Question 1 of 6: Disturbance rejection and phase margin of a cruise-control loop

Question 1: Disturbance rejection and phase margin of a cruise-control 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. A unity-feedback cruise controller in which the road-grade torque is subtracted at the plant input, so the vehicle dynamics receive the actuator command less the disturbance.

Find. The steady-state speed error caused by a step grade, the steady-state speed error caused by the undulating road, the phase margin of the loop, and one modification of $C(s)$ that raises the margin while leaving the steady-state tracking performance intact.

Approach. Write the error in terms of both inputs directly from the block diagram, evaluate it at $s = 0$ for the step and at $s = j0.5$ for the sinusoid, then solve the magnitude condition $|L(j\omega)| = 1$ for the gain crossover and read the phase there.

1. Part (a) — write the error in terms of both inputs. The plant receives $u - d$ and the controller sees $r - y$, so $$y = P(s)\bigl[u - d\bigr], \qquad u = C(s)\bigl[r - y\bigr].$$ Eliminating $u$ gives $y\bigl[1 + C(s)P(s)\bigr] = C(s)P(s)\,r - P(s)\,d$, and subtracting from $r$ leaves the error in the compact form $$\boxed{\;e = r - y = \frac{r + P(s)\,d}{1 + L(s)}, \qquad L(s) = C(s)P(s)\;}$$ The disturbance is weighted by the whole plant because it is injected upstream of it; had the grade torque been subtracted at the plant output, the weight would have been unity instead.
2. Evaluate the loop gain at DC. Both first-order factors reduce to unity at $s = 0$, so $$L(0) = \frac{1000}{(1)(1)} = 1000, \qquad P(0) = 100 .$$ The loop has no integrator, so it is Type 0: a constant disturbance leaves a permanent, but small, speed offset.
3. Apply the final-value theorem to the step grade. With $r$ held constant the reference contributes nothing further to the change in error, and setting $d(s) = 1/s$, $$e_{ss} = \lim_{s \to 0} s\,\frac{P(s)\,\tfrac{1}{s}}{1 + L(s)} = \frac{P(0)}{1 + L(0)} = \frac{100}{1 + 1000},$$ $$\boxed{\;e_{ss} = \frac{100}{1001} = 0.0999\ \text{speed units}\;}$$ Physically the loop absorbs all but about one part in a thousand of the grade torque: the vehicle loses roughly a tenth of a unit of speed on the incline and holds that loss indefinitely.
4. Part (b) — move to the frequency domain. A sinusoidal disturbance is not a case for the final-value theorem; the steady-state error is itself a sinusoid at the same frequency, scaled and shifted by the transfer function from $d$ to $e$ evaluated on the imaginary axis. At $\omega = 0.5\ \text{rad/s}$, $$P(j0.5) = \frac{100}{1 + j5} = 19.612\,\angle{-78.690^\circ},$$ $$L(j0.5) = \frac{1000}{(1+j5)(1+j2.5)} = \frac{1000}{-11.5 + j7.5} = -61.008 - j39.788 .$$
5. Form the sensitivity at that frequency. Adding one to the loop gain, $$1 + L(j0.5) = -60.008 - j39.788, \qquad |1 + L(j0.5)| = 72.000, \qquad \angle\bigl(1+L\bigr) = -146.454^\circ .$$ The ratio that multiplies the disturbance is therefore $$\left|\frac{P}{1+L}\right|_{\omega = 0.5} = \frac{19.612}{72.000} = 0.27238, \qquad \angle\frac{P}{1+L} = -78.690^\circ + 146.454^\circ = +67.764^\circ .$$
6. Scale by the disturbance amplitude. The forcing has amplitude 3, so $$\boxed{\;e_{ss}(t) = 3 \times 0.27238\,\sin\!\bigl(0.5t + 67.76^\circ\bigr) = 0.817\,\sin\!\bigl(0.5t + 67.76^\circ\bigr)\;}$$ The undulating road produces a speed ripple of amplitude $0.817$ — more than eight times the steady offset of part (a), because at $0.5\ \text{rad/s}$ the loop gain has already fallen from 1000 to about 73 and the plant itself is rolling off more slowly than the loop can compensate.
7. Part (c) — locate the gain crossover. Setting the loop magnitude to unity and squaring, $$|L(j\omega)|^2 = \frac{10^6}{(1 + 100\omega^2)(1 + 25\omega^2)} = 1 \;\Longrightarrow\; 2500\,\omega^4 + 125\,\omega^2 - 999\,999 = 0 .$$ Solving the quadratic in $\omega^2$ and keeping the positive root, $$\omega^2 = 19.9750, \qquad \boxed{\;\omega_{gc} = 4.4693\ \text{rad/s}\;}$$
8. Read the phase at the crossover. Each first-order lag contributes its own arctangent, $$\angle L(j\omega_{gc}) = -\arctan(10\omega_{gc}) - \arctan(5\omega_{gc}) = -88.718^\circ - 87.438^\circ = -176.156^\circ,$$ so the phase margin is $$\boxed{\;\mathrm{PM} = 180^\circ + \angle L(j\omega_{gc}) = 3.84^\circ\;}$$ Both lags are already deep into their $-90^\circ$ asymptotes by the time the magnitude reaches unity, which is exactly why so little margin survives. A loop with under four degrees of margin is violently under-damped: it is stable on paper but will ring badly and tolerate almost no additional delay.
9. Part (d) — identify what may and may not be changed. Both steady-state answers above are fixed by $L(0) = 1000$: part (a) scales as $P(0)/(1+L(0))$ and part (b) as the low-frequency sensitivity. Simply reducing the controller gain would indeed buy margin, but it would raise every steady-state error in proportion, so it is precisely the change the question forbids. What is needed is extra phase near $4.5\ \text{rad/s}$ with no change in gain at DC — that is the definition of a lead network with unity DC gain, $$C_{\text{new}}(s) = C(s)\cdot\frac{1 + \alpha T s}{1 + T s}, \qquad \alpha \gt 1 .$$
10. Size the lead network. A lead contributes a maximum phase of $\phi_{\max} = \arcsin\frac{\alpha - 1}{\alpha + 1}$ at the geometric-mean frequency $\omega_m = 1/(T\sqrt{\alpha})$. Choosing $\alpha = 12$ gives $\phi_{\max} = 57.80^\circ$, and placing that peak at $\omega_m = 10\ \text{rad/s}$ — comfortably above the old crossover, because the lead also lifts the magnitude and pushes the crossover out — fixes $$T = \frac{1}{\omega_m\sqrt{\alpha}} = \frac{1}{10\sqrt{12}} = 0.028868\ \text{s},$$ $$\boxed{\;C_{\text{new}}(s) = \frac{10}{5s+1}\cdot \frac{1 + s/2.887}{1 + s/34.641}\;}$$
11. Confirm the redesign. Recomputing the crossover of the compensated loop numerically gives $$\omega_{gc}' = 7.289\ \text{rad/s}, \qquad \mathrm{PM}' = 58.9^\circ,$$ while $C_{\text{new}}(0) = C(0) = 10$ leaves $L(0) = 1000$ and therefore leaves the answers to (a) and (b) exactly as computed. The margin has risen by a factor of fifteen at no cost in steady-state accuracy, which is the whole point of the question.