Lead compensator for k/[s(1+0.1s)(1+0.001s)], Kv = 1000

A lead compensator $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$) is designed on the Bode plot: it supplies its maximum lead $\phi_m$ ($\sin\phi_m=\frac{1-\alpha}{1+\alpha}$) at the new gain crossover $\omega_m=\frac{1}{T\sqrt\alpha}$, where it adds $10\log(1/\alpha)$ dB of gain.

Step 1: K from the velocity error constant

$$K_v=\lim_{s\to0}sG(s)=K=1000\ \text{s}^{-1}$$ $$G(s)=\frac{1000}{s(1+0.1s)(1+0.001s)}$$

Step 2: Uncompensated Bode plot

Corners at 10 and 1000 rad/s. Asymptotic magnitude: 60 dB at $\omega=1$, $-20$ dB/dec to 10 (40 dB), $-40$ dB/dec to 1000 ($-40$ dB), then $-60$ dB/dec.

$$\phi(\omega)=-90^\circ-\tan^{-1}0.1\omega-\tan^{-1}0.001\omega$$

From $|G|=1$: $\omega_{gc}=99.5$ rad/s, $\phi=-179.94^\circ$, so

$$PM_{unc}\approx0^\circ$$

(and $\omega_{pc}=100$ rad/s, GM ≈ 0 dB). The system is on the verge of instability.

Step 3: Required phase lead

$$\phi_m=PM_d-PM_{unc}+\epsilon=45^\circ-0^\circ+10^\circ=55^\circ$$

A margin of $\epsilon=10^\circ$ is taken because the crossover will move to a higher frequency where the plant phase is lower (a 5° margin gives only 44° and fails the check).

Step 4: α

$$\alpha=\frac{1-\sin55^\circ}{1+\sin55^\circ}=\frac{1-0.819}{1+0.819}=0.0994\approx0.1$$

With $\alpha=0.1$: $\phi_m=\sin^{-1}(0.9/1.1)=54.9^\circ$.

Step 5: New gain crossover frequency

The lead adds $10\log(1/0.1)=10$ dB at $\omega_m$, so place $\omega_m$ where $|G(j\omega)|=-10$ dB:

$$\frac{1000}{\omega\sqrt{1+0.01\omega^2}\sqrt{1+10^{-6}\omega^2}}=0.316\Rightarrow\omega_m\approx176\ \text{rad/s}$$

Step 6: Time constants

$$\begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{176\times0.316}=0.018\ \text{s}\\ \alpha T&=0.0018\ \text{s} \end{aligned}$$

Zero at $1/T=55.6$ rad/s, pole at $1/\alpha T=556$ rad/s.

$$G_c(s)=\frac{1+0.018s}{1+0.0018s}$$

An amplifier of gain $1/\alpha=10$ makes up the lead network's attenuation, so $K_v$ stays 1000.

Step 7: Compensated system and check

$$G_c(s)G(s)=\frac{1000(1+0.018s)}{s(1+0.1s)(1+0.001s)(1+0.0018s)}$$

At $\omega_{gc}'=176.8$ rad/s:

$$\begin{aligned} \phi&=-90^\circ-\tan^{-1}17.68-\tan^{-1}0.177+\tan^{-1}3.18-\tan^{-1}0.318\\ &=-90^\circ-86.8^\circ-10.0^\circ+72.5^\circ-17.6^\circ=-131.9^\circ\\ PM&=180^\circ-131.9^\circ=48.1^\circ\ (\ge45^\circ) \end{aligned}$$

GM ≈ 17.5 dB at $\omega_{pc}\approx696$ rad/s.

 dB 60 |\  uncompensated  -20
    37 |  \__ 10    compensated
     0 |------\----x----------- w
       |    100 \  176.8
       |         \__  -40/-60
 lead adds phase between 55.6 and 556 rad/s

Answer: $G_c(s)=\dfrac{1+0.018s}{1+0.0018s}$ (α = 0.1, T = 0.018 s) with K = 1000 gives Kv = 1000 s⁻¹ and PM ≈ 48° (≥ 45°).