eyeprolog

Sizing an RC low-pass filter

Which frequencies does this resistor and capacitor let through?

A resistor and a capacitor in series form a first-order low-pass filter: slow signals pass, fast ones are damped. Two formulas describe it, the time constant τ = R·C and the cutoff frequency f꜀ = 1 / (2πτ). This example turns them into rules over component facts.

Run it in the playground · Source

The question

Filter 1 uses a 10 kΩ resistor and a 1 µF capacitor. What kind of filter is it, what is its time constant, and what is its cutoff frequency?

The components

component(filter1, resistor_ohm, 10000.0).
component(filter1, capacitor_f, 0.000001).
constant(pi, 3.141592653589793).

Components are stored as filter, attribute, value triples, with units in the attribute name. More filters, or more attributes, need no change to the rules.

The rules

time_constant(Filter, Tau) :-
  component(Filter, resistor_ohm, R),
  component(Filter, capacitor_f, C),
  (Tau is R * C).

cutoff_frequency(Filter, Frequency) :-
  time_constant(Filter, Tau),
  constant(pi, Pi),
  (Twopi is 2.0 * Pi),
  (Denominator is Twopi * Tau),
  (Frequency is 1.0 / Denominator).

type(Filter, first_order_low_pass) :-
  component(Filter, resistor_ohm, _r),
  component(Filter, capacitor_f, _c).

Each formula is one rule, and each arithmetic step is its own goal, so the proof shows the intermediate values.

The answer

eyeprolog examples/electrical-rc-filter.pl

type(filter1, first_order_low_pass).
timeConstant_s(filter1, 0.01).
cutoffFrequency_Hz(filter1, 15.915494309189533).

The time constant is 10 ms and the cutoff is about 15.9 Hz: signals well below 16 Hz pass almost unchanged, faster ones are attenuated.

The reason

eyeprolog --proof examples/electrical-rc-filter.pl

step(time_constant(filter1, 0.01),
     rule(4),
     ['Filter' = filter1, 'Tau' = 0.01, 'R' = 10000.0, 'C' = 0.000001],
     [component(filter1, resistor_ohm, 10000.0),
      component(filter1, capacitor_f, 0.000001),

The 10 ms time constant comes from the two component facts by clause 4, multiplying 10000 Ω by 0.000001 F. The cutoff frequency builds on it.

The check

eyeprolog --proof examples/electrical-rc-filter.pl > electrical-rc-filter.why.pl
eyeprolog --check-proof electrical-rc-filter.why.pl examples/electrical-rc-filter.pl

condition('C1', resolution, ok, 8).
condition('C4', coverage, ok, 15).
condition('C7', relevance, ok, 15).
verdict(checked).

All 8 rule steps replay and the 4 arithmetic steps are recomputed. Nothing is taken on trust. The proof certifies the arithmetic from the stated component values; it cannot certify that the real parts have those values.

Try it

Which capacitor gives a 50 Hz cutoff with the same resistor? Add it as filter2.
Add a rule that classifies a signal frequency as passed or attenuated relative to the cutoff.
Add a tolerance attribute to the capacitor and compute the lowest and highest possible cutoff.

Learn more in The Art of EyeProlog, Chapter 29 (numerical models).

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