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Convolution Integral

  1. Mar 18, 2013 #1
    Boas Ch. 8, Sec. 10 #3

    Use the convolution integral to find the inverse transforms of:
    [tex]
    \frac{p}{(p^{2}-1)^{2}} = \frac{p}{p^{2}-1} \frac{1}{p^{2}-1}
    [/tex]

    I'm completely confused with these things. Are we supposed to figure out the inverse Laplace transform and then use that within our convolution integral? I am completely lost. Just looking for some advice, thanks.

    I know the convolution integral takes the form of:
    [tex]
    \int_{-\infty}^\infty\,f(x)g(z-x)dx
    [/tex]
     
  2. jcsd
  3. Mar 18, 2013 #2

    Ray Vickson

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    For functions defined on [0,∞) the convolution is, instead:
    [tex] \int_0^z f(x) g(z-x) \, dx \text{ for }z \geq 0.[/tex]
    And yes, you are supposed to figure out the inverse Laplace of p/(p^2-1), then do a convolution, exactly as it says.
     
  4. Mar 18, 2013 #3
    Just to clarify, you mean '..inverse Laplace of
    [tex]
    \frac{p}{(p^{2}-1)^{2}}[/tex]
    right?
     
  5. Mar 18, 2013 #4
    So far...
    [tex]
    \int \frac{p}{(p^{2}-1)^{2}}dp[/tex]
    getting...
    [tex]
    \frac{-1}{2(p^{2}-1)}[/tex]

    Now do I look for the inverse transform? Because If I do that my answer slightly varies from the book's. Boas shows:
    [tex]
    \frac{tsinht}{2}[/tex]
    whereas I'm getting
    [tex]\frac{sinht}{2}[/tex]

    ..hmmm?

    (thanks for the help)
     
  6. Mar 18, 2013 #5
    However, if I actually paid attention to what you had said maybe I'd be doing something different. I'll try the 'real' method now...
     
  7. Mar 18, 2013 #6
    After finding my inverse transform, being [itex]cosh(at)sinh(at)[/itex] I used their exponential forms to integrate.
    [tex]
    \int_0^\infty \left(\frac{e^{at}+e^{-at}}{2}\right)\frac{e^{at}-e^{-at}}{2}dt[/tex]
    ...which I end up getting
    [tex]
    \frac{1}{2}\left(cosh(at)-t\right)[/tex]
    which still isn't right.

    Clearly, I'm confused.
     
  8. Mar 18, 2013 #7

    vela

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    That's not a convolution integral. You want to convolve sinh and cosh.
     
  9. Mar 18, 2013 #8
    Then I'm even more lost...
     
  10. Mar 18, 2013 #9

    vela

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    Look at Ray's integral in post #2 and yours. Yours doesn't have a ##z##. I'm not sure how you came up with your integral.
     
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