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How would you find the inverse Laplace transformation of $\displaystyle \frac{3s+4}{s^2-16}$ when $s>4$? Thanks!!

I dont really understand what we need to do for this question. Please help

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  • $\begingroup$ See here for different techniques. $\endgroup$ – Mhenni Benghorbal Mar 25 '13 at 2:47
  • $\begingroup$ You really do not want to restrict $s$ to avoid the pole. $\endgroup$ – Ron Gordon Mar 25 '13 at 2:49
  • $\begingroup$ @MichaelRametta: if you have an issue with what you posted above and the DEQ, that sounds like a new question. You can post the ODE and your solution and we can figure out where it went wrong. Regards $\endgroup$ – Amzoti Mar 27 '13 at 21:49
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Hint:

Write the partial fraction fraction expansion as:

$$\displaystyle \tag 1 \frac{3s+4}{s^2-16} = \frac{1}{s+4} + \frac{2}{s-4}$$

Now, take the inverse Laplace of each of the terms on the right-hand-side (RHS) of $(1)$.

We have for $s \gt a$:

$$\mathcal{L}^{-1}\left(\frac{1}{s-a}\right) = e^{at}$$

Clear?

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  • $\begingroup$ +1. Have you seen this one? math.stackexchange.com/q/341507/8581 $\endgroup$ – mrs Mar 26 '13 at 14:49
  • $\begingroup$ I hadn't, but now it has my curiosity up! Thanks for the reference! $\endgroup$ – Amzoti Mar 26 '13 at 15:19
  • $\begingroup$ so what would this answer be $\endgroup$ – Michael Rametta Mar 27 '13 at 19:46
  • $\begingroup$ $e^{-4t}$ for the positive term and $e^{4t}$ for the negative term. $\endgroup$ – Amzoti Mar 27 '13 at 19:48
  • $\begingroup$ Great i got i got the same answers thanks $\endgroup$ – Michael Rametta Mar 27 '13 at 20:09

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