PID Control

How the model works

The plant is a mass on a spring with friction: m x'' + c x' + k x = u + d with m = c = k = 1. The controller measures the position x, compares it with the setpoint r, and sets the force

u = K_p e + K_i ∫e dt − K_d dx/dt, e = r − x

Proportional action pushes harder the further away the mass is. Alone it leaves a steady-state error, because the spring needs a constant force to hold the mass away from its rest point: with a P controller the mass settles at K_p / (k + K_p) of the target. Higher K_p shrinks the error but makes the response more oscillatory.

Integral action accumulates past error and keeps pushing until the error is zero, removing steady-state error, at the cost of more overshoot. Derivative action opposes motion, adding damping. Here it acts on the measured velocity rather than the error, so a sudden setpoint change does not cause a "derivative kick".

Real actuators have limits. When the force saturates, the integral keeps growing (windup) and causes extra overshoot; anti-windup stops integrating while the output is saturated.

Linear plant, continuous control simulated with 1 ms steps, ideal sensor apart from the optional noise (which is smoothed by a 5 ms filter). Overshoot is measured relative to the final value; settling uses a ±2% band. The disturbance force is applied at t = 10 s.

Guided experiments

1. Proportional only. K_p = 10. Prediction: a fast but very oscillatory response with about 62% overshoot, settling at 0.91 instead of 1: a 9.1% steady-state error, 1/(1 + 10). P only
2. Add derivative. K_p = 10, K_d = 4. Prediction: the extra damping cuts overshoot to about 3% and settling time to under 2 s, but the 9.1% error remains. PD
3. Add integral. K_p = 10, K_i = 5, K_d = 4, then push the mass with a −5 N disturbance at 10 s. Prediction: the error goes to zero, with about 14% overshoot. The disturbance knocks the mass down to about 0.6, and the integral brings it back to 1. With PD alone the same push would leave it at about 0.45. PID + disturbance PD + disturbance
4. Integral windup. PID with the force limited to ±3 N, without and with anti-windup. Prediction: without anti-windup the stored integral causes about 21% overshoot; with anti-windup it drops to about 4%. Windup Anti-windup

Challenge

With P control only on this plant (k = 1), what gain K_p keeps the steady-state error below 2%? What happens to the oscillation as you raise K_p that far?

References

• Åström, K. J. and Murray, R. M., Feedback Systems: An Introduction for Scientists and Engineers, 2nd ed. (2021), ch. 11.
• Åström, K. J. and Hägglund, T., Advanced PID Control (2006), ch. 3.
• Ogata, K., Modern Control Engineering, 5th ed. (2010), ch. 5 and 8.