# what is a quarter wave decay response after Ziegler–Nichols tuning?

Ziegler–Nichols tuning creates a "quarter wave decay". What does that mean? I cannot find anything online!

• Maybe they are referring to a damping coefficient, $\zeta$, of 1/4 (from $1/(s^2+2\zeta\omega s+\omega^2)$), since I did a quick test tuning a PID with this method for a simple system and noticed that the step response had a similar oscillation and decay structure. – fibonatic May 13 '16 at 23:59
• The phrase you are looking for is "quarter amplitude decay ratio", which somehow became associated with acceptable control performance. – OldUgly May 16 '16 at 6:30

The response of a system can be seen as a sum of it's stationary response and it's transient response.

The transient response disappears after some time. It has an oscilatory behavior, even if it is just an exponential (see complex Fourier series: every periodic function is a sum of exponentials).

By "quarter wave decay" we mean that, after doing Ziegler-Nichols' tuning, the transient response will (usually) decay four times faster. Thus, it's period is a quarter of the untuned response.

In other words: you increase the frequency of the transient response, also increasing the rate at which it's energy is dissipated. This makes the period of existence of the transient response a quarter of the original one. The downside is that this creates a high overshoot.

Edit:

Ziegler-Nichols aims to reduce the time that it takes to reach the permanent (set point) response, while reducing the overshoot to 1/4 of the untunned response. This is done by adjustments in the proportional gains.

I found this GIF in wikipedia that illustrates well that concept.