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i am trying to understand the behaviour of an RC integrator for \$f>>1/RC\$, i.e. in the integration domain. I know that at the capacitor i have \$V = \int V_{in}/RC dt\$; experimentally i find that when i use as input square waves with tunable duty cycle (but fixed frequency such that \$f>>1/RC\$) i have an approximate constant voltage at about the mean of the input signal (that obviously increases as the duty cycle increases).

I can't really understand why. I can approximate the integral with the mean, but then i have \$ V= \bar{V_{in}} T/RC \$ with T the period; this however tends to zero as T < RC!!

Thank you very much, i am using this to have a "tunable" voltage generator from Arduino, i know it works but i want to understand why it is working!

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You are generating an input waveform to the integrator that has two components. We might describe it as a rectangular wave between 0 and (I'm guessing) +5V.

  • It has a DC component - its mean value.

  • It has an AC component that has a +ve peak and a -ve peak.

Your integrator equation doesn't apply to the DC component because its frequency is zero, which is below the required frequency for "integration" operation. This DC component passes through the integrator unmolested.

Only the AC component is integrated, because its frequency components are at (or even better, above) the required frequency for "integration" operation.

enter image description here

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  • \$\begingroup\$ You were probably a more direct and to the point than I'd have been. Nice. \$\endgroup\$ – jonk Dec 18 '18 at 3:22
  • \$\begingroup\$ Thats exactly what I needed. Thank you! \$\endgroup\$ – Lenz Dec 19 '18 at 8:38

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