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
\(M_p \ \alpha \ \frac{1}{ζ}\)