Triple-Lag Loop with a Dead-Time Sensor — Stability Limit and Margins: 23-Chem-A6 Process Dynamics and Control, December 2017, Q3

Problem 3: Triple-Lag Loop with a Dead-Time Sensor — Stability Limit and 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. A unity-feedback loop with three identical first-order lags in the process and, in case (ii), a pure transport lag in the measurement:

Element	Transfer function
Controller	$G_c=K_c$
Process	$G_p=1/(s+1)^3$
Sensor, case (i)	$H=1$
Sensor, case (ii)	$H=e^{-0.7s}$ (0.7 s dead time)

Find. (a) ultimate gain $K_{cu}$ for each sensor; (b) gain and phase margins at $K_c=1$.

Approach. For the rational case use the Routh test on the characteristic polynomial; for the dead-time case the delay is transcendental, so work directly with the frequency response — find the phase-crossover frequency where $\angle L=-180^\circ$ and set $|L|=1$ there.

(a-i) Rational case — Routh. With $H=1$ the characteristic equation is $1+K_cG_p=0\Rightarrow(s+1)^3+K_c=0$, i.e. $s^3+3s^2+3s+(1+K_c)=0$. The Routh $s^1$ entry $\dfrac{3\cdot3-(1+K_c)}{3}=\dfrac{8-K_c}{3}$ must stay positive: $$\boxed{K_{cu}=8}\qquad(\text{at }K_c=8:\ 3s^2+9=0\Rightarrow\omega_u=\sqrt3=1.73\ \text{rad/s}).$$
(a-ii) Dead-time case — frequency response. A delay cannot be handled by Routh, so use $L(j\omega)=K_c\,e^{-0.7j\omega}/(1+j\omega)^3$. The phase-crossover frequency solves $\angle L=-0.7\omega-3\tan^{-1}\omega=-\pi$: $$0.7\omega+3\tan^{-1}\omega=\pi\ \Rightarrow\ \omega_{co}=1.039\ \text{rad/s}.$$ Setting the amplitude ratio to unity there, $\dfrac{K_{cu}}{(1+\omega_{co}^2)^{3/2}}=1$: $$\boxed{K_{cu}=(1+1.039^2)^{3/2}=3.00}.$$ The 0.7 s dead time slashes the stable gain from 8 to 3.
Frequency-response picture (case ii). The normalised amplitude ratio $|G_pH|=1/(1+\omega^2)^{3/2}$ is flat at $0$ dB below the corner $\omega=1$ (three coincident poles) and falls at $-60$ dB/decade above it. The phase $\phi=-0.7\omega-3\tan^{-1}\omega$ starts at $0^\circ$, and because the dead-time term $-0.7\omega$ grows without bound the phase decreases past $-180^\circ$ (at $\omega_{co}=1.039$) toward $-\infty$ — unlike the pure triple lag, whose phase would asymptote at $-270^\circ$.

Problem 3: Bode plot of $K_cG_pH$ for case (ii). Normalised AR (blue) with the $0$ dB and $-60$ dB/dec asymptotes (dashed); corner at $\omega=1$. The phase (lower panel) is driven below $-180^\circ$ at $\omega_{co}=1.04$ by the dead time and continues toward $-\infty$.

(b-i) Margins at $K_c=1$, $H=1$. Phase crossover: $3\tan^{-1}\omega=\pi\Rightarrow\omega=\tan60^\circ=\sqrt3$. There $|L|=1/(1+3)^{3/2}=1/8$, so $$\boxed{\text{GM}=8\ (18.1\ \text{dB})}.$$ Since $|L(0)|=1$ and $|L|$ decreases monotonically, the gain-crossover frequency is $\omega=0$ where $\phi=0$, giving $\text{PM}=180^\circ+0=180^\circ$.
(b-ii) Margins at $K_c=1$, $H=e^{-0.7s}$. Phase crossover is $\omega_{co}=1.039$ (step 2); there $|L|=1/(1+1.039^2)^{3/2}=1/3.00$, so $$\boxed{\text{GM}=3.00\ (9.5\ \text{dB})}.$$ The dead time does not change $|L|$, so again $|L(0)|=1$, the gain crossover is at $\omega=0$ and $\text{PM}=180^\circ$. The two cases share the phase margin but the delay cuts the gain margin from 8 to 3.

Quantity	Case (i) $H=1$	Case (ii) $H=e^{-0.7s}$
Ultimate gain $K_{cu}$	$8$	$3.00$
Phase-crossover $\omega_{co}$	$\sqrt3=1.73$	$1.04$ rad/s
Gain margin at $K_c=1$	$8$ (18.1 dB)	$3.0$ (9.5 dB)
Phase margin at $K_c=1$	$180^\circ$	$180^\circ$

Concept & Theory

Topic. Routh test versus direct frequency-response stability; ultimate gain; gain and phase margins; the destabilising effect of dead time.

Key relations. Routh array for $s^3+3s^2+3s+(1+K_c)$; phase crossover $\angle L(j\omega_{co})=-180^\circ$; $\text{GM}=1/|L(j\omega_{co})|$; $\text{PM}=180^\circ+\angle L(j\omega_{gc})$ with $|L(j\omega_{gc})|=1$.

Why this works. A rational characteristic polynomial admits the algebraic Routh test, but $e^{-0.7s}$ is transcendental, so only the frequency response (which handles the delay exactly as a phase $-\theta\omega$) locates the stability limit — hence the “no Padé” instruction. Because the dead time adds phase lag without touching magnitude, it lowers $\omega_{co}$ and therefore the gain margin, while leaving the (already very large) phase margin unchanged here.

Common pitfall. Trying to apply Routh to the delay via Padé when forbidden; reading $\omega_{co}$ in degrees (the $0.7\omega$ term is in radians); or forgetting that at $K_c=1$ the DC loop gain is exactly $1$, so the gain crossover sits at $\omega=0$.

Source. Seborg et al., Process Dynamics and Control, Ch. 11 & 14 (Routh test, Bode stability, gain/phase margins); Stephanopoulos, Ch. 17–18 (dead-time frequency response).

Question 3 of 8: Triple-Lag Loop with a Dead-Time Sensor — Stability Limit and Margins

Problem 3: Triple-Lag Loop with a Dead-Time Sensor — Stability Limit and Margins (20%)