Proportional design to a specified gain margin for a plant with transport delay: 22-Elec-B2 Advanced Control Systems, December 2013, Q6

Question 6: Proportional design to a specified gain margin for a plant with transport delay (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 type-1 plant with one lag and a pure transport delay, to be closed with a proportional gain against a gain-margin specification.

Given data — Question 6
Quantity	Value
Plant	$P(s) = \dfrac{e^{-0.2s}}{s(2s+1)}$
Transport delay	$\theta = 0.2$ s
Lag time constant	$T = 2$ s (pole at $s=-0.5$)
Controller	$C(s) = K$, proportional only
Specification	gain margin $= 8$ dB
Feedback	unity negative

Find. (a) the gain $K$ meeting the 8 dB gain margin; (b) the phase margin that results; (c) the steady-state output to a unit step set-point.

Approach. Locate the phase crossover from the phase equation alone — the delay contributes phase but no attenuation, so the crossover frequency does not depend on $K$ — then scale the magnitude there to the required margin, and finally find the new gain crossover to read the phase margin.

Write the frequency response. A pure delay has unit magnitude at every frequency and a phase that grows linearly with frequency, so $$|K P(j\omega)| = \frac{K}{\omega\sqrt{1+4\omega^{2}}},\qquad \angle KP(j\omega) = -\,\theta\omega - 90^\circ - \arctan(2\omega),$$ with $\theta\omega$ expressed in radians before conversion. Because the delay is transcendental, the closed-loop characteristic equation $s(2s+1)+Ke^{-0.2s}=0$ is not a polynomial and the Routh test is unavailable — the frequency-domain route is the only closed-form one.
Locate the phase crossover. Setting the phase to $-180^\circ$ and working in radians, $$0.2\,\omega + \arctan(2\omega) = \frac{\pi}{2}.$$ The left side rises monotonically from 0, so the root is unique; solving numerically, $$\omega_{c\varphi} = 1.5553\ \text{rad/s}.$$ Crucially this frequency is set by the phase alone and is therefore independent of $K$ — a proportional gain slides the magnitude curve up and down but never moves the $-180^\circ$ crossing.
Evaluate the plant magnitude there. $$|P(j\omega_{c\varphi})| = \frac{1}{1.5553\sqrt{1+4(1.5553)^{2}}} = \frac{1}{1.5553 \times 3.2674} = 0.19679,$$ so the uncompensated loop ($K=1$) already has a gain margin of $-20\log_{10}(0.19679) = 14.12$ dB. Meeting an 8 dB specification therefore means increasing the gain.
Solve for the required gain. A gain margin of 8 dB means the loop magnitude at the phase crossover must equal $10^{-8/20}$: $$K\,|P(j\omega_{c\varphi})| = 10^{-8/20} = 0.39811 \quad\Longrightarrow\quad K = \frac{0.39811}{0.19679}$$ $$\boxed{K = 2.0230}$$ Checking, $20\log_{10}(2.0230 \times 0.19679) = -8.00$ dB exactly.
Find the new gain crossover. Setting $|KP(j\omega)|=1$, $$\omega\sqrt{1+4\omega^{2}} = K = 2.0230 \quad\Longrightarrow\quad \omega_{cg} = 0.9456\ \text{rad/s}.$$
Read the phase margin. Evaluating the phase at that frequency, in degrees, $$\angle KP(j\omega_{cg}) = -\frac{180}{\pi}(0.2)(0.9456) - 90^\circ - \arctan\!\big(2(0.9456)\big) = -10.84^\circ - 90^\circ - 62.13^\circ = -162.97^\circ,$$ $$\boxed{\text{PM} = 180^\circ - 162.97^\circ = 17.0^\circ \ \text{at } \omega_{cg}=0.946\ \text{rad/s}}$$ The delay costs almost 11° of that phase on its own. A 17° margin is lightly damped — roughly $\zeta \approx 0.17$ by the usual $\zeta \approx \text{PM}/100$ rule of thumb — which is the price of insisting on an 8 dB rather than a larger gain margin on a delayed plant.
Determine the steady-state output. The plant contains a free integrator, so the loop is type 1 and the position error constant is infinite: $$K_{pos} = \lim_{s\to0} KP(s) = \lim_{s\to0}\frac{K e^{-0.2s}}{s(2s+1)} = \infty \quad\Longrightarrow\quad e_{ss} = \frac{1}{1+K_{pos}} = 0 .$$ Therefore the output tracks the unit step exactly in steady state: $$\boxed{y(\infty) = 1, \qquad e_{ss} = 0}$$ This holds for any stable $K$, so it is a property of the plant structure rather than of the design just completed.

