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- 1. Lecture Notes Lecture Notes Introduction to Control Systems Introduction to Control Systems Instructor: 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/ 20 September 2011 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1
- 2. Chapter 2 Chapter 2 MATHEMATICAL MODELS OF MATHEMATICAL MODELS OF CONTINUOUS CONTROL SYSTEMS CONTINUOUS CONTROL SYSTEMS 20 September 2011 2 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 3. The concept of mathematical model Content Content The concept of mathematical model Transfer function Block diagram algebra Block diagram algebra Signal flow graph State space equation State space equation Linearized models of nonlinear systems Nonlinear state equation Nonlinear state equation Linearized state equation 20 September 2011 3 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 4. The concept of mathematical models The concept of mathematical models 20 September 2011 4 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 5. If you design a control system what do you need to know about the Question Question If you design a control system, what do you need to know about the plant or the process to be control? What are the advantages of mathematical models? 20 September 2011 5 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 6. Practical control systems are diverse and different in nature Why mathematical model? Why mathematical model? Practical control systems are diverse and different in nature. It is necessary to have a common method for analysis and design of different type of control systems ⇒ Mathematics d e e t type o co t o syste s ⇒ at e at cs The relationship between input and output of a LTI system of can be described by linear constant coefficient equations: y q Linear Time- I i t S t u(t) y(t) = + + + + − ) ( ) ( ) ( ) ( 1 1 1 1 0 t y a t dy a t y d a t y d a n n L ) ( ) ( ) ( ) ( 1 1 1 1 0 t u b t du b t u d b t u d b m m + + + + − L Invariant System + + + + − − ) ( 1 1 1 0 t y a dt a dt a dt a n n n n ) ( 1 1 1 0 t u b dt b dt b dt b m m m m + + + + − − n: system order, for proper systems: n≥m. ai, bi: parameter of the system 20 September 2011 6 ai, bi: parameter of the system © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 7. Example: Car dynamics Example: Car dynamics ) ( ) ( ) ( t f t Bv t dv M = + ) ( ) ( t f t Bv dt M + M: mass of the car B friction coefficient: system parameters M: mass of the car, B friction coefficient: system parameters f(t): engine driving force: input v(t): car speed: output 20 September 2011 7 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 8. Example: Car suspension Example: Car suspension ) ( ) ( ) ( ) ( 2 2 t f t Ky dt t dy B dt t y d M = + + dt dt M: equivalent mass B friction constant, K spring stiffness f(t): external force: input 20 September 2011 8 f(t): external force: input y(t): travel of the car body: output © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 9. Example: Elevator Example: Elevator M t K M t dy B t y d M + + + ) ( ) ( ) ( 2 g M t K g M dt y B dt y M T L B + = + + ) ( ) ( ) ( 2 τ ML Cabin & MB Counter- balance M f bi d l d M b l Cabin & load ML: mass of cabin and load, MB: counterbalance B friction constant, K gear box constant τ(t): driving moment of the motor: input 20 September 2011 9 ( ) g p y(t): position of the cabin: output © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 10. Difficult to solve differential equation order n (n>2) Disadvantages of differential equation model Disadvantages of differential equation model Difficult to solve differential equation order n (n>2) = + + + + − − − ) ( ) ( ) ( ) ( 1 1 1 1 0 t y a dt t dy a dt t y d a dt t y d a n n n n n n L dt dt dt ) ( ) ( ) ( ) ( 1 1 1 1 0 t u b dt t du b dt t u d b dt t u d b m m m m m m + + + + − − − L System analysis based on differential equation model is difficult. System design based on differential equation model is almost impossible in general cases. It is necessary to have another mathematical model that makes the It is necessary to have another mathematical model that makes the analysis and design of control systems easier: transfer function 20 September 2011 10 state space equation © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 11. T f f ti T f f ti Transfer functions Transfer functions 20 September 2011 11 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 12. The Laplace transform of a function f(t) defined for all real numbers Definition of Laplace transform Definition of Laplace transform The Laplace transform of a function f(t), defined for all real numbers t ≥ 0, is the function F(s), defined by: +∞ where: { } ∫ − = = 0 ). ( ) ( ) ( dt e t f s F t f st L where: − s : complex variable (Laplace variable) − L : Laplace operator p p − F(s) Laplace transform of f(t). The Laplace transform exists if the integral of ƒ(t) in the interval [0,+∞) is convergence. 