Find the inverse Laplace transform of $X(S)= \frac{2+2s e^{-2s}+4e^{-4s}}{s^2+4s+3}$ with $\Re(s)>-1$

Question

Find the inverse Laplace transform of

$$X(S)= \frac{2+2s e^{-2s}+4e^{-4s}}{s^2+4s+3} \qquad \Re(s)>-1$$

I never learned how to use this in class and so I've seen a couple youtube videos however they are too easy and not as complex as this one.

I am aware that this is a step function so therefore it will involve this property: $$e^{-cs}F(s)= f(t-c)u(t-c)$$

So far I factored the denominator and separated the numerator and have:

$$\frac{2}{(s+1)(s+3)} + \frac{2se^{-2s}}{(s+1)(s+3)} + e^{4s}\frac{4}{(s+1)(s+3)}$$

Anything like references to problems similar as this one would help me.

Answer 1

Hint: Rewrite your last equation as

$$\frac{2}{(s+1)(s+3)} + e^{-2s}\frac{2s}{(s+1)(s+3)} + e^{-4s}\frac{4}{(s+1)(s+3)}$$

and then do partial fraction decomposition on all three terms. This will form

$$\left(\frac{1}{s+1}-\frac{1}{s+3}\right)+e^{-2s}\left(\frac{3}{s+3}-\frac{1}{s+1}\right)+e^{-4s}\left(\frac{2}{s+1}-\frac{2}{s+3}\right)$$

from which you can now use the property $e^{-cs}F(s)= \mathcal{L}\{f(t-c)u(t-c)\}(s)$.