control system, open and closed loop engineering

PPT on Control system

1. Control Systems

2. A control system consisting of interconnected components is designed to achieve a desired purpose. To understand the purpose of a control system, it is useful to examine examples of control systems through the course of history. These early systems incorporated many of the same ideas of feedback that are in use today. Modern control engineering practice includes the use of control design strategies for improving manufacturing processes, the efficiency of energy use, advanced automobile control, including rapid transit, among others. We also discuss the notion of a design gap. The gap exists between the complex physical system under investigation and the model used in the control system synthesis. The iterative nature of design allows us to handle the design gap effectively while accomplishing necessary tradeoffs in complexity, performance, and cost in order to meet the design specifications.

3. Introduction System – An interconnection of elements and devices for a desired purpose. Control System – An interconnection of components forming a system configuration that will provide a desired response. Process – The device, plant, or system under control. The input and output relationship represents the cause- and- effect relationship of the process.

4. Introduction Multivariable Control System Open-Loop Control Systems utilize a controller or control actuator to obtain the desired response. Closed- Loop Control Systems utilizes feedback to compare the actual output to the desired output response.

5. History Watt’s Flyball Governor (18 th century) Greece (BC) – Float regulator mechanism Holland (16 th Century)– Temperature regulator

6. History Water- level float regulator

7. History

8. History 18th Century James Watt’s centrifugal governor for the speed control of a steam engine. 1920s Minorsky worked on automatic controllers for steering ships. 1930s Nyquist developed a method for analyzing the stability of controlled systems 1940s Frequency response methods made it possible to design linear closed- loop control systems 1950s Root- locus method due to Evans was fully developed 1960s State space methods, optimal control, adaptive control and 1980s Learning controls are begun to investigated and developed. Present and on- going research fields. Recent application of modern control theory includes such non- engineering systems such as biological, biomedical, economic and socio- economic systems ???????????????????????????????????

9. Automobile steering control system. The driver uses the difference between the actual and the desired direction of travel to generate a controlled adjustment of the steering wheel. Typical direction- of- travel response. Examples of Modern Control Systems

10. Examples of Modern Control Systems

11. Examples of Modern Control Systems

12. Examples of Modern Control Systems

13. Examples of Modern Control Systems

14. Examples of Modern Control Systems

15. Examples of Modern Control Systems

16. Examples of Modern Control Systems

17. Examples of Modern Control Systems

18. The Future of Control Systems

19. The Future of Control Systems

20. Control System Design

21. Design Example

22. ELECTRIC SHIP CONCEPT Ship Service Power Main Power Distribution Propulsion Motor Motor Drive Generator Prime Mover Power Conversion Module Electric Drive Reduce # of Prime Movers Fuel savings Reduced maintenance Technology Insertion Warfighting Capabilities Vision Integrated Power System All Electric Ship Electrically Reconfigurable Ship Reduced manning Automation Eliminate auxiliary systems (steam, hydraulics, compressed air) Increasing Affordability and Military Capability Design Example

23. CVN(X) FUTURE AIRCRAFT CARRIER Design Example

24. Design Example

25. Design Example

26. Design Example

27. Design Example

28. Design Example

29. Design Example

30. Sequential Design Example

31. Sequential Design Example

32. We use quantitative mathematical models of physical systems to design and analyze control systems. The dynamic behavior is generally described by ordinary differential equations. We will consider a wide range of systems, including mechanical, hydraulic, and electrical. Since most physical systems are nonlinear, we will discuss linearization approximations, which allow us to use Laplace transform methods. We will then proceed to obtain the input–output relationship for components and subsystems in the form of transfer functions. The transfer function blocks can be organized into block diagrams or signal- flow graphs to graphically depict the interconnections. Block diagrams (and signal- flow graphs) are very convenient and natural tools for designing and analyzing complicated control systems Mathematical Models of Systems Objectives

33. Introduction Six Step Approach to Dynamic System Problems Define the system and its components Formulate the mathematical model and list the necessary assumptions Write the differential equations describing the model Solve the equations for the desired output variables Examine the solutions and the assumptions If necessary, reanalyze or redesign the system

34. Differential Equation of Physical Systems T a (t)  T s (t) T a (t) T s (t)  (t)  s (t)   a (t) T a (t) = through - variable angular rate difference = across- variable

35. Differential Equation of Physical Systems v 21 dt Describing Equation L  d i E 1  L  i 2 2 v 21 1 d k dt  F E 2 k 2 1  F  21 1 d k dt  T 1 T 2 E  2 k P 21 I  d Q dt 2 2 E 1  I  Q Electrical Inductance Translational Spring Rotational Spring Fluid Inertia Energy or Power

36. Differential Equation of Physical Systems Electrical Capacitance Translational Mass Rotational Mass Fluid Capacitance Thermal Capacitance 21 21 2 i C  d v E 1  M  v dt 2 2 F M  d v dt E 1 2 2 2  M  v 2 T J  d  dt E 1 2 2 2  J  21 f dt Q C  d P 2 f 21 2 E 1  C  P dt q C t  d T 2 E C t  T 2

37. Differential Equation of Physical Systems Electrical Resistance Translational Damper Rotational Damper Fluid Resistance Thermal Resistance F b  v 21 P b  v 2 21 21 21 2 i 1  v P 1  v R R T b  21 21 2 P b  R f 21 Q 1  P R f 21 2 P 1  P R t 21 q 1  T R t 21 P 1  T

38. Differential Equation of Physical Systems dt 2 dt M  d 2 y(t)  b  d y(t)  k  y(t) r(t)

39. Differential Equation of Physical Systems v(t) R d  C  v(t)  dt 1  t L   v(t) dt r(t)   1  t y(t) K 1  e  sin   1  t   1 

40. Differential Equation of Physical Systems

41. Differential Equation of Physical Systems K 2  1  2  .5  2  10   2  t y(t)  K 2  e  sin   2  t   2   2  2   2  t y1(t)  K 2  e   2  t y2(t)   K 2  e 1 1 2 3 4 5 6 7 1 y(t) y1(t) y2(t) t

42. Linear Approximations

43. Linear Approximations Linear Systems - Necessary condition Principle of Superposition Property of Homogeneity Taylor Series http://www.maths.abdn.ac.uk/%7Eigc/tch/ma2001/notes/node46.ht ml

44. Linear Approximations – Example 2.1 M  200gm T  M  g  L  sin    g  9.8 m s 2 L  100cm   0rad 16     15    T 1     M  g  L  sin    T 2     M  g  L  cos           T 4 3 2 1 1 2 3 4 10 5 5 T 1 (  ) 10  T 2 (  ) Students are encourag ed to investigate linear approximation accuracy for different va  lu

45. The Laplace Transform Historical Perspective - Heaviside’s Operators Origin of Operational Calculus (1887)

46. p d dt 1 p  t   1 du i v Z(p) Z(p) R  L  p i 1 R  L  p  H(t) 1 R  L  p   L  p  1   H(t)  R  2  L  1 p 2  1   R 1 R  L p  R  3  L  1        p 3    .....  H(t) 1 p n  H(t) t n n  2 2 2   L  3 3   R   t 3    L    ..    R  L i 1   R  t    R   t i 1 R e   R   t   L       1   Expanded in a power series v = H(t) Historical Perspective - Heaviside’s Operators Origin of Operational Calculus (1887) (*) Oliver Heaviside: Sage in Solitude, Paul J. Nahin, IEEE Press 1987 .

47. The Laplace Transform Definition   L(f(t))  s  t f(t)  e dt   = F(s) Here the complex frequency is s   j  w dt    s  t f(t)  e The Laplace Transform exists when     this means that the integral converg

48. The Laplace Transform Determine the Lap lace transform for the functions a) f 1 (t)  1 for t    dt  s  t  F 1 (s)   e = 1  (s  t)   e s 1 s b) f 2 (t)   e  (a  t) F 2 (s) dt e  (a  t)  (s  t)  e   = 1 s  1   e  [(s  a)  t] 2 F (s) 1 s  a

49. note that the initial condition is included in the transform = sF(s) - f(0+)  dt  L   d f(t)    dt  (s  t)  s   f(t)  e = - f(0+) +   (s  t)    f(t)    s  e  dt     (s  t) f(t)  e =     u dv du  s  e  (s  t)  dt we obtain v f(t) and and, from which  u  v   v du  dv df(t) u e  (s  t) where =   u dv  by the use of L d f(t)  dt      dt  (s  t) d f(t)  e dt    Evaluate the laplace transform of the derivative of a function The Laplace Transform

50. The Laplace Transform Practical Example - Consider the circuit. The KVL equation is 4  i(t)  2  d i(t) assum e i(0+) = 5 A dt Applying the Laplace Transform , we have   dt    4  i(t)  2  d i(t)   e  ( s  t) dt        dt    4   i(t)  e  ( s  t) dt  2   d i(t)  e  ( s  t) dt 4  I(s)  2  (s  I(s)  i(0)) 4  I(s)  2  s  I(s)  10 I(s)  5 s  2 2 4 1 t transform ing back to the time domain, with our present knowledge of Laplace transform , we may say that t  (0  0.01  2) 6 i( t) 2 i(t)  5  e  ( 2  t)

51. The Partial- Fraction Expansion (or Heaviside expansion theorem) Suppose that The partial fraction expansion indicates that F(s) consists of a sum of terms, each of which is a factor of the denominator. The values of K1 and K2 are determined by combining the individual fractions by means of the lowest common denominator and comparing the resultant numerator coefficients with those of the coefficients of the numerator before separation in different terms. F ( s ) s  z1 ( s p1 ) ( s p2 )    or F ( s ) K1 s  p1 K2 s p2   Evaluation of Ki in the manner just described requires the simultaneous solution of n equations. An alternative method is to multiply both sides of the equation by (s + pi) then setting s= - pi, the right- hand side is zero except for Ki so that Ki   ( s pi )  ( s z1 ) ( s p1 ) ( s p2 )    s = - pi The Laplace Transform

52. The Laplace Transform s - > t - > infinite s - > infinite Lim(s  F(s)) t - > 7. Final- value Theorem Lim(f(t)) Lim(s  F(s)) 6. Initial- value Theorem Lim(f(t)) f(0)     F(s) ds f(t) t 5. Frequency Integration F(s  a) f(t)  e  (a  t) 4. Frequency shifting  d F(s) ds differentiation t  f(t) 3. Frequency f(at ) 2. Time scaling 1  F  s  a  a  f(t  T)  u(t  T) 1. Time delay Frequency Doma e  ( s  T)  F(s) Property Time Domain

53. Useful Transform Pairs The Laplace Transform

54. The Laplace Transform y(s)  M   s  b    y o  k  M  n s  2    s 2  2  n   n 2  2  1  s 2   b   s    M  s1    n    n   n k M  b  2  k  M  n 2 s2    n       1 Roots Real Real repeated Imaginary (conjug ates) Complex (conjugates) 1   2 s1    n   j  n  2 s2    n   j  n  1   Consider the mass- spring- damper system Y(s) (Ms  b)  yo Ms 2  bs  k equation 2.2

55. The Laplace Transform

56. The Transfer Function of Linear Systems 1 V (s)  R 1 Cs      I(s) 1 Z (s) R 2 Z (s) 1 Cs V 2 (s ) 1    Cs   I(s) V 2 (s ) V 1 (s ) 1 Cs R  1 Cs Z 2 (s) Z 1 (s)  Z 2 (s)

57. The Transfer Function of Linear Systems Example 2.2 dt 2 dt d 2 y(t)  4  d y(t)  3  y(t) 2  r(t) Initial Conditions: Y(0) 1 d dt y(0) r(t) 1 The Laplace transform yields:  s 2  Y(s)  s  y(0)   4  (s  Y(s)  y(0))  3 Y(s) Since R(s)=1/s and y(0)=1, we obtain: 2  R(s) Y(s) (s  4) 2  s 2  4s  3  s   s 2  4s  3   The partial fra ction expansion yields: Y(s) 3 2  1 2      1 1 3          2 3  (s  1) (s  3)   (s  1) (s  3)  s    Therefore the transient response is:  2 2    3 3 y(t)  3  e  t  1  e  3  t     1e  t  1  e  3t   2 The steady- state response is: lim y(t) t  2 3

58. The Transfer Function of Linear Systems

59. The Transfer Function of Linear Systems

60. The Transfer Function of Linear Systems

61. The Transfer Function of Linear Systems

62.  K f  i f T m K 1  K f  i f (t)  i a (t) field controled motor - Lapalce Transfo T m (s)  K 1  K f  I a   I f (s) V f (s)  R f  L f  s   I f (s) T m (s) T L (s)  T d (s) T L (s) J  s 2  (s)  b  s  (s) rearranging equations T L (s) T m (s)  T d (s) T m (s) K m  I f (s) I f (s) Vf(s) R f  L f  s The Transfer Function of Linear Systems Td(s)  (s) f V (s) K m s  (J  s  b)   L f  s  R f 

63. The Transfer Function of Linear Systems

64. The Transfer Function of Linear Systems

65. The Transfer Function of Linear Systems

66. The Transfer Function of Linear Systems

67. The Transfer Function of Linear Systems V 2 (s) V 1 (s) RCs  1 V 2 (s)  RCs V 1 (s)

68. The Transfer Function of Linear Systems V 2 (s) V 1 (s) R 2  R 1  C  s  1  R 1 V 2 (s) V 1 (s)   R 1  C 1  s  1   R 2  C 2  s  1  R 1  C 2  s

69. The Transfer Function of Linear Systems  (s) V f (s) K m s  (J  s  b)  L f  s  R f   (s) V a (s) K m s   R a  L a  s  (J  s  b)  K b  K m

70. The Transfer Function of Linear Systems V o (s) V c (s) K    R c  R q   s  c  1    s  q  1   c L c R c  q L q R q For the unloaded case: i d  c  q 0.05s   c  0.5s V 12 V 34 V q V d  (s) V c (s) K m s   s  1   J (b  m) m = slope of linearized torque-speed curve (normally negative)

71. The Transfer Function of Linear Systems Y(s) X(s) s(Ms  B) K K A  k x k p d g A 2    k p  k x dx k p B  b   d g dP g g(x  P) flow A = area of piston Gear Ratio = n = N1/N N 2  L N 1  m  L n  m  L n  m

72. The Transfer Function of Linear Systems V 2 (s) R 2 V 1 (s) R R 2 R R 2 R 1  R 2   max k s V 2 (s) k s   1 (s)   2 (s)  V 2 (s) k s  erro ( r s) V ba ttery  ma x

73. The Transfer Function of Linear Systems V 2 (s) K t  (s) K t  s  (s) K t constant V 2 (s) V 1 (s) k a s   1 Ro = output resistance Co = output capacitanc  R o  C o   1s and is often negligibl for controller amplifie

74. The Transfer Function of Linear Systems T(s) q(s) 1 t  R  C  s   Q  S  1  T T o  T e = temperature difference due to thermal proc = thermal capacitance = fluid flow rate = constant = specific heat of water = thermal resistance of insulation C t Q S R t q(s) = rate of heat flow of heating element x o (t) y(t)  x in (t) X o (s) X in (s)  s 2  M  s 2   b   s  k M For low frequency oscillations, where    n X o  j   X in  j    2 k M

75. The Transfer Function of Linear Systems x r  converts radial motion to linear mo

76. Block Diagram Models

77. Block Diagram Models

78. Block Diagram Models Original Diagram Equivalent Diagram Original Diagram Equivalent Diagram

79. Block Diagram Models Original Diagram Equivalent Diagram Original Diagram Equivalent Diagram

80. Block Diagram Models Original Diagram Equivalent Diagram Original Diagram Equivalent Diagram

81. Block Diagram Models

82. Block Diagram Models Example 2.7

83. Block Diagram Models Example 2.7

84. Signal- Flow Graph Models For complex systems, the block diagram method can become difficult to complete. By using the signal- flow graph model, the reduction procedure (used in the block diagram method) is not necessary to determine the relationship between system variables.

