Exam help! A loop-gain greater than one

[zorro]

Hi, zorro,

Why not necessarily the output has to grow? If signal at some frequency is increasing while traveling over the loop and remains in phase (G>1 at zero phase implies just this), then the output does necessarily have to grow.
---
The output always have to grow for a G>1, P(G)=0. Why for the same G>1, P(G)=0 it does grow in some cases and does not grow in the other cases? Why?

Jasmine

Hi Jasmine,

Some preliminaries (variations on my previous post):
The output maintains stable if the output is exactly the same as the input (G=1 exactly). As I said, If G is real and G>1 for some frequency, the output is no more the same as the input, and that is not an equilibrium point.
The equilibrium point is at that complex frequency at which the output is exactly the same at the input at any time. If for some frequency it happens G>1, it is not G=1 at this is not an equilibrium point. Then, where is there an equilibrium point? If there is one for some growing complex exponential (poles of closed loop in RHP) the circuit is unstable, otherwise it is stable.
The fact that G>1 for some sinusoidal waveform guarantees that an equilibrium point is not there, but it does not guarantee that the equilibrium point is in the RHP.

Now, let’s try to solve the paradox: why not necessarily the ouput has to grow?

This example will be helpful:
Once a student asked me, astonished, how an amplifier with gain –2 at all frequencies could be stable when connecting the ouput to the input. His reasoning was as follows:
Imagine that the input is 1. Then the ouput is –2. As it is connected to the input, the input becomes –2 and the output becomes +4, then –8, etc……..

The problem is that he was thinking in discrete time, not in continuous time. Implicitly he put a delay in the loop.

In continuous time, the only stable point is with output 0 (solution of output=input at all times).
In discrete time with one-sample delay (also in continuous time but including a pure delay), the situation would be really unstable. Plot the root lucus of such a simple system in discrete time with one-sample delay: in open loop there is a pole at z=0. Increasing the gain, the pole moves to left, an for any G>1 (A<-1) the pole goes outside the unit circle => unstability.

You are making that kind of reasoning. You think something like this: “if some cycle has amplitude 1, then the following one has amplitude G, then G^2, and so on”.
You place an additional delay in the loop. That makes you think that the output does necessarily have to grow. It is very intuitive, but incorrect.

Is it clear?

Regards

Z

 

[jasmin_123]

Dear, Zorro,

I like and understand your explanations, but I am not sure that they are complete enough.

This week, I was thinking about the hint (see my message of 12 Feb 2007 23:01) that G has nothing to do with the oscillator behavior. This has led me to the following conclusion (not sure yet whether it is OK).

The loop gain, G(s), is a network function defined for ZSR only. The oscillator response is a ZIR one. Does the G(s), defined in ZSR only, have meaning in ZIR? It seems that not.

Consider, for example, a zero-order amplifier with the gain A=10 and a feedback network with the feedback transmission B(jwo)=0.2. When you measure G in Fig. 1, you get G(jwo)=A*B(jwo)=2. But in ZIR (see Fig. 2), B(jwo)=2 has no meaning, because if A=10, then B(jwo) should be 0.1! and G(jw0) should be 1!

Conclusion: A, B, and G has no meaning in ZIR, and, hence, a single value of G says nothing about the oscillator stability.

Does it make sense?

Jasmine

 

[eternal_nan]

Hi,

This is a pretty tough question. Here is a couple of very contrived attempts:

Attempt 1:
Quote from earlier posting by OP: 1. Consider a noise-free positive-feedback oscillator at almost zero initial conditions, which means a 100% linear case.

If there is absolutely no noise (like what was said in above quote) the oscillator will never start due to lack of initial excitation to get amplified around the positive feedback loop.

Attempt 2 (even more contrived):
The Barkhausen condition of >1 loop gain at N*2pi phase:
The phase lag around the loop when (open) loop gain becomes > 1 could be infinite (lim(N->infinity)), in this case the oscillator will theoretically oscillate but will be infinitely low frequency? I guess this is practically equivalent to 'not starting'.

Added after 2 minutes:

Correction to previous posting: Ignore "Attempt 2", another 20sec thinking on the topic made me realise it is stupid .

 

[Mr.MEB]

So what's the answer finally?

 

[jasmin_123]

The right answer is below; at least, it was accepted by the prof.

The loop gain, G(s), is a network function defined for ZSR only. The oscillator response is a ZIR one. Does the G(s), defined in ZSR only, have meaning in ZIR? It seems that not.

Consider, for example, a zero-order amplifier with the gain A=10 and a feedback network with the feedback transmission B(jwo)=0.2. When you measure G in Fig. 1, you get G(jwo)=A*B(jwo)=2. But in ZIR (see Fig. 2), B(jwo)=2 has no meaning, because if A=10, then B(jwo) should be 0.1! and G(jw0) should be 1!

Conclusion: A, B, and G has no meaning in ZIR, and, hence, a single value of G says nothing about the oscillator stability.

