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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 | ||
Latest revision as of 10:22, 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]= 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})u[-n] = 5^{n+1}/4*u[-n] $(Clarkjv 23:35, 31 January 2011 (UTC))
Answer 3
Write it here.
