1. ## Inverse Laplace Transform

Hello,

Find the inverse Laplace Transform

$F(s) = \frac{s}{2s^2 + 2s +1}$

2. Originally Posted by funnyinga
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?