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Please help to find inverse laplace transform : $$ F(s)=\frac{2 \omega^3}{(s^2+\omega^2)^2} $$

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It would be helpful to know what you have tried so far, and also which relevant concepts you do or do not understand. – Michael Albanese Dec 22 '12 at 12:12
Also consider rewriting your question. One may find "plz" offensive – Nameless Dec 22 '12 at 12:15
Use the Convolution formula for Laplace Transforms, applied to $H(s)=G(s)={1\over s^2+\omega^2}$, and multiply both sides of the result by $2\omega^3$ afterwards. See here for details. – David Mitra Dec 22 '12 at 12:26
@ David Mitra Thank you – Amir Alizadeh Dec 24 '12 at 20:33
up vote 3 down vote accepted

We know $L(e^{at})=\frac1{s-a}$

Putting $a=ib,L(e^{ibt})=\frac1{s-ib}$

or $L(\cos bt)+iL(\sin bt)=\frac{s+ib}{s^2+b^2}$

So, $L(\cos bt)=\frac s{s^2+b^2}$ and $L(\sin bt)=\frac b{s^2+b^2}$

Again, as $L(t^n)=\frac{n!}{s^{n+1}}$

and $L\{e^{ctf(t)}\}=F(s-c)$ where $F(s)=L\{f(t)\},$


Putting $n=1,L(te^{ibt})=\frac{(s+ib)^2}{(s^2+b^2)^2}\implies L(t\cos bt)=\frac{s^2-b^2}{(s^2+b^2)^2}$

Let $\frac1{(s^2+b^2)^2}=A\frac1{s^2+b^2}+B\frac{s^2-b^2}{(s^2+b^2)^2}$

or, $1=A(s^2+b^2)+B(s^2-b^2)=s^2(A+B)+b^2(A-B)$

So, $A+B=0, b^2(A-B)=1\implies -B=A=\frac1{2b^2}$

So, $L^{-1}\left(\frac1{(s^2+b^2)^2}\right)=\frac1{2b^2}\left(\frac{\sin bt}b-t\cos bt\right)$

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thanks to David Mitra $$ F(s)=\frac{2w^3}{(s^2+w^2)^2}=2w.(\frac{w}{s^2+w^2}).(\frac{w}{s^2+w^2}) $$ $$ f(t)=2w.\sin(wt)*\sin(wt)=2w \int_0^t \sin(w(t-\tau))\sin(w \tau) d \tau $$ $$ \sin(w(t-\tau))\sin(w \tau)=\frac{1}{2}\left[ \cos(wt-2w \tau) -\cos(w \tau) \right] $$ $$ f(t)=w \int_0^t \cos(wt-2w \tau) -\cos(w \tau) d\tau$$ $$ f(t)=w \left[ \int_0^t\cos(wt)\cos(2w\tau)d\tau + \int_0^t \sin(wt) \sin(2w\tau) d\tau - \int_0^t \cos(wt)d\tau \right] $$ $$ f(t)=w \left[ \frac{\cos(wt)\sin(2wt)}{2w}+\frac{-\sin(wt)\left[\cos(2wt)-1 \right]}{2w} -t\cos(wt)\right] $$ $$ f(t)=\sin(wt)-wt \cos(wt) $$

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