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===Answer 1=== | ===Answer 1=== | ||

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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} | ||

+ | 1, & \mbox{if }k = n-1 \\ | ||

+ | 0, & \mbox{if }k \ne n-1 | ||

+ | \end{cases}</math> | ||

+ | |||

+ | <math>y[n] = \frac{1}{2^{n-1}}</math> | ||

+ | --[[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

## Contents

# 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} \ $

You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!

### 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.