Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

RolyPoly Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. RolyPoly Group Title
    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...

    • one year ago
  2. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  3. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    find A,B,C

    • one year ago
  4. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Ax+B ?! As+B?!

    • one year ago
  5. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    yes, As+B

    • one year ago
  6. hartnn Group Title
    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 ?

    • one year ago
  7. RolyPoly Group Title
    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 \(\frac{-1}{\omega^2(s^2 + \omega ^2)}\) ??

    • one year ago
  8. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    did u forget s ?

    • one year ago
  9. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    As+B

    • one year ago
  10. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  11. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    w^2 is constant

    • one year ago
  12. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh!!! Sorry!!!

    • one year ago
  13. RolyPoly Group Title
    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?

    • one year ago
  14. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i was trying exactly that...using convolution

    • one year ago
  15. hartnn Group Title
    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 ?

    • one year ago
  16. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    and i am getting same answer with convolution also.

    • one year ago
  17. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  18. hartnn Group Title
    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.....

    • one year ago
  19. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    doubts ? i assume you are trying...

    • one year ago
  20. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  21. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  22. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    !!!!! How come...

    • one year ago
  23. hartnn Group Title
    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..

    • one year ago
  24. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  25. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  26. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  27. hartnn Group Title
    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*....

    • one year ago
  28. hartnn Group Title
    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...

    • one year ago
  29. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  30. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    then take it as 1...

    • one year ago
  31. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  32. RolyPoly Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I... think I got that..

    • one year ago
  33. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    then its just a normal definite integration problem

    • one year ago
  34. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    didn't get it ?

    • one year ago
  35. RolyPoly Group Title
    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!!!!

    • one year ago
  36. hartnn Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    welcome ^_^

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.