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#<math>x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k])</math>
 
#<math>x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k])</math>
 
#<math>k'=n-k</math>
 
#<math>k'=n-k</math>
#<math>x[n]*h[n]=\sum_{k=\infty}^{-\infty}(x[n-k']h[k'])</math> from 1 and 2
+
#<math>x[n]*h[n]=\sum_{k'=\infty}^{-\infty}(x[n-k']h[k'])</math> from 1 and 2
#<math>x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k])</math>
+
 
#<math>x[n]*h[n]=h[n]*x[n]</math>
 
#<math>x[n]*h[n]=h[n]*x[n]</math>

Latest revision as of 15:19, 24 June 2008

Given: $ y[n]=x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k]) $

  1. $ x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k]) $
  2. $ k'=n-k $
  3. $ x[n]*h[n]=\sum_{k'=\infty}^{-\infty}(x[n-k']h[k']) $ from 1 and 2
  4. $ x[n]*h[n]=h[n]*x[n] $

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