Inverse Laplace Transformation, Possible Step Function

Question

I have been asked to find the inverse Laplace transform of the following:

\begin{align} F(s)=&\frac{(s-2)e^{-s}}{(s^2 -4s+3)} \end{align} I am having trouble breaking this down into easily transformed fractions. Obviously the denominator can be changed to $(s-3)(s-1)$, but I'm not sure where to go from that.

Any hints or suggestions?

Answer 1

HINT:

$$\frac{s-2}{(s-1)(s-3)}=\frac{1/2}{s-3}+\frac{1/2}{s-1}$$

Answer 2

Well, using the convolution theorem:

$$\mathscr{L}_\text{s}^{-1}\left[\frac{1}{e^\text{s}}\cdot\frac{\text{s}-2}{\text{s}^2-4\cdot\text{s}+3}\right]_{\left(t\right)}=\mathscr{L}_\text{s}^{-1}\left[\frac{1}{e^\text{s}}\right]_{\left(t\right)}\space*\space\mathscr{L}_\text{s}^{-1}\left[\frac{\text{s}-2}{\text{s}^2-4\cdot\text{s}+3}\right]_{\left(t\right)}\tag1$$

Now, we also know:

$$\frac{\text{s}-2}{\text{s}^2-4\cdot\text{s}+3}=\frac{1}{2}\cdot\left\{\frac{1}{\text{s}-1}+\frac{1}{\text{s}-3}\right\}\tag2$$

So, we get:

$$\mathscr{L}_\text{s}^{-1}\left[\frac{1}{e^\text{s}}\right]_{\left(t\right)}\space*\space\left\{\frac{1}{2}\cdot\left(\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}-1}\right]_{\left(t\right)}+\mathscr{L}_\text{s}^{-1}\left[\frac{1}{\text{s}-3}\right]_{\left(t\right)}\right)\right\}\tag3$$