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

Here is my work:


$$s^2+0s−16 = (s+4)(s-4)$$

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

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


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


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



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

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

  • $\begingroup$ what does that symbol represent $\endgroup$ – Michael Rametta Apr 4 '13 at 1:27
  • $\begingroup$ could where would you put the step in this equation $\endgroup$ – Michael Rametta Apr 4 '13 at 1:50
  • $\begingroup$ the $\delta$ symbol? it represents an impulse function. you can learn more about it here. en.wikipedia.org/wiki/Dirac_delta_function $\endgroup$ – franklin Apr 4 '13 at 2:05
  • $\begingroup$ can you represent it as a step by function by putting "step" wherever necessary $\endgroup$ – Michael Rametta Apr 4 '13 at 2:38
  • $\begingroup$ 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. $\endgroup$ – 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?

  • $\begingroup$ Yea i see that but i still am lost $\endgroup$ – Michael Rametta Apr 3 '13 at 23:20
  • $\begingroup$ 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? $\endgroup$ – Amzoti Apr 3 '13 at 23:24
  • $\begingroup$ so was i completly wrong with the way i tried above? $\endgroup$ – Michael Rametta Apr 3 '13 at 23:58
  • $\begingroup$ "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. $\endgroup$ – franklin Apr 4 '13 at 0:21
  • $\begingroup$ @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 $\endgroup$ – Amzoti Apr 4 '13 at 0:30

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