85. Signal- Flow Graph Models Y 1 (s) G 11 (s)  R 1 (s)  G 12 (s)  R 2 (s) Y 2 (s) G 21 (s)  R 1 (s)  G 22 (s)  R 2 (s)

86. Signal- Flow Graph Models a 11  x 1  a 12  x 2  r 1 x 1 a 21  x 1  a 22  x 2  r 2 x 2

87. Signal- Flow Graph Models Example 2.8 Y(s) R(s) G 1  G 2  G 3  G 4   1  L 3  L 4   G 5  G 6  G 7  G 8   1  L 1  L 2  1  L 1  L 2  L 3  L 4  L 1  L 3  L 1  L 4  L 2  L 3  L 2  L 4

88. Signal- Flow Graph Models Example 2.10 Y(s) R(s) 1  G 2  G 3  H 2  G 3  G 4  H 1  G 1  G 2  G 3  G 4  H 3 G 1  G 2  G 3  G 4

89. Signal- Flow Graph Models Y(s) R(s) P 1  P 2  2  P 3  P 1 G 1  G 2  G 3  G 4  G 5  G 6 P 2 G 1  G 2  G 7  G 6 P 3 G 1  G 2  G 3  G 4  G 8  1   L 1  L 2  L 3  L 4  L 5  L 6  L 7  L 8    L 5  L 7  L 5  L 4  L 3  L 4   1  3 1  2 1  L 5 1  G 4  H 4

90. Design Examples

91. Speed control of an electric traction motor. Design Examples

92. Design Examples

93. Design Examples

94. Design Examples

95. Design Examples

96. BLOCK DIAGRAM REDUCTION OF MU L TIP L E SYSTEMS

97. Components of a block diagram for a linear, time-invariant system

98. a. Cascaded subsystems; b. equivalent transfer function

99. a. Parallel subsystems; b. equivalent transfer function

100. Feedback control system; simplified model; equivalent transfer function

101. Block diagram algebra for summing junctions equivalent forms for moving a block to the left past a summing junction; to the right past a summing junction

102. Block diagram algebra for pickoff points equivalent forms for moving a block to the left past a pickoff point; to the right past a pickoff point

103. Block diagram reduction via familiar forms for Example Problem: Reduce the block diagram shown in figure to a single transfer function

104. Steps in solving Example a. collapse summing junctions; form equivalent cascaded system in the forward path form equivalent parallel system in the feedback path; b. c. d. form equivalent feedback system and multiply by cascadedG 1 (s) Block diagram reduction via familiar forms for Example Cont.

105. Problem: Reduce the block diagram shown in figure to a single transfer function Block diagram reduction by moving blocks Example

106. Steps in the block diagram reduction for Example Move G 2 (s) to the left past of pickoff point to create parallel subsystems, and reduce the feedback system of G 3 (s) and H 3 (s) Reduce parallel pair of 1/G 2 (s) and unity, and push G 1 (s) to the right past summing junction Collapse the summing junctions, add the 2 feedback elements, and combine the last 2 cascade blocks Reduce the feedback system to the left finally, Multiple the 2 cascade blocks and obtain final result.

107. Second-order feedback control system The closed loop transfer function is Note K is the amplifier gain, As K varies, the poles move through the three ranges of operations OD, CD, and UD 0<K<a 2 /4 system is over damped K = a 2 /4 K > a 2 /4 system is critically damped system is under damped s 2  as  K T ( s )  K

108. Finding transient response Example Problem: For the system shown, find peak time, percent overshot, and settling time. Solution: The closed loop transfer function is T ( s )  25 s 2  5 s  25 And 1   2 X 10  16 .3 3  e   / n   2 5  5 2  n  5 so  = .5 p  n 1   2 T    .7 2 6 sec % O S s T  4  1 .6 sec n  using values for  and  n and equation in chapter 4 we find

109. Gain design for transient response Example Problem: Design the value of gain K, so that the system will respond with a 10% overshot. Solution: The closed loop transfer function is For 10% OS we find We substitute this value in previous equation to find K = 17.9  5 s  K T ( s )  K s 2  = 5  K an d  5 thu s  2  2 K n n  = .59 1

110. Signal- flow graph components: system; signal; interconnection of systems and signals

111. a. cascaded system nodes cascaded system signal- flow graph; b. d. c. parallel system nodes parallel system signal- flow graph; feedback system nodes feedback system signal- flow graph Building signal- flow graphs

112. Problem: Convert the block diagram to a signal- flow graph. Converting a block diagram to a signal- flow graph

113. Signal- flow graph development: signal nodes; signal- flow graph; simplified signal- flow graph Converting a block diagram to a signal- flow graph

114. Mason’s ƌule - Definitions Loop gain: The product of branch gains found by traversing a path that starts at a node and ends at the same node, following the direction of the signal flow, without passing through any other node more than once. G 2 (s)H 2 (s), G 4 (s)H 2 (s), G 4 (s)G 5 (s)H 3 (s), G 4 (s)G 6 (s)H 3 (s) Forward- path gain: The product of gains found by traversing a path from input node to output node in the direction of signal flow. G 1 (s)G 2 (s)G 3 (s)G 4 (s)G 5 (s)G 7 (s), G 1 (s)G 2 (s)G 3 (s)G 4 (s)G 5 (s)G 7 (s) Nontouching loops: loops that do not have any nodes in common. G 2 (s)H 1 (s) does not touch G 4 (s)H 2 (s), G 4 (s)G 5 (s)H 3 (s), and G 4 (s)G 6 (s)H 3 (s) Nontouching- loop gain: The product of loop gains from nontouching loops taken 2, 3,4, or more at a time. [G 2 (s)H 1 (s)][G 4 (s)H 2 (s)], [G 2 (s)H 1 (s)][G 4 (s)G 5 (s)H 3 (s)], [G 2 (s)H 1 (s)][G 4 (s)G 6 (s)H 3 (s)]

115. Mason’s Rule The Transfer function. C(s)/ R(s), of a system represented by a signal- flow graph is Where K = number of forward paths T k = the k th forward- path gain nontouching-loop gains taken 3 at a time + = 1 -  loop gains +  nontouching- loop gains taken 2 at a time -  nontouching- loop gains taken 4 at a time - ……. that touch the k th forward path. In other words, = - loop gain terms in  k is for  med b  y eliminating from   k those loop gains that touch the k th forward path.  G ( s )  C ( s )   T k  k R ( s ) k   

116. Transfer function via Mason’s rule Now Problem: Find the transfer function for the signal flow graph Solution: forward path G 1 (s)G 2 (s)G 3 (s)G 4 (s)G 5 (s) Loop gains G 2 (s)H 1 (s), G 4 (s)H 2 (s), G 7 (s)H 4 (s), G 2 (s)G 3 (s)G 4 (s)G 5 (s)G 6 (s)G 7 (s)G 8 (s) Nontouching loops 2at a time G 2 (s)H 1 (s)G 4 (s)H 2 (s) G 2 (s)H 1 (s)G 7 (s)H 4 (s) G 4 (s)H 2 (s)G 7 (s)H 4 (s) 3at a time G 2 (s)H 1 (s)G 4 (s)H 2 (s)G 7 (s)H 4 (s)  = 1-[G2(s)H1(s)+G 4 (s)H 2 (s)+G 7 (s)H 4 (s)+ G 2 (s)G 3 (s)G 4 (s)G 5 (s)G 6 (s)G 7 (s)G 8 (s)] + [G 2 (s)H 1 (s)G 4 (s)H 2 (s) + G 2 (s)H 1 (s)G 7 (s)H 4 (s) + G 4 (s)H 2 (s)G 7 (s)H 4 (s)] – [G 2 (s)H 1 (s)G 4 (s)H 2 (s)G 7 (s)H 4 (s)]  1 = 1 - G7(s)H4(s)  [G 1 (s)G 2 (s)G 3 (s)G 4 (s)G 5 (s)] [1- G 7 (s)H 4 (s)]  G ( s )  T 1  1 

117. Signal- Flow Graphs of State Equations x 2   6 x 1  2 x 2  2 x 3  5 r x 3  x 1  3 x 2  4 x 3  7 r y   4 x 1  6 x 2  9 x 3 a. place nodes; b. interconnect state variables and derivatives; c. form dx1/dt ; d. form dx2/dt x P  ro 2 b x lem  : 5 dr x aw  sig 3 n x al- f  low 2 r graph for: 1 1 2 3

118. (continued) form dx 3 /dt; form output Signal- Flow Graphs of State Equations

119. Alternate Representation: Cascade Form C ( s )  24 R ( s ) ( s  2)( s  3)( s  4)

120. Alternate Representation: Cascade Form 1 x   4 x 1  x 2 x 2   3 x 2 3 x  y  c ( t )  x 1  x 3  2 x 3  24 r y   1    4     24   1    X   3 1  X  2     X    r   

121. Al t e rn a t 2 e 4 R e p r e s e n t a ti o n 1 : P 2 a r all e l F 2 o 4 r m C ( s ) R ( s ) ( s  2)( s  3)( s  4) ( s  2) ( s  3) 12 ( s  4)     x 1   2 x 1 x 2   3 x 2 x 3  y  c ( t )  x 1  x 2  12 r  24 r  4 x 3  12 r  x 3 1 1  X  X    24  r    2   12   3     4   y   1 X      12   

122. Alternate Representation: Parallel Form Repeated roots ( s  1) 2 ( s  2) ( s  1) 2 C ( s )  ( s  3)  2 1 1 ( s  1) ( s  2) R ( s ) x 1   x 1  x 2 x 2  x 2 x 3  y  c ( t )  x 1  2 x 3  r  1 / 2 x 2  x 3 + 2 r X     1   1  1   X   2   y   1  1 / 2 1  X    2  r       1  

123. G(s) = C(s)/R(s) = (s 2 + 7s + 2)/(s 3 + 9s 2 + 26s + 24) This form is obtained from the phase-variable form simply by ordering the phase variable in reverse order Alternate Representation: controller canonical form y  7  2 1   x 2  3    x    x 1    2    2  1   x     r    9     x 3     1    3   1   x 1     26     24  x 1    x  x        y   1 7 2   x 2   2    x 2        x     r     x 3        24   x 1   1   3    x 1    x   9  26    1    1  x 1    x 3      

124. Alternate Representation: controller canonical form System matrices that contain the coefficients of the characteristic polynomial are called companion matrices to the characteristic polynomial. Phase-variable form result in lower companion matrix Controller canonical form results in upper companion matrix

125. Alternate Representation: observer canonical form Observer canonical form so named for its use in the design of observers G(s) = C(s)/R(s) = (s 2 + 7s + 2)/(s 3 + 9s 2 + 26s + 24) = (1/s+7/s 2 +2/s 3 )/(1+9/s+26/s 2 +24/s 3 ) Cross multiplying (1/s+7/s 2 +2/s 3 )R(s) = (1+9/s+26/s 2 +24/s 3 ) C(s) And C(s) = 1/s[R(s)- 9C(s)] +1/s 2 [7R(s)- 26C(s)]+1/s 3 [2R(s)-24C(s)] = 1/s{ [R(s)- 9C(s)] + 1/s {[7R(s)- 26C(s)]+1/s [2R(s)-24C(s)]}}

126. Alternate Representation: observer canonical form x 1   9 x 1  x 2  r x 2   26 x 1  x 3 +7 r x 3   24 x 1  2 r y  c ( t )  x 1 1  X   7  r  9 1   1  X     26     24        2   y   1  X Note that the observer form has A matrix that is transpose of the controller canonical form, B vector is the transpose of the controller C vector, and C vector is the transpose of the controller B vector. The 2 forms are called duals.

127. Feedback control system for Example Problem Represent the feedback control system shown in state space. Model the forward transfer function in cascade form. Solution first we model the forward transfer function as in (a), Second we add the feedback and input paths as shown in (b) complete system. Write state equations x 1   3 x 1 x 2   x 2 - 2 x 2  100 ( r - c ) but c  5 x 1  ( x 2  3 x 1 )  2 x 1  x 2

128. Feedback control system for Example x 1   3 x 1  x 2 x 2   200 x 1  102 x 2  100 r y  c ( t )  2 x 1  x 2  y   2 1  X  3 1  X     200  102   100     X   r 

129. State- space forms for C(s)/R(s) =(s+ 3)/[(s+ 4)(s+ 6)]. Note: y = c(t)

130. UNIT-II TIME RESPONSE ANA L Y SIS

131. In this chapter we extend the ideas of modeling to include control system characteristics, such as sensitivity to model uncertainties, steady- state errors, transient response characteristics to input test signals, and disturbance rejection. We investigate the important role of the system error signal which we generally try to minimize. We will also develop the concept of the sensitivity of a system to a parameter change, since it is desirable to minimize the effects of unwanted parameter variation. We then describe the transient performance of a feedback system and show how this performance can be readily improved. We will also investigate a design that reduces the impact of disturbance signals. Feedback Control System Characteristics Objectives

132. Open- And Closed- Loop Control Systems An open- loop (direct) system operates without feedback and directly generates the output in response to an input signal. A closed- loop system uses a measurement of the output signal and a comparison with the desired output to generate an error signal that is applied to the actuator.

