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Math Help - inverse laplace transform

  1. #1
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    inverse laplace transform

    does anybody know the inverse laplace transform of (5+s)/(s^2 +2s) , to revert to a function of t again....
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  2. #2
    Senior Member Peritus's Avatar
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    partial fraction expansion:

    <br />
\frac{{5 + s}}<br />
{{s(s + 2)}} = \frac{5}<br />
{{s(s + 2)}} - \frac{1}<br />
{{s + 2}} = \frac{a}<br />
{s} + \frac{b}<br />
{{s + 2}} + \frac{1}<br />
{{s + 2}}<br />

    I'll assume that you'll be able to find the constants a and b.

    Now all we need to do is look at the transform table (I'll assume that the TF is casual) :

    <br />
\begin{gathered}<br />
  L\left( {e^{at} u(t)} \right) = \frac{1}<br />
{{s - a}} \hfill \\<br />
  L\left( {u(t)} \right) = \frac{1}<br />
{s} \hfill \\ <br />
\end{gathered} <br />

    ...
    Last edited by Peritus; March 19th 2008 at 12:39 PM.
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  3. #3
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    is that meant to be 5/s(s+2) + 1/s+2
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  4. #4
    Member Danshader's Avatar
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    (5+s)/(s^2 + 2s)
    = 5/(s^2 + 2s) + s/(s^2 + 2s)
    = 5/s(s + 2) + s/s(s + 2)
    = 5/s(s + 2) + 1/(s + 2)

    applying partial fraction to the first term we get,

    = 5/2s -5/2(s+2) + 1/(s+2)

    applying inverse Laplace transform (using the table)

    =5/2 -5 exp(-2t)/2 + exp(-2t)
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