Lecture 8-9 block-diagram_representation_of_control_systems
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This document provides an overview of block diagram representations of control systems. It discusses how block diagrams can be used to represent the input-output relationships of system elements and operations like addition and subtraction. Examples are given of drawing block diagrams from mathematical equations and reducing complex block diagrams to canonical form through techniques like combining blocks in series/parallel and eliminating feedback loops. The document also covers how block diagrams can represent multi-input, multi-output systems and provides an example block diagram of an armature-controlled DC motor system.
Lecture 8-9 block-diagram_representation_of_control_systems
- 1. Feedback Control Systems (FCS) Lecture-8-9 Block Diagram Representation of Control Systems Dr. Imtiaz Hussain email: imtiaz.hussain@faculty.muet.edu.pk URL :http://imtiazhussainkalwar.weebly.com/
- 2. Introduction • A Block Diagram is a shorthand pictorial representation of the cause-and-effect relationship of a system. • The interior of the rectangle representing the block usually contains a description of or the name of the element, gain, or the symbol for the mathematical operation to be performed on the input to yield the output. • The arrows represent the direction of information or signal flow. x d dt y
- 3. Introduction • The operations of addition and subtraction have a special representation. • The block becomes a small circle, called a summing point, with the appropriate plus or minus sign associated with the arrows entering the circle. • The output is the algebraic sum of the inputs. • Any number of inputs may enter a summing point. • Some books put a cross in the circle.
- 4. Introduction • In order to have the same signal or variable be an input to more than one block or summing point, a takeoff (or pickoff) point is used. • This permits the signal to proceed unaltered along several different paths to several destinations.
- 5. Example-1 • Consider the following equations in which 𝑥1 , 𝑥2 , 𝑥3 , are variables, and 𝑎1 , 𝑎2 are general coefficients or mathematical operators. x3 a1 x1 a2 x2 5
- 6. Example-1 x3 a1 x1 a2 x2 5
- 7. Example-2 • Draw the Block Diagrams of the following equations. (1) ( 2) dx1 1 x2 a1 x1dt dt b x3 a1 d 2 x2 dt 2 dx1 3 bx1 dt
- 8. Canonical Form of A Feedback Control System
- 9. Characteristic Equation • The control ratio is the closed loop transfer function of the system. C( s ) G( s ) R( s ) 1 G( s ) H ( s ) • The denominator of closed loop transfer function determines the characteristic equation of the system. • Which is usually determined as: 1 G( s )H ( s ) 0
- 10. Example-3 1. Open loop transfer function B( s ) G( s ) H ( s ) E( s ) 2. Feed Forward Transfer function C( s ) G( s ) R( s ) 1 G( s ) H ( s ) 3. control ratio 4. feedback ratio 5. error ratio C (s) G (s) E (s) G(s ) B( s ) G( s ) H ( s ) R( s ) 1 G ( s ) H ( s ) E( s ) 1 R( s ) 1 G( s ) H ( s ) 6. closed loop transfer function H (s ) C( s ) G( s ) R( s ) 1 G( s ) H ( s ) 7. characteristic equation 1 G( s )H ( s ) 0 8. Open loop poles and zeros if 9. closed loop poles and zeros if K=10.
- 11. Reduction techniques 1. Combining blocks in cascade G2 G1 G1G2 2. Combining blocks in parallel G1 G2 G1 G2
- 12. 3. Eliminating a feedback loop G G 1 GH H G H 1 G 1 G
- 13. Example-4: Reduce the Block Diagram to Canonical Form.
- 15. Example-5 • For the system represented by the following block diagram determine: 1. 2. 3. 4. 5. 6. 7. 8. Open loop transfer function Feed Forward Transfer function control ratio feedback ratio error ratio closed loop transfer function characteristic equation closed loop poles and zeros if K=10.
- 16. Example-5 – First we will reduce the given block diagram to canonical form K s 1
- 17. Example-5 K s 1 K G s 1 K 1 GH 1 s s 1
- 18. Example-5 (see example-3) 1. Open loop transfer function B( s ) G( s ) H ( s ) E( s ) 2. Feed Forward Transfer function C( s ) G( s ) E( s ) C( s ) G( s ) 3. control ratio R( s ) 1 G( s ) H ( s ) 4. feedback ratio 5. error ratio G(s ) B( s ) G( s ) H ( s ) R( s ) 1 G ( s ) H ( s ) E( s ) 1 R( s ) 1 G( s ) H ( s ) 6. closed loop transfer function C( s ) G( s ) R( s ) 1 G( s ) H ( s ) 7. characteristic equation 1 G( s )H ( s ) 0 8. closed loop poles and zeros if K=10. H (s )
- 19. Example-6 • For the system represented by the following block diagram determine: 1. 2. 3. 4. 5. 6. 7. 8. Open loop transfer function Feed Forward Transfer function control ratio feedback ratio error ratio closed loop transfer function characteristic equation closed loop poles and zeros if K=100.
