P, PI and PID controllers

Why it matters

More than nine in ten industrial loops — flow, pressure, level, temperature — run on some form of PID controller, usually inside a DCS or a single-loop controller on a panel. An instrumentation engineer is expected to know what each term does to offset, overshoot and stability, to read the tuning parameters on a faceplate (gain or proportional band, integral time, derivative time), and to tune a loop that misbehaves.

Key ideas

Proportional (P) action: u = Kp·e. The controller output is proportional to the present error.

• Raising Kp speeds the response and reduces, but never removes, steady-state error for a type-0 plant (offset = 1/(1 + Kp·Kplant) for a unit step).
• Too much Kp causes oscillation and eventually instability in plants with three or more lags or dead time.
• In process industry the gain is often stated as proportional band PB = 100/Kp % (for normalised signals).

Integral (I) action: u = (Kp/Ti)·∫e dt. Output keeps changing while any error remains.

• Adds a pole at the origin, raising system type by one, so a constant set-point or load disturbance leaves zero offset.
• Adds up to 90° of phase lag at low frequency, so it reduces stability margins; too short Ti gives slow oscillation.
• Integral windup: if the actuator saturates (valve fully open), the integral keeps growing and causes a large overshoot later. Anti-windup (clamping or back-calculation) is standard in real controllers.

Derivative (D) action: u = Kp·Td·de/dt. Output responds to the rate of change of error — it "anticipates".

• Adds phase lead (a zero), increasing damping and allowing higher gain, so overshoot falls and response speeds up.
• Has no effect on steady-state error (a constant error has zero derivative).
• Amplifies high-frequency noise, so it is always filtered (Td·s/(1 + Td·s/N) with N typically 5–20) and often applied to the measurement rather than the error, to avoid a "derivative kick" on set-point steps.

Combinations

• PI: the workhorse for flow, pressure and level loops — zero offset, modest speed.
• PD: improves transient response of servo loops; does not remove offset in type-0 plants.
• PID: used for slower loops with significant lag, such as temperature, where derivative helps damping.

In the s-plane: PI adds a pole at the origin and a zero at −1/Ti; PD adds a zero at −1/Td; ideal PID adds a pole at the origin and two zeros. PI behaves like a lag compensator and PD like a lead compensator.

Tuning: by model (pole-zero cancellation, IMC) or by experiment. The classic Ziegler–Nichols closed-loop method finds the ultimate gain Ku (P-only gain giving sustained oscillation) and its period Pu, then sets: P — Kp = 0.5Ku; PI — Kp = 0.45Ku, Ti = Pu/1.2; PID — Kp = 0.6Ku, Ti = Pu/2, Td = Pu/8. These give an aggressive, roughly quarter-decay response and are a starting point, not a final setting.

Formulas

• u(t) = Kp·[e(t) + (1/Ti)·∫e dt + Td·de/dt] — ideal (ISA standard) PID form.
• u(t) = Kp·e + Ki·∫e dt + Kd·de/dt — parallel form, with Ki = Kp/Ti, Kd = Kp·Td.
• C(s) = Kp·(1 + 1/(Ti·s) + Td·s) — PID transfer function.
• C(s) = Kp·(Ti·s + 1)/(Ti·s) — PI controller (zero at −1/Ti).
• C(s) = Kp + Kd·s — PD controller (zero at −Kp/Kd).
• PB (%) = 100/Kp — proportional band (normalised signals).

Symbols: e error and u controller output (in signal units such as %, mA or V); Kp proportional gain (output unit per error unit); Ti integral (reset) time and Td derivative time, both in s; Ki in s⁻¹ and Kd in s (times the gain units).

Worked examples

Example 1 — P versus PI on a first-order process Given: process Gp(s) = 2/(5s + 1) (gain 2, τ = 5 s), unity feedback, unit step in set point.

1. P control with Kp = 4: loop gain 8/(5s + 1); closed loop T(s) = 8/(5s + 9).
2. Steady state: T(0) = 8/9 = 0.889, so offset = 1 − 0.889 = 0.111 (11.1 %). Closed-loop time constant 5/9 = 0.556 s.
3. PI control with Kp = 4 and Ti = 5 s (zero placed on the process pole): C(s)·Gp(s) = 4·(5s + 1)/(5s) × 2/(5s + 1) = 8/(5s).
4. Closed loop: T(s) = 8/(5s + 8); T(0) = 1, so offset = 0. Time constant 5/8 = 0.625 s.

Result: P-only leaves 11.1 % offset (τ = 0.556 s); PI removes the offset (τ = 0.625 s) at almost the same speed.

Example 2 — choosing PD gains for a servo (GATE level) Given: plant G(s) = 1/[s(s + 1)], unity feedback, PD controller C(s) = Kp + Kd·s. Find Kp, Kd for ωn = 4 rad/s and ζ = 0.7, and compare with P-only control at the same Kp.

1. Characteristic equation: s(s + 1) + Kp + Kd·s = s² + (1 + Kd)s + Kp = 0.
2. Match s² + 2ζωn s + ωn²: Kp = ωn² = 16; 1 + Kd = 2 × 0.7 × 4 = 5.6 → Kd = 4.6 s.
3. P-only with Kp = 16: s² + s + 16, so ωn = 4, ζ = 1/8 = 0.125, overshoot e^(−π × 0.125/√(1 − 0.0156)) = 67 %.
4. With PD: ζ = 0.7 gives about 4.6 % overshoot from the poles alone; the controller zero at −Kp/Kd = −3.48 raises the actual step overshoot to about 14 % (found by simulation).
5. Velocity constant is unchanged: Kv = lim s·C·G = Kp = 16 s⁻¹.

Result: Kp = 16, Kd = 4.6 s — derivative action cuts overshoot from 67 % to about 14 % while keeping Kv = 16 s⁻¹.

Common mistakes

• Saying P control "eliminates" steady-state error; it only reduces it for a type-0 plant.
• Confusing Ki with 1/Ti: in the ideal form Ki = Kp/Ti.
• Expecting derivative action to reduce steady-state error.
• Using unfiltered derivative on a noisy measurement.
• Forgetting integral windup when the actuator saturates.
• Treating a large proportional band as high gain — PB is inversely proportional to gain.

For GATE IN

Expect questions on the effect of P, I and D actions on type, steady-state error, damping and stability; numerical problems that combine a controller with a simple plant to find offset, closed-loop poles or gains for a given ζ and ωn; and controller transfer-function forms and their pole-zero locations. Practise relating proportional band, integral time and derivative time to the transfer function.

Quick check

1. What is the proportional band of a controller with Kp = 2.5?
2. Which action removes offset for a step load disturbance on a type-0 process?
3. A PI controller has Kp = 3, Ti = 6 s. Where is its zero?
4. What problem does a derivative filter solve?

Answers: 1. 40 %; 2. Integral action; 3. At s = −1/6 s⁻¹; 4. It limits the amplification of high-frequency measurement noise.