ase of s9stems
is "rowi!" rai#l9. 032.114 also oerates i! the hi"hl9 oulate# 2.4 @7A %S6 4a!# (2.43 to
2.403$ @7A)> which rovi#es o!l9 03 67A of sectrum to accommo#ate a variet9 of other
ra#iati!" ro#ucts> i!clu#i!" cor#less ho!es> microwave ove!s> other WLANs> a!# erso!al
area !etworks ('ANS). This makes susceti4ilit9 to i!terfere!ce a rimar9 co!cer!. The
occuie# 4a!#wi#th of the srea#+sectrum cha!!el is 22 67A> so the %S6 4a!# accommo#ates
o!l9 three !o!+overlai!" cha!!els sace# 2$ 67A aart. To hel miti"ate i!terfere!ce effects>
032.114 #esi"!ates a! otio!al fre;ue!c9 a"ile or hoi!" mo#e usi!" the three !o!+overlai!"
cha!!els or si- overlai!" cha!!els sace# at 13 67A.
032.114 uses ei"ht+chi comleme!tar9 co#e ke9i!" (CCC) as the mo#ulatio! scheme to
achieve the hi"her #ata rates. %!stea# of the Barker co#es use# to e!co#e a!# srea# the #ata for
the lower rates> CCC uses a !earl9 ortho"o!al comle- co#e set calle# comleme!tar9
se;ue!ces. The chi rate remai!s co!siste!t with the ori"i!al /SSS s9stem at 11 6chi5s> while
the #ata rate varies to match cha!!el co!#itio!s 49 cha!"i!" the srea#i!" factor a!#5or the
To achieve #ata rates of $.$ a!# 11 645s> the srea#i!" le!"th is first re#uce# from 11 to ei"ht
chis. This i!creases the s9m4ol rate from 1 6s9m5s to 1.3.$ 6s9m5s. :or the $.$+645s 4it rate
with a 1.3.$ 67A s9m4ol rate> it is !ecessar9 to tra!smit 4 4its5s9m4ol ($.$ 645s51.3.$
6s9m5s) a!# for 11 645s> a! 0 4its5s9m4ol. The CCC aroach take! i! 032.114> which kees
the N'SC srea#+sectrum si"!al a!# still rovi#es the re;uire# !um4er of 4its5s9m4ol> uses all
4ut two of the 4its to select from a set of srea#i!" se;ue!ces a!# the remai!i!" two 4its to
rotate the se;ue!ce. The selectio! of the se;ue!ce> coule# with the rotatio!> rerese!ts the
s9m4ol co!ve9i!" the four or ei"ht 4its of #ata. :or all 032.114 a9loa# #ata rates> the ream4le
a!# hea#er are se!t at the 1 645s rate.
The 23 67A+wi#e 4a!#wi#th of WLAN si"!als makes ower e!veloe measureme!ts #ifficult
4ecause most sectrum a!al9Aers have resolutio! 4a!#wi#th filters that are limite# to 13 67A or
less. Therefore> the si"!al is co!si#era4l9 atte!uate# 49 the time the ower is measure# withi!
the i!strume!t. ?ector si"!al a!al9Aers are availa4le with i!formatio! 4a!#wi#ths that are
co!si#era4l9 "reater tha! 23 67A> maki!" WLAN si"!al a!al9sis more accurate.
The 032.114 sta!#ar# uses error vector ma"!itu#e (&?6) as a measure of mo#ulatio! ;ualit9.
This measureme!t has 4ecome commo! for most wireless alicatio!s. The u!#erl9i!"
hilosoh9 of &?6 is that a!9 si"!al #eteriorate# 49 a !ois9 cha!!el ca! 4e rerese!te# as the
sum of a! i#eal si"!al a!# a! error si"!al. The test i!strume!t #etermi!es the error si"!al 49
reco!structi!" the i#eal si"!al 4ase# o! #etecte# si"!al i!formatio! a!# su4tracti!" it from the
actual si"!al at each samle oi!t.
Co+,arison of 802.11a and 802.11b
A #raw4ack of the $ @7A 4a!#> which has receive# co!si#era4le atte!tio!> is its shorter
wavele!"th. 7i"her+fre;ue!c9 si"!als will have more trou4le roa"ati!" throu"h h9sical
o4structio!s e!cou!tere# i! a! office (walls> floors> a!# fur!iture) tha! those at 2.4 @7A. A!
a#va!ta"e of 032.11a is its i!tri!sic a4ilit9 to ha!#le #ela9 srea# or multiath reflectio! effects.
The slower s9m4ol rate a!# laceme!t of si"!ifica!t "uar# time arou!# each s9m4ol> usi!" a
tech!i;ue calle# c9clical e-te!sio!> re#uces the i!ter+s9m4ol i!terfere!ce (%S%) cause# 49
multiath i!terfere!ce. (The last o!e+;uarter of the s9m4ol ulse is coie# a!# attache# to the
4e"i!!i!" of the 4urst. /ue to the erio#ic !ature of the si"!al> the =u!ctio! at the start of the
ori"i!al 4urst will alwa9s 4e co!ti!uous.) To co!trast> 032.114 !etworks are "e!erall9 ra!"e+
limite# 49 multiath i!terfere!ce rather tha! the loss of si"!al stre!"th over #ista!ce.
Whe! it comes to #elo9me!t of a wireless LAN> oeratio!al characteristics have 4ee!
comare# to those of cellular s9stems> where fre;ue!c9 la!!i!" of overlai!" cells mi!imiAes
mutual i!terfere!ce suort mo4ilit9 a!# seamless cha!!el ha!#off. The three !o!+overlai!"
fre;ue!c9 cha!!els availa4le for %&&& 032.114 are at a #isa#va!ta"e comare# to the "reater
!um4er of cha!!els availa4le to 032.11a. The a##itio!al cha!!els allow more overlai!" access
oi!ts withi! a "ive! area while avoi#i!" a##itio!al mutual i!terfere!ce.
Both 032.114 a!# 032.11a use #9!amic rate shifti!" where the s9stem will automaticall9 a#=ust
the #ata rate 4ase# o! the co!#itio! of the ra#io cha!!el. %f the cha!!el is clear> the! the mo#es
with the hi"hest #ata rates are use#. But as i!terfere!ce is i!tro#uce# i!to the cha!!el> the ra#io
will fall 4ack to a slower> al4eit more ro4ust> tra!smissio! scheme.
Network la!!i!" is critical to the #evelome!t of a! otimiAe# s9stem. &ach !etwork must 4e
customiAe# to satisf9 the la!!e# alicatio!s a!# the h9sical e!viro!me!t. 8e;uireme!ts must
4e researche# a!# well+#ocume!te#> i!clu#i!" a!ticiate# roami!" a!# #ata rates !ee#e# for
alicatio!s to 4e use# at secific locatio!s. A site surve9 must 4e thorou"h a!# realistic to
a#e;uatel9 characteriAe the 8: e!viro!me!t of the roose# wireless !etwork i! terms of ra!"e>
cha!!el i!terfere!ce a!# #ela9 srea#.
%t woul# 4e u!realistic to e-ect to realiAe the full #ata rate caa4ilit9 ($4 645s) of 032.11a if the
access oi!ts of a! e-isti!" 032.114 !etwork otimiAe# to oerate at full see# (11 645s) XYT
were siml9 relace#. But as has 4ee! show!> 032.11a is faster tha! 032.114 at a!9 ra!"e. Cost
vs. erforma!ce re;uireme!ts !ee# thorou"h a!al9sis #uri!" the !etwork la!!i!" sta"e to arrive
at the aroriate imleme!tatio! #ecisio!.
Testi!" is critical to a!9 ro#uct #evelome!t rocess. WLAN ro#ucts re;uire that secial
atte!tio! 4e "ive! to #esi"! verificatio! a!# characteriAatio! 4ecause sta!#ar#iAe# oeratio!
across multive!#or ro#ucts ma9 4e re;uire#. To rovi#e a! efficie!t #evelome!t e!viro!me!t>
test tools are availa4le to ;uickl9 #ia"!ose ro4lems a!# isolate them throu"hout all #esi"!
se"me!ts. These tools ca! 4e use# withi! the ma!ufacturi!" rocess to "e!erate a!# a!al9Ae
ro#uctio! metrics for rocess a!# ro#uct imroveme!t.
&ve! #uri!" these lea! eco!omic times> whe! there is a re#uce# #ema!# for tech!olo"9
ro#ucts> the !ew> 4ut alrea#9 ro4ust WLAN market is ro=ecte# to "row 49 a! or#er of
ma"!itu#e over the !e-t five 9ears. These wireless !etworks will re;uire i!creasi!" #ata rates to
rovi#e the simulta!eous #istri4utio! of %!ter!et #ata> hi"h+;ualit9 vi#eo a!# au#io i! the office
or at home. %! a##itio! to hi"her #ata rates> it is almost a fore"o!e co!clusio! that e!#+users will
4e #ema!#i!" co!ti!uous imroveme!ts i! fu!ctio!alit9> ease+of+use a!# relia4ilit9.
Wireless !etworki!" has a romisi!" future with 032.11 lea#i!" the wa9 as the sta!#ar# for
a#otio! i! local !etworki!" e!viro!me!ts. 032.11 a##resses mo4ilit9> securit9> relia4ilit9> a!#
the #9!amic !ature of wireless LANS while keei!" comati4ilit9 with 032+t9e le"ac9
!etworks. &-ect to see availa4ilit9 of 032.11 ro#ucts i!crease #ramaticall9 i! the !ear future as
4usi!esses #iscover the i!crease# ro#uctivit9 rovi#e# 49 Hu!tethere#E !etworks.
032.11+4ase# !etworks have see! wi#esrea# #elo9me!t across ma!9 fiel#s> mai!l9 #ue to the
h9sical co!ve!ie!ces of ra#io+4ase# commu!icatio!. This #elo9me!t> however> was
re#icate# i! art o! the user e-ectatio! of co!fi#e!tialit9 a!# availa4ilit9. This aer
a##resse# the availa4ilit9 asect of that e;uatio!. We e-ami!e# the 032.11 6AC la9er a!#
#escri4e# the architecture a!# the mai! fu!ctio!s of the 6AC as art of the rotocol stack also
we ma#e a few comariso!s. W& thi!k that with the comariso!s we have offere# a! i!teresti!"
issue a!# that the #escritio! has more efficie!c9. Securit9 as a art of the rotocol stack has
also 4ee! me!tio! 4ecause the rotocol stack as a art of the software is the first art i! what
eole are i!tereste# i!. The wi#esrea# acceta!ce of WLANs #ee!#s o! i!#ustr9
sta!#ar#iAatio! to e!sure ro#uct comati4ilit9 a!# relia4ilit9 amo!" the various ma!ufacturers.
The %!stitute of &lectrical a!# &lectro!ics &!"i!eers (%&&&) ratifie# the ori"i!al 032.11
secificatio! i! 1<<. as the sta!#ar# for wireless LANs. That versio! of 032.11 rovi#es for 1
64s a!# 2 64s #ata rates a!# a set of fu!#ame!tal si"!ali!" metho#s a!# other services.
The most critical issue affecti!" WLAN #ema!# has 4ee! limite# throu"hut. The #ata rates
suorte# 49 the ori"i!al 032.11 sta!#ar# are too slow to suort most "e!eral 4usi!ess
re;uireme!ts a!# have slowe# a#otio! of WLANs. 8eco"!iAi!" the critical !ee# to suort
hi"her #ata+tra!smissio! rates> the %&&& rece!tl9 ratifie# the 032.114 sta!#ar# (also k!ow! as
032.11 7i"h 8ate) for tra!smissio!s of u to 11 64s. @lo4al re"ulator9 4o#ies a!# ve!#or
allia!ces have e!#orse# this !ew hi"h+rate sta!#ar#> which romises to oe! !ew markets for
WLANs i! lar"e e!terrise> small office> a!# home e!viro!me!ts. With 032.114> WLANs will
4e a4le to achieve wireless erforma!ce a!# throu"hut comara4le to wire# &ther!et.
To#a9Es 4usi!ess e!viro!me!t is characteriAe# 49 a! i!creasi!"l9 mo4ile workforce a!# flatter
or"a!iAatio!s. &mlo9ees are e;uie# with !ote4ook comuters a!# se!# more of their time
worki!" i! teams that cross fu!ctio!al> or"a!iAatio!al> a!# "eo"rahic 4ou!#aries. 6uch of
these workersE ro#uctivit9 occurs i! meeti!"s a!# awa9 from their #esks. 1sers !ee# access to
the !etwork far 4e9o!# their erso!al #esktos. WLANs fit well i! this work e!viro!me!t>
"ivi!" mo4ile workers much+!ee#e# free#om i! their !etwork access. With a wireless !etwork>
workers ca! access i!formatio! from a!9where i! the cororatio!Fa co!fere!ce room> the
cafeteria> or a remote 4ra!ch office. Wireless LANs rovi#e a 4e!efit for %T ma!a"ers as well>
allowi!" them to #esi"!> #elo9> a!# e!ha!ce !etworks without re"ar# to the availa4ilit9 of
wiri!"> savi!" 4oth effort a!# #ollars.
Busi!esses of all siAes ca! 4e!efit from #elo9i!" a WLAN s9stem> which rovi#es a owerful
com4i!atio! of wire# !etwork throu"hut> mo4ile access> a!# co!fi"uratio! fle-i4ilit9. The
eco!omic 4e!efits ca! a## u to as much as J1,>333 er userFmeasure# i! worker ro#uctivit9>
or"a!iAatio!al efficie!c9> reve!ue "ai!> a!# cost savi!"sFover wire# alter!atives.
So at the e!# we hoe that we have offere# a "e!eral overview of the %&&& 032.11> eseciall9
rotocol stack a!# h9sical la9ers> arts that are #efi!e# i! the %&&& Sta!#ar#. The %&&& 032.11
is a hu"e toic a!# the Sta!#ar#s we ca! #ow!loa# free from the i!ter!et.
WLAN D Wireless Local Area Network
LAN + Local Area Network
6AC + 6e#ium Access Co!trol
'7B D 'h9sical La9er
%S6 + %!#ustrial> Scie!tific> a!# 6e#ical
CS6A5C/ + Carrier Se!se> 6ultile Access with Collisio! /etect
%SO + %!ter!atio!al Sta!#ar#s Or"a!iAatio!
/SSS + /irect Se;ue!ce Srea# Sectrum
NoS + Nualit9 Of Service
TC' + Tra!smit ower co!trol
/:S + /9!amic fre;ue!c9 selectio!
6%6O + 6ultile %!ut> 6ultile Outut
LLC + Lo"ical Li!k Co!trol
/S + /istri4utio! S9stem
W&' + Wire# &;uivale!t 'rivac9
NA? + Network Allocatio! ?ector
C8C + C9clic 8e#u!#a!c9 Check
%&&&+SA + %!stitute of &lectrical a!# &lectro!ics &!"i!eers Sta!#ar#s Associatio!
&?6 + &rror ?ector 6a"!itu#e
CCC + comleme!tar9 co#e ke9i!"
C/: + Cumulative /istri4utio! :u!ctio!
CC/: + Comleme!tar9 Cumulative /istri4utio! :u!ctio!
:&C + :orwar# &rror Correctio!
/N'SC + /iffere!tial Nua#rature 'hase Shift Ce9i!"
/B'SC + /iffere!tial Bi!ar9 'hase Shift Ce9i!"
'N D 'seu#o Noise
8: D 8a#io :re;ue!c9
O:/6 + Ortho"o!al :re;ue!c9 /ivisio! 6ultile-i!"
::C + :e#eral Commu!icatio!s Commissio!
/SSS + /irect Se;ue!ce Srea# Sectrum
:7SS + :re;ue!c9 7oi!" Srea# Sectrum
''6 + 'ulse 'ositio! 6o#ulatio!
78+/SSS + 7i"h 8ate /irect Se;ue!ce Srea# Sectrum
''/1 + 'LC' 'rotocol /ata 1!it
'6/ + 'h9sical 6e#ium /ee!#e!t
'LC' + 'h9sical La9er Co!ver"e!ce 'roce#ure
IS& 2 International Standards &r-ani=ation
The %!ter!atio!al Sta!#ar#s Or"a!iAatio! is reso!si4le for a wi#e ra!"e of sta!#ar#s> i!clu#i!"
those releva!t to !etworki!". The %SO #eveloe# the OS% (Oe! S9stem %!terco!!ectio!)
refere!ce mo#el which is a oular !etworki!" refere!ce tool.
(A) 2 (ireless ocal Area )et'ork
This is a "e!eric term coveri!" a multitu#e of tech!olo"ies rovi#i!" local area !etworki!" via a
ra#io li!k. &-amles of WLAN tech!olo"ies i!clu#e (i2;i (Wireless :i#elit9)> 802.11b a!#
802.11a> 7ierLAN> *luetooth> Ir4A (%!frare# /ata Associatio!) a!# 4.CT (/i"ital
&!ha!ce# Cor#less Telecommu!icatio!s) etc.
(i8A6 2 (orld'ide Intero,erability for 8icro'a#e Access
The term Wi6A2 has 4ecome s9!o!9mous with the %&&& 032.1, suite of sta!#ar#s. These
#efi!e the ra#io or air i!terface withi! two 4roa# ra#io 4a!#s 2@7A to 11@7A (%&&& 032.1,a)
a!# 13@7A + ,,@7A (%&&& 032.1,c) althou"h i!itial i!terest is co!fi!e# to the li!e of si"ht
4a!#s + 2.$@7A> 3.$@7A a!# $.0@7A. %t is a!ticiate# that Wi6A2 will 4e use# i!itiall9 as a
4ackhaul co!!ectio! with other tech!olo"ies such as Wi+:i 4ei!" use# to cover the Sfi!al mileT.
(i2;i 2 (ireless ;idelity
Wi+:i is a! i!teroera4ilit9 sta!#ar# #eveloe# 49 W&CA (Wireless &ther!et Comati4ilit9
Allia!ce) a!# issue# to those ma!ufacturers whose %&&& 802.11a a!# 802>11b e;uime!t has
asse# a suite of 4asic i!teroera4ilit9 tests. &;uime!t assi!" these tests carries the Wi+:i
Note* Wi+:i !ot Wi:i.
8AC 2 8ediu+ Access Control
6e#ia Access Co!trol is the lower of the two su4la9ers of the /ata Li!k La9er. %! "e!eral terms>
6AC ha!#les access to a share# me#ium> a!# ca! 4e fou!# withi! ma!9 #iffere!t tech!olo"ies.
:or e-amle> 6AC metho#olo"ies are emlo9e# withi! .thernet> %P:S> a!# 58TS etc.
A 'h9sical Li!k is the co!!ectio! 4etwee! #evices.
4SSS 2 4irect Se<uence S,read S,ectru+
/irect Se;ue!ce Srea# Sectrum is 4ase# o! the multil9i!" of the Base4a!# si"!al #ata with a
4roa#4a!# srea#i!" co#e. The result is terme# the chi rate. The characteristics of the
4roa#4a!# srea#i!" co#e are that of seu#ora!#om !oise. Co!se;ue!tl9 the receiver
s9!chro!iAe# to the co#e will o4tai! the !arrow4a!# si"!al. All other receivers will see the
srea# si"!al as white or colore# !oise.
?oS 2 ?uality of Ser#ice
The erforma!ce of a commu!icatio!s cha!!el or s9stem is usuall9 e-resse# i! terms of NoS
(Nualit9 of Service). /ee!#i!" uo! the commu!icatio! s9stem> NoS ma9 relate to service
erforma!ce> S): (Si"!al to Noise 8atio)> *.: (Bit &rror 8atio)> ma-imum a!# mea!
throu"hut rate> relia4l9> riorit9 a!# other factors secific to each service.
TCP 2 Trans+ission Control Protocol
Tra!smissio! Co!trol 'rotocol is a relia4le octet streami!" rotocol use# 49 the ma=orit9 of
alicatio!s o! the Internet. %t rovi#es a co!!ectio!+orie!te#> full+#ule-> oi!t to oi!t service
C 2 o-ical ink Control
%! the %P:S s9stem the LLC rotocol rovi#es a hi"hl9 relia4le cihere# lo"ical li!k 4etwee!
the 8S (6o4ile Statio!) a!# S%S) (Servi!" @'8S Suort No#e). %t is i!#ee!#e!t of the
u!#erl9i!" ra#io i!terface rotocols e!a4li!" the i!tro#uctio! of alter!ative @'8S ra#io
solutio!s with mi!imal cha!"es to the Network Switchi!" S9stem.
C:C 2 Cyclic :edundancy Code
A li!ear error co#e that is "e!erate# usi!" a ol9!omial fu!ctio! o! the #ata to 4e se!t> the
remai!#er from the rocess 4ei!" the C8C. This is se!t alo!" with #ata so that a arit9 check of
the receive# #ata ca! 4e co!#ucte#.
I... 2 Institute of .lectrical and .lectronics .n-ineers
The %!stitute of &lectrical a!# &lectro!ics &!"i!eers is a rofessio!al or"a!iAatio! whose
activities i!clu#e the #evelome!t of commu!icatio!s a!# !etwork sta!#ar#s.
'art of the I... 802.11 famil9 of secificatio!s> this wireless local area !etwork tech!olo"9 is
comrise# of a hi"h see# h9sical la9er oerati!" i! the $@7A u!lice!se# 4a!# a!# suorts
#ata rates u to $464s. &;uime!t oerati!" i! accor#a!ce with the %&&& secificatio!s a!#
assi!" the Allia!ces i!teroera4ilit9 tests is a4le to #isla9 the Wi+:i lo"o.
Several ma!ufacturers have #eveloe# e;uime!t which is caa4le of oerati!" i! accor#a!ce
with 4oth %&&& 032.11a a!# I... 802.11bsecificatio!s.
'art of the I... 802.11 famil9 of secificatio!s> %&&& 032.114 is curre!tl9 the most oular
wireless !etworki!" tech!olo"9. The e;uime!t oerates i! the 2.4@7A u!lice!se# 4a!# a!#
utiliAes 9:74SSS (7i"h 8ate + /irect Se;ue!ce Srea# Sectrum) e!a4li!" #ata rates of u to
1164s to 4e achieve#. &;uime!t oerati!" i! accor#a!ce with the %&&& secificatio!s a!#
assi!" the (i2;i Alliances i!teroera4ilit9 tests is a4le to #isla9 the Wi+:i lo"o.
Several manufacturers have developed equipment which is capable of
operating in accordance with both I... 802.11a and IEEE 802.11b
I... 802.11 2 (ireless
This is a! I... (%!stitute of &lectrical a!# &lectro!ic &!"i!eers) tech!ical sta!#ar# coveri!"
(A) (Wireless Local Area Network) tech!olo"9. The sta!#ar#s have 4ee! #ivi#e# i!to su4
"rous with 032.114 curre!tl9 the most commo!. This rovi#es a wireless commu!icatio! at u
to 1164s a!# oerates withi! the 2.4@7A IS8 (%!#ustrial Scie!tific a!# 6e#ical) 4a!#. The
032.11a 4ase# e;uime!t is !ow commerciall9 availa4le a!# rovi#es #ata rates u to $464s
a!# oerates i! the $@7A %S6 4a!#. %&&& 032.11 !etworks are comrise# of Stations> Wireless
6e#ium> AP (Access 'oi!ts) a!# a /S (/istri4utio! S9stem).
AP 2 Access Point
A! Access 'oi!t is a #evice fou!# withi! a! %&&& 802.11 !etwork which rovi#es the oi!t of
i!terco!!ectio! 4etwee! the wireless Station (lato comuter> P4A ('erso!!el /i"ital
Assista!t) etc.) a!# the wire# !etwork.
.SS 2 .3tended Ser#ice Set
A! &-te!#e# Service Set is comrise# of a !um4er of %&&& 802.11 BSS (Basic Service Set) a!#
e!a4les limite# mo4ilit9 withi! the (A) (Wireless Local Area Network). Statio!s are a4le to
move 4etwee! *SS withi! a si!"le &SS 9et still remai! Sco!!ecte#T to the fi-e# !etwork a!# so
co!ti!ue to receive emails etc. As a Station moves i!to a !ew BSS> it will carr9 out a
resuscitatio! roce#ure with the !ew AP (Access 'oi!t).
SSI4 2 Ser#ice Set Identifier
The Service Set %#e!tifier or Network Name is use# withi! %&&& 802.11 !etworks to i#e!tif9 a
articular !etwork. %t is usuall9 set 49 the a#mi!istrator setti!" u the (A) (Wireless Local
Area Network) a!# will 4e u!i;ue withi! a *SS (Basic Service Set) or .SS (&-te!#e# Service
Set). The SS%/ ma9 4e 4roa#cast from a! AP (Access 'oi!t) withi! the wireless !etwork to
e!a4le Stations to #etermi!e which !etwork to SAssociateT with. 7owever> this feature shoul#
4e #isa4le# as it ma9 assist Shackers> or war#riversT i! "ai!i!" access to a rivate !etwork.
SAP 2 Ser#ice Access Point
A co!cetual oi!t where a rotocol la9er offers access to its services to the la9er a4ove or
Table of Contents 1. Abstraction 2. Introduction
2.1 Overview 2.2 Wireless Local Area Network 2.3 The Basic Structure of a Wireless LAN 2.4 Comparing a Wireless LAN to a Wired Network 2.5 The IEEE Standard
3. 802.11 Protocol Stack
3.1 Protocol Structure 3.2 Protocol Stack Architecture 3.2.1 Station (STA) Architecture: 3.2.2 Access-Point (AP) Architecture: 3.2.3 Basic Service Set (BSS): 3.2.4 Independent Basic Service Set (IBSS): 3.2.5 Infrastructure 3.2.6 Extended Service Set (ESS): 3.2.7 Service Set Identifier (SSID): 3.2.8 Basic Service Set Identifier (BSSID) 3.3 Protocol Stack for UNIX 3.4 Compare Overall Structure of 802.11b / 802.15.1 Coexistence Mechanism 3.5.1 MIH SAP Reference Model for 802.11 3.5.2 MIH SAP Reference Model for 802.16 3.6 Data Link Layer 3.6.1 Support for Time-Bounded Data 3.7 MAC Functional Description 3.7.1 MAC Architecture 3.8 Security
3.8.1 Preventing Access to Network Resources 3.8.2 Eavesdropping
4. Physical Layer
4.1 The physical layer basics 4.2 PLCP Frame Fields 4.3 Infrared (IR) 4.4 Spread Spectrum 4.5 Frequency Hopping Spread Spectrum (FHSS) 4.6 Direct Sequence Spread Spectrum (DSSS) 4.6.1 DSSS Modulation 4.6.2 Transmit Frequencies 4.7 The IEEE 802.11a 4.7.1 Practice 802.11a 4.8 The IEEE 802.11b 4.8.1 Practice 802.11b 4.9Comparison of 802.11a and 802.11b
5. Conclusion 6. Abbreviation 7. Glossary
The writing of this Research Report was prompted to maintain two main developments of the IEEE 802.11 Standard physical protocol stack and Physical Layer enhanced from the developments in wireless communication in the past decade. First we had to do huge research activities in this topic. This has been a subject study since the sixties, so that during our exploring work we have selected a lot of materials and picked up the most visible things for the student to understand it more easily and clearly. So that we were concentrated to present the issue in modern wireless concepts in a coherent and unified manner and to illustrate the concepts in that way they are applied. The concepts can be structured into these levels: - Listing characteristics and modeling - Application of these concepts But of course there is interplay between these structures. So this Research report is written based on the material for the students in the sixth semester. Also in the end to understand better the terminology and the huge number of abbreviations explained and some definitions.
The past decade has seen many advances in physical-layer communication theory and their implementation in wireless systems. So that in this Research Report we are going to define and view fundamentals of wireless communication and that especially the IEEE 802.11 Standard and explain the advantages at a level that is accessible to our audience with a basic background. Wireless communication is one of the most vibrant areas in the communication field today. This is due to a confluence of several factors. First, there has been an explosive increase in demand for the tether less connectivity, driven so far mainly by cellular telephony but expected to be soon eclipsed by wireless data application. First there has been an explosive increase in demand for tether less connectivity, driven so far mainly by cellular telephony but expected to be soon eclipsed by wireless data applications. Second, the dramatic process in VLSI technology has enabled small area and low power implementation of sophisticated signal processing algorithms and coding techniques. Third, the success of second generation digital wireless standards and provide a concrete demonstration that good ideas from communication theory can have impact in practice. There are two fundamental aspects of wireless communication that make the problem challenging and interesting. These aspects are by and large not as significant in wire line communication. First the phenomenon of fading: the time variation of the channel strengths due to the small-scale effect of multipath fading, as well as large scale effects. Second, unlike the in the wired world where each transmitter-receiver pair can often be thought of as an isolated point-to point link, wireless users communicate over the air and there is significant interface between them. The original 802.11 standard specified three separate physical layers. Two are radiobased and one is infrared light-based. The original radio-based layers are spread spectrum: frequency hopping and direct sequence. These are all in the 2.4 GHz band. An additional 4
the suite is the definition of the protocols. This is specified in both the 2. in the areas of encryption. due process. Strictly speaking. a technology used for many years for the WLAN security. In a typical WLAN configuration. Wireless LAN’s typically augment or replace wired computer networks.11 standard includes a common Medium Access Control (MAC) Layer. and the receive function.11 frames.1a release. The Institute of Electrical and Electronics Engineers Standards Association (IEEESA) is the leading developer of global industry standards in a broad-range of industries. Wireless technologies use radio transmissions as the means for transmitting data. openness. We have discussed about them in this rapport. Transportation. the physical medium dependent layer (PMD) and the physical layer convergence procedure (PLCP). and transmits data between the components of the WLAN. a transceiver—or access point—connects to the wired network from a fixed ilocation using a standard Ethernet cable. the 802. which is also radio-based: orthogonal frequency division multiplexing (OFDM). the transmit function.4 and the 5 GHz bands. authentication and key management. IEEE 802. There are two sub layers in the 802.11 specifications. 802. the IEEE-SA has offered an established standards development program that features balance. is a closely grouped system of devices that communicate via radio waves instead of wires. 802. Nanotechnology. Telecommunications. in the simplest sense. The access point receives. and the stack is the software implementation of them. The PMD is the sub layer lowest on the stack.11 comprise several alternative physical layers that specify the transmission and reception of 802. including: Power and Energy. The standard IEEE 802. added yet another PHY: complimentary code keying orthogonal frequency division multiplexing (CCK-OFDM). which defines protocols that govern the operation of the wireless LAN.11i enhances the WEP (Wireline Equivalent Privacy). to simple devices such as wireless headphones. Wireless technologies range from complex systems. A protocol stack is a particular software implementation of a computer networking protocol suite. enable one or more devices to communicate without physical connections – without requiring network cabling. Security is one of the first concerns of people deploying a Wireless LAN. microphones. Information Assurance. The PHY layer has three basic functions. In addition. For over a century. providing users with more flexibility and freedom of movement within the workplace. Information Technology. they must conform to the same PHY layer.11g. The terms are often used interchangeably. and consensus. It transmits and receives bits over the air. whereas wired technologies use cables.physical layer in the 5 GHz band was added with the 802. These are the carrier sense function. The IEEE (define) 802. buffers. The latest release.11 physical layers. Biomedical and Healthcare. Overview Wireless technologies. and other devices that do not process or store information.11i is designed to provide secured communication of wireless LAN as defined by all the IEEE 802. 5 . such as WLANs and cell phones. Note that for 2 devices to be able to interact. In Computer network a substantial part is the Wireless Local Area Network (WLAN).11 committee has addressed the issue by providing what is called WEP (Wired Equivalent Privacy) Authentication: A function that determines whether a Station is allowed to participate in network communication.
Expect to see availability of 802. is a closely grouped system of devices that communicate via radio waves instead of wires. WLANs have gained strong popularity in a number of vertical markets. A wide variety of industries have discovered the benefits a WLAN can bring—not only to daily tasks but also to the balance sheet.S. including the health-care. Over the last seven years.11 leading the way as the standard for adoption in local networking environments. wireless LAN market is rapidly approaching $1 billion in revenues.11 products increase dramatically in the near future as businesses discover the increased productivity provided by ‘untethered’ networks. Today WLANs are becoming more widely recognized as a general-purpose connectivity alternative for a broad range of business customers. manufacturing. providing users with more flexibility and freedom of movement within the workplace. The U. WLANs combine data connectivity with user mobility. and. through simplified configuration. reliability. Wireless LANs can pump information and data to executives in the boardroom and to employees in the warehouse. retail. security. minimizing the need for wired connections. Wireless Local Area Network In Computer network a substantial part is the Wireless Local Area Network (WLAN). and academic arenas. a wired LAN within a building or campus. Thus. 802. enable movable LANs. and the dynamic nature of wireless LANS while keeping compatibility with 802-type legacy networks. then wireless networks are its heart. Using electromagnetic waves. If information is the lifeblood of today's business environment. Wireless LAN’s typically augment or replace wired computer networks. Users can access the company intranet or even the World Wide Web from anywhere on the company campus without relying on the availability of wired cables and connection.11 addresses mobility. 6 . WLAN’s transmits and receive data over the air.And as conclusion we will say that wireless networking has a promising future with 802. warehousing. A wireless LAN (WLAN) is a flexible data communication system implemented as an extension or as an alternative for. These industries have profited from the productivity gains of using hand-held terminals and notebook computers to transmit real-time information to centralized hosts for processing.
ISA or PCI adapters for desktop computers. a plethora of applications and devices support the 802. or similar devices integrated into handheld units. printers.11b in popularity. Wireless LAN components that use the 802. buffers.11a high data rate standard perform at speeds up to 54 Mbps.The Basic Structure of a Wireless LAN In a typical WLAN configuration. A single access point can support a small group of users and can function within a range of anywhere from 30 to several hundred feet. and transmits data between the components of the WLAN—whether laptops. By working with a wireless vendor well-versed in security issues.4 GHz frequency range. These users—as well as those who hesitate to deploy wireless technology because of security concerns— stand to benefit from understanding the security options currently available. Users equipped with handheld devices or notebook computers can transmit data to the access point when within range. usually in the form of radio network interface cards (NICs). Such factors also affect wired network speeds.11a exceed 802.11. the distance between network components. Of the three main variations of 802.11g and 802. the type of WLAN system in use.11 WLAN infrastructure. industry experts anticipate that it won’t be long before 802. which operates in the 2. The access point can be installed anywhere in the facility as long as good radio coverage is maintained. and the efficiency of the wired network elements all influence the overall speed and performance of a wireless network. even as the industry moves aggressively to provide even more secure protocols. companies can dramatically enhance the security of its wireless communications system. but most commercial LANs operate at speeds from 10 megabits per second (10BaseT) to 100 Mbps (100BaseT). Wireless users recognize the benefits of the technology and need to know how to protect their business-critical data. 7 . The wireless devices communicate with the network operating system via WLAN adapters. The number of users. or any other wireless equipment—and the wired network infrastructure. Although this standard is much more widely implemented than its newer sister technologies. handheld devices. The access point receives. almost a five-fold increase from the performance of the 802.11b standard. as in the case of notebook computers. Almost all mobile applications today lend themselves to deployment of an 802. Comparing a Wireless LAN to a Wired Network The speed at which a WLAN performs depends on the type and configuration of the devices within the network. a transceiver—or access point—connects to the wired network from a fixed location using a standard Ethernet cable.11b standard.
IEEE 802. Scientific.7-GHz band (5725–5850MHz). and the 5.11 today. and the frameworks and means for supporting security and quality of service over a WLAN.5MHz). including: • Power and Energy • Biomedical and Healthcare • Information Technology • Telecommunications • Transportation • Nanotechnology • Information Assurance 8 .4-GHz band (2400–2483.11 consists of a family of standards that defines the physical layers (PHY) and the Medium Access Control (MAC) layer of a WLAN. WLAN network architectures. how a WLAN interacts with an IP core network. and consensus. The Institute of Electrical and Electronics Engineers Standards Association (IEEE-SA) is the leading developer of global industry standards in a broad-range of industries. 2. and Medical (ISM) radio frequency bands.WLANs typically use the unlicensed Industrial.11 standards family includes the following key standards: The IEEE Standard For over a century. due process. The IEEE 802. In the United States. The most widely adopted WLAN standard around the world is 802. openness. the ISM bands include the 900-MHz band (902–928 MHz). the IEEE-SA has offered an established standards development program that features balance.
" Specifies a PHY that operates in the 5 GHz U-NII band in the 802. (See IEEE status page. Commonly referred to as the LLC or Logical Link Control specification. Standards project no longer endorsed by the IEEE.1D incorporates P802. Interfaces with the network Layer 3. 802.1 Overview Bridging Basics of physical and logical networking concepts. The LLC is the top sub-layer in the data-link layer.1j and 802. Click for a list of the "hot" 802. Disbanded The original token-passing standard for twisted-pair.3 technologies. No longer Practices endorsed by the IEEE. g. Often called "IBM Token-Ring. "Granddaddy" of the 802 specifications.) Withdrawn PAR.1p and P802.11 Wi-Fi 802. including MAC-based bridging (Media Access Control)." (See IEEE status page. Current speeds range from 10 Mbps to 10 Gbps. Products that implement 802. Superseded by 802. shielded copper cables. It also incorporates and supersedes published standards 802. twisted-pair copper.8 802.) Fiber Optic Practices Integrated Services LAN Interoperable LAN security Withdrawn PAR.11 standards must pass tests and are referred to as "Wi-Fi certified.10 Broadband LAN Withdrawn Standard.) Superseded **Contains: IEEE Std 802. multiple access with collision detect" (CSMA/CD) over coax. Provides asynchronous networking using "carrier sense.6k.9 802. Standards project no longer endorsed by the IEEE. Withdrawn Date: Feb 07. are amendments to the original 802.6 802.11a 9 . etc.) Wireless LAN Media Access Control and Physical Layer specification.10b-1992.3 Ethernet 802.1D-2004. 802. and fiber media.The following table lists highlights of the most popular sections of IEEE 802 and has links for additional information: 802 802. LAN/MAN bridging and management." "Superseded **Revision of 802.11 standard.) 802. virtual LANs and port-based access control. 2003. OSI Layer 2. (See IEEE status page. Covers management and the lower sub-layers of OSI Layer 2.1D-1990 edition (ISO/IEC 10038).11a. (See IEEE status page.7 802.5 Token Bus Token Ring Distributed queue dual bus (DQDB) 802.2 Logical Link 802. (See IEEE status page. Supports copper and fiber cabling from 4 Mbps to 100 Mbps.4 802. b.12e.
2.11a and 802.85 .35 AND 5. in a fashion that permits interoperation with 802.11m 10 . with fallback speeds that include the "b" speeds Enhancement to 802.725-5. as well as options for key caching and preauthentication Japanese regulatory extensions to 802. Enhancement to 802.4 GHz band. voice.11 family specifications Corrections and amendments to existing documentation 802.11i 802. or 1 Mbps from 11 Mbps max.11b devices Uses OFDM Modulation (Orthogonal FDM) Operates at up to 54 megabits per second (Mbps).5.400 GHz to 2.11g 802. falls back to 5.11a specification Frequency range 4.11d 802.US .11 that added higher data rate modes to the DSSS (Direct Sequence Spread Spectrum) already defined in the original 802.11b that allows for global roaming Particulars can be set at Media Access Control (MAC) layer Enhancement to 802.11h 802. authentication.0 GHz Radio resource measurements for networks using 802.initially 5.11 family specifications Maintenance of 802.9 GHz to 5.11b 802.4835 GHz Beacons at 1 Mbps.11 that includes quality of service (QoS) features Facilitates prioritization of data.11b Enhancement to 802.11 that offers additional security for WLAN applications Defines more robust encryption.11a that resolves interference issues Dynamic frequency selection (DFS) Transmit power control (TPC) Enhancement to 802.11e 802.11j 802.since expanded to additional frequencies Uses Orthogonal Frequency-Division Multiplexing Enhanced data speed to 54 Mbps Ratified after 802. and video transmissions Extends the maximum data rate of WLAN devices that operate in the 2. and key exchange.15-5.11 standard Boosted data speed to 11 Mbps 22 MHz Bandwidth yields 3 non-overlapping channels in the frequency range of 2.11k 802.
4 802.3a 802. TGn Sync.17 802.1 802.15.18 802. and hands-free headset at 2. and 350+ MHz Competing proposals come from the groups. EWC.16 Wireless Metropolitan Area Networks 802.14 802.15.15 802. 5. 2.19 802. high-bandwidth "ultra wideband" link Short range wireless sensor networks Extension of network coverage without increasing the transmit power or the receiver sensitivity Enhanced reliability via route redundancy Easier network configuration .11 family specifications Demand Priority Not used Cable modems Wireless Personal Area Networks Bluetooth UWB ZigBee Mesh network Increases Ethernet data rate to 100 Mbps by controlling media utilization.4. Not used Withdrawn PAR. Products that implement 802. Short range (10m) wireless technology for cordless mouse.8 GHz) or licensed (700 MHz. 2.16 standards can undergo WiMAX certification testing.11x 802.Better device battery life This family of standards covers Fixed and Mobile Broadband Wireless Access methods used to create Wireless Metropolitan Area Networks (WMANs.20 mission and project scope 11 . Standards project no longer endorsed by the IEEE.5 802.4 GHz.5 – 3.20 Resilient Packet IEEE working group description Ring Radio IEEE 802. Short range.15. multiple output) Miss-used "generic" term for 802.under development Several competing and non-compatible technologies.15.802. and WWiSE and are all variations based on MIMO (multiple input. often called "pre-n" Top speeds claimed of 108. 802.18 standards committee Regulatory TAG Coexistence Mobile IEEE 802.11n Higher-speed standards -.19 Coexistence Technical Advisory Group IEEE 802.6 GHz) frequency bands.12 802. 240.) Connects Base Stations to the Internet using OFDM in unlicensed (900 MHz.13 802. Communications specification that was approved in early 2002 by the IEEE for wireless personal area networks (WPANs). keyboard.
and the stack is the software implementation of them.11 protocol stack.22 Media Independent Handoff Wireless Regional Area IEEE 802. The protocols used by all the 802 variants.22 mission and project scope 802. The lowest protocol always deals with "low-level". The physical layer corresponds to the OSI physical layer fairly well. including Ethernet. User applications habitually deal only with the topmost layers (See also OSI model).Broadband Wireless Access 802. the MAC (Medium Access 12 . Every higher layer adds more features.11. This modularization makes design and evaluation easier. physical interaction of the hardware. Strictly speaking.21 802. Individual protocols within a suite are often designed with a single purpose in mind.11 Protocol Stack A protocol stack is a particular software implementation of a computer networking protocol suite. In 802. the suite is the definition of the protocols. In the figure below we see a partial view of the 802. Because each protocol module usually communicates with two others. The terms are often used interchangeably. have a certain commonality of structure. they are commonly imagined as layers in a stack of protocols. but the data link layer in all the 802 protocols is split into two or more sublayers.21 mission and project scope IEEE 802.
11 standard specifies three transmission techniques allowed in the physical layer. and who gets to transmit next. respectively. Radio-controlled garage door openers also use this piece of the spectrum. using techniques called FHSS and DSSS. whose job it is to hide the differences between the different 802 variants and make them indistinguishable as far as the network layer is concerned. These are called OFDM and HRDSSS. The infrared method uses much the same technology as television remote controls do. 13 . Both of these use a part of the spectrum that does not require licensing (the 2. so your notebook computer may find itself in competition with your garage door. Above it is the LLC (Logical Link Control) sublayer. They operate at up to 54 Mbps and 11 Mbps. two new techniques were introduced to achieve higher bandwidth. In 2001. Now we will examine each of them briefly. All of these techniques operate at 1 or 2 Mbps and at low enough power that they do not conflict too much. The other two use short-range radio. Cordless telephones and microwave ovens also use this band. a second OFDM modulation was introduced.4-GHz ISM band). but in a different frequency band from the first one. The 1997 802. We studied the LLC when examining Ethernet earlier in this chapter and will not repeat that material here. In 1999.Control) sublayer determines how the channel is allocated. that is.
More Data indicates that there are more frames buffered to this station. Beacon frame. To DS . Disassociation frame. Duration/ID (ID) Station ID is used for Power-Save poll message frame type.contain up to 4 addresses (source. transmittion and receiver addresses) depending on the frame control field (the ToDS and FromDS bits).is set to 1 when the frame is received from the Distribution System (DS) MF. 802. It is used to represent the order of different fragments belonging to the same frame and to recognize packet duplications.consists of fragment number and sequence number.11. 802.Protocol Structure In the figure below we can see the Wireless LAN by IEEE 802. Reassociation response frame. . Sequence Control .Order indicates that the frame is being sent using the Strictly-Ordered service class.WEP indicates that the frame body is encrypted according to the WEP (wired equivalent privacy) algorithm. (For receiver to recognize duplicate transmissions of frames) Pwr .More Fragment is set to 1 when there are more fragments belonging to the same frame following the current fragment Retry indicates that this fragment is a retransmission of a previously transmitted fragment.indicates the version of IEEE 802. Probe request frame and Probe response frame. Reassociation request frame. W . Association response frame. destination. 802.Power Management indicates the power management mode that the station will be in after the transmission of the frame. 14 . Subtype . Control and Data. Probe frame. Type . Deauthentication frame.11a.11 protocol family MAC frame structure: 02 2 6 6 6 2 6 2312 Frame Address Duration Control 1 Frame Control Structure: 2 Version • • • 2 Type 4 Subtype 1 To DS Address 2 Address 3 Se q Address 4 Data 4 Check sum 1 From DS 1 MF 1 Retry 1 Pwr 1 More 1 W 1 O • • • • • • • • • • • • • Protocol Version .11g. The duration value is used for the Network Allocation Vector (NAV) calculation.802.Frame subtype: Authentication frame.11b.Frame type: Management. More .is set to 1 when the frame is sent to Distribution System (DS) From DS .11 standard. Address fields (1-4) .11n801. Association request frame.
11 Via the Bridge Tunnel encapsulation scheme are encapsulated the Ethernet Types 8137 (Novell IPX) and 80F3 (AARP) All other Ethernet Types: encapsulated via the RFC 1042 (Standard for the Transmission of IP Datagram’s over IEEE 802 Networks) encapsulation scheme Maximum Data limited to 1500 octets Bridging to Ethernet is done transparently 15 . Station (STA) Architecture: The Station Architecture is a device that contains the IEEE conformant MAC and PHY interface in the wireless medium. work-stations and it is implemented in the Avaya Wireless IEEE 802.11 (Wireless Local Area Networks). The difference between a portable and mobile station is that a portable station moves from point to point but is only used at a fixed point. Mobile stations access the LAN during movement . portable or fixed. • • • • • • • • • The most important features and conditions of the Station Architecture are It has a driver interface like the Ethernet All protocol Stacks are virtually supported by the STA The Frame translation is done according to the IEEE STD 802. The Station Architecture is most often available in terminals like laptops.3 frames with this architecture are translated to 802. CRC .is information that is transmitted or received. is referred to as a station in 802. Protocol Stack Architecture Each computer. mobile. but on the other side it does not provide access to a distribution system.contains a 32-bit Cyclic Redundancy Check (CRC).• • Data .11 PC-Card.1H The IEEE 802.
So that in the IBSS architecture takes place the direct Station-to-Station communication system.) and the wired network. and provides access to a distribution system for associated stations. The Access-Point Architecture offers a Power Management support.Access-Point (AP) Architecture: An Access Point is a device found within an IEEE 802. The Access Point Architecture is a device that contains IEEE 802.11 network which provides the point of interconnection between the wireless Station (laptop computer. In a protocol stack architecture the traffic typically flows through the Access-Point. Most often it contains infra-structure products that connect to wired backbones It is implemented in Avaya Wireless IEEE 802. Basic Service Set (BSS): 16 .11 PCCard when it is inserted in an AP-500 or AP-1000 The most important features and conditions of AccessPoint (AP) Architecture: • The Stations select an Access-Point and they associate with that.11 conformant MAC and PHY interface to the wireless medium. PDA (Personnel Digital Assistant) etc. The Access point is a part that supports roaming and also they provide time synchronization functions like beaconing.
Mostly these networks are spontaneous and can be set up rapidly. an IBSS is simply comprised of one or more Stations which communicate directly with each other. The diameter of the cells is twice the coverage-distance between two wireless stations. In this case we have similarity to a “cell” in the pre IEEE terminology also a BSS can have an Access-Point and that both in standalone networks and in building-wide configurations. The BSS may or may not include AP (Access Point) which provides a connection onto a fixed distribution system such as an Ethernet network. The minimum BSS consists of two stations. So that the Basic Service Set (BSS) forms a self-contained network in which no 17 . As such. An ad-hoc network is a network where stations communicate only peer to peer.11 WLAN (Wireless Local Area Network). Ad-Hoc or IBSS networks are characteristically limited both temporally and spatially.11 LANs use the BSS as the standard building block. There is no base and no one gives permission to talk.11 networks in that no network infrastructure is required.A BSS that stands alone and is not connected to a base is called an Independent Basic Service Set (IBSS) or is referred to as an AdHoc Network. 802. Independent Basic Service Set (IBSS): An Independent Basic Service Set also called ad hoc network is the simplest of all IEEE 802. that is the logical function that determines when a station can transmit or receive. In the BSS architecture a set of stations is controlled by a single “Coordination Function”. they form a Basic Service Set (BSS). IBSS (Independent Basic Service Set) and Infrastructure Basic Service Set. Two types of BSS exist. The contraction should not be confused with an Infrastructure BSS (Basic Service Set). or it just can run without and Access-Point but only in standalone networks.The Basic Service Set is a term used to describe the collection of Stations which may communicate together within an 802. When two or more stations come together to communicate with each other.
18 . Creating large and complex networks using BSS's and DS's leads us to the next level of hierarchy. An access point is a station. Infrastructure When BSS's are interconnected the network becomes one with infrastructure.access to a Distribution System is available. Each BSS becomes a component of an extended. or it is also similar to a BSS without an AccessPoint. This concept of DS increases network coverage. The beauty of the ESS is the entire network looks like an independent basic service set to the Logical Link Control layer (LLC). One of the stations in the IBSS can be configured to “initiate” the network and assume the Coordination Function. Extended Service Set (ESS): An Extended Service Set is comprised of a number of IEEE 802. Entry to the DS is accomplished with the use of Access Points (AP). Stations are able to move between BSS within a single ESS yet still remain “connected” to the fixed network and so continue to receive emails etc. thus addressable. larger network. The diameter of the cell is determined by coverage distance between two wireless stations. the Extended Service Set or ESS.11 infrastructures have several elements. With help of the Access-Points data moves then between the BSS and the DS. Infrastructure is established in the network when BSS are interconnected. As a Station moves into a new BSS. Two or more BSS's are interconnected using a Distribution System or DS. it will carry out a reassociation procedure with the new AP (Access Point). This means that stations within the ESS can communicate or even move between BSS’s transparently to the LLC. so that the 802.11 BSS (Basic Service Set) and enables limited mobility within the WLAN (Wireless Local Area Network).
In the wired network there are used cables to interconnect the Access-Points. In the wireless network are used wirelesses to interconnect the Access-Points.It is the same here that the traffic always flows via Access-Point. Example: of Extended Service Set (ESS) with single BSS and integrated DS 19 . and the diameter of the cell is double the coverage distance between two wireless stations Distribution System (DS).There is available a system to interconnect a set of Basic Service Sets and there is integrated a single Access-Point in a standalone network.
The most important things about the SSID are that it is 32 octets long and it is similar to “Domain-ID” in the pre-IEEE Wave LAN systems. So we can conclude that one network independent from that if it is ESS or IBSS it has always one SSID. It is usually set by the administrator setting up the WLAN and will be unique within a BSS (Basic Service Set) or ESS (Extended Service Set).Example: Extended Service Set (ESS) BSS’s with wired Distribution System (DS) Example: Extended Service Set (ESS) BSS’s and wireless Distribution System (DS) Service Set Identifier (SSID): The Service Set Identifier or Network Name is specified within IEEE 802. However.11 networks to identify a particular network. this feature should be disabled as it may assist hackers or wardrivers in gaining access to a private network. The SSID may be broadcast from an AP within the wireless network to enable Stations to determine which network to “Associate” with. 20 .
21 . The figure shows the flow of an incoming request through the various control mechanisms. an adaptation/policy agent installs policies into the kernel via a special API.IEEE Wave LAN systems. the BSSID is the MAC (Medium Access Control) address of the AP and in Independent BSS or ad hoc networks. with the new capabilities utilized either by user-space agents or applications themselves. The policy agent interacts with the kernel via an enhanced socket interface by sending (receiving) messages to (from) special control sockets. In the infrastructure BSS networks. To give a better idea how it looks like the figure below shows the basic components of the enhanced protocol stack architecture. and actions to perform on the selected traffic. that means that it is in the MAC address format. Protocol Stack for UNIX Also there are known developments for architectural enhancements for Unix-based servers to provide a protocol stack for UNIX. The policies specify filters to select the traffic to be controlled.Basic Service Set Identifier (BSSID) The BSSID is a 48bit identity used to identify a particular BSS (Basic Service Set) within one area. the BSSID is generated randomly. The BSSID identifies the cells and it is 6 octets long. The value of the BSSID is the same as the MAC address of the radio in the AccessPoint. This architecture permits control over an application's inbound network traffic via policybased traffic management. There is also visible a similarity to the NWID in the pre.
11 data plane and can encapsulate MIH messages in data frames. The 802. This is a binary signal that gates when the WLAN and WPAN can each transmit packets.1 Baseband MIH SAP Reference Model for 802. It is similar to the 802.15.11 management frames from the 802.11b MAC and 802. The TX Request and TX Confirm are discreet signals exchanged for every packet transmission attempt. The TX Confirm carries a status value that is one of: allowed or denied.1 LM + LC entities provide status information to the MEHTA control entities.11 does not currently support Class 1 data frames.15.11b / 802. The MIH MLME SAP defines the interface between the MIH Function and the MLME. 802.11 MAC Status Status 802. MIH messages can be transported over the 802. where the LLC SAP (LSAP) defines the interface of the MIH Function with the 802. However. Before the association between Mobile Node and access point takes place.11 PLCP + PHY 802. since 802. the L2 transport of MIH messages can rely on 802.1 Coexistence Mechanism An AWMA transmission control entity is integrated with the WLAN MAC layer and provides a Medium Free signal to the Bluetooth Baseband layer.11 data plane only after the Mobile Node has associated with the 802. The MEHTA control entity receives a per-transmission transmit request (TX Request) and issues a per-transmission transmit confirm (TX Confirm) to each stack to indicate whether the transmission can proceed.15.11 Stack Collaborative Coexistence Mechanism Tx Enable 802.15.Compare Overall Structure of 802.11 The logical placement of the MIH Function in the 802.11 access point.1 Stack TDMA Control Tx Enable 802.11 protocol stack for stations and access points is shown in the figure.3. 22 .15.1 LM + LC Tx Request Tx Confirm (status) MEHTA Control Tx Request Tx Confirm (status) 802.11 management plane (MLME).
so that we can compare better what is the difference between the 802.16. Applications 802.16 The logical placement of the MIH Function in the 802.16 Management Plane. With GPCS a more efficient LLC/SNAP encapsulation (8 bytes overhead) could create the needed room for the MIH Ethertype in 802.Layer 3 Mobility Protocol (L3MP).16 frame. The MIH Function and the Network Control and Management System (NCMS) share the C_SAP and M_SAP for access to the mobility-management services of the Mobility Control Entity and Management Entity in the 802.11 and 802. 23 . The only option available for L2 transport would be to first encapsulate the MIH messages into Ethernet frames with an MIH Ethertype value. Alternatively.16 connections that carry the MIH messages. Handover Policy. since Ethernet CS is not ubiquitous.16 standards and WiMAX only enable the encapsulation of IP packets and Ethernet frames. This approach limits both the efficiency of the L2 transport of MIH messages.16g. The Service-Specific Convergence Sublayer instances currently available in the 802. a solution that enables better efficiency and easier accessibility of L2 transport capabilities could become available with the possible standardization of the Generic Packet Convergence Sublayer (GPCS) recently proposed within 802. and that since it imposes the addition of full Ethernet overhead – at least 18 bytes – to the MIH frame and the availability of L2 transport capabilities for MIH. Transport. and then mandate the adoption of Ethernet CS for 802.16 data frames may take multiple forms.16 protocol stack is shown in the figure. The mechanisms for the direct encapsulation of MIH frames into 802.21 Scope MIH_SAP Media Independent Handover (MIH) Function MIH_SME_SAP MIH Event Service MIH Command Service MIH Information Service LSAP SME MLME_SAP MLME_SAP Logical Link Control (LLC) MLME MAC_SAP MAC PHY_SAP MLME_PLME_SAP PLME_SAP PHY PLME MIH SAP Reference Model for 802. Higher-Layer Mobility Protocol.
once successfully received by the sender.2 LLC and 48-bit addressing as other 802 LANs. completes the process. the receiving station issues an ACK frame that. but the MAC is unique to WLANs. For 802.3. which means an ACK packet is sent by the receiving station to confirm that the data packet arrived intact. In an 802. allowing for very simple bridging from wireless to IEEE wired networks. Applications 802.21 Scope MIH_SAP NCMS Media Independent Handover (MIH) Function MIH Event Service MIH Command Service MIH Information Service CS_SAP Service-Specific Convergence Sublayer (CS) MAC_SAP C_SAP MAC Common Part Sublayer (MAC CPS) Security Sublayer PHY_SAP Management Plane M_SAP Physical Layer (PHY) Data Link Layer As an important part of the protocol stack. collision detection is not possible due to what is known as the “near/far” problem: to detect a collision.3 Ethernet LANs.11 MAC is very similar in concept compared to the 802. Higher-Layer Mobility Protocol. Transport. randomly selected period of time and then transmits if the medium is still free. if no activity is detected. but in radio systems the transmission drowns out the ability of the station to “hear” a collision. which is designed to support multiple users on a shared medium by having the sender sense the medium before accessing it. If the ACK frame is 24 . the data link layer within 802. Handover Policy. A station wishing to transmit senses the air. the Carrier Sense Multiple Access with Collision Detection (CSMA/CD) it is regulated from the protocol how Ethernet stations are going to establish access to the wire and how they detect and handle collisions that occur when two or more devices try to simultaneously communicate over the LAN. a station must be able to transmit and listen at the same time. If the packet is received intact.11 use a slightly modified protocol known as Carrier Sense Multiple Access with Collision Avoidance (CSMA/CA) or the Distributed Coordination Function (DCF). the station waits an additional.11 uses the same 802.Layer 3 Mobility Protocol (L3MP).11 consists of two sublayers: Logical Link Control (LLC) and Media Access Control (MAC). and.11 WLAN. To account for this difference. CSMA/CA attempts to avoid collisions by using explicit packet acknowledgment (ACK). CSMA/CA works as follows. The 802. The 802. 802.
for which retransmission would be expensive from a bandwidth standpoint. Support for Time-Bounded Data 25 . Each packet has a CRC checksum calculated and attached to ensure that the data was not corrupted in transit.3 does not have. where higher-level protocols such as TCP handle error checking. Since RTS/CTS adds additional overhead to the network by temporarily reserving the medium. which is useful in very congested environments or when interference is a factor. so that an 802. allowing the sending station to transmit and receive a packet acknowledgment without any chance of collision. Another MAC-layer problem specific to wireless is the “hidden node” issue. RTS/CTS Procedure eliminates the “Hidden Node” Problem To solve this problem. a collision is assumed to have occurred and the data packet is transmitted again after waiting another random amount of time. in which two stations on opposite sides of an access point can both “hear” activity from an access point. usually due to distance or an obstruction. Since all stations in the network can hear the access point.11 LAN will always have slower performance than an equivalent Ethernet LAN. the 802. When this feature is in use. it is typically used only on the largest-sized packets. This technique reduces the need for retransmission in many cases and thus improves overall wireless network performance. Packet fragmentation allows large packets to be broken into smaller units when sent over the air. it does add some overhead to 802. either because the original data packet was not received intact or the ACK was not received intact. This explicit ACK mechanism also handles interference and other radio related problems very effectively. This is different from Ethernet. but not from each other. a sending station transmits an RTS and waits for the access point to reply with CTS. Finally. the CTS causes them to delay any intended transmissions.not detected by the sending station.11 MAC layer provides for two other robustness features: CRC checksum and packet fragmentation. rendering the process transparent to higher level protocols. However. 802. since larger packets have a better chance of being corrupted.11 specify an optional Request to Send/Clear to Send (RTS/CTS) protocol at the MAC layer. The MAC layer is responsible for reassembling fragments received. CSMA/CA thus provides a way of sharing access over the air.11 that 802.
the MAC Layer manages and maintains communications between 802. a maximum latency is guaranteed. The NAV must be zero before a station can attempt to send a frame. This process reserves the medium for the sending station.javvin. which provides a variety of functions that support the operation of 802. A downside to PCF is that it is not particularly scalable.11 frames. 802. a station must first gain access to the medium. the station must wait a random period of time before attempting to access the medium again. During the periods when the system is in PCF mode.11 standard specifies a common medium access control (MAC) Layer.11 stations contend for access and attempt to send frames when there is no other station transmitting. which is a counter resident at each station that represents the amount of time that the previous frame needs to send its frame. As opposed to the DCF. such as 802. and stations receive data from the access point only when they are polled. distributed coordination function (DCF) and point coordination function (PCF).11 stations (radio network cards and access points) by coordinating access to a shared radio channel and utilizing protocols that enhance communications over a wireless medium. The random delay causes stations to wait different periods of time and avoids all of them sensing the medium at exactly the same time. a station calculates the amount of time necessary to send the frame based on the frame's length and data rate. and after a given time move on to the next station. Since PCF gives every station a turn to transmit in a predetermined fashion.11 MAC specifications through the Point Coordination Function (PCF).11b or 802.com/wireless/MACAddress. If another station is sending a frame. Often viewed as the "brains" of the network. The 802. the MAC Layer checks the value of its network allocation vector (NAV). time is spliced between the system being in PCF mode and in DCF (CSMA/CA) mode. they examine this duration field value and use it as the basis for setting their corresponding NAVs. The station places a value representing this time in the duration field in the header of the frame. transmission.html 26 .11 standard defines two forms of medium access. where control is distributed to all stations. and receiving of 802.Time-bounded data such as voice and video is supported in the 802. stations are polite and wait until the channel is free. This ensures that multiple stations wanting to send data don't transmit at the same time. If a BSS is set up with PCF enabled.11 MAC Layer uses an 802.11a. to perform the tasks of carrier sensing. As a condition to accessing the medium. DCF is mandatory and based on the CSMA/CA (carrier sense multiple access with collision avoidance) protocol. MAC Functional Description The 802. 1 http://www. the 802. In general.1 Before transmitting frames. which is a radio channel that stations share.11 Physical (PHY) Layer. An important aspect of the DCF is a random back off timer that a station uses if it detects a busy medium. When stations receive the frame. which can be ineffective in large networks. Prior to transmitting a frame. in that a single point needs to have control of media access and must poll all stations. No station is allowed to transmit unless it is polled. If the channel is in use.11-based wireless LANs. in PCF mode a single access point controls access to the media. finding the channel idle. With DCF. the access point will poll each station for data.
and colliding with each other. With radio-based LANs. mainly because the station can't have it's receiver on while transmitting the frame. especially when the number of active users increases. a transmitting station can't listen for collisions while sending data. The back off timer significantly reduces the number of collisions and corresponding retransmissions. the enhanced MAC layer is constituted of two Convergence sub-layers.e.transmitting. MAC Intermediate Sub-Layer (MIS) and MAC Lower Sub-layer (MLS). As a result. LLC Convergence Sub-Layer (LLCCS) and Segmentation and Re-assembly (SAR). video applications) and asynchronous (i.11 standard defines the optional point coordination function (PCF) where the access point grants access to an individual station to the medium by polling the station during the contention free period.. the 802. If the sending station doesn't receive an ACK after a specified period of time. the sending station will assume that there was a collision (or RF interference) and retransmit the frame.e. The MLS sub-layer is in charge of building 802. then switches to a contention period when stations use DCF. Stations can't transmit frames unless the access point polls them first. MAC Architecture The new MAC access scheme described hereafter enhances the current 802. The MIS also integrates the Error and Flow Control functions. In addition.11 compatible MPDUs from MIS transfer unit and signaling information. The MIS embeds the core transfer function of the MAC layer and is based on short fixed-size transfer units. and two transfer sub-layers. the receiving station needs to send an acknowledgement (ACK) if it detects no errors in the received frame. The MAC SAP consistency is maintained by the LLCCS sub-layer. As shown in . The MAC SAP is kept identical while the PHY SAP may be modified according to the capabilities of the PHY layer. it can implement the 27 . This process enables support for both synchronous (i.11 MAC. and delivers them to the PHY layer. The period of time for PCF-based data traffic (if enabled) occurs alternately between contention (DCF) periods. The SAR sub-layer performs the adaptation between the variable size packet provided by the LLCCS and the transfer units managed by the MIS.. For supporting time-bounded delivery of data frames. The access point polls stations according to a polling list. e-mail and Web browsing applications) modes of operation.
11i enhances the WEP (Wireline Equivalent Privacy). which is a quick fix of the WEB weaknesses.11 specifications. 2. The IEEE 802.11i is designed to provide secured communication of wireless LAN as defined by all the IEEE 802.11i. IEEE 802.11 frame that get protected are known as additional authentication data (AAD). The standard IEEE 802. authentication and key management. this code is called Michael. CCMP uses AES in counter mode. TKIP uses a message integrity code to enable devices to authenticate that the packets are coming from the claimed source. Counter-Mode/CBC-MAC Protocol (CCMP): a data-confidentiality protocol that is responsible for packet authentication as well as encryption. The additional parts of the IEEE 802.11 committee has addressed the issue by providing what is called WEP (Wired Equivalent Privacy) Authentication: A function that determines whether a Station is allowed to participate in network communication. CCMP uses a 128-bit key.11i is based on the Wi-Fi Protected Access (WPA). a technology used for many years for the WLAN security. For authentication and integrity. CCMP uses Cipher Block Chaining Message Authentication Code (CBC-MAC). Also TKIP uses a mixing function to defeat weak-key attacks.11i has the following key components: 1. in the areas of encryption.11 MAC Protocol Stack Comparison Security Security is one of the first concerns of people deploying a Wireless LAN. which enabled attackers to decrypt traffic. Temporal Key Integrity Protocol (TKIP): it is data-confidentiality protocol and it was designed to improve the security of products that were implemented through WEP. For confidentiality. In IEEE 802.LLC MAC Packet Sequence Number Assignment LLC Sequence Number Assignment Fragmentation Encryption MPDU Header + CRC LLCCS Segmentation Segment Sequence Number Assignment SAR Error and Flow Control MAC MIS Encryption MPDU Header Signalling Insertion MLS PHY Extended MAC PHY Legacy 802. AAD includes the packets source and destination and protects against attackers replaying packets to different destinations. IEEE 802. CCMP protects some fields that aren't encrypted. 28 . the 802.
1X ties a protocol called EAP (Extensible Authentication Protocol) to both the wired and wireless LAN media and support multiple authentication methods. The WEP algorithm is a simple algorithm based on RSA?s RC4 algorithm. 4. which is a Pseudo Random Number Generator (PRNG).11i.1x for key exchange. 802. which restarts the PRNG for each frame.11i client card and access point to negotiate which protocol to use during specific traffic circumstances and to discover any unknown security parameters.3. Eavesdropping Eavesdropping is prevented by the use of the WEP algorithm. where packets may get lost (as any LAN). as well as dynamically varying encryption keys. Because IEEE 802.11i has more than one data-confidentiality protocol.1x: offers an effective framework for authenticating and controlling user traffic to a protected network. initialized by a shared secret key. this is needed in order to work on a connectionless environment.11i provides an algorithm for the IEEE 802.11i: WLAN Security Standards Preventing Access to Network Resources This is done by the use of an Authentication mechanism where a station needs to prove knowledge of the current key. EAP encapsulation over LANs (EAPOL)– it is the key protocol in IEEE 802. Self Synchronizing: The algorithm synchronized again for each message. The first is referred to as the 4-way handshake and the second is the group key handshake. on the sense that an intruder needs to enter the premises (by using a physical key) in order to connect his workstation to the wired LAN. IEEE 802. Two main EAPOL-key exchanges are defined in IEEE 802.IEEE 802. This PRNG outputs a key sequence of pseudo-random bits equal in length to the largest possible packet. Protocol Structure . this is very similar to the Wired LAN privacy. which has the following properties: Reasonable strong: Brute-force attack to this algorithm is difficult because of the fact that every frame is sent with an Initialization Vector. which is combined with the outgoing/incoming packet producing the packet transmitted in the air. 29 .IEEE 802.
but the MAC layers are considerably different. an encoding scheme is used in which a group of 4 bits is encoded as a 16-bit codeword containing fifteen 0s and a single 1. 802. the 802.11 standard includes a common Medium Access Control (MAC) Layer.11b. Both the FHSS and DSSS modes are specified for operation in the 2. This code has the property that a small error in time synchronization leads to 30 .11a and 802. respectively.11b products.95 microns. however. scientific and medical (ISM) band..11a. Each of the five permitted transmission techniques makes it possible to send a MAC frame from one station to another. The third physical layer alternative is an infrared system using near-visible light in the 850 nm to 950 nm range as the transmission medium. rather than any absolute measurements of the phase change.11g. The physical layer basics To know the physical layer terminology we need to understand the essential intricacies of 802. will be to compare the physical layer characteristics of 802. not line of sight) transmission at 0. however. which defines protocols that govern the operation of the wireless LAN.11 standard: 802.11 and HIPERLAN/II have similar physical layer characteristics operating in the 5 GHz band and use the modulation scheme orthogonal frequency division multiplexing (OFDM). as well as a European Telecommunications Standards Institute (ETSI) standard. DBPSK is phase modulation using two distinct carrier phases for data signaling providing one bit per symbol. With a ruling from the Federal Communications Commission that will now allow OFDM digital transmission technology to operate in the ISM band and the promise of interoperability with a large installed base of 802. They differ. which has sometimes been jokingly referred to as the interference suppression is mandatory band because it is heavily used by various electronic products.4 GHz industrial. The differential characteristic of the modulation schemes indicates the use of the difference in phase from the last change or symbol to determine the current symbol's value. High Performance LAN (HIPERLAN/II). Although not detailed here. GFSK is a modulation scheme in which the data are first filtered by a Gaussian filter in the Baseband.11 frames.85 or 0. in the technology used and speed achievable.11a. 2 and 4 bit represent the number of frequency offsets used to represent data symbols of one and two bits.Physical Layer The IEEE (define) 802. many of the same issues will apply. At the forefront of the new WLAN options that will enable much higher data rates are two supplements to the IEEE 802. Two speeds are permitted: 1 Mbps and 2 Mbps. The focus here.11g extension to the standard begins to garner the attention of WLAN equipment providers. using what is called Gray code. in quadrature. The infrared option uses diffused (i. to signal two bits per symbol.11 comprise several alternative physical layers that specify the transmission and reception of 802. In addition. At 1 Mbps.e. With HIPERLAN/II sharing several of the same physical properties as 802. it will offer data rates equal to or exceeding 22 Mb/s with products available late in 2002.11.11b and 802. DQPSK is a type of phase modulation using two pairs of distinct carrier phases. and then modulated with a simple frequency modulation. Another standard that warrants mention in this context is IEEE 802. Both 802.
