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Inverse Laplace Transform of this expression?

  1. Aug 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the Inverse Laplace Transform of [itex]\frac{1}{(s^{2} + 1)^{2}}[/itex]

    2. Relevant equations



    3. The attempt at a solution
    I tried using partial fractions but it didn't work. It looks like a cosine transform, but I don't know what else to do. Help please :(
     
  2. jcsd
  3. Aug 16, 2013 #2

    LCKurtz

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    Think of it as$$
    \frac 1 {s^2+1}\cdot \frac 1 {s^2+1}$$What theorem do you know about the product of transforms?
     
  4. Aug 16, 2013 #3
    Ohh the convolution theorem?? I'm working on it
     
  5. Aug 16, 2013 #4
    I got it now. Thank you so much
     
  6. Aug 18, 2013 #5

    Ray Vickson

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    Another way: start with
    [tex] g_a(s) = \frac{1}{(s^2+1)(s^2+a^2)}, \: a \neq 1[/tex]
    expand in partial fractions, find the inverse Laplace transform ##F_a(t)##, then take the limit as ##a \to 1.##
     
  7. Aug 18, 2013 #6
    I expanded as you suggested and applied the transform. I got:

    [itex]\frac{sin t}{a^{2} - 1}[/itex] + [itex]\frac{sin(at)}{a(1 - a^{2} )}[/itex]

    But if I try to take the limit as a goes to 1, they just go to infinity. What am I doing wrong?
     
  8. Aug 18, 2013 #7

    Ray Vickson

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    Set ##a = 1+\epsilon##, and expand things out until you get something in which you can use l'Hospital's rule to evaluate the limit as ##\epsilon \to 0.## I've done it, and it works!

    Or, you can write your result as
    [tex] \frac{\sin(at) - a \sin(t)}{a(1-a^2)}[/tex] and then use l'Hospital.
     
    Last edited: Aug 18, 2013
  9. Aug 18, 2013 #8
    Omg you are right. I made the sum of fractions and used L'Hopital rule... I got the same result. That was amazing!! I learnt a valuable method today, thank you
     
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