# Practice Question on Computing the Output of an LTI system by Convolution

The unit impulse response h[n] of a DT LTI system is

$h[n]= u[n-1]-u[n-101]. \$

Use convolution to compute the system's response to the input

$x[n]= \frac{1}{2^n} \$

$y[n]=x[n]*h[n]=\sum_{k=- \infty}^\infty \frac{1}{2^k} (u[n-k-1]-u[n-k-101])=\sum_{k=- \infty}^\infty \frac{1}{2^k} u[n-k-1] - \sum_{k=- \infty}^\infty \frac{1}{2^k} u[n-k-101]$

but

$u[n-k-1]=\begin{cases} 1, & \mbox{if }[n-k-1] \ge 0 \\ 0, & \mbox{if }[n-k-1] < 0 \end{cases} = \begin{cases} 1, & \mbox{if }k \le n-1 \\ 0, & \mbox{if }k > n-1 \end{cases}$

and

$u[n-k-101]=\begin{cases} 1, & \mbox{if }[n-k-101] \ge 0 \\ 0, & \mbox{if }[n-k-101] < 0 \end{cases} = \begin{cases} 1, & \mbox{if }k \le n-101 \\ 0, & \mbox{if }k > n-101 \end{cases}$

so

$y[n] = \sum_{k=- \infty}^{n-1} \frac{1}{2^k} - \sum_{k=- \infty}^{n-101} \frac{1}{2^k}$

I'm pretty sure it's right up to here, and it doesn't need to be divided into two cases, because there is no condition for k<0. But I'm not sure how to solve this sum.

--Cmcmican 19:28, 28 January 2011 (UTC)

TA's comment: Very good. I suggest that you flip the input signal for this problem and not the impulse response. The convolution will then be much easier. In order to compute summations, you may want to use geometric series' formulas.

--Ahmadi 22:49, 29 January 2011 (UTC)

Note: $\sum_{k=m}^n ar^k=\frac{a(r^m-r^{n+1})}{1-r}.$ (The general formula, more valuable than the specific ones professors tend to give)

y[n] = h[n] * x[n] = x[n] * h[n]

Change the arguments & limits of the sum. Therefore, $\sum_{k=1}^{101} (1/2)^{n-k}=(1/2)^n * (1/2-(1/2)^{102})/(1/2)$

(Clarkjv 22:34, 31 January 2011 (UTC))