hw3sol - EE 376B Information Theory Prof T Cover Handout#9...

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EE 376B Handout #9 Information Theory Tuesday, May 2, 2011 Prof. T. Cover Solutions to Homework Set #3 1. The cooperative capacity of a multiple access channel. p ( y | x 1 , x 2 ) a45 a45 a45 a8 a8 a8 a8 a8 a42 a72 a72 a72 a72 a72 a106 a45 X n 1 X n 2 ( W 1 , W 2 ) ( ˆ W 1 , ˆ W 2 ) Y n Figure 1: Multiple access channel with cooperating senders. (a) Suppose X 1 and X 2 have access to both indices W 1 ∈ { 1 , 2 nR 1 } , W 2 ∈ { 1 , 2 nR 2 } . Thus the codewords X n 1 ( W 1 , W 2 ) , X n 2 ( W 1 , W 2 ) depend on both indices. Find the capacity region. (b) Evaluate this region for the binary erasure multiple access channel Y = X 1 + X 2 , X i ∈ { 0 , 1 } . Compare to the non-cooperative region. Solution: The cooperative capacity of a multiple access channel (a) When both senders have access to the pair of messages to be transmitted, they can act in concert. The channel is then equivalent to a single user channel with the input X = ( X 1 , X 2 ) ∈ X 1 ×X 2 , and the message W = ( W 1 , W 2 ). The capacity of this single user channel is C = max p ( x ) I ( X ; Y ) = max p ( x 1 ,x 2 ) I ( X 1 , X 2 ; Y ). The two senders can send at any combination of rates with the total rate R 1 + R 2 C. 1
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(b) The capacity for the binary erasure multiple access channel was evaluated in class. When the two senders cooperate to send a common message, the capacity is C = max p ( x 1 ,x 2 ) I ( X 1 , X 2 ; Y ) = max H ( Y ) = log 3 , achieved by (for example) a uniform distribution on the pairs, (0,0), (0,1) and (1,1). The cooperative and non-cooperative regions are illustrated in Figure 2. a45 a54 a64 a64 a64 a64 a64 a64 a64 a64 1 2 1 2 R 1 R 2 0 C 2 = 1 C 1 = 1 C coop = log 3 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 Figure 2: Cooperative and non-cooperative capacity for a binary erasure multiple access channel. 2
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2. Square channel. What is the capacity of the following multiple access channel? X 1 {− 1 , 0 , 1 } X 2 {− 1 , 0 , 1 } Y = X 2 1 + X 2 2 (a) Find the capacity region. (b) Describe p * ( x 1 ) , p * ( x 2 ) achieving a point on the boundary of the capacity region. (c) What is the capacity if Y = X 1 X 2 ? Solution: Square channel. (a) First of all note that {− 1 , 1 } are degenerate from the point of view of the receiver Y . Define Z 1 = X 2 1 and Z 2 = X 2 2 then Y = Z 1 + Z 2 and Z i ∈ { 0 , 1 } . Therefore, we are back to the case of binary erasure multiple access channel described in Example 14.3.3 of the text. The capacity region is as given in Figure 14.13. (b) Note that a single distribution can achieve the whole region. (The union and convexification operations are not necessary for this specific channel.) Take p ( x 1 ) = ( α, 1 / 2 , 1 / 2 α ) and p ( x 2 ) = ( β, 1 / 2 , 1 / 2 β ) for any 0 α, β 1 / 2. Then, we have I ( X 1 ; Y | X 2 ) = H ( Y | X 2 ) = 1 I ( X 2 ; Y | X 1 ) = H ( Y | X 1 ) = 1 I ( X 1 , X 2 ; Y ) = H ( Y ) = 3 / 2 . (Note: Putting positive probability masses on both +1 and 1 does no worse than sticking to either +1 or 1.) (c) We can use the same argument as in Question 2. Since Y ∈ {− 1 , 0 , +1 } , R 1 + R 2 log(3) and ( R 1 = log(3) , R 2 = 0) is acheived by the distribution p * 1 = ( 1 3 , 1 3 , 1 3 ) and p * 2 that puts all its mass on { +1 } or {− 1 } . Therefore, the capacity region is the same as the binary multiplier channel with the bound 1 replaced log(3).
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