20 September 2011 12 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 13. Gi th f ti f(t) d (t) d th i ti L l Properties of Laplace transform Properties of Laplace transform Given the functions f(t) and g(t), and their respective Laplace transforms F(s) and G(s): { } ) ( ) ( s F t f = L { } ) ( ) ( s G t g = L { } ) ( ) ( s F t f = L { } ) ( ) ( s G t g = L Linearity { } ) ( . ) ( . ) ( . ) ( . s G b s F a t g b t f a + = + L Time shifting { } ) ( . ) ( s F e T t f Ts − = − L ) ( ⎫ ⎧df Differentiation ) 0 ( ) ( ) ( + − = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ f s sF dt t df L s F t ) ( ⎫ ⎧ Integration s s F d f t ) ( ) ( 0 = ⎭ ⎬ ⎫ ⎩ ⎨ ⎧ ∫ τ τ L Final value theorem ) ( lim ) ( lim s sF t f = 20 September 2011 13 Final value theorem ) ( lim ) ( lim 0 s sF t f s t → ∞ → = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 14. Laplace transform of basic functions Laplace transform of basic functions Unit step function: 1 ⎧ ≥ 0 f 1 i u(t) { } s t u 1 ) ( = L ⎩ ⎨ ⎧ < ≥ = 0 t f 0 0 t f 1 ) ( i i t u t 0 1 Dirac function: t 0 ⎩ ⎨ ⎧ = ∞ ≠ = 0 t f 0 t f 0 ) ( i i t δ δ(t) ⎩ = ∞ 0 t f i ∫ +∞ =1 ) ( dt t δ { } 1 ) ( = t δ L t 0 1 20 September 2011 14 ∫∞ − ) ( t 0 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 15. Laplace transform of basic functions (cont’) Laplace transform of basic functions (cont’) R f i Ramp function: ⎧ ≥ 0 t f i t r(t) 1 ⎩ ⎨ ⎧ < ≥ = = 0 t f 0 0 t f ) ( ) ( i i t t tu t r t 0 1 1 { } 2 1 ) ( . s t u t = L Exponential function t 0 1 ⎧ ≥ − 0 f t i at e f(t) { } at 1 ) ( L ⎩ ⎨ ⎧ < ≥ = = − 0 f 0 0 f ) ( . ) ( t i t i at e t u e t f t 0 1 { } a s t u e at + = − 1 ) ( . L 20 September 2011 15 t 0 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 16. Laplace transform of basic functions (cont’) Laplace transform of basic functions (cont’) Si id l f i Sinusoidal function ⎧ ≥ 0 t f sin i t ω f(t) ⎩ ⎨ ⎧ < ≥ = = 0 t f 0 0 t f sin ) ( ). (sin ) ( i i t t u t t f ω ω f( ) t 0 t 0 { } 2 2 ) ( ) (sin ω ω ω + = s t u t L Table of Laplace transform: Appendix A, Feedback control of f p f pp f dynamic systems, Franklin et. al. 20 September 2011 16 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 17. Consider a system described by the differential equation: Definition of transfer function Definition of transfer function Consider a system described by the differential equation: Linear time i i t t u(t) y(t) = + + + + − ) ( ) ( ) ( ) ( 1 1 1 1 0 t y a t dy a t y d a t y d a n n L invariant system + + + + − − ) ( 1 1 1 0 t y a dt a dt a dt a n n n n ) ( ) ( ) ( ) ( 1 1 1 1 0 t u b d t du b d t u d b d t u d b m m m m m m + + + + − − L Taking the Laplace transform the two sides of the above equation, using differentiation property and assuming that the initial condition ) ( 1 1 1 0 dt dt dt m m m m − g p p y g are zeros, we have: = + + + + − − ) ( ) ( ) ( ) ( 1 1 1 0 s Y a s sY a s Y s a s Y s a n n n n L 20 September 2011 17 ) ( ) ( ) ( ) ( 1 1 1 0 s U b s sU b s U s b s U s b m m m m + + + + − − L © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 18. Transfer function: Definition of transfer function (cont’) Definition of transfer function (cont’) Transfer function: m m m m b s b s b s b s Y s G + + + + = = − − 1 1 1 1 0 ) ( ) ( L Definition: Transfer function of a system is the ratio between the n n n n a s a s a s a s U + + + + − − 1 1 1 0 ) ( ) ( L y Laplace transform of the output signal and the Laplace transform of the input signal assuming that initial conditions are zeros. 20 September 2011 18 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 19. Transfer function of components Transfer function of components Procedure to find the transfer function of a component Procedure to find the transfer function of a component Step 1: Establish the differential equation describing the input- output relationship of the components by: p p p y ¾ Applying Kirchhoff's law, current-voltage relationship of resistors, capacitors, inductors,... for the electrical components. ¾ Applying Newton's law the relationship between friction and ¾ Applying Newton s law, the relationship between friction and velocity, the relationship between force and deformation of springs ... for the mechanical components. A l h f l l f i f f h ¾ Apply heat transfer law, law of conservation of energy,… for the thermal processes. ¾ ... Step 2: Taking the Laplace transform of the two sides of the differential equation established in step 1, we find the transfer function of the component 20 September 2011 19 function of the component. © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 20. Transfer function of some type of controllers Transfer function of some type of controllers Passive compensators First order integrator: Passive compensators R C 1 ) (s G g C 1 ) ( + = RCs s G R C First order differentiator: 1 ) ( = RC RCs s G 1 ) ( + RCs 20 September 2011 20 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 21. Transfer function of some type of controllers (cont’) Transfer function of some type of controllers (cont’) Passive compensators Passive compensators Phase lead compensator: C R 1 + Ts α R1 R2 1 1 ) ( + + = Ts Ts K s G C α R R R 2 1 2 R R R KC + = 2 1 1 2 R R C R R T + = 1 2 2 1 > + = R R R α Phase lag compensator : R1 R2 1 ) ( + = Ts K s G C α C 1 ) ( + Ts C 1 1 < R α 20 September 2011 21 1 = C K C R R T ) ( 2 1 + = 1 2 1 < + = R R α © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 22. Transfer function of some type of controllers (cont’) Transfer function of some type of controllers (cont’) Active controllers Active controllers K G ) ( Proportional Controller (P) P K s G = ) ( 2 R KP − = 1 R KP Proportional Integral controller (PI) p g ( ) s K K s G I P + = ) ( s 2 R KP − = C R KI 1 − = 20 September 2011 22 1 R P C R I 1 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 23. Transfer function of some type of controllers (cont’) Transfer function of some type of controllers (cont’) Active controllers Active controllers Proportional Derivative controller (PD) s K K s G D P + = ) ( 2 R K C R K Proportional Integral Derivative controller (PID) 1 2 R KP − = C R KD 2 − = p g ( ) s K s K K s G D I P + + = ) ( 2 1 2 2 1 1 C R C R C R KP + − = 2 1 1 C R KI − = 20 September 2011 23 1 2C R KD − = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 24. Transfer function of DC motors Transfer function of DC motors Equivalent diagram