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== Part a ==
 
== Part a ==
  
<font size="3">System: <math>X_{k}[n-k] \to Y_{k}[n] = (k+1)^2 \delta [n-(k+1)]</math>
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<font size="3">System: <math>X_{k}[n]=\delta[n-k] \to Y_{k}[n] = (k+1)^2 \delta [n-(k+1)]</math>
  
Time-delay: <math>X_{k}[n-k] \to X_{k}[n-N-k]</math>
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Time-delay: <math>X_{k}[n]=\delta[n-k] \to X_{k}[n-N]=\delta[n-N-k]</math>
  
  

Revision as of 11:29, 11 September 2008

Part a

System: $ X_{k}[n]=\delta[n-k] \to Y_{k}[n] = (k+1)^2 \delta [n-(k+1)] $

Time-delay: $ X_{k}[n]=\delta[n-k] \to X_{k}[n-N]=\delta[n-N-k] $


$ X_{k}[n] \to timedelay \to sys \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $

$ X_{k}[n] \to sys \to timedelay \to Z_{k}[n]=(k+1)^2 \delta [n-N-(k+1)] $


Since $ (k+1)^2 \delta [n-N-(k+1)] $ is equal to $ (k+1)^2 \delta [n-N-(k+1)] $, the system is time-invariant.

Part b

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009