## RolyPoly 2 years ago Find the inverse Laplace transform of $$\frac{1}{s(s^2 + \omega ^2)}$$

1. RolyPoly

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

2. hartnn

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

3. hartnn

find A,B,C

4. RolyPoly

Ax+B ?! As+B?!

5. hartnn

yes, As+B

6. hartnn

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

7. RolyPoly

$(As+B)s + C(s^2 + \omega^2) = 1$ $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 $$\frac{-1}{\omega^2(s^2 + \omega ^2)}$$ ??

8. hartnn

did u forget s ?

9. hartnn

As+B

10. hartnn

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

11. hartnn

w^2 is constant

12. RolyPoly

Oh!!! Sorry!!!

13. RolyPoly

$L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})=\frac{1}{\omega^2}-\frac{cos\omega t}{\omega^2} = \frac{1-cos\omega t}{\omega^2}$No wonder why I couldn't get the answer!!! Thanks!! May I know if how to get the answer using the convolution one?

14. hartnn

i was trying exactly that...using convolution

15. hartnn

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

16. hartnn

and i am getting same answer with convolution also.

17. hartnn

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

18. hartnn

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.....

19. hartnn

doubts ? i assume you are trying...

20. RolyPoly

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

21. hartnn

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

22. RolyPoly

!!!!! How come...

23. hartnn

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..

24. RolyPoly

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

25. hartnn

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

26. RolyPoly

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

27. hartnn

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*....

28. hartnn

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...

29. RolyPoly

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

30. hartnn

then take it as 1...

31. hartnn

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

32. RolyPoly

I... think I got that..

33. hartnn

then its just a normal definite integration problem

34. hartnn

didn't get it ?

35. RolyPoly

$f*g =\int_0^tf(\tau) g(t-\tau) d\tau$ $\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!!!!

36. hartnn

welcome ^_^