Black's formula and feedback system stability

Question

Consider a hypothetical system with open-loop transfer function $G(s)$.
Place it in positive feedback with unit gain. (That is, take its output and directly add it to its input.)
The closed-loop system now satisfies $Y(s) = G(s) (X(s) + Y(s))$, i.e.: $$\frac{Y(s)}{X(s)} = \frac{G(s)}{1 - G(s)}$$

This is, of course, simply Black's formula.

My questions:

If we have $G(s) = 2$, then this system stable? Why?

Intuitively, this system is (obviously!) unstable, because it is in positive unit feedback with gain > 1.
Yet according to Black's formula, it is actually a stable system with gain $2 / (1-2) = -2$!
(Notice that it has no poles in the right half-plane -- in fact, it has no poles or zeros at all!)

What is there a discrepancy between the math and my intuition? Which one is correct and why?
What assumptions am I breaking exactly, if any?

Answer 1

This an an algebraic loop, not a dynamic one. The question of stability doesn't really mean much.

Answer 2

The maths is correct.

If by "intuition" you actually mean "causal reasoning", then I am afraid that it does not work in a straightforward manner on the analysis of feedback systems. As Åström and Murray so aptly put it:

Simple causal reasoning about a feedback system is difficult because the first system influences the second and the second system influences the first, leading to a circular argument. This makes reasoning based on cause and effect tricky and it is necessary to analyse the system as a whole. A consequence of this is that the behaviour of feedback systems is often counter-intuitive and it is therefore necessary to resort to formal methods to understand them.

Having said that, it is still possible to justify your results. For comparisons, we shall consider three cases: (a) unity feedback, (b) feedback with transient response, and (c) feedback with a time delay.

The given plant $G(s)$ is a proportional transfer element; that is, the input-output relationship is time-independent. In other word, the closed loop consisting of only P-elements is a static system, so its transfer function is always equal to some constant, no matter how the plants are interconnected. However, the existence of a system, the output of which follows the input without lag and instantly assumes the value determined by the static gain, is questionable.

So, suppose we have a more realistic system, in which the output sensor behaves like a first-order lag with a time constant $T_{o}$ and steady gain of $1$; that is,

$$F(s) = \frac{1}{T_{o}s + 1}, \qquad T_{o} \in \mathbb{R}_{> 0}.$$

Let the plant be a P-element; that is,

$$G(s) = K_{p}.$$

Then the transfer function of the closed loop with positive feedback is

$$\frac{Y(s)}{X(s)} = \frac{K_{p}(T_{o}s + 1)}{T_{o}s + 1 - K_{p}}.$$

Now we have a dynamical system. It is true that for $G(s) = K_{p} = 2$ the system is unstable, but this is not due to the positive feedback, because there exists a range of values that $G(s)$ could assume such that the closed loop is BIBO stable; that is, by the Hurwitz criterion, the system is necessarily stable for all $K_{p} < 1$, which is counter-intuitive but true.

Finally, consider the same system, but now with an output sensor that is $\tau$ seconds late in reporting the output. We can model this using the transport delay element having the transfer function

$$F(s) = e^{-\tau s}.$$

The transfer function of the corresponding closed-loop with positive feedback is expressible as

$$\frac{Y(s)}{X(s)} = \frac{K_{p}}{1 - K_{p}e^{-\tau s}}.$$

Since the exponential function is transcendental, applying classical control theory to analyse the system is non-trivial. However, in this case, the system response is fully in accord with our intuition. If $K_{p} = 2$ and the input is a unity step function, then we expect the output values at time $0, \tau, 2\tau, 3\tau, \dotsc$ to form the sequence $0, 2, 6, 14, \dotsc$, which is governed by the recurrence relation

$$x_{n + 1} = 2x_{n} + 2, \qquad x_{0} = 0, \quad n = 0, 1, 2, \dotsc,$$

which clearly implies that the system is BIBO unstable, but our gut feeling is only valid for this special case.