ch10 - Chapter 10 Information Theory and Coding 10.1...

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Chapter 10 Information Theory and Coding 10.1 Problem Solutions Problem 10.1 The information in the message is I ( x )= log 2 (0 . 8) = 0 . 3219 bits I ( x log e (0 . 8) = 0 . 2231 nats I ( x log 10 (0 . 8) = 0 . 0969 Hartleys Problem 10.2 (a) I ( x log 2 (52) = 5 . 7004 bits (b) I ( x log 2 [(52)(52)] = 11 . 4009 bits (c) I ( x log 2 [(52)(51)] = 11 . 3729 bits Problem 10.3 The entropy is I ( x 0 . 3log 2 (0 . 3) 0 . 25 log 2 (0 . 25) 0 . 25 log 2 (0 . 25) 0 . 1log 2 (0 . 1) 0 . 05 log 2 (0 . 05) 0 . 05 log 2 (0 . 05) = 2 . 1477 bits The maximum entropy is I ( x )=log 2 5=2 . 3219 bits Problem 10.4 1

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2 CHAPTER 10. INFORMATION THEORY AND CODING 0.7 0.5 0.2 0.2 0.2 0.2 0.1 0.6 0.3 1 y 3 x 2 x 1 x 3 y 2 y Figure 10.1: I ( x )=2 . 471 bits Problem 10.5 (a) The channel diagram is shown is illustrated in Figure 10.1. (b) The output probabilities are p ( y 1 )= 1 3 (0 . 5+0 . 2+0 . 1) = 8 30 =0 . 2667 p ( y 2 1 3 (0 . 3+0 . 6+0 . 2) = 11 30 . 3667 p ( y 3) = 1 3 (0 . . . 7) = 11 30 . 3667 (c) Since [ P ( Y )] = [ P ( X )] [ P ( Y | X )] we can write [ P ( X )] = [ P ( Y )] [ P ( Y | X )] 1 which is [ P ( X )] = [0 . 333 0 . 333 0 . 333] 2 . 5333 1 . 1333 0 . 4000 0 . 8000 2 . 2000 0 . 4000 0 . 1333 0 . 4667 1 . 6000
10.1. PROBLEM SOLUTIONS 3 This gives [ P ( X )] = [0 . 5333 0 . 2000 02 . 667] (d) The joint probability matrix is [ P ( X ; Y )] = [0 . 333 0 . 333 0 . 333] 0 . 5333 0 0 00 . 2000 0 000 . 2667 0 . 50 . 30 . 2 0 . 20 . 60 . 2 0 . 10 . . 7 which gives [ P ( X ; Y )] = 0 . 2667 0 . 1600 0 . 1067 0 . 0400 0 . 1200 0 . 0400 0 . 0267 0 . 0533 0 . 1867 Note that the column sum gives the output possibilities [ P ( Y )] , and the row sum gives the input probabilities [ P ( X )] . Problem 10.6 For a noiseless channel, the transition probability matrix is a square matrix with 1s on the main diagonal and zeros elsewhere. The joint probability matrix is a square matrix with the input probabilities on the main diagonal and zeros elsewhere. In other words [ P ( X ; Y )] = p ( x 1 )0 ••• 0 0 p ( x 2 ) 0 . . . . . . . . . . . . p ( x n ) Problem 10.7 This problem may be solved by raising the channel matrix A = 0 . 999 0 . 001 0 . 001 0 . 999 which corresponds to an error probability of 0 . 001 ,toincreas ingpowers n and seeing where the error probability reaches the critical value of 0 . 08 . Consider the MATLAB program a = [0.999 0.001; 0.001 0.999]; % channel matrix n = 1; % initial value a1 = a; % save a while a1(1,2) < 0.08 n=n+1; a1=a^n; end

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4 CHAPTER 10. INFORMATION THEORY AND CODING n-1 % display result Executing the program yields n 1=87 . Thus we compute (MATLAB code is given) & a^87 ans = 0.9201 0.0799 0.0799 0.9201 & a^88 ans = 0.9192 0.0808 0.0808 0.9192 Thus 87 cascaded channels meets the speci & cation for P E < 0 . 08 . However cascading 88 channels yields P E > 0 . 08 and the speci & cation is not satis & ed. (Note: This may appear to be an impractical result since such a large number of cascaded channels are speci & ed. However, the channel A may represent a lengthly cable with a large number of repeaters. There are a number of other practical examples.) Problem 10.8 The & rst step is to write H ( Y | X ) H ( Y ) .Th i sg iv e s H ( Y | X ) H ( Y )= X i X j p ( x i ,y j )log 2 p ( y j | x i ) X j p ( y j 2 p ( y j ) which is H ( Y | X ) H ( Y X i X j p ( x i j )[log 2 p ( y j | x i ) log 2 p ( y i )] or H ( Y | X ) H ( Y 1 ln 2 X i X j p ( x i j )ln p ( x i j ) p ( x i ) p ( x i ) or H ( Y | X ) H ( Y 1 ln 2 X i X j p ( x i j p ( x i )( y j ) p ( x i j ) Since ln x x 1 , the preceding expression can be written H ( Y | X ) H ( Y ) 1 ln 2 X i X j p ( x i j ) p ( x i ) p ( y j ) p ( x i j ) 1
10.1. PROBLEM SOLUTIONS 5 2 y 1 y 3 x 2 x 1 x Figure 10.2: or H ( Y | X )

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ch10 - Chapter 10 Information Theory and Coding 10.1...

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