In Part I and Part II of this post we looked into the problem of finding a suitable discrete-time controller when the system being controlled is fast to reach steady-state so that it can be modeled as the single delay

$$y(t) = G u(t – \tau), \quad \tau > 0.$$

In Part I we showed that a proportional controller is not a good choice and in Part II we found out that, surprisingly, an integral only controller is an excellent choice but requires the use of a gain $\hat{G}$ that has to match the gain of the system $G$. In the present post we will revisit the design of a dynamic controller with the goal of showing that:

- the Smith predictor from Part II also had a zero;
- the integral controller is indeed an excellent choice;
- the mismatch $\hat{G} \neq G$ is not necessarily catastrophic.

Start with a generic dynamic controller of the form

$$u(t)=-a u(t – \tau) + b (\bar{y}(t) – y(t)).$$

The Smith predictor is a special case of the above controller in which

$$a = -1, \quad b = \hat{G}^{-1}.$$

Now calculate the following quantity

$$y(t) + a y(t – \tau) = G u(t – \tau) + a G u(t – 2 \tau) = G b \, e(t-\tau)$$

and observe that the relationship between $y$ and $e$ is described by the first-order (one delay) difference equation

$$y(t) + a y(t – \tau) = G b \, e(t-\tau).$$

This equation represents the open-loop (series) connection of the controller with the system. Because the system had one delay and the controller had another one, one would expect that the relatioship between $y$ and $e$ would have two delays. The fact that it has only one delay is the result of a pole-zero cancellation, which means that the above controller, and therefore the Smith predictor must have a zero. We will look into that in more detail in a future post.

In order to close the loop we need to substitute for the error to obtain

$$y(t) + a y(t – \tau) = G b \, e(t-\tau) = G b \, (\bar{y}(t-\tau) – y(t-\tau))$$

from which

$$y(t) + (a+Gb) \, y(t – \tau) = G b \, \bar{y}(t -\tau)$$

Proceeding as in Part I one can conclude that this system is asymptotically stable if

$$|a + G b| < 1.$$

In addition if

$$a=-1,$$

that is if the controller is an integrator, then it also displays asymptotic tracking of constant references because

$$G b \, \tilde{y} = \tilde{y} + (-1+Gb) \, \tilde{y} = G b \, \bar{y} \quad \implies \quad \tilde{y} = \bar{y}.$$

This property is independent of the choice of $b$ or the value of $G$ as long as

$$|G b -1| < 1$$

The conclusion is that an integrator, that is $a = -1$, is an excellent choice!

Finally, note that the choice $b=\hat{G}^{-1}$ when $\hat{G}=G$ leads to

$$G b – 1=G \hat{G}^{-1} -1=0$$

so that convergence to the constant reference is as fast as possible. If $\hat{G}\neq G$ nothing catastrophic ensues as long as $|G \hat{G}^{-1} – 1|<1$. In particular, if

$$G \geq \hat{G} > (1/2) G$$

then $0 \leq G \hat{G}^{-1} – 1 < 1$ so that the closed-loop is stable and without oscillations. If

$$\hat{G} > G$$

the closed loop is stable but it now oscillates because $-1 < G \hat{G}^{-1} – 1 < 0$. Values of $\hat{G} \leq (1/2) G$ lead to instability. The conclusion is that as long as $\hat{G} > (1/2) G$ then the closed-loop is still asymptotically stable and that if $\hat{G} \leq G$ then there will be no oscillations. There also seems to be no danger in overestimating $G$, as long as one can tolerate the oscillations. In fact, the onset of oscillations could even be used to estimate a suitable value of $\hat{G}$.