IMC Design for a Non-Minimum-Phase Dead-Time Process: 20-Bio-A2 Process Dynamics and Control, May 2017, Q3

Problem 3: IMC Design for a Non-Minimum-Phase Dead-Time 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. $G_p(s)=\dfrac{10(0.5-s)e^{-10s}}{100s+1}=\dfrac{5(1-2s)e^{-10s}}{100s+1}$ — a first-order-lag process with dead time and a right-half-plane zero at $s=+0.5$; desired closed-loop time constant $\tau_c=5$.

Find. (a) the IMC controller $q(s)$ and its block diagram; (b) the perfect-model closed-loop unit-step response.

Problem 3(a): IMC structure. The controller $q(s)$ drives both the real process and an internal copy of the model in parallel; the difference between the true output and the model's prediction is fed back and added to the set point, so under a perfect model that feedback carries zero signal and the loop reduces to open-loop $q(s)G_p(s)$.

Approach. Factor $G_p$ into a non-invertible part $G_p^-$ (all-pass: the dead time and the RHP zero, normalised to unit steady-state gain) and an invertible, minimum-phase part $G_p^+$. Invert only $G_p^+$ and append a first-order filter $f(s)=1/(\tau_cs+1)$ for properness and robustness — that product is $q(s)$. With a perfect model the closed-loop servo transfer is simply $G_p^-(s)f(s)$; invert its unit-step response by partial fractions.

(a) Factor the model. Write $10(0.5-s)=5(1-2s)$, so $G_p(s)=\underbrace{(1-2s)e^{-10s}}_{G_p^-(s),\ G_p^-(0)=1}\cdot\underbrace{\dfrac{5}{100s+1}}_{G_p^+(s)}$. The RHP zero and the dead time both belong to $G_p^-$ because inverting either would create a non-causal, unbounded controller.
(a) IMC controller. With filter order 1 (enough to make $q$ proper, since $G_p^+$ is first order) and $\tau_c=5$ (the value specified for part b), $$\boxed{q(s)=\big[G_p^+(s)\big]^{-1}f(s)=\frac{100s+1}{5}\cdot\frac{1}{5s+1}=\frac{100s+1}{5(5s+1)}.}$$ The classical feedback-equivalent controller is $G_c=q/(1-G_pq)$, which is not required numerically here but follows the same block algebra shown in the figure.
(b) Perfect-model servo transfer. With $\tilde G_p=G_p$, the internal feedback path in the figure carries zero signal, and the closed loop reduces to $\dfrac{Y}{Y_{sp}}=G_p^+q\cdot G_p^-\big/G_p^+=G_p^-(s)f(s)$: $$\boxed{\frac{Y(s)}{Y_{sp}(s)}=(1-2s)e^{-10s}\cdot\frac{1}{5s+1}=\frac{(1-2s)e^{-10s}}{5s+1}.}$$
(b) Invert the unit step. Ignoring the pure delay for a moment, $\dfrac{1-2s}{s(5s+1)}=\dfrac1s-\dfrac{1.4}{s+0.2}$, so the un-delayed response is $y'(t)=1-1.4e^{-0.2t}$. Re-inserting the $10\ \text{s}$ delay shifts this in time: $$\boxed{y(t)=\begin{cases}0,&t<10\\[2pt]1-1.4\,e^{-0.2(t-10)},&t\ge10.\end{cases}}$$
Interpret. At $t=10^+$, $y=1-1.4=-0.4$ — the RHP zero forces the output to undershoot the moment the dead time clears, before rising smoothly (time constant $5\ \text{s}$) to the set point $y(\infty)=1$ (zero offset, since $G_p^-(0)f(0)=1$).

Result	Value
Model split	$G_p^-=(1-2s)e^{-10s}$, $G_p^+=5/(100s+1)$
IMC controller	$q(s)=(100s+1)/[5(5s+1)]$
Servo transfer (perfect model)	$Y/Y_{sp}=(1-2s)e^{-10s}/(5s+1)$
Step response	$y(t)=0$ for $t<10$; $1-1.4e^{-0.2(t-10)}$ for $t\ge10$
Undershoot at $t=10^+$	$y=-0.4$

Concept & Theory

Topic. Internal Model Control (IMC) of a non-minimum-phase, dead-time process; factorisation into invertible/non-invertible parts; inverse response from a design RHP zero.

Key relations. $G_p=G_p^-G_p^+$, $G_p^-(0)=1$; $q=(G_p^+)^{-1}f$, $f=1/(\tau_cs+1)$; perfect-model servo $Y/Y_{sp}=G_p^-f$.

Why this works. Any attempt to invert the dead time or the RHP zero would require future information or an unbounded, non-causal controller, so IMC design always leaves those factors untouched in the closed loop; the best achievable servo response is therefore the non-invertible part shaped only by the tuning filter, which is exactly $G_p^-f$.

Common pitfall. Trying to invert $(1-2s)$ or $e^{-10s}$ directly (produces an unrealisable controller with a right-half-plane pole or a predictor), or forgetting that the achievable closed-loop gain is fixed at $G_p^-(0)=1$ regardless of $\tau_c$.

Source. Seborg et al., Process Dynamics and Control, Ch. 12 (IMC design, all-pass factorisation, filter selection); Morari & Zafiriou, Robust Process Control, Ch. 3–4.

Question 3 of 8: IMC Design for a Non-Minimum-Phase Dead-Time Process

Problem 3: IMC Design for a Non-Minimum-Phase Dead-Time Process (20% total)