IMC Design for a Process with Dead Time and an RHP Zero: 23-Chem-A6 Process Dynamics and Control, December 2016, Q5

Problem 5: IMC Design for a Process with Dead Time and an RHP Zero (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. $G(s)=\dfrac{10(s-1)e^{-10s}}{100s+1}$ — open-loop stable pole at $s=-0.01$, a right-half-plane zero at $s=+1$, and dead time $\theta=10$. Desired $\tau_c=10$.

Find. (a) the IMC controller $q$ and its block diagram; (b) the servo response $c(t)$ to a unit step.

The IMC loop drives the process $G$ and a parallel model $\tilde G$ with the same signal; the feedback is the difference between plant and model outputs, so with a perfect model the loop is effectively open and the controller $q$ can be designed by direct inversion of the invertible part of $G$.

Approach. Factor $G=G_+G_-$ into an all-pass part $G_+$ (RHP zero + dead time, $G_+(0)=1$) that cannot be inverted and an invertible minimum-phase part $G_-$; set $q=G_-^{-1}f$ with filter $f=1/(\tau_c s+1)$; then the servo transfer function is simply $G_+f$.

Factor the process. Write $10(s-1)=-10(1-s)$ so the all-pass factor has unit DC gain: $$G_+=(1-s)\,e^{-10s}\ \ (G_+(0)=1),\qquad G_-=\frac{-10}{100s+1}.$$
(a) IMC controller. Invert the good part and add a first-order filter with $\tau_c=10$: $$\boxed{q=G_-^{-1}f=-\frac{100s+1}{10\,(10s+1)}.}$$ The realised feedback controller is $G_c=\dfrac{q}{1-\tilde G q}$, but the IMC form above is the design object.
(b) Servo transfer function. With a perfect model the closed loop reduces to $$\frac{C}{R}=G_+f=\frac{(1-s)\,e^{-10s}}{10s+1}.$$
Unit step, partial fractions. For $R=1/s$, ignoring the delay for the moment, $\dfrac{(1-s)}{s(10s+1)}=\dfrac{1}{s}-\dfrac{1.1}{s+0.1}$ (residues: $1$ at $s=0$, $-1.1$ at $s=-0.1$).
Invert & reinstate the delay. $$\boxed{c(t)=\Big[\,1-1.1\,e^{-0.1(t-10)}\,\Big]\,u(t-10).}$$
Interpret. Nothing happens until $t=10$ (dead time); at $t=10^+$ the output jumps to $c=1-1.1=-0.1$, an inverse response (wrong-way kick) caused by the RHP zero, then rises monotonically to the final value $1$ — offset-free tracking with the specified first-order approach.

Quantity	Value
All-pass factor $G_+$	$(1-s)e^{-10s}$
IMC controller $q$	$-(100s+1)/[10(10s+1)]$
Servo response	$c(t)=[1-1.1e^{-0.1(t-10)}]\,u(t-10)$
Initial undershoot	$-0.1$ at $t=10^+$ (inverse response)
Final value	$1$ (no offset)

Concept & Theory

Topic. Internal Model Control; all-pass factorisation; inverse response from a right-half-plane zero.

Key relations. $G=G_+G_-$ with $|G_+(j\omega)|=1$, $G_+(0)=1$; IMC controller $q=G_-^{-1}/(\tau_c s+1)$; servo $C/R=G_+/(\tau_c s+1)$.

Why this works. IMC never tries to invert dead time or an RHP zero (which would be non-causal or unstable); it inverts only the minimum-phase part and lets the all-pass factor pass through, so the best achievable response is the all-pass dynamics filtered by $1/(\tau_c s+1)$. The RHP zero guarantees the output first moves the wrong way.

Common pitfall. Putting the RHP zero into the invertible part (this would make $q$ unstable), or dropping the $(1-s)$ factor from the servo response and thereby missing the inverse-response undershoot.

Source. Seborg, Edgar, Mellichamp & Doyle, Process Dynamics and Control (4th ed., Wiley) Ch. 12 (IMC and direct synthesis); Stephanopoulos, Chemical Process Control (Prentice Hall) (dead-time compensation).

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

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