Lead compensator for K/[s(1+0.2s)], Kv≥20, PM≥44°

Lead network: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$; maximum lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB (after the $1/\alpha$ attenuation is made up by an amplifier).

Step 1: K

$$K_v=\lim_{s\to0}sG(s)=K=20\ \text{s}^{-1}\Rightarrow G(s)=\frac{20}{s(1+0.2s)}$$

Step 2: Uncompensated system

Asymptotes: $-20$ dB/dec through 26 dB at $\omega=1$; corner at 5 rad/s; then $-40$ dB/dec.

$$\frac{20}{\omega\sqrt{1+0.04\omega^2}}=1\Rightarrow0.04\omega^4+\omega^2-400=0\Rightarrow\omega^2=88.3,\ \omega_{gc}=9.40\ \text{rad/s}$$ $$PM=90^\circ-\tan^{-1}(0.2\times9.40)=90^\circ-62.0^\circ=28.0^\circ$$

Step 3: Phase lead needed

$$\phi_m=44^\circ-28.0^\circ+5^\circ=21.0^\circ$$

Step 4: α

$$\alpha=\frac{1-\sin21^\circ}{1+\sin21^\circ}=\frac{1-0.358}{1+0.358}=0.473$$

Step 5: New gain crossover

The network adds $10\log(1/0.473)=3.25$ dB at $\omega_m$. Find where $|G|=-3.25$ dB ($=\sqrt\alpha=0.688$):

$$\frac{20}{\omega\sqrt{1+0.04\omega^2}}=0.688\Rightarrow\omega_m=11.55\ \text{rad/s}$$

Step 6: T

$$T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{11.55\times0.688}=0.126\ \text{s},\qquad\alpha T=0.0595\approx0.06\ \text{s}$$

Zero at $1/T=7.94$ rad/s, pole at $1/\alpha T=16.8$ rad/s.

$$G_c(s)=\frac{1+0.126s}{1+0.06s}$$

Passive R-C network: $\dfrac{E_o}{E_i}=\alpha\dfrac{1+Ts}{1+\alpha Ts}$ with $T=R_1C$, $\alpha=\dfrac{R_2}{R_1+R_2}$; an amplifier of gain $1/\alpha\approx2.1$ restores $K_v$. E.g. $C=1\ \mu$F, $R_1=126\ \text{k}\Omega$, $R_2=113\ \text{k}\Omega$.

Step 7: Check

$$G_c(s)G(s)=\frac{20(1+0.126s)}{s(1+0.2s)(1+0.06s)}$$ $$PM=180^\circ-135.8^\circ=44.2^\circ\approx44^\circ$$

GM = ∞ (phase stays above $-180^\circ$).

Answer: $G_c(s)=\dfrac{1+0.126s}{1+0.06s}$ (α = 0.473) with K = 20 gives Kv = 20 s⁻¹ and PM ≈ 44°.