(New page: ==Problem 5== An LTI system has unit impulse response <math>h[n]=u[n]-u[n-2]</math>. Computer (a) the system's function <math>H(z)</math> and (b) the system's response to the input <math>...)
 
(Problem 5)
Line 1: Line 1:
 
==Problem 5==
 
==Problem 5==
An LTI system has unit impulse response <math>h[n]=u[n]-u[n-2]</math>.  Computer (a) the system's function <math>H(z)</math> and (b) the system's response to the input <math>x[n]=\cos(\pi n)</math>.
+
An LTI system has unit impulse response <math>h[n]=u[n]-u[n-2]</math>.  Compute (a) the system's function <math>H(z)</math> and (b) the system's response to the input <math>x[n]=\cos(\pi n)</math>.
 +
 
 +
===Part A===
 +
First, note this is discrete time.  Doing the problem in continuous time gives a very different result, as I tragically learned on the test...
 +
 
 +
<math>H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k}</math>
 +
 
 +
<math>H(z)=z+\frac{1}{z}</math> due to the two step functions.

Revision as of 03:39, 15 October 2008

Problem 5

An LTI system has unit impulse response $ h[n]=u[n]-u[n-2] $. Compute (a) the system's function $ H(z) $ and (b) the system's response to the input $ x[n]=\cos(\pi n) $.

Part A

First, note this is discrete time. Doing the problem in continuous time gives a very different result, as I tragically learned on the test...

$ H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k} $

$ H(z)=z+\frac{1}{z} $ due to the two step functions.

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin