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Math Help - Inverse Laplace Transform

  1. #1
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    Inverse Laplace Transform

    Hello,

    Find the inverse Laplace Transform

     F(s) = \frac{s}{2s^2 + 2s +1}
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by funnyinga View Post
    Hello,

    Find the inverse Laplace Transform

     F(s) = \frac{s}{2s^2 + 2s +1}
    Note that this is the same as F(s)=\frac{1}{2}\cdot\frac{s}{(s^2+s)+\tfrac{1}{2}  }.

    Completeing the square in the denominator yields F(s)=\frac{1}{2}\cdot\frac{s}{\left(s+\tfrac{1}{2}  \right)^2+\tfrac{1}{4}}.

    Therefore, the inverse Laplace Transform is

    \mathcal{L}^{-1}\left\{\frac{1}{2}\cdot\frac{s}{\left(s+\tfrac{1  }{2}\right)^2+\tfrac{1}{4}}\right\}=\frac{1}{2}\ma  thcal{L}^{-1}\left\{\frac{s+\tfrac{1}{2}}{\left(s+\tfrac{1}{2  }\right)^2+\tfrac{1}{4}}\right\}-\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{\tfrac{1}{2}}{\left(s+\tfrac{1}{2}\  right)^2+\tfrac{1}{4}}\right\} =\tfrac{1}{2}e^{-\frac{1}{2}t}\left[\cos\left(\tfrac{1}{2}t\right)-\sin\left(\tfrac{1}{2}t\right)\right] (via translation theorems).

    Does this make sense?
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