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# SPTPIntro To Robotics CPE 493A

WVU

GPA 3.73

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This 13 page Class Notes was uploaded by Freeda Predovic on Saturday September 12, 2015. The Class Notes belongs to CPE 493A at West Virginia University taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/202667/cpe-493a-west-virginia-university in Computer Engineering at West Virginia University.

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Date Created: 09/12/15

Probability Part 2 Random Variables in Wireless Networks Wireless Networking CPE 493g 1 Common Continuous Random Variables 11 Uniform One of the most basic continuous random variable is the uniform random variable A uniform random variable takes on values over the continuous range a b with equal probability The pdf is as follows equation 224 from textbook i for a lt z lt b 7 12711 7 7 fX 7 0 elsewhere 1 z 7 m H b 7 a lt b 7 a gt where b a 7 2 and the H function is de ned as 7 l for 7 S 1 Ha 7 0 elsewhere 1 12 Gaussian 121 The Central Limit Theorem Let X17X27 XN be a set of N iid continuous random variables eg uni N form Then the pdf of Y ZXn is fYy fX1y 1599 fXNy As N 7gt 007 Y will have a Gaussian distribution my 7 aj expfly22m Where m is the mean of Y and a2 is the variance of Y This is called the Cental Limit Theorem and justi es the use of Gaussian RVs for many en iueeiiu quot 39 eg error receiver noise interference etc A shorthand notation that is often used is X N 77 m 02 This tells us that X has a normal ie Gaussian distribution With mean m and variance 0 122 Properties of Gaussian RVs Gaussian RVs have the following properties 1 A Gaussian RV is completely described by its mean and variance 2 The sum of 2 Gaussian RVs is also Gaussian More speci cally if Y X1 X2 With X2 N nm1a and X2 N nm2a then Y N 77m1 m2a 0 9 If X and Y are Gaussian and uncorrelated ie EXY EXEY then X and Y are independent F The pdf of a Gaussian RV is symmetric about its mean ie PY S m PY gt m 12 123 Example Applying the properties of Gaussian RVs X1 and X2 are iid n23 What is the pdf of Y X1 X2 What is the median of Y lf EX1X2 4 then are X1 and X2 independent 124 Computing the GDP of Gaussian RVs Let X be Gaussian7 then its CDF is FXa fXzdz Where erfz 0 eimzdz is called the error function You can compute this in matlab by using the erf comman 125 Area Under the Tail of a Gaussian RV The complement of the CDF is the area under the tail 00 li FXa Where erfcz lierfz is called the complementary error function7 which can be computed in matlab using erfc The area under the tail can also be put in terms of the Qfunctz39on Qz mHmz 1 f 2 game In particular7 1 FX04 Q a am 126 Properties of the Qfunction The Qfunction has the following properties 1 Q0 121 2 0 3 QH 17 QM Let X N 77021702 then FXW Qltmgagt You can download a rnatlab script for computing the Qfunction from the course webpage7 or use the following Qfunction table 2 l Qz H 2 l Qz H 2 l Qz H 2 l Qz 010 050000 110 0115866 210 0102275 310 0100135 011 0146017 111 0113567 211 0101786 311 0100097 012 0142074 112 0111507 212 0101390 312 0100069 013 0138209 113 0109680 213 0101072 3 3 0100048 014 0134458 114 0108076 214 0100820 3 4 0100034 05 0130854 15 0106681 25 0100621 35 0100023 016 0127425 116 0105480 216 0100466 3 6 0100016 017 0124196 117 0104457 217 0100347 3 7 0100011 018 0121186 118 0103593 218 0100256 318 0100007 019 0118406 119 0102872 219 0100187 319 0100005 127 Examples Let X N 770 1 Compute the following probabilities 1 PX gt 2 2 PX g 2 3 PX gt 71 4 PX lt 71 Let X N 771 4 Compute the following 1 PX gt 2 2 PX g 2 3 PX gt 75 4 PX lt 75 128 Complex Gaussian Random Variables Let Z X jY where j xil is the imaginary number If X and Y are Gaussian7 then Z is complex Gaussian Unless otherwise specified7 the real and imaginary components of a complex Gaussian random variable are assumed to be uncorrelated7 and hence7 independent 13 Rayleigh Let Z be complex Gaussian7 ie Z X jY Let the real and imaginary parts of Z be independent7 have zero mean7 and equal variances 02 De ne the random variable R to be the magnitude of Z7 ie R 7 lZl xX2 Y2 Variable R is said to be Rayleigh and have the following pdf equation 321 in text MT gag um lt2 where is the unit step function The rst two moments of