133. Open- And Closed- Loop Control Systems H(s) 1 Y(s) G(s) 1  G(s )  R(s) E(s ) 1  G(s ) 1  R(s) Thus, to reduce the error, the magnitude of 1  G(s )   1 H(s)  1 Y(s) G(s ) 1  H(s) G(s)  R(s) E(s ) 1 1  H[(s)  G(s )]  R(s) Thus, to reduce the error, the magnitude of 1  G(s ) H(s)   1 Error Signal

134. Sensitivity of Control Systems To Parameter Variations GH(s)  1 1 Y (s)  R (s) H(s) Output affected only by H(s) G(s)   G(s) Open Loop  Y (s)  G(s)  R (s) Closed Loop Y (s)   Y (s) 1   G(s)   G(s)   H(s)  G(s)   G(s)   R (s)  Y (s)  G(s)  1  GH(s)   GH(s)  (1  GH(s))  R (s) GH(s)   GH(s) The change in the output of the closed system is reduced by a factor of 1+GH(s)  Y (s)  G(s)  R (s)  (1 GH(s)) 2 For the closed- loop case if

135. Sensitivity of Control Systems To Parameter Variations T ( s ) Y ( s ) R ( s ) S  T ( s ) T ( s )  G ( s ) G ( s ) S d T d T T d G  d G     G    d     T  d T   G G d     G  d  T T ( s ) 1 1  H [ ( s )  G ( s ) ] S G T   d   T G   d   G  d   d T   G T   d   T G   d   G  d   d T   G T ( 1  GH ) 2 G ( 1  GH ) 1  G S G T 1 ( 1  GH ) T  GH ( 1  GH ) Sensitivity of the closed- loop to G variations reduced Sensitivity of the closed-loop to H variations When GH is large sensitivity approaches 1 Changes in H directly affects the output response S H

136. Example 4.1 Open loop v o  K a  v in Closed loop  R2 R1  K a R p R1  R2 T  k a T 1  K  S Ka T a a 1 1  K  T S Ka 1 If Ka is large, the sensitivity is low. 4 K a  10   0.1 S T Ka  1 3 1  10  4  9.99  10

137. Control of the Transient Response of Control Systems  (s) V a (s) G(s) K 1  1  s  1 where, K 1 K m R a  b  K b  K m  1 R a  J R a  b  K b  K m

138. Control of the Transient Response of Control Systems  (s) K a  G(s) R(s) 1  K a  K t  G(s) K a  K 1  1  s  1  K a  K t  K 1  1 K 1 K a   1   1  K a  K t  K 1     s   

139. Control of the Transient Response of Control Systems

140. Disturbance Signals In a Feedback Control Systems R(s)

141. Disturbance Signals In a Feedback Control Systems

142. G1(s) Ka  Km Ra G2(s) 1 (J  s  b) H(s) Kt  Kb Ka E(s)  (s) G(s)  Td(s) 1  G1(s)  G2(s)  H(s) G 1 G 2 H ( s )  1 E(s) 1 G1(s)  H(s)  Td(s) If G1(s)H(s) very large the effect of the disturba can be minimized G1(s) H(s) Ra Ka  Km  Kt  Kb  Ka   approximately Ka  Km  Kt Ra since Ka >> Kb Strive to maintain Ka large and Ra < 2 ohms Disturbance Signals In a Feedback Control Systems

143. Steady- State Error E o (s) R(s)  Y(s) (1  G(s))  R(s) E c (s) 1  R(s) 1  G(s) Steady State Error H(s) 1 lim e(t) t  lim s  E(s) s  For a step unit input e o (infinite) s lim s  (1  G(s))  1 s  lim (1  G(0)) s  e c (infinite)  1  G(s)  s 1   1 lim s   s  1   lim s   1  G(0) 

144. The Cost of Feedback Increased Number of components and Complexity Loss of Gain Instability

145. Design Example: English Channel Boring Machines Y(s) T(s)  R(s)  Td(s)  D(s) Y(s) K  11  s s 2  12  s  K  R(s)  1 s 2  12  s  K  D(s)

146. Design Example: English Channel Boring Machines Study system for different Values of gain K Steady state error for R(s)=1/s and D(s)=0 e(t) lim t  infinite 1 s 2  s        1 lim s    s   1  K G  11  s  s Steady state error for R(s)=0 D(s)=1/s y(t) lim t  infinite 1 2 s  12  s  K   lim s   s    s   1 1 K T d

147. Transient vs Steady- State The output of any differential equation can be broken up into two parts, •a transient part (which decays to zero as t goes to infinity) and •a steady- state part (which does not decay to zero as t goes to infinity). t  y ( t )  y tr ( t )  y ss ( t ) lim y tr ( t )  Either part might be zero in any particular case.

148. Prototype systems 1 st Order system 2 nd order system Agenda: transfer function response to test signals impulse step ramp parabolic sinusoidal c ( t )  c ( t )  kr ( t ) 1  c ( t )  2 2  c ( t )   c ( t )  kr ( t ) n n

149. 1 st order system Impulse response Step response Ramp response Relationship between impulse, step and ramp Relationship between impulse, step and ramp responses G ( s )  C ( s )  1 T R ( s ) s  1 T c ( t )  1 e  t T 1( t ) T  r ( t )   ( t ), R ( s )  1, r ( t )  1( t ), R ( s )  1 , s r ( t )  t 1( t ), R ( s )  1 , step c ( t )    1  e  t T   1( t ) s 2 c ( t )   t   T  Te  t T   1( t ) ramp

150. 1 st Order system Prototype parameter: Time constant Relate problem specific parameter to prototype parameter. Parameters: problem specific constants. Numbers that do not change with time, but do change from problem to problem. We learn that the time constant defines a problem specific time scale that is more convenient than the arbitrary time scale of seconds, minutes, hours, days, etc, or fractions thereof.

151. Transient vs Steady state Consider the impulse, step, ramp responses computed earlier. Identify the steady state and the transient parts.

152. 1 st order system Impulse response Step response Ramp response Relationship between impulse, step and ramp Relationship between impulse, step and ramp responses G ( s )  C ( s )  1 T , T  R ( s ) s  1 T c ( t )  1 e  t T 1( t ) T  r ( t )   ( t ), R ( s )  1, r ( t )  1( t ), R ( s )  1 , s r ( t )  t 1( t ), R ( s )  1 , step c ( t )    1  e  t T   1( t ) s 2   c ( t )   t  T  Te  t T  1( t ) ramp Consider the impulse, step, ramp responses computed earlier. Identify the steady state and the transient parts. Compare steady- state part to input function, transient part to TF.

153. 2 nd order system Over damped (two real distinct roots = two 1 st order systems with real poles) Critically damped (a single pole of multiplicity two, highly unlikely, requires exact matching) Underdamped (complex conjugate pair of poles, oscillatory behavior, most common) step response 2 G ( s )  C ( s )  n s 2  2  s   2 n n R ( s ) K   1  1   2     1   2  ( t )  K  1  c   sin  t  tan   1( t )   e   n t step d    1   2 e sin  1( t ) n   t  n d   c ( t )  K   t    

154. 2 nd Order System Prototype parameters: undamped natural frequency, damping ratio Relating problem specific parameters to prototype parameters

155. Transient vs Steady state Consider the step, responses computed earlier. Identify the steady state and the transient parts.

156. 2 nd order system Over damped (two real distinct roots = two 1 st order systems with real poles) Critically damped (a single pole of multiplicity two, highly unlikely, requires exact matching) Underdamped (complex conjugate pair of poles, oscillatory behavior, most common) step response 2 G ( s )  C ( s )  n s 2  2  s   2 n n R ( s ) K   1  1   2     1   2  ( t )  K  1  c   sin  t  tan   1( t )   e   n t step d    1   2 e sin  1( t ) n   t  n d   c ( t )  K   t    

157. Use of Prototypes Too many examples to cover them all We cover important prototypes We develop intuition on the prototypes We cover how to convert specific examples to prototypes We transfer our insight, based on the study of the prototypes to the specific situations.

158. Transient- Response Spedifications Delay time, t d : The time required for the response to reach half the final value the very first time. Rise time, t r : the time required for the response to rise from 10% to 90% (common for overdamped and 1 st order systems); 5% to 95%; or 0% to 100% (common for underdamped systems); of its final value Peak time, t p : Maximum (percent) overshoot, M p : Settling time, t s

159. Order Systems r t      d t p  d    100% p M  e   1   2 t  4 T  4  4 2% s n   t  3 T  3  3 5% s n   Derived relations fo  r 2  nd  1   2 d n    n   tan  1   d       See book for details. (Pg. 232) Allowable M p determines damping ratio . Settling time then determines undamped natural frequency . Theory is used to derive relationships between design specifications and prototype parameters . Which are related to problem parameters.

160. Higher order system PFEs have linear denominators. each term with a real pole has a time constant each complex conjugate pair of poles has a damping ratio and an undamped natural frequency.

161. Proportional control of plant w integrator G ( s )  1 G C ( s )  K p , s ( Js  b )

162. Integral control of Plant w disturbance 1 G ( s )  K , G ( s )  s ( Js  b ) C s

163. Proportional Control of plant w/o integrator G ( s )  1 G ( s )  K , Ts  1 C

164. Integral control of plant w/o integrator 1 G ( s )  K , G ( s )  Ts  1 C s

165. UNIT- III S T ABIL I T Y A N A L Y S I S IN S- DOMAIN

166. The issue of ensuring the stability of a closed- loop feedback system is central to control system design. Knowing that an unstable closed- loop system is generally of no practical value, we seek methods to help us analyze and design stable systems. A stable system should exhibit a bounded output if the corresponding input is bounded. This is known as bounded- input, bounded- output stability and is one of the main topics of this chapter. The stability of a feedback system is directly related to the location of the roots of the characteristic equation of the system transfer function. The Routh– Hurwitz method is introduced as a useful tool for assessing system stability. The technique allows us to compute the number of roots of the characteristic equation in the right half- plane without actually computing the values of the roots. Thus we can determine stability without the added computational burden of determining characteristic root locations. This gives us a design method for determining values of certain system parameters that will lead to closed- loop stability. For stable systems we will introduce the notion of relative stability, which allows us to characterize the degree of stability. Chapter 6 – The Stability of Linear Feedback Systems

167. The Concept of Stability A stable system is a dynamic system with a bounded response to a bounded input. Absolute stability is a stable/not stable characterization for a closed- loop feedback system. Given that a system is stable we can further characterize the degree of stability, or the relative stability.

168. The Concept of Stability The concept of stability can be illustrated by a cone placed on a plane horizontal surface. A necessary and sufficient condition for a feedback system to be stable is that all the poles of the system transfer function have negative real parts. A system is considered marginally stable if only certain bounded inputs will result in a bounded output.

169. The Routh- Hurwitz Stability Criterion It was discovered that all coefficients of the characteristic polynomial must have the same sign and non- zero if all the roots are in the left- hand plane. These requirements are necessary but not sufficient. If the above requirements are not met, it is known that the system is unstable. But, if the requirements are met, we still must investigate the system further to determine the stability of the system. The Routh- Hurwitz criterion is a necessary and sufficient criterion for the stability of linear systems.

170. The Routh- Hurwitz Stability Criterion n  1 n  3 n  1 n  1 s n s n  1 s n  2 s n  3 n  2 n  1   1 n  1  1 n  3 n  1 n  1 n  3 n  1 n  1 n  1 b              s a n n s n  1  a a s n  a s n  2    a s  a  1 b b a n  1 a n  3 n  1 n  3 b c a a a a n  2 a a a a a n  2 a n  4   a n  1  a n  2   a n  a n  3    1 b h b b b a n a n  2 a n  4 a n  1 a n  3 a n  5 n  1 n  3 n  5 c n  1 c n  3 c n  5 Characteristic equation, q(s) Routh array 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.

171. The Routh- Hurwitz Stability Criterion Case One: No element in the first column is zero. Example 6.1 Second-order system The Characteristic polynomial of a second- order system is: q(s) a 2  s 2  a 1  s  a The Routh array is written as: w here: b 1 a 1  a  (0)  a 2 a 1 a Theref ore the requirement for a stable second- order system is simply that a l coeff icients be positive or all the coefficients be negative. 1 s 2 s 1 s 1 a 2 a a b

172. The Routh- Hurwitz Stability Criterion Case Two: Zeros in the first column while some elements of the row containing a zero in the first column are nonzero. If only one element in the array is zero, it may be replaced w ith a sma l positiv e number  that is allow ed to approach zero after completing the array. q(s) s 5  2s 4  2s 3  4s 2  11s  10 The Routh array is then: w here: 2  2  1  4 4   2  6  12 6  c 1  10  b 1  c 1 d 1 6 2   c 1 There are two sign changes in the first column due to the large negative number calculated for c1. Thus, the system is unstable because two roots lie in the right half of the plane. s 5 s 4 s 3 s 2 s 1 s 1 1 1 2 11 2 4 10 b 6 c 10 d 1 10

173. The Routh- Hurwitz Stability Criterion Case Three: Zeros in the first column, and the other elements of the row containing the zero are also zero. This case occurs when the polynomial q(s) has zeros located sy metrically about the origin of the s- plane, such as (s+  )(s -  ) or (s+j  )(s-j  ). This case is solved using the auxiliary polynomial, U(s), w hich is located in the row above the row containing the zero entry in the Routh array. q(s) s 3  2  s 2  4s  K Routh array: For a stable system we require that  s  8 For the marginally stable case, K=8, the s^1 row of the Routh array contains all zeros. The auxiliary plynomial comes f rom the s^2 row. U(s) 2s 2  Ks 2  s 2  8 2  s 2  4  2(s  j  2)(s  j  2) It can be proven that U(s) is a factor of the characteristic polynomial: q(s) s  2 U(s) 2 Thus, w hen K=8, the factors of the characteristic polynomial are: q(s) (s  2)(s  j  2)(s  j  2) s 3 s 2 s 1 s 1 4 2 K 8  K 2 K

174. The Routh- Hurwitz Stability Criterion Case Four: Repeated roots of the characteristic equation on the jw- axis. With simple roots on the jw- axis, the system will have a marginally stable behavior. This is not the case if the roots are repeated. Repeated roots on the jw- axis will cause the system to be unstable. Unfortunately, the routh-array will fail to reveal this instability.

175. Example 6.4

176. Example 6.5 Welding control Using block diagram reduction we find that: The Routh array is then: Ka s 4 s 3 s 2 s 1 s c 11 ( K  6) Ka b Ka 3 3 1 6 For the system to be stable both b 3 and c 3 must be positive. Using these equations a relationship can be determined for K a where: b 3 60  K 6 and c 3 b 3 (K  6)  6  Ka b 3

177. The Relative Stability of Feedback Control Systems It is often necessary to know the relative damping of each root to the characteristic equation. Relative system stability can be measured by observing the relative real part of each root. In this diagram r2 is relatively more stable than the pair of roots labeled r1. One method of determining the relative stability of each root is to use an axis shift in the s-domain and then use the Routh array as shown in Example 6.6 of the text.

