Question 5 of 5: Root locus of a driver–train closed loop

Question 5: Root locus of a driver–train closed loop (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.

Find. (a) The root locus of the unity-feedback loop as \(K\) sweeps over the positive reals, with the equivalent open-loop poles and zeros, the real-axis segments, the asymptotes and the resulting stability verdict. (b) A reasoned discussion of the model’s limitations as a description of a real driver stopping a real train.

Approach. Assemble \(G = G_dG_t\) numerically, then re-arrange the characteristic equation into the form \(1 + K\,\hat G(s) = 0\) so that \(K\) — which appears inside \(G_d\), not as a leading gain — becomes the locus parameter. Sketch the locus of \(\hat G\), and read the stability verdict from the sign of the characteristic polynomial’s constant term.

1. Part (a) — reduce the train dynamics to numbers. The numerator and denominator constants are $$k_bfK_p = 142.5 \times 0.24 \times 47.5 = 1624.5, \qquad M(1+k_e) = 8000 \times 1.1 = 8800 ,$$ so $$G_t(s) = \frac{1624.5}{8800\,s(1.2s+1)} = \frac{0.18460}{s(1.2s+1)} .$$ The train is a Type 1 plant with a single lag corner at \(1/\tau = 0.8333\) rad/s.
2. Part (a) (continued) — assemble the open-loop transfer function. With \(h = 0.003\) and \(L/2 = 0.5\), $$G_d(s) = 0.003\left(1+\frac{K}{s}\right)\frac{s-0.5}{s+0.5} = 0.003\,\frac{(s+K)(s-0.5)}{s\,(s+0.5)} ,$$ so $$G(s) = G_d(s)G_t(s) = \underbrace{0.003 \times 0.18460}_{c\;=\;5.538\times10^{-4}} \cdot\frac{(s+K)(s-0.5)}{s^2\,(s+0.5)(1.2s+1)} .$$ Three structural features decide everything that follows: a double pole at the origin (one integrator from the train, one from the driver’s integral action), a right-half-plane zero at \(s=+0.5\) contributed by the first-order Padé factor that stands in for the driver’s reaction delay, and a zero at \(s=-K\) that moves with the parameter being swept.
3. Part (a) (continued) — recast the characteristic equation in \(K\). Because \(K\) sits inside a numerator factor rather than multiplying the whole loop, the locus cannot be drawn from \(G\) directly. Setting \(1+G(s)=0\) and multiplying out, $$s^2(s+0.5)(1.2s+1) + c\,(s+K)(s-0.5) = 0 ,$$ then splitting the \(K\)-free and \(K\)-proportional parts, $$\boxed{1 + K\cdot\frac{c\,(s-0.5)} {s^2(s+0.5)(1.2s+1) + c\,s(s-0.5)} = 0}$$ This is a standard \(1+K\hat G = 0\) problem. Its equivalent open-loop zero is the Padé zero at \(s = +0.5\), and its equivalent poles are the roots of \(1.2s^4 + 1.6s^3 + 0.50055\,s^2 - 2.769\times10^{-4}s\), namely $$s = 0,\quad s = +5.522\times10^{-4},\quad s = -0.50279,\quad s = -0.83110 .$$
4. Part (a) (continued) — sketch the locus. With four equivalent poles and one equivalent zero there are \(4-1 = 3\) asymptotes, at \(\pm60^\circ\) and \(180^\circ\), radiating from the centroid $$\sigma_a = \frac{(0 + 5.522\!\times\!10^{-4} - 0.50279 - 0.83110) - (+0.5)}{3} = \frac{-1.83334}{3} = -0.6111 .$$ The real-axis segments follow the odd-count rule: the interval \([+5.522\times10^{-4},\,+0.5]\) lies between a pole and the zero and is on the locus, as is \([-0.83110,\,-0.50279]\) and the whole of \((-\infty,-0.83110]\). So one branch runs from the small positive pole rightward into the zero at \(+0.5\), two branches break away from the segment near the origin into the complex plane, and one runs left to infinity.

