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Find the inverse Laplace transform

  1. Sep 6, 2011 #1
    Find the inverse Laplace transform


    I tried factoring out the solution, but run into the problem with the imaginary numbers and am still stuck with the s^2+2j, which I have to factor out once more, and that's where the problem gets even messier. What do I do?
  2. jcsd
  3. Sep 6, 2011 #2
    Factor the denominator first and split into partial fractions...
  4. Sep 6, 2011 #3
    I split the denominator into


    How can I split the above into partial fractions since I have the imaginary numbers in both.

    Also, how do I use the Latex in writing these equations, I am trying to use my regular Latex syntax but for some reason it has no effect here.

  5. Sep 6, 2011 #4
    You should be able to factor it without using complex numbers. That is, you should be able to find a,b,c and d such that


    Enclose them in [itex] .... [/ itex] brackets (without the spaces). So instead of writing $s^2$, you write [itex] s^2 [/ itex].
    The equivalent of $$ ... $$ is [tex] ... [/ tex]
  6. Sep 6, 2011 #5
    If I break it out to be


    which is just the [itex] e^{-at}sin(bt) [\itex] unfortunately I have the [itex] s^2 [\itex] terms I need to get rid of. I think I can work it out if I break it down into complex numbers from here but not sure if this is the right move.
  7. Sep 6, 2011 #6
    And looks like my Latex syntax is still not working
  8. Sep 6, 2011 #7
    Have you tried factoring it the way I suggested??

    Also, you have the wrong / in your tex brackets.
  9. Sep 6, 2011 #8
    I didn't see it right away but here is what I have

    [itex] s^4+4=(s^2+2s+2)(s^2-2s+2)=((s+1)^2+1)((s-1)^2+1) [/itex]

    this will also change the numerator so what I ended up getting overall is

    [itex] \frac{(s+1)-\frac{1}{2}s}{(s+1)^2+1}-\frac{(s-1)-\frac{1}{2}s}{(s+1)^2+1} [/itex]

    Now just trying to figure out the right functions to transform this back into the t domain, any suggestions?
  10. Sep 6, 2011 #9
    That's good, so working it out a bit further. We get that we need to take the inverse Laplace transform of





    This shouldn't be too hard. It should be of the form [itex]e^{-at}\sin(\omega t)u(t)[/itex] or [itex]e^{-at}\cos(\omega t)u(t)[/itex]
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