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Inverse Laplace Transform

  1. Dec 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Find [tex] H(s) = \frac{Y(s)}{X(s)} [/tex]
    [tex]\frac {d^2y(t)}{dt^2} + a\frac {dy(t)}{dt} = x(t) + by(t)[/tex]

    2. Relevant equations


    3. The attempt at a solution
    [tex][s^2 + as - b] Y(s) = X(s)[/tex]
    [tex] H(s) = \frac{1}{s^2+as-b} [/tex]

    I assume the inverse is a sign or a cosine but unsure which one.
     
  2. jcsd
  3. Dec 7, 2014 #2

    Ray Vickson

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    If ##Y(s)## is the LT of ##y(t)##, what are the LTs of ##y'(t)## and ##y''(t)##?

    Hint: not what you wrote.
     
  4. Dec 7, 2014 #3
    You can use the Bromwich integral to find the inverse Laplace transform.
    The poles of ##s## are ##s = \frac{-a\pm\sqrt{a^2 + 4b}}{2}##
    \begin{align}
    \mathcal{L}^{-1}\Bigl\{\frac{1}{s^2 + as - b}\Bigr\} &= \frac{1}{2\pi i}\int_{\gamma -i\infty}^{\gamma +i\infty}\frac{e^{st}}{s^2 + as - b}ds\\
    &= \sum\text{Res}\\
    &= \lim_{s\to s_+}(s - s_+)\frac{e^{st}}{s^2 + as - b} + \lim_{s\to s_-}(s - s_-)\frac{e^{st}}{s^2 + as - b}
    \end{align}
    where ##s_+=\frac{-a + \sqrt{a^2 + 4b}}{2}## and ##s_-=\frac{-a - \sqrt{a^2 + 4b}}{2}##
     
  5. Dec 7, 2014 #4
    The LT of ##y'(t)## and ##y''(t)## are sY[s} and ##s^2Y## respectively. Im not sure what you mean exactly.
     
  6. Dec 7, 2014 #5
    The region of convergence then, would it be btw the two values of s?
     
  7. Dec 7, 2014 #6
    You need to find y(t) for the ROC.
     
  8. Dec 7, 2014 #7

    Ray Vickson

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    What I mean is that your statements above are false, in general. Consult a table of Laplace transforms to see what you should have written.
     
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