23-Chem-A6 Process Dynamics and Control · December 2015
Question 5 of 8: IMC Design, Servo Response and IMC–PID Equivalence (FOPDT)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2015 — 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. Most parts are quantitative (dynamic modelling, transfer functions, step/pulse/ramp responses, Routh and Nyquist stability, and IMC design); qualitative sketches are drawn as real figures where 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, Servo Response and IMC–PID Equivalence (FOPDT) (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 $q(s)$ + block diagram; (b) the servo step response; (c) equivalent PID parameters via 1–1 Padé.
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; the perfect-model servo transfer is then just the delay times the filter; finally rearrange the IMC controller into classical form with a Padé-approximated delay to read off PID parameters.
(a) Factor and invert. Split $G_p=G_p^{-}G_p^{+}$ with the all-pass delay $G_p^{+}=e^{-5s}$ (kept, since inverting it is non-causal) and the invertible part $G_p^{-}=\dfrac{10}{100s+1}$. The IMC controller is $q=(G_p^{-})^{-1}f$ with filter $f=\dfrac{1}{\tau_cs+1}=\dfrac{1}{10s+1}$: $$\boxed{q(s)=\frac{100s+1}{10}\cdot\frac{1}{10s+1}=\frac{100s+1}{10\,(10s+1)}}.$$
Problem 5(a): IMC structure. The controller $q$ drives the real process $G_p$ and the internal model $\tilde G_p$ in parallel; only the model mismatch (here zero) is fed back, so the loop behaves open-loop-like and is offset-free at the set point.
(b) Servo transfer (perfect model). With $\tilde G_p=G_p$ the IMC closed loop collapses to $\dfrac{Y}{Y_{sp}}=G_p^{+}f=\dfrac{e^{-5s}}{10s+1}$. Inverting the transform, the response to a unit set-point step is a delayed first-order rise: $$\boxed{y(t)=\Big[1-e^{-(t-5)/10}\Big]\,\mathcal U(t-5)},$$ 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 the tuning parameter $\tau_c$.
Problem 5(b): IMC servo response for $\tau_c=10$. The output waits out the $5$ s dead time, then rises as a first-order lag to the new set point with no overshoot and no offset.
(c) Equivalent PID via 1–1 Padé. Approximate the delay $e^{-5s}\approx\dfrac{1-2.5s}{1+2.5s}$ ($\theta/2=2.5$). The classical controller $G_c=\dfrac{q}{1-\tilde G_pq}$ reduces (after the $(1-2.5s)$ factors cancel) to $$G_c=\frac{(100s+1)(1+2.5s)}{10\,(\tau_c+2.5)\,s}=\frac{1}{K(\tau_c+\theta/2)}\Big[(\tau+\tfrac\theta2)+\tfrac1s+\tfrac{\theta\tau}{2}s\Big],$$ which matches an ideal PID $G_c=K_c\big(1+\tfrac{1}{\tau_Is}+\tau_Ds\big)$ with $$\boxed{K_c=\frac{\tau+\theta/2}{K(\tau_c+\theta/2)}=0.82,\quad \tau_I=\tau+\tfrac\theta2=102.5\ \text{s},\quad \tau_D=\frac{\tau\theta}{2\tau+\theta}=2.44\ \text{s}}.$$
