Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 10 Stability Analysis and Controller Tuning ※ Bounded-input bounded-output (BIBO) stability * Ex. 10.1 A level process with P control.

Similar presentations


Presentation on theme: "Chapter 10 Stability Analysis and Controller Tuning ※ Bounded-input bounded-output (BIBO) stability * Ex. 10.1 A level process with P control."— Presentation transcript:

1 Chapter 10 Stability Analysis and Controller Tuning ※ Bounded-input bounded-output (BIBO) stability * Ex A level process with P control

2 (S1) Models (S2) Solution by Laplace transform where

3 Note: (1)Stable if K c <0 (2)Unstable K c >0 (3)Steady state performance by

4 * Ex A level process without control (1)Response to a sine flow disturbance (2)Response to a step flow disturbance

5 ※ Stability analysis

6 Note: Assume G d (s) is stable.

7 * Stability of linearized closed-loop systems Ex The series chemical reactors with PI controller

8 @ Known values 1.Process 2.Controller

9 @ Formulation& stability Stable

10 ◎ Criterion of stability ※ Direct substitution method

11 The response of controlled output: P1. P2.

12 P3.

13 ﹪ Ultimate gain (K cu ): The controller gain at which this point of marginal instability is reached ﹪ Ultimate period (T u ): It shows the period of the oscillation at the ultimate gain * Using the direct substitution method by in the characteristic equation

14

15 Example A.1 Known transfer functions

16 Find: (1) Ultimate gain (2) Ultimate period S1. Characteristic eqn.

17 S2. Let at K c =K cu

18

19 Example A.2 S1.

20 S2. S3.

21 Example A.3 Find the following control loop: (1) Ultimate gain (2) Ultimate period

22 S1. The characteristic eqn. for H(s)=K T /(  T s+1) S2. G c =-K c to avoid the negative gains in the characteristic eqn.

23 S3. By direct substitution of at K c =K cu

24 * Dead-time Since the direct substitution method fails when any of blocks on the loop contains deadt-ime term, an approximation to the dead-time transfer function is used. First-order Padé approximation:

25 Example A.4 Find the ultimate gain and frequency of first- order plus dead-time process S1. Closed-loop system with P control

26 S2. Using Pade approximation

27 S3. Using direct substitution method

28 Note: 1.The ultimate gain goes to infinite as the dead-time approach zero. 2.The ultimate frequency increases as the dead time decreases.

29 ※ Root locus A graphical technique consists of roots of characteristic equation and control loop parameter changes.

30 * Definition: Characteristic equation: Open-loop transfer function (OLTF): Generalized OLTF:

31 Example B.1: a characteristic equation is given S1. Decide open-loop poles and zeros by OLTF

32 S2. Depict by the polynomial (characteristic equation) K c :1/3

33 S3. Analysis

34 Example B.2: a characteristic equation is given S1. Decide poles and zeros

35 S2. Depict by the polynomial (characteristic equation)

36 S3. Analysis

37 Example B.3: a characteristic equation is given S1. Decide poles and zeros S2. Depict by the polynomial (characteristic equation)

38

39 S3. Analysis

40 @ Review of complex number c=a+ib

41  Polar notations

42 P1. Multiplication for two complex numbers (c, p) P2. Division for two complex numbers (c, p)

43 @ Rules for root locus diagram (1)Characteristic equation (2)Magnitude and angle conditions

44 Since

45 (3)Rule for searching roots of characteristic equation Ex. A system have two OLTF poles (x) and one OLTF zero (o) Note: If the angle condition is satisfied, then the point s 1 is the part of the root locus

46

47

48 Example B.4 Depict the root locus of a characteristic equation (heat exchanger control loop with P control) S1. OLTF

49 S2. Rule for root locus (i)From rule 1 where the root locus exists are indicated. (ii)From rule 2 indicate that the root locus is originated at the poles of OLTF. (iii)n=3, three branches or loci are indicated. (iv)Because m=0 (zeros), all loci approach infinity as K c increases. (v)Determine CG= and asymptotes with angles,  =60°, 180 °, 300 °. (vi)Calculate the breakaway point by