of a DC motor Equivalent diagram of a DC motor − La : armature induction − ω : motor speed R t i t M l d i ti − Ra : armature resistance − Mt : load inertia − Ua : armature voltage − B : friction constant − Ea : back electromotive force − J : moment of inertia of the rotor 20 September 2011 24 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 25. Transfer function of DC motors Transfer function of DC motors Applying Kirchhoff's law for the armature circuit: ) ( ) ( ). ( ) ( t E dt t di L R t i t U a a a a a a + + = (1) dt ) ( ) ( t K t E ω Φ = a where: K : electromotive force constant (2) Φ : excitation magnetic flux Applying Newton’s law for the rotating part of the motor: t d J t B t M t M ) ( ) ( ) ( ) ( ω (3) dt J t B t M t M L ) ( ) ( ) ( ) ( ω + + = where: ) ( ) ( t i K t M a Φ = (3) (4) 20 September 2011 25 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 26. Transfer function of DC motors Transfer function of DC motors T ki h L l f f (1) (2) (3) (4) l d Taking the Laplace transform of (1), (2), (3), (4) leads to: (5) ) ( ) ( ). ( ) ( s E s sI L R s I s U a a a a a a + + = (6) (7) ) ( ) ( s K s E ω Φ = a ) ( ) ( ) ( ) ( s Js s B s M s M L ω ω + + = (8) ) ( ) ( s i K s M a Φ = Denote: a a L T = Electromagnetic time constant a a R B J Tc = Mechanical time constant 20 September 2011 26 B © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 27. Transfer function of DC motors Transfer function of DC motors From (5) and (7) we have: From (5) and (7), we have: ) 1 ( ) ( ) ( ) ( s T R s E s U s I a a a + − = (5’) ) 1 ( s T R a a + ) 1 ( ) ( ) ( ) ( s T B s M s M s L + − = ω ) 1 ( s T B c + From (5’), (6), (7’) and (8) we can develop the block diagram of the DC motor as follow: the DC motor as follow: ) (s Ua ) (s ML R / 1 ) ( a ) (s Ea a sT R + 1 / 1 a 20 September 2011 27 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 28. Transfer function of a thermal process Transfer function of a thermal process Temperature of th u(t) y(t) Electric power supplying t th 100% the oven to the oven 100% (t) (t) y(t) y(t) 20 September 2011 28 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ “Exact” characteristic of the oven Approximate characteristic of the oven
- 29. Transfer function of a thermal process (cont’) Transfer function of a thermal process (cont’) The approximate transfer function of the thermal ) (s Y The approximate transfer function of the thermal process can be calculated by using the equation: ) ( ) ( ) ( s U s Y s G = 1 The input is the unit step signal, then s s U 1 ) ( = The approximate output is: ) ( ) ( 1 T t f t y − = The approximate output is: ) ( ) ( 1 T t f t y where: ) 1 ( ) ( 2 /T t e K t f − − = The Laplace transform of f (t) is: ) ( K F The Laplace transform of f (t) is: ) 1 ( ) ( 2s T s s F + = ) ( 1 Ke s Y s T = − Applying the time delay theorem: ) 1 ( ) ( 2s T s s Y + = Applying the time delay theorem: ) ( ) ( 1 − Ke s Y G s T 20 September 2011 29 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1 ) ( ) ( ) ( 2 + = = s T s U s G ⇒
- 30. Transfer function of a car Transfer function of a car M: car mass B: friction constant B: friction constant f(t): driving force v(t): car speed ) ( ) ( ) ( t f t Bv dt t dv M = + Differential equation: dt Transfer function B M F s V s G = = 1 ) ( ) ( ) ( ⇔ 1 ) ( = T K s G B Ms s F + ) ( 1 + Ts where K 1 = M T = 20 September 2011 30 B K B T © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 31. Transfer function of an suspension system Transfer function of an suspension system M: equivalent car mass q B: friction constant K: spring stiffness f(t): external force f(t): external force y(t): travel of car body Differential equation: ) ( ) ( ) ( ) ( 2 2 t f t Ky dt t dy B dt t y d M = + + dt dt Transfer function s Y s G = = 1 ) ( ) ( 20 September 2011 31 Transfer function K Bs Ms s F s G + + = = 2 ) ( ) ( © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 32. Transfer functions of sensors Transfer functions of sensors y(t) yfb(t) Sensor y(t) yfb(t) Feedback signal yfb(t) is proportional to y(t), so transfer functions of g yfb( ) p p y( ), sensors are usually constant: fb K s H = ) ( Ex: Suppose that temperature of a furnace changing in the range y(t) = 0÷5000C, if a sensor converts the temperature to a voltage in the ( ) 0 5V h h f f i f h i range yfb(t) 0÷5V, then the transfer function of the sensor is: ) / ( 01 . 0 ) ( 500 / ) ( 5 ) ( 0 0 C V C V K s H fb = = = If the sensor has a delay time, then the transfer function of the sensor is: K s H fb = ) ( 20 September 2011 32 s T s H fb + = 1 ) ( © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 33. Transfer functions of control systems Transfer functions of control systems 20 September 2011 33
- 34. Block diagram Block diagram Block diagram is a diagram of a system in which the principal parts Block diagram is a diagram of a system, in which the principal parts or functions are represented by blocks connected by lines, that show the relationships of the blocks. A block diagram composes of 3 components: Function block Summing point Summing point Pickoff point 20 September 2011 34 Function block Pickoff point Summing point © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 35. Block diagram algebra Block diagram algebra Transfer function of systems in series Transfer function of systems in series G U1 (s) Y1 (s) G G Un (s) Yn (s) U(s) Y(s) G1 1 ( ) G2 U2(s) Y2 (s) Gn … ( ) ( ) Gs U(s) Y (s) ∏ = n i s s G s G ) ( ) ( 20 September 2011 35 ∏ = i 1 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 36. Block diagram algebra (cont’) Block diagram algebra (cont’) Transfer function of systems in parallel Transfer