Figure Q6.1 — Bode plot of the compensated loop. The delay contributes phase only, so the −180° crossing sits at ω = 1.555 rad/s regardless of K; choosing K = 2.0230 pulls the magnitude there down to −8 dB, which is the required gain margin.

Check: the delay is handled exactly, not by approximation. All numbers above use $e^{-j0.2\omega}$ directly. A first-order Padé substitution $e^{-0.2s}\approx(1-0.1s)/(1+0.1s)$ would make the characteristic equation a polynomial and allow a Routh solution, but it shifts the phase crossover by a few per cent and with it the design gain; the exact frequency-domain calculation is preferable and is what the mark scheme expects here. The $17^\circ$ phase margin also means the closed-loop step response will be markedly oscillatory: if the specification really is on transient quality rather than on gain margin, this plant wants a lead or a PI-lead compensator rather than a bare proportional gain.

Final results — Question 6
Quantity	Symbol	Value
Phase crossover frequency	$\omega_{c\varphi}$	1.5553 rad/s
Plant magnitude there	$\|P(j\omega_{c\varphi})\|$	0.19679
Uncompensated gain margin	—	14.12 dB
Design gain for GM = 8 dB	$K$	2.0230
Gain crossover frequency	$\omega_{cg}$	0.9456 rad/s
Phase margin	PM	17.0°
Steady-state output, unit step	$y(\infty)$	1 (zero error, type-1 loop)

Topic. Frequency-domain design of a proportional controller to a gain-margin specification, for a plant containing a pure transport delay.

Key relations. $|e^{-j\theta\omega}|=1$ and $\angle e^{-j\theta\omega} = -\theta\omega$; phase crossover from $\theta\omega+\arctan(T\omega)=\pi/2$ for a $1/[s(Ts+1)]$ plant; $\text{GM} = -20\log_{10}|KP(j\omega_{c\varphi})|$; $\text{PM} = 180^\circ + \angle KP(j\omega_{cg})$; a type-1 loop gives zero steady-state step error.

Why this works. Separating magnitude from phase is what makes this design a two-line calculation. The delay affects only the phase, and a proportional gain affects only the magnitude, so the two crossover frequencies decouple: $\omega_{c\varphi}$ can be found once and for all before $K$ is known, and then $K$ follows from a single magnitude evaluation. Only the phase margin requires going back to the phase equation, because $K$ does move the gain crossover. The same separation explains why a delayed plant can never be stabilised by gain alone beyond a fixed limit — the phase deficit at any given frequency is a property of the plant.

Common pitfall. Mixing radians and degrees in the phase equation: the delay term is naturally $\theta\omega$ in radians while the $\arctan$ terms are usually taken in degrees, and forgetting the conversion produces a phase crossover that is wrong by orders of magnitude. A second pitfall is assuming the specification requires reducing the gain; here the uncompensated margin is 14.12 dB, so meeting an 8 dB specification means gain must be added, and a candidate who assumes otherwise will report a $K$ below unity. Third, attempting a Routh analysis on $s(2s+1)+Ke^{-0.2s}=0$ — that equation has infinitely many roots and is not a polynomial.

Source. Franklin, Powell & Emami-Naeini, Feedback Control of Dynamic Systems, 7th ed., §6.4–6.5 (stability margins) and §6.9 (time delay); Åström & Murray, Feedback Systems, 2nd ed., §9.3 and §10.3 (loops with delay).

Question 6 of 6: Proportional design to a specified gain margin for a plant with transport delay

Question 6: Proportional design to a specified gain margin for a plant with transport delay (25 marks)