Basic question on stability of feedback systems

Thread Starter

Pif Ouille

Joined Jun 27, 2019
7
Hi everybody,
I have a very basic question about stability that has been bothering me lately.

Considering a negative feedback system, we know that \[ T_{CL}(s)=\frac{A(s)}{1+A(s)B(s)} \]
Clearly, if \(\ A(s)B(s)=-1\), then \(T_{CL}(s)\) goes to infinity and the system is unstable. However, with this reasonning, we conclude there is no other value of the loop gain that makes the system instable.

However, Nyquist criterion, or Phase/Gain margins, tell us that if \[|A(s)B(s)|\geq 1\] when \[\angle A(s)B(s)=0\], the system is also instable. Therefore, the system is instable for value like \[A(s)B(s)=-2, -3, \dots\].

Can someone explain this apparent paradox?

Thank you so much!
 

Thread Starter

Pif Ouille

Joined Jun 27, 2019
7
Thank you for your answer.

AB = -1 is just an oscillation condition, a subset of unstable systems.
Yes. I know that.
My question is how come the closed-loop transfer function seems to indicate that AB=-1 is the only condition for instability?
 

Papabravo

Joined Feb 24, 2006
17,240
Thank you for your answer.


Yes. I know that.
My question is how come the closed-loop transfer function seems to indicate that AB=-1 is the only condition for instability?
You are considering only the magnitude. A system can also be unstable if the phase is not well behaved. This is like the difference between scalar algebra and vector algebra. You have to consider both things simultaneously. You will hear the terms "gain margin" and "phase margin". They encapsulate how far away the system is from instability for some range of frequencies.
 

anhnha

Joined Apr 19, 2012
885
Thank you for your answer.


Yes. I know that.
My question is how come the closed-loop transfer function seems to indicate that AB=-1 is the only condition for instability?
That is what I mentioned. AB is the loop gain and AB = -1 is only one condition for oscillation. If you want to check stability you have to check if all the poles on the left half plane.
 

tindel

Joined Sep 16, 2012
810
First, your understanding of the Nyquist stability criterion is flawed. The criterion is really easy (at least on the surface - it takes time to appreciate).
Z=N+P
where
Z is the number of unstable closed loop poles
N is the number of clockwise encirclements around (-1,0) in the w-domain
P is the number of open loop poles (P) in the RHP.

Secondly, it's worth keeping in mind that s=jw. It’s an imaginary term. Plug that into A(s)B(s) and solve and you’ll eventually end up with a real and imaginary part, a vector at any given frequency. Also note that the vector changes in both magnitude and phase, as the frequency changes.

With those two things in mind... If at some frequency that open loop vector is -2 at 0 degrees, that’s the same as saying 2 at 180 degrees. If you’re at that point then you potentially have an encirclement around (-1,0) in the w-domain and you’re potentially unstable. You must look at the entire Nyquist plot to know for sure (assuming no open loop poles in the RHP).

I will say that in most applications, if you have a gain greater than 1 with a phase of 180 degrees, you will likely have a clockwise encirclement of -1 and be unstable, but without looking at the entire Nyquist plot, it's difficult to know for sure.
 

LvW

Joined Jun 13, 2013
1,337
However, with this reasonning, we conclude there is no other value of the loop gain that makes the system instable.
That is not correct.
AB=-1 is the oscillation condition with unity loop gain magnitude - and the corresponding loop phase is -180deg.
Together with the inverting sign at the summing junction we have -360 deg (identical to 0 deg).
However, when the mentioned phase shift (-180deg) is reached at a frequency where the magnitude of the loop gain is even larger than unity, the Nyquist stability criterion says that the closed loop will be unstable.
In some cases, the ciruit produces a clipped sinewave or (nearly) a squarewave or the output goes into saturation.

In short: The condition loop gain=-1 is the stability LIMIT.
 

Thread Starter

Pif Ouille

Joined Jun 27, 2019
7
Thank you all for your answers.

I'm beginning to understand how I got completely confused.
Instability means poles with positive real part. period. The magnitude of the transfer function only goes to infinity if these poles are on the jw axis.

Acutally, I was studying RF amplifiers and my confusion came from slides 26/27 of the following lecture
http://rfic.eecs.berkeley.edu/142/pdf/module7.pdf

The reasonning seems perfectly rigorous and but leads to 1+T=0 as the only way for a system to be instable.

anyway, I'll keep thinking :)

Thanks again
 

LvW

Joined Jun 13, 2013
1,337
Thank you all for your answers.

I'm beginning to understand how I got completely confused.
Instability means poles with positive real part. period. The magnitude of the transfer function only goes to infinity if these poles are on the jw axis.

Acutally, I was studying RF amplifiers and my confusion came from slides 26/27 of the following lecture
http://rfic.eecs.berkeley.edu/142/pdf/module7.pdf

The reasonning seems perfectly rigorous and but leads to 1+T=0 as the only way for a system to be instable.

anyway, I'll keep thinking :)

Thanks again
Perhaps this can clarify something:
* In the referenced document, the stability criterion is derived from expressions for a closed-loop system .
It says: We can clearly see that instability implies that T = −1, which is exactly what we learned in feedback system analysis.
My comment: Nothing is said about T<-1 (e.g. T=-1.5).
Hence, this does not mean that instability exlusively would exist for T=-1 only.

* The quoted sentence was derived from expressions which contain small-signal parameters.
Hence, all equations and expressions are valid for a stable closed-loop system (poles in the left half of the s-plane) - until fpr T=-1 the closed-loop gain will reach infinity (poles at the jw-axis). This is the oscillation condition (stability limit).
Because small-signal parameters are not valid for closed-loop poles in the right half of the s-plane, this method could not derive any criterion for T<-1.
 

Thread Starter

Pif Ouille

Joined Jun 27, 2019
7
My comment: Nothing is said about T<-1 (e.g. T=-1.5).
Hence, this does not mean that instability exlusively would exist for T=-1 only.
I have to confess that I don't really follow your logic here.
If "instability implies T=-1", then I'd say that T≠-1 means no instability. No?

Actually, the sentence you quoted is the very one that confused me.
In the end, I guess what it's saying is:
"if T=-1 for some YL/YS, then there exist some YL/YS that make the system unstable"
But even this simple cause/effect statement is not completely obvious to me :)

Again, thanks everybody for your help.
 

LvW

Joined Jun 13, 2013
1,337
I have to confess that I don't really follow your logic here.
If "instability implies T=-1", then I'd say that T≠-1 means no instability. No?
As far as my "logic" is concerned:
Do you know the difference between the two conditions which can be "sufficient" and/or "necessary"?
At T=-1 a closed-loop is unstable. Does this mean that T=-1 is the only condition for instability?
 

Thread Starter

Pif Ouille

Joined Jun 27, 2019
7
I'm sorry if I sounded sarcastic. That wasn't my intention. And I thank you again for taking the time to help me.

However, I still don't agree with you.
"instability implies T=-1" as said in the slides is not the same as your proposition "at t=-1, a closed loop is unstable",
which, in fact, is "T=-1 implies instability"...

So please don't be sarcastic either.

To me, "Instability implies T=-1" means there is no way to find an instability that corresponds to some value of T different from -1.
 

LvW

Joined Jun 13, 2013
1,337
I'm sorry if I sounded sarcastic. That wasn't my intention. And I thank you again for taking the time to help me.

However, I still don't agree with you.
"instability implies T=-1" as said in the slides is not the same as your proposition "at t=-1, a closed loop is unstable",
which, in fact, is "T=-1 implies instability"...

So please don't be sarcastic either.

To me, "Instability implies T=-1" means there is no way to find an instability that corresponds to some value of T different from -1.
It is even more "complicated". When you study the GENERAL Nyquist criterion (not only the simplified form) you will see that there are circuits which are stable - even when there is a frequency where the loop gain is T=-1.
 

