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== Question 6a ==
 
== Question 6a ==
  
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<math>x[n] \rightarrow \mbox{Time Delay} \rightarrow y[n]=x[n-n_0] \rightarrow System \rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)]</math>
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<math>x[n] \rightarrow System \rightarrow Y_k[n]=(k+1)^2 \delta [n-(k+1)] \rightarrow \mbox{Time Delay}\rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)]</math>
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So the system is time invariant.
  
 
== Question 6b ==
 
== Question 6b ==

Revision as of 15:27, 11 September 2008

System

Input: $ X_k[n]=\delta [n-k] $

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

For any non-negative integer k


Question 6a

$ x[n] \rightarrow \mbox{Time Delay} \rightarrow y[n]=x[n-n_0] \rightarrow System \rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)] $

$ x[n] \rightarrow System \rightarrow Y_k[n]=(k+1)^2 \delta [n-(k+1)] \rightarrow \mbox{Time Delay}\rightarrow Y_k[n-n_0]=(k+1)^2 \delta [n-n_0-(k+1)] $

So the system is time invariant.

Question 6b

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett