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Unformatted text preview: ECE 534: Elements of Information Theory, Fall 2010 Homework 12 Solutions (ALL DUE TO Kenneth S. Palacio Baus) December 1, 2010 1. Problem 15.7. Convexity of capacity region of broadcast channel . Let C R 2 be the ca pacity region of all achievable rate pairs R = ( R 1 ,R 2 ) for the broadcast channel. Show that C is a convex set by using a timesharing argument. Specifically, show that if R (1) and R (2) are achievable, R (1) + (1 ) R (2) is achievable for 0 1. Solution: We have two rate pairs that are achievable: R (1) = ( R (1) 1 ,R (1) 2 ) and R (2) = ( R (2) 1 , (2) 2 ) for which, we have two sequences of codes: ((2 nR (1) 1 , 2 nR (1) 2 ) ,n ) and ((2 nR (2) 1 , 2 nR (2) 2 ) ,n ). Like it is done in the proof of Theorem 15.3.2 for the case of the multipleaccess channel in the textbook, here we can apply a similar argument and construct a third codebook of length n at a rate R (1) +(1 ) R (2) using the first codebook for the first n symbols, and the second codebook for the last (1 ) n . We have that the number of X 1 codewords for the new code is given by: 2 nR (1) 1 2 n (1 ) R (2) 1 = 2 n ( R (1) 1 +(1 ) R (2) 1 ) , And the number of X 2 codewords for the new code is given by: 2 nR (1) 2 2 n (1 ) R (2) 2 = 2 n ( R (1) 2 +(1 ) R (2) 2 ) , So, we have obtained a rate: R (1) + (1 ) R (2) . Recalling that the overall probability of error is less that the sum of the probabilities of error for each of the segments: P ( n ) e P ( n ) e (1) + P ((1 ) n ) e (2) We see that the probability of error goes to 0 as n , hence the rate is achievable. 1 2. Problem 15.11. Converse for the degraded broadcast channel . The following chain of inequal ities proves the converse for the degraded discrete memoryless broadcast channel. Provide reasons for each of the labeled inequalities. Setup for converse for degraded broadcast channel capacity: ( W 1 ,W 2 ) indep X n ( W 1 ,W 2 ) Y n 1 Y n 2 Encoding: f n : 2 nR 1 2 nR 2 X n Decoding: g n : Y n 1 2 nR 1 ,h n : Y n 2 2 nR 2 . Let U i = ( W 2 ,Y i 1 . Then nR 2 Fano I ( W 2 ; Y n 2 ) (1) = ( a ) n X i =1 I ( W 2 ; Y 2 i  Y i 1 2 ) (2) = ( b ) X i ( H ( Y 2 i  Y i 1 2 ) H ( Y 2 i  W 2 ,Y i 1 2 )) (3) ( c ) X i ( H ( Y 2 i ) H ( Y 2 i  W 2 ,Y i 1 2 ,Y i 1 1 )) (4) = ( d ) X i ( H ( Y 2 i ) H ( Y 2 i  W 2 ,Y i 1 1 )) (5) = ( e ) X i I ( U i ; Y 2 i ) (6) Solution: Reasons for each of the labeled inequalities: (a) is given by the Chain rule....
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This note was uploaded on 01/19/2012 for the course ECE 534 taught by Professor Natashadevroye during the Fall '10 term at Ill. Chicago.
 Fall '10
 NatashaDevroye

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