Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the information below correct?

Find the inverse Laplace transform of $$ F(s) = \frac{s}{s^2 + 4s + 13}$$

Soln: a) Complete the squares to simplify our denominator $$ s^2 + 4s + 13 = (s+2)^2 + 9 = (s+2)^2 + 3^2$$ $$\mathscr{L}^{-1}\left\{F(s)\right\} = \frac{s}{(s+2)^2 + 3^2}. $$ From the table we can deduce that this is $$\mathscr{L}^{-1}\left\{F(s)\right\} = e^{-2t} \cos(3t).$$

share|improve this question
1  
(+1) for showing your work, among other things. Don't forget to "accept" an answer when it's all said and done! You can also revisit old questions of yours and accept answers, if you're into the whole "closure" thing :) –  The Chaz 2.0 Jul 20 '11 at 4:17

2 Answers 2

HINT: $$ \frac{s}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2 - 2}}{{(s + 2)^2 + 3^2 }} = \frac{{s + 2}}{{(s + 2)^2 + 3^2 }} - \frac{2}{3}\frac{3}{{(s + 2)^2 + 3^2 }}. $$ Now note that the inverse transforms of $\frac{{s + \alpha }}{{(s + \alpha )^2 + \omega ^2 }}$ and $\frac{\omega}{{(s + \alpha)^2 + \omega^2 }}$ are $e^{-\alpha t} \cos(\omega t)$ and $e^{-\alpha t} \sin (\omega t)$, respectively.

share|improve this answer
2  
Shoot!!! I see what you mean. Thanks ! –  user10695 Jul 20 '11 at 3:04

For rational functions you can use decomposition to partial fractions: $$\frac{s}{(s+2)^2+3^2}=\frac{s}{(s+2+3i)(s+2-3i)}=\frac{1/2+i/3}{s+2-3i}+\frac{1/2-i/3}{s+2+3i}.$$ The only thing you need to know now is that the inverse Laplace transform of $1/(s+\alpha)$ is $\exp(-\alpha x)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.