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Inverse laplace transform question

  1. Nov 22, 2006 #1
    How do i find the inverse laplace transform of s/(s-2)^2
  2. jcsd
  3. Nov 22, 2006 #2
    take the transform of [tex] f(t)=(1+t)e^{2t} [/tex]

    using the "shift" property so..

    [tex] F(s+a)=L[e^{-at}f(t)] [/tex] using

    [tex] \frac{s+2}{s^{2}}=s^{-1}+2s^{-2} [/tex]

    the inverse of above is just 1+t then multiply it by exp(2t) and you get it without recalling "residue theorem"...:tongue2:
  4. Jul 21, 2007 #3
    Use partial fractions:

    [tex]\frac{s}{(s-2)^2} = \frac{A}{(s-2)^2} + \frac{B}{s-2}[/tex]

    To find A, multiply both sides by [tex]s-2[/tex] and evaluate at [tex]s=2[/tex]:

    [tex]s = A[/tex]

    [tex]A = 2[/tex]

    Now to find B, go back to the original expression again, and multiply (again) both sides by [tex]s-2[/tex],
    then differentiate with respect to s and evaluate at [tex]s=2[/tex]:

    [tex]\frac{d}{ds}s = \frac{d}{ds}[B(s-2)][/tex]

    [tex]B = 1[/tex]

    Now plug A and B into the original expression:

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

    So the inverse Laplace would be:

    [tex]L^{-1} = [2te^{2t}+e^{2t}] u(t)[/tex]
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