1. سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر حسين بلندي - دکتر سید مجید اسما عیل زاده

2. Recap. Optimal Systems, Index of Performance, Introduction to Ess. 2

3. Steady-State Tracking & System Types 3

4. In General: Unity feedback control: G(s) C(s) + - r(s) e y(s) plant G o.l. (s) + - r(s) e y(s) 4

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10. r(t) t 10

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15. r(t) t 0 15

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21. r(t)=R·1(t) r(s)=R/s r(t)=R·t·1(t) r(s)=R/s 2 r(t)=R·1/2·t 2 r(s)=R/s 3 type 0 (N=0 a 0 ≠0) K p =b 0 /a 0 e ss =R/(1+K p ) K v =0 e ss =∞ K a =0 e ss =∞ type 1 (N=1 a 0 =0 a 1 ≠0 b 0 ≠0 ) K p = ∞ e ss =0 K v =b 0 /a 1 e ss =R/K v K a =0 e ss =∞ type 2, N=2 a 0 =a 1 =0 a 2 ≠0,b 0 ≠0 K p = ∞ e ss =0 K v = ∞ e ss =0 K p =b 0 /a 2 e ss =R/K a type≥3, N ≥ 3 a 0 =a 1 =a 2 =0 b 0 ≠0 K p = ∞ e ss =0 K v = ∞ e ss =0 K a = ∞ e ss =0 sys. type ref. input 21

22. Example of tank H + - C 22

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24. + r(s) K p s+K I s + - r(s) e ω n 2 s(s+2ξ ω n ) 1 Ts+1 e.g. 24

25. example G(s) r(s) e(s)y(s) 25

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28. The stability of linear feedback systems. 28

29. The concept of stability. 29

30. 30 The stability of a cone.

31. 31 Tacoma Narrows Bridge As oscillation begins At catastrophic failure on November 7, 1940.

32. 32 Bounded Input - Bounded Output criterion An unconstrained linear system is stable if the output response is bounded for all bounded inputs. Note: By a bounded input, we mean an input variable that stays within upper and lower limits for all values of time

33. 33 Example: BI-BO criterion Q.If the step function is applied at the input of continuous system and the output remains below a certain level for all time, is the system stable?

34. 34 Example: BI-BO criterion Q.If the step function is applied at the input of continuous system and the output remains below a certain level for all time, is the system stable? A.The system is not necessarily stable since the output must be bounded for every bounded input. A bounded output to one specific bounded input does not ensure stability.

35. 35 Example: BI-BO criterion Q. If the step function is applied at the input of continuous system and the output is of the form y = t, is the system stable or unstable?

36. 36 Example: BI-BO criterion Q. If the step function is applied at the input of continuous system and the output is of the form y = t, is the system stable or unstable? A. This system is unstable since a bounded input produced an unbounded output.

37. 37 Characteristic root locations criterion A system is stable if all the poles of the transfer function have negative real parts. Stability in the s-plane.

38. 38 Stable system A necessary and sufficient condition for a feedback system to be stable is that all the poles of the transfer function have negative real parts. This means that all the poles are in the left-hand s-plane.

39. 39 Determine in each case if the set of roots represents, stable, marginally stable, or unstable systems. a)-1, -2 e) -2 + j, -2 - j, 2j, -2j b)-1, +1 f) 2,-1,-3 c)-3,-2,0g) -6,-4,7 d) -1 + j, -1- j h) -2 + 3j, -2 -3j, -2

40. 40 Determine in each case if the set of roots represents, stable, marginally stable, or unstable systems. a)-1, -2 e) -2 + j, -2 - j, 2j, -2j b)-1, +1 f) 2,-1,-3 c)-3,-2,0g) -6,-4,7 d) -1 + j, -1- j h) -2 + 3j, -2 -3j, -2

41. 41 Marginally stable If there are any poles on the jω-axis (with none on the right hand side), then the system is considered marginally stable.

42. 42 Unstable If there are one or more poles in the right- hand s-plane or there are repeated roots on the jω-axis, the system is unstable.

43. 43 Example: Poles and zeros Q. A system has poles at -1 and -5 and zeros at 1 and -2. Is the system stable?

44. 44 Example: Poles and zeros Q. A system has poles at -1 and -5 and zeros at 1 and -2. Is the system stable? A.The system is stable since the poles are roots of the system characteristic equation which have negative real parts. The fact that the system has a zero with a positive real part does not affect its stability.

45. 45 Example: Characteristic equation Q. Determine if the system with the following characteristic equation is stable: (s+1)(s+2)(s-3) = 0

46. 46 Example: Characteristic equation Q. Determine if the system with the following characteristic equation is stable: (s+1)(s+2)(s-3) = 0 A.This characteristic equation has the roots -1,-2, and 3 and therefore represent an unstable systems since there is a positive real root.

47. 47 Example: integrator Q. The differential equation of an integrator may be written as follows: dy/dt = x Determine if an integrator is stable.

48. 48 Example: integrator

49. 49 Example: integrator A.The characteristic equation of his system is s = 0. Since the root does not have a negative real part, an integrator is not stable. Since it has no roots with positive real parts, integrator is marginally stable

50. 50 Stability in the s-plane. NOT STABLESTABLE

51. The Routh-Hurwitz Stability Criterion. 51

52. 52 Introduction Goal: Determining whether the system is stable or unstable from a characteristic equation in polynomial form without actually solving for the roots. Routh’s stability criterion is useful for determining the ranges of coefficients of polynomials for stability, especially when the coefficients are in symbolic (nonnumerical) form.

