Question 3 of 8: Lead Controller Design from a Steady-State-Error, Overshoot and Settling-Time Specification

Question 3: Lead Controller Design from a Steady-State-Error, Overshoot and Settling-Time Specification (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 Q3.1 (redrawn as a block diagram) — unity-feedback loop with the Lead Controller ahead of the process $G(s)$. See the official exam paper.]

Given. $G(s)=2500/[(s+1)(s+5)(s+50)]$ (Type 0); design targets $e_{ss,c}\le4\%$, $PO_c\le20\%$, $T_{settle(\pm2\%),c}\le0.5$ s.

Find. $K_{pos,u}$, $K_{pos,c}$, $K_c$; $\Phi_{m,u}$, $\omega_{cp,u}$; $\alpha$, $\tau$, $G_c(s)$; the compensated step-response specs.

Approach. The ess spec fixes $K_c$ exactly as in Question 2. The PO and $T_{settle}$ specs together fix a MINIMUM required $\zeta$ and $\omega_n$ (via $T_{settle}=4/(\zeta\omega_n)$), which in turn fixes a minimum crossover frequency $\omega_{cp,c}$ through the standard gain-crossover relation. A single lead stage is designed to provide unity gain and the required extra phase EXACTLY at that target crossover.

1. Part 1) — position constants. $K_{pos,u}=G(0)=\dfrac{2500}{(1)(5)(50)}=\boxed{10}$, so $e_{ss,u}=1/11=9.09\%$. Meeting $e_{ss,c}=4\%$ needs $K_{pos,c}=1/0.04-1=\boxed{24}$, so $$\boxed{K_c=\frac{24}{10}=2.4}.$$
2. Part 2) — uncompensated PM/crossover and the design targets. With $K_c=2.4$ applied, $K_cG(j\omega)$ crosses 0 dB at $\boxed{\omega_{cp,u}\approx10.25\ \text{rad/s}}$, where $\boxed{\Phi_{m,u}\approx20.0^\circ}$ — far short of what a 20% overshoot needs. From $PO_c\le20\%$: $\zeta=-\ln(0.2)/\sqrt{\pi^2+\ln^2(0.2)}\approx0.456$, so $\Phi_{m,c}\approx100\zeta\approx \boxed{45.6^\circ}$. From $T_{settle(\pm2\%)}\le0.5$ s: $\omega_n\ge4/(\zeta\times0.5)\approx \boxed{17.5\ \text{rad/s}}$, which via the gain-crossover relation $\omega_{cp}=\omega_n\sqrt{\sqrt{4\zeta^4{+}1}-2\zeta^2}$ requires $\boxed{\omega_{cp,c}\approx14.3\ \text{rad/s}}$ — HIGHER than the uncompensated crossover, so a LEAD network (which adds phase and gain at high frequency) is the right choice.
3. Part 3) — lead controller design and closed-loop estimate. At $\omega_{cp,u}=10.25$ rad/s the uncompensated phase margin was only $20^\circ$; pushing the crossover out further to $14.3$ rad/s makes the uncompensated phase there even more negative ($\Phi_{available}\approx7.2^\circ$), so a single lead stage would need an unrealistically large phase boost to hit $45.6^\circ$ exactly at that frequency. Choosing the practical, comfortably-achievable maximum phase $\boxed{\phi_{max}=45^\circ}$ instead (a standard single-stage design choice): $$\alpha=\frac{1-\sin45^\circ}{1+\sin45^\circ}\approx\boxed{0.172},\qquad \tau=\frac{1}{\omega_m\sqrt{\alpha}},\ \ \omega_m=\omega_{cp,c}=16.2\ \text{rad/s}\ \Rightarrow\ \boxed{\tau\approx0.149}.$$ $$\boxed{G_c(s)=2.4\cdot\frac{0.149s+1}{0.0256s+1}}\qquad(a_1\approx0.357,\ a_0=2.4,\ b_1\approx0.0256).$$ Re-checking the FULL compensated loop confirms the actual crossover sits at $\omega_{cp,c}\approx16.2$ rad/s with an achieved $\boxed{\Phi_{m,c}\approx47.7^\circ}$ — comfortably above the $45.6^\circ$ target, since the higher achieved crossover more than compensates for the modest phase shortfall. On the sketch, the compensated magnitude sits ABOVE the uncompensated curve near the new crossover (the lead's mid-band gain boost $1/\sqrt{\alpha}\approx2.4$, i.e. $+7.6$ dB) while the DC gain stays fixed at $K_c=2.4$.
4. Part 3, cont'd — step response estimate. Using $\zeta_{ach}=\Phi_{m,c}/100\approx0.477$ and $\omega_{n,ach}=\omega_{cp,c}/\sqrt{\sqrt{4\zeta_{ach}^4{+}1}-2\zeta_{ach}^2}\approx20.2$ rad/s: $$PO\approx\boxed{18.2\%}\ (\le20\%\ \checkmark),\qquad T_{settle(\pm2\%)}=\frac{4}{\zeta_{ach}\omega_{n,ach}}\approx\boxed{0.415\ \text{s}}\ (\le0.5\ \text{s}\ \checkmark),\qquad T_{rise(0-100\%)}\approx\boxed{0.116\ \text{s}},\qquad e_{ss(step\%)}=\boxed{4.0\%}\ (\text{by design}).$$ Both the overshoot and settling-time specifications are met with a small comfortable margin.