sol7 - ECE 255A Fall 2011 Homework 7 Solutions P 1(a...

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ECE 255A Fall 2011 Homework 7 Solutions 1. (a) Capacity of Gaussian channel is 1 2 log(1 + P N ) and that of BSC is 1 - h ( p e ), where p e = R P 1 2 π e - z 2 / 2 dz . Using the above relations, we can compute the capacities for various values of P . P Gaussian BSC 0.25 0.161 0.108 1 0.50 0.369 8 1.585 0.9762 (b) i. P → ∞ : Capacity of Gaussian channel tends to , while the capacity of the BSC approaches 1, therefore the ratio tends to 0. ii. P 0: We are interested in the quantity, lim P 0 1 - h ( p e ) 0 . 5 log(1+ P ) , where p e = R P 1 2 π e - z 2 / 2 dz . Since both numerator and denomitor tends to 0 applying L’Hpital’s rule twice, we get lim P 0 1 - h ( p e ) 0 . 5 log(1 + P ) = lim P 0 2 ln 2 (log(1 - p e ) - log( p e )) dp e dP = lim P 0 2 ln 2 log(1 - p e ) - log( p e ) 2 π 2 P = lim P 0 8 dp e dP P 2 π = 2 π Solutions to Chapter 8 problem 8. Uniformly distributed noise We can expand the mutual information I ( X ; Y ) = h ( Y ) - h ( Y | X ) = h ( Y ) - h ( Z ) and h ( Z ) = log 2, since Z U ( - 1 , 1). The output Y is a sum a of a discrete and a continuous random variable, and if the probabilities of X are p - 2 ,p - 1 ,...,p 2 , then the output distribution of Y has a uniform distribution with weight p - 2 / 2 for - 3
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sol7 - ECE 255A Fall 2011 Homework 7 Solutions P 1(a...

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