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| − | + | = Practice Question on Computing the Output of an LTI system by Convolution = | |
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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 | The unit impulse response h[n] of a DT LTI system is | ||
| − | <math>h[n]= 5^n u[-n]. \ </math> | + | <math>h[n]= 5^n u[-n]. \ </math> |
| − | Use convolution to compute the system's response to the input | + | Use convolution to compute the system's response to the input |
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| + | <math>x[n]= u[n] \ </math> | ||
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| − | ==Share your answers below== | + | |
| − | 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! | + | == Share your answers below == |
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| + | 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! | ||
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---- | ---- | ||
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| − | + | === Answer 1 === | |
| − | + | <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> | |
| − | + | 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). | |
| − | + | <math>y[n]=x[n]*h[n]=\Bigg(\sum_{k=n}^\infty 5^{n-k}\Bigg)u[n]</math> | |
| − | --[[User:Cmcmican|Cmcmican]] 21:17, 31 January 2011 (UTC) | + | 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 === | ||
| + | Starting from <math>(\sum_{k=-\infty}^n 5^{k})u[-n]</math> | ||
| + | It can be observed that <math>(\sum_{k=-\infty}^n 5^{k}) = (5^-\infty - 5^{n+1})/(1-5) = 5^{n+1}/4 | ||
| + | Therefore, | ||
| + | (\sum_{k=-\infty}^n 5^{k}\Bigg)u[-n] = 5^{n+1}/4*u[-n] </math>([[User:Clarkjv|Clarkjv]] 23:35, 31 January 2011 (UTC)) | ||
| + | === Answer 3 === | ||
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| + | Write it here. | ||
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| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
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| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 19:35, 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
Starting from $ (\sum_{k=-\infty}^n 5^{k})u[-n] $ It can be observed that $ (\sum_{k=-\infty}^n 5^{k}) = (5^-\infty - 5^{n+1})/(1-5) = 5^{n+1}/4 Therefore, (\sum_{k=-\infty}^n 5^{k}\Bigg)u[-n] = 5^{n+1}/4*u[-n] $(Clarkjv 23:35, 31 January 2011 (UTC))
Answer 3
Write it here.
