Question 6 of 6: Proportional control of a delayed integrator to a specified gain margin

Question 6: Proportional control of a delayed integrator to a specified gain margin (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 pure integrator in series with a two-second transport delay, to be closed with a proportional controller in a unity-feedback loop.

Find. The gain that delivers an 8 dB gain margin, the phase margin that results, and the steady-state value and approximate shape of the closed-loop unit-step response.

Approach. Use the fact that a transport delay is all-pass: it contributes phase but no magnitude, so the phase crossover depends only on the delay and the gain crossover only on the gain. Both margins then follow in closed form, and the free integrator settles the steady-state value by inspection.

1. Part (a) — locate the phase crossover, which the gain cannot move. With $C=K$ the loop is $L(j\omega)=K e^{-2j\omega}/(j\omega)$, whose phase is$$\angle L(j\omega)=-90^{\circ}-2\omega\ \text{rad}$$Setting this to $-180^{\circ}$ requires $2\omega=\pi/2$, so$$\boxed{\omega_{pc}=\frac{\pi}{4}=0.7854\ \text{rad/s}}$$The gain $K$ appears nowhere in that equation: a proportional gain shifts the magnitude curve but never the phase curve, so the phase crossover is fixed by the delay alone.
2. Convert the 8 dB specification into a gain. The gain margin is the reciprocal of the loop magnitude at the phase crossover, and a delay has unit magnitude at every frequency, so $|L(j\omega)|=K/\omega$ and$$\text{GM}=\frac{1}{|L(j\omega_{pc})|}=\frac{\omega_{pc}}{K}=10^{8/20}=2.5119$$$$K=\frac{\omega_{pc}}{2.5119}=\frac{0.7854}{2.5119}=\boxed{0.3127}$$Note that Routh’s criterion is unavailable here: $1+Ke^{-2s}/s$ is transcendental, not a polynomial. Frequency response is the natural tool, and a Padé approximation would only inject error into a problem that has an exact closed form.
3. Part (b) — find the gain crossover, which is trivially $K$. Because the delay is all-pass, $|L(j\omega)|=K/\omega=1$ gives$$\omega_{gc}=K=0.3127\ \text{rad/s}$$The magnitude and phase conditions decouple completely on this plant, which is why every margin here has a closed form.
4. Evaluate the phase margin. Substituting $\omega_{gc}$ into the phase expression and converting the delay term from radians to degrees:$$\text{PM}=180^{\circ}+\angle L(j\omega_{gc})=90^{\circ}-\frac{180^{\circ}}{\pi}\,(2\times 0.3127)=90^{\circ}-35.83^{\circ}=\boxed{54.2^{\circ}}$$An 8 dB gain margin and a 54 degree phase margin together describe a well-damped loop, which the simulated response below confirms.
5. Part (c) — get the steady-state output by inspection. The loop contains a free integrator, so the loop gain grows without bound as $s\to 0$ and the sensitivity $S(0)=1/(1+L(0))=0$. For a unit step reference the steady-state error is therefore zero and$$\boxed{y(\infty)=1}$$The transport delay changes when the output gets there, never where it ends up.
6. Sketch the response with its three characteristic features. Simulating $\dot y(t)=K\left[r(t-2)-y(t-2)\right]$ for a unit step gives the trace below, and three features should appear in any hand sketch. First, the output is exactly zero for the first 2 s — nothing can happen until the step has traversed the delay. Second, it then rises with initial slope $K=0.313$ per second and overshoots to a peak of $1.138$ (about 14 percent) near $t=7.6$ s, consistent with a 54 degree phase margin. Third, the residual ringing has a period close to $2\pi/\omega_{gc}\approx 20$ s and decays away, leaving $y=1$; settling to within 2 percent takes roughly 25 s, which is more than ten times the transport delay.