 

[teteamigo]

A single value of G says nothing about the oscillator stability.

Right. Follows the hint of your Teacher, about incomplete information.


But not search reasons in ZIR or ZSR.


Oscillator is like an amplifier with gain at only one frequency.


We can account the greater number of ways that assures the quest above?

Yes.

Realize an Oscillator (linear systems):
loop-gain(jwo)=1 , this condition when true is the intersection of magnitude and phase condition at wo angular frequency.


https://web.mit.edu/klund/www/weblatex/node4.html [GOOD, GOOD, GOOD]


"The history of the Barkhausen Stability Criterion is an unfortunate one. In 1921, during his study of feedback oscillators, Barkhausen developed a ``formula for self-excitation''

k*F(jw)=1

where K is an amplifier gain factor and F(jw) is the frequency dependence of the feedback loop. This equation was originally intended for the determination of the oscillation frequency for use in radio transmitters. However, before conditionally-stable nonlinear systems were understood, it was widely believed that only a single value of K separated stable and unstable regions of behavior. Thus, Barkhausen's Criterion was incorrectly used as a stability criterion, especially in the German literature "


For positive feedback systems:

system1:Loop-Gain=1 at w1 rad/s

We can conclude that the system will oscillate at w1 rad/s

system2:Loop-Gain>1 at w2 rad/s


We can't conclude that the system will oscilate at any other frequency.



We can start search for singularities in RHP with a loop-gain>1 and with a system like L=1/s more unitary positive feedback.


Loop-Gain=4


1/s=4 for s=1/4


If we input a exponential signal in the loop e^4t the output appears multiplied by 4, then 4e^t/4

The system is unstable. Gives gain to no bounded signal input (signals that aren´t close the equilibrium)


Search oscillations for the same system:

No Poles in imaginary axis jw.

1/s=1 <=> s=1 , the only pole of the system is at RHP

 

[jasmin_123]

"Oscillator is like an amplifier with gain at only one frequency."
This is wrong. Oscillator with zero noise is not an amplifier. It does not have any input.


I am aware about https://web.mit.edu/klund/www/weblatex/node4.html
Unfortunately, these pages give no intuitive explanation to the question I asked.

A different site led me to the right answer:
https://www.ee.bgu.ac.il/~paperno/Positive_Feedback_Oscillators.pdf
https://www.ee.bgu.ac.il/~paperno/1._Positive_Feedback_Oscillators_Illustrations.pdf

 

[teteamigo]

"Oscillator is like an amplifier with gain at only one frequency."
This is wrong. Oscillator with zero noise is not an amplifier. It does not have any input.

In physical systems we have noise power at all frequencies given by active, passive devices.


Look that we have noise even before the potential instable circuit was powered on(supply bias).

"I am aware about https://web.mit.edu/klund/www/weblatex/node4.html
Unfortunately, these pages give no intuitive explanation to the question I asked.

Look in depth...

More intuitive (common science, not common sense) than this is impossible.


We can arrange systems, for example with loop-gain G(s)=A>1 (real, positive), and this is not a condition to have s=jw poles, G(jw)=1 (look my example of Loop gain=1/s, the system is unstable, but never can be marginally stable).


G(s)=A>1 Produces this gain for a specific class of signals.
where A is the small-signal gain when the system is powered on (don't forget it).


We need that this class of signals have any sinusoidal dependence with a exponential boost (e^at, a>0)


The system in steady-sate is marginally stable, because the poles of the system are driven by small-signal gain. When the amplitude increases the small signal gain reduces (because gain compression, nonlinear behaviour), then the poles of the system moves to the imaginary axis.

"A different site led me to the write answer:
https://www.ee.bgu.ac.il/~paperno/Positive_Feedback_Oscillators.pdf
https://www.ee.bgu.ac.il/~paperno/1._Positive_Feedback_Oscillators_Illustrations.pdf

From the 1st link:


"D. Initial Pole Location
We now have to decide on the pole location in the initial
state. Let us first suppose that there should be no difference
between the initial and final pole locations; in the both cases
the poles can be located directly on the jω axis.
Magnitude of the natural response (4) in this case [see Fig.
3(b)] is a function of the initial state. It is an undesirablesituation for the three following reasons. First, we do not
intend to supply the circuit with additional means to control
the initial state. Second, any circuit is subject to disturbances
and noise. In the case of an oscillator, the uncontrolled initial
state, disturbances, and noise will make its steady state
unpredictable. Third, it is impossible to provide the initial pole
locations exactly on the jω axis because of the non-zero
tolerances of the circuit components."


We should be exigent with theory. Right.
But we can't be exigent with the physical systems. We accomodate to Nature.

 

[jasmin_123]

Note that an oscillator does not need any noise to oscillate.

"I am aware about https://web.mit.edu/klund/www/weblatex/node4.html
Unfortunately, these pages give no intuitive explanation to the question I asked.

Look in depth...

