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

Re: inverse laplace transform

Re: inverse laplace transform

Quote:

Originally Posted by

**MaxJasper**

How exactly did you get to that solution? I thought you would use partial fractions in some way?

Re: inverse laplace transform

You can find InvLaplace for the following f(s) in any math handbook:

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

where is real and greater than the real part of any singularity of the function . The integral itself is unwieldy but we can use the Cauchy Residue theorem to calculate it. Assuming that the values are real, we see that the singularities of are just

.

Using the formulas for residues from simple poles we have

,

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

Re: inverse laplace transform

Re: inverse laplace transform

Re: inverse laplace transform

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