Unit I Control System Modeling

Transcription

1 Unit I Control System Modeling

2 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-andeffect relationship of the process.

3 The interaction is defined in terms of variables. i. System input ii. System output iii. Environmental disturbances

4 Control System Control is the process of causing a system variable to conform to some desired value. Manual control Automatic control (involving machines only). A control system is an interconnection of components forming a system configuration that will provide a desired system response. Input Signal Control System Output Signal Energy Source

5 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. Multivariable Control System

6 Control System Classification Missile Launcher System Open-Loop Control System

7 Control System Classification Missile Launcher System Closed-Loop Feedback Control System

8 Manual Vs Automatic Control Control is a process of causing a system variable such as temperature or position to conform to some desired value or trajectory, called reference value or trajectory. For example, driving a car implies controlling the vehicle to follow the desired path to arrive safely at a planned destination. i. If you are driving the car yourself, you are performing manual control of the car. ii. If you use design a machine, or use a computer to do it, then you have built an automatic control system.

9 Control System Classification Desired Output Response Controller Process Output Variables Measurement Multi Input Multi Output (MIMO) System

10 Purpose of Control Systems i.power Amplification (Gain) Positioning of a large radar antenna by low-power rotation of a knob ii.remote Control Robotic arm used to pick up radioactive materials iii.convenience of Input Form Changing room temperature by thermostat position iv.compensation for Disturbances Controlling antenna position in the presence of large wind disturbance torque

11 Historical Developments i.ancient Greece ( to 300 BC) Water float regulation, water clock, automatic oil lamp ii.cornellis Drebbel (7 th century) Temperature control iii.james Watt (8 th century) Flyball governor iv.late 9 th to mid 0 th century Modern control theory

12 Human System The Vetruvian Man

13 Human System i.pancreas Regulates blood glucose level ii.adrenaline Automatically generated to increase the heart rate and oxygen in times of flight iii.eye Follow moving object iv.hand Pick up an object and place it at a predetermined location v.temperature Regulated temperature of 36 C to 37 C

14 History 8th Century James Watt s centrifugal governor for the speed control of a steam engine. 90s Minorsky worked on automatic controllers for steering ships. 930s Nyquist developed a method for analyzing the stability of controlled systems 940s Frequency response methods made it possible to design linear closed-loop control systems 950s Root-locus method due to Evans was fully developed 960s State space methods, optimal control, adaptive control and 980s 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???????????????????????????????????

15 Control System Components i.system, plant or process To be controlled ii.actuators Converts the control signal to a power signal iii.sensors Provides measurement of the system output iv.reference input Represents the desired output

16 General Control System Disturbance Set-point or Reference input Error Controlled Signal Manipulated Variable + + Controller Actuator + Process Actual Output Feedback Signal Sensor

17 Control System Design Process If the performance does not meet specifications, then iterate the configuration and actuator

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

19 Mathematical Modelling of Mechanical Systems 9

20 Basic Types of Mechanical Systems Translational Linear Motion Rotational Rotational Motion 0

21 Translational Mechanical Systems

22 Basic Elements of Translational Mechanical Systems i) Translational Spring ii) Translational Mass iii) Translational Damper

23 Translational Spring A translational spring is a mechanical element that can be deformed by an external force such that the deformation is directly proportional to the force applied to it. i) Translational Spring Circuit Symbols Translational Spring

24 If F is the applied force Translational Spring x x Then x is the deformation if x 0 F Or ( x x ) is the deformation. F The equation of motion is given as Where k F k( x ) x is stiffness of spring expressed in N/m

25 Translational Spring Given two springs with spring constant k and k, obtain the equivalent spring constant k eq for the two springs connected in: () Parallel () Series 5

26 Translational Spring The two springs have same displacement therefore: k x k ( k k ) x F x F () Parallel k eq x F k eq k k If n springs are connected in parallel then: k eq k k k n 6

27 7 Translational Spring () Series F x k x k The forces on two springs are same, F, however displacements are different therefore: k F x k F x Since the total displacement is, and we have x x x x k F eq k F k F k F x x x eq

