# Understanding Problem Oscillators

#### sinatra39

Joined Oct 18, 2018
15
Hello folks,

I have an unpleasant understanding problem about the Barkhausen criterion for oscillators.

It states that the amplitude of the closed-loop transfer function has to be 1 and the phase of it an interger multiple of 2*pi.

This means that the resulting complex number has to be real-valued with an amplitude of 1, hence it is on the real axis on the right side (the circle with a absolute value of 1 cuts the real axis on the right side).

But now comes the confusion for me: The poles-zero plot tells us that a system is stable (the amplitude decreases with time) on the left side.
Exactly on the imaginary axis the system would be stable without damping and on the right side of the imaginary axis the system´s amplitude would rise to infinity.

So the Barkhausen criterion has a pole at 1 on the real axis and therefore should not be stable, but with a transfer function amplitude of 1 it doesnt grow......Im confused about this and would appreciate any help to understand this better.

Thank you very much!

#### LvW

Joined Jun 13, 2013
1,279
Sinatra....you have mixed open-loop with closed-loop conditions.
According to Barhausenss condition for a circuit to be able to oscillate, the LOOP GAIN must fulfill the amplitude condition (unity) and phase condition (360 deg resp. 0 deg).
I suppose, you know how the loop gain is defined and how it can be verified/measured/simulated ?
Note that a closed-loop circuit has no gain at all because there is no input for external signals.

#### sinatra39

Joined Oct 18, 2018
15
ah okay, its true. thank you but it still doesnt clarify the fact that the pole is on the right side of the imaginary axis and therefore would be not stable according the the Laplace theory. for the system to be stable the pole would have to be on the imaginary axis but then the Barkhausen criterion would not be fulfilled, since for an amplitude of 1 the phase would have to be pi/2 or 3pi/2 and couldnt be 0. So with the Barkhausen criterion fulfilled, we only get unstable systems (systems that grow exponentially with time) if we think about it from Laplace pole and zero theory. no?

#### LvW

Joined Jun 13, 2013
1,279
Sinatra - your wording is not clear enough....I dont know which "pole" you are speaking of.
The situation is as follows:
When the loop phase is zero (360 deg) and the loop gain is unity, the closed-loop pole pair is directly on the Im axis.
In practice, the loop gain is made slightly larger than unity - and the complex pole pair has a small pos. real part.

#### MrAl

Joined Jun 17, 2014
8,363
Hello folks,

I have an unpleasant understanding problem about the Barkhausen criterion for oscillators.

It states that the amplitude of the closed-loop transfer function has to be 1 and the phase of it an interger multiple of 2*pi.

This means that the resulting complex number has to be real-valued with an amplitude of 1, hence it is on the real axis on the right side (the circle with a absolute value of 1 cuts the real axis on the right side).

But now comes the confusion for me: The poles-zero plot tells us that a system is stable (the amplitude decreases with time) on the left side.
Exactly on the imaginary axis the system would be stable without damping and on the right side of the imaginary axis the system´s amplitude would rise to infinity.

So the Barkhausen criterion has a pole at 1 on the real axis and therefore should not be stable, but with a transfer function amplitude of 1 it doesnt grow......Im confused about this and would appreciate any help to understand this better.

Thank you very much!
Hi,

If you look at the different types of responses in the jw vs real plane you will see that as you say on the left the response decreases with time and on the right it increases with time. If we have complex poles that also means that on the left it is an exponentially decreasing sinusoid, and on the right it is an exponentially increasing sinusoid. So what about directly in the middle, right on the jw axis?
Isnt that interesting, it's a perfect sine wave!

Yes, right on the jw axis the response is a perfect sine wave. So that means that a sine oscillator has to keep the response on the jw axis one way or another. The way this happens in most cases is the circuit has a pole slightly in the right half place but also has some small nonlinearity built in so that it adjusts the amplitude and thus the loop gain. A typical way to get this is to use a clipping action to limit gain.

If you study this in detail, you will find that as you vary the gain K the circuit goes from LHP to RHP and the perfect spot is some constant gain K, which if maintained, would keep it oscillating steadily. But circuit components vary slightly over time, and K changes slightly even with the smallest temperature drift, so there has to be some way to adjust this as the circuit runs, and that is where the nonlinearity comes in. Typically diodes are used to clip the amplitude and thus keep the circuit in check.
Note again that without the diodes the circuit output would climb continuously so it needs the clipping action to keep it real.

Another way to look at this is by way of pure theory. You could never get a gain SO EXACT that it would keep the pole directly on the jw axis for all the time it was running. You need a way to adjust it automatically. Rather than adjust it both ways, it is simpler to adjust it down only, and so the circuit is first built so the sine exponentially increases (RHP) and then the adjustment only has to lower the gain.
It is interesting that if the required gain K is 10, even if K changes to 10.000001 the amplitude will increase indefinitely although slowly. If the gain K changes to 9.999999 the amplitude will decrease indefinitely. This means an auto tune of some sort has to be incorporated into any real circuit.

