hw6_soln - ECE 562 Fall 2011 November17, 2011 SOLUTIONS TO...

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Unformatted text preview: ECE 562 Fall 2011 November17, 2011 SOLUTIONS TO HOMEWORK ASSIGNMENT 6 1. MF Maximizes SNR. We showed in the class that the SNR for the k-th symbol (ignoring the interference from other symbols) when a linear equalizer c k is used is given by: SNR k = E s | c † k h k | 2 N k c k k 2 Using the Cauchy-Shwarz inequality, show that SNR k is maximized by the matched filter equalizer c k, MF = h k . Ans: By the Cauchy-Shwarz inequality SNR k = E s | c † k h k | 2 N k c k k 2 ≤ E s N k c k k 2 k h k k 2 k c k k 2 = E s N k h k k 2 This maximum value is obtained when c k is a scaled version of h k . In particular, we can choose c k = h k . 2. MSE of linear equalizers. The MSE for the k-th symbol when a linear equalizer c k is used was defined in class to be: MSE k = E h | c † k Z- s m k | 2 i Using the fact that Z = h k s m k + X j 6 = k h j s m j + W show that MSE k = E s | c † k h k- 1 | 2 + E s X j 6 = k | c † k h j | 2 + N k c k k 2 Ans: Expanding the definition of MSE gives MSE k = E s m k- c † k Z 2 = E s m k- N X j =1 c † k h j s m j- c † k w 2 = E s m k c † k h k- 1- X j 6 = k c † k h j s m j- c † k w 2 = E " c † k h k- 1 2 | s m k | 2 + X j 6 = k c † k h j 2 | s m j | 2 + | c † k w | 2 # = E s c † k h k- 1 2 + E s X j 6 = k c † k h j 2 + N k c k k 2 where the second to last line follows from the fact that the symbols are independent and zero mean, and independent from the noise....
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hw6_soln - ECE 562 Fall 2011 November17, 2011 SOLUTIONS TO...

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