Find the inverse Laplace transform of $F(s)=\frac{e^{−6s}}{s^2+0s−16}$

Question

Find the inverse Laplace transform of $$F(s)=\frac{e^{−6s}}{s^2+0s−16}$$

Here is my work:

$$F(s)=\frac{e^{−6s}}{s^2+0s−16}$$

$$s^2+0s−16 = (s+4)(s-4)$$

$$\frac{1}{(s+4)(s-4)} = \frac{A}{s+4} + \frac{B}{s-4}$$

$$A=-\frac{1}{8}$$ $$B=\frac{1}{8}$$

$$e^{−6s}(\frac{A}{s+4}+\frac{B}{s-4})$$

$$u(t) = 0 t<6, 1 u>6$$

$$u(t)=(Ae^{-4t}+Be^{4t})$$.

How would i write this as a form $f(t)$=______?

Answer 1

From

$$\mathcal{L}_s\{f(t)\} = F(s) = \frac{e^{-6s}}{s^2 - 16} $$

we can apply, as @Amzoti has explained, two standard Laplace Transforms. Namely:

$$ \begin{equation} \frac{a}{s^2 - a^2} \to \sinh(at) \end{equation}\tag{1}$$ which you will find in the term $\frac{4}{s^2-4}$ i.e. $\sinh(4t)$

The second is the Dirac Delta, i.e.

$$ e^{-cs} \to \delta(t-c) \tag{2}$$

this you will find in the term $\frac{1}{4}e^{-6s}$, i.e. $\frac{1}4 \delta(t-6)$. Thus the entire transform will be $f(t) = \frac{1}4 \delta(t-6) \sinh(4t)$.

Answer 2

Hint: What if you write:

$$F(s)=\frac{e^{−6s}}{s^2+0s−16} = \frac{e^{−6s}}{4} \frac{4}{s^2 - 4^2}$$

Can you see you have a form for the ILT and a form for the shift theorem?