Internal Model Control (IMC) Design: 23-Chem-A6 Process Dynamics and Control, May 2013, Q3

Problem 3: Internal Model Control (IMC) Design (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. $G_p(s)=\dfrac{5(s-1)e^{-5s}}{100s+1}$ — a first-order process ($\tau_p=100$) with a right-half-plane (RHP) zero at $s=+1$ and a dead time $\theta=5$. IMC filter $f=\dfrac{1}{\tau_f s+1}$, $\tau_f=10$.

Find. (a) the IMC controller $q(s)$ and its block diagram; (b) the perfect-model set-point response.

Problem 3(a): IMC structure. The controller $q(s)$ drives the real process $G_p$; an internal model $\tilde G_p$ predicts the output $\tilde y$, and the model error $\hat d=Y-\tilde y$ is fed back. With a perfect model $\tilde G_p=G_p$, $\hat d=0$ and the set-point response reduces to $Y/Y_{sp}=G_p q$.

Approach. Factor the model into an invertible part and a non-invertible all-pass part (the RHP zero and the dead time, which cannot be inverted without instability); invert only the good part and append the filter to get $q$; then use the fact that with a perfect model the IMC set-point response is simply $G_p^{-}f$ times the non-invertible factor.

Factor the model. Split $G_p=G_p^{+}G_p^{-}$ where $G_p^{+}$ holds the non-invertible dynamics (RHP zero + dead time) and is normalised to unity gain: $$G_p^{+}=(1-s)\,e^{-5s},\qquad G_p^{-}=\dfrac{-5}{100s+1}.$$ (Check: $G_p^{+}G_p^{-}=(1-s)e^{-5s}\cdot\tfrac{-5}{100s+1}=\tfrac{5(s-1)e^{-5s}}{100s+1}=G_p$.) Writing $(1-s)$ rather than $(s-1)$ makes $G_p^{+}(0)=1$, guaranteeing zero offset.
Form the IMC controller. Invert only the good part and add the first-order filter: $$q(s)=\big(G_p^{-}\big)^{-1}f=\dfrac{-(100s+1)}{5}\cdot\dfrac{1}{10s+1}=\boxed{\dfrac{-(100s+1)}{5\,(10s+1)}}$$ The RHP zero and dead time are deliberately not inverted (that would create an unstable, non-causal controller).
Equivalent classical controller (optional). If a standard feedback form is wanted, $G_c=\dfrac{q}{1-G_p q}$; with a perfect model $G_p q=G_p^{+}f=\dfrac{(1-s)e^{-5s}}{10s+1}$, which is stable and proper — confirming the design is realisable.
(b) Perfect-model set-point response. For IMC with no model error the closed loop reduces to $$\dfrac{Y(s)}{Y_{sp}(s)}=G_p q=G_p^{+}f=\dfrac{(1-s)\,e^{-5s}}{10s+1}.$$ The filter sets the closed-loop speed ($\tau_f=10$); the all-pass factor $(1-s)e^{-5s}$ is unavoidable — it imposes the process’s dead time and inverse response on the best achievable response.
Qualitative response to a unit step. With $Y_{sp}=1/s$, ignoring the delay the transient is $w(t)=1-1.1\,e^{-0.1t}$, so $w(0)=\boxed{-0.1}$ (an inverse response: the output first moves the "wrong" way because of the RHP zero) and $w(\infty)=1$ (no offset). Including the dead time, the output stays flat until $t=5$, then dips slightly negative and climbs monotonically to the set point with the $\tau_f=10$ filter time constant — see the plot.

Problem 3(b): perfect-model closed-loop response to a unit set-point step. Nothing happens until the dead time $\theta=5$ elapses; the output then shows a small inverse response (dips to $\approx-0.1$ from the RHP zero) before rising monotonically to $1$ with the filter time constant $\tau_f=10$ — no offset.

Result	Expression
IMC controller	$q(s)=-(100s+1)/[5(10s+1)]$
Closed-loop (perfect model)	$Y/Y_{sp}=(1-s)e^{-5s}/(10s+1)$
Initial (inverse) response	$-0.1$ (after the $\theta=5$ delay)
Steady-state offset	$0$ (since $G_p^{+}(0)=1$)

Concept & Theory

Topic. Internal Model Control for a process with a RHP zero and dead time; the invertible/non-invertible factorisation and the IMC filter.

Key relations. $G_p=G_p^{+}G_p^{-}$ with $G_p^{+}$ all-pass, $|G_p^{+}(j\omega)|$ bounded, $G_p^{+}(0)=1$; controller $q=(G_p^{-})^{-1}f$, filter $f=1/(\tau_f s+1)^n$; perfect-model set-point transfer $Y/Y_{sp}=G_p^{+}f$.

Why this works. Inverting the whole model would demand a controller with RHP poles (from the RHP zero) and prediction (from the dead time) — both impossible. Factoring these into an all-pass block that is left untouched keeps the controller stable and causal, while $f$ makes $q$ proper and tunes robustness/speed. Setting $G_p^{+}(0)=1$ forces unity closed-loop DC gain, hence zero offset.

Common pitfall. Cancelling the RHP zero (unstable pole-zero cancellation) or trying to invert $e^{-5s}$; also forgetting the sign so that $G_p^{+}(0)\ne1$, which would introduce steady-state offset.

Source. Seborg et al., Ch. 12 (IMC design, filter selection, RHP zeros/dead time); Morari & Zafiriou, Robust Process Control, Ch. 3–4.

Question 3 of 8: Internal Model Control (IMC) Design

Problem 3: Internal Model Control (IMC) Design (20%)