178. Problem statement: Design the turning control for a tracked vehicle. Select K and a so that the system is stable. The system is modeled below. Design Example: Tracked Vehicle Turning Control

179. The characteristic equation of this system is: 1  G c  G(s) or K(s  a) 1  s(s  1)(s  2)(s  5) Thus, s(s  1)(s  2)(s  5)  K(s  a) or s 4  8s 3  17s 2  (K  10)s  Ka To determine a stable region for the system , we establish the Routh arra where b 3 126  K 8 and c 3 b 3 (K  10)  8Ka b 3 Ka s 4 s 3 s 2 s 1 s c b Ka 3 3 1 8 17 ( K  10) Ka Design Example: Tracked Vehicle Turning Control

180. Ka s 4 s 3 s 2 s 1 s c 17 ( K  10) Ka b Ka 3 3 1 8 Design Example: Tracked Vehicle Turning Control where b 3 126  K 8 and c 3 b 3 (K  10)  8Ka b 3 Therefore, K  126 K  a  (K  10)(126  K)  64Ka 

181. The Root Locus Method In the preceding chapters we discussed how the performance of a feedback system can be described in terms of the location of the roots of the characteristic equation in the s-plane. We know that the response of a closed- loop feedback system can be adjusted to achieve the desired performance by judicious selection of one or more system parameters. It is very useful to determine how the roots of the characteristic equation move around the s-plane as we change one parameter. The locus of roots in the s- plane can be determined by a graphical method. A graph of the locus of roots as one system parameter varies is known as a root locus plot. The root locus is a powerful tool for designing and analyzing feedback control systems and is the main topic of this chapter. We will discuss practical techniques for obtaining a sketch of a root locus plot by hand. We also consider computer- generated root locus plots and illustrate their effectiveness in the design process. The popular PID controller is introduced as a practical controller structure. We will show that it is possible to use root locus methods for design when two or three parameters vary. This provides us with the opportunity to design feedback systems with two or three adjustable parameters. For example the PID controller has three adjustable parameters. We will also define a measure of sensitivity of a specified root to a small incremental change in a system parameter.

182. The Root Locus Method

183. The root locus is a graphical procedure for determining the poles of a closed- loop system given the poles and zeros of a forward- loop system. Graphically, the locus is the set of paths in the complex plane traced by the closed- loop poles as the root locus gain is varied from zero to infinity. In mathematical terms, given a forward-loop transfer function, K G (s) where K is the root locus gain, and the corresponding closed- loop transfer function the root locus is the set of paths traced by the roots of as K varies from zero to infinity. As K changes, the solution to this equation changes. This equation is called the characteristic equation. This equation defines where the poles will be located for any value of the root locus gain, K. In other words, it defines the characteristics of the system behavior for various values of controller gain. The Root Locus Method

184. The Root Locus Method

185. The Root Locus Method

186. The Root Locus Method

187. The Root Locus Method

188. No matter what we pick K to be, the closed-loop system must always have n poles, where n is the number of poles of G(s). The root locus must have n branches, each branch starts at a pole of G(s) and goes to a zero of G(s). If G(s) has more poles than zeros (as is often the case), m < n and we say that G(s) has zeros at infinity. In this case, the limit of G(s) as s - > infinity is zero. The number of zeros at infinity is n-m, the number of poles minus the number of zeros, and is the number of branches of the root locus that go to infinity (asymptotes). Since the root locus is actually the locations of all possible closed loop poles, from the root locus we can select a gain such that our closed-loop system will perform the way we want. If any of the selected poles are on the right half plane, the closed-loop system will be unstable. The poles that are closest to the imaginary axis have the greatest influence on the closed-loop response, so even though the system has three or four poles, it may still act like a second or even first order system depending on the location(s) of the dominant pole(s). The Root Locus Method

189. Example

190. MATLAB Example - Plotting the root locus of a transfer function Consider an open loop system which has a transfer function of G(s) = (s+7)/s(s+5)(s+15)(s+20) How do we design a feedback controller for the system by using the root locus method? Enter the transfer function, and the command to plot the root locus: num=[1 7]; den=conv(conv([1 0],[1 5]),conv([1 15],[1 20])); rlocus (num,den) axis ([- 22 3 - 15 15]) Example

191. Graphical Method

192. Graphical Method

193. Root Locus Design GUI (rltool) The Root Locus Desi gn GUI is an interactive graphical tool to design compensators using the root locus method. This GUI plots the locus of the closed- loop poles as a function of the compensator gains. You c an use this GUI t o add compensator poles an d zeros and analyze how their location af fects the root lo cus and various time and fr equency domain r esponses. Click on the vari ous controls on the GUI to see what they do.

194. UNIT-IV FREQUENCY RESPONSE AN A L Y SIS

195. Frequency Response Methods and Stability In previous chapters we examined the use of test signals such as a step and a ramp signal. In this chapter we consider the steady-state response of a system to a sinusoidal input test signal. We will see that the response of a linear constant coefficient system to a sinusoidal input signal is an output sinusoidal signal at the same frequency as the input. However, the magnitude and phase of the output signal differ from those of the input sinusoidal signal, and the amount of difference is a function of the input frequency. Thus we will be investigating the steady- state response of the system to a sinusoidal input as the frequency varies. We will examine the transfer function G(s) when s =jw and develop methods for graphically displaying the complex number G(j)as w varies. The Bode plot is one of the most powerful graphical tools for analyzing and designing control systems, and we will cover that subject in this chapter. We will also consider polar plots and log magnitude and phase diagrams. We will develop several time- domain performance measures in terms of the frequency response of the system as well as introduce the concept of system bandwidth.

196. Introduction The frequency response of a system is defined as the steady-state response of the system to a sinusoidal input signal. The sinusoid is a unique input signal, and the resulting output signal for a linear system, as well as signals throughout the system, is sinusoidal in the steady- state; it differs form the input waveform only in amplitude and phase.

197. Frequency Response Plots Polar Plots

198. Frequency Response Plots Polar Plots     1000   999  1000 j   1 R  1 C  0.01  1  1 R  C G     1   1   j     1 Negative  1 0.5 0.5 Im(G(  )) 0.5 Re(G(  ))        1 Positive 

199. Frequency Response Plots Polar Plots 20 Im(G1(  )) 500  997.50 6 1000 60  7  4  10 40 Re(G1(  ))  4  2  10  49.875    .1  1000   0.5 K  100 G1     K     j   j   1 

200. Frequency Response Plots Polar Plots

201. Frequency Response Plots Bode Plots – Real Poles

202. Frequency Response Plots Bode Plots – Real Poles   0.1  0.11  1000   j   1 R  1 C  0.01   R  C 1  1    1  100 G     1 j   1 0.1 1 100  1 10 3 30 10  (break frequency or corner frequency) - 3dB 10 20  log  G(  )  20 (break frequency or corner frequency)

203. Frequency Response Plots Bode Plots – Real Poles 1.5 0.1 1 100 1  10 3       atan    0.5  (  ) 1 10  (break frequency or corner frequency)

204. Frequency Response Plots Bode Plots – Real Poles (Graphical Construction)

205. Frequency Response Plots Bode Plots – Real Poles

206. Frequency Response Plots Bode Plots – Real Poles

207. s i  j  i  i   end  10 i  r range variable: i   N range for plot:   start  1   end  N r  log    step size:  start  .01 N  50  end  100 lowest frequency (in Hz): number of p oints: highest frequency (in Hz): Next , choose a frequency rangefor the plots (use powers of 10 for convenient plotting): G(s)  K  3  s  (1  s)   1  s  K  2 Assume  ps  G     180  arg  G  j    360   if  arg  G  j     1   Magnit ude: db  G     20  log  G  j    Phase shift: Frequency Response Plots Bode Plots – Real Poles

208. range for plot: i   N range variable:  i   end  10 i  r s i  j  i 0.01 0.1 10 100 200 100 100 1  i 20  log  G  s i   0.01 0.1 10 100 ps  G   i   180 1  i Frequency Response Plots Bode Plots – Real Poles

209. Frequency Response Plots Bode Plots – Complex Poles

210. Frequency Response Plots Bode Plots – Complex Poles

211. Frequency Response Plots Bode Plots – Complex Poles 2  r  n  1  2    0.707 M p  G   r  1   2  2  1      0.70

212. Frequency Response Plots Bode Plots – Complex Poles 2  r  n  1  2    0.707 M p  G   r  1  2   2  1      0.707

213. Frequency Response Plots Bode Plots – Complex Poles

214. Frequency Response Plots Bode Plots – Complex Poles

215. Performance Specification In the Frequency Domain

216.   .1  .11  2 K  2 j   1 G     K j   j   1    j   2    G      Bode1   20  log 20 .5 1 .5 2 Bode1(  ) 20 1  Open Loop Bode Diagram T    G     1  G    Bode2     20  log  T     .5 1 1 .5 2 20 10 10 Bode2(  )  Closed- Loop Bode Diagram Performance Specification In the Frequency Domain

217. Performance Specification In the Frequency Domain w  4 Finding the Resonance Frequency Given 20  log  T( w)  wr  Find( w) 5.282 wr  0.813 Mpw  1 Given 20  log(Mpw) 5.282 Mpw  Find(Mpw) Finding Maximum value of the frequency resp Mpw  1.837 0.5 1 1.5 2 20 10 Bode2(  ) 10  Closed- Loop Bode Diagram

218. Performance Specification In the Frequency Domain Assume that the system has dominant second-order roo   .1 Given Mpw     2  2  1     1   Find    wn  .1   0.284 Given wr wn  1  2  2 wn  Find( wn) wn  0.888 0.5 1 1.5 2 20 Finding the natural frequency Finding the damping factor 10 10 Bode2(  )  Closed- Loop Bode Diagram

219. Performance Specification In the Frequency Domain

220. Performance Specification In the Frequency Domain GH1    5 6   j   0.5  j   1    j    1 

221. Performance Specification In the Frequency Domain Example

222. Performance Specification In the Frequency Domain Example

223. Performance Specification In the Frequency Domain Example

224. Performance Specification In the Frequency Domain Example

225. Frequency Response Methods Using MATLAB

226. Frequency Response Methods Using MATLAB

227. Frequency Response Methods Using MATLAB

228. Frequency Response Methods Using MATLAB

229. Frequency Response Methods Using MATLAB

230. Frequency Response Methods Using MATLAB

231. Frequency Response Methods Using MATLAB

232. Bode Plots Bode plot is the representation of the magnitude and phase of G(j*w) (where the frequency vector w contains only positive frequencies). To see the Bode plot of a transfer function, you can use the MATLAB bode command. For example, bode(50,[1 9 30 40]) displays the Bode plots for the transfer function: 50 / (s^3 + 9 s^2 + 30 s + 40)

233. Gain and Phase Margin Let's say that we have the following system: where K is a variable (constant) gain and G(s) is the plant under consideration. The gain margin is defined as the change in open loop gain required to make the system unstable. Systems with greater gain margins can withstand greater changes in system parameters before becoming unstable in closed loop. Keep in mind that unity gain in magnitude is equal to a gain of zero in dB. The phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable. The phase margin is the difference in phase between the phase curve and - 180 deg at the point corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, Wgc). Likewise, the gain margin is the difference between the magnitude curve and 0dB at the point corresponding to the frequency that gives us a phase of - 180 deg (the phase cross over frequency, Wpc).

234. Gain and Phase Margin - 180

235. We can find the gain and phase margins for a system directly, by using MATLAB. Just enter the margin command. e This command returns the gain and phase margins, the gain and phase cross over frequencies, and a graphical representation of thes on the Bode plot. margin(50,[1 9 30 40]) Gain and Phase Margin

236. s i  j  i  i   end  10 i  r range variable: i   N range for plot:   start  1   end  N r  log    step size:  start  .01 N  50  end  100 lowest frequency (in Hz): number of p oints: highest frequency (in Hz): Next , choose a frequency rangefor the plots (use powers of 10 for convenient plotting): G(s)  K  3  s  (1  s)   1  s  K  2 Assume  Phase shift: ps  G     180  arg  G  j    360   if  arg  G  j     1   db  G     20  log  G  j    Magnit ude: Gain and Phase Margin

237. gm  6.021 gm   db  G   gm  Now using the p hase angle plot, estimate the frequency at which the phase shift crosses 180 degr  gm  1.8 Solve for  at the phase shift point of 180 degrees:  gm  root  ps  G   gm   180   gm   gm  1.732 Calculate the gain m a r g i n : degrees pm  18.265  c  1.193 Guess for crossover frequency : Solve for the gain crossover frequency:  c  root  db  G   c    c  Calculate the phase m a r g i n : pm  ps  G  c   180 Gain Margin  c  1 Gain and Phase Margin

238. The Nyquist Stability Criterion The Nyquist plot allows us also to predict the stability and performance of a closed- loop system by observing its open- loop behavior. The Nyquist criterion can be used for design purposes regardless of open- loop stability (Bode design methods assume that the system is stable in open loop). Therefore, we use this criterion to determine closed- loop stability when the Bode plots display confusing information. The Nyquist diagram is basically a plot of G(j* w) where G(s) is the open-loop transfer function and w is a vector of frequencies which encloses the entire right- half plane. In drawing the Nyquist diagram, both positive and negative frequencies (from zero to infinity) are taken into account. In the illustration below we represent positive frequencies in red and negative frequencies in green. The frequency vector used in plotting the Nyquist diagram usually looks like this (if you can imagine the plot stretching out to infinity): However, if we have open- loop poles or zeros on the jw axis, G(s) will not be defined at those points, and we must loop around them when we are plotting the contour. Such a contour would look as follows:

239. The Cauchy criterion The Cauchy criterion (from complex analysis) states that when taking a closed contour in the complex plane, and mapping it through a complex function G(s), the number of times that the plot of G(s) encircles the origin is equal to the number of zeros of G(s) enclosed by the frequency contour minus the number of poles of G(s) enclosed by the frequency contour . Encirclements of the origin are counted as positive if they are in the same direction as the original closed contour or negative if they are in the opposite direction. When studying feedback controls, we are not as interested in G(s) as in the closed- loop transfer function: G(s) 1 + G(s) If 1+ G(s) encircles the origin, then G(s) will enclose the point - 1. Since we are interested in the closed- loop stability, we want to know if there are any closed- loop poles (zeros of 1 + G(s)) in the right- half plane. Therefore, the behavior of the Nyquist diagram around the -1 point in the real axis is very important; however, the axis on the standard nyquist diagram might make it hard to see what's happening around this point

240. Gain and Phase Margin Gain Margin is defined as the change in open- loop gain expressed in decibels (dB), required at 180 degrees of phase shift to make the system unstable. First of all, let's say that we have a system that is stable if there are no Nyquist encirclements of -1, such as : 50 s^3 + 9 s^2 + 30 s + 40 Looking at the roots, we find that we have no open loop poles in the right half plane and therefore no closed- loop poles in the right half plane if there are no Nyquist encirclements of -1. Now, how much can we vary the gain before this system becomes unstable in closed loop? The open- loop system represented by this plot will become unstable in closed loop if the gain is increased past a certain boundary.

