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Unit v mmc

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mecanical measurement and control unit -5 (as per rgpv bhopal)

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Unit v mmc

  1. 1. 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. Chapter 1: Introduction to Control Systems Objectives
  2. 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-and-effect relationship of the process.
  3. 3. 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.
  4. 4. History Watt’s Flyball Governor (18th century) Greece (BC) – Float regulator mechanism Holland (16th Century)– Temperature regulator
  5. 5. History Water-level float regulator
  6. 6. History
  7. 7. 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 ???????????????????????????????????
  8. 8. (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. Examples of Modern Control Systems
  9. 9. Examples of Modern Control Systems
  10. 10. Examples of Modern Control Systems
  11. 11. Examples of Modern Control Systems
  12. 12. Examples of Modern Control Systems
  13. 13. Examples of Modern Control Systems
  14. 14. Examples of Modern Control Systems
  15. 15. Examples of Modern Control Systems
  16. 16. Examples of Modern Control Systems
  17. 17. The Future of Control Systems
  18. 18. Block diagram reduction
  19. 19. Block diagram Transfer Function Consists of Blocks Can be reduced )(sR 2G 3G1G 4G 1H 2H )(sY G )(sY)(sR
  20. 20. Reduction techniques 2G1G 21GG 2. Moving a summing point behind a block G G G 1G 2G 21 GG + 1. Combining blocks in cascade or in parallel
  21. 21. 5. Moving a pickoff point ahead of a block G G G G G 1 G 3. Moving a summing point ahead of a block G G G 1 4. Moving a pickoff point behind a block
  22. 22. 6. Eliminating a feedback loop G H GH G 1 7. Swap with two neighboring summing points A B AB G 1=H G G 1
  23. 23. Example 1 Find the transfer function of the following block diagrams 2G 3G1G 4G 1H 2H )(sY)(sR (a)
  24. 24. 1. Moving pickoff point A ahead of block 2G 2. Eliminate loop I & simplify 324 GGG + B 1G 2H )(sY 4G 2G 1H AB 3G 2G )(sR I Solution:
  25. 25. 3. Moving pickoff point B behind block 324 GGG + 1G B )(sR 21GH 2H )(sY )/(1 324 GGG + II 1G B )(sR C 324 GGG + 2H )(sY 21GH 4G 2G A 3G 324 GGG +
  26. 26. 4. Eliminate loop III )(sR )(1 )( 3242121 3241 GGGHHGG GGGG +++ + )(sY )()(1 )( )( )( )( 32413242121 3241 GGGGGGGHHGG GGGG sR sY sT +++++ + == )(sR 1G C 324 12 GGG HG + )(sY 324 GGG + 2H C )(1 3242 324 GGGH GGG ++ + Using rule 6
  27. 27. 2G1G 1H 2H )(sR )(sY 3H (b)
  28. 28. Solution: 1. Eliminate loop I 2. Moving pickoff point A behind block 22 2 1 HG G + 1G 1H )(sR )(sY 3H BA 22 2 1 HG G + 2 221 G HG+ 1G 1H )(sR )(sY 3H 2G 2H BA II I 22 2 1 HG G + Not a feedback loop ) 1 ( 2 22 13 G HG HH + +
  29. 29. 3. Eliminate loop II )(sR )(sY 22 21 1 HG GG + 2 221 3 )1( G HGH H + + 21211132122 21 1)( )( )( HHGGHGHGGHG GG sR sY sT ++++ == Using rule 6
  30. 30. 2G 4G1G 4H 2H 3H )(sY)(sR 3G 1H (c)
  31. 31. Solution: 2G 4G1G 4H )(sY 3G 1H 2H )(sR A B 3H 4 1 G 4 1 G I 1. Moving pickoff point A behind block 4G 4 3 G H 4 2 G H
