Presented at Reno AIChE Meeting – Nov. 6, 2001 Getting More Information From Relay Feedback Tests

1. Presented at Reno AIChE Meeting – Nov. 6, 2001 Getting More Information From Relay Feedback Tests William L. Luyben Lehigh University

2. Scope • Relay-Feedback Test – method, advantages, • normal results = Ku and Pu • Curve Shapes – change with deadtime • Proposed Curvature Factor • Proposed Identification Method  D, Kp and  • Effectiveness • Conclusion

3. Relay-Feedback Test • Insert relay in feedback loop, specify height “h” • (high gain P controller, Hi and Lo limits on OP • OPhi – OPlo = 2 h) • Produces limit cycle in PV signal • with period = ultimate gain = Pu • and amplitude “a” PVmax – PVmin = 2 a • Calculate ultimate gain = Ku = 4h/a

4. Pu=3.5 a=0.17 0.2 0.1 PV 0 -0.1 -0.2 0 5 10 15 Time (minutes) h=1 1 0.5 OP 0 -0.5 -1 0 5 10 15 Time (minutes)

5. Advantages • Fast • Only need to specify “h” • Gives accurate dynamic information at frequency • that is important for feedback controller design. • Closedloop test. Keeps process in linear region. • Detect load changes from asymmetric pulses. • Limitation - only get two parameters. • Reference - Yu, C. C. Autotuning of PID Controllers, • 1999, Springer, London

6. Proposed Ways to Get More Information • Run two tests: conventional and then with dynamic element • inserted to get another point on the Nyquist plot. • (Li et al; Ind. Eng. Chem. Research1991, 30 1530) • Two-channel test: use two relays in parallel, one with • conventional and one with integrator. • (Friman and Waller; Ind. Eng. Chem. Research 1997, 36, 2662)

7. Curve Shapes Deadtime in process affects shape of curve. Mentioned by (1) Astrom and (2) Friman and Waller. Nothing quantitative proposed by these authors. Basic Idea – use curve shape to find a third process parameter.

8. First-Order with Deadtime 0.5 Deadtime = 1 0.4 63% Gain = Kp = 0.5 With OP = 1 0.3 PV 0.2 0.1 Time constant = 2 0 0 2 4 6 8 10 Time (minutes)

9. First-order/deadtime process D=0.1 0.15 0.1 0.05 Y 0 -0.05 -0.1 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time D=1 1 0.5 Y 0 -0.5 -1 0 1 2 3 4 5 6 7 8 9 10 Time D=10 1 0.5 Y 0 -0.5 -1 0 10 20 30 40 50 60 70 80 90 Time

10. Curvature Factor = a Y a/2 0 time b t1 t2 Pu

11. a Pu a Y 0 F=2 time -a a Y a/2 0 time b t1 t2 Pu a Y 0 time F=0.5 Pu

12. Range of Possible F Factors • is from 0.5 to 2. • Can relate F Curvature Factor • to D Deadtime.

13. D/ 0.1 1 10 Relay-feedback Test: Pu 0.382 2.98 21.4 a 0.0952 0.632 1.0 b 0.0488 0.620 9.31 Ku 13.4 2.01 1.27 F=4b/Pu 0.511 0.832 1.74 IMC:  0.2 1.7 17 Kc 5.25 0.882 0.353 I 1.05 1.5 5.5

14. 2 1.5 D = 1 1 Log10(D/) Pure Deadtime (D=) 0.5 0 -0.5 -1 Pure Integrator (D=0) -1.5 0.5 1 1.5 2 Curvature Factor = F = 4b/Pu

16. Proposed Procedure • Run relay-feedback test  a, b and Pu • Calculate Ku=4h/a , u =2/Pu and F=4b/Pu • Calculate D/ from correlation with F. • Define D/ = c (c is known constant) One equation in unknown .

17. 5. Calculate D = c  6. Calculate Kp 7. Now three process parameters are known: D,  and Kp. 8. Use IMC tuning rules.

