Question 5 of 5: Root-locus design on an open-loop unstable, non-minimum-phase plant

Question 5: Root-locus design on an open-loop unstable, non-minimum-phase 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 plant with two right-half-plane poles and two left-half-plane zeros of the same magnitudes, inside unity feedback, together with the printed locus — a circle of radius $\sqrt2$ centred on the origin, plus two real-axis segments.

Find. The design poles for $\zeta=0.707$ and their gain and settling time; the design poles for a settling time three times shorter at the same damping; the open-loop angle sum at that new point; whether a PD zero can supply the deficiency; and a PI controller that removes the steady-state error.

Approach. Read the design point off the printed circle rather than solving for it, then use the magnitude condition for the gain and $4/|\sigma|$ for the settling time. For the faster target, scale the real part by three and keep the $45^\circ$ ray, evaluate the angle condition there to see how far the point is off the locus, and test whether a real zero could close the gap. Part (g) adds the PI and re-solves the angle and magnitude conditions on the compensated locus.

1. Part (a) — read the design point off the circle. The closed-loop polynomial is $$(s-2)(s-1)+K(s+2)(s+1)=(1+K)s^{2}-3(1-K)s+2(1+K) ,$$ so the product of the two roots is $2(1+K)/(1+K)=2$ for every gain: a complex pair therefore satisfies $|s|^{2}=2$, which is the circle of radius $\sqrt2$ the paper prints. A damping ratio of $0.707=1/\sqrt2$ is the $45^\circ$ ray, and the ray meets the circle at the lattice point $$\boxed{s_{d}=-1\pm j}$$ which is why the question says the scaled graph makes numerical work unnecessary. Checking, $-\mathrm{Re}(s_{d})/|s_{d}|=1/\sqrt2=0.7071$.
2. Part (b) — gain from the magnitude condition. $$K=\frac{|(s_{d}-2)(s_{d}-1)|}{|(s_{d}+2)(s_{d}+1)|} =\frac{|{-3+j}|\,|{-2+j}|}{|1+j|\,|j|} =\frac{3.162\times2.236}{1.414\times1}=\boxed{5.00}$$ The coefficient map confirms it: the real part of the pair is $1.5(1-K)/(1+K)$, which equals $-1$ exactly at $K=5$. Note the loop needs $K>1$ merely to be stable, since the $s$ coefficient $-3(1-K)$ must be positive — a consequence of the two unstable open-loop poles.
3. Part (c) — settling time from the real part. $$T_{s}=\frac{4}{|\sigma|}=\frac{4}{1}=\boxed{4.0\ \text{s}}$$
4. Part (d) — a three-times-faster target at the same damping. One third of the settling time means $T_{s}'=4/3=1.333$ s, so $\sigma'=4/T_{s}'=3$; keeping $\zeta=0.707$ keeps the $45^\circ$ ray, so $\omega_{d}'=\sigma'$ and $$\boxed{s_{d}'=-3\pm3j} .$$ This point is at radius $3\sqrt2=4.243$, well outside the circle $|s|=\sqrt2$, so it is certainly not on the uncompensated locus — no gain can reach it.
5. Part (e) — the open-loop angle sum at the new point. Measuring each vector from $s_{d}'=-3+3j$, $$\angle(s_{d}'+1)=123.69^\circ,\quad \angle(s_{d}'+2)=108.43^\circ,\quad \angle(s_{d}'-1)=143.13^\circ,\quad \angle(s_{d}'-2)=149.04^\circ ,$$ so $$\boxed{\textstyle\sum\angle z-\sum\angle p =(123.69+108.43)-(143.13+149.04)=-60.04^\circ} .$$ Since this is not an odd multiple of $180^\circ$, the point is off the locus by $119.96^\circ$, which quantifies what part (d) already suggested.
6. Part (f) — the PD zero cannot supply the deficiency. To put the point on the compensated locus the added zero would have to contribute $$\theta_{c}=-180^\circ-(-60.04^\circ)=\boxed{-119.96^\circ}$$ (or $+240.04^\circ$ on the other branch). But a real zero at $-z$, seen from a point in the upper half-plane, always subtends an angle strictly between $0^\circ$ and $180^\circ$ — positive, and less than a straight angle — whichever side of the point it is placed. In one line: the claim is invalid, because a PD zero can only add a positive angle between $0^\circ$ and $180^\circ$, whereas this design point needs $-120^\circ$. What would work is an extra pole at $s=-1.270$, which contributes exactly $-119.96^\circ$; that is a lag or lead-lag structure, not a PD.
7. Part (g) — PI controller, angle condition first. Adding $C(s)=K(s+0.1)/s$ puts a pole at the origin and a zero at $-0.1$, a near-cancelling pair whose net angle at the old design point is only $\angle(s_{d}+0.1)-\angle s_{d}=-3.01^\circ$ — small by design, so the dominant pair barely moves. Re-solving the angle condition along the $\zeta=0.707$ ray for the compensated loop $K(s+0.1)(s+2)(s+1)/[s(s-2)(s-1)]$ gives $$\boxed{s_{d,\text{PI}}=-1.022\pm1.022j}$$ with the angle sum returning $-180.000^\circ$ there.
8. Part (g) — gain, controller constants and the third pole. The magnitude condition at that point gives $$K=\frac{|s(s-2)(s-1)|}{|(s+0.1)(s+2)(s+1)|}\bigg|_{s_{d,\text{PI}}}=5.249 \;\Longrightarrow\; \boxed{K_{p}=5.249,\qquad K_{i}=0.1K_{p}=0.525}$$ for $C(s)=K_{p}+K_{i}/s$. The third closed-loop pole lands at $s=-0.080$, close to the PI zero at $-0.1$, so its residue is small and the transient is still governed by the designed pair; all three closed-loop poles are in the left half-plane.
9. Part (g) — confirm the steady-state improvement. Without the PI the loop is Type 0 with $K_{p,\text{pos}}=5\times(2)(1)/[(-2)(-1)]=5$, so a unit step leaves $$e_{ss}=\frac{1}{1+K_{p,\text{pos}}}=\frac{1}{6}=0.167 .$$ The PI adds the free integrator, making the loop Type 1, so $$\boxed{e_{\text{step}}=0,\qquad K_{v}=\lim_{s\to0}sL=0.525,\qquad e_{\text{ramp}}=\frac{1}{K_{v}}=1.905}$$ — the step error is eliminated, which is what the question asked for, at the price of a finite ramp error.

The plant in this question is about as awkward as a second-order plant can be: two poles in the right half-plane, so the loop is unstable for small gains, and two zeros mirroring them, so the locus is trapped on the circle $|s|=\sqrt2$ no matter how large the gain grows. That single geometric fact answers parts (a), (d) and (f) together. The bandwidth is bounded above by the zeros' positions, the demand for a three-times-faster response asks for a point the locus can never visit, and the engineer's PD proposal fails not because of arithmetic but because a zero adds phase lead where lag is needed.