Exam help! A loop-gain greater than one

[jasmin_123]

Could somebody give an INTUITIVE(!) explanation why a loop-gain greater
than one does not necessarily causes instability?

My professor has promised a 30-point bonus in the exam for this answer.
I spent almost all the time before the exam to find any hint on the net and in books
with NO success. Help, please!
---
I am talking about positive feedback oscillators and the Barkhausen oscillation criterion.

It is implied that AOL*B is REAL, POSITIVE, and greater than one, and this does not always mean that the oscillator is unstable. It can be stable.

Again, an INTUTIVE explanation is needed, without Nyquist, root locus, and so on.

Feb 12

Well, guys, I felt that we started to make circles and asked the tutor for a hint (at the expense of the 10 bonus points, too bad).
---
The hint was as follows:

"The loop-gain greater than one, at a frequency where the phase is zero, provides no more information about oscillator stability than the color of the oscillator capacitors does!"

Does it say anything to you???

 

[farzad_m]

because having a pole in RHP doesnt necessarily cause instability because you may have a zero that cancels that pole

 

[jasmin_123]

Thank you farzad, but AOL*B>1 does not always shift the poles to RHP.
It can shift them to LHP (see the attachment).

 

[ender84567]

instability is due to a ROC issue in the intersection of the closed loop with the open loop, other than that its PM and GM, long as you are allowed to have a GM greater than one, your loop gain can be >1 for the given circuit. I'm not sure if that helps, but i'm much better with specific examples than a generalization. Explore ROC(rate of closure), PM(phase margin) and GM(gain margin) on a bode plot, i'm sure there is a better resource out there than me about this and it is difficult to describe without pictures.

 

[jasmin_123]

OK, I have to clarify myself. I am talking about positive feedback oscillators and the Barkhausen oscillation criterion.

It is implied that AOL*B is REAL, POSITIVE, and greater than one, and this does not always mean that the oscillator is unstable. It can be stable.

Again, an INTUTIVE explanation is needed, without Nyquist, root locus, and so on.

Help, please!

 

[sheraz.pervaiz]

If loop gain is greater than one.It will be stable if it is negative feedback and unstable if and only if , it is positive feedback

 

[jasmin_123]

You are not right, sheraz.pervaiz, a POSITIVE-feedback oscillator can be both stable and unstable when its AOL*B>1.

 

[alberr]

Let's take common equation of oscillator. It will be something like this:

W(s)= k/(T^2*s^2 + 2*zeta*T*S + 1)

where
k - gain (close loop gain depend on it);
T - time constant (have effect on poles configuration and phase margin);
and, at last, 'zeta' - damping factor. Characterizes irreversible losses in system.
We may have high gain and phase shift may be convenient for good generation, but high damping will vanish our oscillation and we will get stable loop.
For example, a pendulum with high friction in axis will not oscillate even if we will push it.

 

[Mr.MEB]

Theoritically, a pole int he RHP is unstable.
But in the case of electronics, when you're talking about an oscillator, there are always neglected non-linearities, because of those non-linearities the system could remain stable.
-----------
Try Lyapunov to see if a non-linear system is unstable

 

[jasmin_123]

Hi, alberr, I understand your point, but where is loop gain, AOL*B, in your equation?

Mr. MEB, thank you for your input, but a purely linear case is considered.

Any new INTUITIVE ideas?

 

[Mr.MEB]

OK, I have to clarify myself. I am talking about positive feedback oscillators and the Barkhausen oscillation criterion.

It is implied that AOL*B is REAL, POSITIVE, and greater than one, and this does not always mean that the oscillator is unstable. It can be stable.

You are talking about oscillators right?
Have a look at this INTUITIVE explanation:
As the amplitude of the oscillations grows, the gain (A') of the transistor diminush in the region of the peak of the sinus. so A' is less than A at this regions, making A'OL*B<1.

I don't agree with you when you say : "it's a purely linear case" they exist only in Matlab/Scilab. And mathematically they can not be stable.
Hope that this explaination satisfies your examinator.

 

[jasmin_123]

Let me disagree with you, Mr.MEB,

If oscillations had started at AOL*B>1, and were growing, then the circuit was unstable. If no noise is considered, AOL*B will reach exactly 1, with noise it will be
a bit less than one.

Suppose that the circuit starts oscillations from VERY week initial conditions,
almost zero. It now has to 'decide' whether increase or decrease the amplitude,
depending on the real instant AOL*B>1. This is ABSOLUTELY LINEAR and
ABSOLUTELY PRACTICAL, not MATLAB, situation.

To forget about nonlinearity, suppose almost zero initial condition, AOL*B>1,
and now is the question: "Why the oscillator does NOT have to increase its amplitude?" The amplitude will either increase or decrease. Why?

Any INTUITIVE explanations? Please HELP!

 

[Mr.MEB]

I think we have a communication issue. Sorry, my mother tongue isn't English.

If no noise is considered, AOL*B will reach exactly 1, with noise it will be
a bit less than one.

I didn't understood this point, how the input noise could affect the transfer of the closed loop?
You said:

This is ABSOLUTELY LINEAR

and just before

the amplitude,
depending on the real instant AOL*B>1

??

What ever,

ABSOLUTELY PRACTICAL, not MATLAB, situation.

I understood that it was a real circuit (or plant) and not a simulation.
I wanted to say that "purely linear systems" are rare physical phenomenon, but models are linear.

"Why the oscillator does NOT have to increase its amplitude?" The amplitude will either increase or decrease. Why?

I didn't "grip" exactly all the sense of the question, but if you find the answer, could you share it here?

 

[jasmin_123]

Hi, Mr.MEB,

Communication issue #1:
AOL*B is the loop gain, NOT "closed loop."

