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Finding the inverse Laplace transform

  1. Aug 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the inverse Laplace transform of the following function
    1/[s(s2 + 4)2]


    2. Relevant equations
    1/ (s2 + [itex]\omega[/itex]2)2 = (1/ 2[itex]\omega[/itex]3) (sin[itex]\omega[/itex]t -[itex]\omega[/itex]t cos[itex]\omega[/itex]t)


    3. The attempt at a solution
    L-1 = (1/s) (1/ s2 +22)2
    = (1/16) (sin 2t - 2t cos 2t) as the Laplace transform of (1/s) = 1

    Is this what I am meant to do?
     
  2. jcsd
  3. Aug 21, 2011 #2
    The inverse Laplace transform of a product isn't the product of the inverse Laplace transforms in general. You can go for the convolution theorem (might be a mess), or use partial fractions to decompose that, which requires you to use a few other inverse identities once you get the decomposition.
     
  4. Aug 21, 2011 #3
    Right.. how silly of me, so doing it by partial fractions I got 0=B=C and A= 1/16
    giving me F(s) = (1/16)(1/s) does that seem a little more on track?
     
  5. Aug 21, 2011 #4
    Well using partial fractions is definitely on track. But remember that when you have a quadratic term in the denominator, you have to have a linear term in the numerator (plus, it's squared, so there are going to be two.) The form for the expansion should be:

    [tex]\frac{1}{(s)(s^2+4)^2} = {\frac{A}{s}} + \frac{Bs + C}{s^2 +4} + \frac{Ds + E}{(s^2+4)^2}[/tex]

    Give that a shot.
     
    Last edited: Aug 21, 2011
  6. Aug 21, 2011 #5
    Okay after that I am now getting A = 1/16, B = -1/16, C = 0, D = -1/4, E = 0.

    (1/16)(1/s) - (1/16)(s/[s2 + 4]) - (1/4)(s/(s2 + 4)2
     
  7. Aug 21, 2011 #6
    That looks good! Now that you have those, you can invert them term-by-term since the inverse Laplace transform is linear, so [itex]\mathcal{L}^{-1}(af(t)+bg(t)) = a\mathcal{L}^{-1}(f(t)) + b\mathcal{L}^{-1}(g(t))[/itex] for constants [itex]a[/itex] and [itex]b[/itex].
     
  8. Aug 21, 2011 #7
    Thank you so much!!!!! :D
     
  9. Aug 21, 2011 #8
    You're welcome! :biggrin:
     
  10. Aug 22, 2011 #9

    Ray Vickson

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    There is a standard result: if f(t) <---> g(s), then integral_{x=0..t} f(x) dx < ---> g(s)/s.

    RGV
     
  11. Aug 22, 2011 #10
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