this random variable are ER 7a 3 ER2 202 4 Rayleigh variables model the received envelope of signals that are transmitted through a fading channel when there is no direct line of sight LOS between the transmitter and receiver which typically occurs when the transmitter and receiver are far apart eigi receiver is overthehorizoni Incidentally if the magnitude of Z is Rayleigh then the phase of Z will be uniformly distributed over the range 0 2w 14 Exponential Again let Z be complex Gaussian with real and imaginary components that are independent have zero mean and equal variances 02 lZA Let P lZl2 be the instantaneous power of Z Since Z is random so is P In particular P is exponentially distributed with pdf equation 226 in text fPI AE MWI 5 where A is the parameter of the random variable The mean and variance of P is given by m 6 7 a2 HMH Given the above description exponential random variables can be used to model the power of signals received through fading channels with no direct LOSi However exponential random variables also nd other uses For instance the time between phone calls by a given subscriber is often modelled as being exponential as is the time that each call lastsi The spatial distribution of cellular subscribers can be modelled by representing the xy separation of neighboring users exponential in each direction Even the lifetime of lightbulbs can be fairly accurately modelled as exponential random variables Exponential random variables are said to be memorylessi Loosely stated this means that the time from right now until your next call or the time it takes for your lightbulb to blow is independent from how long it has been since your last call or how long it has been since you changed that lightbulbi 15 Rician Let Z XjY be complex Gaussian with real and imaginary components that are independent and have the equal variances 02 Now however assume that while the imaginary component Y has a zero mean but the real component X has a positive mean mX Then R lZl is said to be Rician and has the following pdf equation 32 in text MT 10 um s where Io is the zeroorder modi ed Bessel function of the rst kind de ned as equation 328 in text 1 27f i mcostO 9 2W 0 e 101 Like the error and Q functions the modi ed Bessel function does not have a closed form solution but can be found through a tablelookup ln matlab it can be evaluated using the besseli function Rician random variables are used to model the received amplitude of signals that go through fading channels with a clear line of sight LOS which often occurs when the transmitter and receiver are close The stronger the LOS component the larger the value of On the other hand as B A 0 the random variable becomes more like a Rayleigh In fact when 6 0 the pdf is exactly the same as that of a Rayleigh random variable Rician random variables are sometimes speci ed in terms of the Rician K factor K 10 Usually the K factor is given in dB units ie K013 loglo K Rayleigh fading is merely Rician fading with K 0 in linear units 16 Nakagami m The Rician pdf is inconvenient because of the presence of the Bessel func tion The Nakagamim distribution closely approximates the Rician pdf and signal measurements have shown that it is an excellent model for received sig nal amplitude through a fading channel even better than Rician The pdf of a Nakagami m random variable is equation 329 in text 27271171 Mr W grew 32 um where is the Gamma function 9 ER2 is the average power second moment and m 2 12 is a constant called the fading gure or fading factor When m 1 this is a Rayleigh distribution For other Rician K factors the Rician random variable is approximated by Nakagami m distribution with fading factor equation 330 in text 1H1 m W 11 2 Common Discrete Random Variables 21 Bernoulli The most basic discrete random variable is the Bernouilli random variable Consider an unfair coin that lands on heads with probability p Assign a random variable X according to X i 1 2va The pdf of this random variable is fXI 1 WW 105ltI 1 Since X is discrete we can also specify it in terms of the pmf lip fork0 pxlkl p forkl 22 Geometric Consider the following random experiment An unfair coin lands on heads with probability p Toss the coin until you get heads Let X be the number of tosses until the rst heads The random variable X is