Transfer function from overshoot and settling time

Assume a standard second-order system $\frac{C(s)}{R(s)} = \frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$. Find $\zeta$ from overshoot and $\omega_n$ from settling time.

Damping ratio from $M_p = 0.25$

$$\begin{aligned} M_p &= e^{-\zeta\pi/\sqrt{1-\zeta^2}} = 0.25 \\ \frac{\zeta\pi}{\sqrt{1-\zeta^2}} &= -\ln 0.25 = 1.3863 \\ \zeta &= \frac{1.3863}{\sqrt{\pi^2+1.3863^2}} = \frac{1.3863}{3.4339} = 0.4037 \end{aligned}$$

Natural frequency from $t_s = 6$ s (2% band)

$$\begin{aligned} t_s &= \frac{4}{\zeta\omega_n} = 6 \Rightarrow \zeta\omega_n = 0.6667 \\ \omega_n &= \frac{0.6667}{0.4037} = 1.651\ \text{rad/s} \end{aligned}$$

Transfer function

$\omega_n^2 = 2.727$, $2\zeta\omega_n = 1.333$:

$$\frac{C(s)}{R(s)} = \frac{2.727}{s^2 + 1.333s + 2.727}$$

Peak time

$$\begin{aligned} \omega_d &= \omega_n\sqrt{1-\zeta^2} = 1.651\sqrt{1-0.4037^2} = 1.511\ \text{rad/s} \\ t_p &= \frac{\pi}{\omega_d} = \frac{3.1416}{1.511} = 2.079\ \text{s} \end{aligned}$$

Answer: $\zeta = 0.404$, $\omega_n = 1.651$ rad/s, $T(s) = \dfrac{2.727}{s^2+1.333s+2.727}$, peak time $t_p = 2.08$ s.