recitation9_sol.pdf

# recitation9_sol.pdf - UNIVERSITY OF ILLINOIS AT...

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UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Department of Electrical and Computer Engineering ECE 310 Digital Signal Processing Recitation 8 Profs. Do & Liang March 26-30, 2018 Circular/Cyclic Convolution ( f g )[ n ] = N - 1 X m =0 f [ m ] g [ h n - m i N ] Where f [ n ] and g [ n ] are non-zero only on [0 , N - 1]. Linear convolution is different from circular convolution ; however if we zero-pad f and g to at least length N + N - 1, we can use circular convolution to calculate linear convolution. Example { x n } 3 n =0 = { 1 , 2 , 3 } and { y n } 3 n =0 = { 1 , 1 , 0 } . Solution To compute the cyclic convolution, we flip one signal and shift it to the right, but we’re now assuming that the flipped signal is periodic . So we can use a modified version of the table method, shifting the signal over and determining where the terms overlap: | 1 2 3 | No shift: 0 1 | 1 0 1 | 1 z [0] = 4 Shift by 1: 1 0 | 1 1 0 | 1 z [1] = 3 Shift by 2: 1 1 | 0 1 1 | 0 z [2] = 5 (The bold part is the flipped y n ). Because we’ve reached the length of the original signal, we can stop.

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