(solution)
(solution)
Line 7: Line 7:
 
<math> y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k]</math>
 
<math> y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k]</math>
  
<math> y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]*h[n-k]</math>
+
<math> y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k]</math>
  
 
<math> n[-k] = 1  -k>=0,    k<=0</math>
 
<math> n[-k] = 1  -k>=0,    k<=0</math>
 +
 +
<math> y[n]=\sum^{0}_{k=-\infty}2^{k}*u[n-k]</math>

Revision as of 10:47, 15 October 2008

question

3. An LTI system has unit impulse response $ h[n]=u[-n] $ Compute the system's response to the input $ x[n]=2^{n}u[-n]. $(Simplify your answer until all \sum signs disappear.)

solution

$ y[n]=x[n]*h[n] $

$ y[n]=\sum^{\infty}_{k=-\infty}x[n]*h[n-k] $

$ y[n]=\sum^{\infty}_{k=-\infty}2^{k}u[-k]u[n-k] $

$ n[-k] = 1 -k>=0, k<=0 $

$ y[n]=\sum^{0}_{k=-\infty}2^{k}*u[n-k] $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett