(New page: Category:ECE301Spring2011Boutin Category:problem solving = Practice Question on Computing the Output of an LTI system by Convolution= The unit impulse response h[n] of a DT LTI sys...)
 
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===Answer 1===
 
===Answer 1===
Write it here.
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<math>y[n]=x[n]*h[n]=\sum_{k=-\infty}^\infty \frac{1}{2^k}\delta[n-1-k]</math>
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<math>\delta[n-1-k] = \begin{cases}
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1,  & \mbox{if }k = n-1 \\
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0, & \mbox{if }k \ne n-1
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\end{cases}</math>
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<math>y[n] = \frac{1}{2^{n-1}}</math>
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--[[User:Cmcmican|Cmcmican]] 20:13, 31 January 2011 (UTC)
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===Answer 2===
 
===Answer 2===
 
Write it here.
 
Write it here.

Revision as of 15:13, 31 January 2011

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]= \delta[n-1]. \ $

Use convolution to compute the system's response to the input

$ x[n]= \frac{1}{2^n} \ $


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Answer 1

$ y[n]=x[n]*h[n]=\sum_{k=-\infty}^\infty \frac{1}{2^k}\delta[n-1-k] $

$ \delta[n-1-k] = \begin{cases} 1, & \mbox{if }k = n-1 \\ 0, & \mbox{if }k \ne n-1 \end{cases} $

$ y[n] = \frac{1}{2^{n-1}} $ --Cmcmican 20:13, 31 January 2011 (UTC)


Answer 2

Write it here.

Answer 3

Write it here.


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