Feedback and PID control

Open loop and closed loop

Open loop: you set the pump to 1180 rpm and hope the pressure is right. If the pipes are dirtier than expected, you never find out.

Closed loop: you measure the pressure, compare it with what you want, and correct the speed continuously. The controller does what a person would do by hand, faster and without getting tired.

• SP, set-point (consigne): what you want, for example 25 m of head.
• PV, process variable (mesure): what the sensor reads.
• e = SP − PV, the error.
• CV, control variable: what the controller commands, here the pump speed.
• Disturbance: anything that pushes PV away, for example a valve closing downstream.

The three terms

$$CV(t) = K_p\, e(t) + K_i \int_0^t e\, d\tau + K_d \frac{de}{dt}$$

P, proportional: how far off am I now? Push harder the bigger the error. More $K_p$ reacts faster, but on its own it leaves a steady-state offset.

I, integral: how long have I been off? It adds up the error over time and keeps pushing while any error remains, so it removes the offset completely. Too much $K_i$ overshoots and oscillates.

D, derivative: how fast is it changing? It brakes the response when PV approaches the set-point quickly, which damps overshoot. It amplifies sensor noise, so pump and fan loops usually leave it out and use a PI controller.

FoundationStart here if this is new to you

Think of steering a car to stay in the middle of the lane. P: the further you drift, the more you turn the wheel. I: if a steady crosswind keeps pushing you, you slowly hold a little extra turn until you are centred again. D: if you are swinging back towards the centre quickly, you ease off early so you do not overshoot.

Predict first

A P-only controller (Ki = 0) holds a pump's pressure. Demand changes and the pump must settle at a new speed. Where does the pressure end up?

Exactly on the set-pointA little off the set-point, and it stays thereIt oscillates forever

Kp

1.20

Ki

0.00

Kd

0.00

Overshoot

0.0 %

Settling time

does not settle

Error before disturbance

0.455

Open the full PID playground

Why loops go unstable

Every real loop has lag and delay: the motor takes time to change speed, the sensor filters its reading, the fieldbus adds a few scans. The controller always acts on slightly old news. With a small gain that is harmless. With a large gain it overcorrects on old information, overshoots, overcorrects the other way, and the oscillation grows. Raise $K_p$ above about 9.5 in the widget and it never settles.

Tuning a loop

1. Start with $K_i = K_d = 0$ and a small $K_p$.
2. Raise $K_p$ until the response is quick with little overshoot.
3. Add $K_i$ slowly until the offset disappears in reasonable time.
4. Add $K_d$ only if overshoot is a problem and the signal is clean.

Ziegler-Nichols is the classic systematic method. With P only, raise $K_p$ until the loop oscillates steadily. That gain is $K_u$ and the period is $P_u$. Then use $K_p = 0.45 K_u$ and $K_i = 0.54 K_u / P_u$ for a PI controller. It is known to be aggressive.

The full playground has five real processes, all these effects as switches, and tested presets.

ExplorerGo deeper: derivations and open questions

Frequency view. For a plant $G(s)$ and controller $C(s)$, the loop is stable while the phase of $C(j\omega)G(j\omega)$ stays above −180° where its gain crosses 1. A pure delay $e^{-\theta s}$ adds phase lag $\omega\theta$ with no change in gain, which is why delay, more than anything else, limits how aggressive a loop can be.

Open question. Two first-order lags with a PI controller can never be made unstable by $K_p$ alone. Why? What do you have to add before large $K_p$ can destabilise the loop? (Lab II depends on the answer.)