Correct Answer - Option 4 :
\(\frac{4}{{{{\rm{s}}^2} + 2{\rm{s}} + 4}}\)
Given the output of the second-order system is
\({\rm{y}}\left( {\rm{t}} \right) = 1 - \frac{2}{{\sqrt 3 }}{{\rm{e}}^{ - {\rm{t}}}}\cos \left( {\sqrt {3{\rm{t}}} - \frac{{\rm{\pi }}}{6}} \right)\)
Compare this with the standard equation with unit step input
\({\rm{c}}\left( {\rm{t}} \right) = 1 - \frac{{{{\rm{e}}^{ - {\rm{\zeta }}{{\rm{\omega }}_{{\rm{nt}}}}}}}}{{\sqrt {1 - {{\rm{\zeta }}^2}} }}{\rm{sin}}\left( {{{\rm{\omega }}_{\rm{d}}}{\rm{t}} + \phi } \right)\)
\(\therefore {\rm{\;}}\sqrt {1 - {{\rm{\zeta }}^2}} = \frac{{\sqrt 3 }}{2}\)
By squaring on both sides
\(\Rightarrow 1 - {{\rm{\zeta }}^2} = \frac{3}{4}\)
\(\Rightarrow {\rm{\zeta }} = \frac{1}{2} = 0.5\)
\(\Rightarrow {\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}} = 1 \)
\(\Rightarrow {{\rm{\omega }}_{\rm{n}}} = \frac{1}{{\rm{\zeta }}} = 2\)
The transfer function of the standard second-order system is:
\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)
Where
ζ is the damping ratio = 0.5
ωn is the natural frequency = 2
\({\rm{TF}} = \frac{4}{{{{\rm{s}}^2} + 2{\rm{s}} + 4}}\)