Time-invariant system

  • Thread starter steven-ka
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  • #1
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I would like to have some assistance with two examples of checking times in/variant systems.

1) y(t)=x(-t)


2) y(t)=x(t^2)

I would like to know what's wrong with the following solution of mine(especially the second one):

1) y(t)=x(-t)


x1(t)=x1(t-t0) => y1(t)=x(-t-t0)

y2(t)=y2(t-t0)=> y2(t)=x(-(t-t0))=x(-t+t0)

y1(t)=!y2(t) => time variant system.

2) y(t)=x(t^2)

x1(t)=x1(t-t0)=> y1(t)=x(t^2-t0)

y2(t)=y2(t-t0)=> y2(t)=x((t-t0)^2)

y1(t)=!y2(t)=> time variant system.

I'll appreciate any helpful comment :) thanks.
 
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Answers and Replies

  • #2
haruspex
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I agree with your answers but I am baffled by the proofs.
I assume you are defining x1 as the tIme shifted input and y1 as the corresponding time shifted output. So surely you should write x1(t)=x(t-t0), y1(t)=x1(-t), etc?
 
  • #3
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Yeah yea exactly, so do you think its true?
 
  • #4
haruspex
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Yes, but surely the point is to come up with a working proof.
 

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