Communication issue #2:
For small signals, ALMOST ANY practical circuit is LINEAR.
Many circuits operate ONLY in small-signal mode.

"Purely linear systems" are NOT a rare physical phenomenon,
otherwise small-signal analysis would have no sense.

In steady state, noise of oscillators reduce their AOL*B below one.
Oscillator with noise is actually an amplifier, and its response is solely
forced one; its natural response is zero in steady state.

Communication issue #3:
Natural response of a linear circuit depends on its network function.
AOL*B(s) is a part of the network function, hence, the circuit
natural response does depend on AOL*B(s).

Communication issue #3:
Would I hope to find the answer by myself I would not ask for help.

My professor says that analog engineer must answer ALL elementary
conceptual questions. He drives me mad with this question. But he seems
right, because some engineers I talked to said that an oscillator with
an AOL*B, say, 2 is always unstable. This is so because AOL*B=2 means "the
grows of the signal around the loop." BUT THIS IS NOT TRUE!

Let me repeat the question in the most simple form:

1. Consider a noise-free positive-feedback oscillator at almost zero initial conditions, which means a 100% linear case.
2. AOL*B=2.
3. Why the oscillations do not have to grow?

 

[aomeen]

Hello all,

I just would like to ask about something

you are talking about "Stability of Oscillator" ?

Oscillators are nither stable, nor unstable. osillators are marginallys stable

At start of oscillation loop gain "Open loop gain" MUST be greater than 1 , in the same time the round phase for the desired frequency MUST be 2pi , under postive feedback assumption.

As the oscillator builds, the loop gain drops to 1 This is not due to noise. This is due to amplitude growth. while the phase condition is still applies. That's to say, at start of oscillation some poles are right handed. In steady state All poles are either left handed or pure imaginary.

Now, For your quistion. simply make the loop gain ∞, but make the phase shift for the desired signal 180 "Negative feedback system" This is a typical ideal opamp amplifier which is totally stable.

My point is : Try check for the phase condition. Perhaps the feedback network has phase-amplitude relation "That is linear" such that under higher amplitude the phase shift is no more 360 "Although I dont have anything in mind now that can act as such...

Plz share the answer after your prof. tells it to you . Hope it's something that has a meaning

 

[jasmin_123]

>Oscillators are nither stable, nor unstable. osillators are marginallys stable

It is true ONLY in steady state. I am talking about the transient from the initial conditions to steady state.

>At start of oscillation loop gain "Open loop gain" MUST be greater than 1 , in the same time the round phase for the desired frequency MUST be 2pi , under postive feedback >assumption.

>As the oscillator builds, the loop gain drops to 1 This is not due to noise. This is due to amplitude growth. while the phase condition is still applies. That's to say, at start of oscillation some poles are right handed. In steady state All poles are either left handed >or pure imaginary.


It is not true. An oscillator can start from a loop gain of 0.1 and reach sustained oscillations at the loop gain of 1.

>Now, For your quistion. simply make the loop gain ∞, but make the phase shift for the desired signal 180 "Negative feedback system" This is a typical ideal opamp amplifier >which is totally stable.

This is not a positive-feedback oscillator we are talking about.

>My point is : Try check for the phase condition. Perhaps the feedback network has >phase-amplitude relation "That is linear" such that under higher amplitude the phase shift is no more 360 "Although I dont have anything in mind now that can act as such...

At AOL*B=2, the loop phase shift is zero or 360.

>Plz share the answer after your prof. tells it to you . Hope it's something that has a meaning

Sure! but is here a Guru who can help me to get these damned 30 points in the exam?

The question:

1. Consider a noise-free positive-feedback oscillator at almost zero initial conditions, which means a 100% linear case.
2. AOL*B=2.
3. Why the oscillations do not have to grow?

Could somebody give an INTUITIVE(!) explanation for this ELEMENTARY question?

 

[Mr.MEB]

All right,
I was a student too (not so long ago), and the same question was asked in the electronics lab.
The explanation was nearly what I said.
you can find the same "hint on the net" :
just type "oscillator amplitude" in google and the first link is :

In most oscillator circuits, oscillation builds up from zero when power is first applied, under linear circuit operation. However, limiting amplifier saturation and other non-linear effects end up keeping the oscillator's amplitude from building up indefinitely. Thus, oscillators are not the simplest devices in the world to accurately design, simulate or model. There is a real art to GOOD, STABLE oscillator design. As you learn more about oscillators you will certainly grow to appreciate them!

http://www.sss-mag.com/cosc.html
But seems to me that you are confused or misuse the terms that belong to automatic control field:
- Stability
- Amplitude is not the instantaneous value of the signal, but the diff between peaks of a (pseudo)periodic signal.
- Transfer gain : is not the instantaneous value of (Voutput / Vinput)
- The assumption that oscillators belong to the "small-signal" analysis
- Signal noise vs plant modelistation noise

If this is not the answer for your 30 points, probably I haven't understood the question well again. Do not repost the same question 4th time but instead, try to explain your words:
"osc. stability", "osc. growth", "stability in RHP poles for LINEAR systems"

 

[jasmin_123]

Do you mean, Mr.MEB, that the oscillation in the question above will NOT grow due to misunderstanding, misuse, and confusion?

How can I 'sell' such an answer?

 

[Mr.MEB]

I don't understand oscillation grow.
But in the case you mention:
- Linear system
- RHP pole
There will be no "oscillation grow", the system will go unstable, the output value will go to inf.
So you must admit that the oscillator work in non linear regions.

 

[jasmin_123]

Mr.MEB, please carefully read the question: what asked is why the oscillations DO NOT HAVE TO grow AT START. What will happen after they will either grow or decay is not important and does not affect the situation at start.

By the way, RHP is not mentioned in the question.