PID control — proportional, integral, derivative terms: Exercises

Before we start, the four quantities we will use over and over (from the parent note, restated so this page stands alone):

Figure — PID control — proportional, integral, derivative terms

Level 1 — Recognition

Problem 1.1

Match each symptom to the term whose job it is to cure it. (a) The system reacts too slowly. (b) It settles but always a little off-target. (c) It overshoots and rings.

Recall Solution

Each term has one signature disease-and-cure:

(a) Too slow → boost == $K_{p}$ == (Proportional). P is the present-error push; a bigger push means a faster response.
(b) Steady offset → boost == $K_{i}$ == (Integral). Only the integral remembers a persistent error and keeps ramping until the offset dies.
(c) Overshoot / ringing → boost == $K_{d}$ == (Derivative). D reacts to the rate $\overset{e}{˙}$ , braking before you shoot past.

Answer: (a) P, (b) I, (c) D.

Problem 1.2

Write the error signal $e (t)$ and evaluate it when $r = 3 0^{\circ}$ and $y = 3 4^{\circ}$ .

Recall Solution

$e (t) = r (t) - y (t)$ . Plug in: $e = 3 0^{\circ} - 3 4^{\circ} = - 4^{\circ} .$ The sign is negative, meaning we have overshot the target (we are above it). The controller will command $u$ in the direction that pulls $y$ back down.

Level 2 — Application

Problem 2.1

A reaction wheel holds a satellite's pitch at $r = 0^{\circ}$ . A constant solar torque acts as a disturbance $d = 3$ (output units). Using pure P with $K_{p} = 6$ , find the steady-state error $e_{ss}$ .

Recall Solution

WHAT we need: the resting error once everything stops changing. WHY this formula: at rest the control output must exactly cancel the disturbance, otherwise something would still be moving. With pure P the only output is $u = K_{p} e$ , so at equilibrium $K_{p} e_{ss} = d \Rightarrow e_{ss} = \frac{d}{K _{p}} = \frac{3}{6} = 0. 5^{\circ} .$ A pure-P loop can shrink this but never zero it — zero error would give zero output, and then nothing fights the torque. This is exactly the steady-state error the integral is invented to kill.

Problem 2.2

Same satellite, now P + I. In steady state, what is $e_{ss}$ , and what value does the integral output settle to?

Recall Solution

WHAT: find the resting error and resting integral value. WHY: the integrator changes its output at rate $\overset{u}{˙}_{I} = K_{i} e$ . For the whole system to stop changing, this rate must be zero: $\overset{u}{˙}_{I} = K_{i} e_{ss} = 0 \Rightarrow e_{ss} = 0.$ So the only resting state has zero error — the integral quietly climbs until it does. Its final value must supply the whole cancellation on its own: $u_{I} = d = 3.$ P contributes nothing at rest (because $e_{ss} = 0$ ), so the integrator alone holds off the disturbance.

Problem 2.3

Convert the continuous integral $\int_{0}^{t} e d τ$ into the running sum a flight computer uses, with $Δ t = 0.2$ s and samples $e_{0} = 1.0, e_{1} = 0.6, e_{2} = 0.4$ . Compute the running integral value $I_{2}$ after three samples (rectangle rule).

Recall Solution

WHY a sum: the computer only sees snapshots every $Δ t$ . Each snapshot contributes a thin rectangle of area $e_{j} Δ t$ ; the integral is the total area, i.e. the sum of rectangles. $I_{2} = \sum_{j = 0}^{2} e_{j} Δ t = (1.0 + 0.6 + 0.4) (0.2) = (2.0) (0.2) = 0.4.$ This $I_{2}$ is what gets multiplied by $K_{i}$ to become $u_{I}$ . (See Laplace transforms & transfer functions for the continuous view.)

Level 3 — Analysis

Problem 3.1 (one discrete control update)

Given $K_{p} = 2, K_{i} = 0.5, K_{d} = 1, Δ t = 0.1$ . The previous error $e_{k - 1} = 1.0$ , the current error $e_{k} = 0.8$ , and the stored integral before this step $I_{prev} = 0.3$ . Compute $u_{k}$ .

