Question 1 of 6: Gain limit, phase margin and steady-state tracking of a non-minimum-phase plant

Question 1: Gain limit, phase margin and steady-state tracking of a 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 single-loop system whose forward path is an integrating controller followed by a first-order plant carrying a right-half-plane zero, with the disturbance summed at the plant output.

Find. (a) the gain $K_{max}$ at which the loop oscillates and that oscillation frequency; (b) the phase margin at half that gain; (c) the steady-state error to a unit-slope ramp with a unit step disturbance; (d) the steady-state error to $r(t)=2\sin 3t$.

Approach. Write the error in terms of the sensitivity function, apply the Routh–Hurwitz test to the closed-loop characteristic polynomial for parts (a), then read the margins off the open-loop frequency response for (b), and finish with the final-value theorem for (c) and the sinusoidal-steady-state gain of the sensitivity function for (d).

1. Reduce the loop to one error equation. Around the outer summing junction $e = r - y$, and around the inner one $y = P(s)u - d$ with $u = C(s)e$. Eliminating $u$ and $y$, $$e = r - P(s)C(s)e + d \quad\Longrightarrow\quad E(s) = \frac{R(s)+D(s)}{1+L(s)}, \qquad L(s) = C(s)P(s).$$ Both the reference and the disturbance therefore act on the error through the same sensitivity function $S(s) = 1/[1+L(s)]$; only their spectra differ. The open loop is $$L(s) = \frac{3K(10-s)}{s(10+s)(3+2s)}.$$
2. Form the closed-loop characteristic polynomial. Setting $1+L(s)=0$ and clearing denominators, $s(10+s)(3+2s) + 3K(10-s) = 0$. Expanding $s(10+s)(3+2s) = s(2s^{2}+23s+30)$ gives $$2s^{3} + 23s^{2} + (30-3K)s + 30K = 0.$$ Note how the right-half-plane zero enters: it makes the $s$-coefficient decrease with gain, which is what will eventually destabilise the loop.
3. Apply the Routh test. The array is $$\begin{array}{c|cc} s^{3} & 2 & 30-3K\\ s^{2} & 23 & 30K\\ s^{1} & \dfrac{23(30-3K)-2(30K)}{23} = \dfrac{690-129K}{23} & 0\\ s^{0} & 30K & \end{array}$$ The $s^{0}$ row requires $K \gt 0$ and the $s^{2}$ row is always positive, so the binding condition is the $s^{1}$ row. Sustained oscillation occurs when it vanishes: $$690-129K = 0 \quad\Longrightarrow\quad \boxed{K_{max} = \frac{690}{129} = 5.3488}$$ (The condition $30-3K \gt 0$, i.e. $K \lt 10$, is satisfied at this gain and is therefore not the active limit.)
4. Read the oscillation frequency from the auxiliary polynomial. At $K = K_{max}$ the row above the vanishing row supplies the auxiliary polynomial $23s^{2}+30K_{max}=0$, whose roots are the imaginary-axis pair. Keeping the $s^{2}$ pivot, $$\omega_{osc} = \sqrt{\frac{30K_{max}}{23}} = \sqrt{\frac{30(5.3488)}{23}} = \sqrt{6.9767} \quad\Longrightarrow\quad \boxed{\omega_{osc} = 2.6414\ \text{rad/s}}$$ As a check, the cubic factors at this gain into $2(s+11.5)(s^{2}+6.9767)$, and evaluating the open loop directly gives $L(j2.6414) = -1.000 + j0.000$, exactly the sustained-oscillation condition.
5. Halve the gain and locate the new gain crossover. With $K = K_{max}/2 = 2.6744$ the magnitude of the open loop is $$|L(j\omega)| = \frac{3K\sqrt{100+\omega^{2}}}{\omega\sqrt{100+\omega^{2}}\sqrt{9+4\omega^{2}}} = \frac{3K}{\omega\sqrt{9+4\omega^{2}}},$$ the $\sqrt{100+\omega^{2}}$ factors of the right-half-plane zero and the left-half-plane pole cancelling exactly because they are mirror images. Setting this to unity and solving numerically, $\omega_{cg} = 1.7440$ rad/s.
6. Evaluate the phase there. The numerator factor $10-j\omega$ has a positive real part, so the right-half-plane zero contributes $-\arctan(\omega/10)$ — phase lag, exactly as much as the mirror-image pole at $s=-10$ — and $$\angle L(j\omega) = -\arctan\frac{\omega}{10} - 90^\circ - \arctan\frac{\omega}{10} - \arctan\frac{2\omega}{3}.$$ At $\omega_{cg}=1.7440$ this evaluates to $-159.09^\circ$, hence $$\boxed{\text{PM} = 180^\circ - 159.09^\circ = 20.9^\circ \ \text{at}\ \omega_{cg}=1.744\ \text{rad/s}}$$ The gain margin is exactly $20\log_{10}2 = 6.02$ dB, because halving a gain that was critical leaves the phase crossover where it was and drops the magnitude there by a factor of two.

Parts (c) and (d) are both steady-state questions, and both are answered from the same sensitivity function, evaluated at $s\to 0$ for the polynomial inputs and at $s=j3$ for the sinusoid.

7. Steady-state error to the ramp plus the step disturbance. With $R(s)=1/s^{2}$ and $D(s)=1/s$, $$e_{ss} = \lim_{s\to 0} sE(s) = \lim_{s\to 0}\frac{s\left(\dfrac{1}{s^{2}}+\dfrac{1}{s}\right)} {1+L(s)}.$$ Near the origin $L(s)\to K/s$, so $1+L(s)\to (s+K)/s$ and the two contributions separate: $$\underbrace{\lim_{s\to 0}\frac{1/s}{1+L(s)} = \frac{1}{K_{v}}}_{\text{ramp}} \;+\; \underbrace{\lim_{s\to 0}\frac{1}{1+L(s)} = \lim_{s\to 0}\frac{s}{s+K} = 0}_{\text{step disturbance}},\qquad K_{v}=\lim_{s\to 0}sL(s)=\frac{3K(10)}{(10)(3)}=K.$$ The integrator in $C(s)$ rejects the constant disturbance completely, so only the ramp survives. At $K=K_{max}/2=2.6744$, $$\boxed{e_{ss} = \frac{1}{K} = \frac{1}{2.6744} = 0.3739}$$
8. Steady-state error to the sinusoid. A stable linear loop driven by $r(t)=2\sin 3t$ settles to a sinusoid of the same frequency, scaled and shifted by the sensitivity function evaluated on the imaginary axis: $$e_{ss}(t) = 2\,|S(j3)|\,\sin\!\big(3t + \angle S(j3)\big),\qquad S(j3) = \frac{1}{1+L(j3)}.$$ Substituting $K=2.6744$ gives $L(j3) = -0.3958 + j0.0474$ (magnitude $0.3987$, phase $-186.83^\circ$), hence $S(j3) = 1.6501\angle{-4.49^\circ}$ and $$\boxed{e_{ss}(t) = 3.3003\,\sin\!\big(3t - 4.49^\circ\big)}$$ The error is larger than the reference itself: $|S(j3)| = 1.65 \gt 1$ because $3$ rad/s sits just above the gain crossover and close to the $2.64$ rad/s resonance the $20.9^\circ$ phase margin implies. Feedback amplifies rather than attenuates disturbances in this band, which is exactly what a small phase margin means physically.