28 8 Translational Spring Then we can obtain k k k k k k k eq k F k F k F eq If n springs are connected in series then: n n eq k k k k k k k

29 Translational Spring Exercise: Obtain the equivalent stiffness for the following spring networks. i) k3 ii) k3 9

30 Translational Mass Translational Mass is an inertia element. ii) Translational Mass A mechanical system without mass does not exist. If a force F is applied to a mass and it is displaced to x meters then the relation b/w force and displacements is given by Newton s law. F (t ) M x(t ) F M x

31 Translational Damper When the viscosity or drag is not negligible in a system, we often model them with the damping force. All the materials exhibit the property of damping to some extent. iii) Translational Damper If damping in the system is not enough then extra elements (e.g. Dashpot) are added to increase damping.

32 Common Uses of Dashpots Door Stoppers Vehicle Suspension Bridge Suspension Flyover Suspension

33 Translational Damper F Cx F C( x x ) Where C is damping coefficient (N/ms - ).

34 Translational Damper Translational Dampers in series and parallel. C eq C C C eq C C C C

35 Modelling a simple Translational System Example-: Consider a simple horizontal spring-mass system on a frictionless surface, as shown in figure below. or m x kx mx kx 0 35

36 Example- Consider the following system (friction is negligible) k F M x Free Body Diagram f k F M Where f k and f M are force applied by the spring and inertial force respectively. f M 36

37 Example- f k F M f M F f k f M Then the differential equation of the system is: F Mx Taking the Laplace Transform of both sides and ignoring initial conditions we get kx F ( s ) Ms X ( s ) kx ( s ) 37

38 Example- F ( s ) Ms X ( s ) kx ( s The transfer function of the system is ) X ( s ) F ( s ) Ms k if M 000kg k 000 Nm X ( s ) F ( s ) s 38

39 Example- X ( s ) F ( s ) s The pole-zero map of the system is 40 Pole-Zero Map 30 0 Imaginary Axis Real Axis 39

40 Example-3 Consider the following system k F M x Free Body Diagram f k F M f f C C M F f k f M f C 40

41 Example-3 Differential equation of the system is: F M x C x kx Taking the Laplace Transform of both sides and ignoring Initial conditions we get F ( s ) Ms X ( s ) CsX ( s ) kx ( s ) X ( s ) F ( s ) Ms Cs k 4

42 Example-3 X ( s ) F ( s ) Ms Cs k if Pole-Zero Map M 000kg k 000 Nm C 000 N / ms Imaginary Axis X ( s ) F ( s ) s s Real Axis 4

43 Example-4 Consider the following system Free Body Diagram (same as example-3) f k F M f f B M X ( s ) F ( s ) Ms Bs k F f k f M f B 43

44 Example-5 Consider the following system x F x k M B Mechanical Network k x x F M B 44

45 Example-5 Mechanical Network k x x F M B At node x F k ( x x ) At node x 0 k ( x x ) M x B x 45

46 Example-6 Find the transfer function X (s)/f(s) of the following system. k M M B

47 Example-7 x x k B 3 B 4 f ( t ) M M B B x B 3 x f ( t ) k M B B M B 4 47

48 48 Example-8 Find the transfer function of the mechanical translational system given in Figure-. Free Body Diagram Figure- M ) ( t f k f M f B f B M k f f f t f ) ( k Bs Ms s F s X ) ( ) (

49 Restaurant plate dispenser Example-9 49

50 50 Example-0 Find the transfer function X (s)/f(s) of the following system. Free Body Diagram M k f M f B f M ) ( t F k f M f B f k f k B M k k f f f f t F ) ( B M k f f f 0

51 Example- x k x B 3 x 3 B 4 u(t) B M M k 3 k B B 5 5

52 Example-: Automobile Suspension 5

53 Automobile Suspension 53

54 54 Automobile Suspension ). ( ) ( ) ( 0 eq i o i o o x x k x x b x m eq. i i o o o kx bx kx bx x m Taking Laplace Transform of the equation () ) ( ) ( ) ( ) ( ) ( s kx s bsx s kx s bsx s X ms i i o o o k bs ms k bs s X s X i o ) ( ) (

