Triple-Lag Loop with a Sensor — Ultimate Gain, Bode, Margins: 23-Chem-A6 Process Dynamics and Control, December 2013, Q2

Problem 2: Triple-Lag Loop with a Sensor — Ultimate Gain, Bode, Margins (20%)

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. Open loop $L(s)=K_c\,G_p\,H$ with $G_p=1/(s+1)^{3}$ (a triple first-order lag, corner at $\omega=1$) and either $H=1$ or a pure dead time $H=e^{-0.7s}$.

Find. (a) ultimate gains $K_{cu}$; (b) the case-(ii) Bode plot; (c) gain/phase margins at $K_c=1$.

Approach. Stability at the limit is set by the frequency where the open-loop phase reaches $-180^\circ$ (the phase-crossover frequency $\omega_{pc}$): the ultimate gain makes $|L|=1$ there. For case (i) this is confirmed with a Routh array; the dead time in case (ii) adds phase $-0.7\omega$ without changing magnitude, which is exactly what erodes the gain margin.

(a-i) $H=1$ — Routh and frequency methods agree. The characteristic equation is $(s+1)^{3}+K_c=s^{3}+3s^{2}+3s+(1+K_c)=0$. The Routh $s^{1}$ row is $(9-(1+K_c))/3=(8-K_c)/3$, which stays positive only for $K_c<8$; the $s^{0}$ row $1+K_c>0$ is satisfied. Equivalently the phase $-3\arctan\omega=-180^\circ$ gives $\omega_{pc}=\tan60^\circ=\sqrt3$, where $|G_p|=1/(\sqrt{\omega^{2}+1})^{3}=1/8$, so $$\boxed{K_{cu}^{(i)}=1/|G_p(j\sqrt3)|=8}.$$
(a-ii) $H=e^{-0.7s}$ — dead time adds phase only. Now the phase condition is $3\arctan\omega+0.7\omega=\pi$. Solving numerically gives $\omega_{pc}=1.039\ \mathrm{rad/s}$ (down from $\sqrt3=1.732$). The dead time has unit magnitude, so $|G_pH|=1/(\sqrt{\omega_{pc}^{2}+1})^{3}=0.334$ and $$\boxed{K_{cu}^{(ii)}=1/0.334\approx3.0}.$$ The delay slashes the ultimate gain from 8 to 3.
(b) Bode facts for case (ii). Amplitude ratio: DC value $|G_pH(0)|=1$ (0 dB, slope 0); a single corner at $\omega=1$ (the triple pole); above it the slope is $-3$ (i.e. $-60\ \mathrm{dB/decade}$). Phase: $\phi=-3\arctan\omega-0.7\omega$ — it starts at $0^\circ$, the poles alone would asymptote to $-270^\circ$, but the dead time makes $\phi\to-\infty$ as $\omega\to\infty$; it crosses $-180^\circ$ at $\omega=1.04$ (see figure).

Problem 2(b): Bode plot of $K_cG_pH$ for case (ii). Top — amplitude ratio: flat at 0 dB below the corner $\omega=1$, then $-60\ \mathrm{dB/decade}$ (dashed asymptote); the dead time does not change magnitude. Bottom — phase falls without bound because of the $-0.7\omega$ dead-time contribution, crossing $-180^\circ$ at $\omega_{co}=1.04\ \mathrm{rad/s}$, which fixes the ultimate gain.

(c) Margins at $K_c=1$. With $K_c=1$ the magnitude curve is $|L|=1/(\omega^{2}+1)^{3/2}\le1$, equal to 1 only at $\omega=0$, so the gain-crossover frequency is $\omega_{gc}\to0$ where $\phi=0$; hence in both cases the phase margin is $\approx180^\circ$ (the loop gain never exceeds unity for $\omega>0$). The gain margins are read at the phase crossover: case (i) $\mathrm{GM}=8$ ($+18.1\ \mathrm{dB}$); case (ii) $\mathrm{GM}=3.0$ ($+9.5\ \mathrm{dB}$). The dead time cuts the gain margin nearly threefold while leaving the phase margin untouched — the classic signature of transport lag.

Quantity	Case (i) $H=1$	Case (ii) $H=e^{-0.7s}$
Phase-crossover $\omega_{pc}$	$\sqrt3=1.732$	$1.039$
Ultimate gain $K_{cu}$	$\mathbf{8}$	$\mathbf{3.0}$
Gain margin at $K_c=1$	$8$ ($18.1$ dB)	$3.0$ ($9.5$ dB)
Phase margin at $K_c=1$	$\approx180^\circ$	$\approx180^\circ$

Concept & Theory

Topic. Frequency-response stability of a triple-lag loop with transport delay; ultimate gain, gain/phase margins.

Key relations. Bode stability: at $\omega_{pc}$ where $\angle L=-180^\circ$, $K_{cu}=1/|G_pH(j\omega_{pc})|$; $\mathrm{GM}=1/|L(j\omega_{pc})|$; $\mathrm{PM}=180^\circ+\angle L(j\omega_{gc})$; dead time $e^{-\theta s}$ contributes phase $-\theta\omega$ and magnitude $1$.

Why this works. A pure delay is an all-pass element: it rotates the phasor at every frequency without attenuating it. It therefore leaves the magnitude Bode plot (and the phase margin here) unchanged but drives the phase past $-180^\circ$ at a lower frequency, where the magnitude is still large — so the same loop tolerates far less gain.

Common pitfall. Using a Padé approximation for $e^{-0.7s}$ (the question forbids it) — that would wrongly add a right-half-plane zero and change the magnitude curve. Also, reporting $\omega_{pc}=\sqrt3$ for case (ii); the delay lowers it to $1.04$.

Source. Stephanopoulos, Ch. 17–18 (Bode/Nyquist stability, gain & phase margins); Seborg et al., Ch. 14 (frequency response, dead-time effects).

Question 2 of 8: Triple-Lag Loop with a Sensor — Ultimate Gain, Bode, Margins

Problem 2: Triple-Lag Loop with a Sensor — Ultimate Gain, Bode, Margins (20%)