# Thread: inverse laplace transform

1. ## inverse laplace transform

I have the laplace transform:

x(s) = 1 / ((s + a)^2 + (w^2 - a^2))

what is the inverse laplace if w^2 < a^2 ????

I have tried attempting the question without any luck. Step by step solution showing how it is solved would be very helpful for me to understand

2. ## Re: inverse laplace transform

$x(t)=-\frac{e^{t \left(-a-\sqrt{a^2-w^2}\right)}-e^{t \left(-a+\sqrt{a^2-w^2}\right)}}{2 \sqrt{a^2-w^2}}$

3. ## Re: inverse laplace transform

Originally Posted by MaxJasper
$x(t)=-\frac{e^{t \left(-a-\sqrt{a^2-w^2}\right)}-e^{t \left(-a+\sqrt{a^2-w^2}\right)}}{2 \sqrt{a^2-w^2}}$
How exactly did you get to that solution? I thought you would use partial fractions in some way?

4. ## Re: inverse laplace transform

You can find InvLaplace for the following f(s) in any math handbook:
$f(s)=\mathcal{L}_s^{-1}\left[\frac{1}{(a+s)^2+b^2}\right](t)$

$F(t)=\frac{e^{-a t} \text{Sin}[b t]}{b}$

5. ## Re: inverse laplace transform

Or you can do it the hard way using complex analysis. If you know some complex analysis, read on. The inverse Laplace transform is the following

$\mathcal{L}^{-1} \{x(s)\} = x(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{st}x(s)\,ds$

where $\gamma$ is real and greater than the real part of any singularity of the function $x(s)$. The integral itself is unwieldy but we can use the Cauchy Residue theorem to calculate it. Assuming that the values $a,w$ are real, we see that the singularities of $x(s)$ are just

$s_1 = -a + \sqrt{a^2-w^2},\quad s_2 = -a - \sqrt{a^2-w^2}$.

Using the formulas for residues from simple poles we have

$x(t) = \frac{e^{s_2 t}}{(s_2+a)-\sqrt{a^2-w^2}} + \frac{e^{s_1 t}}{(s_1+a)+\sqrt{a^2-w^2}}$,

which after simplifying leads to the expression given by MaxJasper. If you want I can elaborate on any of the parts.

6. ## Re: inverse laplace transform

Is this for w^2 < a^2?

Yes

8. ## Re: inverse laplace transform

And what about the answer in the case w^2 = a^2??