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Laplace transform heavyside function

  1. Jun 20, 2009 #1
    1. The problem statement, all variables and given/known data
    find the transform of
    sin (wt) u(t - b)

    I have to get it into the form for the time shift but I do not know how to deal with the unit function

    help appreciated
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 20, 2009 #2
    The Laplace transform is defined as:
    [tex]\mathcal{L}\lbrace f(t)\rbrace = \int_0^{\infty} e^{-st}f(t) dt[/tex]
    Now, let's put in
    [tex]f(t) = g(t) u(t-b)[/tex]
    where u(t-b) is zero for t<b and 1 for t>b. Plugging this in:

    [tex] \int_0^{\infty} e^{-st}f(t) dt = \int_0^{\infty} e^{-st}g(t) u(t-b) dt[/tex]

    Now split this integral into two parts:

    [tex]\int_0^{\infty} e^{-st}g(t) u(t-b) dt = \int_0^{b} e^{-st}g(t) u(t-b) dt + \int_b^{\infty} e^{-st}g(t) u(t-b) dt[/tex]

    The first integral vanishes, since u(t-b) is zero on the integration domain. In the second the u-function is simply 1. I'll let you solve the remaining steps.
  4. Jun 20, 2009 #3
    ahhh cool

    now lets say I have
    (t-1)^4 u(t)

    function is 1 from 0 < t < 1 and [tex](t-1)^4[/tex] for t> 1

    I can easily do the transform of t^4 how would i play with it to just do a time shift after the transform or do I have to go the integral route and do parts?
  5. Jun 22, 2009 #4


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    Homework Helper

    If you do not have to work it out from first principles (and I do not think you have to), you could do it by using this L-transform formula:
    [tex]L[f(t-a)u(t-a)] = e^{-as}F(s) \text{ where F(s) is L.T. of } f(t)[/tex].

    Of course in the case of [tex](t-1)^4 u(t)[/tex], it's not in a form which yields itself readily to that formula. In such a case you may have to work out the binomial expansion of (t-1)^4 before applying that formula.
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