Lead compensator for K/[s(1+0.2s)], Kv=10, PM=50°

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

Step 1: K from $K_v$

$$K_v=\lim_{s\to0}sG(s)=K=10\ \text{s}^{-1}\Rightarrow G(s)=\frac{10}{s(1+0.2s)}$$

Step 2: Uncompensated system

Asymptotic magnitude: 20 dB at $\omega=1$ with $-20$ dB/dec, corner at $1/0.2=5$ rad/s, then $-40$ dB/dec. Phase $\phi=-90^\circ-\tan^{-1}0.2\omega$.

Gain crossover: $\dfrac{10}{\omega\sqrt{1+0.04\omega^2}}=1\Rightarrow0.04\omega^4+\omega^2-100=0$

$$\begin{aligned} \omega^2&=\frac{-1+\sqrt{1+16}}{0.08}=39.04\Rightarrow\omega_{gc}=6.25\ \text{rad/s}\\ PM&=180^\circ-90^\circ-\tan^{-1}(1.25)=90^\circ-51.3^\circ=38.7^\circ \end{aligned}$$

Step 3: Required phase lead

$$\phi_m=50^\circ-38.7^\circ+5^\circ=16.3^\circ$$

(5° added because the crossover moves right, where the plant has more lag.)

Step 4: α

$$\alpha=\frac{1-\sin16.3^\circ}{1+\sin16.3^\circ}=\frac{1-0.281}{1+0.281}=0.561$$

Step 5: New gain crossover

Lead gain at $\omega_m$: $10\log(1/0.561)=2.51$ dB. Find where $|G|=-2.51$ dB:

$$\frac{10}{\omega\sqrt{1+0.04\omega^2}}=\sqrt{0.561}=0.749\Rightarrow\omega_m=7.44\ \text{rad/s}$$

Step 6: T

$$\begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{7.44\times0.749}=0.179\approx0.18\ \text{s}\\ \alpha T&=0.561\times0.179=0.10\ \text{s} \end{aligned}$$ $$G_c(s)=\frac{1+0.18s}{1+0.1s}$$

Zero at 5.56 rad/s, pole at 10 rad/s. An amplifier of gain $1/\alpha\approx1.8$ makes up the network's attenuation, keeping $K_v=10$.

Step 7: Compensated system

$$G_c(s)G(s)=\frac{10(1+0.18s)}{s(1+0.2s)(1+0.1s)}$$

Check at the new crossover $\omega_{gc}'=7.47$ rad/s:

$$\begin{aligned} \phi&=-90^\circ+\tan^{-1}1.345-\tan^{-1}1.494-\tan^{-1}0.747\\ &=-90^\circ+53.4^\circ-56.2^\circ-36.8^\circ=-129.6^\circ\\ PM&=180^\circ-129.6^\circ=50.4^\circ \end{aligned}$$

GM = ∞ (phase never reaches −180°).

Bode diagram of the compensated system

Asymptotes: $-20$ dB/dec from 20 dB at $\omega=1$; corner 5 (pole) → $-40$; corner 5.56 (zero) → $-20$; corner 10 (pole) → $-40$ dB/dec.

 dB 40 |\  -20 dB/dec
    20 |  \  (w=1)
       |    \     compensated
     0 |------\---x------------- w
       |     5 \ 7.47 -20 between
       |   5.56 \__   5.56 and 10
       |          \__ -40 after 10
 deg -90|__
     -130|    \___o  PM = 50.4 deg
     -180|-------------\_______

Answer: $G_c(s)=\dfrac{1+0.18s}{1+0.1s}$ (α ≈ 0.56) with K = 10 gives Kv = 10 s⁻¹ and PM ≈ 50° at ω ≈ 7.5 rad/s.