(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>
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<math>  r = -k \!</math>
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<math>=\sum_{r=0}^{\infty}(\frac{1}{2})^{r}u[-n-r]</math> 
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<math>-n-r \geqq  0</math>
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<math>-n\geqq r</math>
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<math>=(\sum_{r=0}^{-n}(\frac{1}{2})^{n})</math>
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if -n>= 0
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else ,  0
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therefore,
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<math> =\frac{1-\frac{1}{2}^{-n+1}}{1-\frac{1}{2}}</math>
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if n=<0
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else,  0

Latest revision as of 11:06, 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] $

$ r = -k \! $

$ =\sum_{r=0}^{\infty}(\frac{1}{2})^{r}u[-n-r] $

$ -n-r \geqq 0 $

$ -n\geqq r $

$ =(\sum_{r=0}^{-n}(\frac{1}{2})^{n}) $

if -n>= 0

else , 0

therefore,

$ =\frac{1-\frac{1}{2}^{-n+1}}{1-\frac{1}{2}} $

if n=<0

else, 0

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Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

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