HW07s09_soln

# HW07s09_soln - ECE-2025 Homework#7 Solutions Spring-2009...

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Unformatted text preview: ECE-2025 Homework #7 Solutions Spring-2009 Problem 7.1: A linear, time-invariant discrete-time system is described by the following difference equation, y[n] = -x[n] + 2x[n-2] + 3x[n-4] - 4x[n-5] ; (a) The impulse response is: ] 5 [ 4 ] 4 [ 3 ] 2 [ 2 ] [ ] [ − − − + − + − = n n n n n h δ δ δ δ (b) The filter coefficients: , 4 , 3 , 2 , 1 5 4 2 = − = = = − = k b b b b b otherwise, for the causal FIR representation: ∑ = − ⋅ = M k k n x k h n y ] [ ] [ ] [ , M=5 ;. (c).Order of the filter: M=5 , and Length of the filter: L=M+1=6 ; Problem 7.2: 1. y[n] = nx[n-1] (Multiplier); (a) Impulse response: ] 1 [ ] [ − = n n n h δ (The output is 1 only when n=1 , and 0 otherwise); (b) If ] [ ] [ ] [ 2 2 1 1 n x n x n x α α + = then ] [ ] [ ] 1 [ ] 1 [ ] 1 [ ] [ 2 2 1 1 2 2 1 1 n y n y n nx n nx n nx n y α α α α + = − + − = − = so h[n] is linear; (c) If ] [ ] [ 1 n n x n x − = then ] [ ] 1 [ ] [ ] 1 [ ] [ 1 n n y n n x n n n y n n nx n y − ≠ − − + − = − − = so h[n] is not time-invariant; (d) The output depends only on past and current inputs, so h[n] is causal; 2. ] 2 [ ] [ | 1 | − = n x n y (Time Distortion); (a) Impulse response: ] 2 [ ] [ | 1 | = = − n n h δ (The output is always zero because the index is never 0) (b) If ] [ ] [ ] [ 2 2 1 1 n x...
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HW07s09_soln - ECE-2025 Homework#7 Solutions Spring-2009...

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