Will a noise-free oscillator produce sustained oscillations?

[jasmin_123]

Will a noise-free oscillator produce sustained oscillations?

Where its poles will be located after adding the noise?

 

[subharpe]

Re: Will a noise-free oscillator produce sustained oscillati

Ideally noise free oscillator cannot start oscillating. Any oscillator needs wideband noise to pick it's tuned frequency for oscillation.

 

[FvM]

Re: Will a noise-free oscillator produce sustained oscillati

Considering a power supply, also a DC offset could source initial energy to the resonator. Examining start-up behaviour of real oscillators, this is actually a more common mechanism than noise, I think.

 

[subharpe]

Re: Will a noise-free oscillator produce sustained oscillati

Hi, sorry if I am wrong. But what I understand is that power supply is an essential requirement for the amplifier section in the oscillator which compensates for the losses in the circuit. But without a input to the tank circuit that contains the frequency of tuning can it start oscillating? Suppose you have a tank circuit with L and C. This is a oscillatory circuit that needs a pulse at the input to start oscillating. The pulse contains all frequencies of same power and the tank circuit picks up the particular tuning frequency for resonance. The other frequencies die outbecause of sharpness of response of the tuner.
But in practice the oscillation will not be sustained because of non-zero resistance in the circuit. Therefore the loss has to be compensated and an amplifier is required or a negative resistance device is required for counter-balancing the resistance in the circuit. But essential thing to start oscillation is a signal having the resonant frequency as one component.

 

[teteamigo]

Re: Will a noise-free oscillator produce sustained oscillati

This is a oscillatory circuit that needs a pulse at the input to start oscillating. The pulse contains all frequencies of same power and the tank circuit picks up the particular tuning frequency for resonance. ... But essential thing to start oscillation is a signal having the resonant frequency as one component.

It was an conclusion that i reached when studied first time oscillators in electronics, after learned systems theory.

But I was impressed when i mounted the first oscillator (practical circuit), and there wasn't input signal. And nobody (teacher, coleagues) talked to me about input signal, noise in oscillators...


But I already had time to test all my doubts, in simulation (ring oscillator), practical circuits (LC pierce, phase shift, ...).

 

[jmachado]

All practical circuits have noise (thermal or flicker noise) this is why you do not require an input source to your oscillator. Then you need to guaranty your circuit as you refer have enough gain to sustain oscillation. Your LC tank will do the rest.

 

[jasmin_123]

To make the long story short:

1. A noise-free oscillator will produce sustained oscillations as a
response to its initial conditions. Its forced response is zero.

2. After adding noise, the oscillator poles will shift to the left of the jω axis.
Its natural response will decay, and its forced response will define the
oscillations.

 

[FvM]

Re: Will a noise-free oscillator produce sustained oscillati

Noise (a voltage or current source) and poles are unrelated properties, I think.

 

[teteamigo]

Re: Will a noise-free oscillator produce sustained oscillati

To make the long story short:

1. A noise-free oscillator will produce sustained oscillations as a
response to its initial conditions. Its forced response is zero.

Correct.

Read some papers about Injection Locking to force the oscillator.

Load, Source Pulling

2. After adding noise, the oscillator poles will shift to the left of the jω axis.
Its natural response will decay, and its forced response will define the
oscillations.

Correct.

Convince to you that the noise don't control the poles as you say above: "shift to the left of jw".



In pendulum we have a natural response equal to forced response.

 

[jasmin_123]

Noise (a voltage or current source) and poles are unrelated properties, I think.

It is assumed that the amp. in the oscillator is not 100% linear.
When noise is added, and the oscillation amplitude grows, the amp. gain becomes lower,
the loop gain also decreases, and the poles move left. The unstable oscillator becomes a
stable, frequency selective amplifier.

You can find this in books.

Added after 4 minutes:

teteamigo, again, I am not able to understand you.
For example: "In pendulum we have a natural response equal to forced response."
Could you please explain what do you mean by that?

