# Laplace transform combination

## Homework Statement

Not a homework problem exactly, but in an EE textbook I saw something to the following effect:

To obtain the time domain behavior for $$\frac{s}{s^2+\frac{\omega_0}{Q} + {\omega_0}^2}$$ the following Laplace transforms are combined to cancel the term in the numerator:

$$\frac{1}{(s-\alpha)^2+b^2} = \frac{1}{b}e^{\alpha t}sin(bt)$$

and

$$\frac{s+\alpha}{(s-a)^2 + b^2} = e^{\alpha t}cos(bt)$$

I may just be tired, but I'm having a lot of trouble seeing how those two Laplace transforms were algebraically combined to get "s" alone in the numerator. Any hint would be appreciated.

## Answers and Replies

gabbagabbahey
Homework Helper
Gold Member
$$\frac{s}{(s-\alpha)^2 + b^2}=\frac{s+\alpha-\alpha}{(s-\alpha)^2 + b^2}$$ Aha! I see it now. Need to remember my "exponential shift" rule.

So we get $$\frac{s+\alpha-\alpha}{(s-\alpha)^2 + b^2} = \frac{s-\alpha}{(s-\alpha)^2 + b^2} + \frac{\alpha}{(s-\alpha)^2 + b^2} = e^{\alpha t}L^{-1}(\frac{s}{s^2+b^2} + \alpha\frac{1}{s^2+b^2})$$

which I believe is the Laplace transform of

$$e^{\alpha t}*(cos(bt) + \frac{\alpha}{b}sin(bt))$$

Thanks!