55 Example-3: Train Suspension Bogie- Car Body Secondary Bogie- Wheelsets Suspension Primary Suspension Bogie Frame 55

56 Example: Train Suspension 56

57 Rotational Mechanical Systems 57

58 Basic Elements of Rotational Mechanical Systems Rotational Spring T k( )

59 Basic Elements of Rotational Mechanical Systems Rotational Damper C T T C ( )

60 Basic Elements of Rotational Mechanical Systems Moment of Inertia J T T J

61 Example- k T J B J 3 k k B 3 T J J k

62 Example- k T J B J 3 B 4 B B 3 k B 3 T J B B J 3 B 4

63 Example-3 J k T J B k

64 Example-4

65 Block Diagram Reduction Technique

66 Block diagram Transfer Function Consists of Blocks Can be reduced G 4 R(s) G G G3 Y (s) H H R(s) G Y(s)

67 Reduction techniques. Combining blocks in cascade or in parallel G G G G G G G G. Moving a summing point behind a block G G G

68 3. Moving a summing point ahead of a block G G G 4. Moving a pickoff point behind a block G G G 5. Moving a pickoff point ahead of a block G G G

69 6. Eliminating a feedback loop G H G GH G G G H 7. Swap with two neighboring summing points A B B A

70 Example Find the transfer function of the following block diagrams (a) G 4 R(s) G G G3 Y(s) H H

71 I R(s) G B G G 4 A G 3 Y(s) H H G Solution:. Moving pickoff point A ahead of block G. Eliminate loop I & simplify B G G 4 G3

72 R(s) G B G G 4 4 A G 3 G G G 3 Y(s) H HG 3. Moving pickoff point B behind block R(s) G B G G 4 G3 G G 4 G3 II C Y(s) H G H /( G4 GG 3)

73 4. Eliminate loop III R(s) ) ( ) ( G G G H GG H G G G G ) Y(s ) ( ) ( ) ( ) ( ) ( ) ( G G G G G G G H H G G G G G G s R s Y s T R(s) G C 3 4 G G G G H Y(s) 3 4 G G G H C ) ( G G G H G G G Using rule 6

74 (b) R(s) G G Y(s) H H H 3

75 Solution:. Eliminate loop I R(s) G H A G G G H H I B Y(s). Moving pickoff point A behind block R(s) A G H H 3 G G H 3 G G H H G G H B H Y(s) II 3 GH H( G Not a feedback loop )

76 3. Eliminate loop II R(s) GG G H Y(s) H 3 H ( G G H ) Using rule 6 T( s) Y( s) R( s) G H GG GG H G H 3 GG H H

77 (c) R(s) G H 4 G G 3 G4 Y(s) H 3 H H

78 Solution:. Moving pickoff point A behind block G 4 I R(s) G H 4 A G G 3 G4 B Y(s) H 3 H H G H 3 4 H G 4 G G 4 4

79 . Eliminate loop I and Simplify R(s) G GG3G4 G G H 3 H G 4 H G II B Y(s) III H II feedback III Not feedback GG3G4 G G H G G H H G G 4 4 H

80 R(s) Y(s) 4 4 G H H G H G G H G G G G G G 3. Eliminate loop II & IIII ) ( ) ( ) ( GG G G H GG G H G G H G G H GG G G R s s Y s T Using rule 6

81 (d) H R(s) G G A G 3 B Y(s) H G 4

82 Solution:. Moving pickoff point A behind block G 3 I H R(s) G G A B G 3 Y(s) H G 3 H G 3 G 4

83 . Eliminate loop I & Simplify H G G 3 B GG 3 B H G 3 II H G H 3 R(s) G GG3 G H G G H 3 Y(s) H G 3 G 4

