Question 5 of 7: Proportional vs. lead control; margins and dominant model

Question 5 — Proportional vs. lead control; margins and dominant model [5 + 10 + 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. $G(s)=\dfrac{1}{s(s+1)(s+2.7)}$ (type 1); measured $G_m=20$ dB at $1.64$ rad/s, $P_m=63.5^\circ$ at $0.347$ rad/s (for $K=1$).

Find. Proportional $K_p$ and its adequacy (A); lead $K_c,\tau,\alpha$ (B); dominant-model specs (C).

[Figure not reproduced: Figure Q5.1 (redrawn) — cascade compensator $G_c(s)$ ahead of the type-1 plant. See the official exam paper.]

Part A — Proportional control [5]

Approach. The ramp error of a type-1 loop is $1/K_v$; find $K_p$, then test stability against the plant's $20$ dB gain margin.

1. Gain for the ramp error. $K_v=\lim_{s\to0}sK_pG(s)=\dfrac{K_p}{(1)(2.7)}$, and $e_{ss,\text{ramp}}=\dfrac{1}{K_v}=\dfrac{2.7}{K_p}=0.27\Rightarrow \boxed{K_p=10}.$
2. Stability check. Raising the gain to $K_p=10$ ($+20$ dB) lifts the whole magnitude curve by exactly the $20$ dB gain margin, so the new gain-crossover moves to $1.64$ rad/s — precisely the phase-crossover ($\angle L=-180^\circ$). The phase margin collapses to $\approx0^\circ$.
3. Adequacy. With $P_m\approx0^\circ$ the loop is only marginally stable, so proportional control is not adequate — it exhausts the entire gain margin and leaves no damping. A dynamic (lead) compensator is required.

Part B — Lead-compensator design [10]

Approach. Keep the DC gain $K_c=10$ for the ramp error, then place the lead so the phase at the new crossover $\omega_{cp}=3$ rad/s gives $\Phi_m=45^\circ$.

1. Compensator DC gain. The ramp-error requirement is unchanged, so $K_v=K_c/2.7=1/0.27\Rightarrow \boxed{K_c=10}$ (the lead's DC gain equals the proportional value).
2. Phase deficit at $\omega_{cp}=3$. $\angle G(j3)=-90^\circ-\tan^{-1}3-\tan^{-1}\tfrac{3}{2.7}=-209.6^\circ$, so to obtain $\angle L(j3)=-180^\circ+45^\circ=-135^\circ$ the lead must supply $\phi_m=-135^\circ-(-209.6^\circ)=\boxed{74.6^\circ}.$
3. Lead ratio. $\alpha=\dfrac{1-\sin\phi_m}{1+\sin\phi_m}=\dfrac{1-\sin74.6^\circ}{1+\sin74.6^\circ}=\boxed{0.018}.$
4. Corner frequencies. Centre the lead's maximum phase at $\omega_{cp}$: $\omega_m=\dfrac{1}{\tau\sqrt{\alpha}}=3\Rightarrow \tau=\dfrac{1}{3\sqrt{0.018}}=2.46$ s. The zero is at $-1/\tau=-0.41$ and the pole at $-1/(\alpha\tau)=-22.1$, giving $$\boxed{G_c(s)=10\,\frac{2.46s+1}{0.045s+1}}.$$

Part C — Dominant second-order estimate [5]

Approach. Use the standard phase-margin–damping correlation and the crossover–$\omega_n$ relation for the compensated $\Phi_m=45^\circ$, $\omega_{cp}=3$ rad/s.

1. Damping from phase margin. For $P_m\le70^\circ$, $\zeta\approx\dfrac{P_m}{100}=\dfrac{45}{100}=\boxed{0.45}.$
2. Natural frequency. $\omega_{cp}=\omega_n\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}\Rightarrow \omega_n=\dfrac{3}{0.821}=\boxed{3.65\ \text{rad/s}}.$
3. Step-response estimates. $e_{ss}=0$ (type 1 to a step); $\text{PO}=e^{-\zeta\pi/\sqrt{1-\zeta^2}}\times100\approx\boxed{20\%}$; $T_{s(\pm2\%)}=\dfrac{4}{\zeta\omega_n}=\dfrac{4}{1.64}\approx\boxed{2.4\ \text{s}}.$