said to be geometric It can be speci ed in terms of the following pmf equation 220 from text pxk plipk71 k 012 The mean and variance of this random variable are the error in the textbook has been corrected 0 em gwe Geometric random variables are useful when analyzing data transmission protocols as illustrated in the following examp e 221 Example Bluetooth uses an automatic repeat request ARQ protocol that works as fol lows Data packets are formed by combining up to 2712 data bits and some overhead information The data packet is transmitted and if received success fully the destination sends back an acknowledgement ACK to the sender If an ACK is received at the sender then it goes on to transmit the next packet but if an ACK is not received due to a transmission error in either the forward or return channel then the same packet is resent The roundtrip time for the data packet and ACK is 375 milliseconds assuming the maximum length packet Let p denote the probability that an ACK is received by the sender Then 4 l 7 p is the probability that an error occurs either on the forward path sender to destination or if the data transmission was successful that the ACK sent over the return channel was received in error First assume errorfree transmission 17 1 Determine the maximum data rate supported by Bluetooth What is the maximum roundtrip error probability 4 for which the average throughput is at least 500 kbps 23 Binomial Let X1X2 XN be a set of iid Bernoulli random variables then the pdf of N Y 2 X7 is 711 fYy fX1yfX2ymfXNy 469 106lty 7 1 469 p6y 1 469 106lty 71 ilt iv gtpkqn k5yi k k0 Where q 1 7 p and lt iv gt is called the binomial coef cient7 de ned by n 7 nl k 7 k n 7 k A fair coin is tossed 5 times Find the probability that 2 or fewer heads were tossed 231 Example 1 232 Example 2 Data is transmitted over a noisy Channel With bit error probability p 0001 Find the probability that there is more than one bit error in a byte of received data 24 Poisson Consider the following random experiment Assume that you have a timer that after being set expires after a random amount of time and this random amount of time is an exponential random variable with parameter A gt 0 The mean length of the timer is therefore lA units where a unit could be a second minute hour etc At time zero set the timer and let it run Once it expires reset it and let it run again Repeat this process until a speci ed amount of time has passed one unit Let X be the number of times the timer goes off during a time unit The discrete random variable X is Poisson and has pmf equation 219 from text k 7 mic A I k 012 12 Like the geometric random variable there is no upper bound on how high k can be Poisson random variables are useful for modelling network traf c If the time between a customer s cell phone calls is exponential then the number of calls heshe makes per day is Poisson 241 Example On average Mary makes two calls on her cell phone per day Her calling plan allows her to make up to 20 calls per week until she gets penalized with an overage charge What is the probability that on a given week she must pay an overage charge Assume that the time between her calls is exponential Errata and Clari cations Probability Part 2 Random Variables in Wireless Networks Wireless Networking CPE 493g 1 Poisson What is right 0 The notes including the example 0 The book 0 My hints except for the numerical answer 0 The of cial homework solutions attached Whats wrong 0 The numerical answer I emailed to you in Another Homework Hintlli Source of confusion 0 Let X be the time between events Assume that X is exponential If mX is the mean time between events then A is the number of events per unit time and mX UK 0 Let Y be the number of events per unit time If the time X between events is exponential then the number of events per unit time Y is Poissoni The mean value of Y is my A and the variance is a i 0 Note that the mean of X and the mean of Y are inversely proportional 2 Geometric Whatls right 0 The pmf PXlkl qu l k 012W where q l 7p Whats wrong 0 The mean and variance given in the book and notes The correct values Proof of m Ele kaqk l 160 10231641 1 160 10 1 7 a l P This implies that we need to correct example 221 le lH i6916 7p0 rD 6 l l H O

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