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== 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]=u[n-5]</math>
+
<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>
+
<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>
  
where z is an input into our system.  Let <math>z = e^{jk\omega_o}</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>
  
<math>F[z]=\sum_{m= - \infty}^{\infty}\delta{[m-5]}e^{-mjk\omega_o}</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]]

Latest revision as of 07:52, 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

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