How to calculate the inverse transform of this function?

Question

I don’t know how to calculate the inverse transform of this function: $z = L^{-1}\{-3s^3/(3s^4+ 16s^2+16)\}$

The solution is:

$$z= \frac12 \cos(\frac{2t}{\sqrt{3}}) - \frac32 \cos(2t)$$

Answer 1

Here $~3s^4+ 16s^2+16=(s^2+4)(3s^2+4)~$ and hence by partial fraction we have, $$\dfrac{3s^3}{3s^4+ 16s^2+16}=\dfrac{3s}{2(s^2+4)}-\dfrac{3s}{2(3s^2+4)}$$ So inverse Laplace transform, $$\mathcal L^{-1}\left\{-\dfrac{3s^3}{3s^4+ 16s^2+16}\right\}=-\mathcal L^{-1}\left\{\dfrac{3s}{2(s^2+4)}\right\}+\mathcal L^{-1}\left\{\dfrac{3s}{2(3s^2+4)}\right\}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\dfrac{1}{2}\cdot\mathcal L^{-1}\left\{\dfrac{3s}{3s^2+4}\right\}-\dfrac{3}{2}\cdot\mathcal L^{-1}\left\{\dfrac{s}{s^2+4}\right\}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\dfrac{1}{2}\cdot\mathcal L^{-1}\left\{\dfrac{s}{s^2+\frac{4}{3}}\right\}-\dfrac{3}{2}\cdot\mathcal L^{-1}\left\{\dfrac{s}{s^2+4}\right\}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\dfrac{1}{2}\cdot\cos\left(\dfrac{2t}{\sqrt 3}\right)-\dfrac{3}{2}\cdot\cos(2t)$$

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Note:

• $$\mathcal{L}=\{\sin(at)\}=\frac{a}{p^2+a^2}\implies \mathcal{L}^{-1}\left\{\frac{a}{p^2+a^2}\right\}=\sin(at)$$
• $$\mathcal{L}=\{\cos(at)\}=\frac{p}{p^2+a^2}\implies \mathcal{L}^{-1}\left\{\frac{p}{p^2+a^2}\right\}=\cos(at)$$