Internal Model Control of a Dead-Time-Dominant Process: 20-Bio-A2 Process Dynamics and Control, May 2018, Q6

Problem 6: Internal Model Control of 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. $G_p(s)=\dfrac{5e^{-10s}}{10s+1}$, IMC filter time constant $\tau_c=20\ \text{s}$, perfect model $\tilde G_p=G_p$, no Padé approximation of the dead time.

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

Problem 6(a): standard IMC structure. $q(s)$ drives both the true process and an internal copy of the model; with a perfect model the feedback path carries zero signal, so the loop reduces to open-loop $q(s)G_p(s)$.

Approach. Factor $G_p$ into a non-invertible part $G_p^-$ (the dead time; the gain is $+5$ so there is no RHP zero, only the delay is non-invertible) and an invertible minimum-phase part $G_p^+$; invert $G_p^+$ and append a first-order filter $f(s)=1/(\tau_cs+1)$ to form $q(s)$; convert to the classical equivalent $G_c=q/(1-G_pq)$ without any Padé approximation; then, with a perfect model, the servo transfer reduces to $G_p^-f$ and can be inverted directly (delay reinserted as a pure time shift).

(a) Factor the model. $G_p(s)=\underbrace{e^{-10s}}_{G_p^-,\ G_p^-(0)=1}\cdot\underbrace{\dfrac{5}{10s+1}}_{G_p^+}$; only the dead time is non-invertible (no RHP zero here).
(a) IMC controller. With a first-order filter (matching the first-order $G_p^+$) and $\tau_c=20$, $$\boxed{q(s)=G_c^{*}=\big[G_p^+\big]^{-1}f(s)=\frac{10s+1}{5}\cdot\frac{1}{20s+1}=\frac{10s+1}{5(20s+1)}.}$$
(a) Classical feedback-equivalent. With a perfect model, $1-G_pq=1-\dfrac{e^{-10s}}{20s+1}$, so $$\boxed{G_c(s)=\frac{q(s)}{1-G_p(s)q(s)}=\frac{10s+1}{5\big[20s+1-e^{-10s}\big]}.}$$ Because $e^{-10s}$ appears transcendentally in the denominator (no Padé approximation applied), $G_c(s)$ is not reducible to a rational $K_c(1+1/(\tau_Is)+\tau_Ds)$ — it is a Smith-predictor-type controller, not classical PID.
(b) Perfect-model servo transfer. Since $\tilde G_p=G_p$, the internal feedback signal is exactly zero and the loop reduces to $Y/Y_{sp}=G_p^-f$: $$\boxed{\frac{Y(s)}{Y_{sp}(s)}=e^{-10s}\cdot\frac{1}{20s+1}.}$$
(b) Invert the unit step. Ignoring the delay momentarily, $\dfrac{1}{s(20s+1)}=\dfrac1s-\dfrac{1}{s+0.05}$, giving the undelayed response $1-e^{-t/20}$. Reinserting the $10\ \text{s}$ delay as a pure time shift, $$\boxed{\delta C(t)=\begin{cases}0,&t\lt10\\[2pt]1-e^{-(t-10)/20},&t\ge10.\end{cases}}$$

Result	Value
Model split	$G_p^-=e^{-10s}$, $G_p^+=5/(10s+1)$
IMC controller	$q(s)=(10s+1)/[5(20s+1)]$
Classical equivalent	$G_c=(10s+1)/\{5[20s+1-e^{-10s}]\}$ — not PID form
Servo transfer (perfect model)	$Y/Y_{sp}=e^{-10s}/(20s+1)$
Step response	$\delta C(t)=0$ 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. $G_p=G_p^-G_p^+$; $q=[G_p^+]^{-1}f$; $G_c=q/(1-G_pq)$; perfect-model servo response $Y/Y_{sp}=G_p^-f$.

Why this works. IMC design never inverts the non-invertible (dead-time/RHP-zero) part of the model, so the best achievable closed-loop response is always that part shaped only by the tuning filter; converting back to the classical feedback form necessarily inherits the same transcendental delay term, which is exactly why $G_c$ cannot be rational/PID unless the delay is first approximated (e.g. by Padé).

Common pitfall. Applying a Padé approximation before deriving $G_c$ (the question explicitly forbids this) or mistakenly claiming $G_c$ is PID because $q(s)$ itself looks like a simple lead/lag.

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

Question 6 of 8: Internal Model Control of a Dead-Time-Dominant Process

Problem 6: Internal Model Control of a Dead-Time-Dominant Process (20% total)