Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly

  • 3 years ago

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

  • This Question is Closed
  1. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  2. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  3. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    find A,B,C

  4. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Ax+B ?! As+B?!

  5. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes, As+B

  6. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  7. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[(As+B)s + C(s^2 + \omega^2) = 1\] \[A+C = 0\]\[B =0\]\[C = \frac{1}{\omega^2}\] \[L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-1}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\]Hmm.. How to do the inverse of Laplace tramsform of \(\frac{-1}{\omega^2(s^2 + \omega ^2)}\) ??

  8. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    did u forget s ?

  9. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    As+B

  10. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  11. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    w^2 is constant

  12. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh!!! Sorry!!!

  13. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[L^{-1}(\frac{1}{s(s^2 + \omega ^2)})=L^{-1}(\frac{-s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})=\frac{1}{\omega^2}-\frac{cos\omega t}{\omega^2} = \frac{1-cos\omega t}{\omega^2}\]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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i was trying exactly that...using convolution

  15. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  16. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and i am getting same answer with convolution also.

  17. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  18. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    doubts ? i assume you are trying...

  20. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  21. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  22. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    !!!!! How come...

  23. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  25. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  26. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  27. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  30. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    then take it as 1...

  31. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  32. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I... think I got that..

  33. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    then its just a normal definite integration problem

  34. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    didn't get it ?

  35. RolyPoly
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  36. hartnn
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    welcome ^_^

  37. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy