Lead compensator for 4/[s(s+2)], Kv=20, PM≥50°

Lead network $G_c(s)=K_c\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$), maximum phase lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$.

Step 1: Gain for $K_v$

$$K_v=\lim_{s\to0}s\,K_c\frac{4}{s(s+2)}=2K_c=20\Rightarrow K_c=10$$ $$G_1(s)=\frac{40}{s(s+2)}=\frac{20}{s(1+0.5s)}$$

Step 2: Bode plot of $G_1$

20log20 = 26 dB at $\omega=1$, slope $-20$ dB/dec, corner 2 rad/s, then $-40$ dB/dec.

$$\frac{40}{\omega\sqrt{\omega^2+4}}=1\Rightarrow\omega^4+4\omega^2-1600=0\Rightarrow\omega_{gc}=6.17\ \text{rad/s}$$ $$PM=180^\circ-90^\circ-\tan^{-1}\frac{6.17}{2}=90^\circ-72.0^\circ=18.0^\circ$$

Step 3: Required lead

$$\phi_m=50^\circ-18^\circ+8^\circ=40^\circ$$

Step 4: α

$$\alpha=\frac{1-\sin40^\circ}{1+\sin40^\circ}=\frac{1-0.643}{1+0.643}=0.217\approx0.22$$

Step 5: New crossover

Gain added at $\omega_m$: $10\log(1/0.22)=6.58$ dB. Solve $|G_1(j\omega)|=-6.58$ dB $=\sqrt{0.22}=0.469$:

$$\frac{40}{\omega\sqrt{\omega^2+4}}=0.469\Rightarrow\omega_m=9.13\ \text{rad/s}$$

Step 6: T

$$T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{9.13\times0.469}=0.233\ \text{s},\qquad\alpha T=0.0513\ \text{s}$$

Zero at $1/T=4.28$ rad/s, pole at $1/\alpha T=19.5$ rad/s.

$$G_c(s)=10\,\frac{1+0.233s}{1+0.0513s}=45.4\,\frac{s+4.28}{s+19.5}$$

Step 7: Check

$$G_c(s)G(s)=\frac{40(1+0.233s)}{s(s+2)(1+0.0513s)}$$

At $\omega_{gc}'=9.11$ rad/s:

$$\begin{aligned} \phi&=-90^\circ-\tan^{-1}\frac{9.11}{2}+\tan^{-1}(2.12)-\tan^{-1}(0.467)\\ &=-90^\circ-77.62^\circ+64.78^\circ-25.05^\circ=-127.89^\circ\\ PM&=180^\circ-127.9^\circ\approx52^\circ\ (\ge50^\circ) \end{aligned}$$

GM = ∞.

 dB 26 |\ -20       uncompensated
       |  \__ 2   /
     0 |------\--x---x------ w
       |    6.17   9.11 compensated
       |           \__ -40

Answer: $G_c(s)=10\,\dfrac{1+0.233s}{1+0.0513s}$ (α ≈ 0.22) gives Kv = 20 s⁻¹ and PM ≈ 52°.