PID tuning (Ziegler–Nichols)
Have the ultimate gain from an oscillation test, or the parameters of the process step response? Enter them to get settings and a chart of how the loop handles a set-point step.
Result
- Overshoot (simulated)
- 65.5%
- Settling time (±2%)
- 74.5 s
- Final value
- 1
| Method | Kp | Ti [s] | Td [s] | Ki [1/s] | Kd [s] |
|---|---|---|---|---|---|
| Ziegler–Nichols | 2.7 | 16.67 | — | 0.162 | — |
| CHR | 1.05 | 36 | — | 0.02917 | — |
| SIMC | 1.5 | 30 | — | 0.05 | — |
Simulation: process K·e^(−Ls)/(Ts+1), ideal controller (Kp, Ti, Td), derivative on the measurement with a Td/10 filter. Ki = Kp/Ti and Kd = Kp·Td are the parallel-form settings.
The Ziegler–Nichols method
The first version is the oscillation test: turn off integral and derivative action and raise the P gain until the loop oscillates with constant amplitude. That gain is Ku and the period is Tu. A classic PID then has Kp = 0.6·Ku, Ti = Tu/2, Td = Tu/8.
The second version uses the open-loop step response. Step the control signal and read the gain K, dead time L and time constant T. Ziegler–Nichols rules aim for quarter-amplitude decay (each oscillation is ¼ of the previous one), so they are aggressive and overshoot a lot. CHR and SIMC are gentler.
Reading the chart
The chart shows the simulated response to a set-point step from 0 to 1 for the model K·e^(−Ls)/(Ts+1). A real process will behave a little differently, so treat the settings as a starting point for tuning on site.
Common questions
What is the difference between ideal and parallel form?
In the ideal (ISA) form, integral and derivative depend on Kp: Kp·(e + 1/Ti∫e + Td·de/dt). In the parallel form each term has its own gain: Kp·e + Ki∫e + Kd·de/dt, where Ki = Kp/Ti and Kd = Kp·Td. Check which one your controller uses.
How do I read L and T from a step response?
Draw a tangent at the steepest point. L is the time from the step to where the tangent crosses the starting value, and T is the time from there to where it crosses the final value.