Question 6 of 6: Gain limits, phase margin and disturbance error for a delayed integrator

Question 6: Gain limits, phase margin and disturbance error for a delayed integrator (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. An integrating plant with a one-third-second transport delay, under proportional control, with a constant disturbance subtracted at the plant input.

Find. The stabilising range of $K$, the phase margin at $K = 1$ with the Nyquist sketch, and the tracking error as a function of $K$.

Approach. The characteristic equation is transcendental, so Routh does not apply; the Nyquist criterion does, and for this loop both the magnitude and phase have closed forms, so the crossover frequencies can be written down. The error follows from the transfer function the diagram actually defines, evaluated with the final-value theorem.

1. Part (a) — write the loop frequency response. With $L(s) = K e^{-s/3}/s$, $$|L(j\omega)| = \frac{K}{\omega}, \qquad \angle L(j\omega) = -90^{\circ} - \frac{\omega}{3}\ \text{rad} = -90^{\circ} - 19.10^{\circ}\,\omega .$$ The delay contributes no attenuation at all — it is all-pass — but its phase lag grows without bound, which is what limits the gain.
2. Explain why Routh is unavailable. The characteristic equation $s + Ke^{-s/3} = 0$ is transcendental and has infinitely many roots, so no finite Routh array exists. The Nyquist criterion, which needs only the frequency response, is the right tool; a Padé approximation would work too, but it would introduce error where an exact answer is available.
3. Locate the phase crossover. The phase reaches $-180^{\circ}$ when the delay alone has contributed a further $90^{\circ}$: $$\frac{\omega_{pc}}{3} = \frac{\pi}{2} \;\Longrightarrow\; \omega_{pc} = \frac{3\pi}{2} = 4.712\ \text{rad/s},$$ and this frequency is independent of $K$, since gain does not shift phase.
4. Convert that into the stability limit. Instability begins when the locus passes through $-1$, i.e. when $|L(j\omega_{pc})| = 1$: $$\frac{K}{\omega_{pc}} = 1 \;\Longrightarrow\; K_{crit} = \omega_{pc} = \frac{3\pi}{2} = 4.712 .$$ $L$ has no right-half-plane poles, so the Nyquist criterion requires no encirclements of $-1$, and $$\boxed{0 \lt K \lt \frac{3\pi}{2} = 4.712} .$$
5. Part (b) — find the gain crossover at $K = 1$. Since $|L| = K/\omega$, $$\omega_{gc} = K = 1\ \text{rad/s}\quad\text{when } K = 1 .$$
6. Evaluate the phase margin there. $$\angle L(j1) = -90^{\circ} - \frac{1}{3}\times 57.296^{\circ} = -109.10^{\circ},$$ $$\text{PM} = 180^{\circ} + \angle L(j\omega_{gc}) = \boxed{\text{PM} = 70.9^{\circ}} .$$ The corresponding gain margin is $\text{GM} = K_{crit}/K = 4.712$, i.e. 13.46 dB, and the delay margin is $\text{PM}/\omega_{gc} = 1.237$ s — almost four times the actual delay, so this design is comfortable.
7. Sketch the Nyquist locus. Writing $L(j\omega) = e^{-j\omega/3}/(j\omega)$ in rectangular form, $$\operatorname{Re} L = -\frac{\sin(\omega/3)}{\omega}, \qquad \operatorname{Im} L = -\frac{\cos(\omega/3)}{\omega} .$$ As $\omega \to 0^{+}$ the locus runs off to $-j\infty$ with real part tending to $-1/3$; the indentation around the pole at the origin closes the contour with a large clockwise semicircle at infinity. As $\omega$ grows the locus spirals clockwise into the origin, cutting the negative real axis first at $-0.2122$ ($\omega = 4.712$ rad/s), then at $-0.0424$ ($\omega = 15\pi/2 = 23.56$ rad/s), and so on. With $K = 1$ every crossing lies well to the right of $-1$, so the point is not encircled: $N = 0$, $P = 0$, hence $Z = 0$ and the loop is stable.
8. Part (c) — derive the error transfer function from the diagram. The plant is driven by $u - d$, so $$y = P\,(u - d), \qquad u = Ke, \qquad e = r - y,$$ which eliminates to $$e = \frac{r + P(s)\,d}{1 + K P(s)} .$$ Note the disturbance appears in the numerator multiplied by the plant — that factor is what makes the answer non-zero.
9. Apply the final-value theorem. With $R(s) = 1/s$ and $D(s) = 0.3/s$, $$e(\infty) = \lim_{s\to 0} s\,\frac{\dfrac{1}{s} + \dfrac{0.3\,e^{-s/3}}{s^{2}}} {1 + \dfrac{K e^{-s/3}}{s}} = \lim_{s\to 0}\frac{s + 0.3\,e^{-s/3}}{s + K e^{-s/3}} = \boxed{e(\infty) = \frac{0.3}{K}} .$$ So the tracking error settles at $0.3/K$: $0.3$ at $K = 1$, $0.075$ at $K = 4$.
10. Interpret the result — and what it costs. The step reference on its own would be tracked perfectly, because the plant integrator makes the loop Type 1. The disturbance, however, enters ahead of that integrator, so a steady offset in $y$ is the only way the loop can generate the control signal that cancels $d$; the error is therefore inversely proportional to the gain and never zero. Since stability caps $K$ below $4.712$, the smallest attainable steady-state error is $0.3/4.712 = 0.064$, and any real design would sit well short of that limit. Driving the error to zero requires integral action in the controller, not more proportional gain.