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<math>x[n] \to \Bigg[ system \Bigg] \to y[n]=x[n-1]+x[1-n] \to \Bigg[ time\ delay\ n_0 \Bigg] \to z[n]=y[n-n_0]=x[(n-n_0)-1]+x[1-(n-n_0)]</math>
 
<math>x[n] \to \Bigg[ system \Bigg] \to y[n]=x[n-1]+x[1-n] \to \Bigg[ time\ delay\ n_0 \Bigg] \to z[n]=y[n-n_0]=x[(n-n_0)-1]+x[1-(n-n_0)]</math>
  
<math>=x[n-1-n_0]+x[1-n+n_0]</math>
+
<math>=x[n-1-n_0]+x[1-n+n_0]\,</math>
  
 
The second term in the last equation has a factor of <math>+n_0</math>, so the two are not equal, therefore this system is time variant.
 
The second term in the last equation has a factor of <math>+n_0</math>, so the two are not equal, therefore this system is time variant.

Revision as of 15:09, 26 January 2011

Practice Question on Time Invariance of a System

The input x[n] and the output y[n] of a system are related by the equation

$ y[n]=x[n-1]+x[1-n]. $

Is the system time invariant (yes/no)? Justify your answer.


Share your answers below

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Answer 1

No, this system is time variant. $ x[n] \to \Bigg[ time\ delay\ n_0 \Bigg] \to y[n]=x[n-n_0] \to \Bigg[ system \Bigg] \to z[n]=y[n-1]+y[1-n]=x[(n-1)-n_0]+x[(1-n)-n_0] $

$ x[n] \to \Bigg[ system \Bigg] \to y[n]=x[n-1]+x[1-n] \to \Bigg[ time\ delay\ n_0 \Bigg] \to z[n]=y[n-n_0]=x[(n-n_0)-1]+x[1-(n-n_0)] $

$ =x[n-1-n_0]+x[1-n+n_0]\, $

The second term in the last equation has a factor of $ +n_0 $, so the two are not equal, therefore this system is time variant.

--Cmcmican 19:07, 26 January 2011 (UTC)

Answer 2

Write it here.

Answer 3

Write it here.


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