Question 1 of 6: Near-instability, input-disturbance rejection, and a 90° phase-margin design

Question 1: Near-instability, input-disturbance rejection, and a 90° phase-margin design (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 loop whose controller carries a free integrator and whose plant is a single fast lag, with a disturbance injected immediately ahead of the plant.

Find. (a) evidence that the nominal loop is only marginally damped; (b) the steady-state error to a unit-step input disturbance; (c) the proportional term $K$ that lifts the phase margin to about $90^\circ$; (d) the steady-state error waveform for a 100 rad/s sinusoidal reference.

Approach. Form the loop transfer function once, read near-instability from both the closed-loop pole damping and the phase margin, then use the same $L(s)$ for the disturbance limit, the $90^\circ$ design and the sinusoidal sensitivity.

1. Part (a) — assemble the loop transfer function and its characteristic polynomial. With $L(s)=C(s)P(s)$,$$L(s)=\frac{10^{4}}{s(s+10)}\cdot\frac{10^{6}}{s+10^{6}}=\frac{10^{10}}{s(s+10)(s+10^{6})}$$so the closed-loop characteristic polynomial is $s(s+10)(s+10^{6})+10^{10}$, i.e.$$\Delta(s)=s^{3}+1{,}000{,}010\,s^{2}+10^{7}s+10^{10}$$
2. Part (a) — Routh alone says “stable”, which is exactly why it is the wrong test here. All four coefficients are positive and the single Routh condition for a cubic, $a_{2}a_{1}\gt a_{3}a_{0}$, gives $1{,}000{,}010\times10^{7}=1.00001\times10^{13}$ against $10^{10}$ — satisfied by a factor of a thousand. A Routh test answers the yes/no question and says nothing about how close the loop is to the boundary, so it must be followed by either the poles or the margins.
3. Part (a) — factor the characteristic polynomial and read the damping. The roots of $\Delta(s)$ are$$s_{1}=-1.000000\times10^{6},\qquad s_{2,3}=-4.9950\pm j\,99.8752$$The complex pair has$$\omega_{n}=|s_{2}|=100.00\ \text{rad/s},\qquad \zeta=\frac{-\operatorname{Re}s_{2}}{\omega_{n}}=\frac{4.9950}{100.00}\;\Rightarrow\;\boxed{\zeta=0.0500}$$A damping ratio of five percent is the quantitative statement of “very nearly unstable”: the pair sits at $5.0/100.0$ of the way in from the imaginary axis.
4. Part (a) — convert the damping into observable behaviour. For a dominant second-order pair,$$M_{p}=100\,e^{-\pi\zeta/\sqrt{1-\zeta^{2}}}=85.5\%,\qquad t_{s(2\%)}\approx\frac{4}{\zeta\omega_{n}}=\frac{4}{4.9950}=0.801\ \text{s}$$The ringing frequency is $99.88$ rad/s, i.e. a period of $0.0629$ s, so the step response executes roughly thirteen visible cycles before it settles. A unit step command overshoots to nearly $1.9$.
Closed-loop unit-step response computed from the roots of $\Delta(s)$: 85.5 percent overshoot and about thirteen cycles of ringing before the 2 percent band is reached.
5. Part (a) — the same conclusion from the frequency response. Below $10^{6}$ rad/s the plant is effectively unity, so the gain crossover solves in closed form from $|10^{4}/[j\omega(j\omega+10)]|=1$:$$\omega^{2}\left(\omega^{2}+100\right)=10^{8}\;\Rightarrow\;\omega_{gc}=\sqrt{\tfrac{-100+\sqrt{100^{2}+4\times10^{8}}}{2}}=99.75\ \text{rad/s}$$At that frequency$$\angle L(j\omega_{gc})=-90^\circ-\arctan\frac{99.75}{10}-\arctan\frac{99.75}{10^{6}}=-174.28^\circ \;\Rightarrow\;\boxed{\text{PM}=5.72^\circ}$$Five and three-quarter degrees of phase margin is the frequency-domain form of $\zeta=0.05$ (the rule of thumb $\zeta\approx\text{PM}/100$ in degrees reproduces it almost exactly). Either statement answers part (a).
Open-loop Bode plot. The magnitude crosses 0 dB at 99.75 rad/s where the phase has already reached $-174.28^\circ$, leaving only $5.72^\circ$ of margin.
6. Part (a) — note the trap in the gain margin. The phase crossover occurs where $\arctan(\omega/10)+\arctan(\omega/10^{6})=90^\circ$, which for two arctangents happens exactly when the product of their arguments is unity:$$\frac{\omega}{10}\cdot\frac{\omega}{10^{6}}=1\;\Rightarrow\;\omega_{pc}=\sqrt{10\times10^{6}}=3162.3\ \text{rad/s},\qquad \text{GM}=\frac{1}{|L(j\omega_{pc})|}=1000\;(60.0\ \text{dB})$$A 60 dB gain margin looks superb and is entirely misleading — the loop is fragile in phase, not in gain, so a few degrees of extra lag (a sensor filter, a transport delay of only $1$ ms) would destabilise it. Quoting the gain margin alone is the classic wrong answer to part (a).