function of systems in parallel U (s) Y ( ) G1 U1 (s) Y1 (s) U (s) Y2 (s) U(s) Y(s) G2 U2(s) Y2 (s) … U(s) ( ) Gp U(s) Y (s) Gn Un (s) Yn (s) ∑ = n i s G s G ) ( ) ( 20 September 2011 36 ∑ = i i p s G s G 1 ) ( ) ( © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 37. Block diagram algebra (cont’) Block diagram algebra (cont’) Transfer function of feedback systems Transfer function of feedback systems Negative feedback Unity negative feedback Y(s) R(s) − G(s) E(s) + Y(s) R(s) − G(s) E(s) + H(s) Yht(s) Yht(s) ) ( ) ( s G s G l = ) ( ) ( s G s G l = ) ( ). ( 1 ) ( s H s G s Gcl + ) ( 1 ) ( s G s Gcl + 20 September 2011 37 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 38. Block diagram algebra (cont’) Block diagram algebra (cont’) Transfer function of feedback systems Transfer function of feedback systems Positive feedback Unity positive feedback Y(s) R(s) + G(s) E(s) Y ( ) + Y(s) R(s) + G(s) E(s) Y ( ) + H(s) Yfb(s) Yfb(s) ) ( ) ( 1 ) ( ) ( s H s G s G s Gcl − = ) ( 1 ) ( ) ( s G s G s Gcl − = ) ( ). ( 1 s H s G ) ( 1 s G 20 September 2011 38 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 39. Block diagram algebra (cont’) Block diagram algebra (cont’) Transfer function of multi loop systems Transfer function of multi-loop systems For a complex system consisting of multi feedback loops, we perform equivalent block diagram transformation so that simple connecting q g p g blocks appears, and then we simplify the block diagram from the inner loops to the outer loops. Two block diagrams are equivalent if their input-output relationship are the same 20 September 2011 39 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 40. Block diagram algebra (cont’) Block diagram algebra (cont’) Moving a pickoff point behind a block Moving a pickoff point behind a block 20 September 2011 40 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 41. Block diagram algebra (cont’) Block diagram algebra (cont’) Moving a pickoff point ahead a block Moving a pickoff point ahead a block 20 September 2011 41 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 42. Block diagram algebra (cont’) Block diagram algebra (cont’) Moving a summing point behind a block 20 September 2011 42 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 43. Block diagram algebra (cont’) Block diagram algebra (cont’) M i i i t h d bl k Moving a summing point ahead a block 20 September 2011 43 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 44. Block diagram algebra (cont’) Block diagram algebra (cont’) Interchanging the positions of the two consecutive summing points 20 September 2011 44 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 45. Block diagram algebra (cont’) Block diagram algebra (cont’) Splitting a summing point 20 September 2011 45 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 46. Block diagram algebra (cont’) Block diagram algebra (cont’) Note Note Do not interchange the positions of a pickoff point and a summing point : p Do not interchange the positions of Do not interchange the positions of two summing points if there exists a pickoff point between them: 20 September 2011 46 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 47. Block diagram algebra Block diagram algebra – – Example 1 Example 1 Find the equivalent transfer function of the following system: Y(s) 20 September 2011 47 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 48. Block diagram algebra Block diagram algebra – – Example 1 (cont’) Example 1 (cont’) I h i h i i c d d Interchanging the summing points c and d, Eliminating GA(s)=[G3(s)//G4(s)] Y(s) ) ( ) ( ) ( 4 3 s G s G s GA − = 20 September 2011 48 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 49. Block diagram algebra Block diagram algebra – – Example 1 (cont’) Example 1 (cont’) G ( ) [G ( ) // unity block] GB(s)=[G1(s) // unity block] , GC (s)= feedback loop[G2(s),GA(s)]: Y( ) ) ( 1 ) ( s G s G + = Y(s) ) ( 1 ) ( 1 s G s GB + = )] ( ) ( ) [ ( 1 ) ( ) ( ) ( 1 ) ( ) ( 2 2 s G s G s G s G s G s G s G s GC − + = + = )] ( ) ( ).[ ( 1 ) ( ). ( 1 4 3 2 2 s G s G s G s G s G A + + Equivalent transfer function of the system: ) ( ). ( ) ( s G s G s G C B eq = ) ( )]. ( 1 [ ) ( 2 1 s G s G s G + = 20 September 2011 49 )] ( ) ( ).[ ( 1 ) ( 4 3 2 s G s G s G s Geq − + = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 50. Block diagram algebra Block diagram algebra – – Example 2 Example 2 Find the equivalent transfer function of the following system: Y(s) 20 September 2011 50 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 51. Block diagram algebra Block diagram algebra – – Example 2 (cont’) Example 2 (cont’) Interchanging the positions of the summing points d and e Moving the pickoff point f behind the block G2(s) Y(s) 20 September 2011 51 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 52. Block diagram algebra Block diagram algebra – – Example 2 (cont’) Example 2 (cont’) GB(s) = feedback loop [G2(s), H2(s)] GC(s) = [GA(s)// unity block] GC(s) [GA(s)// unity block] Y(s) Y(s) 20 September 2011 52 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 53. Block diagram algebra Block diagram algebra – – Example 2 (cont’) Example 2 (cont’) G ( ) d (G ( ) G ( ) G ( )) GD(s) = cascade(GB (s), GC(s), G3(s)) Y(s) GE(s) = feedback loop(GD(s), H3(s)) Y(s) 20 September 2011 53 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 54. Block diagram algebra Block diagram algebra – – Example 2 (cont’) Example 2 (cont’) Detailed calculation: Detailed calculation: 1 * H GA = 2 G GA 2 * G G = 2 2 1 H G GB + = H G H + 2 1 2 2 1 1 1 * G H G G H G G A C + = + = + = 2 2 1 3 3 2 3 2 1 2 2 2 2 3 1 1 . . * H G H G G G G G H G H G G G G G G C B D + + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = = 20 September 2011 54 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 55. Block diagram algebra Block diagram algebra – – Example 2 (cont’) Example 2 (cont’) H G G G + 3 1 3 3 3 2 2 2 1 3 3 2 1 3 3 2 2 2 1 3 3 2 3 1 1 1 1 * H H G H G G H G H G G G H H G G G H G H G G G H G G G D D E + + + + = + + + + = + = 3 1 3 3 3 2 2 2 3 2 2 3 1 1 H H G D + + E i l t t f f ti f th t 1 3 3 2 H G G G G + Equivalent transfer function of the system: 1 3 3 2 1 3 1 3 3 3 2 2 2 1 1 1 1 . 