Closing the loop with P and PI control · Fundamentals of Control Engineering

Prediction

A proportional controller reaches 90% of a constant setpoint. Will waiting longer remove the remaining 10%?

Decide whether the controller can maintain a non-zero effort when the error becomes zero.

Proportional action makes error expensive

With $u=K_P e$ , a larger error commands a larger correction. For a unit-gain PT1 plant under unity feedback, the closed-loop response remains first order. Increasing $K_P$ makes it faster and reduces the steady-state error—but does not eliminate that error at finite gain.

Integral action accumulates error:

u(t)=K_P e(t)+K_I\int_0^t e(\tau)\,d\tau.

If a constant error remained, the integrator would keep changing the control input. In the ideal stable loop, the only steady possibility is zero constant error.

Concept check

Why can PI control hold non-zero actuator effort at zero steady error?

Reactive experiment: trade offset for effort

Worked example

For a plant $P(s)=1/(Ts+1)$ with a P controller $K_P$ , the reference-to-output transfer is

\frac{K_P}{Ts+1+K_P}= \frac{K_P/(1+K_P)}{(T/(1+K_P))s+1}.

The loop is $1+K_P$ times faster, but its unit-step final value is $K_P/(1+K_P)$ . At $K_P=9$ , the final value is 0.9—not 1.

Derivation

The final-value view

For a stable unity-feedback loop with loop transfer L(s), the error is E(s)=R(s)/(1+L(s)). A unit step has R(s)=1/s, so e∞=lim(s→0) sE(s)=1/(1+L(0)). A P-controlled finite-DC-gain plant has finite L(0); an integrator makes L(0) unbounded in the ideal model.

Unofficial TUM ACE exam prep

Separate Führungs- and Störverhalten

Do not infer disturbance rejection from the reference response alone. Write the relevant closed-loop transfer function for each signal path and state the assumptions behind the final-value theorem.

Guided exercise

For $P(s)=2/(3s+1)$ , derive the P-controlled closed-loop gain, time constant, and unit-step offset as functions of $K_P$ . Choose $K_P$ for an offset below 5%, then discuss whether the implied actuator demand is plausible.

Hint

The DC loop gain is (2K_P), so the unit-step error is (1/(1+2K_P)).

Summary

Proportional gain usually speeds the loop and reduces offset, at the cost of effort and robustness margin.
Integral action removes constant offset in the ideal stable model.
Saturation, windup, noise, and delay turn “increase the gain” into a design trade-off.

Retrieve it later

Derive the unit-step offset for a P-controlled PT1 plant and explain how an integrator changes the steady-state argument.

Answer from memory tomorrow, then return to check your reasoning.