7. Part (b) — derive the error transfer function for the input disturbance. With the plant driven by $u-d$ and $u=C(s)e$,$$Y=P\left(Ce-D\right),\qquad e=r-y=-Y\ \ (r=0)$$$$\Rightarrow\;-e=PCe-PD\;\Rightarrow\;e\left(1+PC\right)=PD\;\Rightarrow\;E(s)=\frac{P(s)}{1+L(s)}D(s)$$
8. Part (b) — apply the final-value theorem. For $D(s)=1/s$,$$e_{ss}=\lim_{s\to0}s\cdot\frac{P(s)}{1+L(s)}\cdot\frac{1}{s}=\lim_{s\to0}\frac{P(s)\,s(s+10)(s+10^{6})}{s(s+10)(s+10^{6})+10^{10}}=\frac{1\times0}{10^{10}}\;\Rightarrow\;\boxed{e_{ss}=0}$$The sensitivity numerator carries the loop’s free integrator, which sits upstream of the injection point, so the controller can wind up to whatever constant output cancels $d$ exactly. Physically the controller output settles at $u(\infty)=d=1$ and the plant input returns to zero. Had the disturbance entered at the plant output the same integrator would still give zero error, but had $C$ been a static gain the answer would have been $P(0)/[1+P(0)K]$ instead — which sub-part of the loop the integrator occupies is the whole question.
9. Part (c) — translate “PM $\approx 90^\circ$” into two equations. A $90^\circ$ phase margin means $\angle L(j\omega_{gc})=-90^\circ$, i.e. $L(j\omega)$ is purely negative-imaginary at the crossover. Neglecting the $10^{6}$ rad/s pole (its phase contribution at a few tens of rad/s is under $0.003^\circ$), write $A=10^{4}$ and $b=10$ and split $L$:$$L(j\omega)=K+\frac{A}{j\omega(j\omega+b)}=\underbrace{\left[K-\frac{A}{\omega^{2}+b^{2}}\right]}_{\operatorname{Re}L}\;-\;j\underbrace{\frac{Ab}{\omega\left(\omega^{2}+b^{2}\right)}}_{-\operatorname{Im}L}$$
10. Part (c) — solve them. Setting $\operatorname{Re}L=0$ and $|\operatorname{Im}L|=1$ gives an equation in $\omega$ alone,$$\omega^{3}+b^{2}\omega=Ab=10^{5}\;\Rightarrow\;\omega_{gc}=45.70\ \text{rad/s}$$and then the gain follows directly:$$K=\frac{A}{\omega_{gc}^{2}+b^{2}}=\frac{10^{4}}{45.6978^{2}+100}=\frac{10^{4}}{2188.1}\;\Rightarrow\;\boxed{K\approx4.57}$$Re-evaluating the full $L(s)$ with the $10^{6}$ rad/s pole retained gives $\omega_{gc}=45.6978$ rad/s and a phase margin of $89.997^\circ$, so the approximation costs three thousandths of a degree.
11. Part (c) — check that no other crossover is worse. The parallel $K$ flattens $|L|$ to $K=4.57$ above about 100 rad/s, so the magnitude returns through 0 dB twice more, at $49.08$ rad/s and at $4.46\times10^{6}$ rad/s. The phase at those two crossovers is $-54.3^\circ$ and $-77.4^\circ$, i.e. margins of $125.7^\circ$ and $102.6^\circ$. The smallest of the three is the $90^\circ$ we designed for, so the answer stands. This is worth ten seconds of checking whenever a proportional path is placed in parallel with an integrator: a lag–lead controller of this shape rarely has a single crossover.
12. Part (d) — evaluate the sensitivity function at the driving frequency. With $K=0$ the controller is back to $10^{4}/[s(s+10)]$ and $E(s)=S(s)R(s)$ with $S=1/(1+L)$. At $\omega=100$ rad/s,$$j100\,(j100+10)\,(j100+10^{6})=-1.00001\times10^{10}+j\,9.99\times10^{8}$$$$L(j100)=\frac{10^{10}}{-1.00001\times10^{10}+j\,9.99\times10^{8}}=-0.99011-j\,0.09891$$so $1+L(j100)=0.00989-j\,0.09891$, whose magnitude is $0.09940$ and whose angle is $-84.29^\circ$.
13. Part (d) — write the error as a time function. A stable linear loop driven by a sinusoid settles to a sinusoid of the same frequency, scaled and shifted by $S(j\omega)$:$$|S(j100)|=\frac{1}{0.09940}=10.06,\qquad \angle S(j100)=+84.29^\circ$$$$\boxed{e_{ss}(t)=30.18\,\sin\!\left(100t+84.29^\circ\right)}$$The error is ten times larger than the command. That is not an arithmetic slip: $100$ rad/s is essentially the resonant frequency found in part (a), the sensitivity peak sits there, and a loop with $5.7^\circ$ of phase margin necessarily has $|S|_{\max}\approx1/(2\sin(\text{PM}/2))\approx10$. Parts (a) and (d) are the same fact stated twice, and saying so is the mark-earning observation.