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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

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  • $\begingroup$ 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 $F(s)e^{...}$? $\endgroup$
    – Simon S
    Sep 18, 2015 at 5:00
  • $\begingroup$ won't it just be $$f(s)*e$$ ? $\endgroup$
    – Rickz0rz
    Sep 18, 2015 at 5:46
  • $\begingroup$ 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/… $\endgroup$
    – Simon S
    Sep 18, 2015 at 12:39

1 Answer 1

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

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    $\begingroup$ There are a couple of sign errors here $\endgroup$
    – Simon S
    Sep 18, 2015 at 12:40
  • $\begingroup$ @simon +1 for the good catch. I've edited. $\endgroup$
    – Mark Viola
    Sep 18, 2015 at 14:01

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