Electronic – Are oscillators always non linear

control theoryoperational-amplifieroscillatorsignal-theory

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?

Phase-shift Oscillator

Thanks!

Best Answer

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\$.