  32. 32. 2. Eliminate loop I and Simplify II III 443 432 1 HGG GGG +1G )(sY 1H B 4 2 G H )(sR 4 3 G H II 332443 432 1 HGGHGG GGG ++ III 4 142 G HGH − Not feedbackfeedback
  33. 33. )(sR )(sY 4 142 G HGH − 332443 4321 1 HGGHGG GGGG ++ 3. Eliminate loop II & IIII 143212321443332 4321 1)( )( )( HGGGGHGGGHGGHGG GGGG sR sY sT −+++ == Using rule 6
  34. 34. 3G1G 1H 2H )(sR )(sY 4G 2G A B (d)
  35. 35. Solution: 1. Moving pickoff point A behind block 3G I 1H 3 1 G )(sY 1G 1H 2H )(sR 4G 2G A B 3 1 G 3G
  36. 36. 2. Eliminate loop I & Simplify 3G 1H 2G B 3 1 G 2H 32GG B 2 3 1 H G H + 1G )(sR )(sY 4G 3 1 G H 23212 32 1 HGGHG GG ++ II
  37. 37. )(sR )(sY 12123212 321 1 HGGHGGHG GGG +++ 3. Eliminate loop II 12123212 321 4 1)( )( )( HGGHGGHG GGG G sR sY sT +++ +== 4G
  38. 38. 2G1G 4G R Y 1H 3G N Determine the effect of R and N on Y in the following diagram Example 2
  39. 39. NTRTYYY 2121 +=+= If we set N=0, then we can get Y1: RTYY N 101 == = The same, we set R=0 and Y2 is also obtained: NTYY R 202 == = Thus, the output Y is given as follows: 0021 == +=+= RN YYYYY In this linear system, the output Y contains two parts, one part is related to R and the other is caused by N:
  40. 40. Solution: 1. Swap the summing points A and B 2. Eliminate loop II & simplify Y 12 2 1 HG G +1G 4G R N AB 3G 12 21 31 1 HG GG GG + + R Y N 4G II
  41. 41. 3. Let N=0 R HGGGGGGGHG HGGGGGGG Y 1321312112 13213121 1 1 ++++ ++ = 12 21 31 1 HG GG GG + + R Y 12 21 31 1 HG GG GG + + R Y N 4G o o 1YWe can easily get Rewrite the diagram:
  42. 42. 5. Break down the summing point M: 12 421 431 1 HG GGG GGG + + YN 12 21 31 1 HG GG GG + + 4. Let R=0, we can get: 12 21 31 1 HG GG GG + + 4G YN M
  43. 43. ])1( )[( 1 1 1432143142112 13213121 1321312112 21 NHGGGGGGGGGGHG RHGGGGGGG HGGGGGGGHG YYY +++++ ++ ++++ = += 7. According to the principle of superposition, and can be combined together, So: 1Y 2Y 6. Eliminate above loops: YN 12 421 431 1 1 HG GGG GGG + ++ 12 21 31 1 1 1 HG GG GG + ++ N HGGGGGGGHG HGGGGGGGGGGHG Y 1321312112 1432143142112 2 1 1 ++++ ++++ =
  44. 44. Signal Flow Graph
  45. 45. 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.
  46. 46. Comparison of BD and SFG )(sR )(sG )(sC )(sG )(sR )(sC block diagram: signal flow graph: 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.
  47. 47. SFG
  48. 48. Node: It is a point representing a variable. x2 = t 12 x1 +t32 x3 X 2 X1 X2 X3 t12 t32 X1 Branch : A line joining two nodes. Input Node : Node which has only outgoing branches. X1 is input node. In this SFG there are 3 nodes. Definition of terms required in SFG
  49. 49. 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. t12 X1 t23 X3 X4 X2 t34 t43
  50. 50. • 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 1 = G1 G2 G3 G4, P 2 = G5 G6 G7 G8
  51. 51. 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 1 = G2 H2 L 2 = H3 L3= G7 H7 Non-touching loops are L1 & L2, L1 &
  52. 52. SFG terms representation input node (source) b1x a 2x c 4x d 1 3x 3x mixed node mixed node forward path path loop branch node transmittance input node (source)
  53. 53. 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.
