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| − | + | = Practice Question on Computing the Output of an LTI system by Convolution = | |
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| − | = Practice Question on Computing the Output of an LTI system by Convolution= | + | |
The unit impulse response h[n] of a DT LTI system is | The unit impulse response h[n] of a DT LTI system is | ||
| − | <math>h[n]= \frac{1}{5^n}u[n]. \ </math> | + | <math>h[n]= \frac{1}{5^n}u[n]. \ </math> |
| − | Use convolution to compute the system's response to the input | + | Use convolution to compute the system's response to the input |
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| + | <math>x[n]= u[-n-3] \ </math> | ||
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---- | ---- | ||
| − | ==Share your answers below== | + | |
| − | 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! | + | == Share your answers below == |
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| + | 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! | ||
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---- | ---- | ||
| − | ===Answer 1=== | + | |
| − | + | === Answer 1 === | |
| − | ===Answer 2=== | + | |
| − | Write it here. | + | <span class="texhtml">''x''[''n''] * ''h''[''n''] = ''h''[''n''] * ''x''[''n'']</span> |
| − | ===Answer 3=== | + | |
| − | Write it here. | + | <math>=\sum_{k=-\infty}^\infty h[k]x[n-k]</math> <math>=\sum_{k=-\infty}^\infty 1/5^k*u[k]*u[k-n-3]</math> <math>=\sum_{k=0}^\infty 1/5^k * u[k-n-3]</math> |
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| + | There are two cases. Case 1: <math>=\sum_{k=0}^\infty 1/5^k</math> {iff n+3<0 or n<-3} <span class="texhtml"> = ((1 / 5)<sup>0</sup> − (1 / 5)<sup> / </sup>''i''''n''''f''''t''''y'') / (1 − 1 / 5) = 1 / (4 / 5) = 5 / 4</span> | ||
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| + | Case 2: <math>=\sum_{k=n+3}^\infty 1/5^k</math> {iff n+3>0 or n>-3} <math>=((1/5)^{n+3} - (1/5)^{\infty + 1})/(1-1/5) = 1/(4*5^{n+2})</math> ([[User:Clarkjv|Clarkjv]] 01:02, 3 February 2011 (UTC)) | ||
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| + | === Answer 2 === | ||
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| + | Write it here. | ||
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| + | === Answer 3 === | ||
| + | |||
| + | Write it here. | ||
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---- | ---- | ||
| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
| + | |||
| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Revision as of 21:02, 2 February 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]= \frac{1}{5^n}u[n]. \ $
Use convolution to compute the system's response to the input
$ x[n]= u[-n-3] \ $
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
x[n] * h[n] = h[n] * x[n]
$ =\sum_{k=-\infty}^\infty h[k]x[n-k] $ $ =\sum_{k=-\infty}^\infty 1/5^k*u[k]*u[k-n-3] $ $ =\sum_{k=0}^\infty 1/5^k * u[k-n-3] $
There are two cases. Case 1: $ =\sum_{k=0}^\infty 1/5^k $ {iff n+3<0 or n<-3} = ((1 / 5)0 − (1 / 5) / i'n'f't'y) / (1 − 1 / 5) = 1 / (4 / 5) = 5 / 4
Case 2: $ =\sum_{k=n+3}^\infty 1/5^k $ {iff n+3>0 or n>-3} $ =((1/5)^{n+3} - (1/5)^{\infty + 1})/(1-1/5) = 1/(4*5^{n+2}) $ (Clarkjv 01:02, 3 February 2011 (UTC))
Answer 2
Write it here.
Answer 3
Write it here.
