Problem 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] \! $
$ y[n] = \sum^{\infty}_{k = -\infty} x[k] h[n-k] \! $
$ y[n] = \sum^{\infty}_{k = -\infty} 2^ku[-k] u[-(n-k)] \! $
We know $ u[-k] = 1 \! $ when $ K \leq 0 $, else $ u[-k]=0 $, therefore
$ y[n] = \sum^{0}_{k = -\infty} 2^k u[-n+k] \! $
Let $ r = -k \! $, then
$ y[n] = \sum^{\infty}_{r = 0} (\frac{1}{2})^k u[-n-r] \! $
We know $ -n-r \geq 0 $ when $ -n \geq r $
$ y[n] = \left\{ \begin{array}{ll} \sum^{-n}_{r = 0} (\frac{1}{2})^k \! & \text{ when } -n \geq 0, \\ 0 & \text{ else. } \end{array}\right. $
So
$ y[n] = \left\{ \begin{array}{ll} \frac{1- (\frac{1}{2})^{-n+1}}{1 - \frac{1}{2}} \! & \text{ when } -n \geq 0, \\ 0 & \text{ else. } \end{array}\right. $
