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Inverse Laplace Transform

  1. Apr 9, 2013 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations



    3. The attempt at a solution

    I completed the square for the denominator and got:

    (a= -2, k=1)

    But how do I get rid of the s in the numerator? Or do I have to break this up into separate functions?
  2. jcsd
  3. Apr 9, 2013 #2
    Say we have:
    [tex]F(s) = \frac{1}{(s+2)^2 + 1}[/tex]
    so you need to find [itex]\mathcal{L}^{-1}\left\{s F(s)\right\}[/itex]. Have you seen something like that in your transform tables?
  4. Apr 9, 2013 #3


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    The Laplace transform of cos(t} is [tex]\frac{s}{s^2+ 1}[/tex]. You can find that in any table of Laplace transforms.
  5. Apr 9, 2013 #4


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    Gold Member

    Also you could write$$\frac s {(s+2)^2+1}=
    \frac {(s+2)}{(s+2)^2+1}+\frac{-2}{(s+2)^2+1}$$and use the shifting theorem.
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