Recall Solution

Build the three pieces, each answering its own question. Integral (past): add the newest rectangle, then scale. $I_{k} = I_{prev} + e_{k} Δ t = 0.3 + (0.8) (0.1) = 0.38, u_{I} = K_{i} I_{k} = 0.5 (0.38) = 0.19.$ Proportional (present): react to the error right now. $u_{P} = K_{p} e_{k} = 2 (0.8) = 1.6.$ Derivative (future): the backward difference estimates $\overset{e}{˙}$ ; error is falling, so this is negative — a brake. $u_{D} = K_{d} \frac{e _{k} - e _{k - 1}}{Δ t} = 1 \cdot \frac{0.8 - 1.0}{0.1} = - 2.0.$ Total: $u_{k} = 1.6 + 0.19 - 2.0 = - 0.21.$ Interpretation: even though the error is positive (still below target), the error is dropping so fast that D dominates and commands a slight reversal — this is the anti-overshoot instinct in action.

Problem 3.2 (sign census — cover every case)

For each situation give the sign of $u_{D} = K_{d} \overset{e}{˙}$ (take $K_{d} > 0$ ) and say in words what D is doing: (a) $y$ rushing up toward $r$ from below; (b) $y$ overshot and coming back down; (c) $y$ frozen (steady); (d) $y$ drifting away from $r$ .

Recall Solution

D watches the rate of error, $\overset{e}{˙}$ . Recall $e = r - y$ , so if $y$ rises, $e$ falls.

(a) $y$ rising toward target ⇒ $e$ falling ⇒ $\overset{e}{˙} < 0$ ⇒ $u_{D} < 0$ : brakes the approach to prevent overshoot.
(b) $y$ overshot, now falling back ⇒ $e$ rising back toward 0 ⇒ $\overset{e}{˙} > 0$ ⇒ $u_{D} > 0$ : helps the recovery.
(c) $y$ steady ⇒ $\overset{e}{˙} = 0$ ⇒ $u_{D} = 0$ : D says nothing at rest — which is why D alone can never fix a steady offset.
(d) $y$ drifting away ⇒ $∣ e ∣$ growing ⇒ $\overset{e}{˙}$ has the same sign as $e$ ⇒ $u_{D}$ reinforces the correction: fights the runaway early. The zero case (c) is the key insight: D contributes exactly nothing in steady state, so it can neither cause nor cure offset.

Level 4 — Synthesis

Problem 4.1 (diagnose and prescribe)

A drone altitude loop shows three faults across three test flights:

Flight A: reaches target but ~4% below it, forever.
Flight B: fast, but overshoots 30% and rings for several seconds.
Flight C: after adding heavy D, the motors buzz/chatter even when hovering still.

For each, name the term to change and the direction of change, and explain the mechanism.

Recall Solution

Flight A — persistent offset. Increase == $K_{i}$ ==. A constant residual error means the loop rests where $u = K_{p} e_{ss}$ , which is nonzero only if $e_{ss} \neq = 0$ . Adding/boosting integral action forces the resting condition $K_{i} e_{ss} = 0$ , so the accumulator ramps the offset to zero. (This is type elevation — the integrator raises the system type by one.)
Flight B — overshoot & ringing. Increase == $K_{d}$ ==. Ringing means the approach is under-damped. D reacts to $\overset{e}{˙}$ , subtracting command exactly when the error is collapsing fastest, adding damping without slowing the initial response.
Flight C — motor chatter. Decrease == $K_{d}$ == and/or filter it. Differentiation multiplies high-frequency sensor noise: tiny fast wiggles have huge slope, so $\overset{e}{˙}$ explodes and D commands violent, useless twitches. Fix: small $K_{d}$ plus a low-pass filtered derivative.

Problem 4.2 (windup budget)

An actuator saturates at $u_{m a x} = 5$ . During a large maneuver the error sits at $e = 2$ for $T = 3$ s while the output is already pinned at $5$ . With $K_{i} = 0.8$ , how much extra integral output $Δ u_{I}$ accumulates during saturation, and why is this dangerous?

