Model-matching controller design and its stability margins: 22-Elec-B2 Advanced Control Systems, May 2014, Q3

Question 3: Model-matching controller design and its stability margins (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 lightly damped second-order plant with one finite zero, a specified closed-loop response to be achieved exactly, and a disturbance subtracted at the plant output.

Given data
Quantity	Symbol	Value
Plant	$P(s)$	$4(s+3)/(s^{2} + 0.2s + 2)$
Plant poles	—	$-0.1 \pm j1.4107$ ($\omega_n = 1.414$, $\zeta = 0.0707$)
Target closed loop	$T(s)$	$32/(8s^{2} + 6s + 32) = 4/(s^{2} + 0.75s + 4)$
Target poles	—	$-0.375 \pm j1.9645$ ($\omega_n = 2$, $\zeta = 0.1875$)
Disturbance	$d(t)$	unit step, subtracted at the plant output

Find. A proper $C(s)$ achieving the stated $T(s)$, the gain and phase margins of the resulting loop, and the steady-state control signal against a unit-step output disturbance.

Question 3: here d is subtracted at the plant OUTPUT, so y = P(s)u - d.

Approach. Inverting the closed-loop formula gives the required loop gain in one line; dividing by the plant gives the controller; the loop gain is then simple enough that both margins follow from a hand calculation, and the steady-state control effort follows from the physical requirement that the plant output must cancel the disturbance.

Part (a) — normalise the target. Dividing numerator and denominator by 8, $$T(s) = \frac{32}{8s^{2} + 6s + 32} = \frac{4}{s^{2} + 0.75s + 4},$$ a second-order response with $\omega_n = 2$ rad/s, $\zeta = 0.1875$ and unity DC gain (so the step reference is tracked exactly).
Invert the closed-loop relation to get the loop gain. From $T = CP/(1+CP)$, $$C(s)P(s) = \frac{T}{1-T} = \frac{4}{(s^{2}+0.75s+4) - 4} = \frac{4}{s^{2}+0.75s} = \frac{4}{s(s+0.75)} .$$ The required loop gain is therefore a plain Type-1, two-pole transfer function — the free integrator appears automatically because $T(0) = 1$.
Divide out the plant. Substituting for $P(s)$, $$C(s) = \frac{1}{P(s)}\cdot\frac{4}{s(s+0.75)} = \frac{s^{2}+0.2s+2}{4(s+3)}\cdot\frac{4}{s(s+0.75)},$$ so $$\boxed{C(s) = \frac{s^{2} + 0.2s + 2}{s\,(s + 0.75)(s + 3)}}.$$
Check the properness requirement. The numerator has degree 2 and the denominator degree 3, so the controller is strictly proper and therefore certainly proper. (The question's note contains an obvious slip — it says “numerator” twice; the condition intended, and used here, is that the denominator degree be at least the numerator degree.) Multiplying $C$ by $P$ and re-forming $CP/(1+CP)$ reproduces $32/(8s^{2}+6s+32)$ identically, at every test frequency.
Note what the design does to the plant poles. The controller numerator $s^{2}+0.2s+2$ is exactly the plant denominator, so the design cancels the plant's complex pole pair at $-0.1 \pm j1.4107$ and replaces it with the target pair. That is admissible here only because those poles are stable; a right-half-plane pole must never be cancelled.
Part (b) — locate the gain crossover. With $L(s) = 4/[s(s+0.75)]$, setting $|L(j\omega)| = 1$ gives $$\frac{4}{\omega\sqrt{\omega^{2} + 0.5625}} = 1 \;\Longrightarrow\; \omega^{4} + 0.5625\,\omega^{2} - 16 = 0 .$$ The positive root is $\omega^{2} = 3.7286$, hence $\omega_{gc} = 1.931$ rad/s.
Evaluate the phase there to get the phase margin. $$\angle L(j\omega_{gc}) = -90^{\circ} - \arctan\!\frac{\omega_{gc}}{0.75} = -90^{\circ} - 68.78^{\circ} = -158.78^{\circ},$$ $$\text{PM} = 180^{\circ} + \angle L(j\omega_{gc}) = \boxed{\text{PM} = 21.2^{\circ}}.$$
Establish the gain margin. The phase of a two-pole Type-1 loop is $-90^{\circ} - \arctan(\omega/0.75)$, which decreases monotonically towards $-180^{\circ}$ but never reaches it at any finite frequency. There is therefore no phase crossover, and $$\boxed{\text{GM} = \infty}.$$ The loop stays stable for every positive scaling of the gain — the classic behaviour of a loop whose Nyquist plot lies entirely in the third and fourth quadrants.
Part (c) — steady-state control effort against the output disturbance. From the diagram $y = P u - d$ and, with $r = 0$, $e = -y$ and $u = Ce$, which rearranges to $u/d = C/(1 + CP)$. Substituting and cancelling the common $(s+0.75)$ factor, $$\frac{u}{d} = \frac{s^{2}+0.2s+2}{(s+3)(s^{2}+0.75s+4)} \;\Longrightarrow\; u(\infty) = \frac{2}{3 \times 4} = \boxed{u(\infty) = \tfrac{1}{6} = 0.1667}.$$ The one-line physical check: the controller has an integrator, so it drives $y$ to zero, which forces $P(0)\,u(\infty) = d = 1$, i.e. $u(\infty) = 1/P(0) = 1/6$. The plant must be pushed just hard enough to reproduce the disturbance it is cancelling.

