Calculating Laplace inverse

Question

I'm having difficulties calculating a simple Laplace inverse :

$$ \frac{S-4}{S^2-2S-11} $$

I'm new at this and couldn't find good examples for this case. could you please guide me ?

Answer 1

When Laplace transforms are rational functions, the typical line of attack is to put it into the form of partial fractions. Note that $S^2-2S-11=(S-1)^2-12$ has roots $1\pm\sqrt{12}=1\pm2\sqrt{3}$, so:

$$\frac{S-4}{S^2-2S-11}=\frac{S-4}{(S-1-2\sqrt3)(S-1+2\sqrt3)}$$

So, we try to write it in the form $\frac{A}{S-1-2\sqrt3}+\frac{B}{S-1+2\sqrt3}$, with $A,B$ constants. One way of solving for these constants is the cover-up method. We apply this here:

$$A=\frac{S-4}{S-1+2\sqrt3}\vert_{S=1+2\sqrt(3)}=\frac{-3+2\sqrt3}{4\sqrt3}=\frac{2-\sqrt3}{4}$$

$$B=\frac{S-4}{S-1-2\sqrt3}\vert_{S=1-2\sqrt(3)}=\frac{-3-2\sqrt3}{-4\sqrt3}=\frac{2+\sqrt3}{4}$$

Thus, our function's Laplace transform can be written as $\frac{1}{4}(\frac{2-\sqrt3}{S-1+2\sqrt3}+\frac{2+\sqrt3}{S-1-2\sqrt3})$.

We now recall that $e^{at}$ has Laplace transform $\frac{1}{s-a}$, and this allows us to recover our function as:

$$f(t)=\frac{1}{4}((2-\sqrt3)e^{(1-2\sqrt3)t}+(2+\sqrt3)e^{(1+2\sqrt3)t})$$