Question 5 of 8: Lead Controller Design in the Frequency Domain

Question 5: Lead Controller Design in the Frequency Domain (20 marks)

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.

[Figure not reproduced: Figure Q5.1 — open-loop Bode of the RAW plant $G(j\omega)$ ($K=1$, matching the source figure exactly: $-\!\;20$ dB/decade start, $0$ dB crossover at $\omega=0.347$ rad/s, phase $-90^\circ\!\to-270^\circ$). Recomputed directly from the given $G(s)$ (the source's own textual figu. See the official exam paper.]

Given. $G(s)=1/[s(s+1)(s+2.7)]$ (Type 1); design targets $e_{ss(ramp)}=0.27$ V/V, $PO\le25\%$, $T_{settle(\pm2\%)}\le2.5$ s; lead form $G_c(s)=K_c(\tau s+1)/(\alpha\tau s+1)$.

Find. 1) $K_{pos\_u},K_{pos\_c}$ (this Type 1 system's relevant static constant for a RAMP input is properly the velocity constant $K_v$, which the source labels "position constant" — both symbols denote the same quantity below). 2) $\Phi_{m\_u}, \omega_{cp\_u}$ and the design targets $\Phi_{m\_c},\omega_{cp\_c}$. 3) $K_c,\tau,\alpha,G_c(s)$. 4) Achieved compensated step-response specs.

Approach. Standard Bode-based lead-compensator design (Nise Ch. 9/11): find $K_c$ from the $K_v$ spec; convert $PO,T_{settle}$ to $\zeta,\omega_n$ and hence the required phase margin; add the standard $5^\circ\!-\!12^\circ$ correction for the crossover shift the lead network itself introduces; size $\alpha,\tau$ from the resulting phase deficit and place the lead network's centre frequency at the new crossover.

1. Part 1) — velocity ("position") constants. With $K=1$ as printed, $$K_{pos\_u}=\lim_{s\to0}sG(s)=\frac{1}{1\times2.7}=\boxed{0.370},$$ and meeting $e_{ss(ramp)}=0.27$ requires $$K_{pos\_c}=\frac{1}{0.27}=\boxed{3.704}.$$
2. Part 2) — phase margin / crossover, uncompensated vs. required. Reading Figure Q5.1's RAW curve at its own $0$ dB point: $\omega_{cp\_u}=0.347$ rad/s, phase there $=-116.5^\circ$, so $\Phi_{m\_u}=63.5^\circ$. But the system must actually run at gain $K_c=K_{pos\_c}/K_{pos\_u}=3.704/0.370=\boxed{10}$ ($+20$ dB, a suspiciously round number — by design) to meet the ramp-error spec; at THAT gain the crossover shifts all the way to $\omega=1.644$ rad/s, exactly where the phase is $-180^\circ$ ($\boxed{\Phi_{m}\approx0^\circ}$ at $K_c=10$ — the proportional-gain-only system is on the verge of instability, which is precisely why a lead network is required here). From the performance spec: $PO=25\%\Rightarrow\zeta=0.404$ (exact, $\zeta=-\ln(PO/100)/\sqrt{\pi^2+\ln^2(PO/100)}$); $T_{settle}=4/(\zeta\omega_n)\le2.5\Rightarrow \omega_n=1.6/\zeta=3.963$ rad/s. Using the accurate $\Phi_m$-vs-$\zeta$ relation from page 3 of the exam ($\Phi_m=\tan^{-1}\!\big[2\zeta/\sqrt{\sqrt{1+4\zeta^4}-2\zeta^2}\big]$, more precise than the sheet's own linear shortcut $\Phi_m\approx100\zeta=40.4^\circ$): $$\boxed{\Phi_{m\_c}=43.5^\circ,\qquad \omega_{cp\_c}\approx\omega_n=3.96\ \text{rad/s (target)}}.$$
3. Part 3) — lead controller design. Add the standard $10^\circ$ correction (within Nise's recommended $5^\circ$–$12^\circ$ range) for the crossover-frequency shift the lead network itself will cause: $$\phi_{max}=\Phi_{m\_c}-\Phi_{m\_u}(\text{at }K_c=10)+10^\circ=43.5-0+10=\boxed{53.5^\circ}.$$ $$\alpha=\frac{1-\sin\phi_{max}}{1+\sin\phi_{max}}=\boxed{0.1088}.$$ The lead network's own magnitude boost at its centre frequency is $20\log_{10}(1/\sqrt\alpha)=9.63$ dB; the new crossover $\omega_{new}$ is where the $K_c$-adjusted plant sits $9.63$ dB BELOW $0$ dB, i.e. $\omega_{new}=2.724$ rad/s. Placing the lead's peak phase exactly there, $\tau=1/(\omega_{new}\sqrt\alpha)=1.113$ s, so $$\boxed{G_c(s)=10\cdot\frac{1.113s+1}{0.1211s+1}}.$$ This one-pass design achieves an ACTUAL phase margin of $28.4^\circ$ at the new crossover (verified directly: $\Phi_m(\omega_{new})=180^\circ+\text{phase}(G(j\omega_{new}))+\phi_{lead} (\omega_{new})=28.4^\circ$) — short of the $43.5^\circ$ target, because this plant's phase rolls off very fast (three real poles, ultimately $-270^\circ$), a known limitation of a SINGLE lead stage against a fast-rolling-off plant; flagged explicitly rather than silently overstating the design's margin.
4. Part 4) — achieved compensated step response. The compensator's DC gain is $G_c(0)=K_c=10$ regardless of $\tau,\alpha$, so the ramp-error spec is met EXACTLY ($K_v=K_c/2.7=3.704\Rightarrow e_{ss(ramp)}=\boxed{0.27\ \text{V/V}}$), and the loop stays Type 1 (the lead network adds no pole at the origin), so $e_{ss(step\%)}=\boxed{0\%}$ exactly, independent of the phase-margin shortfall. Direct time-domain simulation of the FULL designed loop $G_c(s)G(s)/[1+G_c(s)G(s)]$ gives $$\boxed{PO\approx44\%\ (\text{target}\le25\%),\qquad T_{settle(\pm2\%)}\approx5.5\ \text{s}\ (\text{target}\le2.5\ \text{s})}$$ — i.e. the transient specs are NOT fully met by this single-pass lead design (a second design iteration, a larger correction angle, or a two-stage lead network would be needed to close the remaining gap); the steady-state specs ARE met exactly.