Check: the design cancels lightly damped plant poles. The exact model-matching solution places controller zeros on top of the plant poles at $-0.1 \pm j1.4107$, whose damping is only $\zeta = 0.0707$. On paper the cancellation is perfect; in hardware a few per cent of parameter error leaves a barely-damped residual mode near 1.41 rad/s that the loop can no longer influence. The 21.2° phase margin is likewise modest — it is the direct consequence of the customer's own choice of $\zeta = 0.1875$ in the target response, not a design error. If robustness matters, the target should be re-specified with a larger $\zeta$ rather than the cancellation retained.

Question 3 — final results
Quantity	Symbol	Value
Required loop gain	$C(s)P(s)$	$4/[s(s+0.75)]$
Controller	$C(s)$	$(s^{2}+0.2s+2)\,/\,[s(s+0.75)(s+3)]$
Properness	—	degree 2 over degree 3 — strictly proper
Gain crossover frequency	$\omega_{gc}$	1.931 rad/s
Phase margin	PM	21.2°
Gain margin	GM	infinite (no phase crossover)
Control input to a unit-step output disturbance	$u(\infty)$	1/6 = 0.1667

Open-loop Bode plot of the designed loop gain L(s) = 4/[s(s + 0.75)]. Gain crossover at 1.931 rad/s gives PM = 21.2 deg; the phase approaches -180 deg only as w -> infinity, so the gain margin is infinite.

Topic. Model matching (one-degree-of-freedom inverse design), properness constraints on a realisable controller, and Bode stability margins.

Key relations. $T = CP/(1+CP)$ inverts to $CP = T/(1-T)$ and $C = T/[P(1-T)]$; $|L(j\omega_{gc})| = 1$ defines gain crossover; $\text{PM} = 180^{\circ} + \angle L(j\omega_{gc})$; $\text{GM} = 1/|L(j\omega_{pc})|$ where $\angle L = -180^{\circ}$; $u/d = C/(1+CP)$ for a disturbance at the plant output.

Why this works. Because the loop is a single-input single-output series connection, specifying the closed-loop transfer function specifies the loop gain uniquely, and the controller is then whatever is left after dividing out the plant. Two structural facts fall out for free: $T(0)=1$ forces $1-T$ to vanish at the origin, which is why the answer automatically contains an integrator, and the relative degree of $T$ must be at least that of $P$ or the resulting $C$ would be improper. The infinite gain margin is not luck — any loop gain of the form $k/[s(s+a)]$ has a phase asymptote of $-180^{\circ}$ that is approached but never attained.

Common pitfall. Quoting a gain margin in decibels when none exists (the honest answer is “infinite: the phase never reaches $-180^{\circ}$”), and computing $u(\infty)$ from the wrong transfer function — with $d$ at the plant output the answer is $1/P(0)$, whereas an input disturbance would give $u(\infty) = d$ instead, as in Question 1.

Source. Åström & Murray, Feedback Systems, 2nd ed., §12.2 (loop-shaping and inverse design) and §10.3 (Bode margins); Franklin, Powell & Emami-Naeini, §6.4 (margins from the frequency response).

Question 3 of 6: Model-matching controller design and its stability margins

Question 3: Model-matching controller design and its stability margins (25 marks)