The output of a standard second-order system for a unit-step input is given as $y(t)=1-\frac{2}{\sqrt{3}}{e}^{-t}\cos (\sqrt{3}t-\frac{\pi }{6})$. What is the transfer function of the system?

The correct option is 4

Concept:

Given the output of the second-order system is 

$\mathrm{y}\left(\mathrm{t}\right)=1-\frac{2}{\sqrt{3}}{\mathrm{e}}^{-\mathrm{t}}\cos \left(\sqrt{3\mathrm{t}}-\frac{\pi }{6}\right)$

Compare this with the standard equation with unit step input

$\mathrm{c}\left(\mathrm{t}\right)=1-\frac{{\mathrm{e}}^{-\zeta {\omega }_{\mathrm{n}\mathrm{t}}}}{\sqrt{1-{\zeta }^2}}\mathrm{s}\mathrm{i}\mathrm{n}\left({\omega }_{\mathrm{d}}\mathrm{t}+\varphi \right)$

$\therefore \sqrt{1-{\zeta }^2}=\frac{\sqrt{3}}{2}$

By squaring on both sides

$\Rightarrow 1-{\zeta }^2=\frac{3}{4}$

$\Rightarrow \zeta =\frac{1}{2}=0.5$

$\Rightarrow \zeta {\omega }_{\mathrm{n}}=1$

$\Rightarrow {\omega }_{\mathrm{n}}=\frac{1}{\zeta }=2$

The transfer function of the standard second-order system is:

$TF=\frac{C\left(s\right)}{R\left(s\right)}=\frac{{\omega }_n^2}{s^2+2\zeta {\omega }_{n}s+{\omega }_n^2}$

Where

ζ is the damping ratio = 0.5

ωn is the natural frequency = 2

$\mathrm{T}\mathrm{F}=\frac{4}{{\mathrm{s}}^2+2\mathrm{s}+4}$