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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- Mar 19th 2008, 07:32 AMjohnbarkwithinverse laplace transform
does anybody know the inverse laplace transform of (5+s)/(s^2 +2s) , to revert to a function of t again....

- Mar 19th 2008, 08:09 AMPeritus
partial fraction expansion:

$\displaystyle

\frac{{5 + s}}

{{s(s + 2)}} = \frac{5}

{{s(s + 2)}} - \frac{1}

{{s + 2}} = \frac{a}

{s} + \frac{b}

{{s + 2}} + \frac{1}

{{s + 2}}

$

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) :

$\displaystyle

\begin{gathered}

L\left( {e^{at} u(t)} \right) = \frac{1}

{{s - a}} \hfill \\

L\left( {u(t)} \right) = \frac{1}

{s} \hfill \\

\end{gathered}

$

... - Mar 19th 2008, 09:37 AMjohnbarkwith
is that meant to be 5/s(s+2) + 1/s+2

- Mar 29th 2008, 11:41 PMDanshader
(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)