Question 2 of 8: Internal Model Control of a Process with a RHP Zero and Dead Time

Problem 2: Internal Model Control of 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. A first-order-plus-dead-time process carrying a right-half-plane (inverse-response) zero:

Find. (a) the IMC controller $q(s)$ and its block diagram; (b) the unit-step set-point response for $\tau_c=10$.

Approach. Split the model into an invertible (minimum-phase) factor and a non-invertible all-pass factor (RHP zero + dead time, normalised to unity DC gain); invert only the good factor and multiply by an IMC filter of the target time constant. The servo response is then the all-pass factor times the filter.

1. (a) Factor the model. Write $\tilde G_p=\tilde G_{p+}\tilde G_{p-}$, where $\tilde G_{p+}$ holds the parts that cannot be inverted (the RHP zero and the delay), normalised so $\tilde G_{p+}(0)=1$: $$\tilde G_{p+}=(1-s)\,e^{-10s},\qquad \tilde G_{p-}=\frac{10}{100s+1}.$$ Indeed $\tilde G_{p+}(0)=1\cdot1=1$, so the steady-state gain lives entirely in $\tilde G_{p-}$.
2. (a) IMC controller. Invert the minimum-phase factor and append the first-order filter $f=1/(\tau_c s+1)=1/(10s+1)$: $$\boxed{q(s)=\tilde G_{p-}^{-1}\,f=\frac{100s+1}{10\,(10s+1)}.}$$ This is proper (equal numerator and denominator order), so it is physically realisable. Inverting the RHP zero was deliberately avoided — doing so would place an unstable pole at $s=+1$ in the controller.
3. (a) Block diagram. In the IMC structure (figure above) the controller $q$ drives both the real process $G_p$ and the internal model $\tilde G_p$ in parallel; only the mismatch $(G_p-\tilde G_p)q$ is fed back. With a perfect model that signal is zero, so the loop behaves open-loop-like and is inherently offset-free.
4. (b) Closed-loop servo transfer function. For IMC with a perfect model, $Y/R=\tilde G_{p+}f$: $$\boxed{\frac{Y(s)}{R(s)}=\frac{(1-s)\,e^{-10s}}{10s+1}.}$$ The delay and the RHP zero survive (they cannot be cancelled); the filter sets the closed-loop speed at $\tau_c=10$.
5. (b) Unit step response. With $R(s)=1/s$, ignore the pure delay (a $10$-unit time shift) and invert $W(s)=\dfrac{1-s}{s(10s+1)}$. Partial fractions give $W(s)=\dfrac{1}{s}-\dfrac{1.1}{s+0.1}$, so $$\boxed{y(t)=1-1.1\,e^{-0.1\,(t-10)}\quad(t\ge10),\qquad y(t)=0\ (t<10).}$$
6. (b) Interpret. Just after the dead time the response jumps the wrong way to $y(10^+)=1-1.1=-0.1$ — the inverse response forced by the RHP zero — then climbs monotonically to $y(\infty)=1$, i.e. it tracks the set point with no offset and an effective time constant of $10$.