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
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 $\mathscr{L}^{-1}\left(G(s)\right)=g(t)$.