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− | Case 1 (iff n+3 | + | 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> |
− | <math>=\sum_{k=0}^\infty 1/5^k | + | |
− | + | ||
− | Case 2(iff n+3 | + | 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)) |
− | <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 21:25, 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.