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

The correct option is 4

Concept:

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

\({\rm{TF}} = \frac{4}{{{{\rm{s}}^2} + 2{\rm{s}} + 4}}\)