1 1 . 1 * H H G H G G H G H G G G G H H G H G G H G G G G G G G E E eq + + + + + + + + = + = 3 1 3 3 3 2 2 2 1 H H G H G G H G + + + 1 3 1 3 2 1 1 H G G G G G H H G H G G H G H G G G G G Geq + = ⇒ 20 September 2011 55 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ 1 3 1 3 2 1 3 1 3 3 3 2 2 2 1 H G G G G G H H G H G G H G eq + + + + +
- 56. Block diagram algebra Block diagram algebra – – Example 3 Example 3 Find the equivalent transfer function of the following system: Y(s) 20 September 2011 56 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 57. Block diagram algebra Block diagram algebra – – Example 3 (cont’) Example 3 (cont’) Hint to solve example 3 Hint to solve example 3 Move the summing point e ahead the block G1(s), then interchange the position of the summing points d ande Hint to solve example 3 Hint to solve example 3 g p g p Move the pickoff point f behind the block G2(s) Y(s) 20 September 2011 57 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 58. Block diagram algebra Block diagram algebra – – Example 3 (cont’) Example 3 (cont’) S l ti t E l 3 S l ti t E l 3 Solution to Example 3 Solution to Example 3 Students calculate the equivalent transfer function themselves using the hints in the previous slide. the hints in the previous slide. 20 September 2011 58 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 59. Remarks on block diagram algebra Remarks on block diagram algebra Bl k di l b i l ti l i l th d t l l t th Block diagram algebra is a relatively simple method to calculate the equivalent transfer function of a control system. The main disadvantage of block diagram algebra is its lack of systematic procedure to perform the block diagram transformation; each particular block diagram can be transformed by different heuristic ways. heuristic ways. When calculating the equivalent transfer function, it is necessary to manipulate many calculations on algebraic fractions. This could be a potential source of error if the system is complex enough potential source of error if the system is complex enough. ⇒ Block diagram algebra is only appropriate for finding equivalent transfer function of simple systems. To find equivalent transfer function of complex systems, signal flow graph method (to be discussed later) is more effective. 20 September 2011 59 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 60. Definition of signal flow graph Definition of signal flow graph Y(s) Y(s) Y(s) Y(s) Si l fl h t k i ti f d d b h Block diagram Signal flow graph Signal flow graph: a networks consisting of nodes and branches. Node: a point representing a signal or a variable in the system. Branch: a line directly connecting two nodes, each branch has an arrow showing the signal direction and a transfer function representing the relationship between the signal at the two nodes of the branch the branch Source node: a node from which there are only out-going branches. Sink node: a node to which there are only in-going branches. Hybrid node: a node which both has in going branches and out 20 September 2011 60 Hybrid node: a node which both has in-going branches and out- going branches. © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 61. Definition of signal flow graph (cont’) Definition of signal flow graph (cont’) Forward path: is a path consisting of continuous sequence of Forward path: is a path consisting of continuous sequence of branches that goes in the same direction from a source node to a sink node without passing any single node more than once. Path gain is the product of all transfer functions of the branches Path gain is the product of all transfer functions of the branches belonged to the path. Loop: is a closed path consisting of continuous sequence of branches that goes in the same direction without passing any single node more than once. Loop gain is the product of all transfer functions of the branches Loop gain is the product of all transfer functions of the branches belonged to the loop. Y(s) Y(s) Y(s) 20 September 2011 61 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ Loop Forward path
- 62. Mason’s formula Mason’s formula The equivalent transfer function from a source 1 The equivalent transfer function from a source node to a sink node of a system can be found by using the Mason’s formula: ∑Δ Δ = k k k P G 1 Pk: is the gain of kth forward path from the considered source node to the considered drain node. Δ: is the determinant of the signal flow graph K + − + − = Δ ∑ ∑ ∑ g nontouchin m j i m j i g nontouchin j i j i i i L L L L L L , , , 1 Δ: is the determinant of the signal flow graph. g nontouchin g nontouchin i L : is the gain of the ith loop Δk: is the cofactor of the kth path . Δk is inferred from Δ by removing all the gain(s) of the loop(s) touching the forward path Pk Note: Nontouching loops do not have any common nodes A loop and 20 September 2011 62 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/ Note: Nontouching loops do not have any common nodes. A loop and a path touch together if they have at least one common node.
