Internal Model Control of a Dead-Time-Dominant Process: 20-Bio-A2 Process Dynamics and Control, December 2019, Q4

Problem 4: Internal Model Control of a Dead-Time-Dominant Process (20%)

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. $G_p(s)=\dfrac{5e^{-10s}}{20s+1}$, IMC filter $\tau_c=20\ \text{s}$, perfect model $\tilde G_p=G_p$, no Padé approximation of the dead time.

Find. (a) the block diagram, $G_c^{*}(s)$, the classical feedback-equivalent $G_c(s)$, and whether $G_c$ is PID form; (b) the unit-step closed-loop response $C(t)$.

Problem 4(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 a non-invertible dead-time part $G_p^-=e^{-10s}$ (gain $+5$, no RHP zero, so only the delay is non-invertible) and an invertible minimum-phase part $G_p^+=5/(20s+1)$; invert $G_p^+$ and append the first-order filter $f(s)=1/(\tau_cs+1)$ to form $G_c^{*}$; convert to the classical equivalent $G_c=G_c^{*}/(1-G_pG_c^{*})$ without any Padé approximation; then, with a perfect model, the servo transfer reduces to $G_p^-f$ and inverts directly (delay reinserted as a pure time shift).

(a) Factor the model. $G_p(s)=\underbrace{e^{-10s}}_{G_p^-}\cdot\underbrace{\dfrac{5}{20s+1}}_{G_p^+}$.
(a) IMC controller. With $\tau_c=\tau_p=20\ \text{s}$ (filter time constant matches the process time constant exactly), $$\boxed{G_c^{*}(s)=\big[G_p^+\big]^{-1}f(s)=\frac{20s+1}{5}\cdot\frac{1}{20s+1}=\frac15=0.2\ \ (\text{a pure constant gain}).}$$
(a) Classical feedback-equivalent. With a perfect model, $$\boxed{G_c(s)=\frac{G_c^{*}}{1-G_p G_c^{*}}=\frac{0.2}{1-\dfrac{e^{-10s}}{20s+1}}=\frac{0.2(20s+1)}{20s+1-e^{-10s}}.}$$ Because the transcendental delay term $e^{-10s}$ appears in the denominator (no Padé approximation applied), $G_c(s)$ cannot be reduced to a rational $K_c\!\left(1+\tfrac1{\tau_Is}+\tau_Ds\right)$ — $\boxed{G_c\ \text{is NOT of PID form}}$ (it is a Smith-predictor-type dead-time compensator).
(b) Perfect-model servo transfer. With $\tilde G_p=G_p$ the internal feedback signal is exactly zero, so the loop reduces to $Y/Y_{sp}=G_p^-f\cdot(\text{gain from }G_p^+ G_c^{*}=1)$: substituting, $$\boxed{\frac{Y(s)}{Y_{sp}(s)}=G_p(s)\,G_c^{*}(s)=\frac{5e^{-10s}}{20s+1}\times0.2=\frac{e^{-10s}}{20s+1}.}$$
(b) Invert the unit step. Ignoring the delay momentarily, $\mathcal L^{-1}\!\left[\dfrac{1}{s(20s+1)}\right]=1-e^{-t/20}$; reinserting the $10\ \text{s}$ delay as a pure time shift, $$\boxed{C(t)=\begin{cases}0,&t\lt10\ \text{s}\\[2pt]1-e^{-(t-10)/20},&t\ge10\ \text{s}.\end{cases}}$$ At $t=30\ \text{s}$: $C=1-e^{-1}=0.632$; as $t\to\infty$, $C\to1$ (zero offset, as expected for a step in set point with a perfect model).

Result	Value
Model split	$G_p^-=e^{-10s}$, $G_p^+=5/(20s+1)$
IMC controller	$G_c^{*}(s)=0.2$ (constant, since $\tau_c=\tau_p$)
Classical equivalent	$G_c=0.2(20s+1)/(20s+1-e^{-10s})$ — not PID form
Servo transfer (perfect model)	$Y/Y_{sp}=e^{-10s}/(20s+1)$
Step response	$C(t)=0$ for $t\lt10$; $1-e^{-(t-10)/20}$ for $t\ge10$

Concept & Theory

Topic. Internal Model Control (IMC) design and its exact (non-Padé) classical feedback equivalent for a first-order-plus-dead-time process.

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

Why this works. IMC design never inverts the non-invertible (dead-time) part of the model, so the best achievable closed-loop response is the dead time acting on the filter response; when the chosen filter time constant equals the process lag time constant exactly, the invertible first-order dynamics cancel perfectly, leaving a pure gain for $G_c^{*}$ — a special (and here exam-convenient) case.

Common pitfall. Applying a Padé approximation before deriving $G_c$ (the question explicitly forbids it) or mistaking the constant $G_c^{*}=0.2$ for evidence that $G_c$ itself must be simple — the classical equivalent still carries the full transcendental delay.

Source. Seborg et al., Process Dynamics and Control, Ch. 12 (IMC design, filter selection, classical-equivalent controller).

Question 4 of 8: Internal Model Control of a Dead-Time-Dominant Process

Problem 4: Internal Model Control of a Dead-Time-Dominant Process (20%)