(No difference)
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Revision as of 09:29, 21 November 2008
Given:
$ y[n]=x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k]) $
- $ x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k]) $
- $ k'=n-k $
- $ x[n]*h[n]=\sum_{k'=\infty}^{-\infty}(x[n-k']h[k']) $ from 1 and 2
- $ x[n]*h[n]=h[n]*x[n] $
