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convolution_extra

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

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ECE 5650/4650 Modern DSP Notes 1 Additional Notes on Convolution Sum Computation Example : Consider two finite length se- quences x(n) n n 1 n 2 h(n) n n 3 n 4 ( ) ( ) ( ) k y n x k h n k ± Ç The sum must be evaluated for each value of n x(k) k n 1 n 2 h(n-k) n-n 3 n-n 4 k h(n-k) n-n 3 n-n 4 k h(n-k) n-n 3 n-n 4 k The first nonzero value occurs when n - n 3 = n 1 or n = n 1 + n 3 The last nonzero value occurs when n - n 4 = n 2 or n = n 2 + n 4 N h = n 4 - n 3 + 1 N x = n 2 - n 1 + 1

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ECE 5650/4650 Modern DSP Notes 2 Additional Notes on Convolution Sum Computation Summary : y ( n ) is nonzero for at most 1 3 2 4 1 y x h n n n n n N N N + + = + Example : Two finite duration sequences given by ( ) ( ) ( ) ( ), 0 x n h n u n u n N N = = > . . . . . . n x(n) = h(n) 0 1 2 N-1 . . . -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: . . . . . . k x(k) 0 1 2 N-1 . . . -1 . . . k h(n-k) n n-(N-1) . . . . . . Case 1 : n < 0 which implies no overlap, so y ( n ) = 0 Case 2 : 0 1 n N 0 ( ) (1)(1) 1 n k y n n = = = +
ECE 5650/4650 Modern DSP Notes 3

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

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