53. 53 A necessary condition for Routh Stability A necessary condition for stability of the system is that all of the roots of its characteristic equation have negative real parts, which in turn requires that all the coefficients be positive. A necessary (but not sufficient) condition for stability is that all the coefficients of the polynomial characteristic equation be positive.

54. 54 A necessary and sufficient condition for Stability Routh’s formulation requires the computation of a triangular array that is a function of the coefficients of the polynomial characteristic equation. A system is stable if and only if all the elements of the first column of the Routh array are positive

55. 55 Characteristic Equation Consider an nth-order system whose the characteristic equation (which is also the denominator of the transfer function) is:

56. 56 Determining the Routh array Consider the characteristic equation: First arrange the coefficients of the characteristic equation in two rows, beginning with the first and second coefficients and followed by the even-numbered and odd-numbered coefficients:

57. 57 Routh array: method (cont’d) Then add subsequent rows to complete the Routh array: ?!

58. 58 Routh array: method (cont’d) Compute elements for the 3 rd row:

59. 59 Routh-Hurwitz Criterion The Routh-Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array.

60. 60 Three distinct cases of the first column array 1.No element in the first column is zero. 2.There is a zero in the first column and in this row. 3.There is a zero in the first column and the other zero in this row.

61. 61 Case1. No element in the first column is zero

62. 62 Second-order system The requirement for a stable second-order system is that all the coefficients be positive or all coefficients be negative.

63. 63 Third-order system

64. 64 Third-order system For system to be stable, it is necessary and sufficient that the coefficients a 0,a 1, a 2, a 3 be positive and a 2 a 1 > a 0 a 3.

65. 65 Unstable system Two roots of q(s) lie in the right-hand s-plane.

66. 66 Two changes in sign

67. 67 Case 2: There is a zero in the first column and in this row. The system is unstable for all values of gain K.

68. 68 Case 3:A zero in the first column and the other zero in this row. For a stable system: U(s) = 2s 2 + 8 = 2 (s + j2)(s - j2) For a marginal stability: K=8 0 < K < 8

69. 69 Ex. A six-legged micro robot It is equipped with sensor network that includes 150 sensors of 12 different types. The legs are instrumented so that the robot can determine the lay of the terrain, the surface texture, hardness, and color.

70. 70 Micro robot

71. 71 Micro robot The auxiliary polynomial:

72. 72 The auxiliary polynomial The equation that immediately precedes the zero entry in the Routh array.

73. 73 Micro robot The robot is using flexible legs with high-gain controllers that may become unstable and oscillate.

74. 74 Example : Given the characteristic equation, is the system described by this characteristic equation stable? Answer: One coefficient (-2) is negative. Therefore, the system does not satisfy the necessary condition for stability.

75. 75 Example : Given the characteristic equation, is the system described by this characteristic equation stable? Answer: All the coefficients are positive and nonzero. Therefore, the system satisfies the necessary condition for stability. We should determine whether any of the coefficients of the first column of the Routh array are negative.

76. 76 Example (cont’d): Routh array

77. 77 Example (cont’d): Routh array

78. 78 Example (cont’d): Routh array

79. 79 Example (cont’d): Routh array The elements of the 1 st column are not all positive: the system is unstable

80. 80 Example : Stability versus Parameter Range Consider a feedback system such as: The stability properties of this system are a function of the proportional feedback gain K. Determine the range of K over which the system is stable.

81. 81 Example (cont’d)

82. 82 Example (cont’d) Expressing the characteristic equation in polynomial form, we obtain:

83. 83 Example (cont’d) The corresponding Routh array is: Therefore, the system is stable if and only if

84. 84 Example (cont’d) Solving for the roots gives: -5 and ±1.22j for K=7.5 => The system is unstable (or critically stable) for K=7.5 -4.06 and –0.47 ±1.7j for K=13 -1.90 and –1.54 ±3.27j for K=25 => The system is stable for both K=13 and K=25

85. 85 Example : Stability versus Two Parameter Range Consider a Proportional-Integral (PI) control such as: Find the range of the controller gains so that the PI feedback system is stable.

86. 86 Example (cont’d) The characteristic equation for the system is given by:

87. 87 Example (cont’d) Expressing the characteristic equation in polynomial form, we obtain:

88. 88 Example 4 (cont’d) The corresponding Routh array is: For stability, we must have:

89. 89 The Routh-Hurwitz criterion

90. 90 Relative Stability Further characterization of the degree of stability of a stable closed loop system. Can be measured by the relative real part of each root or pair of roots.

91. 91 Root r 2 is relatively more stable than roots r 1 and r 1 ’.

92. 92 Axis shift

93. 93 EX. Flow graph diagram

94. 94 Ex. Block diagram model.

95. 95 Tracked vehicle turning control. Select K and a so that the system is stable and e ss ≤ 24% of a ramp command.

96. 96 Characteristic equation:

97. 97 Routh array

98. 98 Ramp response for a=0.6 and K=70

99. 99 The steady-state error

100. 100 The closed-loop T(s)

101. 101 The stable region

102. 102 Absolute Stability A closed loop feedback system which could be characterized as either stable or not stable.

103. 103 Stable System A stable system is a dynamic system with a bounded response to a bounded input.

104. 104 Stability of S.S. Eqn.

105. 105 Example: Closed epidemic system

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