# Black's formula and feedback system stability

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?

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

• Would you mind expanding on this? (What's the difference? What would make it dynamic? etc.) – Mehrdad Nov 20 '14 at 19:11
• Dynamic system means that the value of output depends on the previous values of it. This means some kind of derivative or difference should be involved, e.g. there should be $s$ term in your transfer function. When $G(s)=2$ this is just a gain. This means $y(t)=2x(t)$. – obareey Nov 22 '14 at 7:37
• Right! The system y(t)=2x(t) is BIBO-stable: bounded input implies bounded output, trivially. – Pait Nov 22 '14 at 14:58

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.

• When you say you find the existence of such a system questionable, are you really saying that we cannot reason intuitively about any system unless it contains at least one pole? (That's the only way to introduce delay...) – Mehrdad Nov 20 '14 at 19:34
• I mean the existence of such a physical system, if application is of concern. Real systems have for example inertia. – PYK Nov 20 '14 at 19:38
• Even a system has poles, we still in general cannot reason intuitively about it, as the demonstrated by the first example. If we have a system with a pure time delay element, as described in the second example, then our intuition works. – PYK Nov 20 '14 at 20:04
• What about (linear) systems that don't have time involved at all, though? Like image processing, for example. It should be possible to still reason about them intuitively even if there is no delay, right? – Mehrdad Nov 20 '14 at 20:37
• Sorry, but so far my studies mainly involves dynamical systems defined over a single independent variable that is an element of a time set, which is a subgroup of either $(\mathbb{R}, +)$ or $(\mathbb{N}, +)$, in which certain properties such as causality hold. Multidimensional systems defined over spatial coordinates like image processing are beyond my area of expertise. However, I have a hunch that if some facts do not hold for a subclass, they usually will not hold for the superclass. – PYK Nov 20 '14 at 23:54