Exercise set

PID Controller Tuning and Anti-Windup Exercises

Solved PID controller exercises for tuning, steady-state error, derivative filtering, saturation, anti-windup, sampling, margins and validation.

Branch: Automation and Control Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.1 · published

These exercises practise PID controller tuning as an engineering calculation, not as a table lookup. The goal is to connect proportional, integral and derivative action to steady-state error, transient response, actuator limits, sampling, filtering, anti-windup and validation evidence.

All examples use simplified single-loop models. Real PID commissioning also requires plant identification, sensor calibration, actuator checks, sign verification, safety limits, operating-point coverage and test data. A controller that gives a clean spreadsheet result can still fail if the loop sign is wrong, the actuator saturates, the sample time is too slow, the derivative term amplifies noise or the validation test omits disturbances.

How to use these exercises

Solve each problem in two passes. First, compute the requested PID gain, output, margin, saturation effect or sampled update. Second, decide whether the result would support release on real hardware or a real process.

Keep controller form explicit. A dependent form, an independent parallel form, a velocity form and a vendor PLC block may use different gain units. Copying gains between forms without conversion is one of the fastest ways to create a dangerous loop.

Release Evidence Notes

PID evidence should state the controlled variable, manipulated variable, error sign, controller form, gain units, sample time, actuator limits, anti-windup method, derivative filter, tuning method, operating case, disturbance case and acceptance criteria.

Validation should include setpoint steps, disturbance tests, saturation recovery, manual-to-auto transfer, sensor-noise checks, sample-time and delay review, actuator-rate review, gain and phase margin evidence, and a clear record of what changed after tuning.

Engineering Boundary Notes

These exercises use simplified single-loop models. They do not replace plant identification, safety review, actuator sizing, sensor calibration, interlock testing, nonlinear process review, multi-loop interaction study or field commissioning. A PID calculation is acceptable only when controller form, units, sign, limits and operating envelope match the installed implementation.

Common Release Mistakes

copying PID gains between vendor forms without unit conversion;
tuning on a small setpoint step while disturbances or saturation control the real loop;
enabling integral action without anti-windup or manual-to-auto transfer checks;
adding derivative action without sensor-noise and filter review;
ignoring sample time, dead time, actuator rate limit and output clamps;
releasing a loop because it is stable at one operating point while other ranges were not tested.

Scenario Map

Scenario	Main calculation	Engineering decision
Controller form	gain conversion, units and signs	Avoid copying PID gains into the wrong implementation.
Tuning	IMC, Ziegler-Nichols, gain and time constants	Produce a defensible starting point, not a final unchecked setting.
Steady-state error	system type and integral action	Decide when integral action is required and when it adds risk.
Frequency margin	crossover, phase margin and delay	Check whether tuning leaves room for sampling, filtering and dead time.
Saturation	output limits, windup and recovery	Test whether the loop recovers when the actuator cannot follow the command.
Digital implementation	sampled integrator, derivative filter and update timing	Confirm that firmware timing matches the control assumption.
Release validation	multi-criterion pass/fail screen	Hold release when any margin, limit or evidence item is missing.

Validation Package Checklist

controlled variable, manipulated variable and loop sign are verified;
controller form, gain units and sample time are documented;
actuator limits, rate limits and anti-windup behavior are tested;
derivative filter, sensor noise and quantization are reviewed;
setpoint, disturbance, saturation and manual-to-auto tests are recorded;
gain margin, phase margin, delay and operating-point coverage are checked;
final release decision states accept, retune, restrict range or hold for commissioning.

Exercise 1: Convert dependent PID settings to parallel gains

A controller vendor gives the dependent PID form:

\displaystyle u(t)=K_c\left(e(t)+\frac{1}{T_i}\int e(t)\,dt+T_d\frac{de(t)}{dt}\right)

with:

K_c=3.2,\qquad T_i=4.0\ \text{s},\qquad T_d=0.35\ \text{s}.

Convert the settings to the parallel form:

\displaystyle u(t)=K_p e(t)+K_i\int e(t)\,dt+K_d\frac{de(t)}{dt}.

Solution

Match coefficients term by term:

K_p=K_c

\displaystyle K_i=\frac{K_c}{T_i}

K_d=K_cT_d

Substitute the values:

K_p=3.2

\displaystyle K_i=\frac{3.2}{4.0}=0.80\ \text{s}^{-1}

K_d=3.2(0.35)=1.12\ \text{s}.

Engineering Comment

The numbers are not interchangeable labels. K_i in the parallel form has inverse-time units relative to the error-unit basis, while T_i is an integral time. Using T_i=4.0 as if it were K_i=4.0 would make the integral action five times stronger than intended.

Plausibility Check

The integral time is several seconds, so the parallel integral gain should be smaller than the proportional gain. The derivative time is less than one second, so the derivative gain should be smaller than the proportional gain in this unit convention. The converted values match that expectation.

Exercise 2: Compute a proportional output and check sign

A temperature loop uses:

e=r-y

where $r$ is setpoint and $y$ is measured temperature. At one instant:

r=80\ \text{deg C},\qquad y=73\ \text{deg C},\qquad K_p=4.5.

The output increases heater power when $u$ increases. Compute the proportional contribution and state whether the sign is sensible.

Solution

The error is:

e=80-73=7\ \text{deg C}.

The proportional contribution is:

u_P=K_p e=4.5(7)=31.5.

The error is positive because the measured temperature is below the setpoint. A positive output increases heater power, so the sign is sensible for heating.

Engineering Comment

This is a loop-sign test. If the output increase instead opened a cooling valve, the same algebra would drive the process away from the setpoint. PID tuning should not begin until the error sign and actuator direction are verified.

Plausibility Check

The measured temperature is low, so the heater command should increase. A positive proportional output satisfies that physical expectation.

Exercise 3: Estimate steady-state error with proportional control

A unity-feedback loop uses proportional control:

C(s)=K_p

and a first-order plant:

\displaystyle G(s)=\frac{2}{5s+1}.

For $K_p=3$ , estimate the final error for a unit step input.

Solution

The loop transfer function is:

\displaystyle L(s)=C(s)G(s)=\frac{6}{5s+1}.

For a type 0 unity-feedback system, the position error constant is:

K_{\text{pos}}=\lim_{s\to0}L(s)=6.

The step steady-state error is:

\displaystyle e_\infty=\frac{1}{1+K_{\text{pos}}} =\frac{1}{1+6} =0.1429.

The final error is about $14.3\%$ of the unit step.

Engineering Comment

Proportional gain reduces offset but does not eliminate it for this type 0 loop. Increasing $K_p$ would reduce offset but can also reduce stability margin, increase overshoot and demand more actuator effort.

Plausibility Check

The plant static gain is $2$ , and the proportional gain is $3$ , so the DC loop gain is $6$ . An error near one seventh is plausible.

Exercise 4: Add integral action and check the characteristic equation

The same plant is controlled by:

\displaystyle C(s)=K_p+\frac{K_i}{s}

with:

K_p=3,\qquad K_i=0.8.

Find the closed-loop characteristic equation and determine whether the second-order denominator is stable.

Solution

The loop transfer function is:

\displaystyle L(s)=\left(3+\frac{0.8}{s}\right)\frac{2}{5s+1} =\frac{6s+1.6}{s(5s+1)}.

The unity-feedback characteristic equation is:

1+L(s)=0.

Therefore:

s(5s+1)+6s+1.6=0.

Expand:

5s^2+s+6s+1.6=0

5s^2+7s+1.6=0.

For a second-order polynomial:

a_2s^2+a_1s+a_0

stability requires all coefficients positive. Here:

a_2=5,\qquad a_1=7,\qquad a_0=1.6.

All are positive, so the model is stable.

Engineering Comment

Integral action changes the loop type and can remove steady-state error for a step input, but it also adds a pole at the origin before closure. The stability check is therefore not optional.

Plausibility Check

The integral gain is modest relative to the proportional action. The stable result is plausible, but the damping and actuator demand still need separate checks.

Exercise 5: Choose IMC PI tuning for a first-order-plus-dead-time model

A process identification test gives a first-order-plus-dead-time model:

\displaystyle G_p(s)=\frac{K e^{-Ls}}{\tau s+1}

with:

K=2.5,\qquad \tau=18\ \text{s},\qquad L=3\ \text{s}.

Use the simple IMC-style PI rules:

\displaystyle K_c=\frac{\tau}{K(\lambda+L)},\qquad T_i=\tau

with desired closed-loop time constant:

\lambda=12\ \text{s}.

Find $K_c$ , $T_i$ and the parallel integral gain $K_i=K_c/T_i$ .

Solution

The controller gain is:

\displaystyle K_c=\frac{18}{2.5(12+3)} =\frac{18}{37.5} =0.48.

The integral time is:

T_i=18\ \text{s}.

The parallel integral gain is:

\displaystyle K_i=\frac{K_c}{T_i} =\frac{0.48}{18} =0.0267\ \text{s}^{-1}.

Engineering Comment

The selected $\lambda$ is larger than the dead time, so the tuning is intentionally moderate. A much smaller $\lambda$ would demand faster response but would leave less margin for valve lag, measurement filtering and model error.

Plausibility Check

The process gain is greater than one and the process is relatively slow, so the controller gain is below one and the integral gain is small. That is consistent with a conservative process-loop tuning.

Exercise 6: Compute Ziegler-Nichols closed-loop PID settings

A relay test or cautious ultimate-gain test gives:

K_u=6.0,\qquad P_u=8.0\ \text{s}.

Using the classic closed-loop Ziegler-Nichols PID starting rules:

K_c=0.6K_u,\qquad T_i=0.5P_u,\qquad T_d=0.125P_u,

find $K_c$ , $T_i$ , $T_d$ , and the parallel gains $K_i=K_c/T_i$ and $K_d=K_cT_d$ .

Solution

The dependent-form gain is:

K_c=0.6(6.0)=3.6.

The integral time is:

T_i=0.5(8.0)=4.0\ \text{s}.

The derivative time is:

T_d=0.125(8.0)=1.0\ \text{s}.

The parallel gains are:

\displaystyle K_i=\frac{3.6}{4.0}=0.90\ \text{s}^{-1}

and:

K_d=3.6(1.0)=3.6\ \text{s}.

Engineering Comment

Ziegler-Nichols settings are starting points, not release settings. They often give an aggressive response. A real commissioning procedure should check overshoot, actuator effort, noise sensitivity, safety limits and robustness before accepting the result.

Plausibility Check

The ultimate gain is $6.0$ , and the suggested controller gain is $60\%$ of that value. That leaves some distance from sustained oscillation, but not necessarily enough for a noisy or delay-dominated process.

Exercise 7: Check derivative filter frequency

A derivative term uses the practical filtered form:

\displaystyle C_D(s)=K_d\frac{s}{1+\tau_f s}.

The derivative time is:

T_d=0.40\ \text{s}

and the filter is chosen with:

\displaystyle N=10,\qquad \tau_f=\frac{T_d}{N}.

Find $\tau_f$ and the filter corner frequency:

\displaystyle f_c=\frac{1}{2\pi\tau_f}.

Solution

The filter time constant is:

\displaystyle \tau_f=\frac{0.40}{10}=0.040\ \text{s}.

The corner frequency is:

\displaystyle f_c=\frac{1}{2\pi(0.040)} =3.98\ \text{Hz}.

Engineering Comment

The derivative path will still react strongly to measurement content below and around this frequency. If the sensor is noisy in that band, derivative action can create output chatter even when the model looks well damped.

Plausibility Check

A $0.04\ \text{s}$ filter is fast compared with a $0.40\ \text{s}$ derivative time, so a corner near $4\ \text{Hz}$ is reasonable.

Exercise 8: Estimate phase-margin loss from digital delay

A PID loop has crossover frequency:

\omega_c=4.0\ \text{rad/s}.