 

[teteamigo]

Re: Will a noise-free oscillator produce sustained oscillati

teteamigo, again, I am not able to understand your language.
For example: "In pendulum we have a natural response equal to forced response."

Forgive my english!!!

Natural=Transient

Forced=Steady


Pendulum

Only one frequency of oscillation possible

by impulse

and by drive the pendulum with a force in phase with movement.

 

[jasmin_123]

teteamigo, I am sorry, but what you are writing about the pendulum transient and forced response is not exactly right.

 

[teteamigo]

Re: Will a noise-free oscillator produce sustained oscillati

teteamigo, I am sorry, but what you are writing about the pendulum transient and forced response is not exactly right.

The only thing wrong is about the amplitude.

In forced situation appointed by me, the amplitude of oscillation grows up, up, up,...


In natural response the amplitude remains constant (neglecting loss).

 

[FvM]

Re: Will a noise-free oscillator produce sustained oscillati

Hello,

It is assumed that the amp. in the oscillator is not 100% linear. When noise is added, and the oscillation amplitude grows, the amp. gain becomes lower, the loop gain also decreases, and the poles move left. The unstable oscillator becomes a stable, frequency selective amplifier.

I agree to a great extent, except for the noise, which is not the cause for growing amplitude rather than a >1 loop gain. This was also subharpes point, I think. Noise is usually discussed in literature as an initial impulse cause you need a nonzero initial oscillator amplitude to allow further increase. As I mentioned, watching start-up of a high Q oscillator, e. g. with a crystal, you see that initial amplitude in the observable exponential rise isn't nV scale as it would be from noise but mV or V in many cases.

I agree, that poles move when transitioning from unexcited state (no matter if it would be stable or not) to steady state.

Regards,
Frank

 

[jasmin_123]

"In forced situation appointed by me, the amplitude of oscillation grows up, up,
up,... "

teteamigo, this is right only if the frequency of the input signal (force) equals to
the pendulum resonant frequency, and the pendulum poles are located at the jw
axis.

FvM, let us assume that an oscillator with no noise starts from initial conditions and
reaches a steady state at a loop gain = 1. The poles will be located at the jω axis in this
case. Let us now add a white noise with a component at the resonant frequency of
the oscillator. This component will contribute a linearly increasing sinusoidal signal
to the oscillator output. Depending on the phase difference between this
signal and the osc. natural response, two scenarios are possible. (i) When the
above signals are in phase then the output will grow, the loop gain will drop
below 1, and the poles will shift to the left. (ii) When the above signals are
out of phase, the output will first decrease down to zero, loop gain will become >1,
and the poles will move right. Eventually, however, the much faster (almost
linearly) increasing forced response will increase the oscillator output from zero to
the value exceeding the natural steady state. This will decrease the loop gain
below 1 and shift the poles to the left of the jω axis.

Hope this clear.

 

[shabab]

i think no

 

[jasmin_123]

Re: Will a noise-free oscillator produce sustained oscillati

i think no

So, this means that a zero-input response of a linear circuit cannot
be a sinusoidal waveform with a constant amplitude.

Is this what you say?

 

[arebee]

Someone has probably already answered this, but here's my contribution.

No. Assuming zero IC's, it'll remain at its equilibrium point if there's no input, even if the equilibrium point is unstable.

Adding noise is (usually) modeled as adding an input at some point in the system (assuming its linear). Hence noise doesn't affect the transfer function or poles of it.

If the system is nonlinear, then the amplitude of the oscillation will eventually move the (linearized equivalent) poles (unless the noise itself is large enough to do so).

 

[melc]

Re: Will a noise-free oscillator produce sustained oscillati

Before this looong talk, maybe you define what is a noise-free oscillator ?
In practice there isn't any noise free device. If you like theory then things are different...

 

[jecyhale]

Re: Will a noise-free oscillator produce sustained oscillati

The pole is detemined by the circuit rather than noise, So I don't think it will move the pole.

And any signal can't change it too.