  54. 54. 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 Pk = the kth forward path gain ∆ = 1 – (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 = 1 – (loop-gain which does not touch the forward path)
  55. 55. Ex: SFG from BD
  56. 56. EX: To find T/F of the given block diagram
  57. 57. Identification of Forward Paths P 1 = 1.1.G1 .G 2 . G3. 1 = G1 G2 G3 P 2 = 1.1.G 2 . G 3 . 1 = G 2 G3
  58. 58. Individual Loops L 1 = G 1G 2 H 1 L 2 = - G 2G 3 H 2 L 3 = - G 4 H 2
  59. 59. L 4 = - G 1 G 4 L 5 = - G 1 G 2 G 3
  60. 60. Construction of SFG from simultaneous equations
  61. 61. t21 t 23 t3 1 t32 t33
  62. 62. After joining all SFG
  63. 63. SFG from Differential equations xyyyy =+′+′′+′′′ 253 Consider the differential equation Step 2: 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:-1 -5 - 2 -3 y′ y ′′ y ′′′ y x (a) tep 1: Solve the above eqn for highest order yyyxy 253 −′−′′−=′′′
  64. 64. y ′′′ x y y′ y ′′ 1 -2 -5 -3 1/s 1/s 1/s 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 1/s as shown in fig (b) (b) Step 4: Reverse the sign of a branch connecting y’’’ to y’’, with condition no change in T/F fn.
  65. 65. Step5: Redraw the SFG as shown.
  66. 66. Problem: to find out loops from the given SFG
  67. 67. Ex: Signal-Flow Graph Models
  68. 68. P 1 = P 2 =
  69. 69. Individual loops L 1 = G2 H2 L 4 = G7 H7 L 3 = G6 H6 L 2= G3 H3 Pair of Non-touching loops L 1L 3 L 1L 4 L2 L3 L 2L 4
  70. 70. ∑∑ ∑ −+−−++− ∆ = ..)21(1( LiLjLkiLjLLL P R Y kk Y s( ) R s( ) G1 G2⋅ G3⋅ G4⋅ 1 L3− L4−( )⋅  G5 G6⋅ G7⋅ G8⋅ 1 L1− L2−( )⋅ + 1 L1− L2− L3− L4− L1 L3⋅+ L1 L4⋅+ L2 L3⋅+ L2 L4⋅+
  71. 71. Ex:
  72. 72. Forward Paths
  73. 73. L5 = -G 4 H 4 L1= -G 5 G 6 H 1 L 3 = -G 8 H 1 L 2 = -G2 G 3G 4G 5 H2 L 4 = - G2 G 7 H2 Loops
  74. 74. Loops L 7 = - G 1G2 G 7G 6 H3 L 6 = - G 1G2 G 3G 4G 8 H3 L 8= - G 1G2 G 3G 4G 5 G 6 H3
  75. 75. Pair of Non-touching loops L 4 L 5 L 3 L 7 L 4 L 5L 7 L 4L 5 L 3L 4
  76. 76. Non-touching loops for paths ∆ 1 = 1 ∆ 2= -G 4 H4 ∆ 3= 1
  77. 77. Signal-Flow Graph Models Y s( ) R s( ) P1 P2 ∆2⋅+ P3+ ∆ P1 G1 G2⋅ G3⋅ G4⋅ G5⋅ G6⋅ P2 G1 G2⋅ G7⋅ G6⋅ P3 G1 G2⋅ G3⋅ G4⋅ G8⋅ ∆ 1 L1 L2+ L3+ L4+ L5+ L6+ L7+ L8+( )− L5 L7⋅ L5 L4⋅+ L3 L4⋅+( )+ ∆1 ∆3 1 ∆2 1 L5− 1 G4 H4⋅+
  78. 78. Block Diagram Reduction Example R _+ _ +1G 2G 3G 1H 2H + + C
  79. 79. R
  80. 80. R
  81. 81. R
  82. 82. R _+ 232121 321 1 HGGHGG GGG +− C
  83. 83. R 321232121 321 1 GGGHGGHGG GGG ++− C
  84. 84. Solution for same problem by using SFG
  85. 85. Forward Path P 1 = G 1 G 2 G3
  86. 86. Loops L 1 = G 1 G 2 H1 L 2 = - G 2 G3 H2
  87. 87. P 1 = G 1 G 2 G3 L 1 = G 1 G 2 H1 L 2 = - G 2 G3 H2 L 3 = - G 1 G 2 G3 ∆1 = 1 ∆ = 1- (L1 + L 2 +L 3 )
  88. 88. SFG from given T/F ( ) 24 ( ) ( 2)( 3)( 4) C s R s s s s = + + + )21()2( 1 1 1 − − + = + s s s
  89. 89. Ex:
  90. 90. Example of block diagram Step 1: Shift take off point from position before a block G4 to position after block G4
  91. 91. Step2 : Solve Yellow block. Step3: Solve pink block. Step4: Solve pink block.

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