#### AlbertHall

Joined Jun 4, 2014
11,395
There is always a practical limit to the amplitude and gain.

Consider an amplifier with a gain of 10 and a 100mV input the output will be 1V. What if the power to amplifier is 5V and the input is 600mV? The output cannot be 6V as would be expected as it would clip at the supply giving a 5V output so now the gain is 5V/600mV or 8.33 so you have an automatic gain control. This would of course lead to distortion but it keeps the oscillator going.

#### sinatra39

Joined Oct 18, 2018
15
First of all, thank you so much for the very precise reply and your time! I am aware that practically, we design circuits, so that our poles are on the right side of the diagram and therefore we have an exponentially rising sine until we clip it, as you said for instance by a diode.

But my understanding problem has a theoretical nature: The Barkhausen criterium needs an amplitude of 1 and also requires a phase of n*2*pi (n=0, 1, 2, .....) degrees, hence has to be real-valued.....this means it cannot be on the imaginary axis => it cannot be a perfect sine without exponential growth or damping according to the Barkhausen criterium................if we fulfill the criterium we can only have a pole on the right side.

As I understand the criterium for me it means we only can have theoretically an exponential rising sine but not an undamped or damped sine. where is my mistake?

#### sinatra39

Joined Oct 18, 2018
15
Hi,

If you look at the different types of responses in the jw vs real plane you will see that as you say on the left the response decreases with time and on the right it increases with time. If we have complex poles that also means that on the left it is an exponentially decreasing sinusoid, and on the right it is an exponentially increasing sinusoid. So what about directly in the middle, right on the jw axis?
Isnt that interesting, it's a perfect sine wave!

Yes, right on the jw axis the response is a perfect sine wave. So that means that a sine oscillator has to keep the response on the jw axis one way or another. The way this happens in most cases is the circuit has a pole slightly in the right half place but also has some small nonlinearity built in so that it adjusts the amplitude and thus the loop gain. A typical way to get this is to use a clipping action to limit gain.

If you study this in detail, you will find that as you vary the gain K the circuit goes from LHP to RHP and the perfect spot is some constant gain K, which if maintained, would keep it oscillating steadily. But circuit components vary slightly over time, and K changes slightly even with the smallest temperature drift, so there has to be some way to adjust this as the circuit runs, and that is where the nonlinearity comes in. Typically diodes are used to clip the amplitude and thus keep the circuit in check.
Note again that without the diodes the circuit output would climb continuously so it needs the clipping action to keep it real.

Another way to look at this is by way of pure theory. You could never get a gain SO EXACT that it would keep the pole directly on the jw axis for all the time it was running. You need a way to adjust it automatically. Rather than adjust it both ways, it is simpler to adjust it down only, and so the circuit is first built so the sine exponentially increases (RHP) and then the adjustment only has to lower the gain.
It is interesting that if the required gain K is 10, even if K changes to 10.000001 the amplitude will increase indefinitely although slowly. If the gain K changes to 9.999999 the amplitude will decrease indefinitely. This means an auto tune of some sort has to be incorporated into any real circuit.
First of all, thank you so much for the very precise reply and your time! I am aware that practically, we design circuits, so that our poles are on the right side of the diagram and therefore we have an exponentially rising sine until we clip it, as you said for instance by a diode.

But my understanding problem has a theoretical nature: The Barkhausen criterium needs an amplitude of 1 and also requires a phase of n*2*pi (n=0, 1, 2, .....) degrees, hence has to be real-valued.....this means it cannot be on the imaginary axis => it cannot be a perfect sine without exponential growth or damping according to the Barkhausen criterium................if we fulfill the criterium we can only have a pole on the right side.

As I understand the criterium for me it means we only can have theoretically an exponential rising sine but not an undamped or damped sine. where is my mistake?

#### sinatra39

Joined Oct 18, 2018
15
Sinatra - your wording is not clear enough....I don`t know which "pole" you are speaking of.
The situation is as follows:
When the loop phase is zero (360 deg) and the loop gain is unity, the closed-loop pole pair is directly on the Im axis.
In practice, the loop gain is made slightly larger than unity - and the complex pole pair has a small pos. real part.
is it? because in the complex plane a phase of 0 or 360 degrees means the value is real, so it can not be on the imaginary axis. no?

#### MrAl

Joined Jun 17, 2014
8,363
First of all, thank you so much for the very precise reply and your time! I am aware that practically, we design circuits, so that our poles are on the right side of the diagram and therefore we have an exponentially rising sine until we clip it, as you said for instance by a diode.

But my understanding problem has a theoretical nature: The Barkhausen criterium needs an amplitude of 1 and also requires a phase of n*2*pi (n=0, 1, 2, .....) degrees, hence has to be real-valued.....this means it cannot be on the imaginary axis => it cannot be a perfect sine without exponential growth or damping according to the Barkhausen criterium................if we fulfill the criterium we can only have a pole on the right side.

