Correct Answer - Option 4 : Equal to zero

__Concept:__

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

\(TF = \frac{{C\left( s \right)}}{{R\left( s \right)}} = \frac{{ω _n^2}}{{{s^2} + 2ζ {ω _n}s + ω _n^2}}\)

Characteristic equation: \({s^2} + 2ζ {ω _n} + ω _n^2 = 0\)

ζ is the damping ratio

ωn is the undamped natural frequency

\({M_p} = {e^{\frac{{ - ζ \pi }}{{\sqrt {1 - {ζ ^2}} }}}}\) ----(1)

__Calculation:__

Given:

Mp = 100%

From the above equation,

\({M_p} = {e^{\frac{{ - ζ \pi }}{{\sqrt {1 - {ζ ^2}} }}}}\)

\(ln\;1 = \frac{{ - \zeta \pi }}{{\sqrt {1 - {\zeta ^2}} }}\) ; (ln 1 = 0)

So, **ζ = 0**

__Note:__

Mp is the maximum peak overshoot of the closed-loop transfer function

\(M_p \ \alpha \ \frac{1}{ζ}\)