Lead vs lag; lead design for K/[s²(1+0.25s)]

Lead vs lag compensator

Design

Step 1: K. $K_a=\lim_{s\to0}s^2G(s)=K=10$, so $G(s)=\dfrac{10}{s^2(1+0.25s)}$.

Step 2: Uncompensated system.

$$|G|=\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}},\qquad\phi=-180^\circ-\tan^{-1}0.25\omega$$

$\omega_{gc}=2.85$ rad/s, $PM=-35.5^\circ$: unstable (phase always below $-180^\circ$).

Step 3: Lead required. $\phi=40^\circ-(-35.5^\circ)+\epsilon\approx80^\circ$ or more. One lead stage usefully gives at most about 60°, and for this plant a single stage gives a PM of only about 8° with $\alpha=0.1$ (and under 25° even as $\alpha\to0$). So two identical lead stages are used:

$$G_c(s)=\left(\frac{1+Ts}{1+\alpha Ts}\right)^2$$

Step 4: α per stage. Take $\phi_m\approx50^\circ$ per stage: $\alpha=0.13$, $\sin\phi_m=\frac{0.87}{1.13}\Rightarrow\phi_m=50.3^\circ$ (total 100.7°).

Step 5: New crossover. Two stages add $20\log(1/0.13)=17.7$ dB at $\omega_m$, so

$$\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}}=0.13\Rightarrow\omega_m=6.39\ \text{rad/s}$$

Step 6: T.

$$T=\frac{1}{6.39\sqrt{0.13}}=0.434\ \text{s},\qquad\alpha T=0.0564\ \text{s}$$ $$G_c(s)=\left(\frac{1+0.434s}{1+0.0564s}\right)^2$$

with an amplifier of gain $1/\alpha^2=59.2$ to cancel the networks' attenuation (so $K_a$ stays 10).

Step 7: Check at 6.39 rad/s.

$$\begin{aligned} \phi&=-180^\circ-\tan^{-1}1.60+2(50.3^\circ)=-137.3^\circ\\ PM&=42.7^\circ\ (\ge40^\circ) \end{aligned}$$

GM ≈ 12.9 dB; all closed-loop poles are in the LHP.

Answer: $G_c(s)=\left(\dfrac{1+0.434s}{1+0.0564s}\right)^2$ with K = 10: Ka = 10 s⁻², PM ≈ 42.7°.