LvW

Joined Jun 13, 2013
1,337
Well, it seems that I won't get away without a good review of all these concepts :)

Thanks again!!
A short comment to the "oscillation condition":
The well-known Barkhausen condition (unity loop gain) is considered to be a necessary condition only.
That means: When a circuit oscillates, it must fulfill this condition.
However, does this mean that a circuit will always oscillate when the loop gain is unity at a certain frequency?
This would be a sufficient oscillation criterion. And the answer is: No !
It was just some years ago that a third condition - in addition to Barkhausen`s condition (magnitude and phase) - was found which allows to formulate such a sufficient criterion.
 

tindel

Joined Sep 16, 2012
810
I’d check out Brian Douglas on YouTube. He has a lot of good control theory videos. I also like katkimshow.

Also if you’re working on RF amplifiers I wouldn’t worry about feedback control too much. Understand the basics, as you do, and keep moving forward.

Finally I’d highly suggest Feedback Control of Dynamic Systems if you really want to know about LTI systems. This book is used in a semester long graduate course at my school. https://www.pearson.com/us/higher-e...f-Dynamic-Systems-8th-Edition/PGM1653583.html
 

Thread Starter

Pif Ouille

Joined Jun 27, 2019
7
Great!! I'm gonna have a look at this book I don't know of.

I know I shouldn't worry to much, but I get annoyed with this kind of details :)
And I thought stability was kind of critical in RF amplifiers, isn't it?

I've actually found a similar question on another forum:
https://math.stackexchange.com/questions/1030429/blacks-formula-and-feedback-system-stability

This short article is also interesting on the matter:
http://web.mit.edu/klund/www/weblatex/node4.html

Anyway, with all your help, things are becoming much clearer.
 

LvW

Joined Jun 13, 2013
1,337
Great!! I'm gonna have a look at this book I don't know of.

I know I shouldn't worry to much, but I get annoyed with this kind of details :)
And I thought stability was kind of critical in RF amplifiers, isn't it?

I've actually found a similar question on another forum:
https://math.stackexchange.com/questions/1030429/blacks-formula-and-feedback-system-stability

This short article is also interesting on the matter:
http://web.mit.edu/klund/www/weblatex/node4.html

Anyway, with all your help, things are becoming much clearer.
My recommendation: Forget the last referenced paper.
The authors first sentence is simply stupid. This man has never read the Barkhausen paper in detail.
He does not know that the Barkhausen criterion is a necessary criterion only.
Or he is unable to discriminate between necessary and sufficient criteria.

PS: I am from Germany - and I have Barkhausens original book on my table. And I can read ...and understand.
 
Last edited:

tindel

Joined Sep 16, 2012
810
I know I shouldn't worry to much, but I get annoyed with this kind of details :)
And I thought stability was kind of critical in RF amplifiers, isn't it?
People spend their entire careers specializing in controls. IEEE has a whole subsection just on Controls and Feedback Systems. You can literally spend a lifetime understanding it, and staying up-to-date in the current state of the art.

As you're learning, I've found that sometimes letting concepts simmer in the back of your mind. Having a basic understanding is good enough. As you have these sort of questions - you may not yet comprehend the answers, but after time working in the industry it will become more clear.

Example: I've worked on projects in the past and had questions about how something is working exactly, even though all the testing has verified that it works properly. Years later I'll be working on a different problem, and realize that what I'm working on currently is the solution to the question I had years prior. But it's taken those years to gain the additional experience needed to connect all the dots. This is normal.

I haven't spent much of my career in RF, so I can't really comment on if controls is super important for RF amplifiers. I imagine it's an important part, but is a subtopic. My naive thoughts are that RF amplifiers have little feedback in order to achieve the frequencies needed to transmit/receive, as feedback is generally used to slow down a open loop circuit to make it stable. Most RF (100MHz+) circuits are shielded to keep parasitic feedback from effecting the circuit. I could be completely wrong about that though.
 
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