Question 7 of 8: Lead-controller design in the frequency domain

Question 7 — Lead-controller design in the frequency domain [15 + 5]

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. $G(s)$ is Type 1. At $\omega=10$: $|G(j10)|=0.0632$, $\angle G(j10)=-198.4^\circ$. Targets: $e_{ss(ramp)}=0.1\Rightarrow K_v=10$; $\Phi_m=50^\circ$ at $\omega_{cp}=10$.

Find. $a_0,a_1,b_1$, $G_c(s)$, and the compensated step specs.

Part 1 — Lead-controller calculation [15]

Approach. The DC gain sets the velocity constant; the magnitude and phase conditions at the target crossover ($|G_cG|=1$, $\angle G_cG=-130^\circ$) fix the two time constants.

1. DC gain for the ramp error. $K_v=\lim_{s\to0}sG_c(s)G(s)=G_c(0)\cdot\dfrac{100}{5\cdot10}=2a_0$. Setting $K_v=10$ gives $\boxed{a_0=5}$.
2. Conditions at $\omega_{cp}=10$. $|G_cG(j10)|=1$ with $|G(j10)|=0.0632$ needs $|G_c(j10)|=\dfrac{1}{0.0632}=15.81$. $\angle G_cG(j10)=-180^\circ+50^\circ=-130^\circ$ with $\angle G(j10)=-198.4^\circ$ needs $\angle G_c(j10)=-130^\circ+198.4^\circ=+68.4^\circ$ — a strong lead.
3. Solve for $a_1,b_1$. With $G_c(j10)=\dfrac{a_0+j10a_1}{1+j10b_1}=\dfrac{5+j10a_1}{1+j10b_1}$, imposing the magnitude $15.81$ and phase $+68.4^\circ$ gives $$\boxed{a_1=1.50,\qquad b_1=0.00552.}$$
4. Controller. $$\boxed{G_c(s)=\frac{1.50\,s+5}{0.00552\,s+1}}$$ — zero at $-a_0/a_1=-3.33$ rad/s, pole at $-1/b_1=-181$ rad/s. The zero is nearer the imaginary axis than the pole, confirming a lead network; the pole/zero ratio $\approx54$ provides the large $68^\circ$ phase boost. Check: $G_c(j10)G(j10)$ has magnitude $1.00$ and phase $-130.0^\circ$, so $\Phi_m=50^\circ$ at $\omega_{cp}=10$ exactly.

Part 2 — Compensated step estimates [5]

1. Damping. $\zeta\approx\dfrac{\Phi_m}{100}=\dfrac{50}{100}=\boxed{0.5}$.
2. Overshoot. $PO=100\,e^{-\zeta\pi/\sqrt{1-\zeta^2}}=100\,e^{-0.5\pi/0.866}=\boxed{16.3\%}$.
3. Settling time. With $\omega_{cp}=\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}$ at $\zeta=0.5$, $\omega_n=10/0.786=12.7$ rad/s, so $T_{settle}(\pm2\%)=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{6.36}=\boxed{0.63\ \text{s}}$.
4. Step error. $G_cG$ retains the integrator (Type 1), so $K_{pos}=\infty$ and $\boxed{e_{ss(step)}=0\%}$.