Lag compensator for 2/[s(s+1)(s+2)], Kv = 5, PM 40°

A lag compensator $G_c(s)=K_c\dfrac{1+Ts}{1+\beta Ts}$ ($\beta>1$) is used here: it gives high gain at low frequency (for $K_v$) while its attenuation of $20\log\beta$ at high frequency lowers the gain crossover to a frequency where the plant already has enough phase margin.

Step 1: Gain for the required $K_v$

$$G(s)H(s)=\frac{2}{s(s+1)(s+2)}=\frac{1}{s(1+s)(1+0.5s)}\Rightarrow K_v=\lim_{s\to0}sGH=1$$

To get $K_v=5$ the compensator must give a DC gain $K_c=5$:

$$G_1(s)=\frac{5}{s(1+s)(1+0.5s)}$$

Step 2: Uncompensated (gain-adjusted) system

$$|G_1|=\frac{5}{\omega\sqrt{1+\omega^2}\sqrt{1+0.25\omega^2}},\qquad\phi=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega$$

• $\omega_{pc}=\sqrt2=1.414$ rad/s, GM $=-4.4$ dB; $\omega_{gc}=1.80$ rad/s, PM $=-13^\circ$: unstable.

Step 3: New gain crossover frequency

Required phase at $\omega_{gc}'$: $-180^\circ+40^\circ+\epsilon$ with $\epsilon=10^\circ$ (for the lag of the compensator), i.e. $-130^\circ$:

$$-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega=-130^\circ\Rightarrow\omega_{gc}'=0.49\ \text{rad/s}$$

Step 4: β

At 0.49 rad/s, $|G_1|=18.95$ dB $=8.86$. The lag network must cut this to 0 dB:

$$20\log\beta=18.95\ \text{dB}\Rightarrow\beta=8.86$$

Step 5: T

Place the zero one decade below the new crossover so the compensator lag there is small:

$$\frac1T=\frac{\omega_{gc}'}{10}\approx0.05\ \text{rad/s}\Rightarrow T=20\ \text{s},\quad\beta T=177\ \text{s}$$

Pole at $1/\beta T=0.00565$ rad/s.

$$G_c(s)=5\,\frac{1+20s}{1+177s}=0.565\,\frac{s+0.05}{s+0.00565}$$

Step 6: Check

$$G_c(s)G(s)H(s)=\frac{5(1+20s)}{s(1+s)(1+0.5s)(1+177s)}$$

 dB 40 |\__ uncompensated
       |   \___
     0 |---x---\------------ w
       | 0.49   \ 1.8
       | compensated curve is
       | 19 dB lower above 0.05

All specifications are met (PM ≥ 40°, GM ≥ 10 dB, $K_v=5$). The price is a lower bandwidth (slower response).

Answer: $G_c(s)=5\,\dfrac{1+20s}{1+177s}$ (β ≈ 8.9, T = 20 s): Kv = 5 s⁻¹, PM ≈ 45°, GM ≈ 14 dB.