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

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 $f(t)$=______?

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 $f(t) = \frac{1}4 \delta(t-6) \sinh(4t)$.

• what does that symbol represent – Michael Rametta Apr 4 '13 at 1:27
• could where would you put the step in this equation – Michael Rametta Apr 4 '13 at 1:50
• the $\delta$ symbol? it represents an impulse function. you can learn more about it here. en.wikipedia.org/wiki/Dirac_delta_function – franklin Apr 4 '13 at 2:05
• can you represent it as a step by function by putting "step" wherever necessary – Michael Rametta Apr 4 '13 at 2:38
• not sure what you're asking @MichaelRametta the function is best represented as an impulse function. a step function implies that at certain regions or intervals the function behaves differently. an impulse function implies that at a certain instant the function's amplitude, if you will, dramatically differs. – franklin Apr 4 '13 at 15:24

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?

• Yea i see that but i still am lost – Michael Rametta Apr 3 '13 at 23:20
• What is the inverse ILT for the second item? Now, what is the shift theorem telling you to do with the values of t? Clear? – Amzoti Apr 3 '13 at 23:24
• so was i completly wrong with the way i tried above? – Michael Rametta Apr 3 '13 at 23:58
• "completely wrong" is a pretty strong phrase. the way that @Amzoti is doing it is just easier. it allows you to apply this "shift theorem" or more formally the unit impulse function or Dirac's Delta. – franklin Apr 4 '13 at 0:21
• @MichaelRametta: I agree with franklin. I am not saying this is wrong - we always hope to have multiple ways of solving problems (although that doesn't always work). The approach I took is more suited to what you did with the shift problem you posed earlier - so I thought that is the way one typically learns it first, before exploring other approaches. Regards – Amzoti Apr 4 '13 at 0:30