5. Part (a) (continued) — the stability verdict, which the locus makes unavoidable. Expanding the characteristic polynomial, $$1.2s^4 + 1.6s^3 + (0.5+c)s^2 + c(K-0.5)s - \tfrac12 cK = 0 .$$ The constant term is \(-\tfrac12 cK\), which is negative for every \(K>0\) while the leading coefficient is positive. A polynomial with real coefficients whose leading and constant terms have opposite signs must have at least one positive real root, so $$\boxed{\text{the closed loop has a right-half-plane pole for every } K>0 \text{ — the model is unstable at every driver gain}}$$ Direct root-finding bears this out: at \(K = 0.1,\,1,\,10,\,100\) the positive real root sits at \(0.0076\), \(0.0224\), \(0.0634\) and \(0.1564\) respectively, growing steadily with \(K\). This is not a marginal case that a careful gain choice can rescue — it is the branch from the equivalent pole at \(+5.5\times10^{-4}\) to the Padé zero at \(+0.5\), and it lies wholly in the right half-plane for the entire sweep. The mark-earning statement for part (a) is therefore not just the shape of the locus but the conclusion that no positive \(K\) stabilises this model, which is exactly the observation part (b) asks to be explained.

Part (b) asks for judgement rather than algebra, and the analysis above supplies its strongest evidence: a model that predicts no human driver can ever stop a train stably is plainly missing something, since drivers do it every day.

Part (b) — why this model is not an accurate description of a real driver–train situation. The most telling objection is the one part (a) produced. The model is unstable for every positive driver gain, which contradicts everyday experience; the defect is structural rather than numerical. It arises from the combination of a double integrator — one from the train’s velocity dynamics and one from the driver’s assumed integral action — with a right-half-plane zero at \(+L/2\). A loop with two poles at the origin and a right-half-plane zero cannot be stabilised by adjusting a single gain, so the model forecloses the very behaviour it is meant to reproduce.

The right-half-plane zero itself is an artefact, not physics. The factor \((s-L/2)/(s+L/2)\) is a first-order Padé approximation to the transport delay \(e^{-sT_d}\) with \(T_d = 4/L\); it reproduces a delay’s phase lag near \(\omega \ll L/2\) but replaces the delay’s unit magnitude with a genuine unstable zero, and it produces a non-physical instantaneous wrong-way jump in the step response. A real driver’s reaction time should be modelled as an actual delay, whose effect is to impose a finite bandwidth ceiling rather than to make the loop unconditionally unstable.

Several further idealisations matter for a real stop. A human operator does not behave as a fixed linear PI element: the response is adaptive and nonlinear, with a dead-band near small speed errors, saturation once the brake demand reaches full application, a threshold below which no correction is made at all, and learning across repeated stops. Reaction time itself is variable and state-dependent — fatigue, workload and expectancy change it — whereas the model fixes \(L\). The train side is equally simplified: the first-order lag \(1/(\tau s+1)\) collapses pneumatic brake-pipe propagation, valve dynamics, wheel–rail adhesion (which varies with contamination and weather and sets a hard limit on retardation), brake-pad friction fade with temperature and speed, longitudinal coupler slack and the resulting in-train force oscillations, and grade forces — all into one time constant. Mass \(M\) is taken constant, though loading varies over a wide range and the effective inertia includes rotating components (which is presumably what the \((1+k_e)\) factor is meant to capture, in a very coarse way).

The control structure is also unrepresentative. The model is a pure single-loop, single-input velocity regulator around a reference \(v_r\), whereas a real driver runs a position-and-velocity task — stopping accurately at a mark — with feedforward knowledge of the braking curve, the gradient and the signalling, plus a supervisory automatic train protection system that intervenes independently. Braking is also physically one-sided: brakes can only decelerate, so the loop is inherently asymmetric in a way a linear model cannot express. Finally, the parameter values are quoted to three or four significant figures for an individual driver, which overstates the precision available for a human operator whose gain and delay vary from stop to stop.

A defensible improvement would keep the plant Type 1, replace the Padé factor with an explicit delay \(e^{-sT_d}\) analysed by frequency response (Routh–Hurwitz is unavailable for a transcendental characteristic equation), replace the pure integral term with the lead–lag-plus-delay crossover model standard for human operators, and add rate and amplitude saturation on the brake command. The resulting loop is conditionally stable with a finite gain margin, which matches observed behaviour and lets the model answer the question it was built for.