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(question)
 
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== question ==
 
== 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^{n}u[-n].</math>(Simplify your answer until all \sum signs disappear.)
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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^{n}u[-n].</math>(Simplify your answer until all <math>\sum</math> signs disappear.)
  
 
== solution ==
 
== solution ==
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<math> y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k]</math>
 
<math> y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k]</math>
  
<math> n[-k] = 1  -k>=0,    k<=0</math>
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n[-k] = 1  ,-k>=0
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 +
k<=0
  
 
<math> y[n]=\sum^{0}_{k=-\infty}2^{k}u[n-k]</math>
 
<math> y[n]=\sum^{0}_{k=-\infty}2^{k}u[n-k]</math>
  
<math> \sum^{0}_{k=n}2^{k}, n<=0</math>
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<math> y[n]=\sum^{0}_{k=n}2^{k}, n<=0</math>
 +
 
 +
y[n]=0, else
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 +
let r=-k, k=-r
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 +
<math> y[n]=\sum^{-n}_{r=0}2^{-r}, n<=0 </math>
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 +
y[n]=0, else
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 +
<math> y[n]=\sum^{-n}_{r=0}(1/2)^{r}, n<=0 </math>
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 +
y[n]=0, else
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 +
<math> y[n] =2(1-(1/2)^{-n+1})u[-n] </math>
  
0, else
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<math> y[n]=(2-2^{n})u[-n]</math>

Latest revision as of 09:56, 15 October 2008

question

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

solution

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

$ y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k] $

$ y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k] $

n[-k] = 1 ,-k>=0

k<=0

$ y[n]=\sum^{0}_{k=-\infty}2^{k}u[n-k] $

$ y[n]=\sum^{0}_{k=n}2^{k}, n<=0 $

y[n]=0, else

let r=-k, k=-r

$ y[n]=\sum^{-n}_{r=0}2^{-r}, n<=0 $

y[n]=0, else

$ y[n]=\sum^{-n}_{r=0}(1/2)^{r}, n<=0 $

y[n]=0, else

$ y[n] =2(1-(1/2)^{-n+1})u[-n] $

$ y[n]=(2-2^{n})u[-n] $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman