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

enter image description here

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.


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.

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

enter image description here

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.


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.


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

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


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)\$: -

enter image description here

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: -

enter image description here

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: -

enter image description here

  • 1
    \$\begingroup\$ When the phase margin becomes negative at the unity gain, the poles move to the right-half plane and the transient response diverges. I'll try to post an explanation later on but look at the graph of my APEC 2009 seminar, page 6 and later, you should have a clearer view. \$\endgroup\$ Jul 22 at 16:28
  • \$\begingroup\$ The why comes with the maths and the closed-loop gain TF whose \$Q\$ becomes negative at some point when the phase drops negative at the unity gain. Try to plot the \$Q\$ given in slide 36 and you'll see that it becomes negative at some point, meaning the roots have a positive real part and the time-domain response has positive exponential exponents. \$\endgroup\$ Jul 22 at 16:40

2 Answers 2


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: -

enter image description here

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.

  • \$\begingroup\$ Okay, so the gain is larger than 1 and the phase is \$-180^\circ\$ at the frequency 41.4 rad/s, so what? The system is unstable at this frequency, (thus also unstable for a step input) but it is also unstable for a 1 rad/s sinusoidal input, and a 0.1 rad/s sinusoidal input. The system is in fact unstable to any input waveform. Why? At 1 rad/s the phase shift is miles away from \$-180^\circ\$. These questions cannot be answered simply using intuition, hence why I ask for a mathematical explanation. \$\endgroup\$
    – Carl
    Jul 22 at 14:41
  • \$\begingroup\$ @Carl the system is a closed-loop oscillator (as pointed out) and, any input signal you apply won't affect this fact and, an oscillator is an unstable amplifier. You can't say it won't be unstable at 1 rad/sec because the elephant in the room is the fact that it has turned into an oscillator at 40 odd radians per second. \$\endgroup\$
    – Andy aka
    Jul 22 at 15:00
  • \$\begingroup\$ Wow it is good to see some common sense \$\endgroup\$
    – Autistic
    Jul 23 at 7:58
  • \$\begingroup\$ @Andyaka I investigated your claim of creating an oscillator by having 180° phase shift while the gain > 1 in my "edit" section. As I see it it contradicts this answer. \$\endgroup\$
    – Carl
    Jul 23 at 8:29
  • 1
    \$\begingroup\$ @ Carl Regarding your 1st comment: A closed-loop circuit cannot be unstable for one certain frequency at the input or - as you wrote - to "any input waveform". When a closed loop is unstable, it will be unstable with or without any input. It does not need any input signal to oscillate or to go into saturation. \$\endgroup\$
    – LvW
    Jul 23 at 13:59

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.

  • \$\begingroup\$ The contents of this answer cannot be true. As I have shown in the edited section of my question I have provided a system whose (open) loop gain is larger than unity while the phase shift is \$-180^\circ\$ and yet the closed loop system is stable (all poles has negative real part). This is in contradiction with your upper paragraph. \$\endgroup\$
    – Carl
    Jul 23 at 12:00
  • \$\begingroup\$ @Carl When you fully understand the contents /limitations of Nyquists stability criterion you will see that for the function you have analyzed under EDIT the simplified version of this criterion does NOT apply! The reason: Three poles of the loop gain function in the origin. The simpified criterion - which the basis for the phase margin determination via Bode diagram - allows only two poles for the loop gain in the origin. This is an artificial (non-realistic) example for which the restrictions as mentioned in my Comment 1 apply. As a consequence - the criterion you have applied is not valid. \$\endgroup\$
    – LvW
    Jul 23 at 12:57
  • \$\begingroup\$ No one as been able to provide a bullet-proof description of stability criterion on this Q&A post so far. Okay, the system I initially investigated had 3 poles in the origin. I have modified it such that it now has 3 poles in -0.01, hence all poles are in the LHP. The result is still the same -> the closed loop system is stable. I could have placed the poles in -0.01, -0.02, -0.03 and arrived at a similar result. \$\endgroup\$
    – Carl
    Jul 23 at 13:09
  • \$\begingroup\$ @Carl Do you really expect from us here a "bullet-proof description of the stability criterion"? I only can repeat: Try to become familiar with this criterion (consult a corresponding book) and learn about the restrictions for the applicability of the simplified version of the criterion. Still - my remarks under "Comment 1" apply. When you extend your simulation down to w=1E-4 or so you will notice that the critical phase line will be crossed TWICE from the loop gains phase function. This inhibits the use of the simplified criterion (Bode diagram). \$\endgroup\$
    – LvW
    Jul 23 at 13:34
  • 1
    \$\begingroup\$ The problem is - as I have mentioned several times - that your arguments are solely based on Bode diagram simulations. You must use the Nyquist diagram in the s-domain. That`s all I finally can say . Example: "Signals and Systems" (Oppenheim, Willsky)., 1983: Chapter 11.4/The Nyquist stability criterion, pges 713-721. \$\endgroup\$
    – LvW
    Jul 23 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.