- 20. Reduction techniques 4. Moving a summing point behind a block G G G 5. Moving a summing point ahead a block G G 1 G
- 21. 6. Moving a pickoff point behind a block G G 1 G 7. Moving a pickoff point ahead of a block G G G
- 22. 8. Swap with two neighboring summing points A B B A
- 23. Example-7 • Reduce the following block diagram to canonical form. H2 _ R +_ + + G1 + H1 C G2 G3
- 27. Example-7 H2 G1 _ R +_ + G1G2 1 G1G2 H1 C G3
- 29. Example-7 R +_ G1G2G3 1 G1G2 H 1 G2G3 H 2 C
- 30. Example 8 Find the transfer function of the following block diagram G4 R (s ) Y (s ) G1 G2 G3 H2 H1
- 31. I G4 R(s) B G1 G2 A G3 H2 H1 G2 Solution: 1. Moving pickoff point A ahead of block 2. Eliminate loop I & simplify B G4 G2G3 G2 Y (s )
- 32. G4 R(s) G1 G A G G 4 G2G3 B Y (s ) 3 2 H2 H1G2 G4 G2G3 3. Moving pickoff point B behind block II R(s) G1 B G4 G2G3 H2 H1G2 1 /(G4 G2G3 ) C Y (s )
- 33. 4. Eliminate loop III R(s) G1 G4 4 G2G3 G G2G3 H 1 H 2 (G4 2 G2G3 ) C C Y (s ) G2 H1 G4 G2G3 R(s) G1 (G4 G2G3 ) 1 G1G 2 H1 H 2 (G4 G2G3 ) G1 (G4 G2G3 ) Y (s) R( s ) 1 G1G 2 H1 H 2 (G4 G2G3 ) G1 (G4 G2G3 ) Y (s )
- 34. Example 9 Find the transfer function of the following block diagrams H4 R(s) Y (s ) G1 G3 G2 H3 H2 H1 G4
- 35. Solution: 1. Moving pickoff point A behind block G4 I H4 R(s) Y (s ) G1 G3 G2 H3 H2 H3 G4 H2 G4 H1 1 G4 1 G4 A G4 B
- 36. 2. Eliminate loop I and Simplify R(s) G2G3G4 1 G3G4 H 4 G1 II Y (s ) B H3 G4 H2 G4 III H1 II feedback G2G3G4 1 G3G4 H 4 G2G3 H 3 III Not feedback H 2 G4 H 1 G4
- 37. 3. Eliminate loop II & IIII R(s) G1G2G3G4 1 G3G4 H 4 G2G3 H 3 Y (s ) H 2 G4 H 1 G4 G1G2G3G4 Y (s) R( s ) 1 G2G3 H 3 G3G4 H 4 G1G2G3 H 2 G1G2G3G4 H1
- 38. Example-10: Reduce the Block Diagram.
- 40. Example-11: Simplify the block diagram then obtain the closeloop transfer function C(S)/R(S). (from Ogata: Page-47)
- 42. Superposition of Multiple Inputs
- 43. Example-12: Multiple Input System. Determine the output C due to inputs R and U using the Superposition Method.
- 46. Example-13: Multiple-Input System. Determine the output C due to inputs R, U1 and U2 using the Superposition Method.
- 49. Example-14: Multi-Input Multi-Output System. Determine C1 and C2 due to R1 and R2.
- 51. Example-14: Continue. When R1 = 0, When R2 = 0,
- 52. Block Diagram of Armature Controlled D.C Motor Ra La c Va ia eb T J La s Ra I a(s) K b(s) Va(s) Js c (s) K m I a(s)
- 53. Block Diagram of Armature Controlled D.C Motor La s Ra I a(s) K b(s) Va(s)
- 54. Block Diagram of Armature Controlled D.C Motor Js c (s) K ma I a(s)
- 55. Block Diagram of Armature Controlled D.C Motor
- 56. To download this lecture visit http://imtiazhussainkalwar.weebly.com/ END OF LECTURES-8-9