1 0 1 2 n n 1 from the previous example we

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Unformatted text preview: ) = h(n) ... ... ... -1 0 1 2 n N-1 – From the previous example we know that y(n) will be at most nonzero on the interval [0, 2(N - 1)] – There are 4 cases to consider: x(k) ... ... ... -1 0 1 2 N-1 k h(n-k) ... ... ... n-(N-1) – – n Case 1: n < 0 which implies no overlap, so y(n) = 0 Case 2: 0 ≤ n ≤ N − 1 n y (n) = ∑ (1)(1) = n + 1 k =0 Additional Notes on Convolution Sum Computation k ECE 5650/4650 Modern DSP Notes – 3 Case3: N − 1 ≤ n ≤ 2 N − 2 y ( n) = N −1 ∑ (1)(1) = ( N − 1) − (n − ( N − 1)) + 1 k = n − ( N −1) = 2N − 1 − n – Case 4: n > 2 N − 2 which implies no overlap, so y(n) = 0 – In summary: 0 ≤ n ≤ N...
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This document was uploaded on 02/25/2012.

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