11 frame that a station wishes to transmit and forms what the 802. which changes with different data rates. due to the low bandwidth (and the fact that sunlight swamps infrared signals). also with only a single 1. Don't expect to see the physical layer fields with 802. 00010100 for 2Mbps. Frame Check Sequence. change radio channels. In practice.11 standard reserves it for future use. This field consists of alternating 0s and 1s. PSDU.11b includes Physical Layer Convergence Procedure (PLCP) and Physical Medium Dependent (PMD) sub-layers.11b is nearly always 11 31 . with 1 and 2 bits per baud.4 WIRELESS LANS 295 possible under current conditions of load and noise. receive signals. the field contains the value of 00001010 for 1Mbps.11b but is not a follow-up to 802. 802. and so on..5. Infrared signals cannot penetrate walls. The PLCP fields.11a. Signal.375 Mbaud. with its binary value equal to the data rate divided by 100Kbps.11b are 1. For example.e. however. Length. The receiver begins synchronizing with the incoming signal after detecting the Sync. the operating speed of 802. however. Service. 2. which stands for Physical Layer Service Data Unit. using phase shift modulation (for compatibility with DSSS). Start Frame Delimiter. These are somewhat sophisticated terms that the standard uses to divide the major functions that occur within the Physical Layer. As with other 802. 5. The data rate may be dynamically adapted during operation to achieve the optimum speed SEC. PLCP Frame Fields The PLCP takes each 802. In fact. The PLCP prepares 802. another spread spectrum technique. its standard was approved first and it got to market first.11 frames for transmission and directs the PMD to actually transmit signals.11 frame being sent).11 standard refers to as a PLCP protocol data unit (PPDU). At 2 Mbps. the actual 802. Nevertheless. This field represents the number of microseconds that it takes to transmit the contents of the PPDU. is a fancy name that represents the contents of the PPDU (i. This field identifies the data rate of the 802. The 802. This ensures that the receiver is initially uses the correct demodulation mechanism. that is one of 0001. and the receiver uses this information to determine the end of the frame.11 Physical layers.4-GHz band. Next. are always sent at the lowest rate.11 frame. which uses 11 million chips/sec to achieve 11 Mbps in the 2. The resulting PPDU includes the following fields in addition to the frame fields imposed by the MAC Layer: Sync. the standard defines this field for containing 16-bit cyclic redundancy check (CRC) result.only a single bit error in the output. 4. 0100. we come to HR-DSSS (High Rate Direct Sequence Spread Spectrum). The MAC Layer also performs error detection functions on the PPDU contents as well.11 radio card removes these fields before the resulting data is processed by the MAC Layer and offered to the analyzer for viewing. This field is always set to 00000000 and the 802. which is 1Mbps.11 analyzers from AirMagnet and Wildpackets. respectively. using Walsh/Hadamard codes. The two slow rates run at 1 Mbaud. the encoding takes 2 bits and produces a 4-bit codeword. It is called 802. Data rates supported by 802. or 1000. The PSDU. In order to detect possible errors in the Physical Layer header. so cells in different rooms are well isolated from each other. The two faster rates run at 1. respectively. This field is always 1111001110100000 and defines the beginning of a frame. with 4 and 8 bits per baud. and 11 Mbps. and so on. this is not a popular option. 0010. alerting the receiver that a receivable signal is present.
802.5 Mbps data rates (in addition to the 1 and 2Mbps rates) utilizing an extension to DSSS called High Rate DSSS (HR/DSSS).11b also defines a rate shifting technique where 11 Mbps networks may fall back to 5.11 committee has produced three different high-speed wireless LANs: 802. In theory it can operate at up to 54 MBps. Spread Spectrum Spread spectrum is a technique trading bandwidth for reliability.11b is slower than 802.11g. The PHY provides three functions.11 architecture (802. An enhanced version of 802. Thirdly. 802. was approved by IEEE in November 2001 after much politicking about whose patented technology it would use. 2 Mbps. The goal is to use more bandwidth than the system really needs for transmission to reduce the impact of localized interference on the media. the PHY provides a carrier sense indication back to the MAC to verify activity on the media. 802.11b) defines 11 Mbps and 5.11a. its range is about 7 times greater. the PHY provides an interface to exchange frames with the upper MAC layer for transmission and reception of data. Spread spectrum spreads the transmitted bandwidth of the resulting signal.11a) defines different multiplexing techniques that can achieve data rates up to 54 Mbps.11 PHY layers.5 Mbps.11 provides three different PHY definitions: Both Frequency Hopping Spread Spectrum (FHSS) and Direct Sequence Spread Spectrum (DSSS) support 1 and 2 Mbps data rates. Frequency Hopping Spread Spectrum (FHSS) 32 . Although 802. 802. First. The concept of PPM is to vary the position of a pulse to represent different binary symbols. Another extension to the standard (802. For 1 Mbps. or 1 Mps under noisy conditions or to interoperate with legacy 802. Infrared (IR) The Infrared PHY utilizes infrared light to transmit binary data either at 1 Mbps (basic access rate) or 2 Mbps (enhanced access rate) using a specific modulation technique for each. It is not yet clear whether this speed will be realized in practice.11a. Infrared transmission at 2 Mbps utilizes a 4 PPM modulation technique.GHz ISM band along with 802. which is more important in many situations.11 physical layer (PHY) is the interface between the MAC and the wireless media where frames are transmitted and received.11a but operates in the narrow 2. It uses the OFDM modulation method of 802.11b.11b. the PHY uses signal carrier and spread spectrum modulation to transmit data frames over the media. An extension to the 802. One can legitimately ask if this is a good thing for a standards committee The 802.11g (not to mention three lowspeed wireless LANs).Mbps. Secondly.4.11b. reducing the peak power but keeping total power the same. What it does mean is that the 802. the infrared PHY uses a 16-pulse position modulation (PPM). and 802.
FHSS’ randomization provides a fair way to allocate spectrum in the unregulated ISM band. Spread-spectrum transmissions can share a frequency band with many types of conventional transmissions with minimal interference. FSK. For example.11 Frequency Hopping Spread Spectrum (FHSS) PHY uses the 2. Frequency Hopping utilizes a set of narrow channels and "hops" through all of them in a predetermined sequence. The third modulation method.MHz wide. which has only two possible frequencies. A pseudorandom number generator is used to produce the sequence of frequencies hopped to. This is accomplished by multiple frequency. which makes it popular for building-to-building links. Its main disadvantage is its low bandwidth. The spread-spectrum signals add minimal noise to the narrow-frequency communications. The 802. FHSS (Frequency Hopping Spread Spectrum) uses 79 channels. It also provides a modicum of security since an intruder who does not know the hopping sequence or dwell time cannot eavesdrop on transmissions. and FHSS offers good resistance to it. As a result. bandwidth can be utilized more efficiently.In FHSS the total frequency band is split into a number of channels. the dwell time. and vice versa. causing them to recede into the background. operating with at 1 or 2 Mbps data rate. is an adjustable parameter. using a pseudorandom sequence known to both transmitter and receiver. Over longer distances. code selected. The instantaneous frequency output of the transmitter jumps from one value to another based on the 33 . Basically. but must be less than 400 msec. the incoming digital stream is shifted in frequency by an amount determined by a code that spreads the signal power over a wide bandwidth. A Frequency-Hop spread-spectrum signal sounds like a momentary noise burst or simply an increase in the background noise for short Frequency-Hop codes on any narrowband receiver except a Frequency-Hop spread-spectrum receiver using the exact same channel sequence as was used by the transmitter. multipath fading can be an issue. each 1. starting at the low end of the 2.4 GHz frequency band is divided into 70 channels of 1 MHz each. In comparison to binary FSK. A spread-spectrum transmission offers three main advantages over a fixed-frequency transmission: Spread-spectrum signals are highly resistant to noise and interference. It is also relatively insensitive to radio interference. The broadcast data is spread across the entire frequency band by hopping between the channels in a pseudo random fashion. Spread-spectrum signals are difficult to intercept. The amount of time spent at each frequency. The process of recollecting a spread signal spreads out noise and interference. The FHSS transmitter is a pseudo-noise PN code controlled frequency synthesizer.4 GHz radio frequency band. the 2. they will hop to the same frequencies simultaneously. Frequency-hopping spread spectrum (FHSS) is a spread-spectrum method of transmitting radio signals by rapidly switching a carrier among many frequency channels. Frequency hopping relies on frequency diversity to combat interference. Every 20 to 400 msec the system "hops" to a new channel following a predetermined cyclic pattern.4-GHz ISM band. FHSS may have 2*10^20 or more. As long as all stations use the same seed to the pseudorandom number generator and stay synchronized in time.
Another important factor in FHSS systems is the rate at which the hops occur. In contrast. In telecommunications. All other receivers will see the spread signal as white or colored noise. instead of adding pseudo-random noise to the data. the amount of redundancy used. Fig. which results in a uniform frequency distribution whose width is determined by the output range of the pseudo-random number generator. The minimum time required to change frequencies is dependent on the information bit rate. The principle of Direct Sequence is to spread a signal on a larger frequency band by multiplexing it with a signature or code to minimize localized interference and background noise. The characteristics of the broadband spreading code are that of pseudorandom noise. DSSS has the following features: for generating spread-spectrum transmissions by phase-modulating a sine wave pseudo randomly with a continuous string of pseudo noise code symbols. the process gain is directly dependent on the number of available frequency choices for a given information rate. which is the reason that it is called spread spectrum. direct-sequence spread spectrum is a modulation technique where the transmitted signal takes up more bandwidth than the information signal that is being modulated. the original signal is recovered by 34 . each bit is modulated by a code. frequency-hopping spread spectrum pseudo-randomly retunes the carrier. To spread the signal. Consequently the receiver synchronized to the code will obtain the narrowband signal. The result is termed the chip rate. A signal structuring technique utilizing a digital code sequence (PN Sequences) having a chip rate much higher than the information signal bit rate. Each information bit of a digital signal is transmitted as a pseudorandom sequence of chips. Varying the instantaneous frequency results in an output spectrum that is effectively spread over the range of frequencies generated. In the receiver. Direct Sequence Spread Spectrum (DSSS) Direct Sequence Spread Spectrum is based on the multiplying of the baseband signal data with a broadband spreading code. and the distance to the nearest interference source. the number of discrete frequencies determines the bandwidth of the system.pseudo-random input from the code generator. each of duration much smaller than a bit. Hence.1 FHSS Spectrum In this system.
The spectral content of an SS signal is shown in Fig. etc). At 54 Mbps.11 standard. First the PN code is modulated onto the information signal using one of several modulation techniques (eg. It uses phase shift modulation at 1 Mbaud.11b uses DSSS to disperse the data frame signal over a relatively wide (approximately 30MHz) portion of the 2.. The scheme used has some similarities to the CDMA system.S.11a. A complex encoding system is used. to use spread spectrum. which is why the Federal Communications Commission (FCC) (define) deems the operation of spread spectrum systems as license free. The 802. 1. different frequencies are used—52 of them. This results in greater immunity to radio frequency (RF) interference as compared to narrowband signaling. either FHSS or DSSS may be used. that rule was dropped as new technologies emerged. Since transmissions are present on multiple frequencies at the same time. This process causes the RF signal to be replaced with a very wide bandwidth signal with the spectral equivalent of a noise signal.4 GHz radio frequency band. The first of the high-speed wireless LANs. 2002).4GHz frequency band. 802. including better immunity to narrowband interference and the possibility of using noncontiguous bands. but different from both CDMA and FHSS. Note that this is just the spectrum of a BPSK signal with a (sin x / x) 2 form. 48 for data and 4 for synchronization—not unlike ADSL. transmitting 1 bit per baud when operating at 1 Mbps and 2 bits per baud when operating at 2 Mbps. The technique has a good spectrum efficiency in terms of bits/Hz and good immunity to multipath fading. It is also a part of the 802. BPSK. Each bit is transmitted as 11 chips. The correlated signal is then filtered and sent to a BPSK demodulator. This is probably the most widely recognized form of spread spectrum. DSSS (Direct Sequence Spread Spectrum) is also restricted to 1 or 2 Mbps. using what is called a Barker sequence.11 Direct Sequence Spread Spectrum (DSSS) PHY also uses the 2.receiving the whole spread channel and demodulating with the same code used by the transmitter. Splitting the signal into many narrow bands has some key advantages over using a single wide band. For years. Part of the motivation for OFDM is compatibility with the European HiperLAN/2 system (Doufexi et al. Then. The DSSS process is performed by effectively multiplying an RF carrier and a pseudo-noise (PN) digital signal. The output is a signal that is a maximum when the two signals exactly equal one another or are "correlated". a doubly balanced mixer is used to multiply the RF carrier and PN modulated information signal. but differs in other ways. uses OFDM (Orthogonal Frequency Division Multiplexing) to deliver up to 54 Mbps in the wider 5GHz ISM band. 35 . 802. Note that in the original 802. but in May 2002. The wide bandwidth provided by the PN code allows the signal power to drop below the noise threshold without loss of information. the FCC required all wireless communications equipment operating in the ISM bands in the U. this technique is considered a form of spread spectrum. QPSK. 216 data bits are encoded into 288-bit symbols. The demodulation process (for the BPSK case) is then simply the mixing/multiplying of the same PN modulated carrier with the incoming RF signal. based on phase-shift modulation for speeds up to 18 Mbps and on QAM above that.11 b and g standards. As the term FDM suggests. The signals generated with this technique appear as noise in the frequency domain.
This is a clever process that enables the data stream to be sent at 2Mbps while using the same amount of bandwidth as the one sent at 1Mbps. For example with 1Mbps operation.Fig. The modulator merely shifts the phase of the center transmit frequency to distinguish a binary 1 from a binary 0 within the data stream. the bandwidth of a DSSS system is a direct function of the chip rate. The result is that systems in a spectrally quiet environment benefit from the possible increase in data transfer rate. The modulator uses similar methods for the higher. 1 BPSK DSSS Spectrum The bandwidth in DSSS systems is often taken as the null-to-null bandwidth of the main lobe of the power spectral density plot (indicated as 2Rc in Fig. One feature of DSSS is that QPSK may be used to increase the data rate. the bandwidth (which sets the process gain) is halved due to the two-fold increase in information transfer. the PMD uses differential binary phase shift keying (DBPSK). This allows a narrower RF bandwidth to accommodate the received signal with the effect of rounding the received pulses in the time domain. specifically 2Rc/RINFO. It should be noted that the power contained in the main lobe comprises 90 percent of the total power. The process gain is reduced because for a given chip rate. the PMD uses differential quadrature phase shift keying (DQPSK).5Mbps and 11Mbps data rates. For 2Mbps transmission. 1). 36 . depending on which data rate is chosen.2 Rc. DSSS Modulation The modulator converts the spread binary signal into an analog waveform through the use of different modulation types. where Rc is the chip rate. The half power bandwidth of this lobe is 1. This isn't really as complex as it sounds. 5. Therefore. This increase of a factor of two bits per symbol of transmitted information over BPSK causes an equivalent reduction in the available process gain. which is similar to DBPSK except that there are four possible phase shifts that represents every two data bits. This is just an extension of the previous equation for process gain.
The first 100 MHz in the lower frequency portion is restricted to a maximum power output of 50 mW. while the third 100 MHz. OFDM operates by dividing the transmitted data into multiple parallel bit streams. 37 .422 2. each with lower relative bit rates and modulating separate narrowband carriers. only allows the use of channels 1 through 11.457 2.484 Various countries limit the use of these channels. however.452 2. new product development has proceeded much more slowly than 802.467 2. the U. has a maximum of 1. The trip in route to the destination will significantly attenuate (define) the signal. This standard uses 300 MHz of bandwidth in the 5 GHz unlicensed national information infrastructure (UNII) band.S.427 2. This complicates matters when designing international public wireless LANs.447 2. which is mainly intended for outdoor applications. authorizes the use all 14 channels.432 2. The spectrum is divided into three domains. but the receiver at the destination will detect the incoming Physical Layer header and reverse (demodulate and dispread) the process implemented by the transmitter The IEEE 802.11a While 802.S.417 2. Japan.Transmit Frequencies The transmitter's modulator translates the spread signal into an analog form with a center frequency corresponding to the radio channel chosen by the user. For example.462 2.412 2.442 2. The second 100 MHz has a higher 250 mW maximum. In that case. and the U.11b.437 2.472 2. This is due to the cost and complexity of implementation. each having restrictions imposed on the maximum allowed output power (see Figure 1). the transmitter outputs the modulated DSSS signal to the antenna in order to propagate the signal to the destination. After RF amplification takes place based on the transmit power you've chosen (100mW maximum for the U. can use channels 1 through 13. referred to as sub-carriers.K.0 W power output. The following identifies the center frequency of each channel: Channel 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Frequency (GHz) 2.). you need to choose channels with the least common denominator.11a was approved in September 1999.
11a with OFDM signals and operating at 5 GHz will bring new challenges in testing. Practice 802. they require the use of equipment from a single source throughout the entire network. This nearly eliminates the need for retransmissions when packet errors are detected. 802.11a products are expected to begin arriving in the first half of 2002. particularly because the data rate will be increasing by a factor of five and using the same bandwidth (20 MHz) to do it. enables the receiver to identify and correct errors made during transmission by sending additional data along with the primary transmission. Each represents four data bits. Four complex modulation methods are employed.11a Design of devices using 802.11b. as well as a method to characterize them. The data rates available in 802. and 64QAM. FEC.11a are touting the availability of operational modes that exceed the 54 Mb/s stated in the specification. These include BPSK. These issues add to the implementation cost of 802. each of these are divided into 52 sub-carriers (four of which carry pilot data) of 300-kHz bandwidth each. overlapping carriers. 802. Four non-overlapping 20 MHz channels are specified in the upper band. 16-QAM. QPSK. Considering the composite waveform resulting from the combination of 52 sub-carriers. because faster data rates are out of the specification's scope.11a specifies eight non-overlapping 20 MHz channels in the lower two bands.11a includes forward error correction (FEC) as part of the specification. together with the type of modulation and the coding rate. reconstructing the original high-rate data stream.11a are noted in Table 2. better phase noise performance is required because of the closely spaced. In addition. which do not have a constant power envelope. Some of the companies developing chipset solutions for 802. This metric is not useful. Although it adds a degree of complication to the Baseband processing. as the true peak power may not occur often. The receiver processes the 52 individual bit streams. Of course. Transmitted signals such as OFDM. 64-QAM has 16 symbols with each representing four data bits. which does not exist within 802. BPSK modulation is always used on the four pilot sub-carriers. depending on the data rate that can be supported by channel conditions between the transmitter and receiver. The high peak-to-average power ratio representative of multicarrier OFDM signals dictates the need for highly linear and efficient amplifiers. Application-specific measurement tools aid in the design and troubleshooting of OFDM signals and systems. 802. 38 . It is usually more meaningful for OFDM signals to associate a percentage probability with a power level.11a products. Quadrature amplitude modulation is a complex modulation method where data are carried in symbols represented by the phase and amplitude of the modulated carrier. so each can be received without interference from another. are not wellcharacterized by peak-to-average power ratio. 16-QAM has 16 symbols. the format requires more linearity in the amplifiers because of the higher peak-to-average power ratio of the transmitted OFDM signal.The sub-carriers are orthogonal.
microwave ovens.375 Msym/s. coupled with the rotation. 802. The IEEE 8102. Products are now widely available. shows the number of decibels above the average power on the horizontal axis.4835 GHz).11b.40 to 2. To achieve data rates of 5.11b payload data rates. while the data rate varies to match channel conditions by changing the spreading factor and/or the modulation scheme.11b uses eight-chip complementary code keying (CCK) as the modulation scheme to achieve the higher data rates. represents the symbol conveying the four or eight bits of data. The chip rate remains consistent with the original DSSS system at 11 Mchip/s. 802. the spreading length is first reduced from 11 to eight chips.11b also operates in the highly populated 2.375 MHz symbol rate. Instead of the Barker codes used to encode and spread the data for the lower rates. and personal area networks (PANS).11b 802. The CCDF. so the ISM band accommodates only three non-overlapping channels spaced 25 MHz apart.5 Mb/s/1.11b designates an optional frequency agile or hopping mode using the three non-overlapping channels or six overlapping channels spaced at 10 MHz. 39 . For all 802. If time gating were not used. the preamble and header are sent at the 1 Mb/s rate. it is necessary to transmit 4 bits/symbol (5. The CCK approach taken in 802.11b. uses all but two of the bits to select from a set of spreading sequences and the remaining two bits to rotate the sequence. CCK uses a nearly orthogonal complex code set called complementary sequences. This increases the symbol rate from 1 Msym/s to 1.4 GHz ISM band (2. which provides only 83 MHz of spectrum to accommodate a variety of other radiating products. an 8 bits/symbol.5 and 11 Mb/s payload data rates in addition to the original 1 and 2 Mb/s rates. other WLANs. The selection of the sequence. For the 5. is an extension of the 802. including cordless phones. which keeps the QPSK spread-spectrum signal and still provides the required number of bits/symbol.0. In this measurement.11 DSSS system previously mentioned and supports higher 5. which was approved by the IEEE in 1999. which is simply the more common cumulative distribution function (CDF) subtracted from 1. and the installed base of systems is growing rapidly.375 Msym/s) and for 11 Mb/s. To help mitigate interference effects. A CCDF measurement would be made over several bursts to improve the accuracy of the measurement. an instrument with time-gating capability is used to select only the active portion of the burst (see Figure 2 lower trace).5-Mb/s bit rate with a 1. This metric links a percentage probability to a power level.A more meaningful method for viewing OFDM signal power characteristics uses the complementary cumulative distribution function (CCDF).5 and 11 Mb/s. 802. This makes susceptibility to interference a primary concern. The occupied bandwidth of the spread-spectrum channel is 22 MHz. and percent probability on vertical axis (see Figure 2 upper trace). the periods when the burst is off would reduce the average power calculation.
operational characteristics have been compared to those of cellular systems. But as interference is introduced into the channel.11b The 20 MHz-wide bandwidth of WLAN signals makes power envelope measurements difficult because most spectrum analyzers have resolution bandwidth filters that are limited to 10 MHz or less. If the channel is clear. An advantage of 802.11b networks are generally rangelimited by multipath interference rather than the loss of signal strength over distance. Both 802. making WLAN signal analysis more accurate. where frequency planning of overlapping cells minimizes mutual interference support mobility and seamless channel handoff.11b and 802. Therefore. A site survey must be thorough and realistic to 40 .4 GHz. albeit more robust. 802. Requirements must be researched and well-documented. Comparison of 802. then the modes with the highest data rates are used. transmission scheme.11b A drawback of the 5 GHz band.11b are at a disadvantage compared to the greater number of channels available to 802. Due to the periodic nature of the signal.Practice 802. the radio will fall back to a slower.11a. (The last one-quarter of the symbol pulse is copied and attached to the beginning of the burst. When it comes to deployment of a wireless LAN. The additional channels allow more overlapping access points within a given area while avoiding additional mutual interference. The 802. reduces the inter-symbol interference (ISI) caused by multipath interference. including anticipated roaming and data rates needed for applications to be used at specific locations. Network planning is critical to the development of an optimized system.) To contrast. the junction at the start of the original burst will always be continuous. The three non-overlapping frequency channels available for IEEE 802. which has received considerable attention. The slower symbol rate and placement of significant guard time around each symbol. floors. Vector signal analyzers are available with information bandwidths that are considerably greater than 20 MHz.11a is its intrinsic ability to handle delay spread or multipath reflection effects.11a use dynamic rate shifting where the system will automatically adjust the data rate based on the condition of the radio channel. using a technique called cyclical extension. and furniture) than those at 2. the signal is considerably attenuated by the time the power is measured within the instrument.11b standard uses error vector magnitude (EVM) as a measure of modulation quality.11a and 802. is its shorter wavelength. Each network must be customized to satisfy the planned applications and the physical environment. The underlying philosophy of EVM is that any signal deteriorated by a noisy channel can be represented as the sum of an ideal signal and an error signal. Higher-frequency signals will have more trouble propagating through physical obstructions encountered in an office (walls. This measurement has become common for most wireless applications. The test instrument determines the error signal by reconstructing the ideal signal based on detected signal information and subtracting it from the actual signal at each sample point.
That version of 802. It would be unrealistic to expect to realize the full data rate capability (54 Mb/s) of 802. These wireless networks will require increasing data rates to provide the simultaneous distribution of Internet data.11 addresses mobility. This deployment. 802. mainly due to the physical conveniences of radio-based communication. The widespread acceptance of WLANs depends on industry standardization to ensure product compatibility and reliability among the various manufacturers. The data rates supported by the original 802.11-based networks have seen widespread deployment across many fields.11 products increase dramatically in the near future as businesses discover the increased productivity provided by ‘untethered’ networks. This paper addressed the availability aspect of that equation.adequately characterize the RF environment of the proposed wireless network in terms of range. Even during these lean economic times. The most critical issue affecting WLAN demand has been limited throughput. But as has been shown. Security as a part of the protocol stack has also been mention because the protocol stack as a part of the software is the first part in what people are interested in. Cost vs. security. ease-of-use and reliability.11 standard are too slow to support most general business requirements and have slowed adoption of WLANs. WLAN products require that special attention be given to design verification and characterization because standardized operation across multivendor products may be required. 802. it is almost a foregone conclusion that end-users will be demanding continuous improvements in functionality. Recognizing the critical need to support 41 . Expect to see availability of 802. Conclusion Wireless networking has a promising future with 802. however.11 MAC layer and described the architecture and the main functions of the MAC as part of the protocol stack also we made a few comparisons.11 specification in 1997 as the standard for wireless LANs. and the dynamic nature of wireless LANS while keeping compatibility with 802-type legacy networks. but already robust WLAN market is projected to grow by an order of magnitude over the next five years. These tools can be used within the manufacturing process to generate and analyze production metrics for process and product improvement. test tools are available to quickly diagnose problems and isolate them throughout all design segments. high-quality video and audio in the office or at home.11 leading the way as the standard for adoption in local networking environments. 802. To provide an efficient development environment. WE think that with the comparisons we have offered an interesting issue and that the description has more efficiency. was predicated in part on the user expectation of confidentiality and availability. the new.11a if the access points of an existing 802.11b at any range. when there is a reduced demand for technology products. We examined the 802. The Institute of Electrical and Electronics Engineers (IEEE) ratified the original 802.11a is faster than 802. channel interference and delay spread. reliability. Testing is critical to any product development process.11 provides for 1 Mbps and 2 Mbps data rates and a set of fundamental signaling methods and other services. performance requirements need thorough analysis during the network planning stage to arrive at the appropriate implementation decision. In addition to higher data rates.11b network optimized to operate at full speed (11 Mb/s) — were simply replaced.
11b standard (also known as 802. revenue gain. giving mobile workers much-needed freedom in their network access. Businesses of all sizes can benefit from deploying a WLAN system. or a remote branch office. and configuration flexibility.11.11 High Rate) for transmissions of up to 11 Mbps. Users need access to the network far beyond their personal desktops. allowing them to design. So at the end we hope that we have offered a general overview of the IEEE 802. Today’s business environment is characterized by an increasingly mobile workforce and flatter organizations. which provides a powerful combination of wired network throughput. The IEEE 802.000 per user—measured in worker productivity. With a wireless network. and geographic boundaries. WLANs will be able to achieve wireless performance and throughput comparable to wired Ethernet. the IEEE recently ratified the 802. saving both effort and dollars. parts that are defined in the IEEE Standard. small office. With 802. which promises to open new markets for WLANs in large enterprise. Much of these workers’ productivity occurs in meetings and away from their desks. 42 . and cost savings—over wired alternatives. the cafeteria. and enhance networks without regard to the availability of wiring. workers can access information from anywhere in the corporation—a conference room.higher data-transmission rates. mobile access. Employees are equipped with notebook computers and spend more of their time working in teams that cross functional. Wireless LANs provide a benefit for IT managers as well. organizational. organizational efficiency. especially protocol stack and physical layers. deploy. Global regulatory bodies and vendor alliances have endorsed this new high-rate standard. WLANs fit well in this work environment.11b. The economic benefits can add up to as much as $16.11 is a huge topic and the Standards we can download free from the internet. and home environments.
Local Area Network MAC .Physical Layer Convergence Procedure 43 .Pulse Position Modulation HR-DSSS .Dynamic frequency selection MIMO .Differential Binary Phase Shift Keying PN – Pseudo Noise RF – Radio Frequency OFDM .Cumulative Distribution Function CCDF . Multiple Access with Collision Detect ISO .Differential Quadrature Phase Shift Keying DBPSK .Abbreviation WLAN – Wireless Local Area Network LAN .Frequency Hopping Spread Spectrum PPM .Physical Medium Dependent PLCP .Institute of Electrical and Electronics Engineers Standards Association EVM .Quality Of Service TCP .Complementary Cumulative Distribution Function FEC .Carrier Sense.Logical Link Control DS .Error Vector Magnitude CCK .Transmit power control DFS .Direct Sequence Spread Spectrum QoS .Wired Equivalent Privacy NAV .Network Allocation Vector CRC .Medium Access Control PHY – Physical Layer ISM . and Medical CSMA/CD .complementary code keying CDF .Direct Sequence Spread Spectrum FHSS . Multiple Output LLC .Multiple Input.Distribution System WEP .High Rate Direct Sequence Spread Spectrum PPDU .Orthogonal Frequency Division Multiplexing FFC .Federal Communications Commission DSSS .International Standards Organization DSSS . Scientific.Cyclic Redundancy Check IEEE-SA .Industrial.Forward Error Correction QAM DQPSK .PLCP Protocol Data Unit PMD .
MAC methodologies are employed within Ethernet. and can be found within many different technologies.16c) although initial interest is confined to the line of sight bands . All other receivers will see the spread signal as white or colored noise. The ISO developed the OSI (Open System Interconnection) reference model which is a popular networking reference tool. The result is termed the chip rate. Note: Wi-Fi not WiFi. MAC handles access to a shared medium.Wireless Local Area Network This is a generic term covering a multitude of technologies providing local area networking via a radio link.Wireless Fidelity Wi-Fi is an interoperability standard developed by WECA (Wireless Ethernet Compatibility Alliance) and issued to those manufacturers whose IEEE 802.11a. Bluetooth.International Standards Organization The International Standards Organization is responsible for a wide range of standards. IrDA (Infrared Data Association) and DECT (Digital Enhanced Cordless Telecommunications) etc. For example.5GHz. Physical Link A Physical Link is the connection between devices.Glossary ISO .11a and 802.16a) and 10GHz . The characteristics of the broadband spreading code are that of pseudorandom noise. 44 . Equipment passing these tests carries the Wi-Fi logo. MAC .2.11b equipment has passed a suite of basic interoperability tests.16 suite of standards.Direct Sequence Spread Spectrum Direct Sequence Spread Spectrum is based on the multiplying of the Baseband signal data with a broadband spreading code. In general terms. WLAN . 3. including those relevant to networking. and UMTS etc.8GHz.66GHz (IEEE 802. WiMAX .Worldwide Interoperability for Microwave Access The term WiMAX has become synonymous with the IEEE 802.Medium Access Control Media Access Control is the lower of the two sublayers of the Data Link Layer. Examples of WLAN technologies include Wi-Fi (Wireless Fidelity). Consequently the receiver synchronized to the code will obtain the narrowband signal. HiperLAN. 802. Wi-Fi .5GHz and 5. GPRS. These define the radio or air interface within two broad radio bands 2GHz to 11GHz (IEEE 802.11b and 802. DSSS . It is anticipated that WiMAX will be used initially as a backhaul connection with other technologies such as Wi-Fi being used to cover the “final mile”.
Depending upon the communication system. IEEE 802.11a Part of the IEEE 802.Transmission Control Protocol Transmission Control Protocol is a reliable octet streaming protocol used by the majority of applications on the Internet. this wireless local area network technology is comprised of a high speed physical layer operating in the 5GHz unlicensed band and supports data rates up to 54Mbps. point to point service between hosts. IEEE .11 family of specifications. It is independent of the underlying radio interface protocols enabling the introduction of alternative GPRS radio solutions with minimal changes to the Network Switching System.QoS . This is sent along with data so that a parity check of the received data can be conducted. reliably.11a and IEEE 802. CRC . LLC .11bspecifications. It provides a connection-oriented.Logical Link Control In the GPRS system the LLC protocol provides a highly reliable ciphered logical link between the MS (Mobile Station) and SGSN (Serving GPRS Support Node). 45 . full-duplex.Cyclic Redundancy Code A linear error code that is generated using a polynomial function on the data to be sent. SNR (Signal to Noise Ratio).Quality of Service The performance of a communications channel or system is usually expressed in terms of QoS (Quality of Service).Institute of Electrical and Electronics Engineers The Institute of Electrical and Electronics Engineers is a professional organization whose activities include the development of communications and network standards. Several manufacturers have developed equipment which is capable of operating in accordance with both IEEE 802. TCP . the remainder from the process being the CRC. maximum and mean throughput rate. BER (Bit Error Ratio). QoS may relate to service performance. priority and other factors specific to each service. Equipment operating in accordance with the IEEE specifications and passing the Alliances interoperability tests is able to display the Wi-Fi logo.
PDA (Personnel Digital Assistant) etc. Stations are able to move between BSS within a single ESS yet still remain “connected” to the fixed network and so continue to receive emails etc.Extended Service Set An Extended Service Set is comprised of a number of IEEE 802.11 network which provides the point of interconnection between the wireless Station (laptop computer.11 BSS (Basic Service Set) and enables limited mobility within the WLAN (Wireless Local Area Network). The 802.IEEE 802. It is usually set by the administrator setting up the WLAN (Wireless Local Area Network) and will be unique within a BSS (Basic Service Set) or ESS (Extended Service Set).Service Access Point A conceptual point where a protocol layer offers access to its services to the layer above or below.11 .11b is currently the most popular wireless networking technology. The standards have been divided into sub groups with 802.) and the wired network. 46 .11b specifications.11b Part of the IEEE 802.11 family of specifications. IEEE 802.11a based equipment is now commercially available and provides data rates up to 54Mbps and operates in the 5GHz ISM band. it will carry out a resuscitation procedure with the new AP (Access Point).Access Point An Access Point is a device found within an IEEE 802.Direct Sequence Spread Spectrum) enabling data rates of up to 11Mbps to be achieved. or wardrivers” in gaining access to a private network. AP (Access Points) and a DS (Distribution System). AP .11b currently the most common. The equipment operates in the 2. Wireless Medium.Wireless This is an IEEE (Institute of Electrical and Electronic Engineers) technical standard covering WLAN (Wireless Local Area Network) technology. However.4GHz ISM (Industrial Scientific and Medical) band.4GHz unlicensed band and utilizes HR/DSSS (High Rate .11 networks are comprised of Stations. IEEE 802. The SSID may be broadcast from an AP (Access Point) within the wireless network to enable Stations to determine which network to “Associate” with.Service Set Identifier The Service Set Identifier or Network Name is used within IEEE 802. SSID . this feature should be disabled as it may assist “hackers. As a Station moves into a new BSS. Equipment operating in accordance with the IEEE specifications and passing the Wi-Fi Alliances interoperability tests is able to display the Wi-Fi logo. This provides a wireless communication at up to 11Mbps and operates within the 2. SAP . IEEE 802.11a and IEEE 802.11 networks to identify a particular network. Several manufacturers have developed equipment which is capable of operating in accordance with both IEEE 802. ESS .
