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Here is a link of an article I've been reading. http://www.analoguehaven.com/moog/thereminblack/hotrod.pdf

In page 4 there is a circuit schematic. Notice there is a part called variable pitch oscillator and a part called fixed pitch oscillator. At the output of each of the two oscillators mentioned there is a capacitor. At page 6, the article states that:

C2 and C6 combine the pitch oscillator signals, while D4, R23, R24, and C23 extract the difference frequency. C2 and C6 also provide weak coupling between the two pitch oscillators, which has the effect of synchronizing the pitch oscillators when their frequencies get very close together. This has the desirable effect of providing a stable 'zero beat', so that the instrument, once properly tuned, is silent when the player steps away from it.

Question:

What I would like to understand is how the two output signals of the two oscillators are combined with the use of the capacitors C2 and C6.

Is it correct to say that the two capacitors work as modulator? I'm saying this because the produced signal as you can see goes through a detector. Could you explain what exactly happens at the capacitors or how they combine the signal?

Thank you.

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C2 and C6 are AC coupling capacitors. An AC signal on their input appears on the output. If you just connected these two capacitors together, they would act as a voltage divider - the voltage at the connecting node is the mean of the voltages at the other side of the capacitors. The "weak coupling" comes about from the fact that since the point connecting them is not pinned, there will be a current at the input of C2 when the input to C6 changes and vice versa.

Finally, there is a diode to ground, and a pair of resistors that divide the voltage generated. This diode is a nonlinear element that will result in some rectification and mixing. Your description in the question describes what it does.

R23, R24 and C23 obviously provide a low pass filter of the voltage on the diode - and that is the input to the VCA. The time constant is fairly short - RC = 15 us (since you have to consider the 4.7k and 10k resistors in parallel for AC purposes).

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  • $\begingroup$ Thanks for the answer. In the end you describe the R23,R34,C23 circuit as a low pass filter. I would like to ask why, because from my search I understood that a detector extracts the envelope of a modulated signal(two signals combined). So here do we have the extraction of the envelope or a low-pass filter(or could it be somehow the same?). Also, could you explain the term pinned. You say in the first paragraph: " The "weak coupling" comes about from the fact that since the point connecting them is not pinned...". I'm not familiar with the term in English. Thank you. $\endgroup$ – Constantine Black Jul 24 '15 at 8:49
  • $\begingroup$ The low pass filtering follows the extraction of the envelope - there are some "bumps" in the envelope that are smoothed out. Pinning as I used it means forcing the voltage to be fixed (for example if the two capacitors were joined at the inverting input of a current follower (virtual earth) they would be "pinned" as the voltage of the common point would not change ("pinned down" to a fixed value). $\endgroup$ – Floris Jul 24 '15 at 11:09
  • $\begingroup$ Excuse me,but where the envelope extraction happens ? $\endgroup$ – Constantine Black Jul 24 '15 at 11:57
  • $\begingroup$ When you feed the sum of two sine waves into a nonlinear element, the sum and difference frequencies appear. A low pass filter ensures only the difference signal makes it through to the next stage. That is the "envelope extraction". $\endgroup$ – Floris Jul 24 '15 at 12:26

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