Overshoot and settling time from pole plot

The open-loop poles at $s = -2 \pm j1$ give

$$G(s) = \frac{K}{(s+2-j)(s+2+j)} = \frac{K}{s^2 + 4s + 5}$$

No gain is marked on the plot, so take $K = 1$ (and show the general result).

Closed-loop transfer function (unity feedback)

$$\frac{C(s)}{R(s)} = \frac{K}{s^2 + 4s + 5 + K}$$

Closed-loop poles: $s = -2 \pm j\sqrt{1 + K}$, so $\sigma = \zeta\omega_n = 2$ for every $K$.

For $K = 1$:

$$\begin{aligned} \omega_n^2 &= 6 \;\Rightarrow\; \omega_n = 2.449\ \text{rad/s} \\ 2\zeta\omega_n &= 4 \;\Rightarrow\; \zeta = \frac{2}{2.449} = 0.8165 \end{aligned}$$

Maximum overshoot

$$M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} = e^{-\pi\sigma/\omega_d} = e^{-2\pi/\sqrt{2}} = e^{-4.443} = 0.0118 = 1.18\%$$

In general, $M_p = e^{-2\pi/\sqrt{1+K}}$.

Settling time (2% criterion)

$$t_s = \frac{4}{\zeta\omega_n} = \frac{4}{2} = 2\ \text{s}$$

This is the same for any $K$, because the real part of the closed-loop poles is always $-2$.

Answer: with $K = 1$: $M_p \approx 1.18\%$ and $t_s = 2$ s.

(If the plotted poles $-2 \pm j1$ are taken directly as the dominant poles, $\zeta = 2/\sqrt{5} = 0.894$, $M_p = e^{-2\pi} = 0.19\%$, and $t_s$ is still 2 s.)