eyeprolog

Control system

What signal should each actuator receive?

A small controller combines sensor measurements, a target value, and three classic ingredients of control engineering: proportional gain, feedforward compensation, and nonlinear feedback. Each intermediate quantity is its own named step, so the proof shows how every signal was computed.

Run it in the playground · Source

The question

Two actuators need a control signal.

Actuator 1 is driven by a normalized input, corrected for a measured disturbance (feedforward).
Actuator 2 is driven by the gap between an output and its target (feedback), with a nonlinear term.

What are the two signals, and can we see exactly how each was derived?

Measurements, observations, targets

measurement(input1, [6, 11]).
measurement(input2, true).
measurement(disturbance1, 35766).
measurement(output2, 24).

observation(state3, 22).

target(output2, 29).

Plain facts: a two-reading input, a switch, a disturbance, an output and its target of 29.

Normalizing and compensating

measurement_normalized(I, M) :-
  measurement(I, [M1, M2]),
  (M1 < M2),
  (Delta is M2 - M1),
  (M is Delta ** 0.5).

control(actuator1, C) :-
  measurement_normalized(input1, M1),
  measurement(input2, true),
  measurement(disturbance1, D1),
  (Proportional is M1 * 19.6),
  numeric_log10(D1, Compensation),
  (C is Proportional - Compensation).

The input pair is normalized to √(11 − 6) = √5. Actuator 1 multiplies that by a gain of 19.6 and subtracts log₁₀ of the disturbance as compensation.

Feedback with a nonlinear term

control(actuator2, C) :-
  observation(state3, P3),
  measurement(output2, M4),
  target(output2, T2),
  (Error is T2 - M4),
  (Differentialerror is P3 - M4),
  (Proportional is 5.8 * Error),
  (Nonlinearfactor is 7.3 / Error),
  (Differential is Nonlinearfactor * Differentialerror),
  (C is Proportional + Differential).

The error is target minus output, 29 − 24 = 5. The proportional part scales it, and a nonlinear factor 7.3 / error weights the difference between an observed state and the output.

The answer

eyeprolog examples/control-system.pl

controlSignal(actuator1, 39.27346198678276).
controlSignal(actuator2, 26.08).
status(actuator1, active).
status(actuator2, active).
normalizedMeasurement(input1, 2.23606797749979).
log10(disturbance1, 4.553470372213121).

The two signals, the status of each actuator, and two intermediate quantities are all reported.

The reason

eyeprolog --proof examples/control-system.pl

step(control(actuator2, 26.08),
     rule(15),
     ['C' = 26.08,
      'P3' = 22,
      'M4' = 24,
      'T2' = 29,
      'Error' = 5,
      'Differentialerror' = -2,
      'Proportional' = 29.0,
      'Nonlinearfactor' = 1.46,
      'Differential' = -2.92],

The bindings read like a worked calculation: an error of 5 gives a proportional part of 29.0, the nonlinear factor is 7.3 / 5 = 1.46, and the differential term is 1.46 × (22 − 24) = −2.92. So 29.0 − 2.92 = 26.08.

The check

eyeprolog --proof examples/control-system.pl > control-system.why.pl
eyeprolog --check-proof control-system.why.pl examples/control-system.pl

condition('C1', resolution, ok, 16).
condition('C5', re_decision, ok, 14).
steps(30).
trusted(0).
verdict(checked).

Of the 30 steps, 16 replay program clauses and 14 are arithmetic the checker recomputes itself. Nothing is taken on trust.

Try it

Set the target to target(output2, 24). The error becomes 0, and the run stops with evaluation_error(zero_divisor) from the nonlinear term. What should a real controller do at zero error?
Swap the input readings to measurement(input1, [11, 6]). The second normalization clause now applies, the normalized value becomes 11, and actuator 1 gets a much larger signal (about 211).
Change the gain 19.6 and predict the new actuator 1 signal before running.

Learn more in The Art of EyeProlog, Chapters 6 and 14.

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