Lag compensator for k/[s(1+0.1s)(1+0.2s)], Kv = 30

A phase-lag network $G_c(s)=\dfrac{1+Ts}{1+\beta Ts}$ ($\beta>1$) has unity DC gain and attenuates high frequencies by $20\log\beta$ dB. It lowers the gain crossover frequency to where the plant phase gives the required PM, while $K_v$ is kept by the gain $K$.

Step 1: K from $K_v$

$$K_v=\lim_{s\to0}sG(s)=K=30\Rightarrow G(s)=\frac{30}{s(1+0.1s)(1+0.2s)}$$

Step 2: Uncompensated system

Corners 5 and 10 rad/s. Asymptotes: $-20$ dB/dec (29.5 dB at $\omega=1$) to 5, $-40$ to 10, $-60$ after.

$$\phi(\omega)=-90^\circ-\tan^{-1}0.1\omega-\tan^{-1}0.2\omega$$

$\omega_{gc}=9.77$ rad/s, PM $=-17.2^\circ$; $\omega_{pc}=7.07$ rad/s, GM $=-6$ dB. The system is unstable.

Step 3: New gain crossover

Required phase: $-180^\circ+40^\circ+10^\circ=-130^\circ$ ($10^\circ$ allowance for the lag network's own phase lag).

$$\tan^{-1}0.1\omega+\tan^{-1}0.2\omega=40^\circ\Rightarrow\omega_{gc}'=2.46\ \text{rad/s}$$

Step 4: β

At 2.46 rad/s, $|G|=20.53$ dB $=10.63$:

$$20\log\beta=20.53\Rightarrow\beta\approx10.6$$

Step 5: T

Zero about one decade below $\omega_{gc}'$:

$$\frac1T\approx\frac{2.46}{10}\approx0.25\ \text{rad/s}\Rightarrow T=4\ \text{s},\quad\beta T=42.5\ \text{s}$$

Pole at $1/\beta T=0.0235$ rad/s.

$$G_c(s)=\frac{1+4s}{1+42.5s}$$

Network realisation (passive R-C lag): $T=R_2C$, $\beta=\frac{R_1+R_2}{R_2}=10.6$; e.g. $C=10\ \mu$F, $R_2=400\ \text{k}\Omega$, $R_1=3.84\ \text{M}\Omega$.

Step 6: Compensated system

$$G_c(s)G(s)=\frac{30(1+4s)}{s(1+0.1s)(1+0.2s)(1+42.5s)}$$ $$PM=180^\circ-135.4^\circ=44.6^\circ\ (\ge40^\circ),\qquad GM=13.9\ \text{dB}$$

Answer: $G_c(s)=\dfrac{1+4s}{1+42.5s}$ (β ≈ 10.6, T = 4 s) with K = 30 gives Kv = 30 s⁻¹ and PM ≈ 44.6°.