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Homework Help: Proving time invarianceof a system

  1. Jul 15, 2009 #1
    i need to prove that
    is time invarient

    i know that
    i need to get the same result for a shift in time
    u-x=s -> du=ds


    so what now??
    how does it prove that

    Last edited: Jul 15, 2009
  2. jcsd
  3. Jul 15, 2009 #2
    I think you forgot to change the limits of integration. Anyways I'm not sure what the context is but it looks like we're just dealing with the integrand under a horizontal translation and the corresponding shift in the limits of integration?
  4. Jul 15, 2009 #3

    if you dont know please dont spam
  5. Jul 15, 2009 #4
    All right I'm starting to get a little annoyed. I'm also sure that before you asked to show f(t-x) = f(t) but whatever. Look it's pretty clear you did not switch the limits of integration, which you need to do for definite integrals. Also, if I am interpreting the question correctly, you generally will not get [tex]M_{f(t-x)}=M_{f(t)}[/tex]. If x is the change in time, f(u-x) shifts the graph of f(u) x units to the right, but your limits of integration for [tex]M_{f(t-x)}[/tex] are those of [tex]M_{f(t)}[/tex] minus x units, which doesn't make much sense. Now I doubt I misunderstood this question 100%, but it would help if you actually posed you question more neatly.
  6. Jul 15, 2009 #5
    the correct answer that its time invariant

    i just miss the final step
  7. Jul 15, 2009 #6


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    Science Advisor

    Yes, and Snipez90 told you exactly what you had done wrong. It is unfortunate that you chose to insult the only person who had tried to help you rather than thinking about what he said. You made a change of variable but failed to change the limits of integration as you should.
  8. Jul 15, 2009 #7
    my variable are s
    not t

    this stuff is so weird

    i dont how to change the variable
    because u-x=s

    there is no t there

    how to do that

    i know how to change variable in a normal one variable integral

    bu this is impossible
  9. Jul 16, 2009 #8
    the correct to my question if the first post in this thread is that it time invariant

    here are similar solved questions
    from the first and 4th examples there i got the idea that
    in the [tex]y(t-\tau)[/tex] step we subtract [tex]\tau[/tex] from every t including the intervals of the integral.
    in the [tex]L(x)(t-\tau)[/tex] step we subtract [tex]\tau[/tex] only from the variable inside x function
    then we if we get[tex]y(t-\tau) =L(x)(t-\tau)[/tex] then its time invariant

    but examples 2 and 3 are not following this pattern

    in the second example
    in the [tex]L(x)(t-\tau)[/tex] step they dont substitute the t with t-tau
    i expect it to be [tex]t-2\tau[/tex] inside the x function

    and in the 4th they add T instead of subtracting it
    Last edited: Jul 16, 2009
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