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Time Invariance

  1. Aug 25, 2007 #1
    Note: I posted this a month ago in the homework forum, but never got a reply. It really is an elementary question, and I think someone here might know the answer. (I hope I don't get an infraction for the re-post! If it's inappropriate, please delete.)

    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?
    Last edited: Aug 25, 2007
  2. jcsd
  3. Aug 25, 2007 #2


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    Gold Member

    Infraction?! You optimist! Nay! A banning, at the least! Begone, beast!

    By the way, it's nice to see that you got your username fixed. :biggrin:
  4. Aug 9, 2008 #3
    The first proof is a bit heavy for me at this time of night... but the second seems to me that you have correctly proven it time variant. I'm no genius though... just letting you know i can't see any problem with your reasoning ;)
  5. Aug 9, 2008 #4


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    Homework Helper

    Um, please note when the last post was dated before posting.
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