The measured phase margin before digital implementation is:

\text{PM}_0=55^\circ.

Sampling, computation and output update add an effective delay:

T_d=0.055\ \text{s}.

Estimate the phase lag:

\displaystyle \phi_d=\omega_cT_d\frac{180}{\pi}

and the remaining phase margin.

Solution

The delay phase lag is:

\displaystyle \phi_d=4.0(0.055)\frac{180}{\pi} =12.6^\circ.

The remaining phase margin is:

\text{PM}=55^\circ-12.6^\circ=42.4^\circ.

Engineering Comment

The loop may still be acceptable if the requirement is, for example, at least $40^\circ$ phase margin. It is not acceptable to ignore the delay simply because the continuous-time design looked stable.

Plausibility Check

The delay is small compared with one second, and the crossover is only $4\ \text{rad/s}$ , so a phase penalty of about $13^\circ$ is plausible.

Exercise 9: Compute saturation and windup during a heat-up event

A digital PI controller computes:

u_{\text{raw}}=K_p e+I.

At one sample:

K_p=5,\qquad e=18,\qquad I=42.

The actuator output is limited to:

0\le u\le 100.

Find the raw output, saturated output and saturation error:

e_{\text{sat}}=u-u_{\text{raw}}.

Solution

The raw output is:

u_{\text{raw}}=5(18)+42=132.

The actuator cannot exceed $100$ , so:

u=100.

The saturation error is:

e_{\text{sat}}=100-132=-32.

Engineering Comment

The controller is asking for more heat than the actuator can deliver. If the integrator continues to grow during this condition, the loop can overshoot after the process finally approaches setpoint.

Plausibility Check

The raw output exceeds the limit by $32$ command units, so the saturated output must be exactly the upper limit. The negative saturation error correctly indicates that the actual actuator output is below the requested output.

Exercise 10: Apply back-calculation anti-windup

Use the saturation case from Exercise 9. The previous integrator state is:

I_k=42.

The discrete integral increment from normal PI action is:

K_iT_s e=3.6.

Back-calculation uses:

I_{k+1}=I_k+K_iT_s e+K_{aw}(u-u_{\text{raw}})

with:

K_{aw}=0.20.

Find the next integrator state.

Solution

From Exercise 9:

u-u_{\text{raw}}=-32.

The anti-windup correction is:

K_{aw}(u-u_{\text{raw}})=0.20(-32)=-6.4.

The next integrator state is:

I_{k+1}=42+3.6-6.4=39.2.

Engineering Comment

The normal integral action would have increased the integrator to $45.6$ . Back-calculation instead reduces it to $39.2$ because the actuator is already saturated. This helps the loop recover when the process catches up.

Plausibility Check

The correction magnitude exceeds the normal integral increment, so the integrator should decrease. The computed result does decrease from $42$ to $39.2$ .

Exercise 11: Decide whether conditional integration should hold

A controller uses conditional integration. The rule is:

if the actuator is saturated high and the error would drive the command higher, hold the integrator;
if the actuator is saturated high and the error would drive the command lower, allow integration.

At one sample:

u=100,\qquad u_{\text{raw}}=118,\qquad e=5.

The loop uses positive controller action: positive error increases output. Should the integrator update be held or allowed?

Solution

The actuator is saturated high because:

u=100<u_{\text{raw}}=118.

The error is positive, and positive error increases output. Therefore the normal integral update would drive the command even higher.

The integrator update should be held.

Engineering Comment

Conditional integration is simple and robust when the sign logic is documented. It can be unsafe if copied into a reverse-acting loop without rechecking the error and actuator direction.

Plausibility Check

The actuator is already at its upper limit. Integrating a positive error would ask for still more output. Holding the integrator is the physically sensible decision.

Exercise 12: Check bumpless manual-to-auto transfer

A loop is in manual mode with actual actuator output:

u_{\text{manual}}=46.

At the transfer instant:

K_p e=12,\qquad D= -3.

For bumpless transfer, choose the integrator state $I$ so that:

u=K_p e+I+D=u_{\text{manual}}.

Find $I$ .

