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T-shifting theorem, laplace transforms

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Using the t-shifting theorem, find the laplace transform of

    [itex] f(x) = tu(t-\pi)[/itex]

    2. Relevant equations
    [itex] L[f(t-a)u(t-a)] = F(s)e^{-as}[/itex]


    3. The attempt at a solution
    Now firstly I should state I already know the answer to the problem, the issue is getting to said answer.

    I think if I understood what [itex] f(x) = tu(t-\pi)[/itex] actually looked like it might help, but here's my process.
    [itex] a = \pi[/itex] so using the relationship between [itex]u(t-a) = e^{-as}[/itex] I get [itex]e^{-as} = e^{-\pi s}[/itex].
    f(t) = t, the laplace transformation of t is [itex]\frac{n!}{s^{n+1}} = \frac{1}{s^2}[/itex]
    Now multiplying [itex]e^{-\pi s}[/itex] by F(s) gives:
    [itex]\frac{e^{-\pi s}}{s^2}[/itex]

    So that's as far as I get, the problem is, the answer includes an extra term [itex]\frac{\pi e^{-\pi s}}{s}[/itex] and I have no idea how the get it. So the complete answer is [itex]\frac{e^{-\pi s}}{s^2} + \frac{\pi e^{-\pi s}}{s}[/itex]

    I feel like I've missed something extremely basic, but no matter how many YouTube videos I watch or different text books I read, I can't make heads or tails of it.. thanks!
     
  2. jcsd
  3. Sep 30, 2012 #2

    rude man

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    f(x)?? where's the x in tu(t-π)?

    OK, smart remarks over.

    First thing you do is set up the defining integral for transforming to Laplace. The tables won't be of much help here. So what's the integral?

    Next, I'll tell you that u(t - π) looks like a step function on your t axis with the step occurring not at t = 0 but at t = π.

    (BTW π is a weird choice for a time displacement, but it's OK).

    And BTW your "extra term" can't be right because it conflicts with the units it must have. tu(t-π) has dimension of time T, and a Laplace integration ups that to T2, so your first expression e-sπ/s2 meets that requirement but the second one does not. So someone gave you an incorrect answer.

    I can confirm that your first expression is a correct term in the total correct answer.
     
  4. Sep 30, 2012 #3
    Haha, my apologies, first time using latex (interesting choice of name for it..) and I copied some code and forgot to change the 'x' to a 't'.

    Are you positive the second term is incorrect? It's highly possible given this is an exercise and solution booklet provided by my University created for this semester. It's a little disappointing though when I can't trust the booklet.

    As for defining the integral, I'm not really sure what you're asking in this case. The lecturer rushes over the basics and doesn't give us adequate notes to at least read up on it.
    Most of the time the only way I learn is through worked examples because the majority of text books don't elaborate on things they expect you to already know and so I find it difficult to follow.
    The problem with that approach is that as soon as I'm presented with a question that's in a format I haven't tackled yet (i.e. this one), my brain goes in to meltdown and locks up.

    I may need to enlist the services of a tutor at this point I think!

    I appreciate the response though, it's nice to know I'm not 100% lost haha
     
  5. Sep 30, 2012 #4

    rude man

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    Oops, no, it's right. I said using π was a dumb choice of number to use for this problem. Normally π is dimensionless, just a number. In this case it has the dimension of time, so your 2nd term is OK after all. Sorry I didn't see it sooner.
    The defining integral for transforming a function f(t) to the Laplace domain is F(s) = ∫0f(t)e-stdt.

    So note that if the dimension of f(t) is D, then the dimension of F(s) must be D*T. For example, for a voltage v(t) the dimensions of V(s) are voltage * time. Careful to use different letters for the time and transformed variables. I always use lower case for the time function and upper case for the transformed function, as I've indicated above.

    Your remaining challenge when evaluating this integral will be what to do about the limits of integration. What I gave you, 0 and ∞, are correct, but remember the function u(t-π) is zero until t = π. That's the only hint I'm going to give you there.
    Nah, get a good text instead. My recommendation would be H. H. Skillin'g Electrical Engineering Circuits if it's still available.
     
  6. Oct 2, 2012 #5

    LCKurtz

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    The reason you can't use that relevant equation directly is that your ##f(t)## isn't expressed in terms of ##t-\pi##. Consider rewriting it like this$$
    f(t) = (t -\pi)u(t-\pi) + \pi u(t-\pi)$$and transforming that.
     
  7. Oct 2, 2012 #6

    rude man

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    I have to apologize, I hadn't noticed the question called for using the time-shifting theorem.
     
  8. Oct 2, 2012 #7
    Ahhhh, yes! that's much clearer LCKurtz! That's easy now, thanks!

    Thank you both for the help!
     
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