IMC Design for a First-Order-Plus-Dead-Time Process: 23-Chem-A6 Process Dynamics and Control, May 2015, 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. First-order-plus-dead-time (FOPDT) process $G_p=\dfrac{5e^{-2s}}{10s+1}$: gain $K_p=5$, lag $\tau_p=10\ \mathrm s$, dead time $\theta=2\ \mathrm s$; IMC filter $f=\dfrac{1}{\lambda s+1}$, $\lambda=20\ \mathrm s$.

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

Problem 5(a): IMC structure. Controller $G_c^{*}$ drives the real process $G_p$; the internal model $\tilde G_p$ predicts $\tilde C$, and the model error $\hat d=C-\tilde C$ is fed back. With a perfect model $\tilde G_p=G_p$, $\hat d=0$ and the servo response reduces to $C/R=G_pG_c^{*}$.

Approach. Factor the model into invertible and all-pass (dead-time) parts, invert only the invertible part and append the filter to build $G_c^{*}$, convert to the classical form $G_c=G_c^{*}/(1-\tilde G_pG_c^{*})$, then evaluate the perfect-model servo transfer function and invert the step.

Factor the model. $G_p=G_p^{-}\,G_p^{+}$ with the invertible part $G_p^{-}=\dfrac{5}{10s+1}$ and the all-pass (non-invertible) dead time $G_p^{+}=e^{-2s}$ (unity gain, $G_p^{+}(0)=1$).
(a) IMC controller. Invert the good part and add the first-order filter: $$\boxed{G_c^{*}(s)=\big(G_p^{-}\big)^{-1}f=\frac{10s+1}{5}\cdot\frac{1}{20s+1}=\frac{10s+1}{5(20s+1)}}.$$ The dead time is deliberately not inverted (that would demand prediction).
(a) Equivalent classical controller. $G_c=\dfrac{G_c^{*}}{1-\tilde G_pG_c^{*}}$; with $\tilde G_p=G_p$ and $G_pG_c^{*}=\dfrac{e^{-2s}}{20s+1}$, $$\boxed{G_c(s)=\frac{10s+1}{5\left(20s+1-e^{-2s}\right)}}.$$ Because $e^{-2s}$ appears in the denominator, $G_c$ is not a PID controller (it is a rational-plus-dead-time compensator, exact only without a Padé approximation).
(b) Perfect-model servo transfer function. With no model error the closed loop collapses to $$\frac{C(s)}{R(s)}=G_pG_c^{*}=G_p^{+}f=\frac{e^{-2s}}{20s+1}.$$ The filter $\lambda=20\ \mathrm s$ sets the closed-loop speed; the dead time $e^{-2s}$ is unavoidable.
(b) Step response. For $R=1/s$, $C(s)=\dfrac{e^{-2s}}{s(20s+1)}$. Inverting (first-order step, delayed by $2\ \mathrm s$): $$\boxed{\delta C(t)=\Big[\,1-e^{-(t-2)/20}\,\Big]\,\mathcal U(t-2)}.$$ The output stays at $0$ until $t=2\ \mathrm s$, then rises exponentially with time constant $20\ \mathrm s$ to the set point — no offset (integral-like action of IMC with $G_p^{+}(0)=1$) and no overshoot.

Result	Expression
IMC controller $G_c^{*}$	$(10s+1)/[5(20s+1)]$
Classical $G_c$	$(10s+1)/[5(20s+1-e^{-2s})]$ — not PID
Servo transfer (perfect model)	$e^{-2s}/(20s+1)$
Step response	$\delta C(t)=[1-e^{-(t-2)/20}]\,\mathcal U(t-2)$

Concept & Theory

Topic. Internal Model Control for FOPDT; invertible/all-pass factorisation; the IMC–classical equivalence; offset-free servo design.

Key relations. $G_p=G_p^{-}G_p^{+}$ ($G_p^{+}$ all-pass, $G_p^{+}(0)=1$); $G_c^{*}=(G_p^{-})^{-1}f$, $f=1/(\lambda s+1)$; classical $G_c=G_c^{*}/(1-\tilde G_pG_c^{*})$; perfect-model servo $C/R=G_p^{+}f$.

Why this works. Inverting the dead time would require a non-causal predictor, so it is left in the all-pass factor; the filter makes $G_c^{*}$ proper and tunes robustness through $\lambda$. Setting $G_p^{+}(0)=1$ forces unity DC gain, hence zero steady-state offset. The residual $e^{-2s}$ inside $G_c$ is exactly why an IMC-derived controller for a dead-time process is not reducible to PID without approximating the delay.

Common pitfall. Trying to invert $e^{-2s}$, or approximating it with Padé when the question forbids it; also forgetting that $\lambda$ (not $\tau_p$) governs the closed-loop time constant.

Source. Seborg et al., Ch. 12 (IMC design, filter selection, FOPDT, 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%)