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Time-invariant system

  1. Nov 24, 2015 #1
    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.
     
    Last edited by a moderator: Nov 24, 2015
  2. jcsd
  3. Nov 25, 2015 #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?
     
  4. Nov 26, 2015 #3
    Yeah yea exactly, so do you think its true?
     
  5. Nov 26, 2015 #4

    haruspex

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    Yes, but surely the point is to come up with a working proof.
     
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