Revision as of 15:28, 28 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} \ $

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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)

Answer 2

Write it here.

Answer 3

Write it here.

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