84 R(s) Y(s) 3 3 H G G H G G H G G G G 3. Eliminate loop II ) ( ) ( ) ( GG H G G H G H GG G G R s s Y s T G 4

85 Example Determine the effect of R and N on Y in the following diagram N G 4 R G G Y H G 3

86 In this linear system, the output Y contains two parts, one part is related to R and the other is caused by N: Y Y Y TR T N If we set N=0, then we can get Y: Y YN 0 TR The same, we set R=0 and Y is also obtained: Y YR 0 T N Thus, the output Y is given as follows: Y Y Y Y N Y 0 R0

87 Solution:. Swap the summing points A and B N G 4 R G II G G H B A Y G 3. Eliminate loop II & simplify G 4 N R G G 3 GG G H Y

88 Rewrite the diagram: N R o G G G 4 3 GG G H o Y 3. Let N=0 R G G 3 GG G H Y We can easily get Y Y GG GG 3 GG G3 H G H GG GG GG G H 3 3 R

89 4. Let R=0, we can get: N Y G G 3 GG G H G 4 M 5. Break down the summing point M: N Y G G G 3 4 GG G4 G H G G 3 GG G H

90 ] ) ( ) [( N GG G G H GG G GG G G H R GG G H GG GG GG G H GG GG G H Y Y Y 7. According to the principle of superposition, and can be combined together, So: Y Y 6. Eliminate above loops: Y N H G G G G G G G 3 H G G G G G N GG G H GG GG G H GG G G H GG G GG G G H Y

91 Signal Flow Graph (SFG)

92 Overview What is Signal Flow Graph (SFG)? Definitions of terms used in SFG Rules for drawing of SFG Mason s Gain formula SFG from simultaneous eqns SFG from differential eqns Examples Solution of a problem by Block diagram reduction technique and SFG SFG from a given Transfer function Examples

93 What is Signal Flow Graph? SFG is a diagram which represents a set of simultaneous equations. This method was developed by S.J.Mason. This method does n t require any reduction technique. It consists of nodes and these nodes are connected by a directed line called branches. Every branch has an arrow which represents the flow of signal. For complicated systems, when Block Diagram (BD) reduction method becomes tedious and time consuming then SFG is a good choice.

94 Comparison of BD and SFG block diagram: signal flow graph: R(s) G(s) C (s) G(s) R (s) C(s) In this case at each step block diagram is to be redrawn. That s why it is tedious method. So wastage of time and space. Only one time SFG is to be drawn and then Mason s gain formula is to be evaluated. So time and space is saved.

95 SFG

96 Definition of terms required in SFG Node: It is a point representing a variable. x = t x +t 3 x 3 X X X 3 t t 3 In this SFG there are 3 nodes. Branch : A line joining two nodes. X X Input Node : Node which has only outgoing branches. X is input node.

97 Output node/ sink node: Only incoming branches. Mixed nodes: Has both incoming and outgoing branches. Transmittance : It is the gain between two nodes. It is generally written on the branch near the arrow. t t 3 t 34 X X X 3 X 4 t 43

98 Path : It is the traversal of connected branches in the direction of branch arrows, such that no node is traversed more than once. Forward path : A path which originates from the input node and terminates at the output node and along which no node is traversed more than once. Forward Path gain : It is the product of branch transmittances of a forward path. P = G G G 3 G 4, P = G 5 G 6 G 7 G 8

99 Loop : Path that originates and terminates at the same node and along which no other node is traversed more than once. Self loop: Path that originates and terminates at the same node. Loop gain: it is the product of branch transmittances of a loop. Non-touching loops: Loops that don t have any common node or branch. L = G H L = H 3 L 3 = G 7 H 7 Non-touching loops are L & L, L & L3, L &L3

100 SFG terms representation branch node x transmittance mixed node a x forward path d b loop c input node (source) x 4 x 3 mixed node path x 3 input node (source)

101 Rules for drawing of SFG from Block diagram All variables, summing points and take off points are represented by nodes. If a summing point is placed before a take off point in the direction of signal flow, in such a case the summing point and take off point shall be represented by a single node. If a summing point is placed after a take off point in the direction of signal flow, in such a case the summing point and take off point shall be represented by separate nodes connected by a branch having transmittance unity.