241. and that the Nyquist diagram can be viewed by typing: nyquist (50, [1 9 30 40 ]) The Nyquist Stability Criterion

242. Phase margin as the change in open- loop phase shift required at unity gain to make a closed- loop system unstable. From our previous example we know that this particular system will be unstable in closed loop if the Nyquist diagram encircles the -1 point. However, we must also realize that if the diagram is shifted by theta degrees, it will then touch the -1 point at the negative real axis, making the system marginally stable in closed loop. Therefore, the angle required to make this system marginally stable in closed loop is called the phase margin (measured in degrees). In order to find the point we measure this angle from, we draw a circle with radius of 1, find the point in the Nyquist diagram with a magnitude of 1 (gain of zero dB), and measure the phase shift needed for this point to be at an angle of 180 deg. Gain and Phase Margin

243. w   100   99.9  100 j   1 s( w)  j  w f( w)   1 G( w)  50  4.6 s( w) 3  9  s( w) 2  30  s( w)  40 5 2 1 1 3 4 5 6 5 2 Re(G(w)) Im(G(w)) The Nyquist Stability Criterion

244. Consider the Negative Feedback System Remember from the Cauchy criterion that the number N of times that the plot of G(s)H(s) encircles -1 is equal to the number Z of zeros of 1 + G(s)H(s) enclosed by the frequency contour minus the number P of poles of 1 + G(s)H(s) enclosed by the frequency contour (N = Z - P). Keeping careful track of open- and closed- loop transfer functions, as well as numerators and denominators, you should convince yourself that: the zeros of 1 + G(s)H(s) are the poles of the closed- loop transfer function the poles of 1 + G(s)H(s) are the poles of the open- loop transfer function. The Nyquist criterion then states that: P = the number of open- loop (unstable) poles of G(s)H(s) N = the number of times the Nyquist diagram encircles - 1 clockwise encirclements of -1 count as positive encirclements counter- clockwise (or anti-clockwise) encirclements of -1 count as negative encirclements Z = the number of right half-plane (positive, real) poles of the closed- loop system The important equation which relates these three quantities is: Z = P + N

245. Knowing the number of right- half plane (unstable) poles in open loop (P), and the number of encirclements of -1 made by the Nyquist diagram (N), we can determine the closed- loop stability of the system. If Z = P + N is a positive, nonzero number, the closed-loop system is unstable. We can also use the Nyquist diagram to find the range of gains for a closed- loop unity feedback system to be stable. The system we will test looks like this: where G(s) is : s^2 + 10 s + 24 s^2 - 8 s + 15 The Nyquist Stability Criterion - Application

246. This system has a gain K which can be varied in order to modify the response of the closed- loop system. However, we will see that we can only vary this gain within certain limits, since we have to make sure that our closed- loop system will be stable. This is what we will be looking for: the range of gains that will make this system stable in the closed loop. The first thing we need to do is find the number of positive real poles in our open- loop transfer function: roots([1 - 8 15]) ans = 5 3 The poles of the open- loop transfer function are both positive. Therefore, we need two anti- clockwise (N = -2) encirclements of the Nyquist diagram in order to have a stable closed- loop system (Z = P + N). If the number of encirclements is less than two or the encirclements are not anti-clockwise, our system will be unstable. Let's look at our Nyquist diagram for a gain of 1: nyquist ([ 1 10 24], [ 1 -8 15]) There are two anti- clockwise encirclements of - 1. Therefore, the system is stable for a gain of 1. The Nyquist Stability Criterion

247. MathCAD Implementation w   100   99.9  100 j   1 s( w)  j  w G( w)  s( w) 2  10  s( w)  24 2 s( w)  8  s( w)  15 2 1 Re(G(w)) 1 2 2 2 Im(G(w)) The Nyquist Stability Criterion There are two anti- clockwise encirclements of - 1. Therefore, the system is stable for a gain of 1.

248. The Nyquist Stability Criterion

249. Time- Domain Performance Criteria Specified In The Frequency Domain Open and closed- loop frequency resp onses are relat ed by : T  j   G  j   1  G  j   M pw 1 2   1   2   0.707 G    u  j  v M M    M    G  j   1  G  j   u  jv 1  u  jv u 2  v 2 (1  u) 2  v 2 Squaring and rearrenging which is the equation of a circle on u- v planwe with a M 2 1  M 2     u      2  v 2 M 2 1  M       2 u M cent er at 2 1  M 2 v

250. Time- Domain Performance Criteria Specified In The Frequency Domain

251. The Nichols Stability Method Polar S tability Plot - Nichol s M a t h c ad Implementati on This examp le makes a polar plot of a transfer function and draws one contour of const ant closed- loop magnitude. To draw the plot, enter a definition for the transfer f u n G c t ( i o s ) n : G(s)  45000 s  (s  2)  (s  30) The frequency range defined by the next two equations provides a logarithmic frequency scal running from 1 to 100. You can change this range by editing the d e f i n i t i on s m f o a r nd  m : m   100  m  10 .02  m Now enter a value for M to define the closed- loop magnitude contour t hat will be plotted. M  1.1 Calculate the points on the M- circle: 2  M 2 M 2      exp 2   j  .01  m      M  1 M  1 M C m    The first plot s ho w G s , the contour of constant closed- loop magnit u d M e ,

252. The Nichols Stability Method The first plot s ho w G s , the contour of constant closed-loop magnit u d M e , , and the Nyquist of the open loop system Im  G  j  m  Im  MC m  Re  G  j  m   Re  MC m    1

253. The Nichols Stability Method

254. The Nichols Stability Method 1  j   j   1    0.2  j   1  G     M pw  2.5 dB  r  0.8 The closed- loop phase angle at  r is equal to - 72 degrees and  b = 1.33 The closed- loop phase angle at  b is equal t - 142 degrees Mpw -72 deg wr=0.8 - 3dB -142 deg

255. The Nichols Stability Method G     0.64 j      j   2  j   1   Phase Margin = 30 degrees On the basis of the p hase we estimat  e  0.30 M pw  9 dB M pw  2.8  r  0.88 From equation M pw 1 2   1   2 We are confront ed with comflectin  gs The app arent conflict is caused by the nature of G(j  ) which slopes rap idally toward 180 degrees   0.18 line from the 0-dB axis. The designer must use the frequency- domain- t ime- domain correlation with caution PM GM

256. The Nichols Stability Method PM GM

257. Examples – Bode and Nyquist

258. Examples - Bode

259. Examples - Bode

260. Examples – Bode and Nyquist

261. Examples - Nichols

262. Examples - Nichols

263. The Design of Feedback Control Systems PID Compensation Networks

264. Different Types of Feedback Control On- Off Control This is the simplest form of control.

265. Proportional Control A proportional controller attempts to perform better than the On-off type by applying power in proportion to the difference in temperature between the measured and the set- point. As the gain is increased the system responds faster to changes in set- point but becomes progressively underdamped and eventually unstable. The final temperature lies below the set- point for this system because some difference is required to keep the heater supplying power.

266. Proportional, Derivative Control The stability and overshoot problems that arise when a proportional controller is used at high gain can be mitigated by adding a term proportional to the time- derivative of the error signal. The value of the damping can be adjusted to achieve a critically damped response.

267. Proportional+Integral+Derivative Control Although PD control deals neatly with the overshoot and ringing problems associated with proportional control it does not cure the problem with the steady-state error. Fortunately it is possible to eliminate this while using relatively low gain by adding an integral term to the control function which becomes

268. The Characteristics of P, I, and D controllers A proportional controller (Kp) will have the effect of reducing the rise time and will reduce, but never eliminate, the steady- state error. An integral control (Ki) will have the effect of eliminating the steady-state error, but it may make the transient response worse. A derivative control (Kd) will have the effect of increasing the stability of the system, reducing the overshoot, and improving the transient response.

269. Proportional Control By only employing proportional control, a steady state error occurs. Proportional and Integral Control The response becomes more oscillatory and needs longer to settle, the error disappears. Proportional, Integral and Derivative Control All design specifications can be reached.

270. CL RESPONSE RISE TIME OVERSHOOT SETTLING TIME S- S ERROR Kp Decrease Increase Small Change Decrease Ki Decrease Increase Increase Eliminate Kd Small Change Decrease Decrease Small Change The Characteristics of P, I, and D controllers

271. Tips for Designing a PID Controller Obtain an open- loop response and determine what needs to be improved Add a proportional control to improve the rise time Add a derivative control to improve the overshoot Add an integral control to eliminate the steady-state error Adjust each of Kp, Ki, and Kd until you obtain a desired overall response. Lastly, please keep in mind that you do not need to implement all three controllers (proportional, derivative, and integral) into a single system, if not necessary. For example, if a PI controller gives a good enough response (like the above example), then you don't need to implement derivative controller to the system. Keep the controller as simple as possible.

272. num=1; den=[1 10 20]; step (num,den) Open-Loop Control - Example G(s) 1 s 2  10s  20

273. Proportional Control - Example The proportional controller (Kp) reduces the rise time, increases the overshoot, and reduces the steady- state error. MATLAB Example Kp=300; num=[Kp]; den=[1 10 20+Kp]; t=0:0.01:2; step(num,den,t) Amplitude Step Response From: U(1) 0.2 0.4 0.6 0.8 1 Time (sec.) 1.2 1. 4 1.6 1.8 2 0.2 0.4 0.6 0.8 1 1.2 1.4 To: Y(1) T(s) Kp s 2  10  s  (20  Kp) Amplitude 0.2 0.4 0.6 0.8 1 1.2 Time (sec.) 1.4 1.6 1.8 2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Step Response From: U(1) To: Y(1) K=300 K=100

274. Amplitude Step Response From: U(1) 0.2 0.4 0.6 0.8 1 Time (sec.) 1.2 1.4 1.6 1.8 2 0.2 0.4 0.6 0.8 1 1.2 1.4 To: Y(1) Kp=300; Kd=10; num=[Kd Kp]; den=[1 10+Kd 20+Kp]; t=0:0.01:2; step(num,den,t) Proportional - Derivative - Example The derivative controller (Kd) reduces both the overshoot and the settling time. MATLAB Example T(s) Kd  s  Kp s 2  (10  Kd)  s  (20  Kp) Amplitude 0.2 0.4 0.6 0.8 1 1.2 Time (sec.) 1.4 1.6 1.8 2 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Step Response From: U(1) To: Y(1) Kd=10 Kd=20

275. Proportional - Integral - Example The integral controller (Ki) decreases the rise time, increases both the overshoot and the settling time, and eliminates the steady-state error MATLAB Example Amplitude Step Response From: U(1) 0.2 0.4 0.6 0.8 1 Time (sec.) 1.2 1.4 1.6 1.8 2 0.4 0.6 0.8 1 1.2 1.4 To: Y(1) Kp=30; Ki=70; num=[Kp Ki]; den=[1 10 20+Kp Ki]; 0.2 t=0:0.01:2; step(num,den,t) T(s) Kp  s  Ki s 3  10  s 2  (20  Kp)  s  Ki Amplitude 0.2 0.4 0.6 0.8 1 1.2 Time (sec.) 1.4 1.6 1.8 2 0.2 0.4 0.6 0.8 1 1.2 1.4 Step Response From: U(1) To: Y(1) Ki=70 Ki=100

276. Syntax rltool rltool(sys) rltool(sys,comp) RLTOOL

277. RLTOOL

278. RLTOOL

279. RLTOOL

280. RLTOOL

281. Consider the following configuration: Example - Practice

282. The design a system for the following specifications: · · · · Zero steady state error Settling time within 5 seconds Rise time within 2 seconds Only some overshoot permitted Example - Practice

283. Lead or Phase- Lead Compensator Using Root Locus A first- order lead compensator can be designed using the root locus. A lead compensator in root locus form is given by A phase- lead compensator tends to shift the root locus toward the left half plane. This results in an improvement in the system's stability and an increase in the response speed. When a lead compensator is added to a system, the value of this intersection will be a larger negative number than it was before. The net number of zeros and poles will be the same (one zero and one pole are added), but the added pole is a larger negative number than the added zero. Thus, the result of a lead compensator is that the asymptotes' intersection is moved further into the left half plane, and the entire root locus will be shifted to the left. This can increase the region of stability as well as the response speed. G c (s) where the magnitude of z is less than the magnitude of p . (s  z) (s  p)

284. In Matlab a phase lead compensator in root locus form is implemented by using the transfer function in the form numlead=kc*[1 z]; denlead=[1 p]; and using the conv() function to implement it with the numerator and denominator of the plant newnum=conv(num,numlead); newden=conv(den,denlead); Lead or Phase- Lead Compensator Using Root Locus

285. Lead or Phase- Lead Compensator Using Frequency Response A first- order phase- lead compensator can be designed using the frequency response. A lead compensator in frequency response form is given by In frequency response design, the phase- lead compensator adds positive phase to the system over the frequency range . A bode plot of a phase- lead compensator looks like the following G c (s)  1   s    1   s  p 1  z 1   m z  p sin   m    1   1

286. Lead or Phase- Lead Compensator Using Frequency Response Additional positive phase increases the phase margin and thus increases the stability of the system. This type of compensator is designed by determining alfa from the amount of phase needed to satisfy the phase margin requirements, and determining tal to place the added phase at the new gain- crossover frequency. Another effect of the lead compensator can be seen in the magnitude plot. The lead compensator increases the gain of the system at high frequencies (the amount of this gain is equal to alfa. This can increase the crossover frequency, which will help to decrease the rise time and settling time of the system.

