# inverse laplace transform of a transfer function

So I'm working on this problem but the $$e^{-s}$$ term is throwing me off..

$$G(s) = \frac{100(s+2)}{s(s^{2}+4)(s+1)}e^{-s}$$

Can someone help me out? I tried using partial fraction expansion to get the partial fraction and then use the table but the "e" is messing me up. Thanks

• Hint: If the Laplace transform of $f$ is $F$, what is the Laplace transform of $f(t)u(t-a)$ where $u$ is the Heavyside step function? And hence what is the inverse transform of $F(s)e^{...}$? – Simon S Sep 18 '15 at 5:00
• won't it just be $$f(s)*e$$ ? – Rickz0rz Sep 18 '15 at 5:46
• No. $F(s)e^{\color{red}{\cdots}}$, where the ellipsis $\color{red}{\cdots}$ comes from the previous part. In any case, see MV's answer below or this MIT lecture: ocw.mit.edu/courses/mathematics/… – Simon S Sep 18 '15 at 12:39

We have for the Laplace transform of $f(t-a)u(t-a)$
\begin{align} \mathscr{L}(f(t-a)u(t-a))(s)&=\int_0^{\infty}f(t-a)u(t-a)e^{-st}\,dt\\\\ &=\int_{-a}^{\infty}f(t)u(t)e^{-s(t+a)}\,dt\\\\ &=e^{-sa}\int_0^{\infty}f(t)u(t)e^{-st}\,dt\\\\ &=e^{-sa}\mathscr{L}(f(t)u(t))(s) \tag 1 \end{align}
For the problem of interest, we have $a=1$ in $(1)$ and are asked to find the inverse Laplace Transform of $G(s)$ where $G(s)$ is given by
$$G(s)=\frac{100(s+2)}{s(s+1)(s^2+4)}e^{-s}$$
Therefore, if the inverse Laplace Transform $\mathscr{L}^{-1}\left(G(s)e^{s}\right)=g(t+1)$, for some function $g(t×1)$, then the inverse Laplace Transform $\mathscr{L}^{-1}\left(G(s)\right)=g(t)$.