IMC Design for a First-Order-plus-Dead-Time Process: 23-Chem-A6 Process Dynamics and Control, December 2013, Q5

Problem 5: IMC Design for a First-Order-plus-Dead-Time 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. A first-order-plus-dead-time (FOPDT) model $G=\dfrac{5e^{-2s}}{10s+1}$ ($K_p=5$, $\tau_p=10$, $\theta=2$); IMC filter $f=\dfrac{1}{\tau_c s+1}$ with $\tau_c=20\ \mathrm{s}$.

Find. (a) $G_c^{*}$, the classical equivalent $G_c$, and whether $G_c$ is PID; (b) the perfect-model set-point response.

Problem 5(a): IMC structure. The controller $G_c^{*}$ drives the real process $G$; an internal model $\tilde G$ predicts the output $\tilde C$, and the model error $\hat d=C-\tilde C$ is subtracted from the set point. With a perfect model ($\tilde G=G$), $\hat d=0$ and the set-point response collapses to $C/C_{sp}=G\,G_c^{*}$.

Approach. Factor the model into an invertible minimum-phase part and a non-invertible all-pass part (the dead time); the IMC controller inverts only the good part and appends the filter. The classical equivalent follows from $G_c=G_c^{*}/(1-\tilde G G_c^{*})$, and the perfect-model set-point response is simply the all-pass factor times the filter.

(a) Factor and form $G_c^{*}$. Split $G=\tilde G_+\tilde G_-$ with $\tilde G_+=e^{-2s}$ (all-pass, unity gain, non-invertible) and $\tilde G_-=\dfrac{5}{10s+1}$. Invert the good part and add the filter: $$\boxed{G_c^{*}=\tilde G_-^{-1}f=\dfrac{10s+1}{5}\cdot\dfrac{1}{20s+1}=\dfrac{10s+1}{5(20s+1)}.}$$
Classical equivalent $G_c$. Since $\tilde G G_c^{*}=\dfrac{5e^{-2s}}{10s+1}\cdot\dfrac{10s+1}{5(20s+1)}=\dfrac{e^{-2s}}{20s+1}$, $$G_c=\dfrac{G_c^{*}}{1-\tilde G G_c^{*}}=\dfrac{10s+1}{5\,(20s+1-e^{-2s})}.$$
Is $G_c$ PID? No. The denominator contains the transcendental term $e^{-2s}$, so $G_c$ is not a rational PID controller — it is a dead-time compensator (Smith-predictor-like). Only if one approximated $e^{-2s}\approx1-2s$ would $20s+1-e^{-2s}\approx22s$, giving the PI form $G_c\approx\dfrac{10s+1}{110s}=0.0909\left(1+\dfrac{1}{10s}\right)$; the question forbids that approximation, so the exact $G_c$ is not PID.
(b) Perfect-model set-point response. With $\tilde G=G$ the loop reduces to $\dfrac{C(s)}{C_{sp}(s)}=\tilde G_+f=\dfrac{e^{-2s}}{20s+1}$. For $C_{sp}=1/s$, $C(s)=\dfrac{e^{-2s}}{s(20s+1)}$, whose inverse is $$\boxed{\delta C(t)=\left[1-e^{-(t-2)/20}\right]\mathcal{U}(t-2).}$$
Interpret. The controlled variable stays at zero through the $2\,$s dead time, then rises as a pure first-order approach with time constant $\tau_c=20\,$s to a final value of 1 — no offset and no oscillation. The filter $\tau_c$ is the single knob that sets closed-loop speed; the dead time is simply inherited from the process and cannot be removed.

Problem 5(b): perfect-model IMC set-point response $\delta C(t)=[1-e^{-(t-2)/20}]\mathcal{U}(t-2)$. Flat during the $2\,$s dead time, then a clean first-order rise (63.2% reached one filter time constant later, at $t=22\,$s) to unity with no offset.

Result	Expression
IMC controller	$G_c^{*}=(10s+1)/[5(20s+1)]$
Classical equivalent	$G_c=(10s+1)/[5(20s+1-e^{-2s})]$
PID form?	No (contains $e^{-2s}$); PI only under a Padé/Taylor approximation
Set-point response	$\delta C(t)=[1-e^{-(t-2)/20}]\,\mathcal{U}(t-2)$, no offset

Concept & Theory

Topic. Internal Model Control for an FOPDT process; invertible/all-pass factorisation; IMC–PID connection.

Key relations. $G=\tilde G_+\tilde G_-$ ($\tilde G_+$ all-pass, $\tilde G_+(0)=1$); $G_c^{*}=\tilde G_-^{-1}f$, $f=1/(\tau_c s+1)$; $G_c=G_c^{*}/(1-\tilde G G_c^{*})$; perfect-model set-point transfer $C/C_{sp}=\tilde G_+f$.

Why this works. Inverting the dead time would demand prediction (a non-causal $e^{+2s}$), so it is left in the all-pass factor and simply appears, unavoidable, in the closed loop. Because $\tilde G_+(0)=1$, the closed-loop DC gain is exactly 1, guaranteeing zero offset without integral tuning.

Common pitfall. Trying to cancel or Padé-approximate $e^{-2s}$ inside $G_c^{*}$, or calling the exact $G_c$ a PID controller — it becomes PID only after the delay is approximated.

Source. Seborg et al., Ch. 12 (IMC design, filter selection, IMC–PID equivalence); Morari & Zafiriou, Robust Process Control, Ch. 3–4.

Question 5 of 8: IMC Design for a First-Order-plus-Dead-Time Process

Problem 5: IMC Design for a First-Order-plus-Dead-Time Process (20%)