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<math>y[n]=h[n]*x[n]=\sum_{k=-\infty}^\infty \frac{1}{5^k}u[k]u[n-k]=\sum_{k=0}^\infty \frac{1}{5^k}u[n-k]=\Bigg( \sum_{k=0}^n \frac{1}{5^k} \Bigg)u[n]</math> | <math>y[n]=h[n]*x[n]=\sum_{k=-\infty}^\infty \frac{1}{5^k}u[k]u[n-k]=\sum_{k=0}^\infty \frac{1}{5^k}u[n-k]=\Bigg( \sum_{k=0}^n \frac{1}{5^k} \Bigg)u[n]</math> | ||
| − | + | I'm not totally sure that this is the way to compute this sum... | |
| − | --[[User:Cmcmican|Cmcmican]] 20: | + | <math>y[n]=\Bigg(\frac{1-(1/5)^{n+1}}{1-(1/5)}\Bigg)u[n]</math> |
| + | |||
| + | --[[User:Cmcmican|Cmcmican]] 20:57, 31 January 2011 (UTC) | ||
===Answer 2=== | ===Answer 2=== | ||
Revision as of 16:57, 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]= \frac{1}{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 \frac{1}{5^k}u[k]u[n-k]=\sum_{k=0}^\infty \frac{1}{5^k}u[n-k]=\Bigg( \sum_{k=0}^n \frac{1}{5^k} \Bigg)u[n] $
I'm not totally sure that this is the way to compute this sum...
$ y[n]=\Bigg(\frac{1-(1/5)^{n+1}}{1-(1/5)}\Bigg)u[n] $
--Cmcmican 20:57, 31 January 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