Why to look in depth if the author of
https://web.mit.edu/klund/www/weblatex/node4.html
himself gave up on intuitive explanation?

We should be exigent with theory. Right.
But we can't be exigent with the physical systems. We accomodate to Nature.

To understand a practical oscillator you first have to understand it without noise and then to add noise. With no noise its poles are to the right of the jω axis, and this is an oscillator. With noise, its poles are to the left of the jω axis, and this is an amplifier that amplifies the noise at all the frequencies but does it selectively, depending on the phase response of its feedback network.

 

[teteamigo]

Note that an oscillator does not need any noise to oscillate.

Ooooh!!!

Oscillator Theory
An oscillator is a positive-feedback control system, which does not have an external input signal, but will generate an output signal if certain conditions are met. In practice, a small input is applied to the feedback system from factors such as noise pick-up, or power-supply transients, initiating the feedback process and resulting in sustained oscillations. A block diagram of an oscillator is shown in Figure 6.

**broken link removed**

To understand a practical oscillator you first have to understand it without noise and then to add noise. With no noise its poles are to the right of the jω axis, and this is an oscillator. With noise, its poles are to the left of the jω axis, and this is an amplifier that amplifies the noise at all the frequencies but does it selectively, depending on the phase response of its feedback network.

Wrong. Wrong. Wrong


An amplifier fournish power to a input signal. (even voltage-gain or current-gain) Not Noise.


Oscillator: amplifier+feedback network

Input->Noise (all fortuit things that we reject in an amplifier)

In RF circuits (waves and antennas everywhere in circuit) we have always feedback network.

 

[jasmin_123]

I repeat this once again: "oscillator does not need any noise to oscillate."
Its oscillations in steady state can originate from its natural response.
(If you do not believe, you can simulate a noise-free oscillator in SPICE.)

With noise, the oscillator poles shift to the left, its natural response decays,
and its forced response defines the steady state.

"Oscillator: amplifier+feedback network." Not always.

Oscillator is simply an unstable circuit, with unstable poles. No input.
Amplifier is a stable circuit stable, with stable poles. Noise is the input.

This is elementary. Think of that and then do not forget to write: "Right, right, right!"

Take care!

 

[teteamigo]

I repeat this once again: "oscillator does not need any noise to oscillate."
Its oscillations in steady state can originate from its natural response.
(If you do not believe, you can simulate a noise-free oscillator in SPICE.)

If I talk to you about oscillators is because already made simulations (Spice, Cadence Virtuoso&Spectre), practical circuits.

Initial conditions. Say anything to you????

With noise, the oscillator poles shift to the left, its natural response decays,
and its forced response defines the steady state.

feedback to you:

This is elementary... Take care!

"Oscillator: amplifier+feedback network." Not always.

Oscillator is simply an unstable circuit, with unstable poles. No input.
Amplifier is a stable circuit stable, with stable poles. Noise is the input.

Noise is the input of Jasmin amplifiers.

For me and the other people, the signal is the input.

The Jasmin is the only people that understands its amplifier.

Superb RIGHT!!!

 

[jasmin_123]

I am sorry, but I am unable to understand your point.

I also dislike your misinterpretation of my words.

 

[halls]

As far as i know, an amplifier is a device that takes the input and amplifies it (thus the name of the device), so its output is A times the input.

If the input of that amplifier is noise, what we have in the output is amplified noise, then this will be a noise amplifier.

If we take a noise amplifier and join its output to its input, we have a noise amplifier with feedback, which amplifies its input, which at the same time, is its output, so we now have a closed loop.

And at last, if we put a filter in the middle of this feedback network, what we have is noise amplifier, with filtered feedback... oh my god, it's an oscillator!!

if i am wrong, i think most textbooks on earth that talk about oscillators, should be rewritten.

 

[jasmin_123]

halls, there is no need to rewrite book. It is better to reread them.

You are missing the following: an oscillator with noise can be modeled by a
frequency selective amp. The amp. selectivity depends on the phase response
of the osc. feedback network.
---

To say the truth, I am already sick and tired explaining elementary things and taking break.
If you guys would like to continue then review basics meanwhile. Seriously!

Yours,
Jasmine

 

[jonw0224]

This link may offer some help...

https://web.mit.edu/klund/www/weblatex/node4.html

-jonathan

 

[jasmin_123]

The below is from
https://web.mit.edu/klund/www/weblatex/node4.html

"Yet quoting a counterexample rarely satisfies the student. Dislodging this intuitive misconception is an uphill battle.
Thus, it is worth refuting this argument in every possible way."

Conclusion: the author does not have any intuitive explanation.

 

[jonw0224]

Jasmine,

My apologies. I must admit that yesterday I had missed the fact that the topic had multiple pages. I did not read the posts containing this website already.

-Jonathan

 

[jasmin_123]

That is OK Jonathan, but what is your opinion?
Do you have an explanation? Do you accept my
explanation?

Jasmine