Internal Model Control Design for a Dead-Time-Dominant Process: 20-Bio-A2 Process Dynamics and Control, Undated paper, Q7

Problem 7: Internal Model Control Design for a Dead-Time-Dominant Process (20% total)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $G_p(s)=\dfrac{10e^{-5s}}{200s+1}$, IMC filter time constant $\tau_c=10$, perfect model $\tilde G_p=G_p$.

Find. (a) the block diagram and $G_c^{*}(s)$; (b) the unit-step closed-loop response $C(t)$ for the perfect-model case.

Problem 7(a): standard IMC structure. $G_c^{*}(s)$ drives both the true process and an internal copy of the model; with a perfect model the feedback path carries zero signal, so the closed loop reduces to open-loop $G_c^{*}(s)G_p(s)$.

Approach. Factor $G_p$ into the non-invertible dead-time part $G_p^-=e^{-5s}$ and the invertible minimum-phase part $G_p^+=10/(200s+1)$; invert $G_p^+$ and append the first-order filter $f(s)=1/(\tau_cs+1)$ to form $G_c^{*}$; then, with a perfect model, the servo transfer reduces exactly to $G_p^-f$ and inverts as a delayed first-order step.

(a) Factor the model. $G_p(s)=\underbrace{e^{-5s}}_{G_p^-}\cdot\underbrace{\dfrac{10}{200s+1}}_{G_p^+}$.
(a) IMC controller. With $\tau_c=10\ (\ne\tau_p=200$, unlike the convenient special case where they coincide$)$: $$\boxed{G_c^{*}(s)=\big[G_p^+\big]^{-1}f(s)=\frac{200s+1}{10}\cdot\frac{1}{10s+1}=\frac{200s+1}{10(10s+1)}}$$ — a lead-dominant (PI-like) filtered inverse, not a constant, because the chosen filter time constant is much faster than the process lag.
(b) Perfect-model servo transfer. With $\tilde G_p=G_p$ the internal feedback signal is exactly zero, so the loop reduces to $Y/Y_{sp}=G_p\,G_c^{*}$: substituting and cancelling $(200s+1)$, $$\boxed{\frac{Y(s)}{Y_{sp}(s)}=\frac{10e^{-5s}}{200s+1}\times\frac{200s+1}{10(10s+1)}=\frac{e^{-5s}}{10s+1}}$$ — a pure dead time cascaded with a first-order filter of the chosen $\tau_c=10$, exactly as IMC theory predicts for a perfect model (the achievable closed loop is limited only by the non-invertible dead time).
(b) Invert the unit step. Ignoring the delay momentarily, $\mathcal L^{-1}\!\left[\dfrac1{s(10s+1)}\right]=1-e^{-t/10}$; reinserting the $5\ \text{s}$ delay as a pure time shift, $$\boxed{C(t)=\begin{cases}0,&t<5\ \text{s}\\[2pt]1-e^{-(t-5)/10},&t\ge5\ \text{s}.\end{cases}}$$ At $t=15\ \text{s}$ (one time constant after the delay clears): $C=1-e^{-1}=0.632$; at $t=30\ \text{s}$: $C=1-e^{-2.5}=0.918$; as $t\to\infty$: $C\to1$ (zero offset, as expected for a step in set point with a perfect model).

Result	Value
Model split	$G_p^-=e^{-5s}$, $G_p^+=10/(200s+1)$
IMC controller	$G_c^{*}(s)=(200s+1)/[10(10s+1)]$
Servo transfer (perfect model)	$Y/Y_{sp}=e^{-5s}/(10s+1)$
Step response	$C(t)=0$ for $t<5$; $1-e^{-(t-5)/10}$ for $t\ge5$

Concept & Theory

Topic. Internal Model Control (IMC) design and its achievable closed-loop response for a first-order-plus-dead-time process, filter time constant chosen independently of the process time constant.

Key relations. $G_p=G_p^-G_p^+$; $G_c^{*}=[G_p^+]^{-1}f$; perfect-model servo response $Y/Y_{sp}=G_p^-G_p^+G_c^{*}=G_p^-f$.

Why this works. IMC design never inverts the non-invertible dead-time part of the model, so with a perfect model the best achievable closed-loop response is exactly the dead time acting on the chosen filter, regardless of the process's own (possibly very different) lag time constant — here $\tau_c=10$ is deliberately much smaller than $\tau_p=200$, giving a much faster closed loop than the open-loop process, at the cost of $G_c^{*}$ now being a dynamic (lead/lag) controller rather than a pure gain.

Common pitfall. Assuming $G_c^{*}$ must always reduce to a constant (that only happens in the special case $\tau_c=\tau_p$); here choosing $\tau_c\ne\tau_p$ deliberately is the point of the design freedom IMC provides.

Source. Seborg et al., Process Dynamics and Control, Ch. 12 (IMC design, filter selection, achievable performance).

Question 7 of 8: Internal Model Control Design for a Dead-Time-Dominant Process

Problem 7: Internal Model Control Design for a Dead-Time-Dominant Process (20% total)