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Inverse Laplace Transform

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Division by s Equals integration by t:
    For this problem use the following property (see relevant equations) to find the inverse transform of the given function: [itex] F(s) = \frac{1}{s(s-1)} [/itex]

    2. Relevant equations
    [itex] L^{-1}(\frac{F(s)}{s}) = \int_{0}^{t} f(\tau)\,d \tau [/itex]

    3. The attempt at a solution
    Using the above formula I have: [itex] F(s) = \frac{1}{s(s-1)} [/itex]
    So : [itex] \frac{F(s)}{s} = \frac{\frac{1}{s(s-1)}}{s} = \frac{1}{s^{2}(s-1)} [/itex]
    Next I used a partial fraction decomposition:
    [itex] \frac{1}{s^{2}(s-1)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{(s-1)} [/itex]
    [itex] As(s-1) + B(s-1) + Cs^{2} = 1 [/itex]
    Solving for A, B, and C I found [itex] A=-1 \hspace{1 mm} B=-1 \hspace{1 mm} C = 1 [/itex]
    So the decomposition is:
    [itex] \frac{1}{s^{2}(s-1)} = -\frac{1}{s} - \frac{1}{s^{2}} + \frac{1}{(s-1)} [/itex]
    The inverse Laplace transform of the above equation is:
    [itex] -t - e^{t} - 1 [/itex]
    However, the answer given in my textbook is:
    [itex] e^{t} - 1 [/itex]
    Coincidentally, when I perform the partial fraction decomposition on [itex] F(s) [/itex], I arrive at the answer given in the back of the text, which is really confusing me, do I not need to divide [itex]F(s)[/itex] by [itex] s [/itex]?
     
    Last edited: Nov 14, 2014
  2. jcsd
  3. Nov 14, 2014 #2

    LCKurtz

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    You have stated the formula wrong. The formula you want is$$
    \mathcal L^{-1}\frac 1 s F(s) = \int_0^t f(\tau)~d\tau$$where ##f## is the inverse of ##F##. In your problem ##F(s) =\frac 1 {s-1}##. Informally, this says if you suppress a factor of ##\frac 1 s## you must integrate the result to get the original inverse.
     
  4. Nov 14, 2014 #3
    ah, thank you very much, the problem makes much more sense now. I am blaming my book though, the formula they provided isn't very clear, they formatted it in the same way that I did above...although to be fair I guess I just completely misinterpreted the problem :)
     
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