hw5sol - EE 376B/Stat 376B Handout #15 Information Theory...

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Unformatted text preview: EE 376B/Stat 376B Handout #15 Information Theory Tuesday, May 16, 2006 Prof. T. Cover Solutions to Homework Set #5 1. Gaussian multiple access. A group of m users, each with power P , is using a Gaussian multiple access channel at capacity, so that m X i =1 R i = C mP N , (1) where C ( x ) = 1 2 log(1 + x ) and N is the receiver noise power. A new user of power P wishes to join in. (a) At what rate can he send without disturbing the other users? (b) What should his power P be so that the new users rate is equal to the combined communication rate C ( mP/N ) of all the other users? Solution: Gaussian multiple access. (a) If the new user is not to disturb other users, his message should be decodable at first. Therefore, R = C P mP + N . (b) We need C P mP + N = C mP N , or equivalently, P = mP ( mP + N ) N . 2. Capacity of multiple access channels. Find the capacity region for each of the following multiple access channels: (a) Additive modulo 2 multiple access access channel. X 1 { , 1 } ,X 2 { , 1 } ,Y = X 1 X 2 . (b) Multiplicative multiple access channel. X 1 {- 1 , 1 } ,X 2 {- 1 , 1 } ,Y = X 1 X 2 . 1 Solution: Capacity of multiple access channels. (a) Additive modulo 2 multiple access channel. Quite clearly we cannot send at a total rate of more than 1 bit, since H ( Y ) 1. We can achieve a rate of 1 bit from sender 1 by setting X 2 = 0, and similarly we can send 1 bit/transmission from sender 2. By simple time sharing we can achieve the entire capacity region which is shown in Figure 1....
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This note was uploaded on 06/10/2008 for the course ECE 376B taught by Professor Tomcover during the Spring '05 term at Stanford.

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hw5sol - EE 376B/Stat 376B Handout #15 Information Theory...

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