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Lecture Fundamentals of control systems: Chapter 2 - TS. Huỳnh Thái Hoàng

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Lecture "Fundamentals of control systems - Chapter 2: Mathematical models of continuous control systems" presentation of content: Presentation of content, transfer function, block diagram algebra, signal flow diagram, state space equation, linearized models of nonlinear systems.

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Nội dung Text: Lecture Fundamentals of control systems: Chapter 2 - TS. Huỳnh Thái Hoàng

  1. Lecture Notes Fundamentals of Control Systems Instructor: Assoc. Prof. Dr. Huynh Thai Hoang Department of Automatic Control Faculty of Electrical & Electronics Engineering Ho Chi Minh City University of Technology Email: hthoang@hcmut.edu.vn huynhthaihoang@yahoo.com Homepage: www4.hcmut.edu.vn/~hthoang/ 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1
  2. Chapter 2 Mathematical Models of Continuous Control Systems 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 2
  3. Content  The concept of mathematical model  Transfer function  Block diagram algebra  Signal flow diagram  State space equation  Linearized models of nonlinear systems  Nonlinear state equation  Linearized equation of state 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 3
  4. The concept of mathematical models 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 4
  5. Question  If you design a control system system, what do you need to know about the plant or the process to be controlled?  What are the advantages of mathematical models? 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 5
  6. Why mathematical model?  Practical control systems are diverse and different in nature nature.  It is necessary to have a common method for analysis and design of different type of control systems  Mathematics  The relationship between input and output of a LTI system of can be described by linear constant coefficient equations: u(t) Linear Time- y(t) Invariant System d n y (t ) d n 1 y (t ) dy (t ) a0 n  a1 n 1    an 1  an y (t )  dt dt dt d mu (t ) d m 1u (t ) du (t ) b0 m  b1 m 1    bm 1  bmu (t ) dt dt dt n: system order,order for proper systems: nm. m ai, bi: parameter of the system 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 6
  7. Example: Car dynamics dv (t ) M  Bv (t )  f (t ) dt M: mass of the car, car B friction coefficient: system parameters f(t): engine driving force: input v(t): car speed: output 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 7
  8. Example: Car suspension d 2 y (t ) dy (t ) M 2 B  Ky (t )  f (t ) dt dt M: equivalent mass B friction constant, K spring stiffness f(t): external force: input (t) travel y(t): t l off the th car body: b d output t t 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 8
  9. Example: Elevator ML: mass of cabin and load, MB: counterbalance t b l B friction constant, MB Kg gear box constant Counter- (t): driving moment of the motor ML balance y(t): position of the cabin Cabin & load d 2 y (t ) dy (t ) ML 2 B  M T g  K (t )  M B g dt dt 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 9
  10. Disadvantages of differential equation model  Difficult to solve differential equation order n (n>2) d n y (t ) d n 1 y (t ) dy (t ) a0 n  a1 n 1    an 1  an y (t )  dt dt dt d mu (t ) d m 1u (t ) du (t ) b0 m  b 1 m 1    bm 1  bmu (t ) dt dt dt  System analysis based on differential equation model is difficult.  System design based on differential equations is almost impossible in general cases.  It is necessary to have another mathematical model that makes the analysis and design of control systems easier:  transfer function  state space equation 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 10
  11. T Transfer f functions f ti 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 11
  12. Definition of Laplace transform The Laplace transform of a function f(t), f(t) defined for all real numbers t ≥ 0, is the function F(s), defined by:  L  f (t )  F ( s)   f (t ).e  st dt 0 where:  s : complex variable (Laplace variable)  L : Laplace operator  F(s) Laplace transform of f(t). The Laplace transform exists if the integral of ƒ(t) in the [0 +) is convergence. interval [0,+ convergence 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 12
  13. Properties of Laplace transform Given the functions f(t) and g(t),g(t) and their respective Laplace transforms F(s) and G(s): L  f (t )  F ( s ) L g (t )  G ( s )  Linearity L a. f (t )  b.g (t )  a.F ( s )  b.G ( s )  Time shifting L  f (t  T )  e Ts .F ( s )  df (t )    Differentiation L   sF ( s )  f ( 0 )  dt  t  F ( s)  Integration L  f ( )d   0  s  Final value theorem lim f (t )  lim sF ( s ) t  s 0 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 13
  14. Laplace transform of basic functions  Unit step function: u(t) 1 if t  0 1 L u (t )  1 u(t )   0 if t  0 s 0 t  Dirac function:  0 if t  0 (t)  (t )     if t  0 1  L  (t )  1   (t )dt  1  0 t 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 14
  15. Laplace transform of basic functions (cont’)  R Ramp ffunction: ti t if t  0 r (t )  tu(t )   0 if t  0 r(t) L t.u (t )  2 1 1 s 0 1 t  e  at if t  0  Exponential function f (t )  e at .u(t )   0 if t  0 f(t) 1 L e .u (t )   at 1 t sa 0 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 15
  16. Laplace transform of basic functions (cont’)  Si Sinusoidal id l ffunction ti f(t)) f( sin t if t  0 f (t )  (sin t ).u (t )   0 if t  0 0 t  L (sin t )u(t )  2 s  2  Table of Laplace p transform: Appendix pp A, Feedback control of dynamic systems, Franklin et. al. 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 16
  17. Definition of transfer function  Consider a system described by the differential equation: u(t) Linear time y(t) invariant system d n y (t ) d n 1 y (t ) dy (t ) a0 n  a1 n 1    a n 1  an y ( t )  dt dt dt d m u(t ) d m1u(t ) du(t ) b0 m  b1 m 1    bm1  bmu(t ) d dt d dt d dt  Taking the Laplace transform the two sides of the above equation, q , usingg differentiation p property p y and assuming g that the initial condition are zeros, we have: a0 s nY ( s )  a1s n 1Y ( s )    an 1sY ( s )  anY ( s )  b0 s mU ( s )  b1s m 1U ( s )    bm 1sU ( s )  bmU ( s ) 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 17
  18. Definition of transfer function (cont’)  Transfer function: Y ( s ) b0 s m  b1s m 1    bm 1s  bm G(s)   U ( s ) a0 s n  a1s n 1    an 1s  an  Definition: Transfer function of a system y is the ratio between the Laplace transform of the output signal and the Laplace transform of the input signal assuming that initial conditions are zeros. zeros 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 18
  19. Transfer function of components Procedure to find the transfer function of a component  Step 1: Establish the differential equation describing the input-output p p relationship p of the components p by: y  Applying Kirchhoff's law, current-voltage relationship of resistors, capacitors, inductors,... for the electrical components. components  Applying Newton's law, the relationship between friction and velocity, the relationship between force and d f deformation ti off springs i ... for f the th mechanical h i l elements. l t  Apply heat transfer law, law of conservation of energy, for the thermal process. p  ...  Step 2: Taking the Laplace transform of the two sides of the differential equation established in step 1 1, we find the transfer function of the component. 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 19
  20. Transfer function of some type of controllers Passive compensators  First order integrator: g R 1 C G ((ss )  RCs  1 C RCs  First order differentiator: R G(s)  RC  1 RCs 6 December 2013 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 20
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