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How does one find the Inverse Laplace transform of $$\frac{6s^2 + 4s + 9}{(s^2 - 12s + 52)(s^2 + 36)}$$ where $s > 6$?

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A related problem. –  Mhenni Benghorbal Apr 18 '13 at 3:02

2 Answers 2

up vote 1 down vote accepted

You need to find the poles of the expression; in your case, you have poles at $s=6 \pm 4 i$ and $s=\pm 6 i$. You then find what are called the residues of the LT times $e^{s t}$ at the poles. The residue at a pole $s_k$ is

$$\lim_{s \rightarrow s_k} \left [ (s-s_k) \frac{6 s^2 + 4 s+9}{(s^2-12 s+52)(s^2+36)} e^{s t} \right ]$$

Not sure what the $s>6$ thing is; it is meaningless in this context.

For each pole, the corresponding residues are:

$$s_1=6 i \implies \frac{9 (6 i)^2 + 4 (6 i) + 9}{((6 i)^2 - 12 (6 i) + 52) (12 i)} e^{i 6 t}$$

$$s_2=-6 i \implies \frac{9 (-6 i)^2 + 4 (-6 i) + 9}{((-6 i)^2 - 12 (-6 i) + 52) (-12 i)} e^{-i 6 t}$$

$$s_3=6 + 4 i \implies \frac{9 (6 + 4 i)^2 + 4 (6 + 4 i) + 9}{(8 i) ((6+4 i)^2+36)} e^{(6 + 4 i)t}$$

$$s_3=6 - 4 i \implies \frac{9 (6 - 4 i)^2 + 4 (6 - 4 i) + 9}{(-8 i) ((6-4 i)^2+36)} e^{(6 - 4 i)t}$$

The ILT is then the sum of these residues. I leave the arithmetic/algebra to you; I get as the ILT

$$-\frac{21}{136} \sin (6 t)-\frac{121}{272} \cos (6 t)+e^{6 t} \left(\frac{579}{544} \sin (4 t)+\frac{121}{272} \cos (4 t)\right)$$

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I think that the s > 6 is there to prevent the imaginary values from appearing. I understand your logic, but this s > 6 supposedly is in place to keep the answer in terms of only cos, sin, etc., not i. Thanks though! –  Scottaveus Apr 18 '13 at 3:11
    
Use $e^{i 6 t} = \cos{6 t} + i \sin{6 t}$, etc. –  Ron Gordon Apr 18 '13 at 3:28

Hints:

  1. Write out the partial fraction expansion.

  2. Put the partial fraction into the forms that let you use the inverse table.

Why do they put the restriction on s (you'll see it if you do 1 and 2)?

Clear?

Update

We are given and asked to find the Inverse Laplace Transform of:

$\displaystyle \frac{6s^2 + 4s + 9}{(s^2 - 12s + 52)(s^2 + 36)}$ where $s > 6$.

We have (after some algebra and putting things into forms we can work with):

$\displaystyle \frac{6s^2 + 4s + 9}{(s^2 - 12s + 52)(s^2 + 36)} = -\frac{121}{272}\left(\frac{s}{s^2 + 6^2}\right) - \frac{42}{272}\left(\frac{6}{s^2 + 6^2}\right) + \frac{121}{272} \left(\frac{s - 6}{((s-6)^2 + 4^2}\right) + \frac{147}{2 \times 272} \left(\frac{4}{((s-6)^2 + 4^2}\right)$

Do you see why the condition $s \gt 6$ was given now?

Can you now find the ILT of each of those expressions?

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Nice work and demonstrations! –  amWhy Apr 18 '13 at 4:07

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