- 63. Signal flow graph Signal flow graph – – Example 1 Example 1 Find the equivalent transfer function of the system described by the Find the equivalent transfer function of the system described by the following signal flow graph: Y(s) R(s) Solution Solution Forward paths: Loop: G G G G G P = 1 4 1 H G L − = 5 4 3 2 1 1 G G G G G P = 5 4 6 1 2 G G G G P = 7 2 1 3 G G G P = 1 4 1 H G L 2 7 2 2 H G G L − = 2 5 4 6 3 H G G G L − = 20 September 2011 63 7 2 1 3 G G G P 2 5 4 3 2 4 H G G G G L − = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 64. Signal flow graph Signal flow graph – – Example 1 (cont’) Example 1 (cont’) The determinant of the SFG: The determinant of the SFG: 2 1 4 3 2 1 ) ( 1 L L L L L L + + + + − = Δ The cofactors of the paths 1 1 = Δ 1 2 = Δ 1 3 1 L − = Δ The equivalent transfer function of the system: ) ( 1 Δ + Δ + Δ P P P G ) ( 3 3 2 2 1 1 Δ + Δ + Δ Δ = P P P Geq 1 4 7 2 1 5 4 6 1 5 4 3 2 1 ) 1 ( H G G G G G G G G G G G G G G + + + = 20 September 2011 64 2 7 2 1 4 2 5 4 3 2 2 5 4 6 2 7 2 1 4 1 H G G H G H G G G G H G G G H G G H G Geq + + + + + = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 65. Signal flow graph Signal flow graph – – Example 2 Example 2 Find the equivalent transfer function of the system described by the Find the equivalent transfer function of the system described by the following block diagram: Y(s) R(s) Solution: Y(s) R(s) 20 September 2011 65 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 66. Signal flow graph Signal flow graph – – Example 2 (cont’) Example 2 (cont’) Y(s) R(s) Forward paths: Loop 3 2 1 1 G G G P = 2 2 1 H G L − = H G G L 3 1 1 2 G H G P = 3 3 2 2 H G G L − = 3 2 1 3 G G G L − = 3 1 3 4 H H G L − = 20 September 2011 66 3 1 3 4 H H G L 1 3 1 5 H G G L − = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 67. Signal flow graph Signal flow graph – – Example 2 (cont’) Example 2 (cont’) Th d i f h SFG The determinant of the SFG: ) ( 1 5 4 3 2 1 L L L L L + + + + − = Δ The cofactors of the paths: 1 1 = Δ1 1 2 = Δ Th i l f f i f h The equivalent transfer function of the system: ) ( 1 2 2 1 1 Δ + Δ Δ = P P Geq Δ 1 3 1 3 1 3 3 2 1 3 3 2 2 2 1 3 1 3 2 1 1 H G G H H G G G G H G G H G H G G G G G Geq + + + + + + = 20 September 2011 67 1 3 1 3 1 3 3 2 1 3 3 2 2 2 1 H G G H H G G G G H G G H G + + + + + © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 68. Signal flow graph Signal flow graph – – Example 3 Example 3 Find the equivalent transfer function of the system described by the q y y following block diagram: Y(s) Solution: Y(s) 20 September 2011 68 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 69. Signal flow graph Signal flow graph – – Example 3 (cont’) Example 3 (cont’) Y(s) 3 2 1 1 G G G P = 2 1 1 H G L − = H G G L Forward path Loop 4 2 G P = 1 2 1 2 H G G L − = 3 2 1 3 G G G L − = 3 3 2 4 H G G L − = 20 September 2011 69 3 3 2 4 H G G L 4 5 G L − = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 70. Signal flow graph Signal flow graph – – Example 3 (cont’) Example 3 (cont’) Determinant of the SFG: Determinant of the SFG: 5 4 1 5 4 5 2 5 1 4 1 5 4 3 2 1 ) ( ) ( 1 L L L L L L L L L L L L L L L L − + + + + + + + + − = Δ The cofactor: 1 1 = Δ The equivalent transer function of the system: ) ( ) ( 1 4 1 4 2 1 2 L L L L L + + + − = Δ The equivalent transer function of the system: ) ( 1 2 2 1 1 Δ + Δ Δ = P P G Num Den 20 September 2011 70 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 71. State space equations State space equations 20 September 2011 71 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 72. State: The state of a system is a set of variables whose values State of a system State of a system State: The state of a system is a set of variables whose values, together with the equations described the system dynamics, will provide future state and output of the system. A nth order system has n state variables. The state variables can be physical variables, but not necessary. State vector: n state variables form a column vector called the state vector. [ ]T n x x x K 2 1 = x 20 September 2011 72 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 73. State equations State equations By using state variables, we can transform the n-order differential y g , equation describing the system dynamics into a set of n first order differential equations (called state equations) of the form: ⎧ ⎩ ⎨ ⎧ = + = ) ( ) ( ) ( ) ( ) ( t t y t u t t Cx B Ax x & where ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ n a a a a a a K K 2 22 21 1 12 11 ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ b b 2 1 [ ] where ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ = nn n n n a a a a a a K M M M K 2 1 2 22 21 A ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ = n b b M 2 B [ ] n c c c K 2 1 = C Note: Depending on how we chose the state variables, a system can be described by many different state equations. 20 September 2011 73 y y q © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 74. State equations State equations – – Example 1 Example 1 A suspension system A suspension system A suspension system A suspension system ) ( ) ( ) ( ) ( 2 t f t Ky t dy B t y d M = + + Differential equation: (*) ) ( ) ( 2 t f t Ky dt B dt M = + + ( ) ⎪ ⎨ ⎧ = 1 ) ( ) ( 2 1 B K t x t x & Denote: ⎨ ⎧ = ) ( ) ( 1 t y t x ⎪ ⎩ ⎪ ⎨ + − − = ) ( 1 ) ( ) ( ) ( 2 1 2 t f M t x M B t x M K t x & ⇒ ⎩ ⎨ = ) ( ) ( ) ( ) ( 2 1 t y t x y & 1 0 ) ( 1 0 ) ( 1 1 t x B K t x ⎥ ⎤ ⎢ ⎡ ⎤ ⎡ ⎥ ⎤ ⎢ ⎡ ⎤ ⎡ & ) ( 1 ) ( ) ( . ) ( ) ( 2 1 2 1 t f M t x M B M K t x ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ + ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ & [ ] ⎤ ⎡ ) ( 1 t x ⇔ [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = ) ( ) ( 0 1 ) ( 2 1 t x t x t y ⎨ ⎧ + = ) ( ) ( ) ( t f t t B Ax x & ⇔ ⎥ ⎤ ⎢ ⎡ = B K 1 0 A ⎥ ⎤ ⎢ ⎡ = 1 0 B [ ] 0 1 = C 20 September 2011 74 ⎩ ⎨ = ) ( ) ( ) ( ) ( ) ( t t y f Cx ⇔ ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − − M M A ⎥ ⎥ ⎦ ⎢ ⎢ ⎣M B [ ] © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 75. State equations State equations – – Example 2 Example 2 DC motor DC motor − La : armature induction − ω : motor speed R t i t M l d i ti − Ra : armature resistance − Mt : load inertia − Ua : armature voltage − B : friction constant − Ea : back electromotive force − J : moment of inertia of the rotor 20 September 2011 75 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 76. State equations State equations – – Example 2 (cont’) Example 2 (cont’) Applying Kirchhoff's law for the armature circuit: ) ( ) ( ). ( ) ( t E dt t di L R t i t U a a a a a a + + = (1) dt ) ( ) ( t K t E ω Φ = a where: K : electromotive force constant (2) Φ : excitation magnetic flux Applying Newton’s law for the rotating part of the motor: Applying Newton s law for the rotating part of the motor: (for simplicity, assuming that load torque is zero) t d J t B t M ) ( ) ( ) ( ω (3) dt J t B t M ) ( ) ( ) ( ω + = where: ) ( ) ( t i K t M a Φ = (3) (4) 20 September 2011 76 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 77. State equations State equations – – Example 2 (cont’) Example 2 (cont’) (1) & (2) ⇒ ) ( 1 ) ( ) ( ) ( t U L t L K t i L R dt t di ö ö ö ö ö ö ö + Φ − − = ω (5) L L L dt ö ö ö (3) & (4) ⇒ ) ( ) ( ) ( t J B t i J K dt t d ω ω − Φ = ö (6) Denote: ⎩ ⎨ ⎧ = = ) ( ) ( ) ( ) ( 2 1 t t x t i t x ω ö ⎩ ⎪ ⎪ ⎧ + Φ − − = ) ( 1 ) ( ) ( ) ( 2 1 1 t U L t x L K t x L R t x & ö ö (5) & (6) ⇒ ⎪ ⎪ ⎩ ⎪ ⎪ ⎨ − Φ = ) ( ) ( ) ( 2 1 2 t x J B t x J K t x L L L & ö ö ö 20 September 2011 77 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 78. State equations State equations – – Example 2 (cont’) Example 2 (cont’) ) ( 1 ) ( ) ( 1 1 t U L t x L K L R t x ö ö ö ⎥ ⎤ ⎢ ⎡ + ⎥ ⎤ ⎢ ⎡ ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ Φ − − = ⎥ ⎤ ⎢ ⎡ & ⎤ ⎡ ) ( ⇔ ) ( 0 ) ( ) ( 2 2 t U L t x J B J K t x ö ö ö ö ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ + ⎥ ⎦ ⎢ ⎣ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ − Φ = ⎥ ⎦ ⎢ ⎣ & [ ] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = ) ( ) ( 1 0 ) ( 2 1 t x t x t ω ⎩ ⎨ ⎧ = + = ) ( ) ( ) ( ) ( ) ( t t t U t t Cx B Ax x ω u & ⇔ ⎤ ⎡ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ Φ Φ − − = B K L K L R ö ö ö A [ ] 1 0 = C ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ = 0 1 ö L B where: 20 September 2011 78 ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − J J ⎥ ⎦ ⎢ ⎣ 0 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 79. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #1: The differential equation Case #1: The differential equation Case #1: The differential equation Case #1: The differential equation do not involve the input derivatives do not involve the input derivatives The differential equation describing the system dynamics is: ) ( ) ( ) ( ) ( ) ( 0 1 1 1 1 0 t u b t y a dt t dy a dt t y d a dt t y d a n n n n n n = + + + + − − − L q g y y ) ( ) ( 1 t y t x = Define the state variables as follow: The first state is the system output: ) ( ) ( ) ( ) ( ) ( ) ( 2 3 1 2 1 t x t x t x t x t y t x = = & & The first state is the system output: The i th state (i=2..n) is chosen to be the first derivative of the (i−1)th state : ) ( ) ( ) ( ) ( 1 2 3 t x t x n n − = & M 20 September 2011 79 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 80. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #1 (cont’) Case #1 (cont’) Case #1 (cont ) Case #1 (cont ) State equation: ⎩ ⎨ ⎧ = + = ) ( ) ( ) ( ) ( ) ( t t y t u t t Cx B Ax x & ⎩ ) ( ) ( t t y Cx where: ⎤ ⎡ ⎤ ⎡ 0 0 1 0 ⎤ ⎡ 0 ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ ) ( ) ( ) ( 2 1 t x t x M ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ 0 1 0 0 0 0 1 0 M M M M K K A ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ 0 0 M B ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎣ = − ) ( ) ( ) ( 1 t x t x t n M x ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ − − − − = − − 1 2 1 1 0 0 0 a a a a n n n K K M M M M A ⎥ ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ⎢ = 0 0 b M B ⎥ ⎦ ⎢ ⎣ ) (t xn ⎥ ⎦ ⎢ ⎣ 0 0 0 0 a a a a ⎥ ⎦ ⎢ ⎣ 0 a [ ] 0 0 0 1 K = C 20 September 2011 80 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 81. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #1: Example Case #1: Example Case #1: Example Case #1: Example Write the state equations describing the following system: ) ( ) ( 10 ) ( 6 ) ( 5 ) ( 2 t u t y t y t y t y = + + + & & & & & & ⎪ ⎪ ⎨ ⎧ = = ) ( ) ( ) ( ) ( 1 2 1 t x t x t y t x & Define the state variables as: ⎪ ⎩ = ) ( ) ( 2 3 t x t x & State equation: ⎩ ⎨ ⎧ + = ) ( ) ( ) ( ) ( ) ( t r t t C B Ax x & ⎤ ⎡ ⎤ ⎡ ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ = 0 0 0 0 B ⎩ ⎨ = ) ( ) ( t t y Cx where ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡ = 5 2 3 5 1 0 0 0 1 0 1 0 0 0 1 0 1 2 3 a a a A ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ = ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ = 5 . 