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Unformatted text preview: ECE 255AN Fall 2011 Homework set 5  solutions Solutions to Chapter 7 problems 28. Choice of Channels (a) This is solved by using the very same trick that was used to solve problem 2.10. Consider the following communication scheme: X = ( X 1 Probability X 2 Probability (1 ) Let ( X ) = ( 1 X = X 1 2 X = X 2 Since the output alphabets Y 1 and Y 2 are disjoint, is a function of Y as well, i.e. X Y . I ( X ; Y, ) = I ( X ; ) + I ( X ; Y  ) = I ( X ; Y ) + I ( X ;  Y ) Since X Y , I ( X ;  Y ) = 0. Therefore, I ( X ; Y ) = I ( X ; ) + I ( X ; Y  ) = H ( ) H (  X ) + I ( X 1 ; Y 1 ) + (1 ) I ( X 2 ; Y 2 ) = H ( ) + I ( X 1 ; Y 1 ) + (1 ) I ( X 2 ; Y 2 ) Thus, it follows that C = sup { H ( ) + C 1 + (1 ) C 2 } . Maximizing over one gets the desired result. The maximum occurs for H ( )+ C 1 C 2 = 0, or = 2 C 1 / (2 C 1 + 2 C 2 )....
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This document was uploaded on 01/19/2012.
 Fall '09

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