# inverse laplace transform

• Aug 20th 2012, 08:01 AM
NFS1
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
• Aug 20th 2012, 08:46 AM
MaxJasper
Re: inverse laplace transform
$\displaystyle x(t)=-\frac{e^{t \left(-a-\sqrt{a^2-w^2}\right)}-e^{t \left(-a+\sqrt{a^2-w^2}\right)}}{2 \sqrt{a^2-w^2}}$
• Aug 20th 2012, 05:26 PM
NFS1
Re: inverse laplace transform
Quote:

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

How exactly did you get to that solution? I thought you would use partial fractions in some way?
• Aug 20th 2012, 06:19 PM
MaxJasper
Re: inverse laplace transform
You can find InvLaplace for the following f(s) in any math handbook:
$\displaystyle f(s)=\mathcal{L}_s^{-1}\left[\frac{1}{(a+s)^2+b^2}\right](t)$

$\displaystyle F(t)=\frac{e^{-a t} \text{Sin}[b t]}{b}$
• Aug 21st 2012, 12:53 AM
Vlasev
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

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

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

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

Using the formulas for residues from simple poles we have

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

which after simplifying leads to the expression given by MaxJasper. If you want I can elaborate on any of the parts.
• Aug 21st 2012, 09:07 AM
NFS1
Re: inverse laplace transform
Is this for w^2 < a^2?
• Aug 21st 2012, 10:18 AM
Vlasev
Re: inverse laplace transform
Yes
• Aug 22nd 2012, 09:54 AM
NFS1
Re: inverse laplace transform