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I was browsing the web for the prior works on the PWM demodulator implementations, and I found patent US4408166A, "Pulse width modulation decoder" from 1982.

Circuit from patent

I cound not obtain the vo equations for fig 3 and 4.

Vo equation for fig3 fig3 eq Vo equation for fig4 fig4 eq

I would really appreciate any help to derrive any of these equations.

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  • \$\begingroup\$ Add a circuit image to your post. Do this because if the link dies your question becomes liable to subsidence and, you probably don't want that. In a wider context, questions should be self contained and rely only on information posted by the OP (yourself). \$\endgroup\$ Commented Oct 2 at 21:55
  • \$\begingroup\$ Thank you for the edits and the notices. I will keep them in mind next time I post anything. \$\endgroup\$ Commented Oct 2 at 22:30
  • \$\begingroup\$ your blue assumptions are wrong. A sample and hold is not a short circuit. that's the point here! And the assumption that your multiplicator also is just a constant feedthrough of the reference circuit makes no sense either, because now your assumption completely removes the actual input signal completely from the system. \$\endgroup\$ Commented Oct 2 at 22:49
  • \$\begingroup\$ About the multiplicator (the X sign enclosed with circle), somehow that symbol represents a switch. It is explained in the text of the patent. But about the sample and hold circuit, when it is sampling can’t i model it as a close circuit \$\endgroup\$ Commented Oct 2 at 23:00
  • \$\begingroup\$ yes, that's the same thing. PWM is on (1) or off (0). Multiplying with 1 or 0 is like connecting the reference signal or 0V to R1. This is really simple! \$\endgroup\$ Commented Oct 2 at 23:01

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Well that capacitor is just the central component of the opamp integrator, and an intergrator is what you'd need to convert a pulse width to an amplitude. So, this is really the minimal circuit anyone would expect! (I'm a fair bit surprised you could use that in 1982. This is how "any experienced technical person in the field would do", and that was usually a ground for exclusion from patentability in that era.)

So, yes. It is exactly what does the DC analysis here. You want to look up the basics of Opamp circuits for the theory how that works. If you are interested in how generally the conversion of simple Laplace domain control circuits to opamp circuits with feedbacks work, that's usually taught in courses called "basics of control engineering" or "electronic control systems".

The circuit from that patent is really a pretty typical entry-level exam question for these lectures; it's too much to explain in a stack overflow answer, but if you're an EE student, chances are high you'll encounter such a lecturer sooner or later.

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  • \$\begingroup\$ Thank you for the answer. I am still a bit skeptical about the integrator part. I assumed the sample & hold circuit would act as close circuit when PWM asserts. This would mean that the capacitor and the resistor will be in parallel and for DC the circuit will act as an inverter not an integrator. I will provide some drawing in my next comment \$\endgroup\$ Commented Oct 2 at 22:39
  • \$\begingroup\$ but it's not what it does in this patent. And sample and hold doesn't help you at all decoding PWM; you integrate over the pulse duration, because that's literally what PWM modulates. \$\endgroup\$ Commented Oct 2 at 22:49
  • \$\begingroup\$ I edited the post to include a schmatic about my question. I don’t why the integrator would work with the parallel resistance to the capacitor. I know that the resistor is there to mitigate the DC offset of the amplifier but doesn't it also ruin the integrator for DC or did I do a mistake somewhere? \$\endgroup\$ Commented Oct 2 at 22:55
  • \$\begingroup\$ I already reacted to your analysis in the comments. Your replacements are invalid. \$\endgroup\$ Commented Oct 2 at 23:00

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