23-Chem-A6 Process Dynamics and Control · December 2016
Question 6 of 8: Stability Limit, Bode Plot and Margins with a Sensor Lag
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: D. E. Seborg, T. F. Edgar, D. A. Mellichamp & F. J. Doyle III, Process Dynamics and Control (4th ed., Wiley) — Laplace-domain modelling, linearisation, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Routh test, Nyquist criterion and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first-order thermal/level dynamics, step/ramp/pulse response and block-diagram algebra. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout. This December 2016 sitting is an open-book paper of eight problems of which any five constitute a complete exam; all eight are solved here.
Problem 6: Stability Limit, Bode Plot and Margins with a Sensor Lag (20%)
Given. $G_p=1/(s+1)^3$ (corner $\omega=1$, three equal lags); sensor either $H=1$ or a pure dead time $H=e^{-0.7s}$.
Find. (a) ultimate gains $K_{cu}$; (b) the Bode plot of case (ii); (c) gain and phase margins at $K_c=1$.
Bode plot of $K_cG_pH$ for case (ii) at $K_c=1$. Amplitude starts at $0$ dB and rolls off at $-60$ dB/decade beyond the triple corner $\omega=1$ (slope $0\to-3$). The phase leaves $0^\circ$ at low frequency and, because of the dead time, falls without bound ($\to-\infty$) — it would level at $-270^\circ$ without the delay. Phase crosses $-180^\circ$ at $\omega\approx1.04$, where the gain is $1/K_{cu}$.
Approach. The stability limit is where the open-loop phase hits $-180^\circ$; read $K_{cu}$ as the reciprocal of the magnitude there. For the margins at $K_c=1$, note the magnitude never exceeds unity, so the gain-crossover sits at $\omega=0$.
(a-i) $H=1$: phase condition. $\angle L=-3\arctan\omega=-180^\circ\Rightarrow\arctan\omega=60^\circ\Rightarrow\omega_u=\sqrt3$. Magnitude $|L|=K_c/(1+\omega^2)^{3/2}=K_c/8$; setting it to $1$, $$\boxed{K_{cu}=(1+3)^{3/2}=8.}$$ (Routh on $(s+1)^3+K_c$ gives the same limit.)
(a-ii) $H=e^{-0.7s}$: phase condition. The delay adds $-0.7\omega$ (rad): $3\arctan\omega+0.7\omega=\pi$. Solving numerically $\omega_u=1.039$ rad/s; then $$\boxed{K_{cu}=(1+1.039^2)^{3/2}=3.0.}$$ Dead time sharply lowers the achievable gain — and because it is exact, no Padé approximation is used.
(b) Bode features (case ii). Corner frequency $\omega=1$; low-frequency amplitude asymptote flat at $0$ dB (unity DC gain), high-frequency asymptote $-60$ dB/decade (three poles). Phase $\to0^\circ$ as $\omega\to0$ and $\to-\infty$ as $\omega\to\infty$ (the $-0.7\omega$ term dominates); without the dead time it would asymptote to $-270^\circ$. See figure.
(c) Gain-crossover frequency. At $K_c=1$, $|L(j\omega)|=1/(1+\omega^2)^{3/2}\le1$, equal to $1$ only at $\omega=0$. So the gain crosses unity at $\omega_{gc}=0$, where the phase is $0^\circ$.
(c) Phase margin. $\text{PM}=180^\circ+\angle L(\omega_{gc})=180^\circ+0^\circ=\boxed{180^\circ}$ for both cases — a large, somewhat academic margin because the loop gain only reaches unity at DC.
(c) Gain margin. $\text{GM}=K_{cu}/K_c=K_{cu}$ at $K_c=1$: case (i) $\text{GM}=8$ ($18.1$ dB); case (ii) $\text{GM}=3.0$ ($9.5$ dB). The dead time cuts the usable gain by more than half.
