# Are oscillators always non linear?

From linear systems theory, self-excited sustained oscillations are only possible by means of a marginally stable system, where poles are located exactly on the imaginary axis. However, such a situation is a impossibility in the real world, as tiny parameter deviations would cause the system to go either stable or unstable. And if it goes unstable, than I assume that it would reach any kind of physical limitation (for instance, saturation) that would make it lose its linearity property.

However, reading about oscillators circuits using opamps, I don't really know where is the non linearity. For instance, is this phase-shift oscillator non linear (taken from Opamps for everyone)? How can I know it? Thanks!

The loop is operating in the linear region for most of the time and only transitions into the non-linear region momentarily, at the peaks of the sine wave, to correct for random changes in sine wave amplitude. The quality of the output sine wave (i.e. the extent of transitions into non-linear mode) depends on ensuring the magnitude of the amplifier linear gain is the inverse of the 3rd order phase shift network gain at the critical frequency, thus giving unity loop gain (but in practice the amplifier gain value is slightly higher than this minimum requirement, as discussed later). Any clipping of the sine wave peaks is filtered out by the 3rd order low pass filter that forms the phase shift circuitry.

For the closed loop to maintain a stable sinusoid at a frequency, $\omega_0$, the open loop gain must be unity and the open loop phase angle must be $\small -180^o$ (with the negative feedback amplifier providing the necessary additional $\small -180^o$ of phase shift). Any other condition will not give a steady state oscillation. Therefore each 1st order lag must be contributing $\small -60^o$ of phase shift.

To calculate the frequency, $\small \omega_0$, at which this occurs, let $\small \tau =RC$, then:

$$\small G(j\omega)= \frac{1}{1+j\omega\tau}$$ $$\small G(j\omega_0)= \frac{1}{1+j\omega_0\tau}$$ $$\small \phi=-arctan(\omega_0\tau)=-60^o$$ $$\small\therefore \omega_0\tau=\sqrt{3}$$ $$\small \omega_0=\frac{\sqrt 3}{\tau}={10}^4\:rad\:s^{-1}$$

The corresponding gain at this frequency is:

$$\small \mid G(j\omega_0)\mid=\frac{1}{\sqrt{1+{(\omega_0\tau)^2}}}=\frac{1}{2}$$

Now, there are are three cascaded 1st order lags, so the overall phase and gain are: $$\small\phi=3\times\:(-60^o)= -180^o$$ and $$\small\mid G\mid = {\left(\frac{1}{2}\right)^3}=\frac{1}{8}$$

So, if everything were ideal, we'd arrange for the amplifier to have a gain of $\small K_A=-8$ and the circuit would then oscillate at: $\small \omega_0=\frac{\sqrt 3}{RC}\:rad\:s^{-1}\:$.

Unfortunately(?), things are not ideal, so the amplifier gain magnitude is arranged to be slightly larger, hence, in the circuit, we have $\small K_A=-\large\frac{1.5M\Omega}{180k\Omega}\small\approx -8.3$.

Given a nominal amplifier gain of $\small -8.3$, the nominal closed loop will oscillate with frequency $\small \omega_0$ and the sine wave amplitude will be just sufficient to give unity loop gain. To achieve this the amplifier is slightly saturated thereby reducing its effective gain to $\small -8$. This gain reduction occurs automatically in the saturated region since $\small K_{A_{eff}}=\large \frac{V_o}{V_i}=\frac{V_{sat}}{V_i}$, and as $\small V_i$ increases the effective gain decreases.

If, now, random circuit variations occur to alter the sine wave amplitude, the amplifier gain adjusts itself to compensate, by virtue of the effective gain being inversely proportional to amplitude.

Thus, the resultant sine wave amplitude at the phase shift network output is an ostensibly stable sine wave with amplitude of approximately $\small\frac{2.5\:V}{8}$, where the saturation levels are assumed to be $\small \pm 2.5 V$.

