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Inverse Laplace Transform Help

  1. Jan 27, 2014 #1
    1. The problem statement, all variables and given/known data

    Is there a way to evaluate [itex]L^{-1}(\frac{F(s)}{s + a})[/itex]? I'm sure if it can be evaluate.

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 27, 2014 #2

    pasmith

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    It depends on what [itex]F[/itex] is. If [itex]F[/itex] has no singularities then the inverse transform
    [tex]
    \mathcal{L}^{-1}\left(\frac{F(s)}{s + a}\right) = \frac{1}{2\pi i}\int_{c-i\infty}^{c + i\infty} \frac{F(s)e^{st}}{s+a}\,ds
    [/tex]
    (where [itex]c \in \mathbb{R}[/itex] is such that there are no singularities of [itex]F(s)/(s+a)[/itex] to the right of the line [itex]\mathrm{Re}(s) = c[/itex]) reduces to the residue of [itex]\dfrac{F(s)e^{st}}{s+a}[/itex] at [itex]s = -a[/itex], which is [itex]F(-a)e^{-at}[/itex].
     
  4. Jan 27, 2014 #3

    Ray Vickson

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    Use convolution: if
    [tex] f(t) = L^{-1}[F(s)](t),[/tex]
    then
    [tex] L^{-1} \left( \frac{F(s)}{s+a} \right) (t) = \int_0^t f(t-\tau) e^{-a \tau} \, d \tau.[/tex]
     
  5. Feb 1, 2014 #4
    So if you have no idea what one of the functions in the frequency domain is and you get something like

    inverse Laplace transform( F(s)G(s) )

    and you know what G(s) of is but F(s) is not given then you have no way to evaluate the expression?
     
  6. Feb 1, 2014 #5

    Ray Vickson

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    Of course. If I give you two different F(s), you will get two different answers.
     
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