Inverse laplace Transform of

Question

How to calculate the inverse laplace transform of $\frac{\omega }{\left ( s^{2}+\omega ^{2} \right )( s^{2}+\omega ^{2} )} $ ?

Answer 1

One possible hint may be to use the following idea (assuming you know the convolution of two functions):

$$\mathcal{L}\left(\int_0^{\infty}f(t)g(x-t)dt\right)=F(s)G(s).$$ where $\mathcal{L}(f(x))=F(s)$ and $\mathcal{L}(g(x))=G(s)$. Note that we have here $$\frac{a}{(a^2+s^2)^2}=\frac{1}a\cdot \frac{a}{a^2+s^2}\cdot\frac{a}{a^2+s^2}$$ and of course you know that $\mathcal{L}(\sin(ax))=\frac{a}{a^2+s^2}$.

Answer 2

An alternative to convolution would be the following approach. You need to know the following Laplace transform pairs:

$$\mathcal{L}\{\sin(\omega T)\}=\frac{\omega}{s^2+\omega^2}\\ \mathcal{L}\{\cos(\omega T)\}=\frac{s}{s^2+\omega^2}\\ \mathcal{L}\{tf(t)\}=-F'(s)\\ \mathcal{L}\{f'(t)\}=sF(s)$$

From this it follows that

$$\mathcal{L}\{t\sin(\omega T)\}=\frac{2\omega s}{(s^2+\omega^2)^2}\\ \mathcal{L}\{t\cos(\omega T)\}=\frac{s^2-\omega^2}{(s^2+\omega^2)^2}\tag{1}$$

The original function can now be written as

$$\frac{\omega}{(s^2+\omega^2)^2}=\frac{1}{\omega}\left[\frac{s^2}{(s^2+\omega^2)^2}-\frac{s^2-\omega^2}{(s^2+\omega^2)^2}\right]$$

which can be directly transformed to the time domain using (1):

$$\mathcal{L}^{-1}\left\{\frac{\omega}{(s^2+\omega^2)^2}\right\}=\frac{1}{\omega}\left[\frac{1}{2\omega}(t\sin(\omega t))'-t\cos(\omega t)\right]=\\=\frac{1}{\omega}\left[\frac{1}{2\omega}(\sin(\omega t)+\omega t\cos(\omega t))-t\cos(\omega t) \right]=\\= \frac{1}{2\omega^2}\left[\sin(\omega t)-\omega t\cos(\omega t)\right]$$