PID controllers and tuning

Why it matters

Well over nine in ten industrial loops — motor speed and position drives, temperature zones, pressure and flow loops, robot joints — run a PID controller or one of its subsets (P, PI, PD). Knowing what each term does to stability, speed and accuracy, and how to get starting gains from a quick plant test, is the most directly usable skill in this subject.

Key ideas

The control law. The controller acts on the error e(t) = r(t) − y(t):

• P (proportional) — reacts to the present error. Raising Kp speeds the response and reduces (but for a type-0 plant cannot remove) steady-state error; too much gain reduces damping and can destabilise the loop.
• I (integral) — reacts to the accumulated past error. It adds a pole at the origin, raising the system type by one, so a constant error cannot persist: step steady-state error becomes zero, and constant load disturbances are rejected. Cost: 90° of phase lag at low frequency, more overshoot, and possible integrator wind-up.
• D (derivative) — reacts to the rate of change of error (a prediction of where the error is heading). It adds phase lead and damping, reducing overshoot and allowing a higher Kp. It has no effect on steady-state error. Cost: it amplifies high-frequency noise and gives a large spike ("derivative kick") when the set point steps.

Forms. Parallel (independent gains) u = Kp·e + Ki·∫e dt + Kd·de/dt, or ideal (ISA) form u = Kp[e + (1/Ti)∫e dt + Td·de/dt], with Ki = Kp/Ti and Kd = Kp·Td. Ti (integral time) and Td (derivative time) are in seconds.

What the terms do in the s-plane. PI = Kp(1 + 1/(Ti·s)) adds a pole at s = 0 and a zero at s = −1/Ti. PD = Kp(1 + Td·s) adds a zero at s = −1/Td, which pulls the root locus left. PID adds a pole at the origin and two zeros.

Practical PID. A pure derivative is never implemented; it is filtered, Td·s/(1 + Td·s/N) with N ≈ 5–20. The derivative is often taken on the measured output instead of the error, to avoid set-point kick. Anti-wind-up (stop integrating or back-calculate when the actuator saturates) prevents large overshoot after saturation.

Discrete implementation at sample time Δt: u_k = Kp·e_k + Ki·Σ(e_j·Δt) + Kd·(e_k − e_{k−1})/Δt.

Tuning.

• Ziegler–Nichols ultimate-cycle (closed-loop) method: with I and D off, raise Kp until the output oscillates with constant amplitude. Record the ultimate gain Ku and period Pu, then use the table below. It gives an aggressive start (about quarter-amplitude decay, roughly 25 % overshoot or more) that is then fine-tuned.
• Ziegler–Nichols reaction-curve (open-loop) method: from a step test of an S-shaped plant, read the delay L and time constant T (slope R); for PID, Kp = 1.2T/(K·L), Ti = 2L, Td = 0.5L, where K is the process gain.
• Ku and Pu can also be obtained from a model with the Routh array (marginal gain and auxiliary equation), as below.

Effect summary (increasing each gain): Kp — rise time down, overshoot up, steady-state error down; Ki — rise time down, overshoot up, settling time up, steady-state error eliminated; Kd — overshoot down, settling time down, little change in rise time and steady-state error. These are tendencies, not laws.

Formulas

u(t) = Kp·e(t) + Ki·∫₀ᵗ e dt + Kd·de/dt

• u: controller output; e: error; Kp: dimensionless or output per error unit; Ki in s⁻¹ × Kp units; Kd in s × Kp units.

C(s) = Kp + Ki/s + Kd·s = Kp(1 + 1/(Ti·s) + Td·s)

Ki = Kp/Ti, Kd = Kp·Td

Ziegler–Nichols ultimate-cycle settings: P: Kp = 0.5Ku PI: Kp = 0.45Ku, Ti = Pu/1.2 PID: Kp = 0.6Ku, Ti = 0.5Pu, Td = 0.125Pu

• Ku: ultimate gain (dimensionless for a normalised loop); Pu: ultimate period (s); Pu = 2π/ωu.

Worked examples

Example 1 (standard — Ziegler–Nichols from a model). A plant G(s) = 1/(s(s + 1)(s + 2)) is under unity feedback. Find Ku and Pu, then the Z–N PID settings.

1. With P control: s³ + 3s² + 2s + Kp = 0. Routh s¹ row: (6 − Kp)/3 = 0 → Ku = 6.
2. Auxiliary equation: 3s² + 6 = 0 → ωu = √2 = 1.414 rad/s.
3. Pu = 2π/ωu = 2π/1.414 = 4.44 s.
4. Kp = 0.6 × 6 = 3.6.
5. Ti = 0.5 × 4.44 = 2.22 s; Td = 0.125 × 4.44 = 0.555 s.
6. Parallel gains: Ki = 3.6/2.22 = 1.62 s⁻¹; Kd = 3.6 × 0.555 = 2.0 s.

Example 2 (GATE level — PD pole placement). A motor position plant is G(s) = 1/(s(s + 1)) with unity feedback. (a) With P control only, what damping ratio does Kp = 16 give? (b) Choose PD gains C(s) = Kp + Kd·s so that the closed loop has ωn = 4 rad/s and ζ = 0.7.

1. P only: s² + s + Kp = 0 → ωn = √16 = 4 rad/s, ζ = 1/(2 × 4) = 0.125 (overshoot about 67 %).
2. PD: 1 + (Kp + Kd·s)/(s(s + 1)) = 0 → s² + (1 + Kd)s + Kp = 0.
3. Match with s² + 2ζωn·s + ωn²: Kp = ωn² = 16; 1 + Kd = 2 × 0.7 × 4 = 5.6 → Kd = 4.6 s.
4. Note: the closed loop is (4.6s + 16)/(s² + 5.6s + 16) — the PD zero at s = −16/4.6 = −3.48 adds some overshoot beyond the 4.6 % of a pure ζ = 0.7 system, so check by simulation.
5. Steady-state error to a step is zero in both cases (the plant is type 1); D added damping without changing ωn.

Common mistakes

• Mixing parallel gains (Ki, Kd) with ideal-form times (Ti, Td): Ki = Kp/Ti, not 1/Ti.
• Believing derivative action removes steady-state error — only integral action (or a type increase) does.
• Forgetting Δt in a discrete derivative or integral when the sample time is not 1 s.
• Using unfiltered derivative on a noisy encoder signal.
• Taking Pu in Hz or using ωu where Pu is needed: Pu = 2π/ωu.
• Expecting Z–N gains to be final; they are starting values that usually need detuning.

For GATE ME

Expect: identifying which mode removes offset or adds damping; Z–N settings from Ku and Pu; finding Ku and Pu via Routh for a given plant; closed-loop characteristic equation with PD or PI control and matching ζ and ωn; steady-state error with and without integral action; controller output for given error values. Practise writing the characteristic equation with the controller included before matching coefficients.

Quick check

1. Ku = 6 and Pu = 2 s. Find the Z–N PID Kp, Ti and Td.
2. Which term eliminates steady-state error to a step for a type-0 plant?
3. Kp = 2, Ki = 1 s⁻¹, Kd = 0.5 s, with e = 3, ∫e dt = 4, de/dt = 2. Find u.
4. Why is the derivative usually taken on the measurement rather than the error?
5. What does integral action do to the system type?

Answers: 1. Kp = 3.6, Ti = 1 s, Td = 0.25 s. 2. Integral. 3. 6 + 4 + 1 = 11. 4. To avoid a large derivative kick when the set point steps. 5. Raises it by one.