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Proving Time Invariance

  1. Jul 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that

    [tex] y(t) = \frac{d}{dt}\left[e^{-t}x(t)\right][/tex]

    is time invariant.

    2. Relevant Information

    I don't think this is TI!! I'm told it is TI, but I think I proved that it is not TI! My proof is below. Am I wrong or is the question wrong in assuming that the system is TI?

    3. The attempt at a solution

    Let [itex]y_1[/itex] be the output when [itex]x(t+t_0)[/itex] is the input, then:

    [tex] y_1(t) = \frac{d}{dt}\left[e^{-t}x(t+t_0)\right] [/tex]


    [tex] y(t+t_0) = \frac{d}{dt}\left[e^{-(t+t_0)}x(t+t_0)\right] = e^{-t_0}\frac{d}{dt}\left[e^{-t}x(t+t_0)\right] [/tex]

    Therefore [itex]y_1(t) \neq y(t+t_0)[/itex] and the system is not time invariant.


    Also, just to make sure I wasn't missing some subtlety because of the differentiation I tried to prove this another way.

    Since x(t) is arbitrary, I assumed x(t)=t, so that:

    [tex]y(t)=\frac{d}{dt}[te^{-t}] = e^{-t}-te^{-t}[/tex]

    Now I time shift the system by 2:

    [tex]y(t+2) = e^{-(t+2)}-(t+2)e^{-(t+2)}=e^{-2}\left[e^{-t}-(t+2)e^{-t}\right][/tex]

    Now I let [itex]y_1(t)[/itex] be the output when the input is [itex]x(t+2)=t+2[/itex]:

    [tex]y_1(t) = \frac{d}{dt}[(t+2)e^{-t}] = e^{-t}-(t+2)e^{-t}[/tex]

    Clearly, then, [itex]y(t+2)\neq y_1(t)[/itex] and the system is not TI for x(t)=t, and hence cannot be TI for arbitrary x(t).


    So, if it is TI, what am I doing wrong? And how would I prove that it is TI?
  2. jcsd
  3. Aug 2, 2007 #2
    Anyone know of a forum where this type of question is likely to get answered? I'm still desperate to get an answer to this. This is like basic systems... someone's gotta know.

    Can I move this to the EE forum?
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