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Differential equations: impulse function

  1. Apr 8, 2006 #1
    How do you go about taking the Laplace transform of something like...

    [tex] \delta (t-1)sin(t) [/tex]

    What does this even mean?
     
  2. jcsd
  3. Apr 8, 2006 #2
    This is straight-forward, it comes directly from applying the defintions of the laplace transform and the impulse function (also called the Dirac delta "function"). So the first thing you should do is look up these definitions.
     
    Last edited: Apr 8, 2006
  4. Apr 10, 2006 #3
    So do I have to convolve them? I know the laplace transform of both of the functions, but in order to find the laplace transform of {h(t)x(t)} it seems I'll need convolution.

    And I'll have to use the fact that

    [tex] \int \delta(t-t_0)f(t)dt = f(t_0) [/tex]

    Am I on the right track?
     
  5. Apr 10, 2006 #4

    HallsofIvy

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    No, this problem, at least as you stated it, has nothing to do with convolution: you are given a function, [itex]\delta(t-1)sin(t)[/itex] and are asked to find the Laplace transform. The Laplace transform of that is
    [tex]\int_0^\infty \delta(t-1)sin(t)e^{-st}dt[/tex].
    That's particularly easy to integrate with that delta function in there! Do you know enough about the delta function to know why that is easy?
     
  6. Apr 10, 2006 #5
    I don't think so. It seems like the integral I posted above would help me in some way, but I'm not sure how.

    EDIT: unless, is the laplace transform

    [tex]e^{-s(1)}sin(1)[/tex]

    ?
     
    Last edited: Apr 10, 2006
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