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<span class="texhtml">''x''[''n''] * ''h''[''n''] = ''h''[''n''] * ''x''[''n'']</span> | <span class="texhtml">''x''[''n''] * ''h''[''n''] = ''h''[''n''] * ''x''[''n'']</span> | ||
| − | <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> | + | <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> |
| − | There are two cases. | + | There are two cases. |
| − | Case | + | Case 1 (iff n+3<0 or n<-3): |
| + | <math>=\sum_{k=0}^\infty 1/5^k</math> | ||
| + | <math>=(1-0)/(1-1/5)=5/4</math> | ||
| + | |||
| + | 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)) | ||
=== Answer 2 === | === Answer 2 === | ||
Revision as of 21:22, 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 (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.
