(New page: == Problem 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^nu[-n] </math>. (simplify your answer until...)
 
Line 2: Line 2:
  
 
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^nu[-n] </math>. (simplify your answer until all <math> \sum </math> signs disappear).
 
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^nu[-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[k]  h[n-k] \!</math>
 +
 +
<math> y[n] = \sum^{\infty}_{k = -\infty} 2^ku[-k] u[-(n-k)] \!</math>
 +
 +
 +
We know <math> u[-k] = 1 \!</math> when <math> K \leq 0 </math>
 +
 +
<math> y[n] = \sum^{0}_{k = -\infty} 2^k u[-n+k] \!</math>
 +
 +
<math> r = -k \!</math>
 +
 +
<math> y[n] = \sum^{\infty}_{r = 0} (\frac{1}{2})^k u[-n-r] \!</math>
 +
 +
We know <math> -n-r \geq 0 </math> when  <math> -n \geq r </math>
 +
 +
<math> y[n] = \sum^{-n}_{r = 0} (\frac{1}{2})^k \!</math> when  <math> -n \geq 0 </math>
 +
 +
 +
<math> y[n] = 0 \!</math>, else
 +
 +
<math> y[n] = \frac{1- (\frac{1}{2})^{-n+1}}{1 - \frac{1}{2}} \!</math> when  <math> n \leq 0 </math>
 +
 +
<math> y[n] = 0 \!</math>, else

Revision as of 10:05, 14 October 2008

Problem 3

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

Solution

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

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

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


We know $ u[-k] = 1 \! $ when $ K \leq 0 $

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

$ r = -k \! $

$ y[n] = \sum^{\infty}_{r = 0} (\frac{1}{2})^k u[-n-r] \! $

We know $ -n-r \geq 0 $ when $ -n \geq r $

$ y[n] = \sum^{-n}_{r = 0} (\frac{1}{2})^k \! $ when $ -n \geq 0 $


$ y[n] = 0 \! $, else

$ y[n] = \frac{1- (\frac{1}{2})^{-n+1}}{1 - \frac{1}{2}} \! $ when $ n \leq 0 $

$ y[n] = 0 \! $, else

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood