(New page: == 6A - Is the system time invariant? == System: <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br> Time-delay, then system: <math>T_k[n]=\delta[n-(k+1)]</math><br> <math>Y_k[n]=(k+1)^2\delta...)
 
(6A - Is the system time invariant?)
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== 6A - Is the system time invariant? ==
 
== 6A - Is the system time invariant? ==
 
System: <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br>
 
System: <math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br>
 +
Time-delay: <math>n-n_0</math><br>
  
 
Time-delay, then system:
 
Time-delay, then system:
<math>T_k[n]=\delta[n-(k+1)]</math><br>
+
<math>T_k[n]=\delta[(n-n_0)-(k+1)]</math><br>
<math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br>
+
<math>Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)]</math><br>
  
 
System, then time-delay:
 
System, then time-delay:
 
<math>T_k[n]=(k+1)^2\delta[n-k]</math><br>
 
<math>T_k[n]=(k+1)^2\delta[n-k]</math><br>
<math>Y_k[n]=(k+1)^2\delta[n-(k+1)]</math><br>
+
<math>Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)]</math><br>
  
 
Both <math>Y_k</math> yield the same output; therefore, the system is time invariant.
 
Both <math>Y_k</math> yield the same output; therefore, the system is time invariant.

Revision as of 06:26, 12 September 2008

6A - Is the system time invariant?

System: $ Y_k[n]=(k+1)^2\delta[n-(k+1)] $
Time-delay: $ n-n_0 $

Time-delay, then system: $ T_k[n]=\delta[(n-n_0)-(k+1)] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $

System, then time-delay: $ T_k[n]=(k+1)^2\delta[n-k] $
$ Y_k[n]=(k+1)^2\delta[(n-n_0)-(k+1)] $

Both $ Y_k $ yield the same output; therefore, the system is time invariant.

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009