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.
Best Answer
We know that the Laplace transform of $sin(𝜔t)$ is given by:
$$ \begin{align*} \mathscr{L}[\sin(𝜔t)] = \dfrac{𝜔}{s^2 + 𝜔^2} = F(s) \\ \end{align*} $$
Based on the time multiplication / frequency differentiation rule that Rollen mentioned, the Laplace Transform can be written as:
$$ \begin{align*} \mathscr{L}[t \sin(𝜔t)] & = -\dfrac{d}{ds}(\dfrac{𝜔}{s^2 + 𝜔^2}) \\ & = -\dfrac{0 * (s^2 + 𝜔^2) - 𝜔(2s)}{(s^2 + 𝜔^2)^2} = \dfrac{2𝜔s}{(s^2 + 𝜔^2)^2} \end{align*} $$
Therefore, the provided inverse Laplace transform is as below:
$$ \begin{align*} y(t) = \mathscr{L^{-1}}\left[\dfrac{4s}{(s^2 +1)^2}\right] = 2t \sin(t) \end{align*} $$