Lead compensator for K/[s(s+10)(s+2)], Kv ≥ 800

A lead network $\dfrac{1+Ts}{1+\alpha Ts}$ (unity DC gain) is placed so that its maximum phase lead occurs at the new gain crossover frequency.

Step 1: Gain for $K_v$

$$K_v = \lim_{s\to0} s\,\frac{K}{s(s+10)(s+2)} = \frac{K}{20} = 800 \;\Rightarrow\; K = 16000$$ $$G(s) = \frac{16000}{s(s+2)(s+10)} = \frac{800}{s(0.5s+1)(0.1s+1)}$$

Step 2: Uncompensated margins

$$\begin{aligned} |G(j\omega)| &= 1 \text{ at } \omega_{gc} = 24.5\ \text{rad/s} \\ \angle G &= -90^\circ - \tan^{-1}\frac{24.5}{2} - \tan^{-1}\frac{24.5}{10} = -243.2^\circ \\ PM &= -63.2^\circ,\quad GM = -36.5 \text{ dB at } \omega_{pc} = 4.47\ \text{rad/s} \end{aligned}$$

The system is badly unstable.

Step 3: Phase lead needed

$$\phi_m = 50^\circ - (-63.2^\circ) + 5^\circ \approx 118^\circ$$

One lead network gives at most about $60^\circ$ in practice, so three identical lead stages are used. A first trial of $3 \times 40^\circ$ gives only $PM \approx 43^\circ$ because the crossover shifts, so $3 \times 45^\circ$ is chosen:

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

Step 4: New gain crossover

Three stages add $3\times10\log(1/\alpha) = 22.96$ dB at $\omega_m$, so $\omega_m$ is where $|G| = \alpha^{1.5} = 0.0711$:

$$\frac{16000}{\omega\sqrt{\omega^2+4}\sqrt{\omega^2+100}} = 0.0711 \;\Rightarrow\; \omega_m = 60.6\ \text{rad/s}$$

Step 5: Pole–zero locations

$$\begin{aligned} T &= \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{60.55\times0.4142} = 0.0399\ \text{s} \\ \text{zero: } s &= -\frac{1}{T} = -25.1 \\ \text{pole: } s &= -\frac{1}{\alpha T} = -146.2 \end{aligned}$$ $$G_c(s) = \left(\frac{1+0.0399s}{1+0.00684s}\right)^3 = 198.0\left(\frac{s+25.1}{s+146.2}\right)^3$$

Step 6: Check

$$G_cG = \frac{16000(1+0.0399s)^3}{s(s+2)(s+10)(1+0.00684s)^3}$$

At $\omega = 60.6$: plant phase $= -90^\circ - 88.1^\circ - 80.6^\circ = -258.7^\circ$; lead $= 3\times45^\circ = 135^\circ$; $PM = 180^\circ - 258.7^\circ + 135^\circ = 56.3^\circ$.

Answer: each stage has its zero at $s = -25.1$ and pole at $s = -146.2$ ($\alpha = 0.1716$); three such stages give $PM = 56.3^\circ \ge 50^\circ$ with $K_v = 800\ \text{s}^{-1}$. (Such a large $K_v$ needs a large lead; in practice a lag–lead compensator would give a lower bandwidth and less noise.)