Question 3 of 8: Lead-controller design in the frequency domain

Question 3 — Lead-controller design in the frequency domain [20]

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Type-0 plant, $G(0)=2500/(1\cdot5\cdot50)=10$. Find. the position constants, target margin/crossover, lead network and step specs. Approach. a lead adds phase near crossover (raising $\Phi_m$) and pushes crossover higher (raising bandwidth to meet the fast settling time), while $K_c$ sets the DC gain for the error spec.

Item (1) — position constants

1. Uncompensated. $K_{pos,u}=G(0)=\boxed{10}$, so $e_{ss,u}=1/11=9.1\%$.
2. Compensated. $e_{ss,c}\le4\%\Rightarrow K_{pos,c}\ge1/0.04-1=\boxed{24}$, so $K_c=24/10=\boxed{2.4}$.

Item (2) — target margin and crossover

1. Uncompensated reads. On Fig Q3.1 (flat $+20$ dB $=20\log_{10}10$ at low frequency) the magnitude crosses 0 dB at $\boxed{\omega_{cp,u}\approx6.2\ \text{rad/s}}$, where the phase is $\approx-139^\circ$, so $\boxed{\Phi_{m,u}\approx41^\circ}$ (exact: $6.17$ rad/s, $41.2^\circ$).
2. Overshoot → damping. $PO\le20\%\Rightarrow\zeta\ge0.456$, so $\Phi_{m,c}\approx100\zeta\approx46^\circ$ (exact 2nd-order value $48^\circ$).
3. Settling → bandwidth. $T_{settle(\pm2\%)}=4/(\zeta\omega_n)\le0.5$ needs $\zeta\omega_n\ge8$, i.e. $\omega_n\ge8/0.456=17.5$ rad/s; the gain crossover must then be at least $\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}\approx14$ rad/s. Choose $\boxed{\omega_{cp,c}=18\ \text{rad/s}}$ to leave headroom, and design to $\boxed{\Phi_{m,c}\ge50^\circ}$ (a little above the exact $48^\circ$).
4. Gain-adjusted margin. After applying $K_c=2.4$ the crossover moves to $\omega=10.3$ rad/s with only $\Phi_m=20^\circ$; at $14$ rad/s only $8^\circ$ remains, and at the chosen $18$ rad/s the gain-adjusted margin is $-1.1^\circ$, so the lead must supply roughly $51^\circ$ or more there.

Item (3) — lead network, sketch and step specs

1. Size $\alpha$ from the magnitude condition. At the new crossover the lead contributes gain $1/\sqrt\alpha$, so $|K_cG(j18)|/\sqrt\alpha=1$. With $|K_cG(j18)|=0.335$: $\sqrt\alpha=0.335$, $\boxed{\alpha=0.112}$, and the peak phase is $\phi_{max}=\sin^{-1}\dfrac{1-\alpha}{1+\alpha}=\boxed{52.9^\circ}$. (Sizing $\alpha$ from the phase deficit at the old crossover instead gives $\alpha=0.140$. The lead's own gain then pushes crossover out to $18.8$ rad/s with only $45^\circ$ of margin, and $PO\approx20.4\%$ misses the spec.)
2. Place the peak at crossover. $\tau=\dfrac{1}{\omega_{cp,c}\sqrt{\alpha}}=0.166$, so the zero is at $-1/\tau=-6.03$ and the pole at $-1/(\alpha\tau)=-53.7$ ($\alpha\tau=0.0186$).
3. Controller. $$\boxed{G_c(s)=2.4\,\frac{0.166\,s+1}{0.0186\,s+1}}\qquad(\text{zero }-6.03,\ \text{pole }-53.7).$$ A numerical check puts the compensated crossover at $\omega=18.0$ rad/s with $\Phi_m=-1.1^\circ+52.9^\circ=51.8^\circ$.
4. Step specs. With $\zeta\approx\Phi_m/100=0.52$ and $\omega_n=23.3$ rad/s: $PO=\boxed{14.9\%}$ ($\le20\%$ ✓), $T_{settle(\pm2\%)}=4/(\zeta\omega_n)=\boxed{0.33\ \text{s}}$ ($\le0.5$ ✓), $T_{rise(0-100\%)}=(\pi-\beta)/\omega_d=\boxed{0.11\ \text{s}}$, and $e_{ss(step)}=1/(1+24)=\boxed{4.0\%}$ ($\le4\%$ ✓). A full fourth-order step simulation of $G_cG$ gives $PO=19.5\%$ and $T_{settle}=0.30$ s. Both specs still hold, even though the lead zero adds some overshoot beyond the second-order estimate.