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HW13s[1]

# HW13s[1] - ECE 534 Elements of Information Theory Fall 2010...

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ECE 534: Elements of Information Theory, Fall 2010 Homework 13 Solutions (ALL DUE to Kenneth S. Palacio Baus) December 1, 2010 1. Problem 15.20. Multiple access (a) Find the capacity region for the multiple-access channel Y = X X 2 1 , where: X 1 ∈ { 2 , 4 } , X - 2 ∈ { 1 , 2 } Solution: The output of the channel is given in the following table: X 1 : X 2 1 2 2 2 4 4 4 16 Table 1: Y = X X 2 1 Here we can see that by fixing X 2 = 1 we can obtain a rate R 1 = 1 per transmission since we are able to decode two symbols at the output of the channel. Similarly, by setting X 1 = 2 it is possible to achieve a rate R 2 = 1. These rates can be obtained by the following expressions, considering that X 1 and X 2 are independent and that both have binary alphabets: R 1 I ( X 1 ; Y | X 2 ) (1) I ( X 1 ; X X 2 1 | X 2 ) (2) H ( X 1 | X 2 ) - H ( X 1 | X X 2 1 , X 2 ) (3) H ( X 1 ) (4) 1 (5) R 2 I ( X 2 ; Y | X 1 ) (6) I ( X 2 ; X X 2 1 | X 1 ) (7) H ( X 2 | X 1 ) - H ( X 2 | X X 2 1 , X 1 ) (8) H ( X 2 ) (9) 1 (10) The third boundary is obtained by: R 1 + R 3 I ( X 1 , X 2 ; Y ) (11) H ( Y ) - H ( Y | X 1 , X 2 ) (12) H ( Y ) (13) 1

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From table 1, we have Y ∈ { 2 , 4 , 16 } . In order to have H ( Y ) maximized we can set Y uniformly distributed as follows: p ( y ) = 2 with probability 1 - α 2 4 with probability α 16 with probability 1 - α 2 Then, we know that α = 1 2 to reach the maximum entropy so: H ( Y ) = H 1 4 , 1 2 , 1 4 = 1 . 5 So, R 1 + R 3 H ( Y ) (14) 1 . 5 (15) From the behavior of the channel, shown in table 1, we can observe that this channel behaves like the Binary erasure multiple-access channel . Then for a fixed rate, i.e. R 1 = 1 we have that the transmission of X 1 will look like noise for the transmission of X 2 . Then we have a binary erasure channel with capacity 0.5 bits for the transmission of X 2 . So, we obtain the capacity region shown in the following figure: (b) Suppose that the range of X 1 is { 1 , 2 } . Is the capacity region decreased? Why or why not? The following table shows the behavior of the channel under the new conditions: X 1 : X 2 1 2 1 1 1 2 2 4 Table 2: Y = X X 2 1 Here we notice that if we set X 1 = 1, then the output of the channel remains the same regardless the value of X 2 . So we won’t be able to recover X 2 by fixing X 1 . The rate R 2 will 2
depend on the value given for X 1 so, we must consider the probability of X 1 to be 1 or 2. Let X 1 and X 2 be Bernoulli random variables with the following distributions: p ( x 1 ) = 1 with probability 1 - r 2 with probability r p ( x 2 ) = 1 with probability 1 - s 2 with probability s Then, R 1 I ( X 1 ; Y | X 2 ) (16) I ( X 1 ; X X 2 1 | X 2 ) (17)

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