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 $
$ y[n] = \sum^{0}_{k = -\infty} 2^k u[-n+k] \! $
$ r = -k \! $
$ 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] = \sum^{-n}_{r = 0} (\frac{1}{2})^k \! $ when $ -n \geq 0 $
$ y[n] = 0 \! $, else
$ y[n] = \frac{1- (\frac{1}{2})^{-n+1}}{1 - \frac{1}{2}} \! $ when $ n \leq 0 $
$ y[n] = 0 \! $, else
