Lead compensator for k/[s²(1+0.25s)], Ka = 10, PM 35°

A lead compensator $G_c(s)=\dfrac{1+Ts}{1+\alpha Ts}$ ($\alpha<1$) adds positive phase near the new gain crossover frequency, which raises the phase margin. Its maximum phase lead $\phi_m$ occurs at $\omega_m=\dfrac{1}{T\sqrt\alpha}$, with $\sin\phi_m=\dfrac{1-\alpha}{1+\alpha}$, and there it adds a gain of $10\log(1/\alpha)$ dB. (The design below uses the Bode plot method.)

Step 1: Gain K from the error constant

$$K_a=\lim_{s\to0}s^2G(s)=\lim_{s\to0}\frac{s^2K}{s^2(1+0.25s)}=K=10$$ $$G(s)=\frac{10}{s^2(1+0.25s)}$$

Step 2: Uncompensated system

$$|G(j\omega)|=\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}},\qquad \phi=-180^\circ-\tan^{-1}0.25\omega$$

Asymptotic Bode magnitude: $-40$ dB/dec through $20\log10=20$ dB at $\omega=1$; corner at $\omega=4$, then $-60$ dB/dec.

Solving $|G|=1$: $\omega_{gc}=2.85$ rad/s and

$$PM=180^\circ+(-180^\circ-\tan^{-1}0.713)=-35.5^\circ$$

The uncompensated system is unstable (phase is always below $-180^\circ$).

Step 3: Phase lead required

$$\phi_{req}=PM_d-PM_{unc}+\epsilon = 35^\circ-(-35.5^\circ)+\epsilon\approx 75^\circ\text{–}85^\circ$$

A single lead stage cannot do this: as the gain crossover moves right, the plant phase falls further (the $(1+0.25s)$ lag). A check shows that one lead section gives a PM of only about $8^\circ$ with $\alpha=0.1$, and stays below $25^\circ$ even as $\alpha\to0$. So we use two identical lead sections in cascade, each giving about half the lead (a double-lead compensator):

$$G_c(s)=\left(\frac{1+Ts}{1+\alpha Ts}\right)^2$$

Step 4: Choose α

Trial 1: $\phi_m=45^\circ$ per stage ($\alpha=0.172$) gives PM ≈ 34.7° (just short). Trial 2: $\phi_m\approx50^\circ$ per stage. Take $\alpha=0.13$:

$$\sin\phi_m=\frac{1-0.13}{1+0.13}=0.770\Rightarrow\phi_m=50.3^\circ\ \text{per stage}\ (100.7^\circ\ \text{total})$$

Step 5: New gain crossover frequency

Two stages add $2\times10\log(1/\alpha)=20\log(1/0.13)=17.72$ dB at $\omega_m$. So the new crossover is where the uncompensated magnitude is $-17.72$ dB:

$$\frac{10}{\omega^2\sqrt{1+0.0625\omega^2}}=0.13\Rightarrow\omega_m=\omega_{gc}'=6.39\ \text{rad/s}$$

Step 6: Time constants

$$\begin{aligned} T&=\frac{1}{\omega_m\sqrt\alpha}=\frac{1}{6.39\times\sqrt{0.13}}=0.434\ \text{s}\\ \alpha T&=0.13\times0.434=0.0564\ \text{s} \end{aligned}$$

Corner frequencies: zero at $1/T=2.30$ rad/s, pole at $1/\alpha T=17.7$ rad/s.

$$G_c(s)=\left(\frac{1+0.434s}{1+0.0564s}\right)^2$$

The attenuation $\alpha^2$ of the passive lead networks is made up by an amplifier of gain $1/\alpha^2=59.2$, so the DC gain, and hence $K_a=10$, is unchanged.

Step 7: Compensated system and check

$$G_c(s)G(s)=\frac{10(1+0.434s)^2}{s^2(1+0.25s)(1+0.0564s)^2}$$

At $\omega=6.39$ rad/s:

$$\begin{aligned} \phi&=-180^\circ-\tan^{-1}(1.60)+2\times50.3^\circ\\ &=-180^\circ-58.0^\circ+100.7^\circ=-137.3^\circ\\ PM&=180^\circ-137.3^\circ=42.7^\circ\ (\ge35^\circ) \end{aligned}$$

GM of the compensated system ≈ 12.9 dB (phase crossover at 17.1 rad/s). All closed-loop poles are in the LHP ($-27.0$, $-4.08\pm j7.90$, $-2.14\pm j1.14$).

R(s)->( )->[Amp 59.2]->[(s+2.3)/(s+17.7)]^2 ->[G(s)]-+->C
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Answer: $G_c(s)=\left(\dfrac{1+0.434s}{1+0.0564s}\right)^2$ (two lead stages, α = 0.13 each) with K = 10 gives Ka = 10 s⁻² and PM ≈ 43° (> 35°).