EE_740_HW4 - EE 740 Homework 4 29 October 07 Edmund G....

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EE 740 Homework 4 29 October 07 Edmund G. Zelnio Wright State University —————————————————————————————————————————————————— Problem 7.2, Cover and Thomas Additive noise channel. Find the channel capacity of the following discrete memoryless channel: Y = X + Z were Pr { Z =0 } =Pr { Z = a } =1 / 2. The alphabet for x is X = { 0 , 1 } . Assume that Z is independent of X .O b s e r v e that the channel capacity depends on the value of a . C =m a x p ( x ) I ( X ; Y )( 1 ) I ( X ; Y )= X x X X y Y p ( x, y ) · log p ( x, y ) p ( x ) p ( y ) (2) = H ( Y ) H ( Y/X H ( X ) H ( X/Y 3 ) If a is not equal to either 1 or 0, the channel capacity is one bit as computed in the spreadsheet in Figure 1. For a not equal 0 or 1 X p(x) Z p(z) Y p(y) p(x,y) / p(x) p(y) 0 0.5 0 0.5 0 0.25 2 0 2 0 1 0.5 a 0.5 1 0.25 0 2 0 2 a0 . 2 5 1+a 0.25 Array to Prevent Log of Zero p(x) p(x,y) 01a 1 + a 2121 0.5 0 0.25 0 0.25 0 1 2 1 2 0.5 1 0 0.25 0 0.25 0.25 0.25 0.25 0.25 p(y) p(x/y) 1010 I(X;Y) 1 0101 H(X) 1 H(X/Y) 0 Array to Prevent Log of Zero 1111 Figure 1: Spreadsheet for I ( X ; Y )i fa is not equal to 0 or 1
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If a is equal to 0, the channel capacity is one bit as computed in the spreadsheet in Figure 2. IF a =0, X p(x) Z p(z) Y p(y) p(x,y) / p(x)p(y) 00 . 50 1 . 5 2 0 1 0.5 1 0.5 0 2 p(x) p(x,y) 01 Array to Prevent Log of Zero 0 . 500 . 2 1 0.5 1 0 0.5 1 2 0.5 0.5 p(y) p(x/y) I(X;Y) 1 10 H(X) 1 H(X/Y) 0 Array to Prevent Log of Zero 11 Figure 2: Spreadsheet for I ( X ; Y )i fa is equal to 1 If a is equal to 1, the channel capacity is 1/2 bit as computed in the spreadsheet in Figure 3. IF a =1, X p(x) Z p(z) Y p(y) p(x,y) / p(x)p(y) 0 0.5 0 0.5 0 0.25 2 1 0 1 0.5 1 0.5 1 0.5 0 1 2 20 . 2 5 p(x) p(x,y) 012 Array to Prevent Log of Zero 0.5 0 0.25 0.25 0 2 1 1 0.5 1 0 0.25 0.25 1 1 2 0.25 0.5 0.25 p(y) p(x/y) I(X;Y) 0.5 1 0.5 0 H(X) 1 . 5 1 H(X/Y) 0.5 Array to Prevent Log of Zero . 5 1 . 5 1 Figure 3: Spreadsheet for I ( X ; Y is equal to 1
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—————————————————————————————————————————————————— Problem 7.4, Cover and Thomas Channel capacity. Consider the discrete memoryless channel Y = X + Z (mod 11), where Z = " 123 1 / 31 / / 3 # and X ∈{ 0 , 1 ,..., 10 } . Assume that Z is independent of X . a. Find the capacity. C =m a x p ( x ) I ( X ; Y )( 4 ) I ( X ; Y )= H ( Y ) H ( Y/X H ( X ) H ( X/Y 5 ) H ( X x ∈{ 0 , 1 ,..., 10 } p ( x )log H ( = x ) Since H(Y / X = x) has cardinality 3 independent of x =l o g H ( = x ) X x ∈{ 0 , 1 ,..., 10 } p ( x ) o g 3 , hence I ( X ; Y H ( Y ) H ( H ( Y ) log 3 log | X |− log 3 log 11 log 3 b. What is the maximizing p ( x )? C = log 11 log 3 when p ( x ± 1 11 , 1 11 1 11 ²
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—————————————————————————————————————————————————— Problem 7.8, Cover and Thomas Channel capacity. The Z=channel has binary input and output alphabets and transition probabilities p ( y/x )as shown in Figure 4. Find the capacity of the Z-channel and the maximizing input probability distribution.
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EE_740_HW4 - EE 740 Homework 4 29 October 07 Edmund G....

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