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===Answer 1=== | ===Answer 1=== | ||
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| + | <math>y[n]=h[n]*x[n]=\sum_{k=-\infty}^\infty 5^{k}u[-k]u[n-k] = \sum_{k=-\infty}^0 5^{k}u[n-k] = \Bigg(\sum_{k=-\infty}^n 5^{k}\Bigg)u[-n]</math> | ||
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| + | I'm not sure where to go with this sum. I tried convolving in the other order, but the result wasn't any more helpful (as far as I can tell). | ||
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| + | <math>y[n]=x[n]*h[n]=\Bigg(\sum_{k=n}^\infty 5^{n-k}\Bigg)u[n]</math> | ||
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| + | Am I making some kind of mistake? How do I solve this sum? | ||
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| + | --[[User:Cmcmican|Cmcmican]] 21:17, 31 January 2011 (UTC) | ||
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===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 17:17, 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]= 5^n u[-n]. \ $
Use convolution to compute the system's response to the input
$ x[n]= u[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]=h[n]*x[n]=\sum_{k=-\infty}^\infty 5^{k}u[-k]u[n-k] = \sum_{k=-\infty}^0 5^{k}u[n-k] = \Bigg(\sum_{k=-\infty}^n 5^{k}\Bigg)u[-n] $
I'm not sure where to go with this sum. I tried convolving in the other order, but the result wasn't any more helpful (as far as I can tell).
$ y[n]=x[n]*h[n]=\Bigg(\sum_{k=n}^\infty 5^{n-k}\Bigg)u[n] $
Am I making some kind of mistake? How do I solve this sum?
--Cmcmican 21:17, 31 January 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