287. In Matlab, a phase lead compensator in frequency response form is implemented by using the transfer function in the form numlead=[aT 1]; denlead=[T 1]; and using the conv() function to multiply it by the numerator and denominator of the plant newnum=conv(num,numlead); newden=conv(den,denlead); Lead or Phase- Lead Compensator Using Frequency Response

288. Lag or Phase- Lag Compensator Using Root Locus A first- order lag compensator can be designed using the root locus. A lag compensator in root locus form is given by When a lag compensator is added to a system, the value of this intersection will be a smaller negative number than it was before. The net number of zeros and poles will be the same (one zero and one pole are added), but the added pole is a smaller negative number than the added zero. Thus, the result of a lag compensator is that the asymptotes' intersection is moved closer to the right half plane, and the entire root locus will be shifted to the right. G c (s) (s  p) where the magnitude of z is greater than the magnitude of p. A phase- lag compensator tends to shift the root locus to the right, which is undesirable. For this reason, the pole and zero of a lag compensator must be placed close together (usually near the origin) so they do not appreciably change the transient response or stability characteristics of the system. (s  z)

289. It was previously stated that that lag controller should only minimally change the transient response because of its negative effect. If the phase- lag compensator is not supposed to change the transient response noticeably, what is it good for? The answer is that a phase- lag compensator can improve the system's steady- state response. It works in the following manner. At high frequencies, the lag controller will have unity gain. At low frequencies, the gain will be z0/p0 which is greater than 1. This factor z/p will multiply the position, velocity, or acceleration constant (Kp, Kv, or Ka), and the steady- state error will thus decrease by the factor z0/p0. In Matlab, a phase lead compensator in root locus form is implemented by using the transfer function in the form numlag=[1 z]; denlag=[1 p]; and using the conv() function to implement it with the numerator and denominator of the plant newnum=conv(num,numlag); newden=conv(den,denlag); Lag or Phase- Lag Compensator Using Root Locus

290. Lag or Phase- Lag Compensator using Frequency Response A first- order phase- lag compensator can be designed using the frequency response. A lag compensator in frequency response form is given by The phase- lag compensator looks similar to a phase- lead compensator, except that a is now less than 1. The main difference is that the lag compensator adds negative phase to the system over the specified frequency range, while a lead compensator adds positive phase over the specified frequency. A bode plot of a phase- lag compensator looks like the following G c (s)  1   s    1   s 

291. In Matlab, a phase- lag compensator in frequency response form is implemented by using the transfer function in the form numlead=[a*T 1]; denlead=a*[T 1]; and using the conv() function to implement it with the numerator and denominator of the plant newnum=conv(num,numlead); newden=conv(den,denlead); Lag or Phase- Lag Compensator using Frequency Response

292. Lead- lag Compensator using either Root Locus or Frequency Response A lead- lag compensator combines the effects of a lead compensator with those of a lag compensator. The result is a system with improved transient response, stability and steady- state error. To implement a lead- lag compensator, first design the lead compensator to achieve the desired transient response and stability, and then add on a lag compensator to improve the steady- state response

293. Exercise - Dominant Pole- Zero Approximations and Compensations The influence of a particular pole (or pair of complex poles) on the response is mainly determin by two factors: the real part of the pole and the relative magnitude of the residue at the pole. T real part determines the rate at which the transient term due to the pole decays; the larger the r part, the faster the decay. The relative magnitude of the residue determines the percentage of total response due to a particular pole. Investigate (using Simulink) the impact of a closed- loop negative real pole on the overshoot of system having complex poles. T(s) pr  n 2 (s  pr)    s 2   2  n  s    n 2   Make pr to vary (2, 3, 5) times the real part of the complex pole for different val  ue(0s.3o,f 0.5, 0.7). Investigate (using Simulink) the impact of a closed- loop negative real zero on the overshoot of system having complex poles. T(s) (s  zr)   s 2   2  n  s    n 2   Make zr to vary (2, 3, 5) times the real part of the complex pole for different valu  e ( s . o 3 f , 0.5, 0.7).

294. Exercise - Lead and Lag Compensation Investigate (using Matlab and Simulink) the effect of lead and lag compensations on the systems indicated below. Summarize your observations. Plot the root-locus, bode diagra and output for a step input before and after the compensations. Remember lead compensation: z<p (place zero below the desired root location or to the left of the fi real poles) lag compensation: z>p (locate the pole and zero near the origin of the s- plane) Lead Compensation (use z=1.33, p=20 and K =15).

295. Lag Compensation (use z=0.09 , and p=0.015, K=1/6 ) Summarize your findings

296. Problem 10.36 Determine a compensator so that the percent overshoot is less than 20% and Kv (velocity constant) is greater than 8.

297. Poles and Zeros and Transfer Functions Transfer Function: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial conditions equal to zero. Transfer functions are defined only for linear time invariant systems. Considerations : Transfer functions can usually be expressed as the ratio of two polynomials in the complex variable, s. Factorization : A transfer function can be factored into the following form. G ( s )  K ( s  z 1 )( s  z 2 )...( s  z m ) ( s  p )( s  p ) ... ( s  p ) 1 2 n The roots of the numerator polynomial are called zeros. The roots of the denominator polynomial are called poles . wlg

298. Poles, Zeros and the S- Plane An Example : You are given the following transfer function. Show the poles and zeros in the s- plane. G ( s )  ( s  8)( s  14) s ( s  4)( s  10) S - plane x x o x o - 4 - 8 - 10 - 14 origin  axis j  axis wlg

299. Poles, Zeros and Bode Plots Characterization : Considering the transfer function of the previous slide. We note that we have 4 different types of terms in the previous general form: These are: , ( s / z  1) s ( s / p  1) 1 K , 1 , B Expressing in dB : Given the tranfer function: G ( jw )  K B ( jw / z  1) ( jw )( jw / p  1) 20log | G ( jw |  20log K  20log | ( jw / z  1) |  20log | jw |  20log | jw / p  1| B wlg

300. Poles, Zeros and Bode Plots Mechanics : We have 4 distinct terms to consider: 20logK B 20log|(jw/z +1)| - 20log|jw| - 20log|(jw/p + 1)| wlg

301.  (rad/sec) dB Mag Phase (deg) 1 1 1 1 1 1 wlg This is a sheet of 5 cycle, semi-log paper. This is the type of paper usually used for preparing Bode plots.

302. Poles, Zeros and Bode Plots Mechanics : The gain term, 20logK B, is just so many dB and this is a straight line on Bode paper, independent of omega (radian frequency). The term, - 20log|jw| = - 20logw, when plotted on semi- log paper is a straight line sloping at - 20dB/decade. It has a magnitude of at w = 1. 20 - 20  =1 - 20db/dec wlg

303. Poles, Zeros and Bode Plots Mechanics : The term, - 20log|(jw/p + 1), is drawn with the following approximation: If w < p we use the approximation that –20log|(jw/p + 1 )| = dB, a flat line on the Bode. If w > p we use the approximation of –20log(w/p), which slopes at -20dB/dec starting at w = p. Illustrated below. It is easy to show that the plot has an error of -3dB at w = p and – 1 dB at w = p/2 and w = 2p. One can easily make these corrections if it is appropriate. 2 - 20 - 40  = p - 20db/dec wlg

304. Poles, Zeros and Bode Plots - 20 - 40 20  = z +20db/dec Mechanics : When we have a term of 20log|(jw/z + 1)| we approximate it be a straight line of slop dB/dec when w < z. We approximate it as 20log(w/z) when w > z, which is a straight line on Bode paper with a slope of + 20dB/dec. Illustrated below. wlg

305. Example 1: Given: G ( jw )  50, 000( jw  10) ( jw  1)( jw  500) First : Always , always , always get the poles and zeros in a form such that the constants are associated with the jw terms. In the above example we do this by factoring out the 10 in the numerator and the 500 in the denominator. G ( jw )  50, 000 x 10( jw /10  1) 100( jw /10  1)  500( jw  1)( jw / 500  1) ( jw  1)( jw / 500  1) Second : When you have neither poles nor zeros at 0, start the Bode at 20log 10 K = 20log 10 100 = 40 dB in this case. wlg

306. Example 1: (continued) Third: Observe the order in which the poles and zeros occur. This is the secret of being able to quickly sketch the Bode. In this example we first have a pole occurring at 1 which causes the Bode to break at 1 and slope – 20 dB/dec. Next, we see a zero occurs at 10 and this causes a slope of +20 dB/dec which cancels out the – 20 dB/dec, resulting in a flat line ( db/dec). Finally, we have a pole that occurs at w = 500 which causes the Bode to slope down at – 20 dB/dec. We are now ready to draw the Bode. Before we draw the Bode we should observe the range over which the transfer function has active poles and zeros. This determines the scale we pick for the w (rad/sec) at the bottom of the Bode. The dB scale depends on the magnitude of the plot and experience is the best teacher here. wlg

307. 1 1 1 1 1 1 dB Mag Phase (deg)  (rad/sec) 1 1 1 1 1 1  (rad/sec) dB Mag Phase (deg) Bode Plot Magnitude for 100(1 + jw/10)/(1 + jw/1)(1 + jw/500) 20 40 - 20 60 - 60 - 60 0.1 1 10 100 1000 10000 wlg

308. Using Matlab For Frequency Response Instruction : We can use Matlab to run the frequency response for the previous example. We place the transfer function in the form: 5000( s  10)  [5000 s  50000] ( s  1)( s  500) [ s 2  501 s  500] The Matlab Program num = [5000 50000]; den = [1 501 500]; Bode (num,den) In the following slide, the resulting magnitude and phase plots (exact) are shown in light color (blue). The approximate plot for the magnitude (Bode) is shown in heavy lines (red). We see the 3 dB errors at the corner frequencies. wlg

309. 10 1 10 2 Frequency (rad/sec) Phase (deg); Magnitude (dB) To: Y(1) Bode Diagrams From: U(1) 40 30 20 10 - 10 10 - 1 10 10 3 10 4 - 100 - 80 - 20 - 40 - 60 1 10 100 500 (1  jw )(1  jw / 500) G ( jw )  100(1  jw /10) Bode for: wlg

310. Phase for Bode Plots Commen t: Generally, the phase for a Bode plot is not as easy to draw or approximate as the magnitude. In this course we will use an analytical method for determining the phase if we want to make a sketch of the phase. Illustration : Consider the transfer function of the previous example. We express the angle as follows:  G ( jw )  tan  1 ( w /10)  tan  1 ( w /1)  tan  1 ( w / 500) We are essentially taking the angle of each pole and zero. Each of these are expressed as the tan - 1 (j part/real part) Usually, about 10 to 15 calculations are sufficient to determine a good idea of what is happening to the phase. wlg

311. Bode Plots Example 2 : Given the transfer function. Plot the Bode magnitude. G ( s )  100(1  s /10) s (1  s /100) 2 Consider first only the two terms of 100 jw Which, when expressed in dB, are; 20log100 – 20 logw. This is plotted below. 1 20 40 - 20 The is a tentative line we use until we encounter the first pole(s) or zero(s) not at the origin. - 20db/dec wlg dB  (rad/sec)

312. 1 1 1 1 1 s (1  s /100) 2 G ( s )  100(1  s /10) dB Mag Phase (deg) 40 20 60 - 20 - 40 - 60 1 10  (rad/sec) 100 1000 0.1 Bode Plots Example 2 : (continued) - 20db/dec -40 db/dec s (1  s /100) 2 1 G ( s )  100(1  s /10) wlg The completed plot is shown below.

313. 1 1 1 1  (rad/sec) dB Mag Bode Plots Example 3: Given : 80(1  jw ) 3 G ( s )  ( jw ) 3 (1  jw / 20) 2 1 1 1 0.1 10 100 40 20 60 - 20 . 20log80 = 38 dB -60 dB/dec -40 dB/dec wlg

314. 1 1 1 dB Mag Phase (deg) 20 40 60 - 20 - 40 - 60 1 10  (rad/sec) 100 1000 0.1 Bode Plots -40 dB/dec + 20 dB/dec Given : Sort of a low pass filter Example 4: 2 G ( jw )  10(1  jw / 2) (1  j 0.025 w )(1  jw / 500) 2 1 1 1 wlg

315. 1 1 1 1 1 1 10  (rad/sec) dB Mag Phase (deg) 40 20 60 - 20 - 40 - 60 1 100 1000 0.1 Bode Plots (1  jw / 2) 2 (1  jw /1700) 2 (1  jw / 30) 2 (1  jw /100) 2 G ( jw )  -40 dB/dec + 40 dB/dec Given: Sort of a low pass filter Example 5 wlg

316. Bode Plots Given: problem 11.15 text ( jw ) 2 (0.1 jw  1) 64( jw  1)(0.01 jw  1) ( jw ) 2 ( jw  10) 640( jw  1)(0.01 jw  1) H ( jw )   0.01 0.1 1 10 100 1000 20 40 - 20 - 40 dB mag . . . . . - 40dB/dec - 20db/dec - 40dB/dec - 20dB/dec Example 6 wlg

317. Bode Plots Design Problem: Design a G(s) that has the following Bode plot. dB mag  rad/sec 20 40 0.1 1 10 100 900 1000 30 30 dB +40 dB/dec - 40dB/dec ? ? Example 7 wlg

318. Bode Plots Procedure: The two break frequencies need to be found. Recall: #dec = log 10 [w 2 /w 1 ] Then we have: ( #dec )( 40dB/dec) = 30 d B log 10 [w 1 /30] = 0.75 w 1 = 5.33 rad/sec Also: log 10 [w 2 /900] (- 40dB/dec) = - 30dB This gives w 2 = 5060 rad/sec wlg

319. Bode Plots Procedure: (1  s / 5.3) 2 (1  s / 5060) 2 G ( s )  (1  s / 30) 2 (1  s / 900) 2 Clearing: ( s  5.3) 2 ( s  5060) 2 G ( s )  ( s  30) 2 ( s  900) 2 Use Matlab and conv: N 2  ( s 2  10120 s  2.56 xe 7 ) N2 = [1 10120 2.56e+7] 7.222e+8 s 4 N 1  ( s 2  10.6 s  28.1) N1 = [1 10.6 28.1] N = conv(N1,N2) 1 1.86e+3 2.58e+7 2.73e+8 s 3 s 2 s 1 s wlg

320. Bode Plots Procedure: The final G(s) is given by; Testing : We now want to test the filter. We will check it at  = 5.3 rad/sec And  = 164. At  = 5.3 the filter has a gain of 6 dB or about 2. At  = 164 the filter has a gain of 30 dB or about 31.6. We will check this out using MATLAB and particularly, Simulink. ( s 4  1860 s 3  9.189 e 2 s 2  5.022 e 7 s  7.29 e 8 ) ( s 4  10130.6 s 3  2.571 e 8 s 2  2.716 e 8 s  7.194 e 8 ) G ( s )  wlg

321. Matlab (Simulink) Model: wlg

322. Filter Output at  = 5.3 rad/sec Produced from Matlab Simulink wlg

323. Filter Output at  = 70 rad/sec Produced from Matlab Simulink wlg

324. Reverse Bode Plot Required: From the partial Bode diagram, determine the transfer function (Assume a minimum phase system)  20 db/dec dB 20 db/dec -20 db/dec 30 1 110 850 68 Not to scale wlg Example 8

325. Reverse Bode Plot Not to scale w (rad/sec) 50 dB 0.5 100 dB -40 dB/dec -20 dB/dec 40 10 dB 300 -20 dB/dec -40 dB/dec Required: From the partial Bode diagram, determine the transfer function (Assume a minimum phase system) wlg Example 9

326. Polar Plot

327. Introduction The polar plot of sinusoidal transfer function G(jω) is a plot of the magnitude of G(jω) verses the phase angle of G(jω) on polar coordinates as ω is varied from zero to infinity. as ω is varied from zero to infinity. Therefore it is the locus of As So it is the plot of vector as ω is varied from zero to infinity G ( j  )  G ( j  ) G ( j  )  G ( j  )  Me j  (  ) Me j  (  )

328. Introduction conti… In the polar plot the magnitude of G(jω) is plotted as the distance from the origin while phase angle is measured from positive real axis. + angle is taken for anticlockwise direction. Polar plot is also known as Nyquist Plot.

