(Question 6a)
(Question 6a)
 
(20 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
== Question 6a ==
 
== Question 6a ==
 +
The system is time variant because of the following example:
 +
 +
Dealing with the following system
 +
 +
<math> X_k[n]=Y_k[n] \,</math>
  
I'm assuming k is the variable representing any fo.
 
  
<math> X_k[n]=X_k[n] \,</math>
 
 
where  
 
where  
 +
 
<math> X_k[n]=\delta[n-k]\,</math>     
 
<math> X_k[n]=\delta[n-k]\,</math>     
 +
 +
 
and  
 
and  
 +
 
<math> Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math>
 
<math> Y_k[n]=(k+1)^2 \delta[n-(k+1)] \,</math>
  
  
 +
Consider the input and output of the system when k = 0
 +
 +
<math> X_0[n]=\delta[n]\,</math>   
 +
 +
 +
and
 +
 +
<math> Y_0[n]=\delta[n-1] \,</math>
 +
 +
 +
 +
If I time shift the input by <math> n_0\,</math> , then run it through the system I obtain:
 +
 +
<math>X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1],</math>
 +
 +
 +
but this isn't equal to running the input through the system, then time shifting the output by <math> n_0\,</math> 
 +
 +
<math>X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\,</math>
  
 +
== Question 6b ==
  
Under this assumption the following system cannot possibly be time invariant because of the <math>(k+1)^2</math> term.
+
Assuming this system is linear, an input
 +
<math> X_0[n]=u[n]\,</math> 
 +
would result in an output
 +
<math> Y_0[n]=u[n-1]\,</math>.

Latest revision as of 15:07, 12 September 2008

Question 6a

The system is time variant because of the following example:

Dealing with the following system

$ X_k[n]=Y_k[n] \, $


where

$ X_k[n]=\delta[n-k]\, $


and

$ Y_k[n]=(k+1)^2 \delta[n-(k+1)] \, $


Consider the input and output of the system when k = 0

$ X_0[n]=\delta[n]\, $


and

$ Y_0[n]=\delta[n-1] \, $


If I time shift the input by $ n_0\, $ , then run it through the system I obtain:

$ X_0[n] \longrightarrow X_0[n-n_0] \longrightarrow Y_1[n]=\delta[n-n_0-1], $


but this isn't equal to running the input through the system, then time shifting the output by $ n_0\, $

$ X_0[n] \longrightarrow Y_1[n]=4\delta[n-1] \longrightarrow Y_1[n-n_0]=4\delta[n-n_0-1]\, $

Question 6b

Assuming this system is linear, an input $ X_0[n]=u[n]\, $ would result in an output $ Y_0[n]=u[n-1]\, $.

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett