Find K and T from step response (overshoot 0.254)

From the response curve, peak overshoot $M_p = 0.254$ (25.4%) and peak time $t_p = 3$ s. These give $\zeta$ and $\omega_n$, which are then matched with the closed-loop transfer function.

Closed-loop transfer function

$$\frac{C(s)}{R(s)} = \frac{\frac{K}{s(1+sT)}}{1+\frac{K}{s(1+sT)}} = \frac{K}{Ts^2+s+K} = \frac{K/T}{s^2+\frac{1}{T}s+\frac{K}{T}}$$

Comparing with $\frac{\omega_n^2}{s^2+2\zeta\omega_n s+\omega_n^2}$:

$$\omega_n^2 = \frac{K}{T}, \qquad 2\zeta\omega_n = \frac{1}{T}$$

Damping ratio

$$\begin{aligned} e^{-\zeta\pi/\sqrt{1-\zeta^2}} &= 0.254 \\ \frac{\zeta\pi}{\sqrt{1-\zeta^2}} &= -\ln 0.254 = 1.3704 \\ \zeta &= \frac{1.3704}{\sqrt{\pi^2+1.3704^2}} = 0.400 \end{aligned}$$

Natural frequency

$$\begin{aligned} t_p &= \frac{\pi}{\omega_d} = 3 \Rightarrow \omega_d = 1.0472\ \text{rad/s} \\ \omega_n &= \frac{\omega_d}{\sqrt{1-\zeta^2}} = \frac{1.0472}{\sqrt{1-0.16}} = 1.1425\ \text{rad/s} \end{aligned}$$

K and T

$$\begin{aligned} T &= \frac{1}{2\zeta\omega_n} = \frac{1}{2\times0.4\times1.1425} = 1.094\ \text{s} \\ K &= \omega_n^2 T = 1.3053\times1.094 = 1.428 \end{aligned}$$

Answer: $\zeta = 0.4$, $\omega_n = 1.142$ rad/s, giving $T \approx 1.094$ s and $K \approx 1.43$.