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[[Category: asan]] | [[Category: asan]] | ||
[[Category: Bonus]] | [[Category: Bonus]] | ||
| + | =Proof of the Commutativity property of LTI systems ([[ECE301]])= | ||
Given: <math>y[n]=x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k])</math> | Given: <math>y[n]=x[n]*h[n]=\sum_{k=-\infty}^{\infty}(x[k]h[n-k])</math> | ||
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#<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> | ||
| + | ---- | ||
| + | [[ECE_301_%28SanSummer2008%29|Back to ECE301 Summer 2008]] | ||
Latest revision as of 10:27, 30 January 2011
Proof of the Commutativity property of LTI systems (ECE301)
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] $
