pcshw6_soln - THE STATE UNIVERSITY OF NEW JERSEY RUTGERS...

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Unformatted text preview: THE STATE UNIVERSITY OF NEW JERSEY RUTGERS College of Engineering Department of Electrical and Computer Engineering 332:322 Principles of Communications Systems Problem Set 6 Spring 2004 Reading: Review Probability (appendix and 322:321 notes/text) Web notes on convexity 1. Counting and Sample Space: Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01010. In each code word, a bit is zero with probability 0.8, independent of any other bit. (a) What is the probability of the code word 00111? SOLUTION: Since the probability of a zero is 0 8, we can express the probability of the code word 00111 as two occurences of a 0 and three occurences of a 1. Therefore (b) What is the probability that a code word contains exactly three ones? SOLUTION: The probability that a code word has excatly three 1's is 2. Simple Probability, Simple Application: A source wishes to transmit radio packets to a receiver over a radio link. The receiver uses error detection to identify packets that have been corrupted by radio noise. When a packet is sent error free the receiver sends an acknowledgement(ACK) back to the source. When the receiver gets a packet with errors, it sends back a negative-acknowledgement(NACK). Each time the source receives a NACK, the packet is re-transmitted. We assume that each packet transmission is independently corrupted by errors with probability q. (a) Find the PMF of X , the number of times that a packet is transmitted by the source. SOLUTION: The source continues to transmit packets until one is received correctly. Hence the total number of times a packet is transmitted is X x, if the first x 1 transmissions were in error. Therefore the PMF of X is (b) Suppose each packet takes 1 millisecond to transmit and that the source waits an additional millisecond to receive the acknowledgement message (ACK or NACK) before 1 PX x qx 0 1 1 q x 12 otherwise P three 1 s 5 3 08 2 02 3 P 00111 08 2 02 3 0 00512 0 0512 transmitting. Let T equal the time required until the packet is successfully received. What is the relationship between T and X ? What is the PMF of T ? SOLUTION: The time required to send a packet is 1 millisecond and the time required to send an acknowledgement back takes another millisecond. Thus if X transmissions of a packet are required to send a packet correctly, then the packet is correctly received after T 2X 1 milliseconds. Therefore, for an odd integer t 0, T t iff X t 2 1 . Thus, t 1 t 1 q 2 1 q t 135 PX PT t 2 0 otherwise 3. Joint PMFs: Calls arriving ar a telephone switch are either voice calls (v) or data calls (d). Each call is a voice call with probability p, independent of any other call. Observe calls at a telephone switch until you observe two voice calls. Let M equal the number of calls up to and including the first voice call. Let N equal the number of calls observed up to and including the second voice call. Find the conditional PMF's PM N m n and PN M n m . Interpret your results. SOLUTION: The key to solving this problem is to find the joint PMF of M and N. Note that N M. For n m, the joint event M m N n has probability, A complete expression for the joint PMF of M and N is m 1 Not surprisingly, if we view each voice call as a successful Bernoulli trial, M has a geometric PMF since it is the number of trials up to and including the first success. Also N has a Pascal PMF since it is the number of trials required to see 2 successes. The conditional PMF's are now easy to find 2 PM N m n n 0 1 otherwise PM N m n PN n 1 m 12 n 1 The interpretation of the conditional PMF of N given M is that given M m,N m where N has geometric PMF with mean 1 p. The conditional PMF of M given N is, PN M n m 1 0 p PM N m n PM m n m 1 p n m 1m otherwise PM m Similarly, for m 1 2 , the marginal PMF of M satisfies, m 1 1 p n p2 1 p m 1 1 p m 1 p n 2 p2 1 p PN n 1 p n 1 1 p m For n 2 3 , the marginal PMF of N satisfies, n 1 n 2 PM N m n p2 n 2 p2 2 1 0 p n 2 p2 m 1 2 n otherwise PM mN n 1 1 p p n 2 p1 p2 p 1; n m m 1 n m 1 p 1m 2 N, Given that call N n was the second voice call, the first voice call is equally likely to occur in any of the previous n 1 calls. otherwise 0 otherwise (b) What is the conditional expected value E[W C]? SOLUTION: The conditional expected value of W given C is 0 0 (c) Find the conditional variance, Var[W C]. SOLUTION: 0 We observe that w2 fW w is an even function. Hence Lastly conditional variance of W given C is 5. Functions of random variables: (a) Find the PDF fY y for the random variable Y , where Y X 3 and X is a uniform random variable in the range (0,1). SOLUTION: This transformation is a one-to-one mapping where if Y y then corresponding X y1 3 . CDF of random variable Y can be written as, x 0 3 fY y d FY y dy 1 y 3 2 3 FY y Prob Y y Prob X y 1 3 FX y 1 3 y1 3 1 dx Var W C E W2 C EWC 0 2 16 32 5 81 E W2 C 2 w2 fW w dw w2 fW w dw E W2 E W2 C w2 fW C w dw 2 w2 fW w dw EWC Making substitution v w2 32, we obtain 32 32 exp v dv 32 32 EWC w fW C w dw 2 4 2 wexp w2 32 dw fW C w w C fW w PC 0 2 exp 32 w2 32 w 0 16 y1 3 0. Hence P C (a) What is the conditional PDF fW C w ? SOLUTION: 1 fW w exp w2 32 32 Since W has expected value 0, fW w is symmetric about w P W 0 0 5. 4. Gaussians: W is a Gaussian random variable with expected value 16. Given the event C W 0 , 0 and variance 2 (b) Let X and Y be independent random variables with f X x exp x u x , and fY y 1 u y 1 u y 1 and let Z X Y , where u denotes unit step function. What is 2 the PDF of random variable Z? HINT:Use convolution! SOLUTION: Since Z is the sum of 2 independent random variables, the PDF of Z can be obtained by convolving PDF's of X and Y . 1 1 4 fZ z exp z y dy z 1 exp z 1 1 2 1 1 exp 2 fZ z exp z y dy exp 1 z 1 1 2 z exp fX z y grals as follows z 1, z yuz y For different regions of z values we get different inte- fZ z 0 1 1 2 fZ z fX z y fY y dy z 1 z 1 ...
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