1. Control Theory D action – Tuning

2. When there’s too much oscillation, this can sometimes be solved by adding a derivative action. This action will take into account how fast the error signal is changing. In the time domain this leads to: - the PD controller: - and the PID controller: Derivative Action

3. PD in the s domain: PID in the s domain: Derivative Action

4. Group Task P or PD? 2/(s 2 +2s) r(t) z(t) ProcessController Compare a P controller with K c =10 with a PD controller with K c =10 and T D =0,2 (step response for the servo problem)

5. Group Task

6. Tuning = choosing appropriate parameters for the controller For PID control: choice of K c, T I and T D 2 possibilities: - Based on (open/closed loop) model - Based on experiments 2 pretty much used methods: - Ziegler-Nichols - Cohen-Coon (not discussed) Tuning

7. (one type of) ZIEGLER-NICHOLS: 1)Start with a P controller with low K c and increase this gain slowly until y oscillates with a constant amplitude. (Remark: every time K c is increased, one should bring the closed loop system out of equilibrium, e.g. by slightly changing the reference!) 2) The gain we have now is called the ultimate gain K u and the period of the oscillation is called the ultimate period T u 3) Choose the PID parameters according to the following table: KcKc TITI TDTD P K u /2-- PI 0.45 K u T u /1.2- PID 0.6 K u T u /2T u /8 Tuning

8. Example: A process with the following step response: Tuning

9. Example: K u = 1.45 T u = 4.5s bv. PID: K c = 0.87 T I = 2.25s T D = 0.65s Tuning

10. Example: K u = 1.45 T u = 4.5s bv. PID: K c = 0.87 T I = 2.25s T D = 0.65s Remark: ZN typically leads to pretty much overshoot... (there are revised ZN methods that solve this problem) 5 1015 1 Tuning

11. Example: Suppose a model of the process was given by Can you now tune according to ZN based on this model? Tuning

12. Example: -3.227dB 1.4rad/s Tuning

13. Problem(s) introduced by D-control? A.Control action in servo problem becomes “infinitely” high. B.High frequency behavior/noise is amplified. C.None of the above. D.Both A and B

14. Bode plots! Bode plot PD:

15. We introduce an extra pole to solve the problem at high frequencies: This means we first filter the error with a low pass filter. Off course the time constant of the filtering << T D (Remark: to solve the problem of suddenly changing reference values, often another solution is presented: the derivative action is put on the MEASUREMENT instead of on the ERROR!) Derivative Action with filtering

16. Difference in Bode plot: Remark: The positive effect on the phase shift is now lost at higher frequencies! Blue: D action Green: D action + prefiltering Derivative Action with filtering

17. 1) What is a time delay? Typical example: measurement comes too late, e.g.: TRTR T R,m TRTR tdtd Influence of a time delay

19. Relationship between T R,m (t) and T R (t)? A.T R,m (t) = T R (t) B.T R,m (t) = T R (t-t d ) C.T R,m (t) = T R (t+t d ) D.None of the above TRTR T R,m tdtd

21. Which is true? A.A time delay = a pure phase shift, B is false B.A time delay = a non-linear subsystem, A is false C.Both are true D.Both are wrong

22. In other words: Dead time t d f(t) f(t-t d ) This is of course a pure phase shift. In Laplace (see table): Dead time t d F(s) e -t d s F(s) A dead time is given as e -t d s in the s domain: it’s non-linear! On the Bodeplot? AR = |e -t d jω |=1 φ =  (e -t d jω ) = -t d ω (in rad) Pure phase shift Influence of a time delay

23. A- Bode plot of a time delay: 2) What is the influence on a feedback system? Influence of dead time CHAPTER 3. PID CONTROL

24. φ =  (e -t d jω ) = -t d ω * 180/π (in degrees) This can have severe impact on the stability: The information comes too late. How can we see this in our analysis? - Based on TF: difficult: time delay = non-linear thing - … but it is a pure phase shift… on Bode plot? Influence of a time delay 2) What is the influence on a feedback system?