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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} \ $
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)
Answer 2
Write it here.
Answer 3
Write it here.
