Question 3 of 5: Cascade-compensator design for a double-integrator plant

Question 3: Cascade-compensator design for a double-integrator plant (25 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.

Given. A double-integrator plant inside unity feedback, with a cascade compensator whose zero is fixed by the question and whose pole and gain are the free variables.

Find. The dominant closed-loop pole pair, the compensator pole and gain that place the locus through it, all remaining closed-loop poles, an honest assessment of the second-order approximation, and the steady-state errors to step, ramp and parabolic commands.

Approach. Translate the two transient specifications into a single design point, use the angle condition to solve for the one unknown compensator pole that puts the locus through it, use the magnitude condition for the gain, then factor the resulting cubic to find the third pole and simulate the true response to test the approximation. The steady-state part is a system-type argument on the finished loop.

1. Part (a) — convert the specifications into a design point. Inverting the overshoot relation, $$\zeta=\frac{-\ln(0.163)}{\sqrt{\pi^{2}+\ln^{2}(0.163)}}=0.5001 ,$$ a value the paper has clearly chosen so that $\zeta=0.5$ exactly. The settling time fixes the real part, $\sigma=4/T_{s}=4/1.667=2.400$, so $\omega_{n}=\sigma/\zeta=4.798$ rad/s and $\omega_{d}=\omega_{n}\sqrt{1-\zeta^{2}}=4.155$ rad/s: $$\boxed{s_{d}=-2.400\pm4.155j}$$
2. Part (b) — the angle condition leaves the compensator pole as the only unknown. With one zero at $-1$, the double plant pole at the origin and an unknown pole at $-p_{c}$, $$\angle(s_{d}+1)-2\angle s_{d}-\angle(s_{d}+p_{c})=-180^\circ .$$ Substituting $\angle(s_{d}+1)=108.61^\circ$ and $2\angle s_{d}=2\times120.00^\circ=240.01^\circ$ gives the required pole angle $\angle(s_{d}+p_{c})=108.61^\circ-240.01^\circ+180^\circ=48.61^\circ$, and therefore $$p_{c}=\sigma+\frac{\omega_{d}}{\tan48.61^\circ} =2.400+\frac{4.155}{1.1346}=\boxed{6.062} ,$$ i.e. a compensator pole at $s=-6.062$. Re-evaluating the full angle sum at $s_{d}$ returns $-180.000^\circ$, so the locus does pass through the design point.
3. Part (c) — the magnitude condition gives the gain. $$K=\frac{|s_{d}|^{2}\,|s_{d}+p_{c}|}{|s_{d}+1|} =\frac{(4.7986)^{2}\times5.5394}{4.3850}=\boxed{29.09}$$ so the compensated forward path is $G_{c}(s)G(s)=29.09(s+1)/[s^{2}(s+6.062)]$.
4. Part (d) — factor the closed-loop cubic. The characteristic polynomial is $$s^{2}(s+6.062)+29.09(s+1)=s^{3}+6.062s^{2}+29.09s+29.09 ,$$ and dividing out the designed pair $s^{2}+4.799s+23.03$ leaves a single real root: $$\boxed{s_{3}=-1.263}$$ The other two roots come back as $-2.400\pm4.155j$ to six figures, confirming the design. There are no other closed-loop poles — the third-order loop has exactly three.
5. Part (e) — the second-order approximation is poor, and the reason is geometric. Two tests both fail. The third pole is at $-1.263$, only $1.263/2.400=0.53$ times the real part of the dominant pair, where dominance conventionally needs a factor of at least five; and it is not cancelled by the zero at $-1$ either, since the pole-zero separation is $0.263$, a full 21% of the pole's own distance from the origin. Its step-response residue is $+0.327$, far from negligible. Simulating the true third-order step response against the ideal second-order model, $$\%OS_{\text{actual}}=34.7\%\ \text{vs}\ 16.3\%\ \text{designed}, \qquad T_{s,\text{actual}}=2.41\ \text{s vs}\ 1.667\ \text{s},$$ with the actual peak arriving early at $t=0.711$ s. Both specifications are missed, and missed on the pessimistic side, so the approximation is not merely inaccurate — it is optimistic, which is the dangerous direction.
6. Part (f) — steady-state errors follow from the system type. The compensated loop $L(s)=29.09(s+1)/[s^{2}(s+6.062)]$ retains both poles at the origin, so it is Type 2 and $$K_{p}=\infty,\qquad K_{v}=\infty,\qquad K_{a}=\lim_{s\to0}s^{2}L(s)=\frac{29.09}{6.062}=4.798 .$$ Hence $$\boxed{e_{\text{step}}=0,\qquad e_{\text{ramp}}=0,\qquad e_{\text{parabola}}=\frac{1}{K_{a}}=0.208}$$ for the unit parabola $r=t^{2}/2$. If the parabola is written $r=t^{2}$ instead, the error doubles to 0.417 — worth stating explicitly, because the two conventions differ by a factor of two and the question does not say which it means.

The design is textbook-correct and still unsatisfactory, which is the real lesson of part (e). Forcing a plant with two poles at the origin through a specified point using only one zero and one pole leaves no freedom to push the third closed-loop pole out of the way: the same angle condition that places the dominant pair also determines where the leftover root lands, and here it lands close in. A designer who needed the specifications actually met would either move the compensator zero further left, accept a larger $p_{c}$, or add a second zero — and would check the simulated response rather than trusting the dominant-pole formulas.