23-Chem-A6 Process Dynamics and Control · May 2017
Question 5 of 8: IMC Design for an FOPDT Process — Controller Form and Servo Response
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 5: IMC Design for an FOPDT Process — Controller Form and Servo Response (20%)
Given. A first-order-plus-dead-time (FOPDT) process:
Parameter
Symbol
Value
Gain
$K$
$10$
Time constant
$\tau$
$100$ s
Dead time
$\theta$
$5$ s
IMC filter
$\tau_c$
$10$ s
Find. (a) the IMC controller $G_c^{*}$, the equivalent classical $G_c$ (no Padé), and whether $G_c$ is PID; (b) the servo step response.
Approach. Factor the model into an invertible minimum-phase part and an all-pass delay, invert only the invertible part and add a filter for properness; convert to the classical controller $G_c=G_c^{*}/(1-\tilde G G_c^{*})$ and inspect its form; then read off the perfect-model servo transfer, which is just the delay shaped by the filter.
(a) IMC controller. Split $G=G^{-}G^{+}$ with all-pass delay $G^{+}=e^{-5s}$ (kept, since inverting it is non-causal) and invertible part $G^{-}=\dfrac{10}{100s+1}$. The IMC controller is $G_c^{*}=(G^{-})^{-1}f$ with filter $f=\dfrac{1}{\tau_cs+1}=\dfrac{1}{10s+1}$: $$\boxed{G_c^{*}(s)=\frac{100s+1}{10}\cdot\frac{1}{10s+1}=\frac{100s+1}{10\,(10s+1)}}.$$
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(a) Equivalent classical controller — is it PID? The classical feedback form is $G_c=\dfrac{G_c^{*}}{1-\tilde G\,G_c^{*}}$. With a perfect model $\tilde G=G$, the loop product is $\tilde G\,G_c^{*}=\dfrac{10e^{-5s}}{100s+1}\cdot\dfrac{100s+1}{10(10s+1)}=\dfrac{e^{-5s}}{10s+1}$, so $$\boxed{G_c(s)=\frac{(100s+1)/[10(10s+1)]}{1-e^{-5s}/(10s+1)}=\frac{100s+1}{10\,(10s+1-e^{-5s})}}.$$ Because the dead-time term $e^{-5s}$ survives in the denominator, $G_c$ is not a finite-dimensional PID controller — it is a rational-plus-transport-delay (Smith-predictor-like) compensator. Only if $e^{-5s}$ is replaced by a Padé approximation does $G_c$ collapse to PID form; the question forbids that step, so the honest answer is no.
(b) Servo transfer and step response. With $\tilde G=G$ the IMC closed loop reduces to $\dfrac{C}{C_{sp}}=G^{+}f=\dfrac{e^{-5s}}{10s+1}$. Inverting the unit-step response ($C_{sp}=1/s$) gives a delayed first-order rise: $$\boxed{\delta C(t)=\Big[1-e^{-(t-5)/10}\Big]\,\mathcal U(t-5)},$$ i.e. dead-flat until $t=5\,$s, then approaching $1$ with time constant $\tau_c=10\,$s ($63\,\%$ at $t=15\,$s). There is no offset and no overshoot — the closed-loop speed is set entirely by $\tau_c$.
