Problem 3 a heres the picture of fx x the quantization

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Unformatted text preview: the quantization levels, and the quantization regions. !a !!! aaaaa ! !s ! s s sa 1 2 ^ 2 x4 ^^ 1 x2 x1 x3 2 ^ 1 2 SQNR = E (XE (X () ))2 ] QX The signal power is E (X ) = 2 Z 1 1 x fX (x)dx = 2 2 1 = 2 2x 3 2 =3 3 x 4 4 2 Z2 0 1 x2 4 (2 x)dx 1 = 2 136 0 2 16 4 The quantization noise power is E (X Q(X )) ] = Z 2 1 =2 =1 2 1 =2 1 =2 1 =2 1 =9 Therefore, 1 (x Q(x))2 fX (x)dx 1 (2 x)dx + 2 Z 2 x 5 2 1 (2 x)dx 4 34 0 1 Z1 Z2 x2 2 x + 1 (2 x)dx + x2 130 x + 25 (2 x)dx 39 9 0 1 Z1 Z2 x3 + 8 x2 193 x + 2 dx + x3 + 16 x2 895 x + 50 dx 3 9 3 9 0 1 1 1 x4 + 8 x3 13 x2 + 2 x + 1 1 x4 + 16 x3 85 x2 + 50 x 2 4 9 18 902 4 9 18 91 13 5 36 + 2 36 x1 3 Z1 2 2=3 SQNR = 1=9 = 6 b) The optimal lossless coder to minimize the average bit rate of the coded stream, assuming that we are coding sample-by-sample, is a Hu man coder. To nd the code, we need to calculate the probabilities of each quantized value. 1 1 P (Q(Xn ) = x1) = P (Q(Xn) = x2) = 4 1 + 2 1 1 = 3 ^ ^ 48 1 1 1=1 P (Q(Xn ) = x3) = P (Q(Xn) = x4) = 2 ^ ^ 48 Here's a possible Hu man tree. 0 x1 : 3 ^8 10 x2 : 3 ^8 110 x3 : ^ 0 0 1 8 0 111 x4 : 1 ^8 1 1 4 5 8 1 1 The...
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This document was uploaded on 01/24/2014.

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