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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[L(1*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\] But then, I don't know how to continue...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2i would rather do partial fractions. \(\huge \frac{1}{s(s^2 + \omega ^2)}=\frac{Ax+B}{(s^2 + \omega ^2)}+\frac{C}{s}\)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2i am getting C=1/w^2 , B=0 , A =1/w^2 are u getting the same ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(As+B)s + C(s^2 + \omega^2) = 1\] \[A+C = 0\]\[B =0\]\[C = \frac{1}{\omega^2}\] \[L^{1}(\frac{1}{s(s^2 + \omega ^2)})=L^{1}(\frac{1}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\]Hmm.. How to do the inverse of Laplace tramsform of \(\frac{1}{\omega^2(s^2 + \omega ^2)}\) ??

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2\(L^{1}(\frac{1}{s(s^2 + \omega ^2)})=L^{1}(\frac{s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[L^{1}(\frac{1}{s(s^2 + \omega ^2)})=L^{1}(\frac{s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})=\frac{1}{\omega^2}\frac{cos\omega t}{\omega^2} = \frac{1cos\omega t}{\omega^2}\]No wonder why I couldn't get the answer!!! Thanks!! May I know if how to get the answer using the convolution one?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2i was trying exactly that...using convolution

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2\(\int_0^t u(tu)sin (\omega u)/\omega \:\:du\) did u reach here ?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2and i am getting same answer with convolution also.

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2also its , \(\huge L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2if and after u get this : \(\int \limits_0^t u(tu) \frac{\sin (\omega u)}{\omega} \:\:du\) u just put tu = something and solve the integral, yo'll get the same answer.....

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2doubts ? i assume you are trying...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Why is it \( L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\) ???

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2because L[u(t)] = 1/s and NOT L[1] = 1/s ...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2do 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..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Of course I try first :) Sorry to keep you waiting again!

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2in the definition its 0 to infinity or infinity to infinity ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00 to infinity. That's what I've learnt..

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2okk....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*....

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2so 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...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I haven't learnt \(L (u(t)) = \frac{1}{s}\)....

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2\(\int_0^t 1.sin (\omega (tu))/\omega \:\:du\) u got this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I... think I got that..

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2then its just a normal definite integration problem

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f*g =\int_0^tf(\tau) g(t\tau) d\tau\] \[\int_0^t 1\cdot \frac{sin (\omega (tu))}{\omega} du\]\[=\frac{1}{\omega^2}cos (\omega t\omega u)_0^t\]\[=\frac{1}{\omega^2}[cos(0)  cos\omega t]=\frac{1cos\omega t}{\omega^2}\] Wow!!! Thanks a ton!!!!
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