(New page: == question == 3. An LTI system has unit impulse response <math>h[n]=u[-n]</math> Compute the system's response to the input <math> x[n]=2^{n}u[-n].(Simplify your answer until all \sum sig...) |
(→question) |
||
| (17 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
== question == | == question == | ||
| − | 3. An LTI system has unit impulse response <math>h[n]=u[-n]</math> Compute the system's response to the input <math> x[n]=2^{n}u[-n].(Simplify your answer until all \sum signs disappear.) | + | 3. An LTI system has unit impulse response <math>h[n]=u[-n]</math> Compute the system's response to the input <math> x[n]=2^{n}u[-n].</math>(Simplify your answer until all <math>\sum</math> signs disappear.) |
| + | |||
| + | == solution == | ||
| + | <math> y[n]=x[n]*h[n]</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]u[n-k]</math> | ||
| + | |||
| + | n[-k] = 1 ,-k>=0 | ||
| + | |||
| + | k<=0 | ||
| + | |||
| + | <math> y[n]=\sum^{0}_{k=-\infty}2^{k}u[n-k]</math> | ||
| + | |||
| + | <math> y[n]=\sum^{0}_{k=n}2^{k}, n<=0</math> | ||
| + | |||
| + | y[n]=0, else | ||
| + | |||
| + | let r=-k, k=-r | ||
| + | |||
| + | <math> y[n]=\sum^{-n}_{r=0}2^{-r}, n<=0 </math> | ||
| + | |||
| + | y[n]=0, else | ||
| + | |||
| + | <math> y[n]=\sum^{-n}_{r=0}(1/2)^{r}, n<=0 </math> | ||
| + | |||
| + | y[n]=0, else | ||
| + | |||
| + | <math> y[n] =2(1-(1/2)^{-n+1})u[-n] </math> | ||
| + | |||
| + | <math> y[n]=(2-2^{n})u[-n]</math> | ||
Latest 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] $
