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== Part b ==
 
== Part b ==
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<font size="3">In order for <math>Y[n]=u[n-1]</math> to be true, <math>X[n]=u[n]</math> must also be true.
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Proof:
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<math>u[n]=\delta[n]-\delta[n-1]</math>
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      <math>\delta[n] \to sys \to \delta[n-1] \to</math>
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                                <math>- \to \delta[n-1]-4\delta[n-2]</math>
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  <math>\delta[n-1] \to sys \to 4\delta[n-2] \to</math>

Revision as of 11:42, 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

In order for $ Y[n]=u[n-1] $ to be true, $ X[n]=u[n] $ must also be true.

Proof:

$ u[n]=\delta[n]-\delta[n-1] $

      $ \delta[n] \to sys \to \delta[n-1] \to $
                                $ - \to \delta[n-1]-4\delta[n-2] $
 $ \delta[n-1] \to sys \to 4\delta[n-2] \to $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin