1. 1 Chap 6 The Compensation of the linear control systems P553

2. 2 Chap 6 The Compensation of the linear control systems §6-1 Introduction 6.1.1 definition of compensation 6.1.2 types of compensation §6-2 The basic controller operation analysis 6.2.1 PI D controller ---active compensation 6.2.2 phase-lead controller 6.2.3 phase-lag controller 6.2.4 phase lag-lead controller §6-3 Cascade compensation method of Root loci §6-4 Cascade compensation method of frequency- Domain §6-5 Feedback compensation passive compensation controller

3. 3 6.1 Introduction 6.1.1 What is compensation or correction of a control system ? solution

4. 4 This closed-loop system can be stable. We make the system stable by increasing a component. 6.1 Introduction This procedureis called the compensation or correction. Definition of the compensation: increasing a component,which makes the system’s performance to be improved, other than only varying the system’s parameters, this procedure is called the compensation or correction of the system.

5. 5 Compensator: The compensator is an additional component or circuit that is inserted into a control system to compensate for a deficient performance. 6.1 Introduction 6.1.2 Types of the compensation

6. 6 (1) Cascade(or series) compensation (2) Feedback compensation (3) Both series and feedback compensation (4) Feed-forward compensation (1) Cascade(or series) compensation Features : simple but the effects to be restricted. 6.1 Introduction

7. 7 (2) Feedback compensation R(s)C(s) - - R(s) C(s) - - Features: complicated but noise limiting, the effects are more than the cascade compensation. (3) Both cascade and feedback compensation - R(s)C(s) - Features: have advantages both of cascade and feedback compensation. 6.1 Introduction

8. 8 (4) Feed-forward compensation Features: theoretically we can make the error of a system to be zero and no effects to the transient performance of the system. 6.1 Introduction R(s)C(s) + - F(s) R(s) - + For inputFor disturbance(voice) Demonstration:

9. 9 R(s)C(s) + - For input 6.1 Introduction But no effect to the characteristic equation: 1+ G 10 G 20 = 0 Question: actually the could not be easy implemented especially maybe the G 20 is variable.

10. 10 6.1 Introduction C(s) F(s) R(s) - + For disturbance(voice) Also no effect to the characteristic equation: 1+ G 10 G 20 = 0 Question: actually the could not be easy implemented especially maybe the G 20 is variable. And the F(s) could not be easy measured.

11. 11 example 6.1 Introduction C(s) N(s) R(s) - - E(s) G CR + + Fig.6.1.7 For the system shown in Fig.6.1.7: Solution : Determine G CR and G CN, make E(s) to be zero. Where: Thinking: if r(t) = n(t) = t, Determine G CR and G CN, make e ss to be zero — as a exercise.

12. 12 6.2 Operation analysis of the basic compensators 6.2.1 Active Compensation PID controller － active “compensator”. Transfer function:

13. 13 PD controller C(s)C(s) G(s)G(s) R(s)R(s) － ＋ ＋ ＋

14. 14 Effects of PD controller: 2) PD controller improve the system’s stability (to increase damping and reduce maximum overshoot); 3) PD controller reduce the rise time and settling time; 4) PD controller increase BW(Band Width) and improve GM(K g ),PM(γ c ), and M r. 1) PD controller does not alter the system type; 6.2 Operation analysis of the basic compensators － bring in the noise !

15. 15 PI controller C(s)C(s) G(s)G(s) R(s)R(s) － ＋ ＋ ＋ 6.2 Operation analysis of the basic compensators

16. 16 Effects of PI controller: 1) Increase the system’s type － clear the steady-state error ; 2) reduce BW(Band Width) and GM(K g ), PM(γ c ) and M r ; 6.2 Operation analysis of the basic compensators beneficial to the noise limiting, not beneficial to the system’s stability.

17. 17 G(s) R(s)C(s) - + PID controller 6.2 Operation analysis of the basic compensators Transfer function: PID controller have advantages both of PI and PD.

18. 18 Circuits of PID _ + C R1R1 urur u0u0 PI controller R2R2 _ + C R1R1 urur u0u0 PD controller R2R2 _ + C1C1 R1R1 urur u0u0 PID controller C2C2 R2R2

19. 19 For example: Disk driver control system

20. 20 solution How to get ？ Shown in 6.3 detail.

21. 21 6.2.2 Passive compensation controllers Types of passive compensation controller

22. 22 Zero and pole 6.2.2 Passive compensation controllers

23. 23 Effects are similar to PD. Compensation ideal: make ω m to be ω c ! Bode plot z p

24. 24 Circuit of the phase-lead controller

25. 25 Zero and pole 6.2.2 Passive compensation controllers

26. 26 Effects are similar to PI. Compensation ideal: Make 1/τto be in the lower frequency-band and far from ω c ! Bode plot

27. 27 Circuit of the Phase-lag controller

28. 28 Zero and pole 6.2.2 Passive compensation controllers

29. 29 Effects are similar to PID. Compensation ideal: First make the phase-lag compensation － to satisfy e ss and compensate a part of γ c. second make the phase-lead compensation － to satisfy the transitional requirements. Bode plot

30. 30 Circuit of the Phase lag-lead controller

31. 31 6.2.3 Comparing active compensation controllers and passive compensation controllers 6.2 Operation analysis of the basic compensators

32. 32 6.3 Cascade compensation by Root loci method 6.3.1 Phase-lead compensation (P569) Example 6.3.1: The root loci of the system shown in Fig.6.3.1 Fig.6.3.1 solution Analysis: unstable. phase-lead compensation

