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Inverse Lapalce transform (inre convolution integral)

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data
    trying to determine a general method for doing inverse-Laplace transforms of a product of two Laplace Transform (L.T.) functions f1(s) and f2(s) where f1(s) and f2(s) are the LTs of F1(t) and F2(t).

    The scenario is for calculating the convolution integral for F1(t)*F2(t)

    If I know the inverse-LT for f1(s) and f2(s), but the product f1(s)f2(s) is quite messy and does not lend itself to any easily identifiable inverse-LT (in what LT tables I have; but maybe I just need to look more carefully), is there a way to get the inverse-LT knowing the individual inverse-LTs?

    2. Relevant equations
    G(t)=F1(t)*F2(t) = f1(s)f2(s) = g(s)
    G(t)=inverse-LT of g(s)


    3. The attempt at a solution
    I see a breif reference to a method, but there's no explanation.

    in an old textbook "Mathematical Methods in Chemical Engineering Volume 3; Seinfeld and Lapidus, 1974" on page 62

    Working with the notion of the convolution integral

    below eqn 3.1.36

    "Note that if the inverse of L.T.s of f1(s) and f2(s) are known, the inverse L.T. of the product can be obtained by integration."

    here "the product" is f1(s)f2(s)

    My only guess is I need to employ the "Residue Theorem" on the complex inversion integral of g(s) above.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 30, 2008 #2

    Office_Shredder

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    We have the convolution * is defined by

    [tex]f*g = \int_{0}^{t}f(u)g(t-u)du[/tex]

    And there's a theorem about this and Laplace transforms (call the Laplace operator L)

    [tex]L(f*g) = L(f).L(g)[/tex] where I use . to denote multiplication since * is already used as convolution So

    [tex] f*g = L^{-1}(L(f).L(g))[/tex] and hence given h.k where h and k are laplace transforms of two functions you know, you can find the laplace inverse of h.k by using the convolution of their individual inverses.

    If you're wondering where to read about this, I got this out of 'Differential and Integral Equations' by Peter J. Collins, chapter 14 section 3
     
  4. Nov 3, 2008 #3
    O_Shredder:

    I think I understand your response, but I'm not 100%.

    In short I interpret your response this way:
    ...to get the inverse-LT (L-1) of L(f).L(g), I just convolute the inverse-LT of L(f) by the inverse-LT of L(g). In otherwords, L-1(L(f).L(g)) = f*g

    Is that right?

    If so, I think we talked each other into circles.
    My principle goal is to calculate f*g.
    So I planned to do that by taking the L-1 of L(f).L(g).

    Getting L(f) and L(g) are easy. However, L(f).L(g) produces a more complex function for which I can not easily get the L-1.

    I found a mention in a text (see above) that:
    "if the inverse of L.T.s of f1(s) and f2(s) are known, the inverse L.T. of the product can be obtained by integration."

    I wasn't sure what was meant there "by integration."
    It didn't say specifically anything like "expansion theorem" or "residue theorem".
     
  5. Nov 3, 2008 #4

    Office_Shredder

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    Haha... I got confused by your f's and F's in your OP... that's my bad. Reading what you quoted in the book it sounds like they might want you to find the inverse L.T. of the product via the convolution. I don't have the particular book you cited, but I looked around a bit and couldn't find an alternate method (which doesn't mean there isn't one though)
     
  6. Nov 4, 2008 #5
    I'll likley have to make a serious effort here?!?!
    Go figure.


    thanks for your help
     
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