Parallel Compensator for Continuous and Relay Control Systems with Difficult Plants

1. Parallel Compensator for Continuous and Relay Control Systems with Difficult Plants Ryszard Gessing Silesian University of Technology Gliwice, Poland

2. Outlineof Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

3. Introduction • Smith predictor (compensator), 1958 • Deng, Iwai and Mizumoto 1999 for minimum phase plants • Gessing ACC 2004 for nonminimum phase plants

4. Outline of Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

5. Plant: is stable Parallel Compensator: Replacement Plant:

6. = const It should be: In steady state:

7. Outline of Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

8. Regulator: (for a large k) Closed loop System:

9. Outline of Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

10. Replacement plant: is chosen so that: n – degree ofM(s), T is chosen so that then the CL system is stable for very large k

11. CL system description: Characteristic equation: or which has (n-1)-multiple root:

12. Example 1 Plant: For the replacement plant is: and parallel compensator: Assume k=1000 which gives

13. For r=1(t-1) we obtain: Settling time for t ~1, u ~1000 for t ~1.15, u ~-140 Acceptable responses for

14. Outline of Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

16. Example 2 Data as in Example 1 For insignificant increase of settling time For settling time ~ 5

17. Comparison with regulator PID Settling time ~ 12.2 about 4-times longer.

18. Outline of Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

19. During fast switchings we have:

20. Equivalent systems Relay system Continuous system switched amplitude then both the systems are equivalent, i.e. for the same reference values r(t) we obtain the same outputsy(t) in both the systems.

21. Example 3 The plant and parallel comensator the same as in previous examples. Relay system: h=0.01, H=10 solid line Continuous system: k=1000, dotted line For h=0.005 both the outputs are nondistinguishable.

22. Relay system h=0.005, H=20 Continuous system The outputs y for both the systems are nondinguishable.

23. Comparison Sliding mode control Parallel compensator Ass; minimum phase plant without this assuption Characteristic equation:

24. Outline of Presentation: • Introduction • Parallel Compensator • Approximate Description of the CL System • Choice of the Replacement Plant • Accounting of the Control Saturation • Relay Implementation • Final Conlusions

25. Easy for desinging • Some freedom in choosing the replacement plant and shaping dynamics • Shortening of transients • It has some robustness with respect to change of the plant parameters. • Implementation of sliding mode control without necessity of applying of higher order derivatives • Equivalence of continuous and relay systems • Plant must be stable • Regulator has a high order