I am going to try to explain better my question. Considering a simple control scheme made up of a forward-path amplification\$A\$ and a feedback-path frequency dependent attenuation \$\beta(j\omega)\$, assuming a "positive" feedback, such that the loop gain is \$T(j\omega)=A\beta(j\omega)\$ then an oscillation can be kept constant in time if certain conditions are met for a precise frequency value \$\omega_0\$, namely the Barkhausen criteria: \$T(j\omega_0)=1\$ or equivalently \$|T(j\omega_0)|=1\$ and \$\angle T(j\omega_0)=0\$. In this way, a sinusoidal signal follows the full loop path and comes back unchanged, without any amplitude or phase modifications.

Now, my questions is related to the phase condition: I have understood that the magnitude condition has to be satisfied in order not to have divergent or damped oscillations (no overall amplification or attenuation per each revolution of the loop),whereas why must the phase be \$0\$? I mean, if it is so, the signal is not delayed after each cycle but what happens if it is not so? Is the signal being shifted every cycle? What is then the resulting output waveform, something like a sliding sinusoid?

Since there are no signals being summed, like it happens in a classical feedback-system with an input, one cannot refer to constructive/destructive interference that would occur due to the phase shift between the two signals added; thus the phase shift of the loop gain \$\angle T(j\omega_0)\$ cannot bring the oscillation to grow up or drop if the magnitude is still \$1\$. Am I right? In other words, if the phase condition is not met, then no steady state oscillation can be produced, but anyway it is not possible to have oscillations that tend to increase or decrease.

EDITED SECTION: As additional doubt linked to the previous question, if the Barkhausen criteria are exactly met for a certain frequency, shouldn't one say something more also about the other frequencies, so that all of them are damped (band-pass behaviour)? In other words, how can one be sure that no other frequencies get amplified and enlarged? (even though, saying this I am not taking into account that at such frequencies the phase would not be 0 too, so I cannot figure out which is the effect of loop gain greater or equal than 1 but phase whatever so, the problem is degenerating again into the phase criterion).

  • \$\begingroup\$ My answer to your edited section is (as mentioned already in my detailed answer): Self-excitement is possible only for fa frequency where the loop phase is zero (resp. 360deg). All other frequencies need not to be "damped" because they are not generated from the beginning. \$\endgroup\$
    – LvW
    Jul 19, 2016 at 15:17
  • \$\begingroup\$ Thank you for your patience and sorry for the delay in the answer, this your last statements perfectly aim at the very core of my puzzlement: the system selects only the angular frequency at which the loop gain phase is zero. This is really enlightening! As last step, if you don't mind, can you help me to understand this theoretical passage? i.e. the reason why the frequency suitable to produce oscillation is only the one meeting the phase condition. \$\endgroup\$
    – Vexx23
    Jul 21, 2016 at 19:02
  • \$\begingroup\$ I am sure this passage is trivial but somehow I cannot give it a very persuasive explanation, I'm just continuing to think to a signal which is brought back after a complete revolution of the loop with a phase shift, maybe this 'abstract' way of thinking is right the unfounded passage. \$\endgroup\$
    – Vexx23
    Jul 21, 2016 at 19:04
  • \$\begingroup\$ Vexx23 - I think, I know what you mean. Of course, an answer is possible based on mathematical derivations (for the frequency as well as the time domain using the LAPLACE transformation). However, such explanations are not possible at this place. Perhaps the followinmg helps: A working oscillator can be seen as an amplifier that creates its own input signal. That means (as an example): A 3V output amplitude must be attenuated and appear at the input of a gain of 3 amplifier (as is the case for the WIEN oscillator) with an amplitude of 1V. And this must hold for one single frequency only. \$\endgroup\$
    – LvW
    Jul 21, 2016 at 20:08
  • \$\begingroup\$ @LvW That's just as I feared, the explanation for the frequency with zero phase is absolutely clear. I am not even sure that someone has ever taken the trouble or looked at the issue of developing such an analysis. Anyway you have helped me so much, I am glad this question could be helpful for others. \$\endgroup\$
    – Vexx23
    Jul 26, 2016 at 23:10

1 Answer 1


I think, I can agree to everything you have explained. That means: Yes, the Barkhausen condition may be applied to steady-state oscillations only. With other words - if the amplitude condition is met (|T(jωo)|=1 or even >1) and the phase condition is NOT met, we can have decaying or even rising oscillations immediately after power switch-on. But - at which frequency? In reality we have the following situation:

The system will always try to oscillate at a frequency where the phase condition is met. This is the only possibility to allow self-excitement of a system with positive feedback. And this frequency very often will be NOT exactly the frequency (wo) where the amplitude condition is fulfilled per design (or "over-fulfilled"). Hence, the system will try to oscillate at another frequency (let`s say w1). If the amplitude condition is still fulfilled at this new frequency w1 - we will have rising oscillation amplitudes (until any limiting mechanism will come into play). However, if the loop gain magnitude will be <1 at the frequency w1 we will observe damped oscillations (or even no oscillations).

