Revision as of 18:49, 29 January 2011

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} \ $

Share your answers below

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

Answer 1

$ 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)

Answer 2

Write it here.

Answer 3

Write it here.

Back to ECE301 Spring 2011 Prof. Boutin