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The unit impulse response h[n] of a DT LTI system is | The unit impulse response h[n] of a DT LTI system is | ||
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===Answer 1=== | ===Answer 1=== | ||
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+ | <math>y[n]=h[n]*x[n]=\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n+2-k]+\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n+1-k]+\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n-k]+\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n-1-k]</math> | ||
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+ | <math>y[n]=\frac{1}{3^{n+2}}+\frac{1}{3^{n+1}}+\frac{1}{3^{n}}+\frac{1}{3^{n-1}}</math> | ||
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+ | --[[User:Cmcmican|Cmcmican]] 20:25, 31 January 2011 (UTC) | ||
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===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. |
Latest revision as of 10:21, 11 November 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]= \frac{1}{3^n} \ $
Use convolution to compute the system's response to the input
$ x[n]= \delta[n+2]+\delta[n+1]+\delta[n]+\delta[n-1]. \ $
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]=h[n]*x[n]=\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n+2-k]+\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n+1-k]+\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n-k]+\sum_{k=-\infty}^\infty \frac{1}{3^k}\delta[n-1-k] $
$ y[n]=\frac{1}{3^{n+2}}+\frac{1}{3^{n+1}}+\frac{1}{3^{n}}+\frac{1}{3^{n-1}} $
--Cmcmican 20:25, 31 January 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.