329. Steps to draw Polar Plot the real and imaginary parts Step 6: Put Re [G(jω) ]=0, determine the frequency at which plot intersects the Im axis and ) Step 1: Determine the T.F G(s) Step 2: Put s=jω in the G(s) Step 3: At ω=0 & ω=∞ find by & Step 4: At ω=0 & ω=∞ find by Step 5: Rationalize the function G(jω) and sep & G ar ( at j e   G ( j  ) lim G ( j  )   lim G ( j  )   lim  G ( j  ) calculate intersection value by putting the above calculated frequ   e ncy in G(jω) lim  G ( j  )  

330. Steps to draw Polar Plot conti… Step 7: Put Im [G(jω) ]=0, determine the frequency at which plot intersects the real axis and calculate intersection value by putting the above calculated frequency in G(jω) Step 8: Sketch the Polar Plot with the help of above information

331. Polar Plot for Type System Let Step 1: Put s=jω (1  sT 1 )(1  sT 2 ) G ( s )  K 2 1 2 2 1 2 1 2 T  1  1   tan  T  tan  K K Step 2: Taking t ( h 1 e  li j  m T it )( f 1 o  r m j  a T g ) nitude of G(jω) G ( j  )   1    T  1   T  

332. Type system conti… 2 2 2 1 2 2 1 2 K  1    T  1  j  T  Ste p l i 3 m : Ta G ki ( n j g  t ) he  l imit of the Ph ase Angle o  f G ( j ω )    1    T  1  j  T  lim G ( j  )  K  K   2 1  1  1 T   180 T  tan   G ( j  )    tan  2 1  1  1 T  T  tan   G ( j  )    tan  lim lim    

333. Type system conti… Step 4: Separate the real and Im part of G(jω) 2 1 2 2 2 4 1 2 2 2 1 2 1 2 2 2 2 4 1 2 2 1   T   T   TT K  ( T  T )  j 1 2 1   T   T   TT K (1   TT ) Step 5: G Pu ( t j R  e ) [G  (jω)]=0  G ( j  )    180   90 T 1  T 2 T 1 T 2 T 1 T 2 1 2 1 2 2 2 2 4 2 2 1 2     1 So When &     G ( j  )  K 1 &    1 1   T   T   T T K (1   2 T T )   T T

334. Type system conti… Step 6: Put Im [G(jω)]=0 2 1 2 2 2 2 2 4    G ( j  )  K      G ( j  )   180 1 So When K  ( T 1  T 2 )     &   1   T   T   TT

335. Type system conti…

336. Polar Plot for Type 1 System Let Step 1: Put s=jω s (1  sT 1 )(1  sT 2 ) G ( s )  K 2 1  1 T  tan  1  T 2 2 2 1 j  (1  j  T )(1  j  T ) 1 2   90  tan  G ( j  )   T  1    T  1  j   K K 

337. Type 1 system conti… Step 2: Taking the limit for magnitude of G(jω) 2 2 2 1 2 2 1 2     T  1  j  T   1   St e l p i m 3: T G ak ( i j n  g ) th  e limit of the P K ha se Angle of G  ( j ω )   1    T  1  j  T  lim G ( j  )  K     1 2  1 T  tan  1  T   270  G ( j  )    90  G ( j  )    90 1 2  1 T  tan  1  T   90  tan   tan  lim lim    

338. Type 1 system conti… Step 4: Separate the real and Im part of G(jω) 1 2 2 2 2 1 2 3 2 1 2 2 1 3 2 2 2 2 1 2    ( T  T 2   T T ) j ( K  2 T T  K )  j 1 2    ( T  T 2   T T )   K ( T  T ) Step 5: Put G R ( e j  [G ) (  jω)]=0 1 2 2 1 3 2 2 2 2 So at     G ( j  )    270    ( T  T 2   T T )   K ( T 1  T 2 )     

339. Type 1 system conti… Step 6: Put Im [G(jω)]=0 G ( j  )    T 1  T 2 T 1 T 2 1 2 1 2 2 1 2     1 So When    3 ( T 2  T 2   2 T 2 T 2 ) j ( K  2 T T  K )  G ( j  )   K T 1 T 2 1 &    1       T T

340. Type 1 system conti…

341. Polar Plot for Type 2 System Let Similar to above s 2 (1  sT )(1  sT ) 1 2 G ( s )  K

342. Type 2 system conti…

343. Note: Introduction of additional pole in denominator contributes a constant - 180 to the angle of G(jω) for all frequencies. See the figure 1, 2 & 3 Figure 1+(- 180 Rotation)=figure 2 Figure 2+(- 180 Rotation)=figure 3

344. Ex: Sketch the polar plot for G(s)=20/s(s+1)(s+2) Solution: Step 1: Put s=jω   90  tan  1   tan  1  / 2   2  1  2  4 20 20 j  ( j   1)( j   2) G ( j  )  

345. Step 2: Taking the limit for magnitude of G(jω)   2  1  2  4   2  1  2  4   lim G ( j  )  20     lim G ( j  )  20      lim  G ( j  )    90  tan  1   tan  1  / 2   270 Step 3: Taking the limit of the Phase Angle of G(jω) lim  G ( j  )    90  tan  1   tan  1  / 2   90

346. Step 4: Separate the real and Im part of G(jω) j 20(  3  2  ) (  4   2 )(4   2 )  60  2 (  4   2 )(4   2 ) G ( j  )   j  60  2 (  4   2 )(4   2 ) So at     G ( j  )    270     

347. Step 6: Put Im [G(jω)]=0 3 G ( j  )     2  G ( j  )   10  2 &         (  4   2 )(4   2 ) So for positivevalueof  j 20(  3  2  )    

348. Gain Margin, Phase Margin & Stability

349. Phase Crossover Frequency (ω p ) : The frequency where a polar plot intersects the –ve real axis is called phase crossover frequency Gain Crossover Frequency (ω g ) : The frequency where a polar plot intersects the unit circle is called gain crossover frequency So at ω g G ( j  )  Unity

350. Phase Margin (PM): Phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability (marginally stabile) Φ m =180 +Φ Where Φ= ∠G(jω g ) if if Φ m >0 => +PM Φ m <0 => - PM (Stable System) (Unstable System)

351. Gain Margin (GM): which the phase angle is - 180 . In terms of dB The gain margin is the reciprocal of magnitude G ( j  ) at the frequency at 1  1 | G ( jwc ) | x GM  1 10 | G ( jwc ) | GM in dB  20 log 10 | G ( jwc ) |   20 log ( x ) 10   20 log

352. Stability Stable: If critical point (- 1+j0) is within the plot as shown, Both GM & PM are +ve GM=20log 10 (1 /x) dB

353. Unstable: If critical point (- 1+j0) is outside the plot as shown, Both GM & PM are - ve GM=20log 10 (1 /x) dB

354. Marginally Stable System: If critical point (- 1+j0) is on the plot as shown, Both GM & PM are ZERO GM=20log 10 (1 /1)=0 dB

355. MATLAB Margin

356. Inverse Polar Plot The inverse polar plot of G(jω) is a graph of 1/G(jω) as a function of ω. Ex: if G(jω) =1/jω then 1/G(jω)=jω     lim G ( j  )  1   lim G ( j  )  1 

357. Knowledge Before Studying Nyquist Criterion 1  G ( s ) H ( s ) G ( s ) T ( s )  D G ( s ) G ( s )  N G ( s ) D H ( s ) H ( s )  N H ( s ) unstable if there is any pole on RHP (right half plane)

358. N G ( s ) D H ( s ) G ( s ) T ( s )   1  G ( s ) H ( s ) D G ( s ) D H ( s )  N G ( s ) N H ( s ) D G ( s ) D H ( s ) G ( s ) H ( s )  N G ( s ) N H ( s ) D G D H  D G D H  N G N H D G D H 1  G ( s ) H ( s )  1  N G N H poles of G(s)H(s) and 1+G(s)H(s) are the same Closed- loop system: zero of 1+G(s)H(s) is pole of T(s) Characteristic equation: Open- loop system:

359. ( s  5)( s  6)( s  7)( s  8) G ( s ) H ( s )  ( s  1)( s  2)( s  3)( s  4) 1  G ( s ) H ( s ) G ( s ) G ( s ) H ( s ) 1  G ( s ) H ( s ) Zero – 1,2,3,4 Poles – 5,6,7,8 Zero – a,b,c,d Poles – 5,6,7,8 Zero – ?,?,?,? Poles – a,b,c,d To know stability, we have to know a,b,c,d

360. Stability from Nyquist plot From a Nyquist plot, we can tell a number of closed-loop poles on the right half plane. If there is any closed-loop pole on the right half plane, the system goes unstable. If there is no closed-loop pole on the right half plane, the system is stable.

361. Nyquist Criterion Nyquist plot is a plot used to verify stability of the system. function ( s  p 1 )( s  p 2 ) ( s  z 1 )( s  z 2 ) F ( s )  mapping all points (contour) from one plane to another by function F(s). mapping contour

362. ( s  p 1 )( s  p 2 ) F ( s )  ( s  z 1 )( s  z 2 )

363. Pole/zero inside the contour has 360 deg. angular change. Pole/zero outside contour has deg. angular change. Move clockwise around contour, zero inside yields rotation in clockwise, pole inside yields rotation in counterclockwise

364. Characteristic equation N = P- Z N = # of counterclockwise direction about the origin P = # of poles of characteristic equation inside contour = # of poles of open- loop system z = # of zeros of characteristic equation inside contour = # of poles of closed- loop system Z = P- N F ( s )  1  G ( s ) H ( s )

365. Characteristic equation Increase size of the contour to cover the right half plane More convenient to consider the open- loop system (with known pole/zero)

366. Nyquist diagram of ‘Open- loop system’ Mapping from characteristic equ. to open- loop system by shifting to the left one step Z = P- N Z = # of closed- loop poles inside the right half plane P = # of open- loop poles inside the right half plane N = # of counterclockwise revolutions around - 1 G ( s ) H ( s )

367. Properties of Nyquist plot If there is a gain, K, in front of open- loop transfer function, the Nyquist plot will expand by a factor of K.

368. Nyquist plot example 1 G ( s )  Closed- loop system ha s s  p 2 ole at 1 Open loop system has pole at 2 If we multiply the open- loop with a gain, K, then w G e ( s c ) an  mo 1 ve the closed- loop pole’s 1 p  os G it ( i S o ) n to ( s t  he 1) left- half plane

369. Nyquist plot example (cont.) New look of open- loop system: Corresponding closed- loop system: s  2 G ( s )  K  Evaluate value of K fo 1 r  s G t ( a s b ) ilit s y  ( K  2) G ( s ) K K  2

370. Adjusting an open- loop gain to guarantee stability Step I : sketch a Nyquist Diagram Step II : find a range of K that makes the system stable!

371. How to make a Nyquist plot? Easy way by Matlab Nyquist: ‘ nyquist ’ Bode: ‘ bode ’

372. Step I: make a Nyquist plot Find Starts from an open- loop transfer function (set K=1) Set and find frequency response At dc, s  j  s at  w  hich  the  imaginary part equals zero 

373. (8   2 ) 2  6 2  2 Need the imaginary term = 0,  (15   2 )  8 j   (8   2 )  6 j  (8   2 )  6 j  (8   2 )  6 j   (15   2 )(8   2 )  48  2  j (154   14  3 ) (15   2 )  8 j  (8   2 )  6 j    2  8 j   15   2  6 j   8 s 2 G ( j  ) H ( j  )   6 s  8 s 2  8 s  15 ( s  2)( s  4) ( s  3)( s  5) G ( s ) H ( s )      0, 11 (15  11)(8  11)  48(11)   540   1.31 (8  11) 2  6 2 (11) 412 Substitute And get  ba  ck i n 1 t 1 o the transfer function G ( s )   1.33

374. At dc, s=0, At imaginary part=0

375. Step II: satisfying stability condition P = 2, N has to be 2 to guarantee stability Marginally stable if the plot intersects - 1 For stability, 1.33K has to be greater than 1 K > 1/1.33 or K > 0.75

376. Example Evaluate a range of K that makes the system stable ( s 2  2 s  2)( s  2) G ( s )  K

377. 4(1   2 )  j  (6   2 ) 16(1   2 ) 2   2 (6   2 ) 2 K (( j  ) 2  2 j   2)( j   2) G ( j  )   At   0, 6 the imaginary part = Plug and get G = - 0.05 Step I: find frequency at which imaginary part = s  j  Set   bac 6 k in the transfer function

378. Step II: consider stability condition P = 0, N has to be to guarantee stability Marginally stable if the plot intersects - 1 For stability, 0.05K has to be less than 1 K < 1/0.05 or K < 20

379. Gain Margin and Phase Margin Gain margin is the change in open- loop gain (in dB), required at 180 of phase shift to make the closed- loop system unstable. Phase margin is the change in open- loop phase shift, required at unity gain to make the closed- loop system unstable. GM/PM tells how much system can tolerate before going unstable!!!

