Revision as of 16:53, 31 January 2011 by Cmcmican (Talk | contribs)

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] \ $


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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] $

Right here I run into the same problem that I had with the original problem. I don't know how to compute this sum. If it were an infinite sum I could compute it, but it's not. I can't find any references that tell me how to compute this sum in non-infinite form. I tried convolving in the opposite order, but came up with $ \Bigg( \sum_{k=0}^n \frac{1}{5^{n-k}} \Bigg)u[n] $ which is no better (as far as I can tell). Is there something I am missing? How do I compute this sum??????

--Cmcmican 20:53, 31 January 2011 (UTC)

Answer 2

Write it here.

Answer 3

Write it here.


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