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Im trying to find the inverse laplace transformation of $\frac{(2s^2 + 9s + 5)}{((s^2 - 16s + 73)(s^2 + 25))}$. when $s \gt 8$.

Here is my work: $$\begin{align} &\frac{(2s^2 + 9s + 5)}{((s^2 - 16s + 73)(s^2 + 25))} \\ &= \frac{\left(\frac{1}{272}\right) * (9s + 580)}{(s^2 - 16s + 73) - \frac{(9s + 180)}{(s^2 + 25)}} \\ &= \left(\frac{1}{272}\right) * \frac{(9s + 580)}{((s - 8)^2 + 9) - \frac{(9s + 180)}{(s^2 + 25)}}, \text{completing the square} \\ &= \left(\frac{1}{272}\right) * \frac{(9(s - 8) + 508)}{((s - 8)^2 + 9) - \frac{(9s + 180)}{(s^2 + 25)}} \end{align}$$

Inverting now yields $$\left(\frac{1}{272}\right) * e^{(8t)} (9 \cos(3t) + \left(\frac{508}{3}\right) \sin(3t)) - (9 \cos(5t) + \left(\frac{180}{5}\right) \sin(5t))$$

Please let me know what is the correct answer because this is incorrect. Appreciate it.

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    $\begingroup$ This is your $23$rd (!!) question: you should by now post using LateX! Go to the FAQ section for direction. There's a rather tiny chance somebody will ever try to read what you wrote without having his eyes hurt... $\endgroup$ – DonAntonio Mar 29 '13 at 14:51
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    $\begingroup$ Hint: You have an error in the completing the square (check that numerator of your first term). Compare your completed square and see if they match. Note, it really helps readability to format using MathJax (see FAQ). Regards $\endgroup$ – Amzoti Mar 29 '13 at 14:59
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    $\begingroup$ Kudos to Amzoti for not only getting into reading the above but actually spotting a mistake...! $\endgroup$ – DonAntonio Mar 29 '13 at 15:20
  • $\begingroup$ @DonAntonio: Trust me, my initial reaction was exactly like yours - it hurts my eyes trying to read it! I totally agree with you, he has enough postings that he should really learn how to use MathJax (note that I had the same issue and it is easy to use). Regards $\endgroup$ – Amzoti Mar 29 '13 at 15:26
  • $\begingroup$ can some tell me the correct answer. and how go i use math jax $\endgroup$ – Michael Rametta Mar 29 '13 at 21:32
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We are are asked to find:

$\tag 1 \displaystyle \mathcal{L^{-1}} \frac{2s^2 + 9s + 5}{(s^2 - 16s + 73)(s^2 + 25)}, \text{when} ~~ s \gt 8.$

We start off by doing the partial fraction expansion.

$\tag 2 \displaystyle \frac{2s^2 + 9s + 5}{(s^2 - 16s + 73)(s^2 + 25)} = \frac{1}{272}\left(\frac{9 s+580}{s^2-16 s+73}-\frac{9 (s+20)}{s^2+25}\right)$

Now, we are looking for having specific forms in order to do the ILT, and for this problem, we are going to use the following four forms:

  • $\displaystyle \mathcal{L^{-1}} \frac{a}{s^2+a^2}, s \gt 0 \rightarrow \sin (at)$

  • $\displaystyle \mathcal{L^{-1}} \frac{s}{s^2+a^2}, s \gt 0 \rightarrow \cos (at)$

  • $\displaystyle \mathcal{L^{-1}} \frac{a}{(s-b)^2 + a^2}, s \gt b \rightarrow e^{bt}\sin (at)$

  • $\displaystyle \mathcal{L^{-1}} \frac{s-b}{(s-b)^2 + a^2}, s \gt b \rightarrow e^{bt}\cos (at)$

So, our task now is to take the partial fraction expansion and write so it looks like the four forms above. We get:

$ \tag 3 \displaystyle \frac{1}{272}\left( 9 \frac{s-8}{(s-8)^2+3^2} + \frac{652}{3} \frac{3}{(s-8)^2 + 3^2} -9 \frac{s}{s^2 + 5^2} -36 \frac{5}{s^2+5^2} \right)$

Do you understand where your error is now? Expand $(3)$ and make sure it is exactly what we started with in $(2)$.

Now, since these are in the exact form we need them to be in, we can use the four $\mathcal{L^{-1}}$ above to write:

$$\displaystyle \frac{1}{272}\left(9 e^{8t}\cos (3t) + \frac{652}{3} \left(e^{8t} \sin(3t)\right) -9 \cos(5t) -36 \sin(5t)\right)$$

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  • $\begingroup$ @amWhy: Thanks again - I am always trying to improve my writing and that is hard - so appreciate the feedback! $\endgroup$ – Amzoti Mar 31 '13 at 2:11
  • $\begingroup$ You really are improving your writing, quite nicely, I may add. And I like your "conversing" with the OP as you have here. Just trust yourself ... You are doing well ;-) $\endgroup$ – Namaste Mar 31 '13 at 2:13

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