Gain factor K1 raising overshoot from 25% to 50%

The closed-loop system is second order; increasing the gain lowers the damping ratio and raises the overshoot.

Closed-loop characteristic equation

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

Comparing with $s^2 + 2\zeta\omega_ns + \omega_n^2$:

$$\omega_n = \sqrt{\frac{K}{T}}, \qquad 2\zeta\omega_n = \frac{1}{T} \;\Rightarrow\; \zeta = \frac{1}{2\sqrt{KT}}$$

So $\zeta \propto 1/\sqrt{K}$ (for fixed $T$).

Damping ratio from overshoot

$$M_p = e^{-\pi\zeta/\sqrt{1-\zeta^2}} \;\Rightarrow\; \zeta = \frac{-\ln M_p}{\sqrt{\pi^2 + (\ln M_p)^2}}$$

• For $M_p = 0.25$: $\ln 0.25 = -1.3863$

$$\zeta_1 = \frac{1.3863}{\sqrt{9.8696 + 1.9218}} = \frac{1.3863}{3.4339} = 0.4037$$

• For $M_p = 0.50$: $\ln 0.5 = -0.6931$

$$\zeta_2 = \frac{0.6931}{\sqrt{9.8696 + 0.4805}} = \frac{0.6931}{3.2172} = 0.2155$$

Gain factor

With gain $K$: $\zeta_1 = \dfrac{1}{2\sqrt{KT}}$. With gain $K_1K$: $\zeta_2 = \dfrac{1}{2\sqrt{K_1KT}}$. Dividing:

$$\frac{\zeta_1}{\zeta_2} = \sqrt{K_1} \;\Rightarrow\; K_1 = \left(\frac{\zeta_1}{\zeta_2}\right)^2 = \left(\frac{0.4037}{0.2155}\right)^2 = (1.8738)^2$$

Answer: $K_1 \approx 3.51$. The gain must be multiplied by about 3.51 to raise the peak overshoot from 25% to 50%.