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Inverse Laplace Transform

  1. May 1, 2008 #1
    My goodness.. I have not come across an inverse Laplace transform like this. My teacher lets us just use a chart to figure them out, but this is definitely not on there. How do I find the inverse Laplace transform of:

    { (1/2)+[(5e^-6s)/(4s^2)] } / (s+5)

    I already used partial fractions to split up the denominator, so there is one more inverse laplace that I need to do on top of this one, but I figure if I get this one, then I can get the other one as well. Any help??!
  2. jcsd
  3. May 1, 2008 #2


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    you will need to use the dirac function for the exponential, the rest is pretty standard.
  4. May 2, 2008 #3
    The e^-6s part means that this part of your signal is delayed in time by 6 seconds. so you can do the inverse transform without it and then when you get your time signal for this part, delay it by 6 seconds. So replace t with t-6 in your answer for the time equation and you should be good.
  5. May 2, 2008 #4
    Do the inverse transform without it? Like... just take that whole term out and treat it like it is zero?
  6. May 2, 2008 #5


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    You'll need to use this:

    [tex]L(f(t-a)u(t-a)) = e^{-as}F(s)[/tex]

    I don't think dirac delta function comes into play here.
  7. May 2, 2008 #6
    Or i also think that applying the laplace transforms of integrals would work here.
  8. Jun 1, 2008 #7
    An easy way to take the inverse Laplace Transform (if you have some knowledge of Complex Calculus) is to take the sum of all the residues of the function e^(zt) f(z), where you are taking the inverse Laplace transform of f(z) and z is the complex variable.
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