Lead compensator for 5/[(s+2)(s+20)], Kv ≥ 20

Assumption: as printed, $G(s)=\frac{5}{(s+2)(s+20)}$ is type 0, so $K_v=\lim_{s\to0}sG(s)=0$ for any gain and no lead network can give $K_v=20$. The intended plant must have an integrator, so the design is done for $G(s)=\dfrac{K}{s(s+2)(s+20)}$, with the gain chosen to give $K_v=20$.

Lead network: $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$, $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, gain added there $10\log(1/\alpha)$.

Step 1: Gain

$$K_v=\lim_{s\to0}sG(s)=\frac{K}{2\times20}=20\Rightarrow K=800$$ $$G(s)=\frac{800}{s(s+2)(s+20)}=\frac{20}{s(1+0.5s)(1+0.05s)}$$

Step 2: Uncompensated system

Corners at 2 and 20 rad/s; asymptotes $-20$, $-40$, $-60$ dB/dec; 26 dB at $\omega=1$.

$$\phi=-90^\circ-\tan^{-1}0.5\omega-\tan^{-1}0.05\omega$$

$\omega_{gc}=6.03$ rad/s, PM $=1.6^\circ$; $\omega_{pc}=6.32$ rad/s, GM $=0.8$ dB. Barely stable.

Step 3: Phase lead

$$\phi_m=35^\circ-1.6^\circ+\epsilon$$

Because of the pole at 20 rad/s, the plant phase falls quickly as the crossover moves right. Trials with $\epsilon=5^\circ$–$12^\circ$ ($\phi_m=38^\circ$–$45^\circ$) give PM of only 28°–33°. A larger margin is needed: take $\phi_m\approx55^\circ$.

Step 4: α

$$\alpha=\frac{1-\sin55^\circ}{1+\sin55^\circ}\approx0.1\quad(\phi_m=54.9^\circ)$$

Step 5: New crossover

The network adds $10\log10=10$ dB, so $\omega_m$ is where $|G|=-10$ dB:

$$\frac{20}{\omega\sqrt{1+0.25\omega^2}\sqrt{1+0.0025\omega^2}}=0.316\Rightarrow\omega_m=10.5\ \text{rad/s}$$

Step 6: T

$$T=\frac{1}{10.5\times\sqrt{0.1}}=0.30\ \text{s},\qquad\alpha T=0.03\ \text{s}$$ $$G_c(s)=\frac{1+0.3s}{1+0.03s}$$

(zero at 3.33 rad/s, pole at 33.3 rad/s; an amplifier of gain $1/\alpha=10$ restores the DC gain).

Step 7: Check

$$G_c(s)G(s)=\frac{800(1+0.3s)}{s(s+2)(s+20)(1+0.03s)}$$ $$PM=180^\circ-141.9^\circ=38.1^\circ\ (\ge35^\circ),\qquad GM=11.9\ \text{dB}\ (\ge10\ \text{dB})$$

Answer (for $G=K/[s(s+2)(s+20)]$): K = 800 and $G_c(s)=\dfrac{1+0.3s}{1+0.03s}$ give Kv = 20 s⁻¹, PM ≈ 38°, GM ≈ 12 dB.