18. Effectiveness First-order with various deadtimes Third-order Inverse response Openloop unstable

19. First-Order; D=0.05 1.6 ZN 1.4 1.2 TL 1 Y005 0.8 IMC 0.6 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time 20 15 ZN 10 TL M005 5 IMC 0 -5 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time Fig. 5

20. First-Order; D=0.1 1.5 ZN IMC 1 TL Y01 0.5 0 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time 10 ZN 8 IMC 6 M01 4 TL 2 0 -2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time Fig. 6

21. First-Order; D=1 1.4 1.2 IMC 1 0.8 ZN Y1 0.6 TL 0.4 0.2 0 0 1 2 3 4 5 6 7 8 9 10 Time 1.8 1.6 IMC 1.4 1.2 M1 1 ZN 0.8 TL 0.6 0.4 0 1 2 3 4 5 6 7 8 9 10 Time Fig. 7

22. First-Order; D=10 1 IMC 0.8 ZN 0.6 Y10 0.4 TL 0.2 0 0 10 20 30 40 50 60 Time 1 IMC 0.9 0.8 0.7 M10 ZN 0.6 0.5 TL 0.4 0.3 0 10 20 30 40 50 60 Time Fig. 8

23. Now test on third-order process. Actual process: Approximate process:

24. Third-Order; D=0.1 0.03 0.02 0.01 0 Y -0.01 -0.02 -0.03 0 2 4 6 8 10 12 14 16 18 20 Time Third-Order; D=1 0.1 0.05 0 Y -0.05 -0.1 0 5 10 15 20 25 30 Time Third-Order; D=10 0.2 0.1 0 Y -0.1 -0.2 0 10 20 30 40 50 60 70 80 90 Time Fig. 9

25. Third-Order; D=0.1 1.4 ZN 1.2 1 IMC 0.8 Y01 0.6 0.4 0.2 0 0 5 10 15 20 25 30 Time 25 ZN 20 15 M01 10 IMC 5 0 -5 0 5 10 15 20 25 30 Fig. 10 Time

26. Third-Order; D=1 1.4 1.2 ZN 1 0.8 IMC Y1 0.6 0.4 0.2 0 0 5 10 15 20 25 30 Time 11 ZN 10 9 8 M1 7 IMC 6 5 4 0 5 10 15 20 25 30 Time Fig. 11

27. Third-Order; D=10 1.4 1.2 IMC 1 0.8 Y10 0.6 ZN 0.4 0.2 0 0 10 20 30 40 50 60 70 80 90 Time 10 9 IMC 8 7 M10 6 ZN 5 4 3 0 10 20 30 40 50 60 70 80 90 Time Fig. 12

28. D=0.1; Inverse response 1.5 Tauz=1.6 1 0.5 Y 0 -0.5 Tauz=0.8 -1 -1.5 0 2 4 6 8 10 12 14 16 18 20 Time D=1; Inverse response 2 Tauz=1.6 1 Y 0 -1 Tauz=0.8 -2 0 2 4 6 8 10 12 14 16 18 20 Time D=10; Inverse response 2 Tauz=1.6 1 0 Y -1 Tauz=0.8 -2 0 10 20 30 40 50 60 70 80 90 Time Fig. 13

29. First-Order Unstable; D=0.1 0.2 0.1 Y1 0 -0.1 -0.2 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 Time First-Order Unstable; D=0.2 0.4 0.2 Y2 0 -0.2 -0.4 0 0.5 1 1.5 2 2.5 3 3.5 4 Time First-Order Unstable; D=0.3 0.4 0.2 Y3 0 -0.2 -0.4 0 1 2 3 4 5 6 Time Fig. 14

30. First-Order Unstable; D=0.1 2 IMC 1.5 TL Y1 1 ZN 0.5 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time 8 6 4 IMC TL 2 M1 0 -2 ZN -4 -6 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Time Fig. 15

31. First-Order Unstable; D=0.2 4 IMC 3 TL 2 Y2 1 ZN 0 -1 -2 0 1 2 3 4 5 6 7 8 9 10 Time 4 3 2 TL 1 M2 0 -1 -2 ZN IMC -3 -4 0 1 2 3 4 5 6 7 8 9 10 Time Fig. 16

32. First-Order Unstable; D=0.3 3.5 3 TL 2.5 2 1.5 Y3 1 ZN 0.5 0 -0.5 -1 0 2 4 6 8 10 12 14 16 18 20 Time 3 2 ZN 1 0 M3 -1 -2 -3 TL -4 -5 0 2 4 6 8 10 12 14 16 18 20 Time Fig. 17

33. Conclusion Only one simple test required. Use shape of curve to deduce deadtime. Works for first and higher-order systems. Does not work for inverse response or unstable.