33. 33 6.3 Cascade compensation by Root loci method Fig.6.3.2

34. 34 6.3 Cascade compensation by Root loci method Fig.6.3.3 There are two approaches to determine z c and p c. (1) Maximum α method (2) Method based on the open-loop gain

35. 35 6.3 Cascade compensation by Root loci method For this example we choose the Maximum α method: Fig.6.3.3 In terms of the sine’s law:

36. 36 Root locus of the compensted system 6.3 Cascade compensation by Root loci method The root locus of the compensated system is shown in Fig.6.3.4 Fig.6.3.4 Steps of the cascade phase-lead Compensation: (1) Determine the dominant roots based on the performance specifications of the system: (2) plot the root locus of the system and analyze what compensation device should be applied.

37. 37 6.3 Cascade compensation by Root loci method (3) Determine the angle φ c to be compensated: (4) calculate θ andγ: (6) plot the root locus of the compensated system and make validity check. (5)calculate z c and p c In terms of the sine’s law :

38. 38 Example 6.3.2: Solution: The root locus of the system is shown in Fig.6.3.5. - 10 Fig.6.3.5 6.3.2 Phase-lag compensation using the root locus (P577)

39. 39 6.3.2 Phase-lag compensation using the root locus (P577) Fig 6.3.6 The detail of the root-loci is shown in Fig 6.3.6.

40. 40 Fig 6.3.6 6.3.2 Phase-lag compensation using the root locus (P577)

41. 41 Validate…… 6.3.2 Phase-lag compensation using the root locus (P577) Steps of the cascade phase-lag Compensation: (1) Determine the dominant roots based on the performance specifications of the system: (2) plot the root locus of the system and analyze what compensation device should be applied. If the phase-lag Compensation be applied:

42. 42 6.3.2 Phase-lag compensation using the root locus (P577) (5) plot the root locus of the compensated system and make validity check. 6.3.3 Phase lag-lead compensation by the root locus method Basic ideal:

43. 43 First: make the phase-lead compensation － to satisfy the transitional requirements. Second: make the phase-lag compensation － to satisfy e ss requirements. Make compensation using PD and PI for example 6.3.1 and example 6.3.2 Exercise: 6.3.3 Phase lag-lead compensation by the root locus method

44. 44 6.4.1 Phase-Lead Compensation using Bode diagram 6.4 Cascade compensation by frequency response method

45. 45 6.4 Cascade compensation by frequency response method

46. 46 6.4 Cascade compensation by frequency response method

47. 47 Example: 6.4 Cascade compensation by frequency response method solution: － － 40dB/dec Fig.6.4.1

48. 48 6.4 Cascade compensation by frequency response method － － 40dB/dec Fig.6.4.1 1.Make validity check for this example. 2. Make compensation using PD for this example. Exercise:

49. 49 6.4.2 Phase-Lag Compensation using Bode diagram Example: 2 6.32 － － 20dB/dec － － 40dB/dec － － 90 0 － － 180 0 － － 20lgβ solution: Fig.6.4.2

50. 50 6.4.2 Phase-Lag Compensation using Bode diagram 2 6.32 － － 20dB/dec － － 40dB/dec － － 90 0 － － 180 0 － － 20lgβ Validate…… Fig.6.4.2

51. 51 6.4.2 Phase-Lag Compensation using Bode diagram 1. Make validity check for this example. 2. Make phase-lag compensation for γ c =50 o and K v =20. Steps of the phase-lag compensation: Exercise:

52. 52 6.4.4 Compensation according to the desired frequency response Example solution 10 100 － 20 － 20dB/dec － 40 － 40dB/dec First: make the phase-lag compensation － to satisfy e ss and compensate a part of γ c. Second: make the phase-lead compensation － to satisfy the requirementsγ c and ω c etc. 6.4.3 Phase-Lag-lead Compensation using Bode diagram Fig.6.4.3

53. 53 6.4.4 Compensation according to the desired frequency response 10 100 － 20 － 20dB/dec － 40 － 40dB/dec In terms of the desired frequency response we have: Fig.6.4.3

54. 54 10 100 － 20 － 20dB/dec － 40 － 40dB/dec 1 Exercise: Fig. 6.4.4

55. 55 6.5 Feedback compensation R(s)R(s)C(s)C(s) － － G’ 0 (s) R(s)R(s) C(s)C(s) － － G’ 20 (s) 6.5.1 The configuration of the Feedback compensation 6.5.2 The basic Feedback compensators Fig. 6.5.1

56. 56 2. Impair(weaken) the influences of the disturbance to the encircled elements. 3. make the performance of the encircled elements to be desired. 1.Decrease the time constant of the encircled elements → Quicken the response of the encircled elements － may be; For example 6.5.3 Function of the feedback:

57. 57 6.5.4 The design procedure of the feedback compensator 1. Design the desired characteristics, such as the desired Bode diagram, of the encircled elements in terms of the system’s analysis. 2. Choose the appropriate feedback compensators to get the desired characteristics. Example R(s)R(s) C(s)C(s) － － G’ 20 (s) Fig. 6.5.2 For the system shown in Fig. 6.5.2, G 20 =10/s 2, the desired G’ 20 (jω) shown in Fig. 6.5.3. determine the G c. 1 10 － 20 － 20dB/dec － 40 － 40dB/dec 0.1 － 40 － 40dB/dec Fig. 6.5.3 solution

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