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Inverse Laplace transform

  1. Sep 12, 2014 #1
    1. The problem statement, all variables and given/known data

    Find the inverse Laplace transform of the expression:

    F(S) = [itex]\frac{3s+5}{s^2 +9}[/itex]


    2. Relevant equations



    3. The attempt at a solution

    From general Laplace transforms, I see a pattern with laplace transforming sin(t) and cos(t) because:

    L{sin(t)+cos(t)} = [itex]\frac{s+1}{s^2 +1}[/itex]

    All I'm missing here is a couple of constants(?).

    I know that the laplace transform works like this:

    L{Asin(Bt)+Ccos(Dt)} = [itex]\frac{Cs}{s^2 +D}[/itex] + [itex]\frac{A*B}{s^2 +B^2}[/itex]

    Looking at my original problem, I can see that I need B to equal D, ##D^2## + ##B^2## = 9, C = 3 and A*B = 5.

    If I set A = 5, B = 3, C = 9 and D = 3 I get ##3^2## + ##3^2## = 18, 5
    3 = 15 (putting numbers into my equations above).

    Therefore I need to divide the whole expression by 3 to get my correct answer:

    L{[itex]\frac{1}{3}([/itex]5sin(3t)+9cos(3t))} = [itex]\frac{3s+5}{s^2 +9}[/itex]

    Is there a simpler way to do this? Or do I just have to use the trial and error method until I find the correct factors/constants?

    Any input on this will be greatly appreciated. Thanks
     
  2. jcsd
  3. Sep 12, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What? One of the equations above said A*B= 5 which is NOT satisfied by A= 5, B= 3. You have only three equations for the four values A, B, C, and D. And there is no reason to assume that A, B, C, and D must be integers.

    Use the fact that the Laplace transform, and so its inverse, is linear.
    [tex]\frac{3s+ 5}{s^2+ 9}= 3\frac{s}{s^2+ 9}+ 5\frac{1}{s^2+ 9}[/tex]
    and look up their inverse transforms separately.

    The laplace transform of [itex]cos(\omega x)[/itex] is [itex]\frac{s}{s^2+ \omega^2}[/itex] so the inverse transform of [itex]3\frac{s}{s^2+ 9}[/itex] is [itex]3 cos(3x)[/itex]

    The Laplace transform of [itex]sin(\omega x)[/itex] is [itex]\frac{\omega}{s^2+ \omega^2}[/itex] so the inverse transform of [itex]5\frac{1}{s^2+ 9}= (5/3)\frac{3}{s^2+ 9}[/itex] is [itex](5/3)sin(3x)[/itex].
     
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