Find K and P from step response

Find the closed-loop transfer function with feedback $(1 + sP)$. Then use the measured overshoot and peak time to get $\zeta$ and $\omega_n$.

Closed-loop transfer function

$$\frac{C(s)}{R(s)} = \frac{G}{1+GH} = \frac{\frac{K}{s(s+2)}}{1 + \frac{K(1+sP)}{s(s+2)}} = \frac{K}{s^2 + (2 + KP)s + K}$$

Its DC gain is 1, which agrees with the final value of 1 in Fig (ii). Comparing:

$$\omega_n^2 = K, \qquad 2\zeta\omega_n = 2 + KP$$

Step 1: $\zeta$ from $M_p = 0.343$

$$\zeta = \frac{-\ln 0.343}{\sqrt{\pi^2 + (\ln 0.343)^2}} = \frac{1.0700}{\sqrt{9.8696 + 1.1449}} = \frac{1.0700}{3.3188} = 0.3224$$

Step 2: $\omega_n$ from $t_p = 3$ s

$$\omega_n = \frac{\pi}{t_p\sqrt{1-\zeta^2}} = \frac{3.1416}{3\sqrt{1 - 0.1040}} = \frac{3.1416}{3(0.9466)} = 1.106\ \text{rad/s}$$

Step 3: $K$

$$K = \omega_n^2 = 1.106^2 = 1.224$$

Step 4: $P$

$$\begin{aligned} 2 + KP &= 2\zeta\omega_n = 2(0.3224)(1.106) = 0.713 \\ P &= \frac{0.713 - 2}{1.224} = -1.05\ \text{s} \end{aligned}$$

Answer: $K \approx 1.22$ and $P \approx -1.05$ s.

Remark: $P$ comes out negative. With these values of $t_p$ and $M_p$, the response is less damped than $K/[s(s+2)]$ with unity feedback would give ($\zeta = 2/(2\sqrt{1.224}) = 0.90$). So the $sP$ term must act as positive rate feedback. If the figure values in your paper differ, use the same four steps with them.