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s3://commoncrawl/crawl-data/CC-MAIN-2016-30/segments/1469258943369.84/warc/CC-MAIN-20160723072903-00060-ip-10-185-27-174.ec2.internal.warc.gz
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CC-MAIN-2016-30
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https://brainmass.com/economics/econometrics/introductory-economics-questions-410214
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math
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Please see attachment for the fully formatted solution.
1) An economics department at a large state university keeps tract of its majors' starting salaries. Does taking econometrics effect starting salary? Let Sal = starting salary in dollars, GPA = grade point average on a 4.0 scale, Metrics = 1 if the student took econometrics, and 0 otherwise. Using a sample size of 50, we obtain
Sal(hat) = 24200 + 1643GPA + 5033METRICS, R-squared=0.74
(se) (1078) (352) (456)
The variance-covariance matrix of the estimated coefficients is
Intercept GPA Metrics
Intercept 116299737 -370463 -124114
GPA -370463 124108 22428
Metrics -124114 22428 208216
a) Show how you can obtain the standard error of the coefficient on GPA, Metrics from the variance-covariance matrix.
b) At the 95% level, testing the testing the overall significance of joint hypotheses.
(Hint: H0: B1 = 0, B2 = 0)
c) At the 95% level, test the hypothesis that the marginal effect of GPA on starting salary is equal to the marginal effect of Metrics on starting salary. (Hint: H0: B1 = B2)© BrainMass Inc. brainmass.com March 21, 2019, 10:15 pm ad1c9bdddf
a) Recall that the i,j-th element on the variance covariance matrix represents cov(xi, xj). Also, cov(xi, xi) = var(xi).
Thus the estimated variances are var(GPA) = 124108 and var(METRIC) = 208216. Standard error (which is the estimated standard deviation) are se(GPA) = 352.29 and var(METRIC) = 456.31. These are the same as the ones calculated by the computer.
b) To test β1 = β2 = 0, we need to use the Bonferroni adjusted t test. To begin, we compute 2 t test statistics
let t1 = (β1_hat - 0)/se(β1_hat) and t2 = (β2_hat - 0)/se(β2_hat)
thus, simplifying gives t1 = 1643/352 = ...
This solution explains how to solve various problems related to economics.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514574409.16/warc/CC-MAIN-20190921104758-20190921130758-00011.warc.gz
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CC-MAIN-2019-39
| 1,782 | 19 |
http://www.icesculptingtools.com/home/homepage/hand-tools/ice-picks-and-chippers/
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math
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Pickle Fork Medium
6 Prong Pick Deluxe$9.98
Consider leaving an ice pick behind when delivering a larger sculpture.
If your customer can break the sculpture into smaller pieces & haul it out, you may not have to return. We have ours trained pretty well.
Ice Picks (5 Pack)$9.98
Ice Pick Pro$4.98
1584 picks $2.59/ea
1 line of text (Company name or website)
Customized Ice Picks$4,102.56
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s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948530841.24/warc/CC-MAIN-20171213201654-20171213221654-00430.warc.gz
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CC-MAIN-2017-51
| 386 | 9 |
https://s1dd.com/2016/01/addition-practice-for-kids/
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math
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Click the gear icon at the top for settings. You can change the maximum and minimum number of digits, whether the problem should have carry-overs or not and how many numbers to add. If carry-overs are disabled, only two numbers will be shown and the minimum digits will be ignored for the second number.
The answer is automatically checked as soon as you are done. If the answer is wrong on the first try, you need to click the "Check Answer" button when you are done.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600402124756.81/warc/CC-MAIN-20201001062039-20201001092039-00493.warc.gz
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CC-MAIN-2020-40
| 468 | 2 |
https://www.hometheatershack.com/tags/cinemascope/
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math
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General Screen Discussion
I am wonder if this is right or I am missing something.
I heard a youtube conversation with a THX guy that the viewing angle you want for cinemascope is 50deg.
Using this calculator
If I put the 100" width I get a distance of...
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s3://commoncrawl/crawl-data/CC-MAIN-2021-17/segments/1618038078021.18/warc/CC-MAIN-20210414185709-20210414215709-00079.warc.gz
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CC-MAIN-2021-17
| 254 | 5 |
https://queenscrap.blogspot.com/2009/05/katzs-conflicts-of-interest.html
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math
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From the NY Observer:
On Wednesday, May 13, Melinda Katz strode into the second floor of the National Arts Club on Gramercy Park South, where a suited crowd of mostly white, male real estate people sat expectantly. It was a luncheon for the Associated Builders and Owners of Greater New York. She was the guest speaker.
Ms. Katz moved easily in this crowd, shaking hands, touching arms, assuming the swagger and bravado peculiar to the real estate set. At times, she seemed much like a beloved younger sister bantering with older brothers.
As chair of the land-use committee, Ms. Katz has become a major player in the real estate community. She has presided, with relatively little controversy, over enormously important land-use decisions, ranging from the redevelopment of Willets Point to the rezoning of 125th Street in Harlem.
But she has done so with a marked disregard for appearances. A swift glance at her campaign finance records reveals contributions from pretty much every single major player in the New York real estate scene, including many with business before her committee: the Rudins, the Toll Brothers and the Walentases, to name a very few.
Friday, May 29, 2009
Posted by Queens Crapper at 12:44 AM
Labels: comptroller, conflict of interest, Melinda Katz
"Grease my palm. No, no, slather it on. That's right, between the fingers and around the wrist."
"Make snide comment. No, no, don't be too specific. That's right, keep it real general and widely applicable to conceal any ignorance of the topic at hand."
Hows this for specific? She's corrupt and is a parasite and gives nothing to the people for the salary she collects.-
Besides "strode" is a good description. She walks like Ellen Degeneris.
She is giving a speech Monday to The Cement League - they are in charge of building hi-rises
What kind of insensitive people are you? She has a child (and a nanny) now; she doesn't have time to attend a grassroots pancake breakfast to raise money for her absolutely no chance bid for comptroller. Taking these bribes, I mean campaign contributions, is the only way for her to raise these needed funds. Come on now anything she does has to be forgiven she falls under the banner of “ SIGLE MOM”. How dare you all question her motives, I said “SINGLE MOM”!
What a hateful blog this is becoming with the select few making everone look like redneck swill
First it's anti-immigrant (really anti-Hispanic regardless if they are legal or even citizens). Then it was anti-black. Now its anti-gay.
Good job guys.
I read because I hate what is being built and how my tax money is spent. But I get embarassed reading this hateful crap sometimes.
ignore the last blog! has to be katz..
she lines her pockets everyday with developers and the real estate. isn't it ashame that at one time her area thought she would be good for the community. We need to vote all these assholes out of their seats and bring in new fresh inspiring people for all communities. TIME FOR CHANGE, and it starts in NOV...
Sometime ago, I offered a looseleaf binder full of printouts regarding all of C.M. Katz's campaign contributors for anyone to peruse at a public hearing chaired by Jessica Lapin which was held at Queensborough Hall.
It included scads of real estate developers (major & minor), contractors, their various subs and affiliated trade groups, building management and maintenance companies, etc.
It depleted my ink cartridge!
I encourage EVERYONE to look up those same statistics for yourselves on line.
They are very "specific" and extremely damning!
One can accurately conclude from these that Melinda Katz is, indeed, the unashamed poster child of NYC's real estate industry!
Anonymous (Katz, actually) said:
"First it's anti-immigrant (really anti-Hispanic regardless if they are legal or even citizens). Then it was anti-black. Now its anti-gay."
Talk about vague. Some specifics, please, Melinda-San.
oh someone playing the race card, it isn't hate when you are exposing corruption. damn whiners. do you need a hanky?
Army recruitment article was the recent one that was anti-Hispanic and anti-immigrant (when the recruitment was for documented immigrants still alot fo people ranted)
The MTA diversity article was recent for Anti-black (Claimed they all eat McD's and smoke on the trains, no one ever seen any MTA employee smoke on a train)
Now anti-gay (Ellen Degeneres, everytime Katz is mentioned, etc)
And I didn't evenpost the complaint earlier and I know it's getting bad on this site
she voted for wilets point to and must of got a lot of $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ to
When all the old yentas pass who will vote for her?
Oh please Melinda Katz is NOT gay- she kind sorta plays that way to hide the fact that in her 16 years of feeding at the public trough she must have been knocking boots with some body(ies). No one is single 16 years unless maybe their boyfriend(s) is(are) unavailable?
When she rails about fighting the old boys club I have to Laugh. Are those the old boys who HANDED her her assembly seat at age 28?Chartier Morris Hevesi? No girls eh?
And no I like Black people and immigrants and gays and lesbians. Melinda does walk like Ellen Degeneres, she's just kind of not feminine. And outside of politics no one would look twice at her, but with the land use committee in her pocket...well that's better than a pair of 4 inch Jimmy Choos.
She believes her own press. She's not all that, she just thinks she is.
I love this blog with all the racist and anti immigrant comments!
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s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948858.7/warc/CC-MAIN-20230328104523-20230328134523-00446.warc.gz
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CC-MAIN-2023-14
| 5,589 | 42 |
http://www.wyzant.com/geo_Inglewood_College_Algebra_tutors.aspx?d=20&pagesize=5&pagenum=5
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math
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Pacific Palisades, CA 90272
Grade and high school math tutor
...They are not integers, they are not fractions (with repeating decimal expansions like 1/3), and they are not irrational numbers (like the square root of 2). It took the field of mathematics a long time to figure out that they are in a separate category of number...
including algebra 2
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s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609535775.35/warc/CC-MAIN-20140416005215-00371-ip-10-147-4-33.ec2.internal.warc.gz
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CC-MAIN-2014-15
| 349 | 4 |
https://www.mathway.pro/fractions/what-is-1.8-as-a-fraction
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math
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What is 1.8 as a fraction?
Converting decimals to fractions is an essential skill. It bridges the gap between two numeric representations. In this guide, we will detail the process of converting 1.8 to a fraction.
You can select other values to familiarize yourself with the conversion guide.
Understanding the decimal: “1.8”
A decimal number has an integer and fractional parts. The decimal point separates the two. The integer is to the left, the fraction to the right. For example, in 1.8, 1 is the integer and 8 is the fraction.
- For the numerator:
- We start with the number 1.8.
- By removing the decimal point, we derive the numerator as 18.
- For the denominator:
- Each position after the decimal represents a division by 10.
- Thus, having 1 positions after the decimal equates to 10 or 101.
- The factors for the numerator and the denominator are numbers that can evenly divide each of them.
- For instance, the factors of 18 include 1, 2, 3, 6, 9, and 18.
- The factors of 10 comprise 1, 2, 5, and 10.
- Greatest Common Divisor (GCD):
- It's the largest number that can evenly divide both the numerator and the denominator.
- In this instance, the GCD for 18 and 10 is 2.
The factors of 18 are:
The factors of 10 are:
Conversion formula (equation):
What is a decimal?
A decimal is a numeral system with a point. This point divides the integer from its fractional part. It provides a straightforward way to express and work with values less than one.
What is a fraction?
A fraction is a mathematical expression of two parts: the numerator on top and the denominator below. It represents partial values, showcasing relationships or comparisons.
A fast guide to the conversion:
Numbers exist in different formats: decimals, fractions, and percentages. Today, we'll convert the decimal 1.8 into its fractional form. This conversion shows the core essence of the number. And making mathematical operations more intuitive.
- Step 1: Laying the groundwork. Before diving into conversion, let's frame our decimal as a fraction over 1. This offers a clean slate, making the subsequent steps systematic. 1.8 =
- Step 2: Gauging the decimal's depth. Decimals vary. Some are short; others are long. Our focus? The number of digits after the decimal point. For 1.8, we have three digits. This insight nudges us to elevate our fraction by a factor of 10 for each digit. Factor = 101 = 10
- Step 3: Amplifying the fraction. Taking our coefficient, it's time to strengthen both the numerator and denominator. The intent? To equalize the fraction by the depth of the decimal. =
- Step 4: Simplifying. Elegance lies in simplicity. Our mission now is to shorten the . This requires seeking common divisors. Here, 2 is our ally, dividing both parts seamlessly. =
The decimal 1.8, when unfurled and explored, translates seamlessly into the fraction 1.8/1. This manual provides an introduction to numbers and deepens number skills. Such conversions, though seemingly elementary, pave the way for advanced mathematical prowess.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100112.41/warc/CC-MAIN-20231129141108-20231129171108-00375.warc.gz
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CC-MAIN-2023-50
| 3,021 | 31 |
https://hcmop.wordpress.com/2012/04/04/relating-areas-and-angles-using-cevas-theorem-and-menalaus-theorem/
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math
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Ceva’s theorem and Menalus’ theorem are widely applied in geometry problems in various ways. As long as the shape that Ceva’s theorem and Menalaus’ theorem is associated with appears in a geometric diagram, it is often useful to write down the expression of these theorem applied to the question and see if it of any use. However, these theorems are not only useful in relating the ratios of lengths. Consider the following problem from the recent APMO.
(APMO 2012 P1) Let be a point in the interior of a triangle , and let be the points of intersection of the line and the side of the triangle, of the line and the side , and of the line and the side respectively. Prove that the area of the triangle must be 6 if the area of each of the triangles and is 1.
This is a very popular diagram that often comes out in MO problems. There are several ratios involved in this diagram that makes it so accessible to problem setters. In particular, this triangle can form 6 Menalaus theorem relationships. To solve the problem above, we shall use the following three equations:
However, it is not very useful to find the ratios of lengths since we are not given any details about the segments in the triangles. However, we can use Menalaus’ theorem to relate the area of triangles. We let be the area of triangle , be the area of triangle and be the area of triangle . Replacing the ratio of lengths using the ratios of areas, we have the following system of simultaneous equations:
There are many ways to solve this simultaneous equation. It is possible to solve the system of equations using inequalities too. By rearranging and adding the equations in the system, we obtain . Using Ceva’s theorem to relate areas, we also have . Using AM-GM inequality, we know that . Since the equality case is only established when all terms are equal, we have and hence the area of the large triangle is 6.
Ceva’s theorem can also be used to relate angles too. From the above diagram, we also have:
Using this form of Ceva’s theorem, we can easily prove the three angle bisectors of a triangle are concurrent i.e. intersect at a common point. The following problem form IMO Shortlist 2001 yields a very beautiful application of Ceva’s theorem in its angular form:
Let be the center of the square inscribed in acute triangle with two vertices of the square on side . Thus one of the two remaining vertices of the square lies on side and the other on segment . Points and are defined in a similar way for inscribed squares with two vertices on sides and , respectively. Prove that lines are concurrent.
Sine rule helps in solving the above problem. Do give it a try.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-22/segments/1526794867841.63/warc/CC-MAIN-20180526170654-20180526190654-00592.warc.gz
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CC-MAIN-2018-22
| 2,662 | 9 |
https://www.jiskha.com/display.cgi?id=1255043547
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math
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posted by Anonymous .
b) Suppose you want to cover the backyard with decorative rock and plant some trees as the first phase of the project. You need 30 tons of rock to cover the area. If each ton cost $60 and each tree is $84, what is the maximum number of trees you can buy with a budget for rock and trees of $2,500? Write an inequality that illustrates the problem and solve. Express your answer as an inequality
I am not good at word problems can someone help me?
Math Inequalities -
You cannot copy and paste to this post. Please type out your question.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889473.61/warc/CC-MAIN-20180120063253-20180120083253-00321.warc.gz
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CC-MAIN-2018-05
| 559 | 5 |
https://www.onlinemathlearning.com/fractions-same-numerator.html
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math
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LaTonya has 2 equal sized hotdogs. She cut the first one into thirds at lunch. Later she cut the second hotdog to make double the number of pieces. Draw a model of LaTonya’s hotdogs.
a. How many pieces is the second hotdog cut into?
b. If she wants to eat 2/3 of the second hotdog, how many pieces should she eat?
Look again at your models of LaTonya’s hotdogs. Let’s change the problem slightly.
What if LaTonya eats 2 pieces of each hotdog?
Figure out what fraction of each hotdog she eats.
She eats 2/3 of the first one and 2/6 of the second one.
Did LaTonya eat the same amount of the first hotdog and the second hotdog?
No. But she ate 2 pieces of each hotdog.
Why is the amount she ate different?
The number of pieces is the same, but the size of each piece is different.
The more you cut up a whole, the smaller the pieces get.
So eating 2 pieces of thirds is more hotdog than 2 pieces of sixths.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141187753.32/warc/CC-MAIN-20201126084625-20201126114625-00527.warc.gz
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CC-MAIN-2020-50
| 1,247 | 17 |
https://medhaavee.com/iq-riddles-with-answers/
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math
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Grow your knowledge with the help of the below easy, hard IQ riddles with answers.
IQ Riddles With Answers For Kids And Students
1. What number gives the same amount when it is added to 3 as when it is multiplied by 3? Answer: 11⁄2. We need 3x = x + 3, so 2x = 3, or x = 2⁄2 (or 11⁄2).
2. What do math teachers like to eat? Answer: Pi(e).
3. Ten copycats were sitting on a boat when one jumped out. How many were left? Answer: None. They were copycats so they all jumped out.
4. If you were born in an odd-numbered year, will you celebrate your 50th birthday in an odd or even year? Answer: Odd.
Also get: Interview Riddles
5. If you multiply me by any other number, the answer will always remain the same. What number am I? Answer: 0
6. If there are three cups of sugar and you take one away, how many do you have? Answer: One cup.
7. I have two coins equaling fifteen cents. One of them is not a nickel, so what are the two coins? Answer: A dime and a nickel; one of them is not a nickel but one of them is.
8. I am a three-digit number. My second digit is four times bigger than the third digit. My first digit is three less than my second digit. Who am I? Answer: 141.
9. You are participating in a race. If you overtake the last person in a race, what position are you in? Answer: You cannot overtake the last person in a race. You can only overtake a person who is ahead or in front of you in a race.
10. Which two numbers make a one-digit number when they are multiplied but a two-digit number when they are added together? Answer: One and nine.
11. When you add eight 8’s, the result you get will be 1,000. When solely using addition, how is this possible? Answer: 888 + 88 + 8 + 8 + 8 = 1,000
12. When does 11 + 3 = 2? Answer: On a clock.
13. If today is Sunday, what day of the week will it be 4400 days from today? Answer: Thursday. The division 4400 / 7 gives 628 remainders 4. This means that 628 weeks go by and there are 4 extra days. This puts us on Thursday.
14. If there are 3 apples and you take 2 away, how many apples do you have? Answer: You will have 2 apples since you took 2 apples.
15. Find the odd one out: First, Second, Third, Fourth, Fifth, Sixth, Seventh, Eighth. Answer: Forth. It should be the Fourth.
16. What is two days after the day after the day before yesterday? Answer: Tomorrow. The day before yesterday was two days ago; the day after the day before yesterday was yesterday; two days after that (yesterday) is tomorrow.
17. What is special about the number 854,917,632? Answer: It’s the numbers from one to nine in alphabetical order.
18. Six glasses are in a single row. The first three glasses are filled with water and the rest are empty. Move only one glass so that empty and filled glasses are aligned alternately. Answer: Pour the water from the 2nd glass into the 5th glass.
19. Only mental calculation is allowed. Take 1000 and add 40 to it. Now add another 1000. Now add 30. Add another 1000. Now add 20. Now add another 1000. Now add 10. What is the total? Answer: 4100. A lot of people will get 5000 which is wrong.
20. If you were running a race, and you passed the person in 2nd place, what place would you be in now? Answer: You would be in 2nd place. You thought first place, right? Well, you passed the guy in second place, not first.
21. If you had one match and entered a dark room in which there were a kerosene lamp, an oil burner, and a wood-burning stove, what do you light first? Answer: The match.
22. If you are eight feet away from a door and with each move you advance half the distance to the door. How many moves will it take to reach the door? Answer: It will never be reached because you will always be half-distance.
Image by Freepik
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CC-MAIN-2023-40
| 3,717 | 26 |
https://www.sciencesnail.com/science/category/physical%20chemistry/2
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math
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How to dilute a solution to Avogadro’s limit
First, we must estimate the mass of a grain of salt. An average grain of regular table salt has dimensions of roughly 0.3 mm × 0.3 mm × 0.3 mm . This corresponds to a volume of 2.7 × 10^-5 mL. The density of NaCl is 2.16 g/mL and so the grain contains roughly 5.83 x 10^-5 g. The molecular weight of NaCl is 58.44 g/mol, giving 9.98 × 10^-7 moles (n). Now we calculate what volume (V) we’d have to dissolve the salt grain in to reach Avogadro’s limit concentration (C).
Reaching Avogadro’s limit by serial dilution
Suppose we instead dilute our initial 1 L salt grain solution by adding 100 mL of it into a fresh 900 mL of water. Each dilution now achieves a 10-fold dilution factor.
Avogadro’s limit in homeopathy
The principles of homeopathy hopefully surprise and confuse you, as it goes against our most basic understanding of physics, chemistry and biology. All scientifically understood drugs exert their effects through physical or chemical interactions with the biological system. Surely a drug cannot interact with/modulate a biological system if none is administered. The most commonly proposed mechanism for homeopathy is the concept of “water memory,” that water molecules are permanently imprinted with the memory of compounds they’ve previously interacted with. Proponents even claim that this information is transmitted to new water molecules upon dilution. Of course, there are serious problems with these ideas. The structural features of water have been heavily studied and are well understood. Water molecules dynamically associate in a hydrogen-bonding network. Presumably, it is through this that the water could “remember.” Unfortunately for homeopathy though, it seems that water is rather forgetful. Experimentally, this network’s has been shown to be remarkably short-lived with all memory gone after only 50 femtoseconds .
The concept of water memory does not even require a scientific rebuttal to expose its absurdity, however. If water is imprinted with all chemicals it has ever meet, then just about all water on Earth would already have memory of near every conceivable medicine or toxin. Furthermore, recall the central premise that potency increases with dilution. It follows that the addition and subsequent dilution of a specific medicine would be ineffective at developing that particular activity over all others already present.
The probability of finding a molecule below Avogadro’s limit
An interesting aspect of Avogadro’s limit is that here the law of large numbers no longer applies. Calculations of moles or concentrations near or below Avogadro’s limit must consider probability. Under these conditions, finding a molecule of interest in a given volume of a solution is probabilistic. Not only this, there is even uncertainty in the concentration of a prepared solution, as long as it is made by serial dilution. I am unaware of any attempts to account for the probabilistic nature of this situation, and so I conclude this article by developing a numerical solution below.
Suppose we have a solution of a molecule of interest at concentration C with volume VT. We withdraw a sample of volume VS. What is the probability of the sample containing nS molecules?
Broadly, the probability of an event is given by the ratio of how many ways there are for the desired outcome to occur over the total number of possible outcomes. We can calculate the probability of the stated event by developing expressions for these two quantities. In doing so, it is helpful to divide the total solution into imaginary Vs-sized portions, with the molecules randomly distributed among them.
We can evaluate the total number of possible outcomes from our random sample by noticing that this is equivalent to a classic combinatorics problem: sorting x identical objects into r distinct groups. The number of possible outcomes is given by the following combination :
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CC-MAIN-2023-50
| 3,964 | 12 |
https://www.hindawi.com/journals/jps/2009/716364/
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math
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Computational Procedures for a Class of GI/D/k Systems in Discrete Time
A class of discrete time GI/D/ systems is considered for which the interarrival times have finite support and customers are served in first-in first-out (FIFO) order. The system is formulated as a single server queue with new general independent interarrival times and constant service duration by assuming cyclic assignment of customers to the identical servers. Then the queue length is set up as a quasi-birth-death (QBD) type Markov chain. It is shown that this transformed GI/D/1 system has special structures which make the computation of the matrix R simple and efficient, thereby reducing the number of multiplications in each iteration significantly. As a result we were able to keep the computation time very low. Moreover, use of the resulting structural properties makes the computation of the distribution of queue length of the transformed system efficient. The computation of the distribution of waiting time is also shown to be simple by exploiting the special structures.
In most communication systems studied in discrete time, the interarrival times of packets usually follow independent general distribution. Hence, this arrival of packets in an ATM switch can be represented as independent general distribution. On the other hand, multiple packets (53 byte cells) are transmitted simultaneously through identical transmission lines in an ATM switch. Therefore, an ATM switch can be considered as a multiserver system with deterministic service time as the packets are of same size. Hence, the performance of an ATM switch can be analyzed by studying a GI/D/ system with finite support on interarrival times.
To the best of our knowledge GI/D/ systems in discrete time have never been analyzed with focus on computational efficiency in the literature. In this paper, matrix-geometric method is used to analyze a class of GI/D systems in discrete time with focus on computational efficiency. The class of GI/D systems considered in this paper has finite support for the interarrival times and the service is first-in first-out (FIFO) order. The idea used here to analyze the waiting time distribution of this multiserver system is due to Crommelin . This idea is also used in analyzing the distribution of waiting time of multiserver systems with constant service time duration for both continuous and discrete times such as [2–20]. The idea is that the assignment of customers to servers in cyclic order does not change the distribution of waiting time in a multiserver system where customers are served in FIFO order in the identical servers with same constant service time. In this case, only an arbitrary server can be studied to compute the waiting time distribution. Crommelin analyzed the waiting time distribution of an M/D/ system by converting the system to an /D/1 system. The work of Crommelin is followed by some studies on deterministic service time multiserver systems in both continuous time and discrete time. Iversen decomposed an M/D/ queue with FIFO into /D/ queues with FIFO for analyzing waiting time distribution. Franx developed expression for the waiting time distribution of the M/D/ queue by a full probabilistic analysis requiring neither generating functions nor Laplace transforms. Wittevrongel and Bruneel used transform-based method to analyze multiserver system with constant service time and correlated arrival process. Nishimura studied a MAP/D/ system in continuous time using spectral technique. Takine analyzed a MAP/D/ system and computed the distribution of queue length by first characterizing the sojourn time distribution. Kim and Chaudhry also computed the distribution of queue length of a discrete time GI/D/ queue after computing the distribution of waiting time by using distributional Little's law and transform-based method. Exact elapsed time from the time of last arrival is approximated with time average in . Chaudhry et al. analyzed the distribution of waiting time of a MAP/D/ system in discrete time. Later Alfa gave a simple and more efficient computational scheme for the MAP/D/ system in discrete time. Alfa also carried out algorithmic analysis for a discrete time BMAP/D/ system by exploiting its structural properties. Chaudhry et al. and Alfa [2, 3] used matrix geometric method for analysis. There is some work on the GI/D/ queues such as Wuyts and Bruneel and Gao et al. [10–14] which used transform-based method. There are other algorithms for multiserver queues with constant service times such as GI/D/1 and GI/D/c queues by Chaudhry , /D/ queue by Chaudhry and Kim , /D/ queue by Chaudhry et al. and Franx , Ph/D/ queues by van Hoorn , and an /D/ queue by Xerocostas and Demertzes .
In this paper, the cyclic assignment of customers to servers is used to model a class of GI/D/ systems in discrete time as a single server system with the assumption of first-in first-out (FIFO) service order. The modeling as single server system sets up the distribution of queue length as a quasi-birth-death (QBD) which has some structural properties. Analysis of the GI/D/ system is carried out efficiently by exploiting these structural properties in this paper. This paper has three contributions—reductions in the computational complexities of (i) the matrix , (ii) the distribution of queue length of the transformed system, and (iii) the distribution of waiting time. The first contribution is that the time complexity of the computation of the matrix is reduced by decreasing the number of multiplications in each iteration from to while requiring the same number of iteration as the natural iteration. Here, is the support of general interarrival times, is the number of servers, and is the duration of service. The second contribution is that the distribution of queue length of the transformed system is computed efficiently by exploiting the special structures of the system. The third and most important contribution is that the computation of the distribution of waiting time is simplified. Chaudhry et al. and Alfa [2, 3] computed the distribution of waiting time using the technique for phase type service time distribution. However, it is shown in this paper that the computation of the distribution of waiting time for deterministic service time distribution in discrete time does not need the complicated steps of phase type service time distribution.
The rest of the paper is organized as follows. Section 2 introduces the GI/D/ system and explains the modeling of this multiserver system into a single server system. Section 3 discusses the special structures of the matrices of our model and exploitation of those structures for efficient computation of the matrix . The computation of the distributions of queue length and waiting time is explained in Sections 4 and 5, respectively. Some numerical examples are provided in Section 6 to show the suitability of our proposed method to compute the matrix . Section 7 concludes the paper.
2. The GI/D/ System
The class of GI/D/ system in discrete time considered here has finite support for interarrival time which is of general distribution. Moreover, the customers are served in FIFO service order in the identical servers which have constant service time.
The interarrival times, , of this multiserver system are general, independent, and identically distributed (iid) with distribution , .
We let be the mean arrival rate of customers to the system with . The interarrival times can be represented as a Markov chain where each state represents the elapsed time using the approach followed in the analysis of discrete time GI/G/1 and /G/1 systems by Alfa . General independent interarrival times can be represented as an absorbing Markov chain with a transition matrix where is a square matrix of order , is a column vector of order , , is an matrix of zeros, and is a column vector of ones of order . The case of infinite interarrival times can be captured for . However, the interarrival times are finite if data is collected from practical systems. Therefore, general independent interarrival times with finite support () are considered in this paper. The probabilities of interarrival times can be represented by a vector . Here, and . This general arrival process can be defined by phase type distribution of dimension in elapsed time representation. The parameters , , and are given as , , for , , , for all and where for and is the transpose of matrix .
The general independent arrival process can be represented by two square matrices and of order where represents a transition from phase to phase with arrivals. Now, and can be represented in terms of phase type distribution . Here, and . The first column of is nonzero and this column is the vector . The matrices and are similar to the matrices of zero or one arrival for MAP distributed interarrival times. However, they are special cases because the matrices and only capture general and independently distributed interarrival times.
A discrete time Markov chain can be considered to represent this general arrival process where is the number of arrivals up to and including time and is the phase of the next arrival at time .
The matrix is stochastic. If is the invariant probability vector for arrival in different phases then and . The arrival rate is given by .
There are identical servers in a GI/D/ system. The service times are of constant duration units (). Service time can be represented as phase type distribution in elapsed time form with representation of dimension where and , is an identity matrix of order . Another column vector can be defined for this phase type distribution .
The condition that ensures the stability of GI/D/ system is where . We assume that the system is stable (i.e., ).
2.1. The Model of Arrivals to an Arbitrary Server
Let us assume that customers are served in first-in first-out (FIFO) order in this GI/D/ system. It can also be assumed that the servers are assigned with customers in a cyclic manner particularly during the idle times. This assignment of customers to servers does not affect the waiting times experienced by the customers as the servers are identical. Using the same approach of Alfa [2, 3] we can consider the case of the th () server. In this approach if the first customer is assigned to the first server then the th server will be assigned with , , th customers. In this case we only need to study one server, an arbitrary server (). This arbitrary server sees the arrival of customers to be according to another independent general process which we call . This arrival process is a -fold convolution of original arrival process to the multiserver system. can be defined by two square matrices and of order . The element ( and ) represents transition from phase to phase with arrivals. Both and can be considered to comprise of square block matrices of order . In this representation for , for and for and or . Similarly for and in all other cases for . The first column and the last row of are zero as the first column and last row of are zero. Only the last rows of are nonzero and only the first element of each of these rows is nonzero.
The matrices and exhibit behavior similar to the matrices and except that () is a new matrix representing the arrival of supercustomers where a supercustomer is the last customer to form a group of customers or the arrival of regular customers to an arbitrary server. On the other hand, () represents arrival of regular customers to the GI/D/ system. Now, this new arrival process to an arbitrary server can be represented by a discrete time Markov chain where is the number of supercustomers that have arrived by time and is the phase of the next arrival at time .
The matrix is stochastic. If , a matrix, is the invariant probability vector for arrival in different phases then and . can also be represented as where is of size . The supercustomer arrival rate is given by .
2.2. The New Model /D/1 System
A model can be developed for a single server queue with arrival and deterministic service time of units following the techniques of Alfa . This single server queue is called /D/1 system and the resulting model is of QBD process type for which standard methods can be used . Let us assume that , , and represent the number of supercustomers, the arrival phase of the next supercustomer, and the elapsed time of the supercustomer in service, respectively, at time in this QBD model. The state space can be represented by for and the state space is for . The total state space is . The transition matrix for this /D/1 system is
where , , , , , and .
Let be the stationary distribution of where , and () is the probability vector representing supercustomers in the /D/1 system. During the computation of the stationary distribution of queue length , a matrix is computed first which is followed by the computations of and . Then is computed using the matrix-geometric result . Here, the matrix is of order which is the minimal nonnegative solution of . The entry of represents the expected number of visits into , starting from , before the first return to level (). Here, is the number of supercustomers in the system and and are any of the phases of the system.
3. Structures of the Matrices
This section explains the structures of the block-matrices , , , , , and and how their structures are exploited for efficient computation of the matrix . The structures of these matrices are also fully exploited in computing the stationary distribution of the queue length in Section 4.
Here, and are matrices of order and , respectively. The first column and the last row of the matrix are zero. On the other hand, only the last rows of the matrix are nonzero and among these last rows only the first column is nonzero which is . The matrix can be represented as where .
The matrix is of order which can be represented as where
is a matrix of order . Moreover, has nonzero columns and each column contains only one nonzero element. has nonzero elements in th () positions. is also a matrix of order matrix. Furthermore, only the first column of is nonzero which contains nonzero elements. has nonzero elements in th () positions.
The matrices , , and are of size . The matrix can be represented as
where and are both matrices of order , and . Hence,
There are nonzero columns in and each column contains only one nonzero element. has these nonzero elements in th () positions. Only the first column of is nonzero with nonzero elements in th () positions.
The matrix can be represented as where
Here, and are both matrices of order . and . There are nonzero columns, nonzero rows, and nonzero elements in . Moreover, each nonzero row or each nonzero column of contains only one nonzero element. On the other hand, there are nonzero columns in from column to each having nonzero elements.
The matrix can be represented as . has only nonzero rows which are in its last rows and nonzero columns from column to column .
3.1. Computation of the Matrix
The matrix can be computed using any of the three linearly convergent formulae from Neuts —, , and which are known as natural, traditional, and -based iterations, respectively. Here, is the th iteration value of , , is a square matrix of the same order as the matrix and . The matrix is the minimal nonnegative solution of . The entry represents the taboo probability of starting from for the eventual visit of the Markov chain to level by visiting under the taboo of level . Any of these three methods terminate in the th () iteration for where is a very small positive constant. In each step, natural itearation requires two matrix multiplications whereas traditional iteration requires three matrix multiplications and one matrix inversion. On the other hand, -based iteration requires two matrix multiplications and one matrix inversion in each step. Traditional iteration requires less number of iterations than natural iteration does while the number of iterations required for -based iteration is less than the numbers of iterations for other two methods. Alfa and Xue developed efficient inversion technique for -based iteration for the matrix in analyzing the queue length of a GI/G/1 system in discrete time.
