Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do I compute the following transform?
$$\frac {s-1}{2s^2+s+6}$$

I've gotten this far:

$$\frac {1}{2}\cdot \frac {s-1}{\left(s+\frac{1}{4}\right)^2 + \frac{47}{16}}$$

share|cite|improve this question
You mean inverse transform, right? – Pedro Tamaroff Nov 21 '12 at 3:21
Yep sorry about that, I'll edit it. – Shankar Kumar Nov 21 '12 at 3:22
up vote 1 down vote accepted

You can now use: $$ \mathcal{L}_s\left(\mathrm{e}^{-\lambda t} \mathrm{e}^{i t \omega}\right) = \int_0^\infty \mathrm{e}^{-s t} \mathrm{e}^{-\lambda t} \mathrm{e}^{i t \omega} \mathrm{d} t = \frac{1}{s + \lambda - i \omega} $$ valid as long as $s+\lambda > 0$, and $\omega \in \mathbb{R}$. From here, taking real and imaginary parts you conclude: $$ \mathcal{L}_s\left(\mathrm{e}^{-\lambda t} \cos\left(\omega t\right)\right) = \frac{s+\lambda}{(s+\lambda)^2 + \omega^2}, \qquad \mathcal{L}_s\left(\mathrm{e}^{-\lambda t} \sin\left(\omega t\right)\right) = \frac{\omega}{(s+\lambda)^2 + \omega^2} $$ Now you the decomposition you obtained and read off the coefficients, keeping in mind that the inverse Laplace transform has the form $\mathrm{e}^{-\lambda t} \left( \alpha \cos(\omega t) + \beta \sin(\omega t) \right)$ for some $\lambda, \omega, \alpha$ and $\beta$.

share|cite|improve this answer

$$ F(s)=\frac {s-1}{2s^2+s+6}=\frac {s-1}{2(s^2+\frac{s}{2} +3)} $$ $$ ax^2+bx=a\left[ (x+\frac{b}{2a})^2-(\frac{b}{2a})^2 \right] $$ So: $$ 2(s^2+\frac{s}{2} +3)=2 \left[ (s+\frac{1}{4})^2-(\frac{1}{4})^2 +3 \right]=2 \left[ (s+\frac{1}{4})^2+(\frac{47}{15}) \right] $$ therefor: $$ F(s)=(\frac{1}{2}) \frac{s-1 {\color{red} { +\frac{5}{4} -\frac{5}{4}} } }{(s+\frac{1}{4})^2+(\frac{47}{15})} = (\frac{1}{2}) \left[ \frac{s+\frac{1}{4}}{(s+\frac{1}{4})^2+(\frac{47}{15})}-\frac{\frac{5}{4}}{(s+\frac{1}{4})^2+(\frac{47}{15})} \right] $$ We know: $$ \mathcal{L}\left[ e^{-at}cos(bt) \right]=\frac{s+a}{(s+a)^2+b^2} $$ $$ \mathcal{L}\left[ e^{-at}sin(bt) \right]=\frac{b}{(s+a)^2+b^2} $$ therefor: $$ f(t)=(\frac{1}{2})e^{\frac{-t}{4}} \left[ cos(\frac{\sqrt{47}}{4}t) - \frac{5}{\sqrt{47}} sin(\frac{\sqrt{47}}{4}t)\right] $$

share|cite|improve this answer

Your Answer


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.