23-Chem-A6 Process Dynamics and Control · May 2014
Question 8 of 8: Bode Diagram and Gain Margin of a Dead-Time Process
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2014 — 04-Chem-A6 Process Dynamics & Control. Three-hour open-book examination; any non-communicating calculator is permitted. Eight problems are printed and any five constitute a complete paper (each worth 20%); all eight are solved below for completeness. The parts are quantitative throughout — tank and reactor modelling, transfer functions, step/impulse responses, IMC design, Nyquist and Bode stability — and every requested plot (IMC response, Nyquist locus, thermocouple response, Bode diagram) is drawn as a real figure.
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, transfer functions, feedback stability, frequency response and IMC design; G. Stephanopoulos, Chemical Process Control: An Introduction to Theory and Practice (Prentice Hall) — Nyquist criterion, Bode stability and dead-time systems; D. R. Coughanowr & S. E. LeBlanc, Process Systems Analysis and Control (3rd ed., McGraw-Hill) — first- and second-order tank dynamics, step/impulse response and linearisation of non-linear balances. Standard control conventions (deviation variables; unity valve/sensor gains unless stated) are used throughout.
Problem 8: Bode Diagram and Gain Margin of a Dead-Time Process (20%)
Given. $G_p(s)=\dfrac{e^{-\theta s}}{s+1}$ with dead time $\theta=2$ and lag time constant $\tau_{lag}=1$; proportional gain $K_c$. Open loop $L(j\omega)=\dfrac{K_c e^{-j2\omega}}{j\omega+1}$.
Find. (a) the Bode magnitude and phase diagrams with corners, asymptotes and slopes; (b) the gain margin at $K_c=1$.
Approach. Separate the amplitude ratio and phase into the first-order-lag and dead-time contributions; sketch the Bode asymptotes; then, for the margin, find the phase-crossover frequency where $\angle L=-180^{\circ}$ and take the reciprocal of the amplitude ratio there.
Amplitude ratio and phase. The dead time has unit magnitude, so $$\mathrm{AR}=|L|=\dfrac{K_c}{\sqrt{\omega^2+1}},\qquad \angle L=-\theta\omega-\tan^{-1}\omega=-2\omega-\tan^{-1}\omega\ \text{(rad)}.$$
(a) Magnitude diagram. Low-frequency asymptote: flat at $\mathrm{AR}=K_c$ (0 dB for $K_c=1$), slope $0$. Corner (break) frequency at the pole, $\omega=1/\tau_{lag}=\boxed{1\ \mathrm{rad/s}}$, where $\mathrm{AR}$ is $3\ \mathrm{dB}$ below the asymptote. High-frequency asymptote: slope $\boxed{-20\ \mathrm{dB/decade}}$ (the dead time does not change magnitude).
(a) Phase diagram. As $\omega\to0$, $\angle L\to0^{\circ}$. The first-order lag contributes $0\to-90^{\circ}$ (through $-45^{\circ}$ at the corner $\omega=1$); the dead time adds an unbounded linear lag $-2\omega$ (rad). Hence the total phase falls through $-180^{\circ}$ and keeps decreasing without limit — the hallmark of dead time.
(b) Gain margin at $K_c=1$. $\mathrm{AR}(\omega_c)=\dfrac{1}{\sqrt{1.14^2+1}}=\dfrac{1}{\sqrt{2.30}}=0.659$, so $$\mathrm{GM}=\dfrac{1}{\mathrm{AR}(\omega_c)}=\boxed{1.52}\ (\approx3.6\ \mathrm{dB}).$$ The gain may be raised by a factor $1.52$ before $\mathrm{AR}$ reaches unity at $\omega_c$; equivalently the ultimate gain is $K_{cu}\approx1.52$, so $K_c=1$ is stable with a modest margin.
Problem 8(a): Bode diagram of $K_cG_p$ at $K_c=1$. Magnitude (top): flat at $0\ \mathrm{dB}$, corner at $\omega=1$ ($-3\ \mathrm{dB}$), then $-20\ \mathrm{dB/decade}$ (asymptotes dashed). Phase (bottom): $0^{\circ}\to$ decreasing without bound; it crosses $-180^{\circ}$ at $\omega_c\approx1.14$, where the magnitude is $0.66$, giving $\mathrm{GM}=1.52$.
