23-Chem-A6 Process Dynamics and Control · May 2017
Question 2 of 8: Frequency Response of an FOPDT Loop — Bode Diagram and Gain-Margin Design
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2017 — 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 paper is entirely quantitative — dynamic modelling, transfer functions, step/ramp responses, Routh and Nyquist stability, frequency-response (Bode) design and IMC — and qualitative sketches are drawn as real figures wherever the paper asks for them.
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.
Problem 2: Frequency Response of an FOPDT Loop — Bode Diagram and Gain-Margin Design (20%)
Given. A first-order-plus-dead-time (FOPDT) plant under proportional control:
Parameter
Symbol
Value
Process gain
$K$
$1$
Time constant
$\tau$
$0.5$ s
Dead time
$\theta$
$0.1$ s
Target gain margin
$\mathrm{GM}$
$1.7$
Find. (a) the asymptotic Bode diagram of $k_cG_p$; (b) the proportional gain $k_c$ giving a gain margin of $1.7$.
Approach. Read the amplitude ratio and phase of $G_p(j\omega)$ directly from its factors (a first-order lag plus a pure delay), sketch the two asymptotic diagrams, then locate the phase-crossover frequency (phase $=-180^\circ$), evaluate the amplitude ratio there, and set $k_c$ so that $k_c\,\mathrm{AR}=1/\mathrm{GM}$.
(a) Amplitude ratio and phase. With $s=j\omega$, $$\mathrm{AR}=|k_cG_p|=\frac{k_c}{\sqrt{1+(\tau\omega)^2}}=\frac{k_c}{\sqrt{1+0.25\,\omega^2}},\qquad \angle G_p=-\theta\omega-\tan^{-1}(\tau\omega)=-0.1\,\omega-\tan^{-1}(0.5\,\omega).$$ The delay contributes no gain (unit magnitude) but a phase that grows linearly and without bound; the lag contributes the roll-off.
Asymptotes and corner. The corner is at $\omega_c=1/\tau=2\,$rad/s. For $\omega\ll2$ the gain is flat at $k_c$ ($0\,$dB when $k_c=1$) and the phase $\to0^\circ$; for $\omega\gg2$ the gain falls at $\boxed{-20\ \text{dB/dec}}$ and the phase, dominated by the delay, tends to $-\infty$. At the corner the lag alone gives $-3\,$dB and $-45^\circ$ (plus a small $-0.1\cdot2=-0.2\,$rad $=-11.5^\circ$ from the delay). See the figure.
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(b) Phase-crossover frequency. The gain margin is measured where the phase equals $-180^\circ=-\pi$: solve $0.1\,\omega_{co}+\tan^{-1}(0.5\,\omega_{co})=\pi$. Iterating gives $$\boxed{\omega_{co}=16.9\ \text{rad/s}}\quad(0.1\cdot16.9+\tan^{-1}(8.45)=1.69+1.453=3.14\approx\pi).$$
Amplitude ratio at $\omega_{co}$. $\mathrm{AR}_{k_c=1}=\dfrac{1}{\sqrt{1+(0.5\cdot16.9)^2}}=\dfrac{1}{\sqrt{1+71.4}}=\dfrac{1}{8.50}=0.1176.$ The ultimate gain (GM$=1$) is therefore $k_{cu}=1/0.1176=8.50$.
Gain for GM $=1.7$. By definition $\mathrm{GM}=\dfrac{1}{k_c\,\mathrm{AR}(\omega_{co})}$, so $$k_c=\frac{1}{\mathrm{GM}\cdot\mathrm{AR}}=\frac{1}{1.7\times0.1176}=\boxed{k_c\approx5.0}.$$ (Equivalently $k_c=k_{cu}/\mathrm{GM}=8.50/1.7$.) At this gain the corresponding phase margin is about $45^\circ$ (gain crossover near $\omega\approx9.8\,$rad/s), a comfortably stable design.
