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. $

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010