Recall Solution

While saturated, the integrator keeps eating error even though the actuator can do no more. Over $3$ s at constant $e = 2$ : $Δ u_{I} = K_{i} e T = (0.8) (2) (3) = 4.8.$ This $4.8$ is stored on top of whatever was already there — a phantom command the actuator never actually delivered. When $y$ finally reaches target, the integrator must first unwind all $4.8$ before its output drops, so $y$ sails far past the target. This is integrator windup. Fix: anti-windup — freeze (or clamp) the integrator whenever the actuator is saturated, so no phantom charge builds. (Related tuning wisdom: Ziegler–Nichols tuning keeps $K_{i}$ modest partly for this reason.)

Level 5 — Mastery

Problem 5.1 (design from a spec)

You must control a slow thermal plant. Requirements: (i) zero steady-state error under a constant heat-loss disturbance; (ii) no more than mild overshoot; (iii) the sensor is noisy. Starting from a working P controller with $K_{p} = 4$ , decide which terms to add, in what order, and justify each choice against the requirements. Then compute the steady-state integral output if the constant disturbance is $d = 10$ .

Recall Solution

Reason term-by-term against the spec:

Requirement (i) zero steady-state error ⇒ we must add I. Only the integral gives the resting condition $K_{i} e_{ss} = 0 \Rightarrow e_{ss} = 0$ . Keep $K_{i}$ modest to avoid windup on this slow plant.
Requirement (ii) mild overshoot ⇒ add a small D for damping — but requirement (iii) noisy sensor means heavy D would amplify noise, so use a small, filtered derivative. If overshoot is already acceptable with P+I, you may even skip D (a PI controller is common on slow, noisy thermal systems).
Keep $K_{p} = 4$ for baseline speed. So the design is PI (optionally a filtered D), chosen because each requirement maps to exactly one term's job.

Steady-state integral output: at rest $e_{ss} = 0$ , so P contributes $K_{p} \cdot 0 = 0$ and D contributes $K_{d} \cdot 0 = 0$ . The integral must therefore supply the entire cancellation of the disturbance: $u_{I} = d = 10.$ The integrator's stored value settles to whatever makes $K_{i} I = 10$ ; the exact $I$ depends on $K_{i}$ , but the output $u_{I} = 10$ is fixed by the physics.

Problem 5.2 (predict the full command)

A satellite pitch loop uses $K_{p} = 3, K_{i} = 0.5, K_{d} = 2$ . At one instant: $e_{k} = 0.4$ , previous error $e_{k - 1} = 0.7$ , stored integral $I_{k} = 1.2$ (already updated), sample time $Δ t = 0.05$ . Compute $u_{P}$ , $u_{I}$ , $u_{D}$ , and the total $u_{k}$ . Then state which term dominates and what it physically means.

Recall Solution

$u_{P} = K_{p} e_{k} = 3 (0.4) = 1.2,$ $u_{I} = K_{i} I_{k} = 0.5 (1.2) = 0.6,$ $u_{D} = K_{d} \frac{e _{k} - e _{k - 1}}{Δ t} = 2 \cdot \frac{0.4 - 0.7}{0.05} = 2 \cdot (- 6) = - 12.$ $u_{k} = 1.2 + 0.6 - 12 = - 10.2.$ Dominant term: D, and strongly negative. Even though there is still a positive error and a positive stored history, the error is collapsing very fast ( $- 6$ per second), so the derivative slams the brakes — the loop is aggressively preventing an overshoot. This is the "start un-turning early" instinct from the parent note, quantified.

Recall Self-test checklist (click to reveal)

Can you, without looking:

State $e = r - y$ and why the sign matters (P1.2)?
Compute $e_{ss} = d / K_{p}$ for pure P and explain why I zeroes it (P2.1, 2.2)?
Run one discrete PID update by hand (P3.1, 5.2)?
Give the sign of $u_{D}$ in all four motion cases including the frozen case (P3.2)?
Explain windup with a number and its fix (P4.2)?
Design a PI/PID from a spec and find the resting integral output (P5.1)? If any is shaky, reopen that problem before moving on.