Balance scales and integral action. Part II

In a previous post we discussed how the principles of operation of a balance scale could be understood as a feedback loop with integral action. We left off after analyzing the impact of measurement disturbances in the loop and mentioned that input disturbances could be used to model “operator errors.”

Fig. 1.13: Feedback loop with input disturbance

The motivation for considering “operator errors” is that it is very likely that a human operating the scale will not be able to produce weights, that is the signal $u$, that are exactly proportional to the error signal $e$. In that case an input disturbance $w$ entering the loop as in Fig. 1.13 can be used to model the “operator error” in producing the “control signal” $u$.

With the addition of the input disturbance, and following the same reasoning as in Part I of this post, it is possible to describe the loop in Fig. 1.13 by the recursive equations

$$y(k+1)=y(k) + u(k) + w(k), \quad u(k) = K e(k), \quad e(k) = \bar{y} – y(k).$$

As before, it is possible to obtain a recursion to describe the error signal $e$ as in

$$
\begin{aligned}
e(k+1) &=\bar{y} – y(k+1) \\
&= \bar{y} – (y(k) +u(k) + w(k)) \\
&= \bar{y} – (\bar{y}-e(k)) -K e(k)- w(k) \\
&= (1 – K) e(k) – w(k)
\end{aligned}
$$

Note how assuming that the reference signal $\bar{y}(k)=\bar{y}$ is constant has the effect of cancelling $\bar{y}$ from the recursion. This is integral action once again at work!

A reasonable assumption on the disturbance $w(k)$ might be that it is small when compared with $u(k)$. For example, that $w(k)=\delta(k) \, u(k)$ and that $\delta(k)$ is bounded, say $|\delta(k)| \leq \gamma$. Note that we are not saying how small $\gamma$ should be! That’s a question that we will seek to answer based on what we know about the system. Indeed, because

$$
\begin{aligned}
|e(k+1)| &= |(1 – K) e(k) – w(k)| \\
&= |(1 – K) e(k) + \delta(k) u(k)| \\
&= |1 – K – \delta(k) K| |e(k)|,
\end{aligned}
$$

the error $e(k)$ converges to zero even in the presence of operator errors as long as

$$|1-K+\delta(k) K|\leq|1-K|+|K||\delta(k)|\leq |1-K|+|K|\gamma<1.$$

In other words, if

$$\gamma<\frac{1-|1-K|}{|K|}.$$

One conclusion is that, since $K > 0$ and $|1-K|<1$, it will always be possible to tolerate some operator error. Better yet, if the operator is cautious, that is if $0<K<1$ then

$$\gamma<\frac{1-|1-K|}{|K|} = \frac{1 – 1 + K}{K} = 1$$

no matter the value of $K$. That is, the balance scale can tolerate up to 100% of operator error, and it will still converge!

The type of analysis performed above is covered in much of Chapter 8. Indeed, the above condition in terms of the $\gamma$ can be obtained via the application of a discrete-time counterpart for the small gain argument provided in Section 8.3, in which one can think of the above model for the operator error as been a particular case of the block-diagram in Fig. 8.3 in which the block $\Delta$ is the $\delta(k)$ above.

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