380. GM and PM via Nyquist plot

381. GM and PM via Bode Plot  G M M The frequency at which the magnitude equals 1 is called the gain crossover frequency   The frequency at which the phase equals 180 degrees is called the phase crossover frequency M  G gain crossover frequency phase crossover frequency

382. Example Find Bode Plot and evaluate a value of K that makes the system stable The system has a unity feedback with an open- loop transfer function ( s  2)( s  4)( s  5) G ( s )  K First, let’s find Bode Plot of G(s) by assuming that K=40 (the value at which magnitude plot starts from dB)

383. At phase = -180, ω = 7 rad/sec, magnitude = -20 dB

384. GM>0, system is stable!!! Can increase gain up 20 dB without causing instability (20dB = 10) Start from K = 40 with K < 400, system is stable

385. Closed- loop transient and closed- loop frequency responses ‘ 2 nd system’ 2 R ( s ) C ( s ) n n s 2  2  2  s    T ( s )  n

386. Magnitude Plot of closed- loop system Damping ratio and closed- loop frequency response 1 2  1   2 p M   p   n 1  2  2

387.  (1  2  2 )  4  4  4  2  2  (1  2  2 )  4  4  4  2  2 BW s BW   n (1  2  2 )  4  4  4  2  2 4  BW    T p 1   2 T   BW = frequency at which magnitude is 3dB down from value at dc (0 rad/sec), or . Response speed and closed- loop frequency response 2 1 M 

388. Find from Open- loop Frequency Response  BW Nichols Charts From open- loop frequency response, we can find  BW at the open- loop frequency that the magnitude lies between -6dB to -7.5dB (phase between - 135 to - 225)

389. Relationship between damping ratio and phase margin of open- loop frequency response   tan  1 2   2  2  1  4  4 M Phase margin of open- loop frequency response Can be written in terms of damping ratio as following

390. Open- loop system with a unity feedback has a bode plot below, approximate settling time and peak time  BW = 3.7 PM=35

391.   0.32  2  2  1  4  4   tan  1 2  M Solve for PM = 35  1.43 4  4  4  2  2 (1  2  2 )   1   2  5.5 4  4  4  2  2 (1  2  2 )  4     BW p BW s T  T

392. UNIT- 5 STATE SPACE REPRESENTATION

393. Objectives How to find mathematical model, called a state-space representation, for a linear, time- invariant system How to convert between transfer function and state space models How to find the solution of state equations for homogeneous &non homogeneous systems

394. Plant Mathematical Model : Differential equation Linear, time invariant Frequency Domain Technique Time Domain Technique

395. Two approaches for analysis and design of control system Classical Technique or Frequency Domain Technique Modern Technique or Time Domain Technique

396. Some definitions system System variable : any variable that responds to an input or initial conditions in a State variables : the smallest set of linearly independent system variables such that the values of the members of the set at time t0 along with known forcing functions completely determine the value of all system variables for all t ≥ t0 State vector : a vector whose elements are the state variables State space : the n- dimensional space whose axes are the state variables State equations : a set of first- order differential equations with b variables, where the n variables to be solved are the state variables Output equation : the algebraic equation that expresses the output variables of a system as linear combination of the state variables and the inputs. For n th- order, write n simultaneous, first- order differential equations in terms of the state variables ( state equations ). If we know the initial condition of all of the state variables at as well as the system input for , we can solve the equations

397. Graphic representation of state space and a state vector

398. For a dynamic system, the state of a system is described in terms of a set of state variables [x 1 (t) x 2 (t) … x n (t)] The state variables are those variables that determine the future behavior of a system when the present state of the system and the excitation signals are known. Consider the system shown in Figure 1, where y 1 (t) and y 2 (t) are the output signals and u 1 (t) and u 2 (t) are the input signals. A set of state variables [x 1 x 2 ... x n ] for the system shown in the figure is a set such that knowledge of the initial values of the state variables [x 1 (t ) x 2 (t ) ... x n (t )] at the initial time t , and of the input signals u 1 (t) and u 2 (t) for t˃=t , suffices to determine the future values of the outputs and state variables. System Input Signals u 1 (t) u 2 (t) Output Signals y 1 (t) y 2 (t) System u(t) Input x(0) Initial conditions y(t) Output Figure 1. Dynamic system.

399. In an actual system, there are several choices of a set of state variables that specify the energy stored in a system and therefore adequately describe the dynamics of the system. The state variables of a system characterize the dynamic behavior of a system. The engineer’s interest is primarily in physical, where the variables are voltages, currents, velocities, positions, pressures, temperatures, and similar physical variables. The State Differential Equation: The state of a system is described by the set of first- order differential equations written in terms of the state variables [x 1 x 2 ... x n ]. These first- order differential equations can be written in general form as x ˙ 1  a 11 x 1  a 12 x 2  … a 1n x n  b 11 u 1   b 1m u m x ˙ 2  a 21 x 1  a 22 x 2  … a 2n x n  b 21 u 1   b 2m u m ⁝  a n1 x 1  a n 2 x 2  … a nn x n  b n1 u 1   b nm u m x ˙ n

400. Thus, this set of simultaneous differential equations can be written in matrix form as follows:     1 11 2n   2    21 22  n   n1 n 2 nn   n   2      u  b nm     u m     b n1   x  a 1n   x 1   a 11 a 12  x 1  d  x   a   ⁝   b  b 1m ⁝  ⁝  ⁝   ⁝    x   ⁝  x   a  a  a   a  a dt  ⁝   n  2   x  n: number of state variables, m: number of inputs. The column matrix consisting of the state variables is called the state vector and is written as  x 1  x    ⁝   x 

401. The vector of input signals is defined as u. Then the system can be represented by the compact notation of the state differential equation as x ˙  A x  Bu This differential equation is also commonly called the state equation. The matrix A is an nxn square matrix, and B is an nxm matrix. The state differential equation relates the rate of change of the state of the system to the state of the system and the input signals. In general, the outputs of a linear system can be related to the state variables and the input signals by the output equation y  C x  D u Where y is the set of output signals expressed in column vector form. The state- space representation (or state- variable representation) is comprised of the state variable differential equation and the output equation.

402. General State Representation D = derivative of the state vector with respect to time = output vector = input or control vector x ˙  Ax  Bu y  Cx  Du x = state vector x ˙ y u A = system matrix B C = input matrix = output matrix = feedforward matrix State equation output equation

403. AN EXAMPLE OF THE STATE VARİABLE CHARACTERİZATİON OF A SYSTEM u(t) Current source L C R V c V o i L i c The state of the system can be described in terms of a set of variables [x 1 x 2 ], where x 1 is the capacitor voltage v c (t) and x 2 is equal to the inductor current i L (t). This choice of state variables is intuitively satisfactory because the stored energy of the network can be described in terms of these variables.

404. Therefore x 1 (t ) and x 2 (t ) represent the total initial energy of the network and thus the state of the system at t=t . Utilizing Kirchhoff’s current low at the junction, we obtain a first order differential equation by describing the rate of change of capacitor voltage L  C dv c c  u(t)  i dt i Kirchhoff’s voltage low for the right- hand loop provides the equation describing the rate of change of inducator current as   R i L  v c L di L dt The output of the system is represented by the linear algebraic equation v  R i L (t)

405. We can write the equations as a set of two first order differential equations in terms of the state variables x 1 [v C (t)] and x 2 [i L (t)] as follows: 2 1 dt dx 2 dx 1 1 1   x 2  u(t) C C  1 x  R x dt L L L C dv c  u(t)  i dt di L L   R i L  v c dt The output signal is then y 1 (t)  v (t)  R x 2 Utilizing the first- order differential equations and the initial conditions of the network represented by [x 1 (t ) x 2 (t )], we can determine the system’s future and its output. The state variables that describe a system are not a unique set, and several alternative sets of state variables can be chosen. For the RLC circuit, we might choose the set of state variables as the two voltages, v C (t) and v L (t).

406.  C     1  C  x    u(t)  1 R      L L   1   x ˙   We can write the state variable differential equation for the RLC circuit as and the output as y   R  x

407. RLC network i  t  q  t  1. State variables

408. 2. L di  Ri  1  idt  v  t  dt C Using i  t   dq dt L R q  v  t  dt 2 dt C d 2 q  dq  1 dq  i dt di   1 q  R i  1 v  t  dt LC L L q  t  i  t  Can be solved using Laplace Transform Other network variables can be obtained (1) C L v  t    1 q  t   Ri  t   v  t  (2) 5. (1),(2) : state- space representation

409. Other variables v R  t  v C  t  L L dv R C R   R v  R v  R v  t  L R dt RC dt dv C 1 v  Each variables : linearly independent

410. In vector- matrix form x ˙  Ax  Bu where   dq dt   x ˙   di dt    R / L    A    1/ LC  1    i  x   q     1 L  B    u  v  t  (1) (2) where y  v L  t  y  Cx  Du C    1 C  R  D  1

411. State space representation using phase variable form dy dt n  1 d n  1 y    a 1 dt  a y  b u dt n d n y  a n  1 x  y 1 dy x 2  dt choose d 2 y x 3  dt 2 x n  dt n  1 d n  1 y diferensiasikan dt 1 x ˙  dy d 2 y x ˙ 2  dt 2 d 3 y x ˙ 3  dt 3 d n y ⁝ x ˙ n  dt n 2 3 ⁝ x ˙ n  1  x n x ˙  x 2 1 x ˙  x x  b u 1 1 2 n  1 n   a x  a x   a n x ˙

412.  x ˙ 1                  b    x 3                       a 2  x 1  a n  1     x n  x n  1        u  x        ⁝  ⁝    1  a 1  a 2  a 3  a 4  a 5  2   1    x ˙ n  x ˙ n  1   ⁝  x ˙ 3  x ˙ 1  1  ⁝         x n  x n  1  2  x 1   x      ⁝  y   1    x 3

413. Example : TF to State Space Inverse Laplace ˙ c ˙˙  9 c ˙˙  2 6 c ˙  24 c  24 r 1. x 3  c ˙ ˙ 2. Select state variables x 1  c x 2  c ˙ x ˙ 1  x 2 x ˙ 2  x 3   24 x 1  26 x 2 3  9 x  24 r 3 x ˙ y  c  x 1 N  s  G  s   D  s  N  s  numerator D  s  denominator

414.    1  x     r  9     x 3     24   2   x 1       2    x ˙ 3      24 1  x ˙     26  x ˙ 1   3  1     x  y   1  x 2   x 

415. Decomposing a transfer function

416. 2 2 Y  s   C  s    b s  b s  b  X  s  1 1 dt 2 d 2 x b 2 1 y  t   dt dx b 1 1  b x 1  y  t   b x 1  b 1 x 2  b 2 x 3

417. Example

418. y  t   2 x 1  7 x 2  x 3 y   2  2    x 3   7 1   x   x 1 

419. State Space to TF x ˙  Ax  Bu y  Cx  Du Laplace Transform sX  s   AX  s   BU  s  Y  s   CX  s   DU  s  X  s    sI  A   1 BU  s  Y  s    C  sI  A   1 B  D  U  s  U  s  T  s   Y  s   C  sI  A   1 B  D

420.     3       x ˙    x  1   10   1  x    u    1  2 y   1     1  s  1  sI  A    s  1   2 s  3    sI  A   1  adj ( sI  A ) det( sI  A ) s 3  3 s 2  2 s  1  ( s 2  3 s  2) s  3    1 s ( s  3)    s  (2 s  1) 1   s 2   s  

421. T  s   C  sI  A   1 B  D T  s   s 3  3 s 2  2 s  1 10( s 2  3 s  2)

422. Solution of homogeneous state equation differential equation is The solution of the state differential equation can be obtained in a manner similar to the approach we utilize for solving a first order differential equation. Consider the first-order differential equation x ˙  ax ; x (0)  dx  axdt Where x(t) and u(t) are scalar functions of time. We expect an exponential solution of the form e at. Taking the Laplace transform of both sides, we have x ˙ on integrating above equation logx=at+c X= e at. e c x=x(0)= e c on substituting the intial condition ,the solution of homogeneous state equation of first order x ˙  e at x (0) x ˙  AX ( t ), x ( o )    k! A k t k 2!  I  At     A 2 t 2 e At

423. Solution of homogeneous state equation 2! k! e  I  At       The solution of the state differential equation can be obtained in a manner similar to the approach we utilize for solving a first order differential equation. Consider the first-order differential equation x ˙  ax  bu ; x (0)  Where x(t) and u(t) are scalar functions of time. By taking laplace transform s X ( s )  x  a X ( s )  bU ( s ) The inverse Laplace transform of X(s) results in the solution t x(t)  e at x(0)   e a ( t  ) b u(  ) d  We expect the solution of the state differential equation to be similar to x(t) and to be of differential form. The matrix exponential function is defined as A 2 t 2 A k t k At

424. which converges for all finite t and any A. Then the solution of the state differential equation is found to be t x(t)  e At x(0)   e A( t  ) Bu(  ) d  X(s)   sI  A   1 x(0)   sI  A   1 BU(s) where we note that [sI-A] - 1 =ϕ(s), which is the Laplace transform of ϕ(t)=e At . The matrix exponential function ϕ(t) describes the unforced response of the system and is called the fundamental or state transition matrix . t x(t)   (t) x(0)    (t   ) Bu(  ) d 

425. THE TRANSFER FUNCTION FROM THE STATE EQUATION The transfer function of a single input- single output (SISO) system can be obtained from the state variable equations. x ˙  A x  Bu y  C x where y is the single output and u is the single input. The Laplace transform of the equations sX(s)  AX(s)  BU(s) Y(s)  CX(s) where B is an nx1 matrix, since u is a single input. We do not include initial conditions, since we seek the transfer function. Reordering the equation

426. [sI  A]X(s)  BU(s) X(s)   sI  A   1 BU(s)   (s)BU(s) Y(s)  C  (s)BU(s) Therefore, the transfer function G(s)=Y(s)/U(s) is G(s)  C  (s)B Example: Determine the transfer function G(s)=Y(s)/U(s) for the RLC circuit as described by the state differential function , y   R  x  u  C     1  C  x    1 R      L L   1   x ˙  

427.    R    s    L L   s 1 C  sI  A    1 1 R L LC s   (s)  s 2  s    R 1   1  L  (s)   sI  A  (s)     1  1  s  L  C  Then the transfer function is G(s)   L LC 1  R s  s 2 R / LC  (s) G(s)  R / LC  s      1  1 1       (s) C  (s)   C     L    L  (s) R    (s)  s  R 

428. CONSIDER THE SYSTEM 24 s 3 y  c  x 1  s 3  9 s 2  26 s  24  C ( s )  24 R ( s ) R ( s )  9 s 2  26 s  24 C ( s )  x ˙ 1  x 2 x ˙ 2  x 3 x ˙ 3  - 24 x 1 - 26 x 2  9 x 3  24 r ˙ c ˙˙  9 c ˙˙  2 6 c ˙  24 c  24 r x 1  c x 2  c ˙ x 3  c ˙ ˙ State variables Output equation System equations

429. y   1   24    9     x 3    1     24  26 x ˙ 2  x 3 x ˙ 2  - 24 x 1 - 26 x 2  9 x 3  24 r y  c  x 1   x 3   2  2    x ˙ 2   2  x ˙ 1  x 2    x   x 1     1  x     r  x 1      x ˙     x ˙ 1 

430. THANK YOU