sol4 - ECE 5620 Spring 2011 Homework #4: Solutions 1. 1 2...

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ECE 5620 Spring 2011 Homework #4: Solutions 1. 1
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4. Since the continuous-time noise is Gaussian and the filter is linear, the discrete-time noise must also be Gaussian. Here is a time-domain approach for showing that the discrete-time nosie is white. One can also work in the frequency domain. Let h ( t ) denote the impulse response of the filter. Then we have ˜ N ( t ) = Z h ( t - τ ) N ( τ ) dτ. So for any n and m , E h ˜ N ± n 2 W ² ˜ N ± m 2 W ²i = E ³Z h ± n 2 W - τ ² N ( τ ) Z h ± m 2 W - s ² N ( s ) ds ´ = Z Z h ± n 2 W - τ ² h ± m 2 W - s ² N 0 2 δ ( τ - s ) dτds = N 0 2 Z h ± τ - n 2 W ² h ± τ - m 2 W ² dτ. Using Parseval’s relation Z h 1 ( t ) h 2 ( t ) dt = Z H 1 ( f ) H * 2 ( f ) df, we obtain E h ˜ N ± n 2 W ² ˜ N ± m 2 W ²i = N 0 2 Z H ( f )exp ± - n W f ² H * ( f )exp ± m W f ² df = N 0 2 Z W - W exp ± - n W f ² exp ± m W f ² df = ( N 0 W if n = m 0 otherwise . Since the samples are uncorrelated, the noise is white. The average power is given by N 0 W . 5. (a) The capacity is given by C a = 1 2 log(1 + S ) . This is shown in the figure. (b) Note that we can no longer use a Gaussian input. We will show that a uniform distribution over {- P, P } is optimal. Suppose X is not uniform, and Pr( X = - P ) = p Pr( X = + P
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sol4 - ECE 5620 Spring 2011 Homework #4: Solutions 1. 1 2...

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