| Line 1: | Line 1: | ||
== Define a DT LTI System == | == Define a DT LTI System == | ||
Let the DT LTI system be: | Let the DT LTI system be: | ||
| − | <math>y[n]= | + | <math>y[n]=x[n-5]</math> |
==Obtain the Unit Impulse Response h[n] and the System Function F[z] of the system== | ==Obtain the Unit Impulse Response h[n] and the System Function F[z] of the system== | ||
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First to obtain the unit impulse response h[n] we plug in <math>\delta{[n]}</math> into our y[n]. | First to obtain the unit impulse response h[n] we plug in <math>\delta{[n]}</math> into our y[n]. | ||
| − | <math>h[n]=\delta{[n-5 | + | <math>h[n]=\delta{[n]-5}</math> |
Then the system function F[z] is obtained by | Then the system function F[z] is obtained by | ||
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<math>F[z]=\sum_{m= - \infty}^{\infty}h[m]z^{-m}</math> | <math>F[z]=\sum_{m= - \infty}^{\infty}h[m]z^{-m}</math> | ||
| − | + | <math>F[z]=\sum_{m= - \infty}^{\infty}\delta{[m-5]}z^{-m}</math> | |
| − | So when z^n is input into our system, we should get <math>F[z]z^n</math> back out. | + | since <math>x[m]\delta{[m-5]}=x[-5]</math> by the sifting property then: |
| + | |||
| + | <math>F[z]=z^{-5}</math> | ||
| + | |||
| + | where z is an input into our system. | ||
| + | |||
| + | So when z^n is input into our system, we should get <math>F[z]z^n=z^{-5}z^n</math> back out. | ||
| + | |||
| + | == Response of the system to the signal defined in Question 1 == | ||
| + | |||
| + | <math>x(t)=(5+3j)cos(4t)+(1+2j)sin(3t)</math> | ||
| − | |||
Go back to [[Homework 4_ECE301Fall2008mboutin]] | Go back to [[Homework 4_ECE301Fall2008mboutin]] | ||
Latest revision as of 08:52, 27 September 2008
Define a DT LTI System
Let the DT LTI system be: $ y[n]=x[n-5] $
Obtain the Unit Impulse Response h[n] and the System Function F[z] of the system
First to obtain the unit impulse response h[n] we plug in $ \delta{[n]} $ into our y[n].
$ h[n]=\delta{[n]-5} $
Then the system function F[z] is obtained by
$ F[z]=\sum_{m= - \infty}^{\infty}h[m]z^{-m} $
$ F[z]=\sum_{m= - \infty}^{\infty}\delta{[m-5]}z^{-m} $
since $ x[m]\delta{[m-5]}=x[-5] $ by the sifting property then:
$ F[z]=z^{-5} $
where z is an input into our system.
So when z^n is input into our system, we should get $ F[z]z^n=z^{-5}z^n $ back out.
Response of the system to the signal defined in Question 1
$ x(t)=(5+3j)cos(4t)+(1+2j)sin(3t) $
Go back to Homework 4_ECE301Fall2008mboutin
