(New page: Given: <math>y[n]=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>x[n]*h[n]=\sum_{k=\infty}^-\...) |
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Given: | Given: | ||
| − | <math>y[n]=x[n]*h[n]=\sum_{k=-\infty}^\infty(x[k]h[n-k])</math> | + | <math>y[n]=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>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]=h[n]*x[n]</math> | #<math>x[n]*h[n]=h[n]*x[n]</math> | ||
Latest revision as of 14:19, 24 June 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] $
