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:
![F(t)=\frac{e^{-a t} \text{Sin}[b t]}{b}](http://latex.codecogs.com/png.latex?F(t)=\frac{e^{-a t} \text{Sin}[b t]}{b})
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
\} = x(t) = \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma-iT}^{\gamma+iT}e^{st}x(s)\,ds)
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??