convolution_extra - ECE 5650/4650 Modern DSP Notes 1...

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Unformatted text preview: ECE 5650/4650 Modern DSP Notes 1 Example: Consider two finite length sequences x(n) h(n) Nx = n2 - n1 + 1 Nh = n4 - n3 + 1 n n1 n n3 n2 n4 W Ç x ( k ) h( n  k )  W y ( n) k – The sum must be evaluated for each value of n x(k) k n2 n1 h(n-k) k n-n4 n-n3 h(n-k) k n-n4 n-n3 h(n-k) k n-n4 n-n3 – The first nonzero value occurs when n - n3 = n1 or n = n1 + n 3 – The last nonzero value occurs when n - n4 = n2 or n = n2 + n4 Additional Notes on Convolution Sum Computation ECE 5650/4650 Modern DSP Notes – • 2 Summary: y(n) is nonzero for at most n1 + n3 ≤ n ≤ n2 + n4 ⇒ N y = N x + N h − 1 Example: Two finite duration sequences given by x(n) = h(n) = u (n) − u (n − N ), N > 0 x(n) = 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 −1 n + 1, y ( n ) = 2 N − 1 − n, N − 1 ≤ n ≤ 2 N − 2 0, otherwise N y(n) ... ... ... ... -1 0 1 N-1 Additional Notes on Convolution Sum Computation 2N-1 n ECE 5650/4650 Modern DSP Notes 4 Example: Two finite duration sequences in sequence explicit representation: h(n) = {1, 2, 1, − 1}, ↑ x(n) = {1, 2, 3,1} ↑ – In the above notation the arrows indicate where n = 0 We need to evaluate the convolution sum for −1 ≤ n ≤ 5 – To evaluate construct the following table: – x(k) vs k h(n-k) 0 0 1 2 3 1 0 0 n = -1 -1 1 2 1 0 0 0 0 0 0 1 n=0 0 -1 1 2 1 0 0 0 0 0 4 n=1 0 0 -1 1 2 1 0 0 0 0 8 n=2 0 0 0 -1 1 2 1 0 0 0 8 n=2 0 0 0 0 -1 1 2 1 0 0 3 n=3 0 0 0 0 0 -1 1 2 1 0 -2 n=4 0 0 0 0 0 0 -1 1 2 1 -1 – y(n) The final output is thus y (n) = { 0,1, 4, 8, 8, 3, − 2, − 1, 0, 0 } ↑ – Is this reasonable? The output should start at (-1 + 0) = -1 and should stop at (3 + 2) = 5, which is indeed the case Additional Notes on Convolution Sum Computation ECE 5650/4650 Modern DSP Notes 5 Example: Two infinite duration sequences, x(n) = u (n), h(n) = a nu (n) – By direct substitution into the convolution sum formula we have y ( n) = ∞ ∑ a u ( k )u ( n − k ) k k =−∞ – The term u(k) sets the lower sum limit to zero while the term u(n - k) sets the upper sum limit to n, hence n < 0 1 − a n+1 0, y ( n) = n k u ( n) = ∑ k = 0 a , n ≥ 0 1 − a 1 1-a y(n) ... ... 1 -1 0 1 2 Additional Notes on Convolution Sum Computation ... n ...
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This document was uploaded on 02/25/2012.

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