There are several efficient methods for computing the matrix for a QBD such as logarithmic reduction algorithm of Latouche and Ramaswami , cyclic reduction technique of Bini and Meini , and invariant subspace approach of Akar and Sohraby . These quadratically convergent algorithms compute a matrix which is the minimal nonnegative solution to . Here, is an matrix for matrices , , and where () represents the probability that starting from state () the Markov chain eventually visits the level () and does so by visiting the state (). The matrices , , and are related and these relations can be found in . Similar to the computation of the matrix , the matrix can be computed using three different iterations—natural, traditional, and -based. Generally a zero matrix is used for the initial value for iterations of the matrix (i.e., ). Latouche and Ramaswami showed that if iterative steps are needed to converge for a very small positive constant such that for the -st natural iteration , then logarithmic reduction requires not more than iterations and overall complexity of their algorithm is . They also developed logarithmic reduction technique for the matrix . However, the quadratic convergent algorithms [26–28] are not considered here for computing the matrix as these algorithms involve inversion of matrices which cannot exploit the special structures of the matrices , , and .
The complexity of computing the matrix can be reduced by using the structure of . The first rows are zero in . Therefore, the first rows are zero in . Among the last rows of , rows are zero which are rows , , . The corresponding rows are zero for . The matrix can be represented as follows:
Here, is a matrix of order which contains th rows as zero rows.
The computations of and require and block-matrix multiplications, respectively, where each block matrix is of order . Similarly, the computation of requires multiplication of and which involves block-matrix multiplications. Inversions of both the matrices and generate block matrices each of order . Inversion of the matrix needs to be computed only once for traditional iteration whereas inversion of the matrix needs to be computed in each iteration for -based iteration. On the other hand, the multiplication of and in traditional iteration and the multiplication of and in -based iteration involve and block-matrix multiplications, respectively, where each block matrix is of order . Each iteration of natural method requires substantially less multiplications than each step of traditional and -based methods do. Therefore, we use natural iteration for developing our algorithm for computing the matrix in this paper.
The matrix of (3.7) can be computed using the natural iteration with the following equation:
where is the th iteration value of with the initial value . The terminating condition for this iteration is same as the terminating condition for natural, traditional, and -based methods but needs to consider only nonzero rows of .
The number of multiplications required to compute is . Similarly, the number of multiplications required to compute both and for is . In the same way, the number of multiplications required to compute is . On the other hand, the numbers of multiplications required to compute by block-matrix multiplication of and and by block-matrix multiplication of and are and , respectively. Hence, the numbers of multiplications required for computing from and from in one iteration are and , respectively. Therefore, the total number of multiplications required to compute all block matrices 's () of the matrix in one iteration is . By exploiting the special structures of the matrices the computational complexity is decreased in each iteration from to which is substantial reduction in computations. Moreover, the memory requirement for our method is also less than that of natural, traditional, -based methods and quadratic convergent algorithms.
4. Stationary Distribution of P for /D/1 System
Two methods for computing stationary distribution for the Markov chain represented by the matrix are briefly presented here which are followed by Section 4.1. Section 4.1 explains how the second method can be made efficient by utilizing the special structures of the matrices.
In the first method an invariant probability vector needs to be computed for a stochastic matrix . The row vector is computed from and . Then and can be expressed as and where .
In the second method, we need to compute , the invariant probability vector of a stochastic matrix defined as . Next a positive real number is computed as follows . On the other hand, is computed using the relations and . and can be computed as , .
4.1. Exploiting Structures for Stationary Distribution
The stationary distribution of can be computed by first computing the probability vector which can be made efficient by exploiting the special structures of the matrices , and . The special structures of the matrices , , , , , , and are due to the special structures of the matrices , , , and . Here, how the computations of these matrices can be carried out efficiently is explained.
Computing requires multiplications by using block-matrix inversion where , , , and . can be expressed as where .
can be represented as
where , , and . The computation of requires instead of multiplications by utilizing the special structure of .
The matrices and can be represented as
where , , for , and . Here, and for are matrices of order and , respectively. Here, multiplying by and by both require multiplications due to utilizing special structures of matrices. If there were no special structures of , and then the multiplying by and by would require and multiplications, respectively.
The matrix of can be represented in the following form:
where is a matrix of order . Here,
We can recursively define , for , for , , for , and .
Let us define another invariant probability vector of the stochastic matrix where is a row vector having scalar elements. Here, is computed from , and for . Next a positive real number is computed as where and . and are both of size . The column vectors and can be expressed as and , respectively. Both and are of size . and for . There are only nonzero elements in and nonzero elements in for . On the other hand, for and .
can be computed as using multiplications instead of multiplications using the special structure of . Let us partition and as and , respectively, for efficient computation of . Here, and for are of size and , respectively. is computed as and these vector-matrix multiplications can be made efficient by considering only nonzero rows of .
can be computed as for () using multiplications instead of multiplications as has only nonzero rows.
5. Waiting-Time Distribution of /D/1 System
The waiting time distribution of an arbitrary customer in /D/1 system is the same as the waiting time distribution of a customer in GI/D/ system. Let be the probability that a customer (i.e., the th customer of a supercustomer group) has to wait units of time. is computed after obtaining the probability vectors 's where is the stationary probability vector that a supercustomer sees supercustomers ahead of him. can be expressed as . Similarly, can be expressed as and can be expressed as .
Two approaches for computing the distribution of waiting time are presented in Sections 5.1 and 5.2, respectively. The first method uses the approach for phase type service time distribution and this approach is used in the analysis of Alfa and Chaudhry et al. . In the second method we develop a computationally simpler approach. This is a simplified version of Method 1. It simplifies the computations of the probability vectors 's for and , the probability mass function of the distribution of waiting time.
5.1. Method 1
The expressions for the stationary distribution of an entering supercustomer to find different number of supercustomers ahead of him in the system can be written as
In this method, the probability that the waiting time is units of time is computed as
where is a square matrix of order that represents the probability distribution of unit of work in system at the arrival of a supercustomer who finds that supercustomers are ahead of him. For example, the element represents the probability that the service of the arbitrary supercustomer, who arrives to find supercustomers in the system with remaining workload of units begins in phase , given that service of the supercustomer who was in service at the arrival of the arbitrary supercustomer was in phase . The matrix is computed in the following way:
5.2. Method 2
This method simplifies the computation of and . Moreover, this method does not need to compute the matrix which simplifies the computation of the probability mass function of the distribution of waiting time.
The special structures of , , , and can be exploited to reduce the number of operations to compute and . Only the first columns of the matrices and are nonzero. Therefore, the first element of is nonzero and the remaining elements of are equal to zero. This phenomenon is quite intuitive as the arriving supercustomer does not find anybody ahead of him, he can start his service in phase instantly and the arrival process for next supercustomer starts at phase 1 of the phases. The first element of is computed as
columns of are nonzero which are column 2 to column . On the other hand, only the first column of is nonzero. Therefore, the first elements of are nonzero. It is quite intuitive that the first elements of are nonzero (i.e., is nonzero and ) as the arriving supercustomer finds supercustomers ahead of him, the arrival process of next supercustomer starts at phase 1 of phases and the ongoing service is in the start of phase . The nonzero elements 's of the vector can be computed using
The special structures of and for result in
If a supercustomer sees supercustomers ahead of him then he has to wait or amount of time for (or units of time for ) before getting service.
First we consider the case of . In this case, , , and . For , (5.3) can be written as
For we can assume and . In this case, .
For , we have from (5.7).
For we can assume and . In this case, . Therefore, we can express for ,
Now, we consider the case . Let us define as the th column of an identity matrix of order . It is found that for and
which are proved by induction in . If the matrix is nonzero then it contains nonzero elements in the first column. The nonzero first column in nonzero is due to elapsed time representation of service time where service starts at phase 1.
Now, . Therefore, we can express for ,
Alfa [2, 3] pointed out that for . However, he did not provide any explicit expression for and his method needs to compute the first column of the matrix recursively as our method does not need to compute. Use of (5.10) to compute can significantly reduce the complexity of computing the distribution of waiting time for any multiserver queueing system in discrete time with deterministic service time such as BMAP/D/ system of Alfa and MAP/D/ systems of Alfa and Chaudhry et al. . Equation (5.10) reduces the usage of memory as there is no need to compute and store the matrix for different values of and .
Now, if an arriving customer has to wait units of time () then there are supercustomers ahead of him and the customer, who is in service in the server, is in phase of service time. Therefore, the probability of a customer waiting () units of time can be expressed by using value of of (5.8) and (5.10) in (5.2)
6. Numerical Examples
This section compares our proposed method of computing the matrix with natural, traditional, and U-based methods and Logarithmic Reduction technique for the matrices and . Three different matrices , , and are computed in each of these methods. The values of the convergence constant used are , , , and . Three identical servers (i.e., ) are considered with duration of service as eight time units (i.e., ) and the support for interarrival times is assumed to be four (i.e., ). Three different probability vectors , , and are used for interarrival times for numerical examples. A program is written in C programming language for different methods to compute the matrices , , and . The program was executed in SunOS 5.10 on an i386 processor which operates at 2660 MHz and has an i387 compatible floating point processor. The number of iterations and time in seconds is recorded in Tables 1–6 for each method from the execution of the program. “LR” is used to denote logarithmic reduction technique in those tables.
Tables 1 and 2 show the number of iterations and time required to compute the matrices , , and for the interarrival times having probability vector . The means of interarrival time and arrival rate for this probability vector are 3.1 time slots and 0.3226/time slot, respectively, and . Tables 3 and 4 show the number of iterations and time required for the interarrival times having probability vector . The means of interarrival time and arrival rate for the probability vector are 3.0 time slots and 0.3333/time slot, respectively, and . Our methods of computing the matrix are found to require least time for these two examples among the methods used here though logarithmic reduction technique and U-based methods require less number of iterations. Tables 5 and 6 show the number of iterations and time required for the interarrival times having probability vector . The means of interarrival time and arrival rate for this probability vector of interarrival times are 2.7 time slots and 0.3704/time slot, respectively, and . Table 6 shows that for the interarrival times probability vector , Logarithmic Reduction techniques for the matrices and require less time to converge than our method of exploiting the structure of the matrices , , and . The relative advantage of our method of exploiting the structures to Logarithmic Reduction techniques decreases with respect to time with increasing value of .
In this paper, a class of GI/D/ systems in discrete time is analyzed by converting it to a single server queue problem with a convoluted arrival process. Then the stationary distribution of the length of the queue of the transformed system is analyzed using QBD approach. The special structures of the matrices make the computation of the matrix and the distribution of queue length of the transformed system efficient. Numerical examples show that our proposed method for computing the matrix can be as efficient as quadratic convergent algorithms such as logarithmic reduction technique. It is also shown that for deterministic service time the waiting time distribution does not need any cumbersome computation like phase type service time distribution.
This research is partially supported by grants from NSERC to A. S. Alfa.
C. D. Crommelin, “Delay probability formulae when the holding times are constant,” Post Ofiice Electrical Engineers Journal, vol. 25, pp. 41–50, 1932.View at: Google Scholar
M. L. Chaudhry, B. K. Yoon, and K. C. Chae, “Waiting-time distribution of a discrete-time multiserver queue with correlated arrivals and deterministic service times: system,” Operations Research Letters, vol. 30, no. 3, pp. 174–180, 2002.View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
P. Gao, S. Wittevrongel, and H. Bruneel, “Delay analysis for a discrete-time GI-D-c queue with arbitrary-length service times,” in Proceedings of the 1st European Performance Engineering Workshop (EPEW '04), pp. 184–195, Toledo, Spain, September 2004.View at: Google Scholar
P. Gao, S. Wittevrongel, and H. Bruneel, “Analysis of discrete-time multiserver queues with constant service times and correlated arrivals,” in Proceedings of the 12th International Conference on Analytical and Stochastic Modelling Techniques and Applications (ASMTA '05), pp. 1–8, Riga, Latvia, May 2005.View at: Google Scholar
P. Gao, S. Wittevrongel, J. Walraevens, and H. Bruneel, “Analytic study of multiserver buffers with two-state Markovian arrivals and constant service times of multiple slots,” Mathematical Methods of Operations Research, vol. 67, no. 2, pp. 269–284, 2008.View at: Publisher Site | Google Scholar | Zentralblatt MATH | MathSciNet
S. Nishimura, “A spectral analysis for MAP/D/N queue,” in Advances in Algorithmic Methods for Stochastic Models, G. Latouche and P. Taylor, Eds., pp. 279–294, Notable, Neshanic Station, NJ, USA, 2000.View at: Google Scholar
M. F. Neuts, Matrix-Geometric Solutions in Stochastic Models, vol. 2, Johns Hopkins University Press, Baltimore, Md, USA, 1981.View at: MathSciNet
G. Hadley, Linear Algebra, Addison-Wesley, Reading, Mass, USA, 1969.
M. M. Rahman and A. S. Alfa, “Computational procedures for a class of GI/D/ systems in discrete time—an extended analysis,” http://www.ee.umanitoba.ca/~mmrahman/gidkreport.pdf.View at: Google Scholar
P. A. Watters, Solaris 10: The Complete Reference, McGraw-Hill, New York, NY, USA, 2005.
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CC-MAIN-2023-50
| 33,302 | 97 |
http://fixunix.com/hewlett-packard/510424-wtb-hp-75-math-pac-1-a-print.html
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math
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WTB HP 75 Math Pac 1
If anyone here has the HP 75 Math Pac module that adds matrix ops
they'd be willing to sell, I'd certainly be interested in it!
I have a copy of the instructions but an original would be nice, too.
Information Systems Administrator
St. Regis Mohawk Tribe Environment Division
412 State Route 37
Akwesasne, NY 13655
(518) 358-5937 Ext. 117
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s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982292975.73/warc/CC-MAIN-20160823195812-00266-ip-10-153-172-175.ec2.internal.warc.gz
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| 359 | 9 |
http://www.jetradar.co.nz/routes/trn/prg?dont_redirect_please=true&marker=direct
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math
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Best airfares from Turin to PragueJetradar compares airfares from Turin to Prague with hundreds of airlines (including Air China, Lufthansa, British Airways) and dozens of online travel agencies in one single place. We provide you with all the options to find the cheapest flights to Prague, you choose where to book. Also there you can find airlines special offers from Turin to Prague. The cheapest round-trip price found within the last week was NZ$279. The cheapest oneway price found within the last week was NZ$95.
Cheapest Flights from Turin to Prague
Airline One Way One Way / Round-Trip Round-Trip Air China NZ$2,878 NZ$400 Lufthansa NZ$859 NZ$216 British Airways NZ$247 NZ$358 Turkish Airlines NZ$230 NZ$435 Swiss International Air Lines NZ$492 Find
Airline One Way One Way / Round-Trip Round-Trip Vietnam Airlines NZ$3,157 Find KLM NZ$237 NZ$211 Air Berlin NZ$185 NZ$397 Air France NZ$276 NZ$230 Austrian Airlines NZ$382 NZ$464
Airline One Way One Way / Round-Trip Round-Trip Alitalia NZ$209 NZ$217 S7 Airlines NZ$963 Find Scandinavian Airlines NZ$270 NZ$654 Hahn Air Find Find Brussels Airlines NZ$220 NZ$245
Airline One Way One Way / Round-Trip Round-Trip BMI Regional Find NZ$255 Lufthansa CityLine NZ$345 NZ$230 TAM Linhas Aereas Find NZ$2,090 Iberia NZ$210 NZ$339 Air Serbia NZ$235 NZ$445
Flights from Turin to Prague
Travelling from Turin to Prague advices and facts
- The distance between Turin and Prague is 467 miles (or 752 kilometers).
- There is no time difference between Turin and Prague. Check this again shortly before your flight.
- The currency exchange rate between Turin (TRN) and Prague (PRG) is 1 EUR = 27.0614 CZK.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315936.22/warc/CC-MAIN-20190821110541-20190821132541-00222.warc.gz
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CC-MAIN-2019-35
| 1,648 | 11 |
https://passemall.com/question/cfa-level-1/consider-the-following-information-regarding-a-convertible-bondpar-value--1000conversion-ratio-5868872611659776/
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math
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Consider the following information regarding a convertible bond:
Par value = $1,000
Conversion ratio = 30:1
Convertible bond price = $1,020
Given that the conversion premium is $30, the current share price is closest to:
Given the convertible bond price of $1,020, a conversion premium of $30 implies that the conversion value is $990. Because the conversion ratio is 30:1, the current price per share will be $990/30 = $33.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00387.warc.gz
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CC-MAIN-2022-21
| 424 | 6 |
http://knowledge.giize.com/free-energy/how-much-energy-is-in-a-magnet-work-function-in-thermodynamics/
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math
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Well, in fact there are a number of ways of looking at that. One common way of estimating how many times energy is stored is simply counting how many times the magnet is held in place by a magnet in the material it is used in. This number, known as the total number of times energy is stored, is often greater than the number of times the magnet is held by it. For example, suppose that at one end of the magnetic core of a magnet, there is a steel core consisting of a single layer of iron, with the iron core just a few millimeters thick, with two thin layers of copper, and with a layer of nickel. Then the total amount of energy stored at the end of the magnet is (x)2/3 x 2x 2 = 7.5, or 3.1 million joules. The total number of times energy is stored is 2x(x)2, so 7.5 x 3.1, or about 5.6 million joules. It is useful to have a rough idea of this in order to compare how much energy is stored in different types of magnetic materials.
If the magnet is held by some material more than once, you can count the number of times energy is stored in the different materials that are held in the same way. The total number of times energy is stored in each material is then (x)2/3 x x x + 7/6 = 2.8 million, or 0.6 million. So we can see now that the two metals used to hold the magnet are the same, but the nickel used on the other side of the magnet is used to store the additional energy of the iron core. This is the process of “multiplication.” The energy stored in the iron cores is added to the energy stored in the copper and nickel cores to form the combined total of energy that is stored. The energy stored in the nickel cores is used to make the nickel cores less attractive to the iron cores. The amount of stored energy in each material is then also (x)2/3 x x x + (5/2)(x)2/3 = 0.3 million, or 0.15 million. We know that the total amount of energy stored in the magnets is (2xx)(2x(x)2/3), which means we need to know how many times the cores of the different materials are held by the different magnets to get the total amount of energy in the material. But how do we
free energy magnet, energy rich compounds, work function in thermodynamics, free energy principle reddit, advantages of free energy generator
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s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243989814.35/warc/CC-MAIN-20210513142421-20210513172421-00136.warc.gz
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CC-MAIN-2021-21
| 2,226 | 3 |
http://tex.stackexchange.com/questions/tagged/starred-version+text-decorations
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math
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s3://commoncrawl/crawl-data/CC-MAIN-2014-35/segments/1408500832032.48/warc/CC-MAIN-20140820021352-00292-ip-10-180-136-8.ec2.internal.warc.gz
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| 2,282 | 55 |
http://catdir.loc.gov/catdir/samples/prin031/91004396.html
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math
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Bibliographic record and links to related information available from the Library of Congress catalog
Copyrighted sample text provided by the publisher and used with permission. May be incomplete or contain other coding.
There is no smallest among the small and no largest among the large; But always something still smaller and something still larger.--Anaxagoras (ca. 500-428 B.C.)
Infinity has many faces. The layman often perceives it as a kind of "number" larger than all numbers. For some primitive tribes infinity begins at three, for anything larger is "many" and therefore uncountable. The photographer's infinity begins at thirty feet from the lens of his camera, while for the astronomer--or should I sad the cosmologist--the entire universe may not be large enough to encompass infinity, for it is not at present known whether our universe is "open" or "closed," bounded or unbounded. The artist has his own image of the infinite, sometimes conceiving it, as van Gogh did, as a vast, unending plane on which his imagination is given free rein, at other times as the endless repetition of a single basic motif, as in the abstract designs of the Moors. And then there is the philosopher, whose infinity is eternity, divinity, or the Almighty Himself. But above all, infinity is the mathematician's realm, for it is in mathematics that the concept has its deepest roots, where it has been shaped and reshaped innumerable times, and where it finally celebrated its greatest triumph.
Mathematical infinity begins with the Greeks. To be sure, mathematics as a science had already reached quite an advanced stage long before the Greek era, as is clear from such works as the Rhind papyrus, a collection of 84 mathematical problems written in hieratic script and dating back to 1650 B.C.1 But the ancient mathematics of the Hindus, the Chinese, the Babylonians, and the Egyptians confined itself solely to practical problems of daily life, such as the measurement of area, volume, weight, and time. In such a system there was no place for as lofty a concept as infinity, for nothing in our daily lives has to deal directly with the infinite. Infinity had to wait until mathematics would make the transition from a strictly practical discipline to an intellectual one, where knowledge for its own sake became the main goal. This transition took place in Greece around the sixth century B.C., and it thus befell the Greeks to be the first to acknowledge the existence of infinity as a central issue in mathematics.
Acknowledge--yes, but not confront! The Greeks came very close to accepting the infinite into their mathematical system, and they just might have preceded the invention of the calculus by some two thousand years, were it not for their lack of a proper system of notation. The Greeks were masters of geometry, and virtually all of classical geometry--the one we learn in school--was formulated by them. Moreover, it was the Greeks who introduced into mathematics the high standards of rigor that have since become the trademark of the profession. They insisted that nothing should be accepted into the body of mathematical knowledge that could nor be logically deduced from previously established facts. It is this insistence on proof that is unique to mathematics and distinguishes it from all other sciences. But while the Greeks excelled in geometry and brought it to perfection, their contribution to algebra was very meager. Algebra is essentially a language, a collection of symbols and a set of rules by which to operate with these symbols (just as a spoken language consists of words and of rules by which to combine these words into meaningful sentences). The Greeks did not possess the algebraic language, and consequently were deprived of its main advantages--the generality it offers and its ability to express in an abstract way relations between variable quantities. It is this fact, more than anything else, which brought about their horror infiniti, their deeply rooted suspicion of the infinite. "The infinite was taboo," said Tobias Dantzig in his classic work, Number--the Language of Science, "it had to be kept out, at any cost; or, failing this, camouflaged by arguments ad absurdum and the like."
Figure 1.1. The runner's paradox.
Nowhere was this fear of the infinite better manifested than in the famous paradoxes of Zeno, a philosopher who lived in Elea in the fourth century B.C. His paradoxes, or "arguments," as they were called, deal with motion and continuity, and in one of them he proposed to show that motion is impossible. His argument seems quite convincing: in order for a runner to move from one point to another, he must first cover half the distance between the two points, then half of the remaining distance, then half of what remains next, and so on ad infinitum (Fig. 1.1) Since this requires an infinite number of steps, Zeno argued, the runner would never reach his destination. Of course, Zeno knew full well that the runner would reach the end point after a finite lapse of time. Yet he did not resolve the paradox; rather, he left it for future generations. In this at least he was humble, admitting that the infinite was beyond his and his generation's intellectual reach. Zeno's paradoxes had to wait another twenty centuries before they would be resolved.
But while the Greeks were unable to grasp the infinite intellectually, they still made some good use of it. They were the first to devise a mathematical method to find the value of that celebrated number which we denote today by pi, the ratio of the circumference of a circle to its diameter. This number has intrigued laymen and scholars alike since the dawn of recorded history. In the Rhind Papyrus (ca. 1650 B.C.), we find its value to be (4/3)4, or very nearly 3.16049, which is within 0.6% of the exact value. It is indeed remarkable that the ancient Egyptians already possessed such a degree of accuracy. The Biblical value of pi, by comparison, is exactly 3, as is clear from a verse in 1 Kings vii 23: "and he made a molten sea, ten cubits from brim to brim, and his height was five cubits; and a line of thirty cubits did encompass him round about." Thus the error in the Biblical value is more than 4.5% of the true value of pi!
Figure 1.2. Regular inscribed
and circumscribing polygons.
Now all the ancient estimations of pi were essentially empirical--they were based on an actual measurement of the circumference and diameter of a circle. The Greeks were the first to propose a method which would give the value of pi to any degree of accuracy by a mathematical process, rather than by measurement. The inventor of this method was Archimedes of Syracuse (ca. 287-212 B.C.), the great scientist who achieved immortal fame as the discoverer of the laws of floating bodies and of the mechanical lever. His method was based on a simple observation: take a circle and circumscribe it by a series of regular polygons of more and more sides. (In a regular polygon, all the sides and angles are equal.) Each polygon has a perimeter slightly in excess of the circumference of the circle; but as we increase the number of sides, the corresponding polygons will encompass the circle more and more tightly (Fig. 1.2). Thus if we can find the perimeters of these polygons and divide them by the diameter of the circle, we will get a fairly close approximation to pi. Archimedes followed this procedure for polygons having 6, 12, 24, 48, and 96 sides, for which he was able to calculate the perimeters by using known methods. For the 96-sided polygon he found the value 3.14271 (this is very close to the 22/7 approximation often used in school). He then repeated the same procedure with polygons touching the circle from within--inscribed polygons--giving values that are all short of the true value. Again by using 96 sides, Archimedes found the value 3.14103, or very nearly 3-10/71. Since the actual circle is "squeezed" between the inscribed and circumscribed polygons, the true value of pi must be somewhere between these values.
Let there be no misunderstanding: Archimedes' method gave an estimation of pi far better than anything before him. But its real innovation was not in this improved value, but in the fact that it enabled one to approximate pi to any desired accuracy, simply by taking polygons of more and more sides. In principle, there is no limit to the degree of accuracy this method could yield--even though for practical purposes (such as in engineering), the above values are more than adequate. In modern language we say that pi is the limit of the values derived from these polygons as the number of sides tends to infinity. Archimedes, of course, did not mention the limit concept explicitly--to do so would have required him to use the language of algebra--but two thousand years later this concept would be the cornerstone around which the calculus would be erected.
1The papyrus is named after the Scottish Egyptologist A. Henry Rhind, who purchased it in 1858. It is now in the British Museum. See The Rhind Mathematical Papyrus by Arnold Buffum Chace, The National Council of Teachers of Mathematics, Reston, Virginia, 1979.
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| 9,180 | 14 |
https://www.criep.eu/en/basili-m-fontini-f-2005-quasi%E2%88%92-option-value-under-ambiguity-economics-bulletin-43-1-10/
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math
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Basili M., Fontini F. (2005).
Quasi−option value under ambiguity. Economics Bulletin, Volume 4(3), pp. 1-10.
Real investments involving irreversibility and ambiguity embed a positive quasi−option value under ambiguity (q.o.v.a.), which modifies the evaluation of an investment decision involving depletion of natural resources by increasing the value of delaying. Q.o.v.a. depends on the specific decision−maker attitude towards ambiguity, expressed by a capacity on the state space. An empirical measure of q.o.v.a. is pointed out. Exploiting the properties of a capacity and its conjugate, the relationship has been established between the upper and lower Choquet integral with respect to a subadditive capacity and the bid and ask price of the underlying asset (output) of the investment decision. The empirical measure of q.o.v.a. is defined as the upper bound of the opportunity value. As an example, q.o.v.a. is applied to evaluate an off−shore petroleum lease under ambiguity.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510412.43/warc/CC-MAIN-20230928130936-20230928160936-00290.warc.gz
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CC-MAIN-2023-40
| 991 | 3 |
https://www.studypool.com/discuss/192017/if-you-know-the-circumference-of-a-circle-what-is-the-r?free
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math
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Divide Circumference by pi (3.1415) to get diameter. Divide diameter by 2 to get radius.
if you are given any one of the following, radius, diameter, circumference, or area. You can find all of the others. Memorize the formulas for Circumference (Pi*diameter) and Area (Pi * Radius squared) and the fact that radius * 2 = diameter, and you can do all of them. so:
Circumference = 2 Pi R
To find R you divide the circumference by (2 Pi)
R = Circumference /(2*3.14159)
Jul 8th, 2014
Studypool's Notebank makes it easy to buy and sell old notes, study guides, reviews, etc.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00642.warc.gz
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CC-MAIN-2017-43
| 570 | 7 |
https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1941708
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math
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Correlated Equilibrium in Games with Incomplete Information
62 Pages Posted: 10 Oct 2011
Date Written: October 10, 2011
We define a notion of correlated equilibrium for games with incomplete information in a general setting with finite players, finite actions, and finite states, which we call Bayes correlated equilibrium. The set of Bayes correlated equilibria of a fixed incomplete information game equals the set of probability distributions over actions, states and types that might arise in any Bayes Nash equilibrium where players observed additional information. We show that more information always shrinks the set of Bayes correlated equilibria.
Keywords: Correlated equilibrium, Incomplete information, Robust predictions, Information structure
JEL Classification: C72, D82, D83
Suggested Citation: Suggested Citation
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s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030337971.74/warc/CC-MAIN-20221007045521-20221007075521-00412.warc.gz
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| 828 | 7 |
https://outdoors.stackexchange.com/questions/11487/how-dependable-is-a-cheap-compass
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math
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Consider the question in the terms of basic physics. Essentially, the magnetic pull of the earth and needle's magnet creates a torque on the compass (wheel / disk / needle etc. ) causing it to rotate. In an ideal compass, the torque meets only dampening resistance - you don't want the needle to move too quickly or you will get an oscillation while hunting for steady state. Thus, the needle will move until it has zero torque placed on it (pointing north)
In a non-ideal compass (a real one), there is some amount of force that counters the rotation of the compass (generally friction). The greater that force, the more potential error that a compass may have. This is true because the force of the earths magnetic pull produced the greatest torque when the needle is at 90 degrees to it. As the needle gets close to pointing north, the exerted torque from the earth reduces. Eventually, the torque exerted by the earth is not enough to overcome the frictional force (s) and the needle no longer moves (even though it is not pointing north). Thus, your reading will be in error.
Cheaper compasses will have a greater frictional component. Often, this component is a function of the angle at which it is held - making the inaccuracies harder to predict.
A good compass will be manufactured with emphasis on reducing frictional forces and will often have a level to help you keep the compass flat while performing readings.
I can't speak to what the price point difference is between good and bad. I can say, though, that a compass has a very specific purpose - to help keep you from getting lost. Do you really want your survival to depend on the lowest bidder?
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00573.warc.gz
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CC-MAIN-2024-10
| 1,662 | 5 |
http://applmathmech.cqjtu.edu.cn/en/article/1991/10
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math
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Abstract: A theoretical analysis is presented for the dynamic plastic behavior of a simply supported rigid, perfectly plastic circular plate in damping medium with finite-deflections subjected to a rectangular pressure pulse. Analytical solutions of every moving stage under both medium and high loads are developed.
Abstract: In this paper we construct the finite-difference scheme for the singularly perturbed boundary value problem for the fourth-order elliptic differential equation on the basis of paper , and prove the uniform convergence of this scheme with respect to the small parameter e in the discrete energy norm. Finally, we give a numerical example.
Abstract: In this paper the complete double-series in the closed region expressing the double-variable functions and their partial derivatives are derived by the H-transforniution and Stockes transformation. Using the double-series, a series solution for the axisyinmetric boundary value problem of the elastic circular cylinder with finite length is presented.In a numerical example, the cylinder subjected to the axisymmetric traellens with various loaded regions is investigated and the distributions of the displacement sand stresses are obtained.It is possible to solve the axisymmetric boundary value problems in the eylinderical coordinates for other scientific fields by use of the method presented in this paper.
Abstract: This paper describes a newly developednon-isotropic multiple-scale turbulence model (MS/ASM) for complex flow calculations. This model focuses on the direct modeling of Reynolds stresses and utilizes split-spectrum concepts to model multiple-scale effects in turbulence. Validation studies on free shear flows, rotating flows and recirculating flows show that the current model performs significantly better than the single-scale k-e model. The present model is relatively inexpensive in terms of CPU time which makes in suitable for broad engineering flow applications.
Abstract: The problem on instability of nonlinear spherical membrane with large axisvmmelric tensile deformations is investigated by using the bifurcation theory. It is proved that all singular points of the nonlinear boundary value problem must be simple limit points. The effect of loading and material parameters on the equilibrium state and its stability is discussed.
Abstract: In this paper, using the differentiability of the solution with respect to the initial value and the parameter, we present a method which, different from Liapunov's direct method. will determine the stability of the non-stationary solution of the initial value problem when the non-stationary solution remains unknown.
Abstract: All the stress components at a rapidly propagating crack-tip in elastic perfectly-plastic material are the functions of only. Making use of this condition and the equations of steady-state motion, plastic stress-strain relations, and Mises yield condition with Poisson ratio, in this paper, we derive the general expression of perfectly plastic field at a rapidly propagating plane-strain crack-tip. Applying this general expression with Poisson ratio to Mode I crack, the perfectly plastic field at the rapidly propagating tip of Mode I plane-strain crack is obtained. This perfectly plastic field contains a Poisson ratio, and thus, we can obtain the effect of Poisson ratio on the perfectly plastic field at the rapidly propagating tip of Mode I plane-strain crack.
Abstract: In this paper we study the stability for equilibrium points of equations in two-population dynamics. We discuss two predator-prey-patch models. Model 1 is described by a differential equation. Model 2 is described by an integral differential equation. We obtain the conditions for the stability of their equilibrium points. The results show that the overall population stability despite local extinction is realizable.
Abstract: This paper makes detailed analyses tor the flexural vibration (frequency) of the hemispherical shell and presents the varying laws of frequency with the rarving boundary angles and the wall thickness of the above shell, It is an important value to develop the instrument, such as hemispherical resonator gyro (HRG), whose sensing component is a hemispherical shell.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233506420.84/warc/CC-MAIN-20230922134342-20230922164342-00057.warc.gz
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CC-MAIN-2023-40
| 4,253 | 9 |
https://www.impan.pl/~piotr/
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math
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Piotr Mankiewicz, professor
Ph. D.:IM PAN 1969, habilitation: IM PAN 1974
Most of my scientific life I worked in Banach space theory. My early research
concern Banach spaces with the Random-Nikodym property, differentiability of
Lipschitz mapping between Banach spaces, uniform and Lipschitz classification
fo Banach spaces and generalizations of the Mazur-Ulam theorem. More
recently, I study random quotients of finite dimensional Banach spaces. This
led to constructions of finite dimensional Banach spaces with relatively
"few" well bounded linear operators and resulted in solving several problems
in both local and structural theory of Banach spaces.
- On the Differentiability of Lipschitz Mapping in Frechet
spaces, Studia Math. 45 (1973), 15-29.
- Fat Equicontinuous Groups Homeomorphism of Linear Topological Spaces
and Their Application to the Problem of Isometries in Linear Metric Spaces,
Studia Math. 64 (1979), 13-23.
- Application of Ultrapowers to the Uniform and Lipschitz
Classification of Banach Spaces, Studia Math. 73 (1982), 225-251.
Jointly with S. Heindrch.
- A Superreflexive Banach Space X with L(X) Admitting a
Homomorphism Onto the Banach Algebra $c(\beta N),$ Israel J. of Math. 65
- A Solution of the Finite-dimensional Homogeneous Banach Space
Problem, Israel J. of Math. 75 (1991), 129-159. Jointly with N.
- Schauder Bases in Quotients of Subspaces of $l_2(X),$ Amer. J.
of Math. 116 (1994), 1341-1363. Jointly with N. Tomczak-Jaegermann.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320303864.86/warc/CC-MAIN-20220122134127-20220122164127-00098.warc.gz
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CC-MAIN-2022-05
| 1,472 | 24 |
http://youtube-downloader-mp3.com/video-id-YZGSEcLlIH0.html
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math
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We could also do it by kirchoff's loop law
Since when is one over four equal to four? In the math that I learned it was equal to zero point two five (one quarter). Serious question here, not a snarky remark--is the math different for electronics?
