| Line 15: | Line 15: | ||
<math>=\sum_{k=-\infty}^{0}2^{k}u[-u+k]</math> | <math>=\sum_{k=-\infty}^{0}2^{k}u[-u+k]</math> | ||
| + | |||
| + | <math> r = -k \!</math> | ||
| + | |||
| + | <math>=\sum_{r=0}^{\infty}(\frac{1}{2})^{r}u[-n-r]</math> | ||
| + | |||
| + | <math>-n-r \geqq 0</math> | ||
| + | |||
| + | <math>-n\geqq r</math> | ||
| + | |||
| + | <math>=(\sum_{r=0}^{-n}(\frac{1}{2})^{n})</math> | ||
Revision as of 09:52, 14 October 2008
Exam1 question 3.
An LTI system has unit impulse response $ h[n] = u[-n] $. compute the system's response to the input $ x[n]=2^nu[-n] $. (Simplify your answer until all $ \sum $ signs disappear.)
Solution
$ y[n] = x[n] * h[n]\! $
$ =\sum_{k=-\infty}^{\infty} x[k]h[n-k] $
$ =\sum_{k=-\infty}^{\infty} 2^{k}u[-k]u[-(n-k)] $
$ =\sum_{k=-\infty}^{0}2^{k}u[-u+k] $
$ r = -k \! $
$ =\sum_{r=0}^{\infty}(\frac{1}{2})^{r}u[-n-r] $
$ -n-r \geqq 0 $
$ -n\geqq r $
$ =(\sum_{r=0}^{-n}(\frac{1}{2})^{n}) $
