does anybody know the inverse laplace transform of (5+s)/(s^2 +2s) , to revert to a function of t again....
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}
$
...
(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)