Answer to your second question: If Barkhausens condition (amplitude and phase) is fulfilled for one single frequency only, there is no chance that oscillations at other frequencies are build up.

EDIT 1(example)

Here is a very illustrative example which exactly meets the subject of your question:

There is a harmonic oscillator consisting of two active opamp integrators in a closed loop (one is inverting and the other one is non-inverting). Hence, under ideal conditions we have a loop phase of 0 deg over a rather broad frequency range. However, for real opamps the loop phase will cross the 0 deg line at one single frequency fo only. We will observe that the loop gain magnitude will cross the 0 dB line at another frequency f1 which is always larger than fo (f1>fo). This is due to the opamps finite open-loop gain and corresponding phase shift.

That means: The loop gain magnitude at the frequency fo will be > 0 dB and the Barkhausen condition is (over-)fulfilled. The circuit oscillates at a frequency fo (where the phase condition is fulfilled). Note that for all frequencies below fo the loop gain is even larger than at f=fo. Nevertheless, the frequency for steady-state oscillations is f=fo.

Final remark: In particular, this oscillator topology is most suitable for explaining why and at which frequency a linear oscillator will oscillate.

EDIT 2 (Comment to the example cicrcuit):

Perhaps the following considerations may further help to answer your (very intelligent) question. If you try to use one of the available simulation programs you will find that the two-integrator oscillator will NOT continuously oscillate in case of simplified first-order models for the opamps. Why not? Because the loop phase will meet the 0 deg line for infinite frequencies only. For all finite frequencies, the loop phase is very close to 0 deg but not exactly zero (second-order system only). For this reason, after power switch-on we will have slowly decaying oscillations.

To allow continuous oscillations, a simple resistor across the feedback capacitor will turn one of the integrators into a two-pole system - and the loop gain now is of 3rd order with zero loop phase at a finite frequency.

  • \$\begingroup\$ As regards the second part of your reasoning, I think it is rather clear, if there exists a frequency such that the phase shift is 0 then such frequency will appear on the output, even if the gain at that frequency is not good to mantain the oscillation, so either damped or divergent. But why, with magnitude 1 and phase not 0, does the oscillation at \$\omega_0\$ "decay"? (shouldn't it be a totally different signal and how can I figure it out what does it look like?) About the last question, shouldn't we guarantee that all the other frequencies are not amplified (e.g. with magnitude > 1)? \$\endgroup\$
    – Vexx23
    Jul 19, 2016 at 14:14
  • \$\begingroup\$ Perhaps my EDIT will answer this question? If not, place another comment. \$\endgroup\$
    – LvW
    Jul 19, 2016 at 14:25
  • \$\begingroup\$ Regarding your last sentence: We are not able to guarantee this because (for practical reasons) we must design the loop gain always >1 at the wanted frequency (tolerances and safe start of oscillations) \$\endgroup\$
    – LvW
    Jul 19, 2016 at 14:31
  • \$\begingroup\$ Sorry if I'm proving unable to make my doubt clear: your example is magnificent in explaining how the oscillations are triggered, naturally selecting a frequency such that the phase is exactly 0 while the magnitude is over-sized so that the oscillations are allowed to grow up till a limitation gets activated. Referring to your example, why does the frequency \$f_1\$ (at which the gain is 1 but the phase is slightly lower) generate anyway an oscillation wich (exponentially?) decay in the very first instants after switch on? I mean, why an oscillation and not else? How can I prove/see it? \$\endgroup\$
    – Vexx23
    Jul 19, 2016 at 14:50
  • \$\begingroup\$ As regards the last part of my question and of your answer, I have added some detail to the question, in the hope that my doubt can prove itself either well-founded or totally unfounded (more likely). \$\endgroup\$
    – Vexx23
    Jul 19, 2016 at 14:59

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