There is a transfer function as below:
$$ \begin{equation} G(s) = \dfrac{4s}{s^2 + 1} \end{equation} $$
Now I would like to find its output inverse laplace transform with input as below:
$$ u(t) = sint $$
Below is how I attempted to find the inverse laplace transformation:
$$ \begin{align*} y(t) & = \mathscr{L^{-1}}[\dfrac{4s}{s^2 + 1} * \dfrac{1}{s^2 + 1}] \\ & = \mathscr{L^{-1}}[\dfrac{4s}{(s^2+1)^2}] \end{align*} $$
However, I am not sure how to take apart the fraction to make it apply the laplace transform sheet that I know. Below is another approach that I found:
$$ \mathscr{L^{-1}}[sF(s)] = f'(t) + \mathscr{L^{-1}}f(0) $$
But, this seems quite weird to me as this still requires me to find the transform of
$$ \mathscr{L^{-1}}[\dfrac{4}{(s^2 + 1)^2}] $$
which seems no much help to me, and I did not see how to apply inverse laplace transform directly on this.
Can anyone point out what I have misunderstood? Thank you for your time and advice.
We know that the Laplace transform of $sin(t)$ is given by:
$$ \begin{align*} \mathscr{L}[\sin(t)] = \dfrac{}{s^2 + ^2} = F(s) \\ \end{align*} $$
Based on the time multiplication / frequency differentiation rule that Rollen mentioned, the Laplace Transform can be written as:
$$ \begin{align*} \mathscr{L}[t \sin(t)] & = -\dfrac{d}{ds}(\dfrac{}{s^2 + ^2}) \\ & = -\dfrac{0 * (s^2 + ^2) - (2s)}{(s^2 + ^2)^2} = \dfrac{2s}{(s^2 + ^2)^2} \end{align*} $$
Therefore, the provided inverse Laplace transform is as below:
$$ \begin{align*} y(t) = \mathscr{L^{-1}}\left[\dfrac{4s}{(s^2 +1)^2}\right] = 2t \sin(t) \end{align*} $$