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<math> y[n]=\sum^{0}_{k=-\infty}2^{k}u[n-k]</math> | <math> y[n]=\sum^{0}_{k=-\infty}2^{k}u[n-k]</math> | ||
| − | <math> \sum^{0}_{k=n}2^{k}, n<=0</math> | + | <math> y[n]=\sum^{0}_{k=n}2^{k}, n<=0</math> |
| − | 0, else | + | y[n]=0, else |
let r=-k, k=-r | let r=-k, k=-r | ||
| − | <math> \sum^{-n}_{r=0}2^{-r}, n<=0 </math> | + | <math> y[n]=\sum^{-n}_{r=0}2^{-r}, n<=0 </math> |
| − | 0, else | + | y[n]=0, else |
| − | <math> \sum^{-n}_{r=0}(1/2)^{r}, n<=0 </math> | + | <math> y[n]=\sum^{-n}_{r=0}(1/2)^{r}, n<=0 </math> |
| − | 0, else | + | y[n]=0, else |
<math> y[n] =2(1-(1/2)^{-n+1})u[-n] </math> | <math> y[n] =2(1-(1/2)^{-n+1})u[-n] </math> | ||
| − | <math> (2-2^{n})u[-n]</math> | + | <math> y[n]=(2-2^{n})u[-n]</math> |
Revision as of 10:56, 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] $
$ y[n]=\sum^{0}_{k=n}2^{k}, n<=0 $
y[n]=0, else
let r=-k, k=-r
$ y[n]=\sum^{-n}_{r=0}2^{-r}, n<=0 $
y[n]=0, else
$ y[n]=\sum^{-n}_{r=0}(1/2)^{r}, n<=0 $
y[n]=0, else
$ y[n] =2(1-(1/2)^{-n+1})u[-n] $
$ y[n]=(2-2^{n})u[-n] $
