Compensator for 4/[s(s+2)]: Kv, PM, GM

The uncompensated system has a low phase margin once the gain is raised for $K_v = 20$, so a phase-lead compensator is designed using the Bode plot.

Take $G_c(s) = K_c\,\alpha\,\dfrac{Ts+1}{\alpha Ts+1}$, $0 < \alpha < 1$.

Step 1: Gain for $K_v$

$$K_v = \lim_{s\to0} s\,G_c(s)G_f(s) = K_c\alpha\cdot\frac{4}{2} = 2K_c\alpha = 20 \;\Rightarrow\; K = K_c\alpha = 10$$

Gain-adjusted system: $G_1(s) = \dfrac{40}{s(s+2)} = \dfrac{20}{s(0.5s+1)}$.

Step 2: Phase margin of $G_1$

Gain crossover where $|G_1(j\omega)| = 1$:

$$\begin{aligned} \frac{40}{\omega\sqrt{\omega^2+4}} &= 1 \;\Rightarrow\; \omega^4 + 4\omega^2 - 1600 = 0 \\ \omega^2 &= 38.05,\quad \omega_{gc} = 6.17\ \text{rad/s} \\ \angle G_1 &= -90^\circ - \tan^{-1}(6.17/2) = -162.0^\circ \\ PM &= 180^\circ - 162.0^\circ = 18.0^\circ \end{aligned}$$

The phase never reaches $-180^\circ$ (second-order type-1 system), so $GM = \infty$; the gain margin requirement of 10 dB is automatically met.

Step 3: Phase lead needed

Add about $8^\circ$ for the extra lag caused by the shift of the crossover frequency:

$$\phi_m = 50^\circ - 18^\circ + 8^\circ = 40^\circ$$ $$\alpha = \frac{1-\sin 40^\circ}{1+\sin 40^\circ} = \frac{0.357}{1.643} = 0.217$$

Step 4: New gain crossover frequency

The compensator adds $10\log(1/\alpha) = 6.63$ dB at $\omega_m$. So $\omega_m$ is where $|G_1| = -6.63$ dB, i.e. $|G_1| = \sqrt{\alpha} = 0.466$:

$$\frac{40}{\omega\sqrt{\omega^2+4}} = 0.466 \;\Rightarrow\; \omega_m = 9.16\ \text{rad/s}$$

Step 5: Corner frequencies

$$\begin{aligned} T &= \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{9.16\times0.466} = 0.234\ \text{s} \\ \text{zero: } \frac{1}{T} &= 4.27\ \text{rad/s}, \qquad \text{pole: } \frac{1}{\alpha T} = 19.66\ \text{rad/s} \\ K_c &= \frac{10}{\alpha} = 46.1 \end{aligned}$$

Step 6: Compensator and check

$$G_c(s) = 46.1\,\frac{s+4.27}{s+19.66} = 10\,\frac{0.234s+1}{0.0508s+1}$$ $$G_c(s)G_f(s) = \frac{40(0.234s+1)}{s(s+2)(0.0508s+1)}$$

Answer: $G_c(s) = 46.1\dfrac{s+4.27}{s+19.66}$ gives $K_v = 20\ \text{s}^{-1}$, $PM = 52.3^\circ\ (\ge 50^\circ)$ and $GM = \infty\ (\ge 10$ dB$)$.