Register to reply

Inverse Laplace Transform(s/((s^2)+1)^2

by fysiikka111
Tags: inverse, laplace, transforms or s2
Share this thread:
fysiikka111
#19
Feb4-11, 06:04 PM
P: 41
So I know that differentiating both sides of the sin(at) LT will give me the square that I need in the denominator, but then I'm stuck with the d/ds on the sin side. I'm not sure what to do after this.
Thanks
hunt_mat
#20
Feb4-11, 08:15 PM
HW Helper
P: 1,583
when you differentiate the integral you take in differentiation under the integral sign:
[tex]
\frac{d}{ds}\int_{0}^{\infty}e^{-st}\sin tdt=\int_{0}^{\infty}\frac{d}{ds}\left( e^{-st}\sin t\right) dt
[/tex]
Differentiating the integrand:
[tex]
\frac{d}{ds}e^{-st}\sin t=-te^{-st}\sin t
[/tex]
As now we regard s as variable and t is fixed. So the above can be written as:
[tex]
(-t\sin t)e^{-st}
[/tex]
and so you're taking the Laplace transform of something different now. That something is the answer to your question. Is this making sense now?
fysiikka111
#21
Feb5-11, 05:25 AM
P: 41
I think I got it now.
After differentiating both sides and dividing by -2:
s/(s^2+a^2)^2=L{tsin at/2}
Hence,
L-1{s/(s^2+1)^2}=tsint/2
However, how do we know that "a" isn't -1, since (-1)^2=1.
hunt_mat
#22
Feb5-11, 12:57 PM
HW Helper
P: 1,583
well done, got there in the end.


Register to reply

Related Discussions
Finding an inverse Laplace Transform for a function - solving IVPs with Laplace Calculus & Beyond Homework 2
Inverse Laplace transform Calculus 2
Inverse Laplace Transform of ... Calculus 0
Inverse Laplace Transform Introductory Physics Homework 5
Laplace inverse transform Calculus 7