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===Answer 1=== | ===Answer 1=== | ||
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+ | <math>y[n]=x[n]*h[n]=\sum_{k=-\infty}^\infty \frac{1}{2^k}\delta[n-1-k]</math> | ||
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+ | <math>\delta[n-1-k] = \begin{cases} | ||
+ | 1, & \mbox{if }k = n-1 \\ | ||
+ | 0, & \mbox{if }k \ne n-1 | ||
+ | \end{cases}</math> | ||
+ | |||
+ | <math>y[n] = \frac{1}{2^{n-1}}</math> | ||
+ | --[[User:Cmcmican|Cmcmican]] 20:13, 31 January 2011 (UTC) | ||
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===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. |
Revision as of 16:13, 31 January 2011
Contents
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]= \delta[n-1]. \ $
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}\delta[n-1-k] $
$ \delta[n-1-k] = \begin{cases} 1, & \mbox{if }k = n-1 \\ 0, & \mbox{if }k \ne n-1 \end{cases} $
$ y[n] = \frac{1}{2^{n-1}} $ --Cmcmican 20:13, 31 January 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.