U guys don't have this in 12th standard?
Just in time, im on this in college
Best explanation sir
Its funny just did this in school today :)
Thanku very much sir, this is very helpful
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s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125948738.65/warc/CC-MAIN-20180427002118-20180427022118-00010.warc.gz
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CC-MAIN-2018-17
| 430 | 7 |
https://pwntestprep.com/2020/08/in-question-10-of-the-backsolving-chapter/
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math
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In question #10 of the backsolving chapter : in the xy plane, a line containing the points (a, a^3) and (10,40) passes through the origin. Which of the following could be the value of a?
I found the explanation in the answer key to be too time-consuming if I were to solve the equation with backsolving. Can you explain how to solve this question algebraically instead?
Sure thing. Algebraically you need to key into the fact that the line passes through the origin. When a line passes through the origin, any other point on the line tells you its slope! The fact that we have tells us that the slope is 4:
So, when is it true that ?
Choice B is , so that’s the answer.
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CC-MAIN-2023-50
| 671 | 5 |
http://hru.cachlambanh.co/lewis-carroll-research-paper.html
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math
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Carroll was, Heath states, not particularly distinguished as either a mathematician or as a logician. Speaking of his mathematical work, Heath states, “Dodgson’s mathematical outlook was, in general, conservative and provincial, aiming no higher than the improvement of elementary teaching or routine calculation”. His work in logic was likewise not on the cutting edge. There are, however, some interesting aspects of his work in logic. For one thing he devised a system of logical diagramming that is similar to Venn’s diagrams and tree diagrams. This involved the use of rectangular boxes to display binary categorical data. This innovation, however, did not supplant Venn diagrams which remain in use today in almost every elementary discussion of the notion of sets. It is to be noted that Carroll never took himself too seriously and that the spirit of whimsy seemed to enter into his scholarly papers. What was perhaps his most important logical insight—that a rule permitting a conclusion to be drawn from a premise in any given system of logic cannot itself be considered to be a premise because, if it is, an infinite regress is generated—was brought forth by Carroll in a paper entitled “What the Tortoise Said to Achilles”. Another of his logical publications involved a discussion of the paradoxes involved in rules for employees entering and exiting a barber Carroll’s was a most interesting mentality. We may contrast him with Charles Babbage who was undoubtedly a more formidable genius, but who wore the sour aspect of a misanthrope all his life.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221210133.37/warc/CC-MAIN-20180815141842-20180815161842-00575.warc.gz
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CC-MAIN-2018-34
| 1,581 | 1 |
https://www.mi.fu-berlin.de/en/math/groups/ag-c/dates/28_05_14Munoz.html
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math
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We recall the definition of the Grothendieck group of an exact category and describe the Grothendieck groups of schemes and rings. We define the first K-theory group of a ring via a short exact sequence of groups and prove Whitehead's lemma. If time permits, we define K_0 and K_1-regularity of a ring and prove that a regular ring is K_0 and K_1-regular.
Further details: http://userpage.fu-berlin.de/hoskins/seminar.html [userpage.fu-berlin.de]
Time & Location
May 28, 2014 | 04:00 PM
SR 005, Arnimallee 3
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817073.16/warc/CC-MAIN-20240416062523-20240416092523-00312.warc.gz
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CC-MAIN-2024-18
| 507 | 5 |
http://www.chegg.com/homework-help/questions-and-answers/conductor-placed-external-electrostatic-field-theexternal-field-uniform-conductor-placed---q208930
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math
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0 pts pendingThe asker has
chosen a Best Answer and the points will be awarded within 24 hours.
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely
isolated from any source of current orcharge.
Which of the following describes the electricfield inside this conductor?
It is in thesame direction as the original
It is in theopposite direction from that of the
It has adirection determined entirely by the
charge on itssurface.
It is alwayszero.
The charge density inside theconductor is:
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeand is directed toward the surface of the conductor. Whatis
the charge density on the surface of
the conductor at that point?
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s3://commoncrawl/crawl-data/CC-MAIN-2013-48/segments/1387345774525/warc/CC-MAIN-20131218054934-00051-ip-10-33-133-15.ec2.internal.warc.gz
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CC-MAIN-2013-48
| 816 | 14 |
https://www.jiskha.com/display.cgi?id=1284077100
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math
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posted by Shadow .
There is something wrong with this definition for a pair of vertical angles: “If AB and CD intersect at point P,then APC and BPD are a pair of vertical angles.” Sketch a counterexample to show why it is not correct. Can you add a phrase to correct it?
- Umm, for the counterexample, would it be correct if I draw an intersecting line but for APC and BPD I make the angle different?
Fuq u Loner
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s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084891980.75/warc/CC-MAIN-20180123151545-20180123171545-00501.warc.gz
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CC-MAIN-2018-05
| 416 | 4 |
http://www.ub.edu/javaoptics/docs_applets/Doc_PolarEn.html
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math
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Light polarization and Fresnel laws
In this applet we study polarization of light and its reflection and refraction in isotropic media. The first window shows how to achieve the different polarization states of light by superposing two plane waves with special characteristics. The second and third windows show the behavior of reflected and refracted waves depending on the properties of an incident wave on a surface separating two media. The first medium is a dielectric while the second one can be either a dielectric or a conductor. Both the amplitude ratios (Fresnel coefficients) and the polarization ellipse changes are studied.
In this window we study the superposition of two plane waves of the same frequency, traveling in the same direction and that have perpendicular electric field directions, different amplitudes and with a phase delay between them.
The amplitudes of the p (Ap) wave and the s (As) wave and the phase delay δ between them can be changed with the sliders. The plot at the bottom represents these two waves. The corresponding electric fields Ep and Es can be expressed as:
As the waves propagate, the endpoint of the resultant electric field vector traces an ellipse, which is plotted on the upper left graph. We can also see the characteristics of this polarization ellipse: the semimajor and semiminor axes, and the angle with the p-axis, in degrees. The Stokes vector S describing the polarization state of light is indicated as well. The Stokes parameters are computed in the following way:
The sign of the S parameter shows whether light is right-handed (S>0), in other words it turns clockwise, or left-handed (S<0), when it turns counter-clockwise. This is indicated in the black window by the letter R or L, respectively.
In this window we study light reflection and refraction in dielectric isotropic media, taking into account Fresnel equations. These equations specify the values of the electric field vector amplitude when light changes from one medium to another one, or when it is reflected, as a function of the incident amplitude. We suppose that we are working at a frequency (wavelength) range for which the media are non-absorbent.
Let us consider an incident wave on a surface separating two transparent media with indices of refraction n and n'. If the incident angle is φ, the wave is reflected with a reflection angle equal to the incident one and is transmitted to the second medium with a refraction angle φ' that can be calculated as:
The indices of refraction of the two media and the incident angle can be changed with the sliders. The program calculates the refraction angle and plots the incident (red), reflected (green) and refracted (blue) rays making the corresponding angles with the surface normal. It also computes the Fresnel coefficients, which are the ratios of the reflected (A'') or refracted (A') amplitudes to the incident amplitude (A):
The p and s subscripts refer to the field that vibrates parallel (p) or perpendicularly (s) to the plane of incidence. This plane is determined by the incident ray and the surface normal (on the plot it corresponds to the plane of the screen itself). The p component is represented as a vector with different orientations on this plane, while the s component is plotted as a vector pointing towards the inside or outside of the plane of the screen.
For normal incidence, the formulae become:
When the "Fresnel coefficients" button is pressed (default option active when starting the program) we can see a plot of the Fresnel coefficients variation with the angle of incidence, in different colors. A yellow dot over the graph indicates the selected incident angle. There are some particularly interesting cases:
This angle is shown on the plot. In this case, we have:
Both reflection coefficients have modulus 1 and there is only a phase delay 2(β-α)-π between the two components (the angle π refers to the minus sign in rp), with
Pressing the "Polarization" button opens a new window. In it we can see the polarization ellipses for incident, reflected and transmitted light, in the same colors as the corresponding rays. The axes scale is between 0 and 1, and if the scale changes values, a new position is indicated. We can also see the characteristics of the incident wave, which are the ones chosen in the "Polarization" window and can be modified by accessing that window.
When total internal reflection occurs, there is no transmitted polarization ellipse (in blue). The values of the angles 2α (angp) and 2β (angs) are also shown, as well as the total phase retardance between the s and p components of the reflected light. This will be the same as the incident light phase retardance plus (angs-angp-π). We can see how this value changes with the incident angle and the indices, and how by total reflection it is possible to obtain elliptically polarized light from incident linearly polarized light.
On the other hand, we can see that when the incident angle equals the Brewster angle, the reflected light is linearly polarized in the normal direction to the incident plane.
In this window we study light reflection on isotropic and conducting media. In this case the index of refraction of the second medium is a complex number,
where n' and k (extinction coefficient) are real and positive constants related to the dielectric constant and the conductivity of the medium. In the applet n' can range between 0.1 and 3.5, and k can range between 0 ("dielectrics" case) and 5.
The Fresnel formulae still apply, but with consideration of the complex index of refraction for the second medium. The reflection coefficients, after some calculation, become:
where all are real numbers except the index of refraction of the second medium and the product
We have taken into account that Snell's law is still valid:
Here φ' has a formal meaning with no geometrical interpretation.
As the Fresnel coefficients are complex numbers, the reflected waves might suffer phase changes (ξp and ξs for the parallel and perpendicular components, respectively), which depend both on the indices of refraction of the two media and on the incident angle. These parameters can be modified with the sliders. The program also shows the total phase retardance between the s and p components for the reflected light, which is the phase delay occasioned by the incident wave plus (ξs-ξp). We must point out that when no additional phase delay is introduced in the reflection, the incident and reflected waves rotate in opposite directions (if one is right-handed, the other one is left-handed, and vice versa). This is because when describing the light polarization state of these two waves they are viewed from opposite sides (each wave is viewed from the direction which the wave is approaching).
For normal incidence, the formulae for dielectric media in this particular case apply, with the complex index of refraction for the second medium:
Transmitted wave behavior is more complex. Here we only study the case of normal incidence, for which the transmissivity T, defined as the ratio between the transmitted and incident intensity, becomes:
(R is the reflectivity). The transmitted wave can be written as a plane wave propagating in the conducting medium in the z direction (perpendicular to the surface separating the two media) and having a complex index of refraction. Its electric field vector can be described as:
This corresponds to a plane wave that propagates in a medium of index n' (real) but with an amplitude that decays exponentially with the distance traveled in the medium (z=0 is taken in the boundary surface). This dependency is shown on a plot, as well as the distance traveled by the wave into the conducting medium for which its amplitude decays in a factor 1/e. This is called skin depth (zp) and characterizes the medium:
λ is the wavelength of light and can be modified with a slider.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823621.10/warc/CC-MAIN-20181211125831-20181211151331-00382.warc.gz
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CC-MAIN-2018-51
| 7,923 | 29 |
https://oz.minisport.com/mini-body-repair-panels/mini-body-trim/roof-gutter-trim-mini.html
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math
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Roof Gutter Trim
Roof Gutter Trim including Genuine and OE Spec Gutter Trim Finishers in Black along with Finisher Clips in Black or Grey.
Please browse through our comprehensive sections, if you cannot find what you are looking for please feel free to contact us on +44 (0) 1282 778731 or email [email protected] and our friendly staff will be more than happy to assist you.
- Genuine Roof Finisher Clip - BlackSpecial Price A$10.34 A$9.40 Regular Price A$11.50
- Black Roof Finisher Gutter TrimSpecial Price A$31.06 A$28.24 Regular Price A$36.21
- Chrome Roof Finisher Gutter TrimSpecial Price A$33.40 A$30.36 Regular Price A$39.75
- Electric Fan & MountingsSpecial Price A$94.74 A$86.13 Regular Price A$125.39
- Mini Detailing Kit by Cooper Car CompanySpecial Price A$58.70 A$53.36 Regular Price A$106.26
- Cooper Rocker Cover Buttons - Black EngravedSpecial Price A$67.49 A$61.35 Regular Price A$84.34
- Door Handle Outer LH - GreySpecial Price A$60.72 A$55.20 Regular Price A$67.46
- Cooper Mini Tyre & Trim by Auto Finesse - 500mlSpecial Price A$9.92 A$9.02 Regular Price A$20.24
- Cooper 8.4" Std 4 pot Alloy Caliper and Pad Kit - Black -VDSpecial Price A$632.50 A$575.00 Regular Price A$840.54
- Brake Servo Unit - MK1 Mini Cooper SSpecial Price A$616.65 A$560.59 Regular Price A$711.41
- Bronze Valve Guides - Competition - Set of 8Special Price A$70.40 A$64.00 Regular Price A$80.96
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s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00609.warc.gz
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CC-MAIN-2023-14
| 1,392 | 14 |
http://mathhelpforum.com/trigonometry/25891-precal-test-tomorrow.html
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math
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Seems that others have similar problems that I do. I've finished up most of my review packet, just a few things I don't quite get yet. Wondering if someone here could clarify these two questions. Then all I have left is to face the daunting task of saving myself from my own stupidity.
1. Find the exact value of the expression. Do not use a calculator.
csc^2 70 degrees - tan^2 20 degrees
a) 2 b) 0 c) -1 d) 1
I think the others I am figuring out right now.
ok, here's the trick. it might take a while for us to go through this with me giving you hints, so i'll just give you a possible solution.
things to know:
(4) the sine of an angle is equal to the cosine of its compliment (complimentary angles are angles that sum to 90 degrees).
so, for instance: , , , etc
particularly what we will use here is:
there are other solutions here. this is just the one that occurred to me first. maybe sixstringartist can come up with another one
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s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541896.91/warc/CC-MAIN-20161202170901-00144-ip-10-31-129-80.ec2.internal.warc.gz
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CC-MAIN-2016-50
| 935 | 11 |
https://smartindia.net.in/ask-a-teacher/question/954/
|
math
|
Ask a Teacher
what is molality?
|Molarity and molality are both measures of concentration of solutions.
One is the ratio of moles to volume of the solution and the other is the ratio of moles to the mass of the solution.
Molality is the number of moles of solute per kilogram of solvent. It is important the mass of solvent is used and not the mass of the solution. Solutions labelled with molal concentration are denoted with a lower case m. A 1.0 m solution contains 1 mole of solute per kilogram of solvent.The important part of remembering the difference is
molarity - M ? moles per liter solution
molality - m ? moles per kilogram solvent.
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s3://commoncrawl/crawl-data/CC-MAIN-2024-18/segments/1712296817043.36/warc/CC-MAIN-20240416031446-20240416061446-00802.warc.gz
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CC-MAIN-2024-18
| 644 | 7 |
https://books.google.com/books/about/Applied_business_statistics.html?id=afkbAQAAMAAJ&hl=en
|
math
|
What people are saying - Write a review
We haven't found any reviews in the usual places.
y Classification of Data
Types of Averages
26 other sections not shown
adjusted annual answer arithmetic mean binomial brands Bureau calculation Census Chapter chi-square class interval class limits coefficient of correlation coefficient of determination computed confidence interval constructed Consumer Price Index correlation analysis data in Exercise degrees of freedom dependent determine difference dollar economic employees error of estimate estimating equation example Exercise Table figures fluctuations forecast formula frequency distribution harmonic mean illustrate independent variables index numbers large number logarithm measure median method Metro Company midpoint Milwaukee metropolitan area moving averages multiple regression normal curve normal distribution null hypothesis ordinate percent probability problem quartile question questionnaire ratio regression and correlation regression coefficients regression equation relationship represent result sample data sample means sample percentages scatter diagram seasonal index selected shown shows simple random sample skewness standard deviation standard error Substituting survey trend origin trend values U.S. Department unexplained variance United universe mean variance variation
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s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189471.55/warc/CC-MAIN-20170322212949-00455-ip-10-233-31-227.ec2.internal.warc.gz
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CC-MAIN-2017-13
| 1,343 | 6 |
http://sau.int/admissions/admission-notice-2016/48-admissions/2016-session/435-msc-applied-maths-entrance-exam-2016.html
|
math
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MSc in Applied Mathematics
MSc Applied Mathematics is a two year (four semester) academic programme.
12 years of schooling + a 3 or 4 year Bachelor’s degree (with Mathematics as a subject for at least two years) from an institution recognized by the government of any of the SAARC countries, with the following minimum eligibility marks criteria:
For candidates from the annual system: 55% and above
For candidates from the semester system (percentage): 60% and above
For candidates from the GPA semester system, the formula for determining percentage is as follows:
CGPA obtained X 100 / Maximum GPA obtainable.
Such candidates are required to have a minimum of 65% marks or above.
Format of the Entrance Test Paper
The duration of the Entrance Test will be 3 hours and the question paper will consist of 100 multiple choice questions in two parts.
PART A : will have 40 questions on Basic Mathematics.
PART B : will have 60 questions on Undergraduate Level Mathematics.
Calculators will not be allowed. However, Log Tables may be used.
The combined syllabus for both Part A and Part B is as follows:
Calculus and Analysis:Limit, continuity, uniform continuity and differentiability; Bolzano Weierstrass theorem; mean value theorems; tangents and normal; maxima and minima; theorems of integral calculus; sequences and series of functions; uniform convergence; power series; Riemann sums; Riemann integration; definite and improper integrals; partial derivatives and Leibnitz theorem; total derivatives; Fourier series; functions of several variables; multiple integrals; line; surface and volume integrals; theorems of Green; Stokes and Gauss; curl; divergence and gradient of vectors.
Algebra: Basic theory of matrices and determinants; groups and their elementary properties; subgroups, normal subgroups, cyclic groups, permutation groups; Lagrange's theorem; quotient groups; homomorphism of groups; isomorphism and correspondence theorems; rings; integral domains and fields; ring homomorphism and ideals; vector space, vector subspace, linear independence of vectors, basis and dimension of a vector space.
Differential equations:General and particular solutions of ordinary differential equations (ODEs); formation of ODE; order, degree and classification of ODEs; integrating factor and linear equations; first order and higher degree linear differential equations with constant coefficients; variation of parameter; equation reducible to linear form; linear and quasi-linear first order partial differential equations (PDEs); Lagrange and Charpits methods for first order PDE; general solutions of higher order PDEs with constant coefficients.
Numerical Analysis: Computer arithmetic; machine computation; bisection, secant; Newton-Raphson and fixed point iteration methods for algebraic and transcendental equations; systems of linear equations: Gauss elimination, LU decomposition, Gauss Jacobi and Gauss Siedal methods, condition number; Finite difference operators; Newton and Lagrange interpolation; least square approximation; numerical differentiation; Trapezoidal and Simpsons integration methods.
Probability and Statistics:Mean, median, mode and standard deviation; conditional probability; independent events; total probability and Baye’s theorem; random variables; expectation, moments generating functions; density and distribution functions, conditional expectation.
Linear Programming: Linear programming problem and its formulation; graphical method, simplex method, artificial starting solution, sensitivity analysis, duality and post-optimality analysis.
Negative Marks for Wrong Answers
If the answer given to any of the Multiple Choice Questions is wrong, ¼ of the marks assigned to that question will be deducted.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281419.3/warc/CC-MAIN-20170116095121-00549-ip-10-171-10-70.ec2.internal.warc.gz
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CC-MAIN-2017-04
| 3,749 | 22 |
https://www.nagwa.com/en/videos/154172092083/
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math
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Isabella has 𝑚 dollars. She wants to buy a book for 10 dollars and a pencil case for four dollars. Describe the condition on 𝑚 for her to be able to afford both items.
So we know that she has a total of 𝑚 dollars. And the two items she would like to buy cost 10 dollars and four dollars. So their total we would find by adding them together.
So in order to afford something, that means we have to have enough money. So that means we don’t need to have exactly 10 plus four dollars. Or we need to have more than that. So 𝑚 needs to be greater than or equal to 10 plus four.
If Isabella had more than 10 plus four dollars, she’d have some money left over. If she had exactly equal to, meaning 10 plus four dollars, she wouldn’t have any money left to spend. But she would still be able to afford both items. So finally, our final answer will be 𝑚 is greater than or equal to 10 plus four.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323583087.95/warc/CC-MAIN-20211015222918-20211016012918-00683.warc.gz
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CC-MAIN-2021-43
| 907 | 4 |
https://mialobel.com/2010/11/04/pulse-of-the-planet-%E2%80%93-frequency-and-math/
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math
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This next segment I helped write and produce for Jim Metzner’s Pulse of the Planet is about music and math. Experimental musician and instrument builder Bart Hopkin talks about how each musical note vibrates at a certain frequency, and how the ratios between these frequencies create musical intervals. (Or something like that. Best listen to really understand it.)
Sounds – Frequency and Math: They’re playing my favorite equation!
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s3://commoncrawl/crawl-data/CC-MAIN-2024-10/segments/1707947474661.10/warc/CC-MAIN-20240226162136-20240226192136-00633.warc.gz
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CC-MAIN-2024-10
| 438 | 2 |
https://ritzblog.akritz.com/2020/01/27/what-does-pie-stand-for-in-math/
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math
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What Does Pie Stand For in Math?
Polynomial in mathematics is usually a pretty exciting topic to discover in the event you like numbers and math.
This is primarily because of the endless exciting and several applications which are capable to become produced using a PIC!
Most probably, you could have come across book report outline the usual pondering that a “base” of any superior thing will have to be well-defined, and therefore many of the well known routines of a common classical mathematician are surely based on a fantastic understanding of this notion. This makes it quick to check regardless of whether your approach to math is appropriate. Consequently, be sure to know what the which means of “ie” is in general so you don’t get confused.
In fact, in all of the disciplines of Mathematics, there will generally be a “chess game” that may be just as well excellent to prevent. This specific example is generally named PIC.
PIC will not be meant to become analyzed in depth, rather make the application of the term into a form that’s easier. http://cs.gmu.edu/~zduric/day/essay-help-accident-victims.html The equation, also known as “pie”, has a root within the equation of a triplicate of your square or cube. The base of your square may be observed because the smallest of the sets of sides as well as the base of the cube will be the largest in the set of sides.
Actually, PIC is actually not the entire story. There is certainly an extremely potent function named n=3 x 2 y, where n and y will be the second and third terms on the term x y. This can be solved in a lot more than one particular way.
As I stated before, the whole notion behind PIC and the PIC process is very highly effective. 1 can solve any trouble in PIC inside a single step, without even possessing to do a series of operations. The expression is less difficult to visualize than the reverse.PIC can resolve any linear equation.
So, we shouldn’t think with the abstract know-how in terms of the complex 1. PIC basically has an answer that will be a https://buyessay.net/book-report lot more accurate than any mathematical formula. Let us see how. The complex kind of complex function is called as a “complex number”.
According for the benefits of this basic formula, we are able to find out the worth of the “term” with the expression ln(x) at any given time. Applying this strategy, we are able to make use of n=2 or n=3 x 2 y to discover the worth from the expression y=xln(x). If we can make use of this technique to find out the value from the actual expressions, we are able to discover the genuine worth from the expression and thus, we are able to do a genuine calculation that is definitely determined by the true values in the variables.
Even even though we can’t seriously learn the precise worth of the complex function when it comes to a genuine value, the PIC technique can generally do the trick. The derivation might be performed with just a bit of further information and facts concerning the worth from the complex function. The derivation also gives us the complete thought on the complex numbers. This isn’t pretty uncomplicated, but not very hard either.
Also, if we keep going with all the concept of PIC, we can just find out all the implications of each of the combinations of your functions and then use it to derive all of the other formulas. Now, the way this is carried out is basic, but nonetheless it is actually possible for any novice student to perform it with out an issue. However, at the finish on the day, a very good idea has to be utilized.
School is not full without mastering the basics on the “pie” and also the “poly”. If you want to pass your Mathematics Exams, be sure you study both.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-40/segments/1695233510300.41/warc/CC-MAIN-20230927135227-20230927165227-00675.warc.gz
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CC-MAIN-2023-40
| 3,756 | 13 |
https://www.mathworks.com/matlabcentral/fileexchange/86488-independent-linkaxes?s_tid=blogs_rc_6
|
math
|
Ever wanted to link some axes only on x while linking others on y and noticed that only one of them works?
Well, fear no longer, here is the extended linkaxes that lets you do exactly this.
The example shows how to link the subplots above/below each other on the x axis, while linking the subplots left/right of each other on the y axis.
Tested only on matlab R2018a.
Just Manuel (2022). Independent Linkaxes (https://www.mathworks.com/matlabcentral/fileexchange/86488-independent-linkaxes), MATLAB Central File Exchange. Retrieved .
MATLAB Release Compatibility
Platform CompatibilityWindows macOS Linux
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!Start Hunting!
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s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00486.warc.gz
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CC-MAIN-2022-49
| 724 | 9 |
https://betterlesson.com/lesson/595406/polar-distance-formula?from=breadcrumb_lesson
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math
|
Yesterday we worked on finding ways to convert from polar ordered pairs to rectangular (or vice versa). Today's lesson will focus on how to find the distance between two points that are given in polar form. I really like this lesson because it is an extension of yesterday and gives students more practice in visualizing polar points in the plane. Students also use past geometric concepts (distance formula and Law of Cosines) in an entirely new coordinate system.
I begin the lesson by showing students slide #2 of the PowerPoint and asking them to find the distance between each of the three pairs of points. We will remind ourselves that these ordered pairs are all written in polar form and recap what each coordinate represents in polar form. I will have students work in their table groups and I will give them about 10-15 minutes to see how far they can get. I stress that there are multiple ways to find the distance between points; if they find an answer for the first pair of points, they should try to use a different method to find the distance for the second pair.
As I walk around the room, here are some hints I will give if students are stuck.
After students have had ample time to work, I will select students to share their work with the class. Below is the order I will hand pick students to share their work with the entire class. Obviously it will depend on the solution strategies I see, but these are usually bound to be present. If one of the solution strategies is not present, you can always get your students started and see if they can finish it.
1. Using the distance formula incorrectly - This overgeneralization of the distance formula is usually present. A student will simply plug the r and θ in for x and y in the rectangular distance formula. Usually a student will catch their mistake while talking it over with their table, but I still think it is really important to talk about why this does not work. Students will need continuous reinforcement that the first coordinate in rectangular form means something completely different than the first coordinate in polar form.
2. Using the distance formula correctly - This is usually the most common strategy for my students. They will convert each of the two points to rectangular form (like we did yesterday) and then plug the new values into the distance formula. I like how the student whose work is shown here left the ordered pairs in (rcosθ, rsinθ) form. This may help students find a general formula for the distance between any two polar points.
3. Using the Law of Cosines - This strategy requires a diagram and the corresponding algebraic work. Again, if no students used this strategy, simply connecting each of the two points to the origin may get them to see a triangle and think about tools they can use.
Finally, the last pair of polar points from the PowerPoint contains all variables, so we will discuss how we can generalize the techniques we came up with to work for any two points in polar form.
Using the distance formula is pretty straightforward, but the Law of Cosines may have some interesting conversations. I discuss in the video below.
Unable to display content. Adobe Flash is required.
After this discussion, I will give students this worksheet and have them work on it for homework.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585242.44/warc/CC-MAIN-20211019043325-20211019073325-00704.warc.gz
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CC-MAIN-2021-43
| 3,300 | 11 |
https://www.nagwa.com/en/plans/139131019525/
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math
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Lesson Plan: Simplifying Algebraic Fractions Mathematics
This lesson plan includes the objectives, prerequisites, and exclusions of the lesson teaching students how to factor algebraic expressions and simplify algebraic fractions.
Students will be able to
- simplify an algebraic fraction with a single term (i.e., a monomial) in the denominator,
- simplify an algebraic fraction that has been factored by canceling the common factors,
- factor an algebraic fraction and simplify it by canceling the common factors,
- find unknown constants by simplifying an algebraic fraction.
Students should already be familiar with
- factoring algebraic expressions including quadratic ones,
- identifying the common factor in an expression,
- simplifying algebraic expressions using rules of exponents.
Students will not cover
- long division of polynomials,
- factoring cubic expressions unless they have a common factor of .
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100081.47/warc/CC-MAIN-20231129105306-20231129135306-00709.warc.gz
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CC-MAIN-2023-50
| 915 | 14 |
https://olcademy.com/description/6731591256751100000164
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math
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- Clear concepts on Matrix!
- Advance solver of Matrices!
1. Types of Matrix Operation on Matrices
2. Matrix Addition
3. Matrix Subtraction
4. Matrix Multiplication
5. Adjoint of a matrix
6. The inverse of a matrix
7. Transpose of a Matrix
These are the following topics we will be going through in this course.
The course will be completed in 2 months. There will be 10 lectures, 1 day per week. In this course, you will gain full knowledge about Matrix Operation.
You will learn the definition of the matrix, different types of matrices, how to add and subtract matrices, how to multiply matrices, and miscellaneous problems related to addition, subtraction, and multiplication. And as well as Transpose, adjoint, and inverse related matrix problems.
No. of Students
I'm an instructor, my name is Mriganka Adhikari. I'm a professional mathematics teacher with more than 3 years of teaching experience. I'm associated with many reputed tutorial institutes of Kolkata, West Bengal. Over 90 students passed out with excellence under my guidance over 3 years. I work with the students as a team and this teamwork will pay in the near future. I hope the trend of the brilliant results will continue with the bonding of hard work by the students and the teacher. Techniques and way of approach are the two most important characters in mathematics. The fame of any teacher is the result of hard endeavor by the students only.
No documents for this course.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991258.68/warc/CC-MAIN-20210517150020-20210517180020-00633.warc.gz
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CC-MAIN-2021-21
| 1,450 | 15 |
http://benfordonline.net/fullreference/486
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math
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Accepted for publication in the International Journal of Combinatorial Number Theory.
ISSN/ISBN: Not available at this time. DOI: Not available at this time.
Abstract: ABSTRACT: We study the following two-player game involving players A and B. Let n be a positive integer known to both players. The player A (B) chooses an n-digit integer a (b). They do this without the knowledge of the other player’s choice. Then the numbers are revealed and the product ab is computed. If the leading digit (i.e., the first digit from the left) of ab is 1, 2 or 3, then player A wins, else player B wins. Since six digits favor B and only three favor A, a naive reasoning would suggest a 2 to 1 advantage for player B. The reality is nearly the opposite. More precisely, let pn be the probability that A wins when both players play optimally. We show the somewhat surprising result that limn→∞ pn = log104≈0.6020599913. We also determine the odds in favor of the players when they use some non-optimal strategies. Our analysis of optimal play involves the scale invariance of the well-known Benford distribution
Not available at this time.
Reference Type: Journal Article
Subject Area(s): Applied Mathematics
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s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189127.26/warc/CC-MAIN-20170322212949-00608-ip-10-233-31-227.ec2.internal.warc.gz
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CC-MAIN-2017-13
| 1,203 | 6 |
http://8211-how-to-write-a-billion-in-numbers.essayyetbest.accountant/
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math
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Write 3 Billion In Numbers? - Math Discussion This list contains selected positive numbers in increasing order, including counts of things, dimensionless quantity and probabilities. Related Questions. What is 35% of a number if 12 is 15% of a number? How many “learnhub” numbers are there? Which 3 numbers have the same answer whether
English numerals - pedia One billion years may be ed eon (or aeon) in astronomy and geology. Ordinal numbers refer to a position in a series. Another common usage is to write out any number that can be expressed as one or two words, and use.
English Numbers - Cardinal Numbers Working out how b these numbers are is made harder because, traditionally, the United States and the United Kingdom meant different amounts when each one talked about a billion. Cardinal numbers from 1 through 1,000,000; 1 one 11 eleven 21 twenty-one 31 thirty-one 2 two 12 twelve 22. In English this number is a billion.
Scripts - How to round decimals using bc in bash? - Ask Ubuntu There are nine zeros in one billion, so you can either add nine zeros to 7 or move the decimal point over to the rht nine places to find your answer. I'm still looking for a pure bc answer to how to round just one value within a function, but. Is there a command to round decimal numbers in txt files?
Sorting a Billion Numbers with Julia – one This post gives an overview of working with larger-than-memory data in the Julia language. To put it more simply go to the end of the file, write the value to disk there, and update the length and position metadata accordingly. Fast array implementations.
How to write a billion in numbers:
Rating: 100 / 100
Overall: 88 Rates
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s3://commoncrawl/crawl-data/CC-MAIN-2018-26/segments/1529267866358.52/warc/CC-MAIN-20180624044127-20180624064127-00425.warc.gz
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CC-MAIN-2018-26
| 1,674 | 8 |
https://tachash.org/word-search-meet-four-math-geniuses-who-changed-the-world/
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math
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Brilliant minds around the world are responsible for advancing our understanding of mathematics. Here are some mathematicians from our Word Search, whose work has had a lasting impact:
Known as the father of geometry, the Greek mathematician is best known for his treatise titled Elements. Written around 300 BC. AD, it defines how formal mathematics has been practiced for over 2,000 years. Euclid’s main idea was to start from certain axioms assumed to be true and then use purely deductive methods to establish a set of theorems. Among the 465 theorems he proposed, there are many famous results, such as the Pythagorean theorem, the triangle inequality and the fact that there are an infinite number of prime numbers.
One of the most famous mathematical formulas – the Pythagorean theorem – belongs to the Greek mathematician and philosopher Pythagoras. His influence on later philosophers and the advancement of Greek philosophy is considered enormous. Plato makes reference to Pythagoras in a number of his works, and Aristotle was one of his pupils. The elusive philosopher was known for being secretive and carefully guarding his mathematical processes – a mystique that only added to his reputation.
The Indian mathematician Srinivasa Ramanujan was a prodigy in this field and, despite his struggle with poverty at a young age, he rose to the forefront of mathematics in India, and later, in the West. Ramanujan worked out the Riemann series – a paradox that has baffled scholars for decades – and also contributed to elliptic integrals, hypergeometric series and functional equations of the zeta function. All this, without formal university training.
German mathematician Johann Gauss is sometimes referred to as the “princeps mathematicorum” (Latin for “the first of mathematicians”). Gauss is responsible for proving the fundamental theorem of algebra and making significant contributions to number theory. Gauss’s work is also essential to advances in physics – his work led to new insights into magnetism and the discovery of Kirchhoff’s circuit laws in electricity.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662601401.72/warc/CC-MAIN-20220526035036-20220526065036-00492.warc.gz
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CC-MAIN-2022-21
| 2,107 | 5 |
https://forums.opera.com/user/ljlevitan
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math
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I cannot figure out how to open a Private window in Opera on my Android phone (so I can take advantage of the built-in VPN). Could I get some help, please?
Best posts made by ljlevitan
This user hasn't posted anything yet.
Latest posts made by ljlevitan
Private WindowOpera for Android
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s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593655906934.51/warc/CC-MAIN-20200710082212-20200710112212-00208.warc.gz
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CC-MAIN-2020-29
| 285 | 5 |
http://gmjacksonphysics.blogspot.com/2016/08/
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math
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Quantum electrodynamics (QED) is perhaps the most precise and successful theory in all of physics. There is, as I've mentioned in pre...
Monday, August 29, 2016
Saturday, August 27, 2016
The universe is getting bigger and colder. We are told that one day it could suffer a heat death. By contrast, the early universe was smaller and hotter. All this implies an inverse correlation between the universe's average temperature and its size.
We might be tempted to model it with the Ideal Gas equation. We could think of the universe as a big balloon filled with gas. As the volume of the balloon increases, the pressure decreases along with the temperature. Below is the equation (P=pressure; T=temperature; V=volume; R=universal gas constant; n=kilomoles of gas):
Right away the equation bursts our balloon. When volume (V) increases, pressure (P) decreases but the temperature can remain constant. So why isn't our universe the same temperature as it was in its early years? Its volume is increasing; its pressure is decreasing; yet, the temperature isn't remaining constant.
