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<math> y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k]</math> | <math> y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k]</math> | ||
| − | <math> y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k] | + | <math> y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k]</math> |
<math> n[-k] = 1 -k>=0, k<=0</math> | <math> n[-k] = 1 -k>=0, k<=0</math> | ||
| + | |||
| + | <math> y[n]=\sum^{0}_{k=-\infty}2^{k}*u[n-k]</math> | ||
Revision as of 09:47, 15 October 2008
question
3. An LTI system has unit impulse response $ h[n]=u[-n] $ Compute the system's response to the input $ x[n]=2^{n}u[-n]. $(Simplify your answer until all \sum signs disappear.)
solution
$ y[n]=x[n]*h[n] $
$ y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k] $
$ y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k] $
$ n[-k] = 1 -k>=0, k<=0 $
$ y[n]=\sum^{0}_{k=-\infty}2^{k}*u[n-k] $