• Chu, in the end, just the fact that clipping results in a system where the gain $K_{a}$ is not constant ($V_o = K_A(V_i) * V_i$) is already enough for us to state that this a non linear system? Actually, my guess is that every oscillator that has any sort feedback circuit to dynamically adjust effective gain depending on the amplitude will be non linear. And it seems to me, from all the answers in this question, that all oscillators work from this principle - thus the answer to the question is that all oscillators are non linear. Would that be correct? – SuperGeo Aug 2 '17 at 21:57
• An oscillator is not a system in the strict definition, since it doesn't have an input signal. But as it uses a nonlinear element to exhibit its operational features, it is a non linear device. – Chu Aug 2 '17 at 22:20
• Can't we consider it a system with noise as input? – SuperGeo Aug 2 '17 at 22:23
• Not really. Thermal noise starts the process and this can hardly be considered an input. Is it important that something may or may not be classified as a system? It's only a name. There will always be grey areas. – Chu Aug 3 '17 at 7:58

A practical oscillator has to be designed taking into account component tolerances.

In your example the ratio of RG to RF needs to be such that the gain round the loop is unity.

Because of the tolerance in RG, RF and the capacitors there has to be some extra gain designed in so that in the worst case the gain is at least unity.

In the typical case this means the gain will be greater than unity so the oscillations will keep building up until one of the amplifiers saturates, that is becomes non-linear, and the amplitude stabilizes. Since one of the amplifiers is saturating its output will not be a sinewave.

In your example saturation provides the amplitude stabilization but other techniques can be used that produce less distortion.

An old technique is to use a thermal device that varies its characteristics with the amplitude of the oscillation and in so doing alters the loop gain to stabilize the oscillations. The first product from Hewlett Packard in 1939, The HP200, used an incandescent lamp as the control element. Thermistors have been used in a similar fashion.

Electronic means can also be used such as incorporating an Automatic Gain Control (AGC) in the loop.

• In the end, just the fact that clipping results in a system where the gain is not constant ($V_o=K_A(V_i)∗V_i$) is already enough for us to state that this a non linear system? Actually, my guess is that every oscillator that has any sort feedback circuit to dynamically adjust effective gain depending on the amplitude will be non linear. And it seems to me, by using clipping, or AGC, or a thermal device, that all oscillators work from this principle - thus the answer to the question is that all oscillators are non linear. Would that be correct? – SuperGeo Aug 2 '17 at 22:25

Any oscillator needs that little bit extra gain to start. Oscillations build cycle after cycle. Your example circuit is powered with a +5v supply. Oscillation peak-to-peak amplitude certainly cannot exceed that supply voltage, so peaks are clipped - there's your non-linearity that keeps amplitude from growing forever.
It is not a particularly good oscillator, because attaching a load to "Vout" may reduce loop gain enough that oscillations slowly die out. And a load resistance will also change the phase of that last RC stage, affecting frequency. Better to take output from any one of the op-amp outputs.
In designing an oscillator, enough loop gain is required to ensure that worst-case component tolerances, and environmental effects, and accounting for power delivered to a load still yield a loop gain > 1. That may require a loop gain considerably higher than 1, which in turn will require considerable non-linearity to keep amplitude stable once the oscillator has started.

• Regarding the idea of attaching a load, the OP used $\frac{1}{4}$ package devices, so the OP has an extra for the added final buffer! ;) (Perhaps the OP expected you to "see it" added at the end there?) – jonk Aug 1 '17 at 23:23

Usually clipping limits the amplitude, but if you want to build a low-distortion oscillator, then every component must stay in its linear range. Clipping is not allowed. Which means we have a problem:

self-excited sustained oscillations are only possible by means of a marginally stable system

The Wien Bridge oscillator is an example of a solution. It uses an AGC (Automatic Gain Control) which tunes the gain of the circuit to keep the output amplitude constant, well below saturation. Oscillation happens via phase shift.

Back in the day, this was implemented with a lightbulb, whose impedance increases as it is warmed up by the signal.

These days, more modern solutions exist, such as JFETs used as resistor. Here is a rather complex example. The mess in the left side detects the signal envelope, and adjusts the JFET's resistance. The actual oscillator is the two opamps in the top right corner.

In the HP3033 sin generator, producing DC to 13Mhz in 0.1Hz steps, the primary oscillator was differential long-tail pair thus the input power (the tail current) was well controlled. Part of the feedback network was a bleedoff network, with losses increases as the square of the voltage.

Thus that oscillator was totally linear.

• Why the downgrade? have you examined that oscillator? I have. With the purpose of learning what set the amplitude. Re-read the sentences in my first paragraph in the answer. I'll later provide the schematic. – analogsystemsrf Aug 15 '17 at 17:59

Van der Pol oscilator is a nonlinear oscilator with nonlinear damping, cf, khalil (nonlinear control, page 335-337).