All opamps are ideal and Rx = 2R in the figure. The comparator is a nonlinear block which operates as follows: if its input Va < 0, its output is +1V. If Va > 0, its output is -1V. To simplify calculations, we can also assume the comparator output to be a square wave alternating between +1 and -1 at steady state.


How to find the amplitude of oscillation at Va and Vb? I have obtained the transfer function between Va and Vb, but I'm unable to get one between the comparator output and Va.

Also, what is the strength of the 3rd harmonic relative to the fundamental at Va and Vb? Using that result, how to find which of these outputs is “cleaner”?

  • \$\begingroup\$ @SacredMechanic..may I ask you: Where did you find this circuit? It is a very interesting one. The oscillation amplitude depends on the selected frequency, the parts of the inverting lowpass (left) and also on the comparator output. However, the quality of the sinusoidal output is excellent. The frequency is twice the cross-over frequency of the DEBOO integrator. \$\endgroup\$
    – LvW
    Jul 27, 2020 at 14:54
  • \$\begingroup\$ Correction: The amplitude does NOT depend on the frequency of oscillation \$\endgroup\$
    – LvW
    Jul 27, 2020 at 16:43

1 Answer 1


At the moment, all I can say is the following:

  • The circuit is an electronic model of a pendulum clock - that means: Each period it receives (from the comparator) an additional "kick" which can compensate the losses of the analog loop. In the literature, this principle is called "restoration of initial conditions".

  • The transfer function of the closed-analog loop is a second-order lowpass with a pole-frequency of app wp=6.28*160 rad/s and a pole quality factor Qp=10.

  • Hence, the step response of this closed-loop has a decaying function exp[-(wp/2Q)t]

  • Periodically, the comparator output (+- 1V) injects energy into the loop thereby compensating the losses (that means: working against the decaying property of the lossy loop).

  • This state of compensation (a kind of equilibrium) defines the final steady-state amplitude of the circuit. This amplitude strongly depends on the comparator output voltage. This "kick" signal may be as low as 1µVolt - the circuit still oscillates, but with a very ow amplitude.

EDIT: Are you still interested in a formula for the amplitude?

Comment:I must admit that I am rather surprised that - up to now - there is no further comment or answer. For my opinion, the shown circuit is very interesting in its function and opens the question: Is it really true that all oscillators are non-linear circuits?

EDIT2: The considerations for finding the oscillation amplitude are as follows:

  • The high-Q lowpass oscillations will decay (without any "kick") resulting in an amplitude difference

[Vmax- Vmax*exp(-t/tau)] = Vmax[1-exp(-t/tau)].

  • This loss must be compensated by the external "kick" (comparator voltage Vc) each half period. For this reason, we set t=1/2fo ; tau=wo/2Q (filter theory). In the design example Q=10 (Bode diagram for the closed.loop magnitude).

Equating the above expression with the amplified comparator voltage Vc*Ac, we can solve for Vmax and the result is Vmax=12.5 V. This is in full agreement with corresponding simulation.

The full swing comparator voltage is Vc=2V and the gain Ac for the comparator voltage is Ac=Zf/R with feedback impedance Zf=10R||(1/sC). (In our case: Ac=0.91).

  • \$\begingroup\$ Yes, I'm still interested in the formula for the amplitude of oscillation at Va and Vb. :-) \$\endgroup\$ Jul 30, 2020 at 7:10
  • \$\begingroup\$ see EDIT2 in my answer... \$\endgroup\$
    – LvW
    Jul 30, 2020 at 10:34

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