IMC Design for a Process with a RHP Zero and Dead Time: 23-Chem-A6 Process Dynamics and Control, May 2018, Q5

Problem 5: IMC Design for a Process with a RHP Zero and Dead Time (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. $G_p(s)=\dfrac{10(0.5-s)e^{-10s}}{100s+1}$ — first-order ($\tau_p=100$) with a right-half-plane zero at $s=+0.5$ and dead time $\theta=10$; desired closed-loop time constant $\tau_c=5$.

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

Problem 5(a): standard IMC structure. The controller $q$ drives both the plant $G_p$ and the internal model $\tilde G_p$; their difference (model mismatch) is fed back and subtracted from the set point $R$. With a perfect model the feedback signal is exactly the disturbance.

Approach. Factor $G_p$ into a non-invertible all-pass part (RHP zero + dead time, scaled to unity at $s=0$) and an invertible minimum-phase part; invert only the latter and cascade a filter $f=1/(\tau_c s+1)$; then form the perfect-model servo transfer function and invert for a step.

(a) Factor the model. Put the numerator in $(1-\text{const}\cdot s)$ form: $10(0.5-s)=5(1-2s)$, so $G_p=\dfrac{5(1-2s)e^{-10s}}{100s+1}$. Split: $$G_{p+}=(1-2s)\,e^{-10s}\ \text{(non-invertible, }G_{p+}(0)=1),\qquad G_{p-}=\frac{5}{100s+1}\ \text{(invertible).}$$
(a) Build the controller. Invert $G_{p-}$ and add the filter $f=\dfrac{1}{\tau_c s+1}=\dfrac{1}{5s+1}$: $$\boxed{q(s)=G_{p-}^{-1}f=\frac{100s+1}{5}\cdot\frac{1}{5s+1}=\frac{100s+1}{5\,(5s+1)}.}$$ The block diagram above is the realisation.
(b) Perfect-model servo response. With $\tilde G_p=G_p$ the closed loop collapses to $C/R=G_p q=G_{p+}f=\dfrac{(1-2s)e^{-10s}}{5s+1}$. For $R=1/s$, expand $\dfrac{1-2s}{s(5s+1)}=\dfrac1s-\dfrac{1.4}{s+0.2}$ (residue at $s=0$ is $1$; at $s=-0.2$ it is $-1.4$), then apply the 10-unit delay: $$\boxed{c(t)=1-1.4\,e^{-0.2(t-10)}\ \ (t\ge10);\qquad c(t)=0\ \ (t<10).}$$
Interpret. The response is dead for 10 time units, then jumps to $c(10^{+})=1-1.4=-0.4$ (inverse-response undershoot from the RHP zero) before rising monotonically to $1$. There is no offset because $G_{p+}(0)=1$ was enforced.

Problem 5(b): perfect-model set-point response. Dead for $t<10$, an inverse-response undershoot to $-0.4$ at $t=10^{+}$, then a smooth first-order rise ($\tau_c=5$) to the offset-free value 1.

Quantity	Value
(a) IMC controller $q$	$\dfrac{100s+1}{5(5s+1)}$
Non-invertible factor	$G_{p+}=(1-2s)e^{-10s}$
(b) Servo response	$c(t)=1-1.4e^{-0.2(t-10)},\ t\ge10$
Undershoot / offset	$-0.4$ at $t=10^{+}$ / none

Concept & Theory

Topic. Internal Model Control; factoring out non-invertible dynamics (dead time, RHP zeros); the filter time constant as the single tuning knob.

Key relations. $q=G_{p-}^{-1}f$, $f=1/(\tau_c s+1)$; perfect-model servo $C/R=G_{p+}f$; unity-gain scaling of $G_{p+}$ guarantees zero offset.

Why this works. A realisable, stable controller cannot invert dead time (its inverse is a non-causal predictor) or a RHP zero (its inverse is an unstable pole). IMC quarantines both in $G_{p+}$, scales it to unity for zero offset, and inverts only the well-behaved remainder. The best achievable closed loop therefore still contains the original delay and inverse response, smoothed by $\tau_c$.

Common pitfall. Trying to cancel the RHP zero (producing an unstable controller), or forgetting to scale $G_{p+}$ to unity so an offset appears.

Source. Seborg et al., Ch. 12 (IMC design); Stephanopoulos, Ch. on model-based control.

Question 5 of 8: IMC Design for a Process with a RHP Zero and Dead Time

Problem 5: IMC Design for a Process with a RHP Zero and Dead Time (20%)