$\displaystyle \mathcal{L}^{-1} \left[ F(s) \cdot e^{-cs} \right] = f(t-c) \, \alpha(t-c)$ where $\displaystyle f(t)=\mathcal{L}^{-1} [ F(s) ]$ and $\displaystyle \alpha$ is the unit step function.
$\displaystyle \mathcal{L}^{-1} \left[ F(s) \cdot e^{-cs} \right] = f(t-c) \, \alpha(t-c)$ where $\displaystyle f(t)=\mathcal{L}^{-1} [ F(s) ]$ and $\displaystyle \alpha$ is the unit step function.
So is $\displaystyle \mathcal{L}^{-1}[F(s)] = \mathcal{L}^{-1}[\frac{1}{s-3}]= e^{3t}$
where t = (t-3)
$\displaystyle =e^{(t-3)}H_3 (t)$
Last edited by Paymemoney; Apr 9th 2011 at 02:58 AM.