Hi!

Need your help. Can anyone pls help me find the inverse Laplace transform of the following:

<font color=pink>Y(s) = s²/(s⁴-9)
</font>

y(t) = ?

Can be done easily if you have Matlab (I dont :( ). Or if you want to compute it manually go ahead.

Haha, good luck getting help on that here. [img]graemlins/hehe.gif[/img] Thankfully I've not had to deal with laplace transforms for at least 2 years now and hopefully never again :D

I've got matlab but no idea how to get it to do a laplace transform

Whenever I get these sorts of problems I usually go look at a table.

http://eqworld.ipmnet.ru/en/auxiliary/aux-inttrans.htm

Thanks.
I already took look at tables. and tried to solve. But this one is difficult. second power in numerator :( . Tables wont help. Thanks anyway.

[ 06-29-2005, 03:14 PM: Message edited by: ZFR ]

Much the same as Vasky I'm afraid - ever since Uni I have spent my life running a refinery and avoiding the hell out of Laplace and his blimmin transforms. Laplace is up towards the top of my set of most most hated names - along with Jacobi, Wronski, and Fourier. Man I am glad that advanced numerical methods are not actually used much at all by engineers in real life ;) .

<font color = lightgreen>Let me get back home from work this evening and I'll have an answer for you. [img]graemlins/petard.gif[/img]

On second thought I decided not to wait.

Substitute k = s^2 so that the problem becomes k/(k^2 - 9) and a quick check of an online Laplace table shows the transform of <font color = white>cosh (at) for k > abs(a)</font>. Since k = s^2, this becomes s > abs(sqrt(a)) or in this case s > abs(sqrt(3)).


For those who haven't studied engineering, the hyperbolic cosine has the interesting property that the transverse forces acting on a hyperbolic cosine curve turn out to be zero. That is to say, the only sources of stress occur in the same dimension as the line of symmetry so that if an arch is built in the form of a hyperbolic cosine only gravity exerts stress on the arch...like the Gateway Arch in St. Louis.</font>

[ 06-29-2005, 09:16 PM: Message edited by: Azred ]

...Man o man. I wonder when I'll have to know THAT stuff. I'm in Calc IV this fall, and have yet to have to study hyperbolic trig...

But then again, being a Physics major kind of demands it, doesn't it?

<font color = lightgreen>Yes, it does. Don't worry--it's easy. [img]graemlins/petard.gif[/img] </font>

Sorry Azred solution seems wrong...

if we have Lapace

Y(s) = 1/s
y(t) = 1

and

Y(s) = 1/s^2
y(t) = t

so in second case you cant use substitution let k = s^2 and expect it to become Y(k) = 1/k hence solution is again 1.

found answer

(1/6)*(sqrt3)*(sinh(sqrt3*t)) + (1/6)*(sqrt3)*(sin(sqrt3*t))