So what's the problem with the Ideal Gas Law? We know that pressure is equivalent to energy density, so let's plug energy density (E/V) into the equation in lieu of pressure (P):
The second equation above reveals the problem: The volume (V) in the denominator cancels the volume in the numerator. Change in volume in this case has no effect on the temperature.
We know that entropy increases as the universe expands. Perhaps lower temperature is a function of increased entropy? Let's take a look at the entropy equations (E=heat or energy; s=change in entropy; T=temperature; k=Boltzmann's constant; omega=number of different arrangements of particles):
If we equate the two equations above, we can derive a temperature equation:
Temperature appears to be a function of energy (E) over the number of ways particles can arrange themselves (omega). Once again, temperature is not a function of volume, but it seems like it should be. We know from experience that a candle can heat up and maintain the temperature of pickle jar, but a candle in an aircraft hanger can be a much colder space.
Let's explore the omega variable and see if we can make it a function of volume. Imagine a single particle with one state in a single space (see first diagram below). In that case, omega equals one. Now double the space. The particle will have two spaces it can occupy (see second diagram below). Omega now equals two.
Entropy has increased due to increased space. We can define omega in the following way:
Below are some examples that include the above diagrams, a coin toss, and dice:
Consider the coin-toss example above. Normally we think of a coin as having two states: heads or tails, so omega equals two, right? Well that's only true if the coin lands on the same spot on your floor because there is no other spot it can land. But suppose you have a big floor and you divide it into a grid of 20 spaces where the coin can land. Omega would then be 2^2 * 20 = 40 states.
Now suppose you have the same floor space, but you now have two coins. Omega becomes 1520 states! We are now ready to put together a temperature equation:
Notice the temperature is now a function of volume because entropy is now a function of volume. If the volume increases, the temperature decreases and so does the pressure (Boyle's Law) if the other omega variables and E are held constant. Remember pressure is E/V. The difference here is that E/V is no longer canceled by V*E/V.
Taking the temperature of the Universe or any other system can be done by this system of equations. The first equation is for systems that have more space than particles. The second is for singularities, high pressure systems, where the energy density is high. The third equation allows you to make a quick-and-dirty calculation. It is similar to the Ideal Gas equation albeit it includes entropy.
Friday, August 26, 2016
To derive the Planck length we start with the Schwarzschild radius (x=length; m=mass; G=Newton's constant; E=energy; c=light speed). (To find out how the Schwarzschild radius is derived, click here.)
We also need Heisenberg's Uncertainty Principle (p=momentum; x=position; h-bar=Planck's constant). (To find out how to derive the Uncertainty Principle, click here.) For our convenience we convert momentum (p) to energy (E) by multiplying both sides by light speed (c):
Divide both sides by E:
Multiply the above result by the Schwarzschild radius, then do some algebra to get the Planck length (x with a sub p):
To get the Planck time (t with a sub p), take the Schwarzschild radius and divide it by c:
Shall we call the above result the Schwarzschild time? We also need the other Heisenberg Uncertainty Principle, the one with energy (E) and time (t). Now do some more algebra to get the Planck time:
To get the Planck mass, set the Schwarzschild time equal to the Heisenberg time, then solve for E-squared:
Let's assume E is kinetic energy. Square it, make a substitution, do some more algebra, and get the Planck mass (m with a sub p):
Thursday, August 25, 2016
This post will further address the problem discussed in a previous post. To read that post, click here.
How much energy is there in the universe? According to one source, there is zero energy. (To read the related article, click here.) Zero energy would make sense if the escape-velocity energy of matter exactly matched the gravitational potential energy. But the universe is expanding. This implies that the positive energy outweighs the negative gravitational energy. Most importantly, the zero-energy theory doesn't match energy measurements. According to the Wilkinson Microwave Anisotropy Probe, the universe's energy density is E-10 joules per cubic meter. Multiply that figure by the volume of the universe and you don't get zero.
In any case, the universe's non-zero energy is allegedly conserved. We have Einstein's famous energy equation below, and the gravitational-energy equation (E=energy; P=momentum; c=light speed; G=Newton's constant; r=radius m=mass):
We might be tempted to add them together (or subtract one from the other).
The first equation below models a singularity with an infinitesimal radius. The total energy (Et) is infinite (or minus infinity). The second equation below models an expanded universe. It's total energy is finite. Clearly energy is not conserved.
So how can energy be conserved? Let's try this:
The first equation above is not a conserved quantity. Energy increases or decreases when the radius changes. The second equation above is a conserved quantity--changes in the radius and energy always make the equation's right side equal to one.
Let's multiply Einstein's equation with the second equation above.
The universe's total energy is now conserved. Changes in the universe's radius cancel the changes in the gravitational energy. Using the above technique, we can factor in as many inverse-square energies as we like. Just for fun, let's factor in electromagnetism (k is a constant; q=charge):
As you can see, when all is said and done, the universe's total energy is equal to the right side of Einstein's equation--and it doesn't have to be zero, so it agrees with the Wilkinson probe's findings.
Sunday, August 21, 2016
More than a hundred years ago Albert Einstein set forth his relativity theory. He had this weird notion that time and space change in magnitude and shape when mass/energy is present or when your spaceship is traveling at high velocity. Many equations have been put together to model this phenomenon, including the ones listed below (t' and t=change in time; M=mass; G=Newton's constant; v=velocity; ds=spacetime; c=light speed; x=distance; r=radius):
The first equation above is the good old spacetime metric. The middle equation is the Lorentz factor. The third shows how mass affects time. These equations tell us that more mass or more velocity cause the change in time to shrink. Like time, distance (x) also shrinks, so we can replace t and t' with x and x' in the last two equations.
Thanks to E=mc^2 we know that mass, momentum and energy are different aspects of the same thing. Thus, the above equations are telling us that whenever we introduce energy into a system, time and space are affected. What the equations don't tell us is how or why energy causes time and space to shrink or curve. Let's see if we can figure it out.
To simplify the process, let's convert space and time to spacetime (ct and ct'). Doing so will make both space and time one variable:
If we call upon Pythagoras we can model spacetime with some simple trigonometry and even derive the Lorentz factor:
Notice in the diagrams, above and below, how ct' shrinks when vt grows. When vt is zero, ct = ct'.
Now let's change the variables to something even Schrodinger could appreciate:
We can now model spacetime with a wave function:
So what happens when a wave's energy increases? Its wavelength shortens. In the case below, the wavelength is ct, but becomes ct' when energy is increased. Notice how the circles or endpoints (particles) of each wave move closer together. Gravity?
In the diagram below we start with an arbitrary sphere of spacetime. The lines inside the first sphere (A) are straight or flat long waves of spacetime. The only energy present is zero-point energy:(1/2)hf. The second sphere (B) has a small circle at the center. This shall represent, say, a black hole. There is more energy present (nfh)--the spacetime wavelengths grow shorter, making sphere B smaller. (Variables: s=angular distance; r=radius or wavelength; phi=angle.)
Sphere B's spacetime is more curved than sphere A's as evidenced by the formula phi = s/r. Distance s covers a greater angle around B than A.
So far, for the sake of simplicity, we've depicted spacetime as a single one-dimensional wave as the first two diagrams below illustrate. The third diagram below is probably more realistic. It shows a complex hodge-podge of waves whose sum and average yield expectation values of vt and ct'.
The relativity theory is consistent with Heisenberg's uncertainty principle: Et >= h-bar/2. If the change in energy (E) is large, then the change in time (t) is small if their product is to equal Planck's constant (h-bar) divided by two. For spacetime we can modify this to Ect >= ch-bar/2.
If spacetime's energy is one half frequency (f) times Planck's constant (h), it follows there would be less spacetime if energy is increased to nfh (n=any positive integer) or some amount greater than 1/2fh. At high energies, there is simply less ground-state energy or spacetime.
Saturday, August 20, 2016
Maybe you've read somewhere that particles can escape a black hole via quantum tunneling, Hawking radiation or pair production; however, according to classical physics, there is no way light can escape--or is there?
Light escapes a gravitational potential if it has sufficient escape velocity. In terms of energy, a photon's kinetic energy must exceed the gravitational potential energy. A black hole's event horizon, whose distance from the center is the Schwarzschild radius (r), is the place where kinetic energy equals potential energy.
Theoretically, if a photon passes through the event horizon toward the black hole's center, it can't escape. You will note that the second equation above assumes a photon has mass (bar-m). This should be converted to a mass equivalent: fh/c^2 (f=frequency; h=Planck's constant; c=light speed; G=Newton's constant; m=black hole mass). We can now derive the Schwarzschild radius:
Houston, we have a problem! The second equation above assumes that a photon's kinetic energy is (1/2)fh. Electromagnetism has kinetic energy (photons) and potential energy (which involves charge). If photons are kinetic energy, then it only makes sense to use a photon's total energy (fh) when deriving the radius (r).
If (1/2)fh equals the gravitational potential energy, then fh must be greater than the gravitational potential energy--and the photon should be able to escape.
The math above indeed shows that fh is greater than the gravitational potential. There is another consideration: does the photon have more than zero energy? The last equation above shows that fh would equal (1/2)fh if fh is zero. When a photon enters a gravitational field it gains energy or frequency (f). When it tries to leave, it loses energy or frequency (f). The equations below demonstrate this (d=distance the photon travels; g=gravitational acceleration):
The total frequency (ft) of the photon is the sum of the initial frequency (fo) plus or minus the change in frequency (triangle f). How much energy the photon gains or loses depends on its angle of entry or exit relative to the unit normal line (n). If the photon is parallel to the unit normal (n), cos(0) equals 1, the photon will lose a maximum amount of energy when exiting a black hole. If the photon leaves at an angle, it will lose less energy.
So, if a photon enters the event horizon it will gain energy. When it tries to leave, it will lose energy. If it has more than zero energy, will it escape? Let's compare the escape velocity (c) to the gravitational equivalent. As you recall, we derived this inequality:
Let's multiply both sides by C^2/fh, then simplify:
Since c is greater than .707c, the photon will escape. In fact it will escape from any point that is outside the radius of Gm/c^2--one half the Schwarzschild radius.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-30/segments/1531676591296.46/warc/CC-MAIN-20180719203515-20180719223515-00272.warc.gz
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CC-MAIN-2018-30
| 13,457 | 64 |
http://www.usablestats.com/askstats/question/301/
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math
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Question 301:What is the advantage of a t-score over a z-score?
Need more than just an answer? Download the Easy t Excel T-test Calculator Package which gets you the answer with the detailed calculations.
Need help with the concepts in this question ? View the Comparing Two Means: 2 Sample t-test Tutorial.
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s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917119080.24/warc/CC-MAIN-20170423031159-00161-ip-10-145-167-34.ec2.internal.warc.gz
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CC-MAIN-2017-17
| 307 | 3 |
http://www.slideserve.com/telma/8-1-ratios-and-proportions
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math
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8-1 Ratios and Proportions. Four Key Terms Two Properties. Proportion. A statement that two ratios are equal. . Key Terms. and. Extended proportion. Written form of three or more ratios are equal. . Key Terms. Properties of Proportions. Cross-Product Property. (1). (2). (3). (4).
Four Key Terms
A statement that two ratios are equal.Key Terms
Written form of three or more ratios are equal.Key Terms
A drawing in which all lengths are proportional to corresponding lengths.
A comparison of each length in a drawing to the actual length.Key Terms
A photo that is 8 in. wide and 10 in. high is enlarged to a poster that is 2 ft. wide and 2.5 ft. high. What is the ratio of the height of the photo to the height of the poster?
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s3://commoncrawl/crawl-data/CC-MAIN-2017-39/segments/1505818688966.39/warc/CC-MAIN-20170922130934-20170922150934-00469.warc.gz
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CC-MAIN-2017-39
| 724 | 7 |
http://slideplayer.com/slide/4623476/
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math
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Presentation on theme: "Magnetic Circuits and Transformers"— Presentation transcript:
1 Magnetic Circuits and Transformers Magnetic FieldsMagnetic CircuitsInductance andMutual InductanceMagnetic MaterialsIdeal TransformersReal TransformersChapter 15: Magnetic Circuits
2 Magnetic FieldsMagnetic fields can be visualized as lines of flux that form closed paths. Using a compass, we can determine the direction of the flux lines at any point. Note that the flux density vector B is tangent to the lines of flux.
4 Force on Moving Electric Charge A charge moving through a magnetic field experiences a force f perpendicular to both the velocity u and flux density B.
5 Force on Moving Electric Charge A charge q moving through a magnetic field experiences a force f perpendicular to both the velocity u and flux density B. where u is the velocity vector and B is a magnetic field. The magnitude of this force is Current that flows through a conductor are electron charges in motion so the force acting on the wire with current in the magnetic field is and in the straight wire of the length l crossing the field under angle
6 Flux Linkage and Induced Voltage When the flux linking a coil changes, a voltage is induced in the coil. The polarity of the voltage is such that if a circuit is formed by placing a resistance across the coil terminals, the resulting current produces a field that tends to oppose the original change in the field. Faraday Law of magnetic induction: voltage e induced by the flux changes is where total flux linkage is N-number of turns, magnetic flux passing through the surface area A, and B is the magnetic field
7 Induced Voltage in a Moving Conductor A voltage is also induced in a conductor moving through a magnetic field in the direction such that the conductor cuts through magnetic flux lines. The flux linkage of the coil is (with uniform magnetic field B) so according to Faraday’s law the voltage induced in the coil is where
8 Ampère’s LawAmpère’s law (generalization of Kirchhoff's law) states that the line integral of magnetic field intensity H around a closed path is equal to the sum of the currents flowing through the surface bounded by the path where magnetic field intensity H is related to flux density B and magnetic permeability since so if H and dl point in the same direction
9 Ampère’s LawThe magnetic field around a long straight wire carrying a current can be determined with Ampère’s law aided by considerations of symmetry.So the magnetic flux density(*)Using Ampère’s law in the toroidal coil, filed intensity isUsing (*) we get inside the toroidal coil:
10 Reluctance of a Magnetic Path Magnetic circuits are analogue of electrical circuits. The magnetomotive force of N-turn current carrying coil is The reluctance R of a magnetic path depends on the mean length l, the area A, and the permeability μ of the material. Magnetic flux is analogous to current in electrical circuit and is related to F and R in a similar way as Ohm’s law
11 Magnetic CircuitsThe magnetic circuit for the toroidal coil can be analyzed to obtain an expression for flux. Magnetomotive force is Where the reluctance is so and the magnetic flux is
12 Magnetic CircuitsExample Magnetic circuit below relative permeability of the core material is 6000 its rectangular cross section is 2 cm by 3 cm. The coil has 500 turns. Find the current needed to establish a flux density in the gap of Bgap=0.25 T.
13 Magnetic CircuitsExample Magnetic circuit below relative permeability of the core material is 6000 its rectangular cross section is 2 cm by 3 cm. The coil has 500 turns. Find the current needed to establish a flux density in the gap of Bgap=0.25 T. Medium length of the magnetic path in the core is lcore=4*6-0.5=23.5cm, and the cross section area is Acore= 2cm*3cm = 6*10-4 m2 the core permeability is
14 Magnetic CircuitsExample The core reluctance is the gap area is computed by adding the gap length to each dimension of cross-section: thus the gap reluctance is:
15 Magnetic CircuitsExample Total reluctance is based on the given flux density B in the gap, the flux is thus magnetomotive force is thus the coil current must be
16 Coil Inductance and Mutual Inductance Coil inductance is defined as flux linkage divided by the current: since from the Faraday law When two coils are wound on the same core we get from the Faraday law:
17 Magnetic MaterialsIn general, relationship between B and H in magnetic materials is nonlinear.Magnetic fields of atoms in small domains are aligned (Fig b).Field directions are random for various domains, so the external magnetic field is zero.When H is increased the magnetic fields tend to align with the applied field.
18 Magnetic MaterialsDomains tend to maintain their alignment even if the applied field is reduced to zero.For very large applied field all the domains are aligned with the field and the slope of B-H curve approaches m0.When H is reduced to 0 from point 3 on the curve, a residual flux density B remains in the core.When H is increased in the reverse direction B is reduced to 0.Hysteresis result from ac current
19 Energy ConsiderationEnergy delivered to the coil is the integral of the power:Sincewhere l is the mean path length and A is the cross-section area, we getAnd since Al is the volume of the core, the per unit volume energy delivered to the coil is
20 Energy LossEnergy lost in the core (converted to heat) during ac operation per cycle is proportional to the area of hysteresis loop.To minimize this energy loss use materials with thin hysteresisBut for permanent magnet we need to use materials with thicj hysteresis and large residual field.Energy is also lost due to eddy currents in the core materialThis can be minimized with isolated sheets of metal or powdered iron cores with insulating binder to interrupt the current flow.
21 A transformer consists of several coils wound on a common core. Ideal TransformersA transformer consists of several coils wound on a common core.In ideal transformer we have:
22 A transformer consists of several coils wound on a common core. Ideal TransformersA transformer consists of several coils wound on a common core.Power in ideal transformer delivered to the load:
23 Ideal Transformers Impedance transformation. Using and We get the input impedance of the ideal transformer equal to:
24 Ideal TransformersConsider the circuit with ideal transformer and find phasor currents and voltages, input impedance, as well as power delivered to the load.The input impedance isSo the input current isThe input voltage is
25 Ideal TransformersPower delivered to the load is the same as the input powerOr directly
26 Figure 15.28 The equivalent circuit of a real transformer. Real TransformersFigure The equivalent circuit of a real transformer.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363149.85/warc/CC-MAIN-20211205065810-20211205095810-00248.warc.gz
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CC-MAIN-2021-49
| 6,811 | 26 |
https://www.infoplease.com/dictionary/linearequation
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math
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Meaning of linear equation
— Math. Math.
- a first-order equation involving two variables: its graph is a straight line in the Cartesian coordinate system.
- any equation such that the sum of two solutions is a solution, and a constant multiple of a solution is a solution. Cf.
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s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257649095.35/warc/CC-MAIN-20180323220107-20180324000107-00717.warc.gz
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CC-MAIN-2018-13
| 279 | 4 |
https://www.careerride.com/fa-composite-cost-of-capital-explained.aspx
|
math
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What is composite cost of capital? Explain the process to compute it?
Composite cost of capital is also known as weighted average cost of capital which is a measurable unit for it. It also tells about the component costs of common stock, preferred stock, and debt. Each of these components is given weightage on the basis of the associated interest rate and other gains and losses with it. It shows the cost of each additional capital as against the average cost of total capital raised. The process to compute this is first computing the weighted average cost of capital which is the collection of weights of other costs summed together. The formula is given as:-
WACC= Wd (cost of debt) + Ws (cost of stock/RE) + Wp (cost of pf. Stock)
In this the cost of debt is calculated in the beginning and it is used to find out the cost of capital and other weights of cost is been calculated after the calculation each and every individual weight of the component is added and then it gives the final composite cost.
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s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00006.warc.gz
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CC-MAIN-2022-21
| 1,010 | 4 |
https://quran.com/2/61/clearquran-with-tafsir
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math
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ﮛ ﮜ ﮝ ﮞ ﮟ ﮠ ﮡ ﮢ ﮣ ﮤ ﮥ ﮦ ﮧ ﮨ ﮩ ﮪ ﮫ ﮬ ﮭ ﮮ ﮯ ﮰ ﮱﯓ ﯔ ﯕ ﯖ ﯗ ﯘ ﯙ ﯚ ﯛﯜ ﯝ ﯞ ﯟ ﯠ ﯡ ﯢﯣ ﯤ ﯥ ﯦ ﯧ ﯨ ﯩ ﯪ ﯫﯬ ﯭ ﯮ ﯯ ﯰ ﯱ ﯲ ﯳ ﯴ ﯵ ﯶﯷ ﯸ ﯹ ﯺ ﯻ ﯼ
Dr. Mustafa Khattab
And ˹remember˺ when you said, “O Moses! We cannot endure the same meal ˹every day˺. So ˹just˺ call upon your Lord on our behalf, He will bring forth for us some of what the earth produces of herbs, cucumbers, garlic, lentils, and onions.” Moses scolded ˹them˺, “Do you exchange what is better for what is worse? ˹You can˺ go down to any village and you will find what you have asked for.” They were stricken with disgrace and misery, and they invited the displeasure of Allah for rejecting Allah’s signs and unjustly killing the prophets. This is ˹a fair reward˺ for their disobedience and violations.
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s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255943.0/warc/CC-MAIN-20190520101929-20190520123929-00289.warc.gz
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CC-MAIN-2019-22
| 898 | 3 |
http://www.thestudentroom.co.uk/showthread.php?t=1965962
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math
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Here are the questions:
1) The points A(-3,19), B(9,11) and C(-15,1) lie on the circumference of a circle. Find the coordinates of the centre of the circle.
Midpoint = (-3+9/2, 19+11/2) = (-3,15)
11-19/9+3 = -3/4
Did the same for the other one, and got: M.P.(9,10) and 2y=3x-7
(4y=-3x+51) + (2y=3x+7) -> 6y=44 -> y=44/6=22/3
Is this right so far?
2) Find the distance between (-4a,0) and (3a,-2a)
=√[49a2+4a2 = √53a2
Would the answer be √53a or 53a?
It is 53a, because the square root cancels out with the squared. Remember it is .
No it isn't
(Original post by Math12345)
It is 53a, because the square root cancels out with the squared. Remember it is sqrt(53a)^2
Thanks for posting! You just need to create an account in order to submit the post
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© Copyright The Student Room 2016 all rights reserved
The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.
Register Number: 04666380 (England and Wales), VAT No. 806 8067 22
Registered Office: International House, Queens Road, Brighton, BN1 3XE
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s3://commoncrawl/crawl-data/CC-MAIN-2016-36/segments/1471982298551.3/warc/CC-MAIN-20160823195818-00239-ip-10-153-172-175.ec2.internal.warc.gz
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CC-MAIN-2016-36
| 1,156 | 24 |
http://www.rejinpaul.com/2012/02/ec2045-satellite-communication-question.html
|
math
|
2. What is Satellite?
3. State Kepler’s first law.
4. State Kepler’s second law.
5. State Kepler’s third law.
6. Define apogee and perigee.
7. What is line of apsides?
8. Define ascending node and descending node.
9. Define Inclination.
10. Define mean anomaly and true anomaly.
11. Mention the apogee and perigee height.
12. What is meant by azimuth angle and look angles?
13. Give the 3 different types of applications with respect to satellite systems.
14. Mention the 3 regions to allocate the frequency for satellite services.
15. Give the types of satellite services.
16. What is mean by DOMSAT, INTELSAT, SARSAT?
17. Define polar-orbiting satellites.
18. Give the advantage of geostationary orbit.
19. Write short notes on station keeping.
20. What are the geostationary satellites?
21. What is sun transit outage?
(b) Explain about U.S Domsats. (6)
2. Discuss briefly the development of INTELSAT starting from the 1960s through the
3. What is meant by polar orbiting? Explain in detail. (16)
4. State Kepler’s three laws of planetary motion. Illustrate in each case their relevance to artificial satellites orbiting the earth. (16)
5. Explain in detail the geocentric-equatorial coordinate system which is based on the
earth’s equatorial plane. (16)
6. Explain in detail about topocentric-horizon coordinate system which is based on the
observer’s horizon plane. (16)
7. Explain in detail about various measure of time. (16)
2. Write short notes on attitude control system.
3. What is declination?
4. What is meant by payload?
5. What is meant by transponder?
6. Write short notes on station keeping.
7. What is meant by Pitch angle?
8. What is an propellant?
9. What is an Yaw and zero ‘g’?
10. Describe the spin stabilized satellites.
11. What is meant by frequency reuse?
12. What is meant by spot beam antenna?
13. What is meant by momentum wheel stabilization?
14. What is polarization interleaving?
15. Define S/N ratio.
16. What is noise weighting?
17. What is noise power spectral density?
18. What is an inter modulation noise?
19. What is an antenna loss?
20. Define sky noise.
21. Define noise factor.
22. What is TWTA?
23. What is an OMT?
2. Explain about Earth eclipse of satellite and sun transit outage. (16)
3. Explain about launching orbits. (16)
4. Explain what is meant by satellite attitude and briefly describe two forms of attitude
Control . (16)
5. Draw the block diagram of TT&C and explain its blocks. (16)
6. Describe briefly the most common type of high-power amplifying device used aboard a communication satellite. (16)
7. Explain about wideband receiver and advanced Tiros-N spacecraft. (16)
8. Describe briefly the antenna subsystem and Anik-E. (16)
9. Explain in detail about thermal control and Morelos. (16)
2. What are the methods of multiple access techniques?
3. What is an CDMA?
4. What is SCPC?
5. What is a thin route service?
6. What is an important feature of Intelsat SCPC system?
7. What is an TDMA? What are the advantages?
8. What is preamble?
9. Define guard time.
10. What is meant by decoding quenching?
11. What is meant by direct closed loop feedback?
12. What is meant by feedback closed loop control?
13. Define frame efficiency.
14. What is meant by digital speech interpolation?
15. What is meant by telephone load activity factor?
16. What are the types of digital speech interpolation?
17. What is meant by freeze out?
18. What is DSI?
19. What are the advantages of SPEC method over DSI method?
20. Define satellite switched TDMA?
21. What is SS / TDMA?
22. What is processing gain?
23. What is burst code word?
24. What is meant by burst position acquisition?
25. What is an single access?
26. What is an multiple access technique?
27. What is meant by frequency reuse?
28. What is meant by space division multiple access?
29. What is an error detecting code?
30. What are the limitations of FDMA-satellite access?
31. Write about pre-assigned TDMA satellite access.
32. Write about demand assigned TDMA satellite access.
2. Explain in detail about FDMA and show how this differs from FDM. (16)
3. Explain in detail the operation of a preassigned SCPC network. (16)
4. Explain in detail the operation of the spade system of demand assignment. What is the function of the common signaling channel? (16)
5. Describe the general operating principles of a TDMA network. Show how the
transmission bit rate is related to the input bit rate. (16)
6. Explain the need for reference burst and preamble and postamble in a TDMA
System . (16)
7. Explain in detail about network synchronization with neat sketch. (16)
8. Define and explain the terms carrier recovery, bit-time recovery, traffic data, frame efficiency and channel capacity. (16)
9. Explain in detail about speech interpolation and prediction. (16)
10. Explain in detail about satellite switched TDMA. (16)
11. Describe briefly about on board signal processing for FDMA/TDM operation. (16)
12. Describe in your own words how signal acquisition and tracking are achieved in a
DS/SS system . And also derive the expression for maximal sequence. (16)
13. Explain the principle behind spectrum spreading and dispreading and how this is
used to minimize interference in a CDMA system. Also determine the throughput
efficiency of the system. (16)
2. Give the difference between KU-band and the C-band receive only systems.
3. What is mean by ODU and IDU.
4. Explain about MATV system.
5. Write about CATV system.
6. Define S/N ratio.
7. What is noise weighting?
8. What is an EIRP?
9. What is noise power spectral density?
10. What is an inter modulation noise?
11. What is an antenna loss?
12. Define noise factor.
13. A satellite downlink at 12 GHz operates with a transmit power of 6 W and an antenna gain of 48.2 dB. Calculate the EIRP in dBW.
14. The range between a ground station and a satellite is 42000 km. Calculate the free
space loss a frequency of 6 GHz.
15. An antenna has a noise temperature of 35 K and it is matched into a receiver which has a noise temperature of 100 K. Calculate the noise power density and the noise power for a BW of 36 MHz.
16. Define Saturation flux density.
17. Write the equations of losses for clear sky conditions.
18. What are the types of antenna losses?
19. Define sky noise.
20. What is an Apsorptive n/w?
only home TV systems. (16)
2. Describe and compare the MATV and CATV systems. (16)
3. With the relevant expression explain in detail about transmission losses. (16)
4. Explain the classifications of system noise temperature. (16)
5. Explain uplink satellite circuit. (16)
6. Explain downlink satellite circuit. (16)
7. Describe briefly about the rains effects. (16)
8. Explain about inter-satellite link. (16)
2. Mention the 3 regions to allocate the frequency for satellite services.
3. Give the types of satellite services.
4. What is mean by DOMSAT, INTELSAT, SARSAT?
5. What are the applications of Radarsat?
6. What is ECEF?
7. What is dilution of precision?
8. What is PDOP?
8. What is DBS?
9. Give the frequency range of US DBS systems with high power satellites.
10. Give the frequency range of US DBS systems with medium power satellites.
11. What is DTH?
12. Write about bit rates for digital television.
13. Give the satellite mobile services.
14. What is GCC and GEC?
15. What is INMARSAT?
16. List out the regions covered by INMARSAT.
17. What is INSAT?
18. List out the INSAT series.
19. What is GSM?
20. What is GPRS?
21. Define DAB.
22. What is DVB?
23. What is GRAMSAT?
2. Explain about indoor and outdoor unit of home receiver. (16)
3. Explain about frequencies and polarization, transponder capacity and bit rates for
digital television. (16)
4. Explain in detail about satellite mobile services. (16)
5. Describe the operation of typical VSAT system. State briefly where VSAT systems and find widest applications. (16)
6. Describe the main features of Radarsat. Explain what is meant by dawn to dusk orbit and why the Radarsat follows such on orbit. (16)
7. Explain why a minimum of four satellites are visible at an earth location utilizing the GPS system for position determination. What does the term dilution of precision refer to?
8. Describe the main features and services offered by the orbcomm satellite system.
How do these services offered by geostationary satellites and terrestrial cellular systems?
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s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321309.25/warc/CC-MAIN-20170627101436-20170627121436-00087.warc.gz
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CC-MAIN-2017-26
| 8,314 | 168 |
http://lists.admb-project.org/pipermail/users/2012-December/002120.html
|
math
|
[ADMB Users] mildly robust methods in count models
davef at otter-rsch.com
Sat Dec 15 15:53:19 PST 2012
> Where I disagree mildly is that in looking at **this particular case**
>it seems to me that there are not important qualitative differences
>between the fits. Yes, the type I error rate/coverage is definitely
>better with the robust model (5.4% vs 7.7% of the time). In my own
>personal universe, 50% undercoverage (i.e. approx 7.5% type I error rate
>for a nominal rate of 5%) is fairly bad, but not terrible -- in the
>context of all the other things that are always wrong with the model
>that we can't control, I consider that a moderate but not an
Well it just happened to be the first example I could run easily.
Just to finish it off I thought I would compare the robust and nonrobust
type 2 (I think you call it) fit to the data, i.e. Mollie's original
model. Again I added a covariance consisting of N(0,1) random noise.
I increased the number of simulations to 40,000 to try and make it
For the robust model
For the original model
That came as a complete surprise to me.
So the original model formulations only rejects the null hypothesis 3.25% of the time.
That seems pretty bad.
The robut procedure improves on both model formulations. So robustness is robust.
> Robust methods are definitely on the list,
>but they might fall below (e.g.) getting people to stop using stepwise
I can hear this conversation where someone is saying, I will agree to
use (mildly)robust methods, but only of you let me use a stepwise approach.
More information about the Users
|
s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585181.6/warc/CC-MAIN-20211017175237-20211017205237-00570.warc.gz
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CC-MAIN-2021-43
| 1,573 | 27 |
http://stat.snu.ac.kr/board_view.php?id=1765
|
math
|
|제목||[학술세미나] [학과세미나] 5월 28일(월) 17시 특별세미나 안내|
|내용||[학과세미나] 5월 28일(월) 17시 특별세미나 안내
▪ 제목 : Optimal shrinkage estimation in heteroscedastic hierarchical models: beyond empirical Bayes
▪ 연사 : Samuel Kou (Harvard University)
▪ 일시 : 2018년 5월 28일(월) PM 17:00 – 18:00
▪ 장소 : 25동 405호
Hierarchical models are powerful statistical tools widely used in scientific and engineering applications. The homoscedastic (equal variance) case has been extensively studied, and it is well known that shrinkage estimates, the James-Stein estimate in particular, offer nice theoretical (e.g., risk) properties. The heteroscedastic (the unequal variance) case, on the other hand, has received less attention, even though it frequently appears in real applications. It is not clear of how to construct "optimal" shrinkage estimate. In this talk, we study this problem. In particular, we consider hierarchical linear models and models beyond Gaussian. We introduce a class of shrinkage estimates, constructed by minimizing an unbiased risk statistic. We show that this class is asymptotically optimal in the heteroscedastic case. We apply the estimates to real examples and observe competitive numerical results.
세미나 안내_180528_Samuel Kou.hwp [14KB]
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s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823318.33/warc/CC-MAIN-20181210055518-20181210081018-00009.warc.gz
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CC-MAIN-2018-51
| 1,351 | 8 |
http://www.endurance.net/RideCamp/archives/past/9807/msg00925.html
|
math
|
Check it Out!
[Date Prev] [Date Next]
[Thread Prev] [Thread Next] [Date Index] [Thread Index] [Author Index] [Subject Index]
Re: suspensory ligament
Stall rest is the WORST thing.. you need to do as others advised.. stall
and walk every day add more time.
ice legs.. do the shots.
find a holistic vet.. try accupunture get the horse balanced
do herbs you indicated a supplement do it.
give ruta grava a homeopathic
also topical Ruta-Rhus-Arnica.. a few times each day
oh have to say this-- I read about this in a book just passing along
the info from the book you see-- just suggesting this is what I would do
IF it were my horse..
I have the Centurion Electromag boot.. it works great.. you may want to
rent one for 3 months.. they advertise in Equus and many other mags
beats the static magnets 100 time over.. 908 417 0050 i believe it
would be worth the cost..
I used it just for a pulled sp tendon.. fast healing
look at THOR laser also -rental many ads for them also - a very
ti can tell you about healing tendons --get his book The Bowed Tendon..
I bought a muscle stimulator unit --thinking I could use it to help fix
any leg or muscle problem..but it is far to complicated for me.. I put
it on my arm and had my hand jerking all over the place.. over my head..
so its back in the box.. never used
need an expert on this tool.. and i dont have the time to learn--as long
as my gang is sound..
any way thats my advice on the ripped tendon -- get ultasound often to
i read about a really super drug in THE HORSE mag..cant find it now
maybe it was online.
helps to heal the tendon with NO scaring.. you did mention a drug -may
Check it Out!
Back to TOC
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s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600400228998.45/warc/CC-MAIN-20200925213517-20200926003517-00073.warc.gz
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CC-MAIN-2020-40
| 1,675 | 33 |
http://news-antique.com/?id=785849&keys=art-artist-market-investment
|
math
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News-Antique.com - Dec 20,2008 - A tracking report on the year-to-date value of the Mei Moses All Art Index has been issued by Beautiful Asset Advisors. According to the fall art market insight report, as of 10/31/2008, the value of the index was 270.58** which, as one would expect, is down on the 2007 year end value of the index which was 279.42. Although the value of the index is published annually, according to the people at Beautiful Asset Advisors® LLC who calculate the index “We are now getting sufficient incremental data during a year so that at periodic intervals we can recreate our index with the sales that have transpired during the year up to that date. If we assume that these periodic sales have taken place at the end of the year than the index numbers we would get would be the ones that would be achieved at the end of the year if no additional information was collected. We have employed this technique to generate the summer and fall tracking results.
