Lead compensator for 2/[s(1+0.5s)], Kv > 20, PM 50°

Lead compensator: $G_c(s)=K_c\dfrac{1+Ts}{1+\alpha Ts}$, $\alpha<1$, with maximum lead $\sin\phi_m=\frac{1-\alpha}{1+\alpha}$ at $\omega_m=\frac{1}{T\sqrt\alpha}$, where it raises the magnitude by $10\log(1/\alpha)$ dB.

Step 1: Gain for $K_v$

$$K_v=\lim_{s\to0}s\,G_c(s)G(s)=K_c\times2=20\Rightarrow K_c=10$$

Gain-adjusted plant:

$$G_1(s)=\frac{20}{s(1+0.5s)}$$

Step 2: Uncompensated (gain-adjusted) system

Asymptotes: $-20$ dB/dec (26 dB at $\omega=1$), corner at 2 rad/s, then $-40$ dB/dec.

$$\frac{20}{\omega\sqrt{1+0.25\omega^2}}=1\Rightarrow0.25\omega^4+\omega^2-400=0\Rightarrow\omega_{gc}=6.17\ \text{rad/s}$$ $$PM=90^\circ-\tan^{-1}(0.5\times6.17)=90^\circ-72.0^\circ=18.0^\circ$$

Step 3: Phase lead needed

$$\phi_m=50^\circ-18^\circ+\epsilon$$

With $\epsilon=5^\circ$ the check gives PM = 49.8° (just short), so take $\epsilon\approx10^\circ$: $\phi_m\approx42^\circ$.

Step 4: α

$$\alpha=\frac{1-\sin42^\circ}{1+\sin42^\circ}=0.198\approx0.2\quad(\phi_m=41.8^\circ)$$

Step 5: New gain crossover

Lead gain at $\omega_m$: $10\log(1/0.2)=6.99$ dB. Find where $|G_1|=-6.99$ dB $(=\sqrt{0.2}=0.447)$:

$$\frac{20}{\omega\sqrt{1+0.25\omega^2}}=0.447\Rightarrow\omega_m=9.35\ \text{rad/s}$$

Step 6: T

$$T=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{9.35\times0.447}=0.239\approx0.24\ \text{s},\qquad\alpha T=0.048\ \text{s}$$

Zero at 4.17 rad/s, pole at 20.8 rad/s.

$$G_c(s)=10\,\frac{1+0.24s}{1+0.048s}=50\,\frac{s+4.17}{s+20.8}$$

Step 7: Check

$$G_c(s)G(s)=\frac{20(1+0.24s)}{s(1+0.5s)(1+0.048s)}$$ $$PM=180^\circ-126.1^\circ=53.9^\circ\ (\ge50^\circ),\quad GM=\infty$$

 dB 26 |\  -20
       |  \__ 2 (corner)
     0 |------\----x------- w
       |    6.17\  9.37 (new wgc)
       |         \__ -40
  lead adds phase between 4.17 and 20.8

Answer: $G_c(s)=10\,\dfrac{1+0.24s}{1+0.048s}$ (α = 0.2): Kv = 20 s⁻¹, PM ≈ 54° at ω ≈ 9.4 rad/s.