question about sine wave oscillator ?

[muoinhohn]

Hi all
I have a question about sine wave oscillator.
Denote A is amplifier gain and β is feedback gain. In the Barkhausen criterion, if Aβ=-1 then system is oscilator.
- When the system is oscillatory, by some reason, mag(Aβ)=1 but phase(Aβ) ≠-180°, what will happen ?
- When the system is oscillatory, by some reason, phase(Aβ) = -180° but mag(Aβ) ≠1, what will happen ?

Thank you in advance

 

[Fom]

The oscillation frequency will be equal to the frequence where phase(Aβ) = -180°.
If gain (Aβ) at this frequency will be less than 1 the oscillation will be dumping sine wave.
If gain (Aβ)=1 it will pure sine wave.
If gain (Aβ)>1 it will be clipped sine wave (sine with harmonics).

 

[nandu_r]

This is defined by barkhausen criteria

A synstem to become oscilator
Aβ (loop gain) ≥ 1 ( so that system can regenerative)
loop phase shift should be more than 180°

In case to generate a sine wave oscilator keep the loop gain slightly more than 1
this will give a good sine wave with minimum distortion

make a note that in margine for the loop gain takes care of your process variation

 

[muoinhohn]

If i well understand the explains that if mag(Aβ) > 1 and phase(Aβ) ≠ -180° (or ≠ 0°) then the system will be oscilate but with a distorsion is great. Is it right ?

Regards.

 

[VVV]

Actually, no. It will not oscillate at all. A large gain cannot cause oscillation, if the phase is not -180°.
This is the basis for all automatic control systems: have large gain (at low frequency), in order to minimize errors, but ensure the phase is not 180°. When the phase eventually does reach 180°, ensure the gain is less than 1, so an oscillation cannot be sustained.

Bad distortion will happen if the gain is much greater than 1 when the phase is 180°. In that case, the oscillaton will start and the amplitude will build up until some non-linearity in the system will limit the gain to 1 (this always happens, otherwise the amplitude would have to reach infinity, which is impossible). But by the time the gain is limited by nonlinearities, the waveform will be badly distorted, since non-linearities actually cause distortion.

 

[muoinhohn]

Hi
Thanks for yours reponses. So It is not clearly for me. In the case mag(Aβ) > 1, that meas that if at the time 0, on the system output there is a noise, then this noise will be transfer throught A and β and therefore the noise will return to output but its magnitude is greater. The process will be continue again, ....
Is there a mathematical explaination for the oscillator?

Regards

 

[VVV]

Well, you pretty much got it. The noise will get the oscillator started. However, you should remember that there is a filter element in this oscillator, that is, something depends on the frequency, either an LC tank, or and RC network, etc. (Usually β is made to be frequency dependent).

That means that out of the many components of the noise (it's pretty much white noise), only the correct frequency will be returned to the input with enough amplitude and phase to make the oscillator run. So the oscillator will run at that frequency.

Mathematical explanations exist, of course, but they may not be so easy to grasp.
However, here is a simplified one, which you probably know already:

the oscillator can be viewed as an inverting amplifier (gain A), with a feedback network ("gain" β, in fact it's in most cases a loss). Now calculate the output, as a function of the input, for the amplifier:
Vout=A*Vin

But Vin is really Vout multiplied by β: Vin=Vout*β.

For the oscillator to work, the input signal Vin must be sufficient to produce Vout when amplified.
So now we can write:
Vout=A*Vout*β, or 1=A*β, which is the condition you already knew.

If the amplifier is inverting, then its gain is -A and the final condition is -1=A*β, meaning the phase of the signal at the amplifier input has to be -180°, instead of 0°.
Since β is really a loss, you can see that the condition requires that the amplifier make up for it, or else the oscillator will not work.
If you have A*β<1, you can see that that A*Vout*β<Vout, so the oscillation cannot be sustained, since all the time the output amplitude will decrease, instead of building up.

 

[muoinhohn]

I agree with you in the case A*β<1 . But what i want to known is that if A*β> 1 how does the oscillator work ?

Regards

 

[VVV]

As I said, if you have A*β>1, then the amplitude will build up, since the input signal is more than what is required to produce the output signal. So the output signal increases, which in turn increases the input signal, which in turn increases the output signal, which..... and so on.

Thus, the amplitude would keep building up to infinity. That, of course is not possible. Something will limit the oscillation: either a non-linear negative-feedback element that you introduce yourself (such in the case of Wien-bridge oscillators), or the changing parameters of the components (with increased amplitude, the transistor begin operating in the large-signal mode, where the gain is less, or the inductors/ resistors in the circuit begin dissipating more with increased signal, thus increasing losses), or even your finite power supply voltage will clip the signal, resulting in distortion, but also limiting the amplitude.

So eventually, something will limit the amplitude. That is why in many cases A*β is made just slightly higher than 1, so the oscillator is guaranteed to start up, but amplitude limiting will start at a fairly low output, thus preventing significant distortion.
In the case of the Wien-bridge, for example, you actually "measure" the output amplitude and adjust the negative feeback accordingly, so as to maintain constant output amplitude. This makes for very low distortion out of this type of oscillator. The concept is applied to other types where amplitude stabilization and low distortion is important.

 

[muoinhohn]

So it is ok. I found on the article "Start-up and Frequency stability in High-Frequency Oscillators" of M. Nguyen and Meyer. They said that in some oscillators that have Pha(A*β) = 0 , and Mag(A*β) > 1, but they are stable.
I'm studying this article. Thank you very much for yours reponses and advises.

 

[flatulent]

You have left off the first part of the criterion. It starts of with "If and only if there are poles in the right half plane."

There are complex control system that have the gain over one with a 180 degree phase shift but they do not oscillate because there are no poles in the right half plane.

The criterion continues with "the frequency the system will oscillate at is the one where the gain is one and the phase shift is 180 degrees."

What you get at startup is a chirp in frequency while the gain changes to reach steady state by limiting or AGC actions.

This chirp was one problem with amateur transmitters many years ago when the master oscillator was turned off for receive and on for transmitting.

 

[muoinhohn]

I've understood this problem . In the case general, we must regard the poles of (1-A*β) in s-plan. If we only regard Phase and Mag of A*β, we can't conclude exactlly. Because the function A*β (jω) is used to find the steady-state reponse when the circuit is stable.