(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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| − | [[ | + | = [[:Category:Problem_solving|Practice Question]] on Computing the Output of an LTI system by Convolution = |
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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. | ||
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| + | Case 1 (iff n+3<0 or n<-3): <math>=\sum_{k=0}^\infty 1/5^k=(1-0)/(1-1/5)=5/4</math> | ||
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| + | Case 2 (iff n+3>0 or n>-3): <math>=\sum_{k=n+3}^\infty 1/5^k</math> <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 === | ||
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| + | Write it here. | ||
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---- | ---- | ||
| − | [[ | + | |
| + | [[2011 Spring ECE 301 Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
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| + | [[Category:ECE301Spring2011Boutin]] [[Category:Problem_solving]] | ||
Latest revision as of 10:22, 11 November 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 (iff n+3<0 or n<-3): $ =\sum_{k=0}^\infty 1/5^k=(1-0)/(1-1/5)=5/4 $
Case 2 (iff n+3>0 or n>-3): $ =\sum_{k=n+3}^\infty 1/5^k $ $ =((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.
