Question 6 of 6: Delay margin, Nyquist plot and the effect of a sensor bias

Question 6: Delay margin, Nyquist plot and the effect of a sensor bias (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 controller around a first-order plant that carries a pure transport delay of one eighth of a second, with the measurement corrupted by a constant sensor bias before it is fed back.

Find. The range of $K$ for closed-loop stability, the phase margin at $K=1$ with the corresponding Nyquist sketch, and the steady-state tracking error produced by the sensor bias.

Approach. The delay makes the characteristic equation transcendental, so Routh cannot be used; work in the frequency domain instead. Locate the phase crossover to get the critical gain, the gain crossover to get the phase margin, and then use block algebra with the final-value theorem for the bias.

1. Write the loop frequency response. With $L=CP$,$$L(j\omega)=\frac{K\,e^{-j\omega/8}}{j\omega\,(j\omega+1)},\qquad |L(j\omega)|=\frac{K}{\omega\sqrt{\omega^{2}+1}},$$$$\angle L(j\omega)=-90^{\circ}-\arctan\omega-\frac{\omega}{8}\ \text{rad}.$$The delay contributes no attenuation at all — it only subtracts phase, and it does so in proportion to frequency, which is what makes it so destructive at high loop bandwidth.
2. (a) Find the phase crossover. Stability is lost when the phase reaches $-180^{\circ}$, i.e. when$$\arctan\omega_{pc}+\frac{\omega_{pc}}{8}=\frac{\pi}{2}.$$This is transcendental, so it is solved numerically (bisection on the interval where the left side crosses $\pi/2$), giving$$\omega_{pc}=2.7712\ \text{rad/s}.$$Without the delay the left side would only approach $\pi/2$ asymptotically and there would be no crossover at all: the loop would be stable for every positive gain. The delay is the sole source of the instability.
3. Convert to a critical gain. At that frequency the magnitude must be unity, so$$K_{\text{crit}}=\omega_{pc}\sqrt{\omega_{pc}^{2}+1}=2.7712\sqrt{2.7712^{2}+1}=8.169$$$$\boxed{\,0\lt K\lt 8.169\,}$$The open loop has no right-half-plane poles ($P=0$), so by the Nyquist criterion the closed loop is stable precisely while the locus fails to encircle $-1$; negative $K$ inverts the sense of the feedback and is excluded.
4. (b) Find the gain crossover at $K=1$. Setting $|L(j\omega)|=1$ with $K=1$,$$\omega^{2}(\omega^{2}+1)=1\;\Longrightarrow\;\omega^{4}+\omega^{2}-1=0\;\Longrightarrow\;\omega_{gc}^{2}=\frac{\sqrt{5}-1}{2}\;\Longrightarrow\;\omega_{gc}=0.78615\ \text{rad/s}.$$The magnitude is unaffected by the delay, so this crossover is exact and closed-form.
5. Evaluate the phase margin. Adding the three phase contributions at $\omega_{gc}$,$$\mathrm{PM}=180^{\circ}+\angle L(j\omega_{gc})=180^{\circ}-90^{\circ}-\underbrace{38.173^{\circ}}_{\arctan 0.78615}-\underbrace{5.630^{\circ}}_{\omega_{gc}/8\ \text{rad}}$$$$\boxed{\,\mathrm{PM}=46.20^{\circ}\ \text{at }\omega_{gc}=0.78615\ \text{rad/s}\,}$$The matching gain margin is $20\log_{10}(8.169)=18.24$ dB. Note how little the delay costs here (5.6 degrees) precisely because the crossover is slow; at the phase crossover the same delay is worth 19.8 degrees.
6. Sketch the Nyquist plot. As $\omega\to 0^{+}$ the integrator sends $|L|\to\infty$ with a phase approaching $-90^{\circ}$, and expanding $L$ for small $\omega$ shows the locus is asymptotic to the vertical line $\operatorname{Re}\{L\}=-(1+\tau)=-1.125$. As $\omega$ increases the phase falls steadily, the locus crosses the negative real axis at $-0.1224$ when $\omega=2.771$ rad/s, and then spirals into the origin because the delay keeps subtracting phase while the magnitude decays. The standard indentation around the pole at the origin closes the contour with a large clockwise arc on the right. With $P=0$ and no encirclement of $-1$, the Nyquist criterion gives $Z=N+P=0$: no unstable closed-loop poles, in agreement with part (a).

1. (c) Write the error with the bias present. The signal fed back is the measurement $y+n$, so the comparator output is $\varepsilon=r-y-n$ while the quantity asked for is the true tracking error $e=r-y$. With $d=0$, $Y=PC\varepsilon=PC\,(R-Y-N)$, giving $Y=\dfrac{PC\,(R-N)}{1+PC}$ and therefore$$\boxed{\,E(s)=R(s)-Y(s)=\frac{R(s)+P(s)C(s)\,N(s)}{1+P(s)C(s)}\,}$$The reference is filtered by the sensitivity, but the bias is filtered by the complementary sensitivity — the two are not the same, and that difference is the whole point of the question.
2. Take the limit. With $R(s)=1/s$ and $N(s)=0.3/s$, the final-value theorem gives$$e(\infty)=\lim_{s\to 0}\frac{1+L(s)(0.3)}{1+L(s)}.$$Because $C$ contains an integrator, $|L(s)|\to\infty$ as $s\to 0$ for every $K\gt 0$, so the limit is the ratio of the two terms that grow:$$\boxed{\,e(\infty)=n=0.3\ \text{for every stabilising }K,\ 0\lt K\lt 8.169\,}$$and correspondingly $y(\infty)=r-e=0.7$.
3. Interpret the result. The answer is deliberately independent of $K$, and that is the engineering lesson: integral action drives the measured signal to the set-point, and the measured signal is $y+n$. The loop therefore parks the true output at $r-n=0.7$ and reports success. Raising the gain, adding more integrators or retuning the controller cannot help, because no feedback path can distinguish a genuine output change from an offset in the instrument that measures it. The only remedies lie outside the loop: calibrate or re-zero the sensor, use a redundant or differently-principled second sensor and vote, or estimate the bias explicitly as an augmented state. Contrast this with a constant output disturbance $d$, which the same integrator would reject completely.