Question

Find the inverse Laplace transform of
F(S)= (2e^(-3s))/(s(s-1)(s-2))

Answer 1

\[ F(S)= \frac{ 2e^{-3s} }{s(s-1)(s-2) }\]

Answer 2

First thing I would take care of is the exponential factor using the formula,
\[\mathcal{L}\{u(t-c)f(t-c)\}=e^{-cs}\mathcal{L}\{f(t)\}\]

Answer 3

Applying this, you have
\[\mathcal{L}^{-1}\left\{\frac{2e^{-3s}}{s(s-1)(s-2)}\right\}=u(t-3)f(t-3)\]
where \(f(t)=\mathcal{L}^{-1}\{F^*(s)\}\), with \(F^*(s)=\dfrac{1}{s(s-1)(s-2)}\).

As for finding \(f(t)\), break up into partial fractions.

Answer 4

@SithsAndGiggles
I am confused by your last part. How did you get the right side?
\[\mathcal{L}^{-1}\left\{\frac{2e^{-3s}}{s(s-1)(s-2)}\right\}=u(t-3)f(t-3) \]

Answer 5

\[\mathcal{L}\{u(t-c)f(t-c)\}=e^{-cs}\mathcal{L}\{f(t)\}\]
implies
\[u(t-c)f(t-c)=\mathcal{L}^{-1}\left\{e^{-cs}\mathcal{L}\{f(t)\}\right\}\]
In this case, you have \(c=3\), and
\[\color{red}{e^{-cs}}\color{blue}{\mathcal{L}\{f(t)\}}=\frac{\color{blue}{2}\color{red}{e^{-3s}}}{\color{blue}{s(s-1)(s-2)}}\]

Answer 6

Makes much sense now. Thanks