(New page: == Exam1 question 3. == An LTI system has unit impulse response <math>h[n] = u[-n]</math>. compute the system's response to the input <math>x[n]=2^nu[-n]</math>. (Simplify your answer u...)
 
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An LTI system has unit impulse response <math>h[n] = u[-n]</math>. compute the system's response to the  
 
An LTI system has unit impulse response <math>h[n] = u[-n]</math>. compute the system's response to the  
 
input <math>x[n]=2^nu[-n]</math>.  (Simplify your answer until all <math>\sum</math>  signs disappear.)
 
input <math>x[n]=2^nu[-n]</math>.  (Simplify your answer until all <math>\sum</math>  signs disappear.)
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== Solution ==
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<math>y[n] = x[n] * h[n]\!</math>
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<math>=\sum_{k=-\infty}^{\infty} x[k]h[n-k]</math>
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<math>=\sum_{k=-\infty}^{\infty} 2^{k}u[-k]u[-(n-k)]</math>
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<math>=\sum_{k=-\infty}^{0}2^{k}u[-u+k]</math>

Revision as of 09:45, 14 October 2008

Exam1 question 3.

An LTI system has unit impulse response $ h[n] = u[-n] $. compute the system's response to the input $ x[n]=2^nu[-n] $. (Simplify your answer until all $ \sum $ signs disappear.)


Solution

$ y[n] = x[n] * h[n]\! $

$ =\sum_{k=-\infty}^{\infty} x[k]h[n-k] $

$ =\sum_{k=-\infty}^{\infty} 2^{k}u[-k]u[-(n-k)] $

$ =\sum_{k=-\infty}^{0}2^{k}u[-u+k] $

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