As I understand the criterium for me it means we only can have theoretically an exponential rising sine but not an undamped or damped sine. where is my mistake?
Hello again,

One of the things you should know about the Barkhausen is that just because a circuit meets that criterion that does not mean that it will oscillate, it just means that it is capable of oscillation.
So a circuit that does not meet that criterion will not oscillate, but a circuit that does meet that criterion may or may not oscillate.

But we need to get more numerical about this. You should state exactly what you mean when you say that "it" has to be real valued. What exactly are you saying has to be real valued?
The transfer function for a well known phase shift oscillator clearly has a pair of complex poles that do not go away somehow. The pair lie right on the jw axis.
With the right gain, the exponential damping goes away (becomes equal to 1) so there is no rise or falling of the amplitude but is steady at one constant level.

So what is it that you are saying has to be "real valued"?

#### sinatra39

Joined Oct 18, 2018
15
Hello again,

One of the things you should know about the Barkhausen is that just because a circuit meets that criterion that does not mean that it will oscillate, it just means that it is capable of oscillation.
So a circuit that does not meet that criterion will not oscillate, but a circuit that does meet that criterion may or may not oscillate.

But we need to get more numerical about this. You should state exactly what you mean when you say that "it" has to be real valued. What exactly are you saying has to be real valued?
The transfer function for a well known phase shift oscillator clearly has a pair of complex poles that do not go away somehow. The pair lie right on the jw axis.
With the right gain, the exponential damping goes away (becomes equal to 1) so there is no rise or falling of the amplitude but is steady at one constant level.

So what is it that you are saying has to be "real valued"?
Hello and thank you,

for me it clearly doesnt have a pair that lies exactly on the imaginary axis. The transfer function in general is a complex number, that means has a fraction of a real number and imaginary number in the form H(s) = a+ jb. If I now want to condition |H(s)|=1, and tan^-1(b/a) = 0 (phase 0) then it doesnt result in a pair of purely imaginary numbers but a real number with amplitude 1 and completely on the real axis, since the phase shift is 0.

#### LvW

Joined Jun 13, 2013
1,279
is it? because in the complex plane a phase of 0 or 360 degrees means the value is real, so it can not be on the imaginary axis. no?
Sinatra - I am afraid you have not he correct understanding of the term "loop hain".
I repeat: When the loop gain LG is real and fulfills Barkhausens criterion (LG=1) the CLOSED-LOOP has a pole pair on the Im axis !!

#### sinatra39

Joined Oct 18, 2018
15
Sinatra - I am afraid you have not he correct understanding of the term "loop hain".
I repeat: When the loop gain LG is real and fulfills Barkhausens criterion (LG=1) the CLOSED-LOOP has a pole pair on the Im axis !!
Hey!

Please check the attached picture. If I have a closed loop gain Hloop(s) as in the picture and the Barkhausen criterium gives me a real valued 1 for the loop gain, then this is at the same time the pole of Hloop(s) and so the pole is not a pair on the imaginary axis but a real valued 1 instead! no?

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#### MrAl

Joined Jun 17, 2014
8,363
Hey!

Please check the attached picture. If I have a closed loop gain Hloop(s) as in the picture and the Barkhausen criterium gives me a real valued 1 for the loop gain, then this is at the same time the pole of Hloop(s) and so the pole is not a pair on the imaginary axis but a real valued 1 instead! no?
Hello,

The denominator is 1+H1*H2 because there is negative feedback.

Try this...
Set H1=K/(s+1)^3
Set H2=1

Using negative feedback the transfer function is:
K/(K+s^3+3*s^2+3*s+1)

the denominator of the transfer function is:
(K+s^3+3*s^2+3*s+1)

Now set K=8 and see what the poles look like.

To check out the overdamped version set K=7, and to check out the underdamped version set K=9.
K=8 is the magic number forward gain for this oscillator.

If you are questioning |H1*H2|=1 then recall that is the amplitude of a sine wave in sinusoidal analysis which is real.
A*sin(wt)
where A is the amplitude.

Last edited:

#### sinatra39

Joined Oct 18, 2018
15
Hello,

The denominator is 1+H1*H2 because there is negative feedback.

Try this...
Set H1=K/(s+1)^3
Set H2=1

Using negative feedback the transfer function is:
K/(K+s^3+3*s^2+3*s+1)

the denominator of the transfer function is:
(K+s^3+3*s^2+3*s+1)

Now set K=8 and see what the poles look like.

To check out the overdamped version set K=7, and to check out the underdamped version set K=9.
K is the magic number forward gain for this oscillator.

If you are questioning |H1*H2|=1 then recall that is the amplitude of a sine wave in sinusoidal analysis which is real.
A*sin(wt)
where A is the amplitude.
Thank you very much! Now I understand! I just had a short wire in my brain. Thank you so much!!