For those of you who haven’t heard of the Mei Moses Index before it is basically a number that represents the rate of return of fine art for a particular year. The rate of return can basically be defined as the annual percentage return realized on an investment. If you haven’t heard of the Mei Moses Index before or have heard of the index but wondered how it is calculated then read on. The Mei Moses index is calculated using data collected from repeat sales of the same work of art which is then used to determine the annual return of that particular work of art. For example, if a painting was purchased in 1990 for $50,000 and then sold in 2000 for $150,000 the total return would be 200% over a 10 year period which could then be broken down to 20% per year. The problem with using the average increase in value per year calculated from the total increase in value according to the two sale points is that the assumption is being made that the work of art increased (or decreased) in value by the same percentage each year which would not be the case. To combat this problem the Mei Moses index uses a special formula to determine as accurately as possible the yearly increase (or decrease) in value.
A simplified example of how the index is calculated is given by Matthew Spiegel who is a finance professor at Yale university. Spiegel’s example is:
Painting C sells in years 1 and 3 and during that time returns 5% per year (10% total for two years). Painting D sells in years 2 and 3 and returns 3% (3% total for one year). The repeat sales index would start by estimating the return from years 2 to 3 at 3% in order to perfectly fit the return on painting D. Given this it would then estimate the return to art from years 1 to 2 at 7% (3%+7% equals
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s3://commoncrawl/crawl-data/CC-MAIN-2017-17/segments/1492917121165.73/warc/CC-MAIN-20170423031201-00389-ip-10-145-167-34.ec2.internal.warc.gz
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CC-MAIN-2017-17
| 2,745 | 4 |
https://www.studyblue.com/notes/note/n/lecture_75b15dpdf/file/387406
|
math
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Lecture 7 ME300 Thermodynamics II Page 1 Lecture 7: Exergy Analysis Closed System Reading Assignment: Sections 7.1-7.3 Problems: P7, 7.15, 7.29 Any questions on Lecture 6? 7.1 Introducing Exergy Exergy: Property that quantifies potential for use. 7.2 Conceptualizing Exergy How to make best use of the situation when a system at a state different from that of the environment is allowed to interact with the environment and eventually comes to equilibrium state (called dead state) during its transient time? Examples: (a) Spontaneous cooling of hot body or warming of a cold body in a room (b) Braking of vehicles and locomotives in motion N a t ur a l ga s i s us e d f or he a t i ng bui l di ng. B ut hot a i r ( ~ 6 5 - 75 o F ) ca nnot r e pr oduc e t he N G i t us e d. D i d w e f ul l y ut i l i z e t he pot e nt i a l of N G f or he a t i ng t he bui l di ng ? H ow ca n w e m i ni m i z e i r r e v e r s i bi l i t i e s ? Lecture 7 ME300 Thermodynamics II Page 2 (c) Heat being dissipated from the heat exchangers behind the refrigerators (d) Simple gas turbine engine?s exhaust energy leading to co-generation power plant (e) Use of low-grade heat coming out of co-gen power plants (f) On-the-road vehicles with internal combustion engines making use of exhaust gas energy for turbochargers -----. Overall system (includes system + environment), system and environment used for evaluating exergy Exergy of a system (1) E x e r gy r e f e r e nce PE M e cha ni ca l W or k E l e ct r i ci t y E x e r gy i s t he m a x i m um t he or e t i ca l w or k obt a i na bl e f r om a n ov e r a l l s y s t e m co ns i s t i ng of a s y s t e m a nd t he e nv i r on m e nt a s t he s y s t e m c om e s i nt o e qui l i br i um w i t h t he e nv i r onm e nt ( pa s s e s t o de a d s t a t e ) . 25 o C , 1 a t m . Lecture 7 ME300 Thermodynamics II Page 3 The closed overall system boundary exergy reference environment is drawn so that only work crosses its boundary, Q=0 (thermally isolate system ? e.g. Joule?s experiment); V0 is constant and the environment is at 25oC and 1 atm. Neglecting ?PE and ?KE, change in the overall system energy, . E, Uo: Initial and final (dead state) energy of the system S, So: Initial and final (dead state) entropy of the system V, Vo: Initial and final (dead state) volume of the system Total volume of the overall system is constant. ?Ue: Change in internal energy of the environment Apply 1st T dS equation ( to replace ?Ue Rearranging gives: , rearrange for , substitute in Equation 2 to get: Replace E by E=U+KE+PE, giving: , here E, exergy is given by Exergy Aspects: 1. Property of a system and environment combined (overall system). Once environment is specified, then it represents property of the system. 2. The maximum work (exergy) cannot be negative. 3. Exergy is not conserved. It is destroyed by irreversibilities. In the limiting case, the system is allowed to change spontaneously to the dead state without producing any work ? potential to produce work destroyed. 4. Exergy ? maximum theoretical work obtainable. Conversely, it can be regarded as minimum theoretical work required to restore the system from dead state to its initial state. 5. A system in dead state means that it is in thermal and mechanical equilibrium with its environment and its value is 0 and no more useful work. ?Thermo-static implies no more work out, thermodynamics implies converting energy to work until it reaches equilibrium? Mongia?s spin in order to Lecture 7 ME300 Thermodynamics II Page 4 minimize struggle with the use of the word thermodynamics (quasi-static assumption) to describe relationship between W, Q and E. Specific Exergy (2) Please, go through example on page 336 and Example 7.1 Calculate specific exergy of saturated water vapor at 120oC, having a velocity 30 m/s and an elevation of 6 m relative to an exergy reference environment where To = 298 K (25oC), po = 1 atm, and g = 9.8 m/s. Recall for saturated liquids: Use Table A-2 to get the properties. Use Eq. (2) to calculate e: Example 7.1: Exergy of an internal combustion engine: 7 bar, 867oC; reference environment: 300 K, 1.013 bar Ignore PE and KE. Ideal gas properties from Table A-22: Let us look at the individual components of e: (a) (b) (c) Focus on u and s to make the most impact on e. Exergy Exchange: Initially the system state is: Interaction with the surrounding occurs via work and heat transfer leading to different state: Subtracting, we get exergy change in the system: Lecture 7 ME300 Thermodynamics II Page 5 Exergy Transfer Exergy destruction Problem 7.31 Known: 2 lb of ammonia at 20 psi with 80% quality is heated by an electrical heating element until its volume increases by 25% with negligible increase in pressure. Find: Calculate: (a) The energy transferred by electrical work and the accompanying exergy transfer. (b) The amount of energy transfer by work to the piston and the accompanying exergy transfer. (c) The change in exergy of the ammonia. (d) The amount of exergy destruction. Engineering Model: 1. NH3 in a closed system with the environment conditions as shown. 2. Adiabatic system with negligible KE and PE. 3. Negligible mass of the electrical resistor. Analysis: Properties of NH3 from Table A-13E and A-14E Lecture 7 ME300 Thermodynamics II Page 6 Schematic & Given Data: Which are nearly saturated conditions at 20 lbf/in.2 20 lb f / i n 2 T S 1 a tm 1 2 0 60 o F Lecture 7 ME300 Thermodynamics II Page 7 Exergy change is given by: Exergy Transfer Exergy destruction Negligible KE, PE and for piston only the equation reduces to: = 5.25 Btu There is no change in volume. Exergy change: 10.8451778?5201.3687?1.106] Energy destruction Lecture 7 ME300 Thermodynamics II Page 8 Lecture 8: Exergy Analysis Closed System Reading Assignment: Section 7.4 Problems: P8, 7.37, 7.44 Mechanical Engineering
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https://www.pass4success.com/prmia/discussions/exam-8006-topic-1-question-70-discussion
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math
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Using covered interest parity, calculate the 3 month CAD/USD forward rate if the spot CAD/USD rate is 1.1239 and the three month interest rates on CAD and USD are 0.75% and 0.4% annually respectively.
Forward rates can be calculated from spot rates and interest rates using the formula Spot x (1+domestic interest rate)/(1+foreign interest rate), where the 'Spot' is expressed as a direct rate (ie as the number of domestic currency units one unit of the foreign currency can buy). In this case the forward rate will be 1.1239 * (1 + 0.75%*90/360) / (1 + 0.4%*90/360) = 1.1249.
It can be confusing to determine which interest rate should be considered 'domestic', and which 'foreign' for this formula. For that, look at the spot rate. Think of the spot rate as being x units of one currency equal to 1 unit of the other currency. In this case, think of the spot rate 1.1239 as 'CAD 1.1239 = USD 1'. The currency that has the '1' in it is the 'foreign' and the other one is 'domestic'.
It is also important to remember how exchange rates are generally quoted. Most exchange rates are quoted in terms of how many foreign currencies does USD 1 buy. Therefore, a rate of 99 for the JPY means that USD 1 is equal to JPY 99. These are called 'direct rates'. However, there are four major world currencies where the rate quote convention is the other way round - these are EUR, GBP, AUD and NZD. For these currencies, the FX quote implies how many US dollars can one unit of these currencies buy. So a quote of '1.1023' for the Euro means EUR 1 is equal to USD 1.1023 and not the other way round.
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CC-MAIN-2023-40
| 1,589 | 4 |
http://www.pscquestionbank.com/
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math
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81. Among the four options given below find the word which does not have a related meaning as the word ‘Judicious’.
82. Among the four options given below find the word which has opposite meaning of the word ‘combat’.
83. Choose the pair in which the words are differently related.
(A) Canberra - Australia
(B) Algeria - Africa
(C) Norway - Europe
(D) Iran - Asia
84. In the following, find the number that replaces the ‘....?....’
846 : 624 : : 774 : ‘....?....’
85. Examine carefully the following statements and answer the question given below :
A and B play football and cricket.
B and C play cricket and hockey.
A and D play basketball and football
C and D play hockey and basketball.
Who plays football, basketball and hockey ?
86. Arrange the following words as per the order in the dictionary :
(A) (iii), (ii), (i), (iv), (v)
(B) (iii), (ii), (i), (v), (iv)
(C) (i), (ii), (iii), (v), (iv)
(D) (ii), (iii), (i), (iv), (v)
87. If February 1, 2004 is Wednesday, then what day is March 3, 2004 ?
88. In the following number series, one number is wrong. Find the wrong number.
89. The shape of acceleration versus mass graph for constant force is :
(C) Straight line
90. The tank appears shallow than its actual depth, due to :
91. In n-type semiconductor the majority carriers are :
(C) Electrons and Holes
(D) None of these
92. In a given sample there are 10, 000 radio-active atoms of half-life period 1 month. The number of atoms remaining without disintegration at the end of 3 months is :
(B) 5, 000
(C) 2, 500
(D) 1, 250
93. A convex lens is placed in water, its focal length :
(C) Remains the same
94. Principles of Law of Inheritance were enunciated by :
95. The infective stage of malarial parasite plasmodium that enters human body is :
(D) minuta form
96. Respiratory centre in human beings is :
(B) medulla oblongata
97. Artificial ripening of fruits is accomplished by treatment with :
(A) Sodium chloride
(B) Indole-3 acetic acid
(D) Ethelene gas
98. In mammals ammonia produced by metabolism is converted into urea in the :
99. The 17th Laureans World Sports Award, 2015 for Life Time Achievement was given away to Niki Landa for:
(D) Motor racing
100. ‘Ardhanareeswaran’ the famous novel written by :
(A) Sugatha Kumari
(B) Perumal Murugan
(C) Akbar Kakkattil
(D) C. Radhakrishnan
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s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541783.81/warc/CC-MAIN-20161202170901-00322-ip-10-31-129-80.ec2.internal.warc.gz
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CC-MAIN-2016-50
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http://tex.stackexchange.com/questions/tagged/ipad+svg
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math
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TeX - LaTeX
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Recommended workflow? LaTeX -> ePub with math viewable on iPad
This started as a comment in an older thread, but I thought I repost it here as a question to the forum... I'm looking for a workflow that will take me efficiently and reliably from existing LaTeX ...
Jan 27 '13 at 19:09
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s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394011355201/warc/CC-MAIN-20140305092235-00071-ip-10-183-142-35.ec2.internal.warc.gz
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https://www.coursehero.com/tutors-problems/Chemistry/8546085-Hi-Im-trying-to-study-for-a-test-but-cannot-figure-out-how-to-solve-t/
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math
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Hi I'm trying to study for a test but cannot figure out how to solve these.
For GaAs, the autoionization constant at 300 K is 4x10^12 cm^-6.
a) What are the values of p and n, the hole and mobile electron carrier concentrations, respectively, at this temperature?
b) Suggest an element that might serve as a dopant, substituting for As, to increase n. Suggest an element that might serve as a dopant, substituting for Ga, to increase p.
c) If a donor is added that makes n= 2 x10^17 cm^-3, what is the value of p?
d)Propose reasonable concentrations of p and n for GaAs at 200 K and 400 K.
e) Will the autoionization equilibrium constant for GaN be larger or smaller than that for GaAs?
a)There are two samples, labelled A and B. Sample A is a physical mixture of GaP and GaAs in a 1:1 mole ratio. Sample B is a GaP0.5As0.5 solid solution. Are samples A and B distinguishable by elemental analysis, X-ray diffraction, absorption spectroscopy, or emission spectroscopy?
3. Recently, it has been found that many living systems use small gaseous molecules as hormones. A company that has strong interests in the control of fruit ripening has determined that a competitor is treating apples with a naturally occurring plant hormone that is generated at the apple orchards with small chemical reactors (the reactors appear to be small heaters with some sort of metal catalyst). The company has collected samples of the only reagent which goes into the reactor and samples of the product that is produced. You carry out gas chromatography, mass spectrometry, and carbon-13 NMR spectroscopy on each of the compounds.
a) The reactant put into the reactor is a compound that contains only C and H, exhibits one peak in the gas chromatogram, and has a mass spectrum with a peak at 56 amu. Draw the structures of three possible isomers that satisfy the octet rule and the mass spectrometry data; one isomer should have only single bonds.
b) For each of the isomers that you have drawn, identify the ideal bond angles for each carbon center on the drawing above.
c) The carbon-13 NMR exhibits just one peak. Identify which isomer of the three that you drew above is the reagent. Explain how you eliminated the others by identifying the number of peaks expected in the carbon-13 NMR spectra.
d) The product of the reaction yields a single peak in the gas chromatogram, has a peak at 28 amu in the mass spectrum, and exhibits one peak in the carbon-13 NMR. Draw the structure of the product.
e) The conversion of the reactant is found to be exothermic. Using bond strength data from your textbook and/or bond strain considerations, rationalize why the reaction is exothermic.
Recently Asked Questions
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CC-MAIN-2019-13
| 3,127 | 18 |
https://uilis.usk.ac.id/oer/items/show/2459
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math
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This book helps students to master the material of a standard US undergraduate first course in Linear Algebra. The material is standard in that the subjects covered are Gaussian reduction, vector spaces, linear maps, determinants, and eigenvalues and eigenvectors. Another standard is book’s audience: sophomores or juniors, usually with a background of at least one semester of calculus. The help that it gives to students comes from taking a developmental approach — this book’s presentation emphasizes motivation and naturalness, using many examples. The developmental approach is what most recommends this book so I will elaborate. Courses at the beginning of a mathematics program focus less on theory and more on calculating.
Mathematics, Saint Michael’s College
Jim Hefferon, “LINEAR ALGEBRA,” Open Educational Resource (OER) - USK Library, accessed December 7, 2023, http://uilis.usk.ac.id/oer/items/show/2459.
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| 930 | 3 |
https://smart-bookmarking.com/qa/quick-answer-what-does-congruent-look-like.html
|
math
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- What is an example of a congruent shape?
- What does congruent mean *?
- What is AAS congruence rule?
- Is AAS same as SAA?
- Why does AAS congruence not work?
- What is difference between similar and congruent triangle?
- What does it mean if two segments are congruent?
- What does a congruent line look like?
- What are the 3 properties of congruence?
- Are all regular octagons congruent?
- How do you know if something is congruent?
- How do you know if a shape is congruent or similar?
- How do you prove AAS?
- What special marks are used to show that segments are congruent?
- Are all similar figures congruent?
- Can circles be congruent?
What is an example of a congruent shape?
Usually, we reserve congruence for two-dimensional figures, but three-dimensional figures, like our chess pieces, can be congruent, too.
Think of all the pawns on a chessboard.
They are all congruent.
To summarize, congruent figures are identical in size and shape; the side lengths and angles are the same..
What does congruent mean *?
Congruent means the same size and shape.
What is AAS congruence rule?
The AAS Theorem says: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent. Notice how it says “non-included side,” meaning you take two consecutive angles and then move on to the next side (in either direction).
Is AAS same as SAA?
AAS Congruence. A variation on ASA is AAS, which is Angle-Angle-Side. … Angle-Angle-Side (AAS or SAA) Congruence Theorem: If two angles and a non-included side in one triangle are congruent to two corresponding angles and a non-included side in another triangle, then the triangles are congruent.
Why does AAS congruence not work?
Knowing only side-side-angle (SSA) does not work because the unknown side could be located in two different places. Knowing only angle-angle-angle (AAA) does not work because it can produce similar but not congruent triangles.
What is difference between similar and congruent triangle?
In mathematics, we say that two objects are similar if they have the same shape, but not necessarily the same size. If the objects also have the same size, they are congruent. …
What does it mean if two segments are congruent?
Congruent angles are angles that have the same measure. Congruent segments are segments that have the same length. … Two points (segments, rays or lines) that divide a segment into three congruent segments trisect the segment. The two points at which the segment is divided are called the trisection points of the segment.
What does a congruent line look like?
Congruent segments are simply line segments that are equal in length. Congruent means equal. Congruent line segments are usually indicated by drawing the same amount of little tic lines in the middle of the segments, perpendicular to the segments. We indicate a line segment by drawing a line over its two endpoints.
What are the 3 properties of congruence?
The three properties of congruence are the reflexive property of congruence, the symmetric property of congruence, and the transitive property of congruence. These properties can be applied to segment, angles, triangles, or any other shape.
Are all regular octagons congruent?
Regular octagons have sides that are congruent. That means that all of the sides of the octagon are the same measurement. … The interior angles of a regular octagon are all 135 degrees and the central and exterior angles are all 45 degrees.
How do you know if something is congruent?
Two polygons are congruent if they are the same size and shape – that is, if their corresponding angles and sides are equal. Move your mouse cursor over the parts of each figure on the left to see the corresponding parts of the congruent figure on the right.
How do you know if a shape is congruent or similar?
Identifying Triangles as Similar, Congruent, or Neither When two line segments have the same length, they are congruent. When two figures have the same shape and size, they are congruent.
How do you prove AAS?
Angle-Angle-Side (AAS) Rule If two angles and a non-included side of one triangle are equal to two angles and a non-included side of another triangle, then the triangles are congruent. In the diagrams below, if AC = QP, angle A = angle Q, and angle B = angle R, then triangle ABC is congruent to triangle QRP.
What special marks are used to show that segments are congruent?
When we draw congruent segments, we use tic marks to show that two segments are congruent.
Are all similar figures congruent?
All congruent figures are similar, but not all similar figures are congruent. Congruence means two objects (whether two dimensional or three dimensional) are identical in size and shape. Everything about them — their angles, lengths of sides, overall dimensions — are identical.
Can circles be congruent?
All circles of the same size are congruent to one another. “Size” can refer to radius, diameter, circumference, area, etc.
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s3://commoncrawl/crawl-data/CC-MAIN-2020-50/segments/1606141674082.61/warc/CC-MAIN-20201201104718-20201201134718-00157.warc.gz
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CC-MAIN-2020-50
| 5,019 | 51 |
https://www.wikiyy.com/en/Pentacontagon
|
math
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A regular pentacontagon
|Edges and vertices||50|
|Schläfli symbol||, t|
|Symmetry group||Dihedral (D50), order 2×50|
|Internal angle (degrees)||172.8°|
|Properties||Convex, cyclic, equilateral, isogonal, isotoxal|
Regular pentacontagon properties
One interior angle in a regular pentacontagon is 1724⁄5°, meaning that one exterior angle would be 71⁄5°.
The area of a regular pentacontagon is (with t = edge length)
and its inradius is
The circumradius of a regular pentacontagon is
The regular pentacontagon has Dih50 dihedral symmetry, order 100, represented by 50 lines of reflection. Dih50 has 5 dihedral subgroups: Dih25, (Dih10, Dih5), and (Dih2, Dih1). It also has 6 more cyclic symmetries as subgroups: (Z50, Z25), (Z10, Z5), and (Z2, Z1), with Zn representing π/n radian rotational symmetry.
John Conway labels these lower symmetries with a letter and order of the symmetry follows the letter. He gives d (diagonal) with mirror lines through vertices, p with mirror lines through edges (perpendicular), i with mirror lines through both vertices and edges, and g for rotational symmetry. a1 labels no symmetry.
These lower symmetries allows degrees of freedom in defining irregular pentacontagons. Only the g50 subgroup has no degrees of freedom but can seen as directed edges.
Coxeter states that every zonogon (a 2m-gon whose opposite sides are parallel and of equal length) can be dissected into m(m-1)/2 parallelograms. In particular this is true for regular polygons with evenly many sides, in which case the parallelograms are all rhombi. For the regular pentacontagon, m=25, it can be divided into 300: 12 sets of 25 rhombs. This decomposition is based on a Petrie polygon projection of a 25-cube.
- Gorini, Catherine A. (2009), The Facts on File Geometry Handbook, Infobase Publishing, p. 120, ISBN 9781438109572.
- The New Elements of Mathematics: Algebra and Geometry by Charles Sanders Peirce (1976), p.298
- Constructible Polygon
- "Archived copy" (PDF). Archived from the original (PDF) on 2015-07-14. Retrieved 2015-02-19.CS1 maint: archived copy as title (link)
- The Symmetries of Things, Chapter 20
- Coxeter, Mathematical recreations and Essays, Thirteenth edition, p.141
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CC-MAIN-2020-05
| 2,206 | 21 |
https://forum.duolingo.com/comment/25362873/what-is-the-best-way-practicing-reading-questions
|
math
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what is the best way practicing reading questions
Any good advices (:
I am afraid, that I don´t understand your question? Read and translate, that´s all.
But in detail:
First read the question sentence, then try to translate it by yourself without cheating (lookup on a dictionary, use an online translating tool or somehting like that.)
2nd hit on the continue button. You´ll be informed, if there was something wrong. Look at it, memorize it and if you don´t understand well, why your answer was wrong open the discussion to read, if there are any explanations which help you to understand. If not, ask your question there and hope for a good answer.
3rd after the lesson use the review button to review your wrong answers again, see, that you understand or memorize all of them.
4th repeat the lesson as often as you can. Use the hover function on the words of the question sentence to get an idea on what you should answer. Repeat the lessons till you have no errors there (it´s easy, because in the normal circles are often the same questions in the same order).
5th Do this for all circles and then try the strengthen button. Do the same here as for the single circles (1-4), but it is a bit more difficult here, because there are some more questions to proof, if you´ve understood the points.
6th Try out the timed practice. Now you have no time to look and proof, you have only the time to answer as quick as you can. But here are much more and more difficult questions and variabilities and the order isn´t every time the same. But this will really strengthen your abilities.
And: Don´t give up! Go forward and come back later, if it works not so good at the particular moment.
This is at least my way.
Is that, what you asked for?
I don't understand the question either. But guessing at what you mean, I'd say stick with it for now, level 11 has long ways to go. I've only passed the half of the course here the other day, and the lesson that marks half way trough is basically first lesson that isn't in present tense. I must say that feels odd. Once you get most of this course done, you can start reading wikipedia articles in german, perhaps pick topics you know a lot about, for example if you are scientifically inclined, you can read scientific articles, as the words that won't be covered in this course, are pretty much the same in german and english (or any other language for that matter, they are pretty universal).
You might also want to practice listening skills, and for that you can watch movies, listen to songs, or, find youtube channels in german. There's one called easy german, where a girl walks around and interviews people in the street about various topics, in german. Everything is subtitled in both german and english, so you can be sure to understand. Problem with that is that spoken german is apparently way simpler than written one, so you'd get to practice your listening, but perhaps you won't get to the hard core german there. good side of it, you'd get exposed to slang and the language you'd also be exposed to if you were to go to germany, or if you were to talk to germans online.
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s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704832583.88/warc/CC-MAIN-20210127183317-20210127213317-00021.warc.gz
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CC-MAIN-2021-04
| 3,137 | 15 |
https://dl.gi.de/items/0c674157-a67d-4a24-b755-74e87057f88c
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math
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Towards a biometric random number generator – a general approach for true random extraction from biometric samples
ISSN der Zeitschrift
Gesellschaft für Informatik e.V.
Biometric systems are per definition used to identify individuals or verify an identity claim - one difficulty of getting reliable decisions is the inherent noise that makes it difficult to extract stable features from biometric data. This paper describes how biometric samples can be used to generate strong random numbers which form the basis of many security protocols. Independent from the biometric modality, the only requirement of the proposed solution are feature vectors of fixed length and structure. Each element of such a feature vector is analyzed for its reliability - only unreliable positions, that cannot be reproduced coherently from one source, are extracted as bits to form the final random bit sequences. Optionally a strong hash-based random extraction can be used. The practicability is shown testing vascular patterns against the NIST-recommended test suite for random number generators.
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CC-MAIN-2024-18
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http://mengnadu.com/index.php/books/an-elementary-treatise-on-solid-geometry
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math
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By Smith, Charles, 1844-1916
This publication used to be initially released sooner than 1923, and represents a duplicate of an immense old paintings, holding an analogous structure because the unique paintings. whereas a few publishers have opted to follow OCR (optical personality reputation) expertise to the method, we think this ends up in sub-optimal effects (frequent typographical mistakes, unusual characters and complicated formatting) and doesn't safely protect the historic personality of the unique artifact. We think this paintings is culturally very important in its unique archival shape. whereas we try to thoroughly fresh and digitally increase the unique paintings, there are sometimes cases the place imperfections equivalent to blurred or lacking pages, negative photographs or errant marks could have been brought as a result of both the standard of the unique paintings or the scanning approach itself. regardless of those occasional imperfections, we've introduced it again into print as a part of our ongoing worldwide publication renovation dedication, delivering consumers with entry to the very best historic reprints. We have fun with your knowing of those occasional imperfections, and truly desire you take pleasure in seeing the ebook in a layout as shut as attainable to that meant via the unique writer.
Read Online or Download An elementary treatise on solid geometry PDF
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Additional resources for An elementary treatise on solid geometry
Are x = 0, y = 0, z= Find the co-ordinates of the centre of the sphere in 25. scribed in the tetrahedron formed by the planes whose equations =- a. are y + z - 0, z + x - 0, x + y = 0, and x + y + z CHAPTER III. SURFACES OF THE SECOND DEGREE. The most general equation of the second degree, viz. 50. 2 ax* + + cz* + 2fyz + 2^,7? + 2htvy -f 2w + 2vy + 2wz + d = 0, contains ten constants. But, since we may multiply or divide the equation by any constant quantity without altering the relation between as, y, and z which it indicates, there are really only nine constants which are fixed for any particular surface, viz.
Find the equations of the straight lines which bisect the angles x y .. ,, , = z- and, x-. = u = -z . between the lines T = m m , I m n V m n = r. Then be two points, one on each line, such that the co-ordinates of P are li\ mr, nr, and of Q are I r, m r, n r; hence the co ordinates of the middle point of PQ are ^(1 + / ) r, ^ (//* + m ) r, % (n + n ) r. Since , Let P, Q S. S. G. . OPOQ 2 T&E STRAIGHT 18 middle the l m+m +l n on is point . By + ri. , 1 the equations ; = required of the equations are of the bisector n- the preceding Article cos 6 therefore sin = A/ therefore sin 26.
Closed polyhedron on any plane is zero. Find the co-ordinates of the centre of the sphere in 24. scribed in the tetrahedron formed by the planes whose equations and x + y + x=l. are x = 0, y = 0, z= Find the co-ordinates of the centre of the sphere in 25. scribed in the tetrahedron formed by the planes whose equations =- a. are y + z - 0, z + x - 0, x + y = 0, and x + y + z CHAPTER III. SURFACES OF THE SECOND DEGREE. The most general equation of the second degree, viz. 50. 2 ax* + + cz* + 2fyz + 2^,7?
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CC-MAIN-2017-51
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https://edoc.hu-berlin.de/handle/18452/3182
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math
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2005-10-21Buch DOI: 10.18452/2530
A Shooting Method for Fully Implicit Index-2 Differential-Algebraic Equations
A shooting method for two-point-boundary value problems for fully implicit index-1 and -2 differential-algebraic equations is presented. A combination of the shooting equations with a method of the calculation of consistent initial values leads to a system of nonlinear algebraic equations with nonsingular Jacobian. Examples are given.
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s3://commoncrawl/crawl-data/CC-MAIN-2023-50/segments/1700679100146.5/warc/CC-MAIN-20231129204528-20231129234528-00361.warc.gz
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https://www.coursehero.com/tutors-problems/Math/482263-A-women-has-9-close-friends-a-In-how-many-ways-can-she-invite-f/
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math
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(a) In how many ways can she invite five of these to dinner? Explain/Show work.
(b) Repeat part (a) with the added stipulation that two of her friends do not like each other so that if she invites one of them she cannot invite the other. Explain/Show work.
(c) Repeat part (a), assuming that the nine friends consist of five single people and 2 married couples. If she invites a husband or wife she must invite the spouse. Explain/Show work.
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http://www.mathisfunforum.com/post.php?tid=19214&qid=262787
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math
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Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °
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Let O be the point at the centre of the circle.
They have one common angle and one right angle so the other angle must be the same.
In ΔABC, the perpendicular from vertices A and B on their opposite sides meet (when produced) the circumcircle of the triangle at points D and E respectively.
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s3://commoncrawl/crawl-data/CC-MAIN-2014-10/segments/1394021587780/warc/CC-MAIN-20140305121307-00005-ip-10-183-142-35.ec2.internal.warc.gz
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CC-MAIN-2014-10
| 523 | 8 |
https://www.brainkart.com/article/Solved-Problems--Ideal-And-Real-Gases,-Thermodynamic-Relations_5475/
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math
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1. A mixture of ideal gases consists of 7kg ofand 2kg ofat a pressure of 4bar and a temperature of 27°C. Determine:
i. Mole fraction of each constituent,
ii. Equivalent molecular weight of the mixture,
iii. Equivalent gas constant of the mixture,
iv. The partial pressure and partial volumes,
v. The volume and density of the mixture
= 2kg p = 4bar T = 27°C
i. Mole fraction of ,
Describe Joule Kelvin effect with the help of T-p diagram
The Joule Kelvin effect or Joule Thomson effect is an efficient way of cooling gases. In this, a gas is made to undergo a continuous throttling process. A constant pressure is maintained at one side of a porous plug and a constant lower pressure at the other side. The apparatus is thermally insulated so that the heat loss can be measured.
Joule –Thomson co –efficient is defined as the change in temperature with change in pressure, keeping the enthalpy remains constant. It is denoted by,
It is defined as the fluid expansion through a minute orifice or slightly opened valve. During this process, pressure and velocity are reduced. But there is no heat transfer and no work done by the system. In this process enthalpy remains constant.
Joule Thomson Experiment:
The figure shows the arrangement of porous plug experiment. In this experiment, a stream of gas at a pressure and temperature is allowed to flow continuously through a porous pig. The gas comes out from the other side of the porous pig at a pressure and temperature .
The whole apparatus is insulated. Therefore no heat transfer takes place. Q = 0.
The system does not exchange work with the surroundings.
So, W=0 from steady flow energy equation we know that
Since there is no considerable change velocity, and , Q=0,W=0, are applied in steady flow energy equation. Therefore, It indicates that the enthalpy is constant for throttling process.
It is assumed that a series of experiments performed on a real gas keeping the initial pressure p1 and temperature T1 constant with various down steam pressures ( p2,p3..... ). It is found that the down steam temperature also changes. The results from these experiments can be plotted as enthalpy curve on T-p plane.
The slope of a constant enthalpy is known as Joule Thomson Coefficient. It is denoted by µ.
For real gas, µ may be either positive or negative depending upon the thermodynamic state of the gas.
2. A mixture of 2kg oxygen and 2kg Argon is in an insulated piston cylinder arrangement at 100kPa, 300K. The piston now compresses the mixture to half its initial volume. Molecular weight of oxygen is 40. Ratio of specific heats for oxygen is 1.39 and for argon is 1.667.
= 32 = 40
= =1.39 =1.667
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CC-MAIN-2023-50
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https://macaulay2.com/doc/Macaulay2/share/doc/Macaulay2/OldToricVectorBundles/html/index.html
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math
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Using the descriptions of Kaneyama and Klyachko this package implements the construction of equivariant vector bundles on toric varieties.
Note that this package implements vector bundles in Kaneyama's description only over pure and full dimensional fans.
uses the OldPolyhedra
package by René Birkner
. At least version 1.1 of OldPolyhedra
must be installed via installPackage
to use OldToricVectorBundles
Each vector bundle is saved either in the description of Kaneyama or the one of Klyachko. The first description gives the multidegrees (in the dual lattice of the fan) of the generators of the bundle over each full dimensional cone, and for each codimension-one cone a transition matrix (See ToricVectorBundleKaneyama
). The description of an equivariant vector bundle given by Klyachko consists of filtrations of a fixed vector space for each ray in the fan of the base variety. Furthermore, these filtrations have to satisfy a certain compatibility condition (See ToricVectorBundleKlyachko
For the mathematical background see
Tamafumi Kaneyama,On equivariant vector bundles on an almost homogeneous variety, Nagoya Math. J. 57, 1975.
Alexander A. Klyachko,Equivariant bundles over toral varieties, Izv. Akad. Nauk SSSR Ser. Mat., 53, 1989.
Markus Perling,Resolution and moduli for equivariant sheaves over toric varieties, PhD Thesis, 2003.
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CC-MAIN-2024-18
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https://www.teacherspayteachers.com/Product/6NS1-6MP1-Identifying-Division-Stories-Strategy-Activity-3747230
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math
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This is a strategic problem solving activity that can be conducted in groups. I performed this as a carousel activity and visited the groups to join in on their fruitful discussion.
When we are working on problem solving, it's important for students to correctly identify the operation and apply it correctly to the given figures. When fractions come into the picture, this can become confusing. One strategy students found helpful was to substitute the story with whole numbers to develop the number sentence. Also, it's useful for students to write out the number sentence for each option first and then look at the problem.
It's easier to share an example, I've included how I would discuss with students below:
Problem: Which of the following problems can be solved using the expressions 3 divided by 1/3 ?
Option 1: 3 people equally share 1/3 of pizza. How much pizza does each person eat?
Discussion Point – Each is sharing part of the pizza, one-third of the whole pie that is. So the division story would be one-third divided by one-third. Each person would have one-ninth of the whole pie.
Option 2: A family ate 1/3 of a 3 foot sandwich. How much did they eat?
Discussion Point – The family collectively ate one-third of this three-foot sandwich. The operative word here being “of”. This means the family ate one-third times three meaning they ate one foot of the sandwich.
Option 3: A pie requires 1/3 pounds of apples. How many apples are needed for the 3 pies?
Discussion Point – Each pie needs one-third of a pound of apples. Meaning one-third for Pie One, one-third for Pie Two and one-third for Pie Three. This is repeated addition (or multiplication), one-third plus one-third plus one-third, or one-third times three. Either way, you’ll need one pound of apples to make the three pies.
Option 4: How many 1/3 lbs of bacon can you get from a 3 lb package?
Discussion Point – If we are splitting up a three pound packet of bacon, we want to know how many groups of one-third pounds there are in the three pounds. Therefore, it would be three divided by one-third, and there would be nine segments of the whole package that are one-third pounds each.
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