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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Where is -ve resistance here? Nov 20^th, 2012, 6:18am Dear all, I know every oscillator built from -ve resistance concept, i was able to find -ve resistance in all oscillator except integrator and Schmitt trigger oscillator.I am sure there is some way to find, can any one point me how to find it in the following fig. Thanks, Raj.
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buddypoor

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Re: Where is -ve resistance here?
Reply #1 - Nov 20^th, 2012, 6:43am

raja.cedt wrote on Nov 20^th, 2012, 6:18am:

Dear all,
I know every oscillator built from -ve resistance concept, i was able to find -ve resistance in all oscillator except integrator and Schmitt trigger oscillator.I am sure there is some way to find, can any one point me how to find it in the following fig.
Thanks,
Raj.

Raj, you strictly must discriminate between
(a) linear/harmonic oscillators of at least 2nd order (pole pair in the right s-plane), and
(b) relaxation generators (I avoid the term "oscillator"), which are of 1st order only.

Your circuit belongs to type (b) and there is no equivalence with a negative-resistance concept. OK?

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nrk1

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Re: Where is -ve resistance here?
Reply #2 - Nov 20^th, 2012, 9:15am

buddypoor wrote on Nov 20^th, 2012, 6:43am:

Raj, you strictly must discriminate between
(a) linear/harmonic oscillators of at least 2nd order (pole pair in the right s-plane), and
(b) relaxation generators (I avoid the term "oscillator"), which are of 1st order only.

Your circuit belongs to type (b) and there is no equivalence with a negative-resistance concept. OK?

This distinction is correct. But just want to point out that a schmitt trigger is a second order system in the sense that there are two state variables-the capacitor voltage and the state of the schmitt trigger. I think it is not possible to make an oscillator with a single state variable.

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buddypoor

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Re: Where is -ve resistance here?
Reply #3 - Nov 20^th, 2012, 10:52am

nrk1 wrote on Nov 20^th, 2012, 9:15am:

This distinction is correct. But just want to point out that a schmitt trigger is a second order system in the sense that there are two state variables-the capacitor voltage and the state of the schmitt trigger. I think it is not possible to make an oscillator with a single state variable.

According to my knowledge - the order of a system is determined by the degree of the transfer functions denominator; and this requires at least two capacitors. I was not referring to the number of state variables.

B.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Where is -ve resistance here? Reply #4 - Nov 20^th, 2012, 11:05am Hello buddypoor and nrk1, thanks for the reply. So i guess relaxation generators don't obey Barkhason condition. �Correct me if i am wrong. Thanks, Raj.
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buddypoor Community Fellow Offline Posts: 529 Bremen, Germany	Re: Where is -ve resistance here? Reply #5 - Nov 20^th, 2012, 11:16am raja.cedt wrote on Nov 20^th, 2012, 11:05am: Hello buddypoor and nrk1, thanks for the reply. So i guess �relaxation generators don't obey Barkhason condition. �Correct me if i am wrong. Thanks, Raj. Of course, you are correct. Barkhausen applies to sinusoidal oscillators only.
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Lex

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Re: Where is -ve resistance here?
Reply #6 - Nov 21^st, 2012, 12:46am

nrk1 wrote on Nov 20^th, 2012, 9:15am:

buddypoor wrote on Nov 20^th, 2012, 6:43am:

Raj, you strictly must discriminate between
(a) linear/harmonic oscillators of at least 2nd order (pole pair in the right s-plane), and
(b) relaxation generators (I avoid the term "oscillator"), which are of 1st order only.

Your circuit belongs to type (b) and there is no equivalence with a negative-resistance concept. OK?

This distinction is correct. But just want to point out that a schmitt trigger is a second order system in the sense that there are two state variables-the capacitor voltage and the state of the schmitt trigger. I think it is not possible to make an oscillator with a single state variable.

You are correct. At least two state variables are needed. It's a mathematical principle that goes beyond EE and can be found in mechanical engineering, civil engineering, etc..

About the relaxation oscillator: its poles are moving back and forth on the s-plane. At the moment the sign changes, the poles are complex, and at that moment in time you could apply the Barkhausen criterion. If it doesn't suffice, the relaxation oscillator will never switch, and the oscillation is stopped.