102 Mason s Gain Formula A technique to reduce a signal-flow graph to a single transfer function requires the application of one formula. The transfer function, C(s)/R(s), of a system represented by a signal-flow graph is k = number of forward path P k = the kth forward path gain = (Σ loop gains) + (Σ non-touching loop gains taken two at a time) (Σ non-touching loop gains taken three at a time)+ so on. k = (loop-gain which does not touch the forward path)

103 Ex: SFG from BD

104 EX: To find T/F of the given block diagram

105 Identification of Forward Paths P =..G.G. G 3. = G G G 3 P =..G. G 3. = G G 3

106 Individual Loops L = G G H L = - G G 3 H L 3 = - G 4 H

107 L 4 = - G G 4 L 5 = - G G G 3

108 Construction of SFG from simultaneous equations

109 t t 3 t 3 t 3 t 33

110

111 After joining all SFG

112 SFG from Differential equations Consider the differential equation y 3 y 5 y y x Step : Solve the above eqn for highest order y x 3 y 5 y y Step : Consider the left hand terms (highest derivative) as dependant variable and all other terms on right hand side as independent variables. Construct the branches of signal flow graph as shown below:- x y y - -5 y y -3 (a)

113 Step 3: Connect the nodes of highest order derivatives to the lowest order der.node and so on. The flow of signal will be from higher node to lower node and transmittance will be /s as shown in fig (b) (b) y /s /s x y y /s y Step 4: Reverse the sign of a branch connecting y to y, with condition no change in T/F fn.

114 Step5: Redraw the SFG as shown.

115 Problem: to find out loops from the given SFG

116 Ex: Signal-Flow Graph Models

117 P = P =

118 Individual loops L = G H L = G 3 H 3 Pair of Non-touching loops L L 3 L L 4 L L 3 L L 4 L 3 = G 6 H 6 L 4 = G 7 H 7

119 Y R P ( ( L L ) LiLj LiLjLk.. k k Ys ( ) Rs ( ) G G G 3 G 4 L 3 L 4 G 7 G 8 L L G 5 G 6 L L L 3 L 4 L L 3 L L 4 L L 3 L L 4

120 Ex:

121 Forward Paths

122 Loops L = -G 5 G 6 H L = -G G 3 G 4 G 5 H L 3 = -G 8 H L 4 = - G G 7 H L 5 = -G 4 H 4

123 Loops L 6 = - G G G 3 G 4 G 8 H 3 L 7 = - G G G 7 G 6 H 3 L 8 = - G G G 3 G 4 G 5 G 6 H 3

124 Pair of Non-touching loops L 5 L 4 L 4 L 5 L 7 L 4 L 3 L 5 L 7 L 3 L 4

125 Non-touching loops for paths = = -G 4 H 4 3 =

126 Signal-Flow Graph Models Y( s) R( s) P P P 3 P G G G 3 G 4 G 5 G 6 P G G G 7 G 6 P 3 G G G 3 G 4 G 8 L L L 3 L 4 L 5 L 6 L 7 L 8 L 5 L 7 L 5 L 4 L 3 L 4 3 L 5 G 4 H 4

127 Block Diagram Reduction Example H R + C G G G 3 H

128 R

129

130 R

131 R

132 R + _ GG G3 G G H G G H 3 C

133 R GG G3 G G H G G H G G G 3 3 C

134 Solution for same problem by using SFG

135 P = G G G 3 Forward Path

136 Loops L = G G H L = - G G 3 H

137 L 3 = - G G G 3 P = G G G 3 L = G G H L = - G G 3 H L 3 = - G G G 3 = = - (L + L +L 3 ) T.F= (G G G 3 )/ [ -G G H + G G G 3 + G G 3 H ]

138 SFG from given T/F C ( s ) 4 R ( s ) ( s )( s 3)( s 4) s ( s) ( s )

139 Ex:

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145 Example of block diagram Step : Shift take off point from position before a block G 4 to position after block G 4

146 Step : Solve Yellow block. Step3: Solve pink block. Step4: Solve pink block.

147