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Inverse Laplacetransform, can't find fitting formula

  1. Dec 30, 2011 #1
    1. The problem statement, all variables and given/known data

    Inverse Laplacetransform (3s+7)/(s^2+2s-2)


    2. Relevant equations

    (3s+7)/(s^2+2s-2)



    3. The attempt at a solution

    Split into 3s/(s^2+2s-2) and 7/(s^2+2s-2).

    I cant find a fitting transformpair(I use tables/formulas).
     
  2. jcsd
  3. Dec 30, 2011 #2
    Let's complete the square of the polynomial equation to get it into a form that we can use.

    s^2 + 2s -2 = (s+1)^2 -3

    Now, we have

    3s/[(s+1)^2 -3] + 7/[(s+1)^2 -3]
    = 3(s+1)/[(s+1)^2 -3] + 4/[(s+1)^2 -3]
    invLaplace => 3(e^-t)hcos(rad3*t) + (4/rad3)(e^-t)hsin(rad3*t)
     
  4. Dec 30, 2011 #3
    I don't have hyperbolicus functions in my transformtable, is there any other way?
     
  5. Dec 30, 2011 #4
    Yes, we can use the definition of sinh and cosh.

    Hyperbolic sine: (e^2x - 1)/(2e^x)
    Hyperbolic cosine: (e^2x + 1)/(2e^x)

    3(e^-t)[(e^2rad3(t) + 1)/(2e^rad3(t))] + (4/rad3)(e^-t)[(e^2rad3(t) - 1)/(2e^rad3(t))]
    =3(e^2rad3(t) + 1)/(2e^(rad3(t)+1)) + (4/rad3)(e^2rad3(t) - 1)/(2e^(rad3(t)+1))
    =[(3rad3+4)/(2rad3)](e^2rad3(t) - (8/rad3))/(e^(rad3(t)+1))
     
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