50 s= – and –0.063 S3. Depict the possible root locus with ω u =0.22 (direct substitution method) and K cu =24

51 Example B.5 Depict the root locus of a characteristic equation (heat exchanger control loop with PI control) S1. OLTF

52 S2. Following rules

53 S3. Depict root locus

54 * Exercises

55 Ans. 8.1

56

57 Ans. 8.2

58

59

60

61 * Dynamic responses for various pole locations

62 ◎ * Which is good method for stability analysis

63 ※ Bode method A brief review: (1)OLTF (2)Frequency response

64 ◎ Stability criterion

65

66 Ex. C.1 Heat exchanger control system (Ex. A.1) * Frequency response stability criterion Determining the frequency at which the phase angle of OLTF is –180°(–π) and AR of OLTF at that frequency

67 S2. Find MR and θ S3. Bode plot in Fig to estimate ω=0.219 by θ= –180° and decide MR= S1. OLTF

68

69 S4. Decide K c as AR=1 # * Stability vs. controller gain In Bode plot, as θ= –180° both ω and MR are determined. Moreover, ω = ω u and K cu can be obtained.

70 Ex. C.2 Analysis of stability for a OTLF S1. MR and θ

71 S2. Show Bode plot (MR vs.  &  vs.  )

72 S3. Find ω u and K cu ω u =0.16 by  = –π K cu =12.8 Ex. C.3 The same process with PD controller and  =0.1 (S1) OLTF

73 (S2) By Fig  u =0.53 and MR=0.038 K cu =33 and  u =0.53

74

75 Ex (S1) Bode plot (AR vs.  &  vs.  ) for K c =1

76 (S2) Stability vs. controller gain K c Ex Determine whether this system is stable.

77 (S1) Bode plot for K c =15 and T I =1 (S2) Since the AR>1 at, the system is unstable.

78 P1. Bode plot for the first-order system

79 P2. Bode plot for the second-order system

80 Ex Determine AR and  of the following transfer function at

81

82 * Controller tuning based on Z-N closed-loop tuning method S1. Calculating  c by setting K c =1 and then determine K u and P u where AR c =

83 S2. Controller tuning constants Ex Calculate controller tuning constants for a process, G p (s)=0039/(5s+1) 3, by uning the Z-N method S1.

84 S2. Bode plot

85 S3. Tuning constants

86 S4. Closed-loop test

87 Ex Integral mode tend to destabilize the control system

88 (1)Gain margin (GM): Total loop gain increase to make the system just unstable. The controller gain that yields a gain margin * Typical specification: GM  2 If P controller with GM=2 is the same as the Z-N Effect of modeling errors on stability

89 (2) Phase margin (PM): * Typical specification: PM>45° Ex. D.1 Consider the same heat exchanger to tune a P controller for specifications (Ex. C.2) (a) While GM=2

90 (b) PM= 45°  θ= –135°. By Fig. in Ex. C.2, we can find and

91 ※ Polar plot The polar plot is a graph of the complex-valued function G(i  ) as  goes from 0 to . Ex. E.1 Consider the amplitude ratio and the phase angle angle of first-order lag are given as

92

93 Ex. E.2 Consider the amplitude ratio and the phase angle angle of second-order lag are given as

94

95 Ex. E.3 Consider the second-order system with tuning K c

96 Ex. E.4. Consider the amplitude ratio and the phase angle angle of pure dead time system are given as

97 ※ Conformal mapping

98 ※ Nyquist stability criterion (Nyquist plot) Ex. E.5 Consider a closed-loop system, its OLTF is given as

99

100 Marginal stable K c =23.8 stable K c <23.8 Unstable stable K c >23.8

101 Exercises: (1)Q (2)Q.10.15


Download ppt "Chapter 10 Stability Analysis and Controller Tuning ※ Bounded-input bounded-output (BIBO) stability * Ex. 10.1 A level process with P control."

Similar presentations


Ads by Google