Question 6 of 6: Transport delay under proportional control: margins, step error and redesign

Question 6: Transport delay under proportional control: margins, step error and redesign (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 lag with a large transport delay in a unity-feedback loop under pure proportional control.

Find. (a) the admissible range of $K$ and the phase margin that goes with it; (b) the steady-state step error as a function of $K$; (c) a controller that removes the step error while keeping at least 6 dB of gain margin.

Approach. A transport delay is all-pass, so magnitude and phase decouple completely: the phase crossover comes from a transcendental phase equation solved numerically, while the magnitude condition stays in closed form. Routh is unavailable throughout, because $1+3Ke^{-4s}/\left(s+1\right)$ is not a polynomial.

1. Part (a) — locate the phase crossover. The delay contributes $-\theta\omega$ radians and the lag $-\arctan\omega$, so$$\angle L(j\omega)=-4\omega-\arctan\omega=-\pi\;\Rightarrow\;4\omega+\arctan\omega=\pi$$This has no closed-form solution; solving numerically (bisection on $0\lt\omega\lt1$) gives$$\omega_{pc}=0.64261\ \text{rad/s}$$Note that $\omega_{pc}$ is independent of $K$, because a proportional gain shifts the magnitude curve without touching the phase.
2. Part (a) — impose the gain-margin specification. A delay has unit magnitude at every frequency, so$$\left|L(j\omega_{pc})\right|=\frac{3K}{\sqrt{1+\omega_{pc}^{2}}}=\frac{3K}{1.18866}=2.52382\,K$$$$\text{GM}=\frac{1}{2.52382K}\ge10^{6/20}=1.99526\;\Rightarrow\;\boxed{0\lt K\le0.19858}$$For reference the loop is marginally stable at $K_{crit}=1/2.52382=0.39622$, and the common shorthand of reading 6 dB as a 2:1 ratio gives the slightly tighter $K\le0.19811$; either is acceptable provided the convention is stated.
3. Part (a) — determine the phase margin, and notice that there is no gain crossover. The magnitude$$\left|L(j\omega)\right|=\frac{3K}{\sqrt{1+\omega^{2}}}$$is monotonically decreasing in $\omega$, so its largest value is at DC:$$\left|L\right|_{\max}=3K\le3\times0.19858=0.59575\quad(-4.50\ \text{dB})$$which is below unity at every frequency. The magnitude curve therefore never reaches 0 dB, no gain crossover frequency exists, and$$\boxed{\text{PM}=\infty}$$This is the graded answer, and it is a statement about the loop rather than a dodge: the Nyquist plot of $L(j\omega)$ is a spiral of radius $3K\le0.596$ that never leaves the unit disc, so no amount of added phase lag can rotate it onto the $-1$ point. Evaluating $-180^\circ-\left(-4\omega-\arctan\omega\right)$ at some arbitrary frequency and quoting the result is the classic wrong answer.
With $K$ at the 6 dB limit the open-loop magnitude peaks at $-4.50$ dB and never reaches 0 dB, so the phase margin is infinite while the gain margin is exactly 6 dB at $\omega_{pc}=0.6426$ rad/s.
4. Part (b) — apply the final-value theorem to the error. With $E(s)=R(s)/\left[1+L(s)\right]$ and $R(s)=1/s$,$$e_{ss}=\lim_{s\to0}\frac{s}{s\left[1+L(s)\right]}=\frac{1}{1+L(0)}$$The delay contributes $e^{0}=1$ at DC, so $L(0)=3K$ and$$\boxed{e_{ss}=\frac{1}{1+3K}}$$Evaluated at the largest gain the specification allows, $e_{ss}=1/\left(1+0.59575\right)=0.6267$: the loop can only remove about 37 percent of a step command. That is the real cost of the dead time — it caps the gain, and a Type-0 loop’s accuracy is set entirely by its gain.
5. Part (c) — choose the controller structure. Zero steady-state step error requires $L(0)=\infty$, i.e. a free integrator in the controller. The minimal choice is$$C(s)=\frac{K_{i}}{s},\qquad L(s)=\frac{3K_{i}e^{-4s}}{s\left(s+1\right)}$$(A full PI, $K_{p}+K_{i}/s$, also works and is faster, but the integral term alone is enough to satisfy both requirements and is easier to defend under the margin constraint.)
6. Part (c) — relocate the phase crossover. The integrator adds a constant $-90^\circ$, so$$-90^\circ-4\omega-\arctan\omega=-180^\circ \;\Rightarrow\;4\omega+\arctan\omega=\frac{\pi}{2}\;\Rightarrow\;\omega_{pc}=0.31615\ \text{rad/s}$$roughly half its previous value — the integrator has cost bandwidth, which is the price of the accuracy.
7. Part (c) — size the integral gain. At the new phase crossover$$\left|L(j\omega_{pc})\right|=\frac{3K_{i}}{\omega_{pc}\sqrt{1+\omega_{pc}^{2}}}=9.04783\,K_{i}\;\Rightarrow\;K_{i}\le\frac{1}{1.99526\times9.04783}=0.055393$$Taking the round value $K_{i}=0.05$ leaves margin in hand:$$\boxed{C(s)=\frac{0.05}{s}}\qquad \text{GM}=\frac{1}{9.04783\times0.05}=2.2105\ (6.89\ \text{dB})$$
8. Part (c) — confirm the redesign is well behaved. Unlike part (a), the integrator makes $\left|L\right|=3K_{i}/\left(\omega\sqrt{1+\omega^{2}}\right)$ frequency dependent, so a gain crossover now exists:$$\omega_{gc}=0.14838\ \text{rad/s},\qquad \text{PM}=180^\circ+\left(-90^\circ-4\omega_{gc}\tfrac{180}{\pi}-\arctan\omega_{gc}\right)=47.6^\circ$$So the redesigned loop has $\boxed{e_{ss}=0,\ \text{GM}=6.89\ \text{dB},\ \text{PM}=47.6^\circ}$ — both requirements met with a conventional and comfortable phase margin. The closed-loop bandwidth is about $0.15$ rad/s, i.e. a settling time of roughly 30 s, which is the honest consequence of a 4 s dead time and should be stated rather than hidden.