Negative phase margin and misconception of the point of instability in negative feedback control loops

Question

Prelude

Consider a canonical block diagram of a negative feedback loop with gain \$K\$, plant transfer function \$G(s)\$ and feedback transfer function \$H(s)\$: -

Regardless of what functions K, G, and H are, the closed loop transfer function for this system is

$$T_\text{cl}(s) = \frac{KG(s)}{1+KG(s)H(s)} \tag{1}$$

The term \$KG(s)H(s)\$ is called the open loop gain (in some literature just "loop gain") and this is of particular interest. Only signals that can be physically made are of interest so we substitute \$s = j\omega\$ and arrive at

$$T_\text{ol}(j\omega) = KG(j\omega)H(j\omega) = |T_\text{ol}(j\omega)|e^{j\theta_\text{ol}(j\omega)} $$

\$T_\text{ol}(j\omega)\$ is a complex function that varies with frequency.

The frequency at which \$T_\text{ol}(j\omega) = -1\$, i.e. \$|T_\text{ol}(j\omega)| = 1 \$ and \$\theta_\text{ol}(j\omega) = \pm180^\circ \$ is special, because it causes the denominator in \$(1)\$ to become \$0\$ and \$T_\text{cl}(j\omega)\$ becomes undefined.

At this point, all textbooks I've read about this subject define the frequency at which \$|T_\text{ol}(j\omega)| = 1 = 0 \: \text{dB}\$ as the crossover frequency \$\omega_c\$. Incidentally, another parameter used to characterize the stability of a closed loop system is defined:

$$\phi_m = \theta_\text{ol}(j\omega_c) + 180^\circ $$

The importance of \$\phi_m\$ is that it reveals how close to instability the closed loop system will be. So a system with \$\phi_m = 5^\circ\$ will be close to instability, while a system with \$\phi_m = 60^\circ\$ will robust.

This all makes perfect sense. So far so good.

Problem

The question is: what happens when \$\phi_m\$ becomes negative?

The textbooks claim that the closed loop system will be unstable, but with no mathematical argument. A simulation supports this claim but with no satisfactory explanation. Let's take an example. I am only interested in the unity feedback case, i.e. \$H(s) = 1\$:

Consider a system with \$H(s) = 1\$ and \$KG(s) = \frac{10^5}{s^3+60s^2+1100s+6000} \$, whose open loop system has negative phase margin.

$$T_\text{ol}(s) = \frac{10^5}{s^3+60s^2+1100s+6000} \Rightarrow$$ $$T_\text{cl}(s) = \frac{10^5}{s^3+60s^2+1100s+106000} $$

But why is the closed loop system unstable? The answer is obviously that \$T_\text{cl}(s)\$ has poles in the right half plane (in this case \$3.56 \pm j39.5\$) but why does a negative phase margin imply that?

Another thing I have seen mentioned in some of TI's documents is that \$\phi_m\$ should be measured with respect to the phase at DC. So if \$\theta_\text{ol}(0) = 180^\circ\$ and \$\theta_\text{ol}(j\omega_c) = 30^\circ \$ then \$\phi_m = 180^\circ - 150^\circ = 30^\circ\$. Where they get this from I don't know.

Question

I am looking for a mathematical explanation of why a negative phase margin in a negative feedback loop implies an unstable closed loop system.

I guess a mathematical description of the question would be:

Let \$T_\text{ol}(s)\$ be a continuous and differentiable function \$T_\text{ol} : \mathbb{C} \to \mathbb{C}\$. Define \$T_\text{cl}(s) = \frac{T_\text{ol}(s)}{1+T_\text{ol}(s)} = \frac{N(s)}{D(s)}, \: s \in \mathbb{C} \$.

Show that if there exists an \$\omega_c \in \mathbb{R}\$ such that \$T_\text{ol}(j\omega_c) =|T_\text{ol}(j\omega_c)|e^{j\theta_\text{ol}(j\omega_c)}\$ where \$|T_\text{ol}(j\omega_c)| \geq 1\$ and \$ \theta_\text{ol}(j\omega_c) = \pm \pi \$ that the polynomial \$D(s)\$ will have at least one root with positive real part.

Sources

Texas instruments claim: https://i.stack.imgur.com/nZhh8.png

Block diagram: Signal processing and linear systems - B.P.Lathi

Edit

I investigated the transfer function \$T_\text{ol}(s) = 10^4 \frac{s^3+3s^2+3s+1}{s^3+60s^2+1100s+6000} \Rightarrow T_\text{cl}(s) = 10^4\frac{s^3+3s^2+3s+1}{10001s^3+30060s^2+31100s+16000}\$.

Have a look at the Bode Plot for \$T_\text{ol}(s)\$ and the impulse response for \$T_\text{cl}(s)\$: -

What do we see? There are in this case two frequencies where the gain is larger than \$1\$ and the phase shift is \$180^\circ\$ so the closed loop system should be unstable, right? But no. Looking at the impulse response for the closed loop system we see that the system is stable.

Investigating \$T_\text{ol}(s) = 10\frac{(s+1)^2}{s^3} \Rightarrow T_\text{cl}(s) = 10\frac{(s+1)^2}{s^3+10s^2+20s+10} \$ shows this equally interesting result: -

Again, the gain is larger than \$1\$ while the phase shift now is \$-180^\circ\$. The closed loop system is still stable.

Edit 2

For completeness sake, consider the system \$T_\text{ol}(s) = 10\frac{(s+1)^2}{(s+0.01)^3} \Rightarrow T_\text{cl}(s) = 10\frac{(s+1)^2}{s^3+10.03s^2+20s+10}\$ to avoid having poles on the imaginary axis. We still arrive at a system that is closed loop stable: -

Answer 1

I am looking for a mathematical explanation of why a negative phase margin in a negative feedback loop implies an unstable closed loop system.

If you look at your specific bode plot, there is a frequency where the phase shift is 180 degrees and, this corresponds with a gain greater than unity hence, the circuit will oscillate: -

Basically it meets the Barkhausen criteria for an oscillator.

Sorry, but I don't think this needs a mathematical explanation.

The question is: what happens when \$\phi_M\$ becomes negative?

If it becomes negative then it must have been non-negative hence, at exactly \$\phi_M = 0\$ then, if the gain is greater than unity, it will oscillate.

Answer 2

I think the answer to your question is rather simple:

When the loop gains phase shift at the "cross-over frequency w_c" (unity loop gain magnitude) is more negative than -180deg, there must be a lower frequency w_o where the phase shift is -180deg with a loop gain which is LARGER than unity. Hence, the stability criterion is violated and the closed loop will be unstable.

Comment 1: There are some special circuits with loop gain zeros and/or poles in the right half of the s-plane for which the above explanation is not applicable. In these cases, the complete Nyqist criterion has to be applied for a stability check.

Comment 2: For the above explanations the loop gain was defined as LG(s)=KG(s)*H(s).

However, in most cases, the minus sign at the summing node is considered to be also a part of the loop gain LG(s)=-KG(s)*H(s). As a consequence, the "critical phase shift" for finding the phase margin is 360deg (0deg) instead of -180deg.

For my opinion, this is the "best" definition because it is in accordance with Barkhausens oscillation condition. More than that, all circuit simulations of the loop gain are using automatically this definition.