Question

Find the inverse Laplace transform of \(\frac{1}{s(s^2 + \omega ^2)}\)

Answer 1

\[L(1*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\]
But then, I don't know how to continue...

Answer 2

i would rather do partial fractions.
\(\huge \frac{1}{s(s^2 + \omega ^2)}=\frac{Ax+B}{(s^2 + \omega ^2)}+\frac{C}{s}\)

Answer 3

find A,B,C

Answer 4

Ax+B ?! As+B?!

Answer 5

yes, As+B

Answer 6

i am getting C=1/w^2 , B=0 , A =-1/w^2
are u getting the same ?

Answer 7

\[(As+B)s + C(s^2 + \omega^2) = 1\]
\[A+C = 0\]\[B =0\]\[C = \frac{1}{\omega^2}\]

\[L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-1}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\]Hmm.. How to do the inverse of Laplace tramsform of \(\frac{-1}{\omega^2(s^2 + \omega ^2)}\) ??

Answer 8

did u forget s ?

Answer 9

As+B

Answer 10

\(L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\)

Answer 11

w^2 is constant

Answer 12

Oh!!! Sorry!!!

Answer 13

\[L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})=\frac{1}{\omega^2}-\frac{cos\omega t}{\omega^2} = \frac{1-cos\omega t}{\omega^2}\]No wonder why I couldn't get the answer!!! Thanks!!

May I know if how to get the answer using the convolution one?

Answer 14

i was trying exactly that...using convolution

Answer 15

\(\int_0^t u(t-u)sin (\omega u)/\omega \:\:du\)
did u reach here ?

Answer 16

and i am getting same answer with convolution also.

Answer 17

also its ,
\(\huge L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\)

Answer 18

if and after u get this :
\(\int \limits_0^t u(t-u) \frac{\sin (\omega u)}{\omega} \:\:du\)
u just put t-u = something
and solve the integral, yo'll get the same answer.....

Answer 19

doubts ? i assume you are trying...

Answer 20

Why is it \( L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\) ???

Answer 21

because L[u(t)] = 1/s
and NOT L[1] = 1/s ...

Answer 22

!!!!! How come...

Answer 23

do u want me to prove that ?
i suggest you try first...using definition of Laplace, with u(t), u get 1/s
with 1, you won't get 1/s..

Answer 24

Of course I try first :) Sorry to keep you waiting again!

Answer 25

in the definition its 0 to infinity or -infinity to infinity ??

Answer 26

0 to infinity. That's what I've learnt..

Answer 27

okk....then u get same answer for 1 and u(t)
because when t>=0, u(t) = 1
then u'll get same answer for both
u(t)*... and 1*....

Answer 28

so suit yourself...i have habit of taking u(t)
because for some transforms, the definition is -infinity to infinity
where u(t)and 1 are not same...

Answer 29

I haven't learnt \(L (u(t)) = \frac{1}{s}\)....

Answer 30

then take it as 1...

Answer 31

\(\int_0^t 1.sin (\omega (t-u))/\omega \:\:du\)
u got this?

Answer 32

I... think I got that..

Answer 33

then its just a normal definite integration problem

Answer 34

didn't get it ?

Answer 35

\[f*g =\int_0^tf(\tau) g(t-\tau) d\tau\]
\[\int_0^t 1\cdot \frac{sin (\omega (t-u))}{\omega} du\]\[=\frac{1}{\omega^2}cos (\omega t-\omega u)|_0^t\]\[=\frac{1}{\omega^2}[cos(0) - cos\omega t]=\frac{1-cos\omega t}{\omega^2}\] Wow!!! Thanks a ton!!!!

Answer 36

welcome ^_^