Considering the naming (generator/oscillator): to my opinion, the circuit produces a stable period and is therefore rightfully called an oscillator. For some 'so called' oscillators on the other hand that can have chaotic states, I'd be more reluctant.

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buddypoor

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Re: Where is -ve resistance here?
Reply #7 - Nov 21^st, 2012, 1:21am

Lex wrote on Nov 21^st, 2012, 12:46am:

...................
About the relaxation oscillator: its poles are moving back and forth on the s-plane. At the moment the sign changes, the poles are complex, and at that moment in time you could apply the Barkhausen criterion.

Hi Lex, I have some problems with the last sentence. Please, can you help and explain a bit?

At first, if it would be true that we really could apply Barkhausen�s rule "at the moment the sign changes" - I ask myself what could be gained by such an application.
But, in general I have some doubts if this is even possible. Barkhausen�s condition is based on the loop gain of a system with feedback.
And the definition of loop gain is based on a linear (or linearized) system in the steady-state mode. And I doubt if these conditions are fulfilled at the moment of "switching".

Thanks and regards
B.

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nrk1

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Re: Where is -ve resistance here?
Reply #8 - Nov 21^st, 2012, 3:52am

buddypoor wrote on Nov 20^th, 2012, 10:52am:

According to my knowledge - the order of a system is determined by the degree of the transfer functions denominator; and this requires at least two capacitors. I was not referring to the number of state variables.

B.

I think a transfer function as a rational polynomial in s can be determined only for a linear time invariant system. In that case, the order of the denominator is equal to the number of state variables which equals the number of (independent) capacitors. The present circuit, with its severe nonlinearity, cannot be described by a transfer function of that sort at all. So counting the number of state variables gives a more general notion of order.

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buddypoor

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Re: Where is -ve resistance here?
Reply #9 - Nov 21^st, 2012, 5:09am

nrk1 wrote on Nov 21^st, 2012, 3:52am:

I think a transfer function as a rational polynomial in s can be determined �only for a linear time invariant system. In that case, the order of the denominator is equal to the number of state variables which equals the number of (independent) capacitors. The present circuit, with its severe nonlinearity, cannot be described by a transfer function of that sort at all. So counting the number of state variables gives a more general notion of order.

OK - I can follow your argumentation and I agree.
However, it is to be mentioned that linear and time-invariant operation of the circuit is also possible if we restrict the positive feedback to a very small value. But that`s not the case, of course, for the dimensioning, which leads to the waveform as shown in post#1.

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raja.cedt Senior Fellow Offline Posts: 1516 Germany	Re: Where is -ve resistance here? Reply #10 - Nov 21^st, 2012, 10:30am hello lex, I didn't understand your post, because of my limited knowledge in the field of non-linear dynamics? Can you explain little brief what is chaotic states. Thanks, Raj.
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Lex

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Re: Where is -ve resistance here?
Reply #11 - Nov 22^nd, 2012, 7:33am

buddypoor wrote on Nov 21^st, 2012, 1:21am:

Lex wrote on Nov 21^st, 2012, 12:46am:

...................
About the relaxation oscillator: its poles are moving back and forth on the s-plane. At the moment the sign changes, the poles are complex, and at that moment in time you could apply the Barkhausen criterion.

Hi Lex, I have some problems with the last sentence. Please, can you help and explain a bit?

At first, if it would be true that we really could apply Barkhausen�s rule "at the moment the sign changes" - I ask myself what could be gained by such an application.
But, in general I have some doubts if this is even possible. Barkhausen�s condition is based on the loop gain of a system with feedback.
And the definition of loop gain is based on a linear (or linearized) system in the steady-state mode. And I doubt if these conditions are fulfilled at the moment of "switching".

Well Barkhausen criterion is a mathematical condition to have linear circuits oscillate, right? I agree that if the switching cannot be modeled linearly it becomes difficult. But a Schmitt trigger opamp can act fairly linear in such a regime. But in general I agree, it depends on the implementation. On the other hand, the theory is not completely covering the harmonic oscillator either, to my opinion. E.g. for a harmonic LC oscillator with some decent amplitude, the transistor's gm changes continuously (modulating the 'steady state' behavior). But in summary, in practice I agree with you. ;-)

@raja: there is quite some text out there on chaotic oscillators. I would say, start at wikipedia1, wikipedia2?

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