| Line 18: | Line 18: | ||
<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
Contents
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.
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
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.
