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RolyPoly
Group Title
Find the inverse Laplace transform of \(\frac{1}{s(s^2 + \omega ^2)}\)
 one year ago
 one year ago
RolyPoly Group Title
Find the inverse Laplace transform of \(\frac{1}{s(s^2 + \omega ^2)}\)
 one year ago
 one year ago

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RolyPoly Group TitleBest ResponseYou've already chosen the best response.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

hartnn Group TitleBest ResponseYou've already chosen the best response.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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Ax+B ?! As+B?!
 one year ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
did u forget s ?
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
\(L^{1}(\frac{1}{s(s^2 + \omega ^2)})=L^{1}(\frac{s}{\omega^2(s^2 + \omega ^2)}+\frac{1}{\omega^2s})\)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
w^2 is constant
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Oh!!! Sorry!!!
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i was trying exactly that...using convolution
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
and i am getting same answer with convolution also.
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
if 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.....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
doubts ? i assume you are trying...
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Why is it \( L(u(t)*\frac{sin\omega t}{\omega }) = \frac{1}{s(s^2+\omega ^2)}\) ???
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
because L[u(t)] = 1/s and NOT L[1] = 1/s ...
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
!!!!! How come...
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Of course I try first :) Sorry to keep you waiting again!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
in the definition its 0 to infinity or infinity to infinity ??
 one year ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
0 to infinity. That's what I've learnt..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.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

hartnn Group TitleBest ResponseYou've already chosen the best response.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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I haven't learnt \(L (u(t)) = \frac{1}{s}\)....
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
then take it as 1...
 one year ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
I... think I got that..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
then its just a normal definite integration problem
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
didn't get it ?
 one year ago

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
welcome ^_^
 one year ago
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