Solution

Set the controller output equal to the manual output:

12+I-3=46.

So:

I+9=46

and:

I=37.

Engineering Comment

Tracking the manual output before automatic transfer prevents a command bump. Without this initialization, the loop could jump immediately to an actuator value unrelated to the operator’s current safe setting.

Plausibility Check

The proportional and derivative terms already contribute $9$ command units. The integrator must supply the remaining $37$ to match the manual output of $46$ .

Exercise 13: Compute an incremental PI update

A digital velocity-form PI controller updates the command as:

\Delta u_k=K_p(e_k-e_{k-1})+K_iT_s e_k.

Given:

K_p=2.4,\qquad K_i=0.8\ \text{s}^{-1},\qquad T_s=0.10\ \text{s},

e_{k-1}=4,\qquad e_k=3,

and previous output:

u_{k-1}=52,

find $\Delta u_k$ and $u_k$ before saturation.

Solution

The proportional increment is:

K_p(e_k-e_{k-1})=2.4(3-4)=-2.4.

The integral increment is:

K_iT_s e_k=0.8(0.10)(3)=0.24.

The command increment is:

\Delta u_k=-2.4+0.24=-2.16.

The new command is:

u_k=52-2.16=49.84.

Engineering Comment

The output decreases because the error is falling. The integral term still pushes upward because the current error remains positive. This is typical PI behavior near a setpoint.

Plausibility Check

The error dropped by one unit, so a modest output decrease is expected. The final value is close to the previous command, not a large jump.

Exercise 14: Choose a sample time from bandwidth guidance

A speed loop is designed for approximate closed-loop bandwidth:

f_b=3\ \text{Hz}.

Use the rule that controller sampling frequency should be at least $20$ times the bandwidth:

f_s\ge 20f_b.

Find the maximum sample time:

\displaystyle T_s\le \frac{1}{f_s}.

Solution

The minimum sampling frequency is:

f_s\ge 20(3)=60\ \text{Hz}.

The maximum sample time is:

\displaystyle T_s\le \frac{1}{60}=0.0167\ \text{s}.

So:

T_s\le 16.7\ \text{ms}.

Engineering Comment

This is only a starting rule. Encoder resolution, computation jitter, PWM update timing, sensor filtering and zero-order-hold delay can require a faster sample rate or a lower bandwidth.

Plausibility Check

A loop with a few-hertz bandwidth should not be updated only a few times per second. A sample time below about $17\ \text{ms}$ is consistent with the stated rule.

Exercise 15: Estimate setpoint weighting effect

A PID loop is at steady operation with:

r_0=80,\qquad y=80.

The setpoint is then changed to:

r_1=100.

The proportional gain is:

K_p=2.0.

For the proportional setpoint-kick contribution:

\Delta u_P=K_p\beta\Delta r,

compare the kick for $\beta=1.0$ and $\beta=0.4$ .

Solution

The setpoint change is:

\Delta r=r_1-r_0=100-80=20.

For $\beta=1.0$ :

\Delta u_P=2.0(1.0)(20)=40.

For $\beta=0.4$ :

\Delta u_P=2.0(0.4)(20)=16.

The reduced weighting cuts the proportional setpoint kick by:

\displaystyle \frac{16}{40}=0.40.

Engineering Comment

Setpoint weighting reduces proportional kick on setpoint changes while retaining disturbance response through measured output feedback. The exact formula depends on the controller structure, so commissioning records should state the vendor form, not only the value of $\beta$ .

Plausibility Check

Reducing $\beta$ from $1.0$ to $0.4$ should reduce the setpoint kick to $40\%$ of its original value. The kick comparison gives $16/40=0.4$ .

Exercise 16: Check actuator headroom after tuning

A tuned controller produces a maximum expected output of:

u_{\max,\text{test}}=82

on a design disturbance. The actuator limit is:

u_{\max}=100.

The release rule requires at least $15\%$ actuator headroom relative to full scale:

u_{\max,\text{allowed}}=100-15=85.

Does the test pass?

Solution

The tested maximum output is:

82.

