Lead compensator for K/[s²(0.2s+1)], Ka = 10, PM 35°

A type-2 plant has $-180^\circ$ phase at low frequency, so the plant pole makes it unstable for any $K$. A phase-lead network is designed on the Bode plot: $G_c(s) = \dfrac{1+Ts}{1+\alpha Ts}$, $\alpha < 1$ (unity DC gain, so $K_a$ is set by $K$).

Step 1: Gain for $K_a$

$$K_a = \lim_{s\to0} s^2\,\frac{K}{s^2(0.2s+1)} = K = 10, \qquad G(s) = \frac{10}{s^2(0.2s+1)}$$

Step 2: Uncompensated phase margin

$$\begin{aligned} |G(j\omega)| &= \frac{10}{\omega^2\sqrt{1+0.04\omega^2}} = 1 \;\Rightarrow\; \omega_{gc} = 2.94\ \text{rad/s} \\ \angle G &= -180^\circ - \tan^{-1}(0.2\times2.94) = -210.4^\circ \\ PM &= -30.4^\circ \quad (\text{unstable}) \end{aligned}$$

Step 3: Phase lead required

$$\phi_m = 35^\circ - (-30.4^\circ) + 5^\circ = 70.4^\circ$$

This is more than about $60^\circ$, which a single lead network cannot give with a sensible $\alpha$. Use two identical lead stages in cascade. Also, the plant magnitude falls at $-40$ to $-60$ dB/decade, so the crossover moves a lot and its phase falls quickly; a first trial with $2 \times 35^\circ$ gives only $PM = 24.5^\circ$. So take $2 \times 45^\circ$:

$$\alpha = \frac{1-\sin45^\circ}{1+\sin45^\circ} = 0.1716 \text{ per stage}$$

Step 4: New gain crossover

The two stages together add $2 \times 10\log(1/\alpha) = 15.3$ dB at $\omega_m$. So $\omega_m$ is where $|G(j\omega)| = \alpha = 0.1716$ ($-15.3$ dB):

$$\frac{10}{\omega^2\sqrt{1+0.04\omega^2}} = 0.1716 \;\Rightarrow\; \omega_m = 6.08\ \text{rad/s}$$

Step 5: Corner frequencies

$$\begin{aligned} T &= \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{6.08\times0.4142} = 0.397\ \text{s} \\ \text{zero: } 1/T &= 2.52\ \text{rad/s},\qquad \text{pole: } 1/(\alpha T) = 14.69\ \text{rad/s} \end{aligned}$$

Step 6: Compensator and check

$$G_c(s) = \left(\frac{1+0.397s}{1+0.0681s}\right)^2 = 34.0\left(\frac{s+2.52}{s+14.69}\right)^2$$ $$G_c(s)G(s) = \frac{10(1+0.397s)^2}{s^2(0.2s+1)(1+0.0681s)^2}$$

At $\omega = 6.08$: plant phase $= -180^\circ - \tan^{-1}(1.216) = -230.6^\circ$, compensator phase $= +90^\circ$, so $PM = 180^\circ - 230.6^\circ + 90^\circ = 39.4^\circ$.

Answer: two cascaded lead stages, $G_c(s) = \left(\dfrac{1+0.397s}{1+0.0681s}\right)^2$, give $K_a = 10$ and $PM = 39.4^\circ \ (\ge 35^\circ)$.