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<math> y[n] = x[n] * h[n] , where * is convolution/,</math> | <math> y[n] = x[n] * h[n] , where * is convolution/,</math> | ||
| − | <math> \sum^{\infty}_{k=-\infty} 2^{k}u[-k]u[-n+k]</math> | + | <math> = \sum^{\infty}_{k=-\infty} 2^{k}u[-k]u[-n+k]</math> |
| − | <math> \sum^{0}_{k=-\infty} 2^{k}u[-n+k]</math> | + | <math> = \sum^{0}_{k=-\infty} 2^{k}u[-n+k]</math> |
| − | <math> \sum^{0}_{k=n} 2^{k}</math> | + | <math> = \sum^{0}_{k=n} 2^{k}</math> |
| − | <math> \sum^{-n}_{0} \frac{1}{2}^{k}, for n<0</math> | + | <math> = \sum^{-n}_{0} \frac{1}{2}^{k}, for n<0</math><br> |
| − | <math> 0 , for n>0</math> | + | <math> = 0 , for n>0</math> |
| − | <math> \ | + | <math> \frac{1-\frac{1}{2}^{-n+1}}{1-\frac{1}{2}}</math> |
Revision as of 13:27, 10 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.)
Answer
$ y[n] = x[n] * h[n] , where * is convolution/, $
$ = \sum^{\infty}_{k=-\infty} 2^{k}u[-k]u[-n+k] $
$ = \sum^{0}_{k=-\infty} 2^{k}u[-n+k] $
$ = \sum^{0}_{k=n} 2^{k} $
$ = \sum^{-n}_{0} \frac{1}{2}^{k}, for n<0 $
$ = 0 , for n>0 $
$ \frac{1-\frac{1}{2}^{-n+1}}{1-\frac{1}{2}} $