The allowed maximum output is:

85.

Because:

82<85,

the actuator-headroom screen passes.

The remaining headroom is:

100-82=18

or $18\%$ of full scale.

Engineering Comment

The loop has enough output margin for the tested disturbance, but this does not prove stability. Headroom should be checked together with phase margin, rate limits, saturation recovery and disturbance coverage.

Plausibility Check

The requirement is $15\%$ headroom and the test leaves $18\%$ , so the pass margin is only $3$ percentage points. This is acceptable but not generous.

Exercise 17: Screen derivative noise amplification

A speed estimate has high-frequency noise with approximate RMS amplitude:

n_{\text{rms}}=0.8\ \text{rpm}.

The discrete derivative uses:

\displaystyle D_k=K_d\frac{y_{k-1}-y_k}{T_s}

with:

K_d=0.015,\qquad T_s=0.005\ \text{s}.

Estimate the derivative-output RMS caused by one-sample noise difference of $0.8\ \text{rpm}$ .

Solution

The noise slope estimate is:

\displaystyle \frac{0.8}{0.005}=160\ \text{rpm/s}.

The derivative output is:

D_{\text{noise}}=0.015(160)=2.4.

Engineering Comment

A few output units of derivative noise may be acceptable for a wide actuator, but it may be unacceptable for a small valve, motor duty cycle or precision servo. The derivative filter and speed-estimation method should be validated with real data.

Plausibility Check

The sample time is very short, so even small one-sample noise produces a large numerical slope. The computed output chatter is therefore plausible.

Exercise 18: Decide whether a PID loop is ready for release

A commissioning test gives the following evidence for a PID temperature loop:

Check	Requirement	Measured
Overshoot	$\le 10\%$	$8.5\%$
Settling time	$\le 90\ \text{s}$	$84\ \text{s}$
Phase margin	$\ge 40^\circ$	$43^\circ$
Maximum output	$\le 85\%$	$88\%$
Saturation recovery test	required	not performed

Should the loop be released?

Solution

Check each requirement.

Overshoot passes:

8.5\%<10\%.

Settling time passes:

84\ \text{s}<90\ \text{s}.

Phase margin passes narrowly:

43^\circ>40^\circ.

Maximum output fails:

88\%>85\%.

The saturation recovery test is missing, so the anti-windup and recovery behavior are not validated.

The loop should not be released. It should be held for actuator-headroom correction, saturation recovery testing and review of whether the phase-margin result remains valid after retuning.

Engineering Comment

Release is controlled by the weakest required evidence, not by the average impression. A loop that tracks setpoints well can still fail in operation if it has no actuator headroom or if saturation recovery is untested.

Plausibility Check

Three checks pass, one numerical check fails and one required test is missing. Holding release is the conservative and technically defensible decision.

Review Checklist

Before accepting a PID controller, check:

controlled variable, manipulated variable and feedback sign;
controller form and gain units;
setpoint weighting, derivative-on-measurement and derivative filter settings;
sample time, computation delay, output update timing and jitter;
actuator output limit, rate limit, deadband and fail state;
anti-windup method and saturation recovery evidence;
setpoint response, disturbance response and operating-point coverage;
phase margin, gain margin or other robustness evidence;
manual-to-auto transfer and startup/shutdown behavior;
final settings, firmware or PLC version, and validation data.

Common Mistakes

Copying PID gains between dependent and parallel forms without conversion.
Tuning setpoint tracking while ignoring disturbance rejection.
Adding integral action without checking stability and saturation.
Using derivative action on a noisy measurement without filtering.
Treating Ziegler-Nichols settings as final release values.
Ignoring sample time, zero-order hold and computation delay.
Accepting a simulation without actuator limits, rate limits or deadband.
Failing to test manual-to-auto transfer.
Reporting gains without units, controller form or sign convention.
Releasing a loop without saturation and anti-windup recovery evidence.

The central habit is to treat PID tuning as a controlled engineering change. A good final answer is not only a set of gains; it is a documented loop boundary, tested implementation and defensible release decision.

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