sol5 - Boston University Department of Electrical and...

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Boston University Department of Electrical and Computer Engineering ENG EC515 Digital Communication ( Fall 2009 ) Solution to Problem set 5 Posted: Mon 26 Oct Suggested reading: Class-notes, handouts, and Ch. 4 textbook Problem 5.1 (Irrelevant data, dimensionality reduction) One of three ( K = 3) messages is transmitted over a vector (real, zero-mean) AWGN channel with noise variance σ 2 . The messages need not be equally likely. Specifically, M ∈ M = { 1 , 2 , 3 } and M p M where p M need not be uniform. The channel codewords (or symbols) are given by C 3 = { x 1 , x 2 , x 3 } where x 1 = (0 . 5 , 1 , 10 , 2) T , x 2 = (0 . 5 , 2 , 10 , 3) T , and x 3 = (0 . 5 , 3 , 10 , 4) T . Notice that all three codewords agree in their first and third components. Let X = x M be the transmitted vector, Y = X + Z be the received vector, and Z be real, zero-mean, AWGN with variance σ 2 . (a) Show that the optimum receiver can ignore Y (1) and Y (3) which are the first and third com- ponents of Y . (b) Now suppose that the noise is not white. Specifically, let all 4 components of the noise be equal to Z where Z ∼ N (0 , σ 2 )( z ) is independent of the message M . Thus, Z (1) = Z (2) = Z (3) = Z (4) = Z . (i) [1pt] Can the optimum receiver ignore Y (1) and Y (3)? (ii) [1pt] Why? (iii) Design an optimum receiver for this situation; (iv) Compute the probability of error of the optimum receiver. Solution: (a) [2pts] b m MAP ( y ) = arg max m ∈M ± p Y | X ( y | x m ) p M ( m ) ² = arg max m ∈M { p Z ( y - x m ) p M ( m ) } = arg min m ∈M {- ln( p Z ( y - x m )) - ln( p M ( m )) } = arg min m ∈M K X i = 1 1 2 σ 2 | y ( i ) - x m ( i ) | 2 - ln( p M ( m )) = arg min m ∈M X i = 2 , 4 1 2 σ 2 | y ( i ) - x m ( i ) | 2 - ln( p M ( m )) + X i = 1 , 3 1 2 σ 2 | y ( i ) - x m ( i ) | 2 = arg min
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This note was uploaded on 11/09/2010 for the course ECE 515 taught by Professor Venkateshsaligrama during the Fall '09 term at BU.

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sol5 - Boston University Department of Electrical and...

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