0 0 0 0 0 a b B 20 September 2011 81 ⎥ ⎦ ⎢ ⎣ − − − ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − − − 5 . 2 3 5 0 1 0 2 0 3 a a a [ ] 0 0 1 = C © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 82. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #2: The differential equation involve the input derivatives Case #2: The differential equation involve the input derivatives Case #2: The differential equation involve the input derivatives Case #2: The differential equation involve the input derivatives Consider a system described by the differential equation: − ) ( ) ( ) ( ) ( 1 t dy t y d t y d n n = + + + + − − ) ( ) ( ) ( ) ( 1 1 1 0 t y a dt t dy a dt t y d a dt t y d a n n n n L ) ( ) ( ) ( ) ( 1 2 2 1 1 0 t u b t du b t u d b t u d b n n − − + + + + L ) ( 1 2 1 1 1 0 t u b dt b dt b dt b n n n n − − − − + + + + Define the state variables as follow: The first state is the system output: The i th state (i=2 n) is equal to the ) ( ) ( ) ( ) ( ) ( 1 1 2 1 t r t x t x t y t x − = = β & The i state (i 2..n) is equal to the first derivative of the (i−1)th state minus a quantity proportional to the i t ) ( ) ( ) ( 2 2 3 1 1 2 t r t x t x − = β M & 20 September 2011 82 input: ) ( ) ( ) ( 1 1 t r t x t x n n n − − − = β & © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 83. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #2 (cont’) Case #2 (cont’) ⎩ ⎨ ⎧ = + = ) ( ) ( ) ( ) ( ) ( t t y t r t t Cx B Ax x & Case #2 (cont ) Case #2 (cont ) State equation: ⎩ = ) ( ) ( t t y Cx where: ⎤ ⎡ 0 0 1 0 ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ ) ( ) ( 2 1 t x t x M ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ 0 1 0 0 0 0 1 0 M M M M K K ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ β β 2 1 ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎣ = − ) ( ) ( ) ( 1 t t x t n M x ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ − − − − = − − 1 2 1 1 0 0 0 a a a a n n n K M M M M A ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎣ = − n β β 1 M B ⎥ ⎦ ⎢ ⎣ ) (t xn ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ 0 0 0 0 a a a a K [ ] C ⎥ ⎦ ⎢ ⎣ n β 20 September 2011 83 [ ] 0 0 0 1 K = C © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 84. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #2 (cont’) Case #2 (cont’) Case #2 (cont ) Case #2 (cont ) The coefficients β in the vector B are calculated as follow: b 1 1 1 0 0 1 a b a b β β − = 1 2 2 1 2 0 1 1 1 2 a a b a a b β β β β β − − = 0 1 2 2 1 2 3 a β β β = M 0 1 1 2 2 1 1 1 a a a a b n n n n n β β β β − − − − − − − − = K 20 September 2011 84 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 85. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #2: Example Case #2: Example Case #2: Example Case #2: Example Write the state equations describing the following system: ) ( 20 ) ( 10 ) ( 10 ) ( 6 ) ( 5 ) ( 2 t u t u t y t y t y t y + = + + + & & & & & & & Define the state variables: ⎪ ⎩ ⎪ ⎨ ⎧ − = = ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 1 1 2 1 t r t x t x t y t x β β & & The state equation: ⎩ ⎨ ⎧ + = ) ( ) ( ) ( ) ( ) ( t r t t C B Ax x & ⎪ ⎩ − = ) ( ) ( ) ( 2 2 3 t r t x t x β & ⎩ ⎨ = ) ( ) ( t t y Cx ⎤ ⎡ ⎤ ⎡ where: ⎥ ⎤ ⎢ ⎡ 1 β ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡ = 5 2 3 5 1 0 0 0 1 0 1 0 0 0 1 0 1 2 3 a a a A ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣ = 3 2 β β B 20 September 2011 85 [ ] 0 0 1 = C ⎥ ⎦ ⎢ ⎣ − − − ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ − − − 5 . 2 3 5 0 1 0 2 0 3 a a a © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 86. Method for establishing state equations from differential equations Method for establishing state equations from differential equations Case #2: Example (cont’)) Case #2: Example (cont’)) Case #2: Example (cont )) Case #2: Example (cont )) The elements of vector B are calculated as follow: ⎪ ⎪ ⎪ ⎧ × = = = 0 5 10 0 2 0 0 0 1 b a b β β ⎪ ⎪ ⎪ ⎪ ⎨ − = × − × − = − − = = × − = − = 15 0 6 10 5 20 5 2 0 5 10 1 2 2 1 2 0 1 1 1 2 a a b a a b β β β β β ⎪ ⎩ = = = 15 2 0 3 a β ⎤ ⎡ 0 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − = 15 5 0 B ⇒ 20 September 2011 86 ⎦ ⎣ © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 87. State State- -space equations in controllable canonical form space equations in controllable canonical form Consider a system described by the differential equation: Consider a system described by the differential equation: = + + + + − − − ) ( ) ( ) ( ) ( 1 1 1 1 0 t y a dt t dy a dt t y d a dt t y d a n n n n n n L dt dt dt ) ( ) ( ) ( ) ( 1 1 1 1 0 t u b dt t du b dt t u d b dt t u d b m m m m m m + + + + − − − L or equivalently by the transfer function: m m m m b s b s b s b G + + + + − 1 1 1 0 ... ) ( n n n n m m a s a s a s a s G + + + + = − − − 1 1 1 0 1 1 0 ... ) ( Th t ll bl i l t t ti f th t i t d The controllable canonical state equations of the system is presented in the next slide. 20 September 2011 87 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 88. State State- -space equations in controllable canonical form (cont’) space equations in controllable canonical form (cont’) ⎧ ) ( ) ( ) ( t t t B A & ⎩ ⎨ ⎧ = + = ) ( ) ( ) ( ) ( ) ( t t y t r t t Cx B Ax x State equations: Where: ⎥ ⎤ ⎢ ⎡ 0 0 1 0 K ⎥ ⎤ ⎢ ⎡0 ⎤ ⎡ ) (t ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ = 0 1 0 0 M M M M K A ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ = 0 M B ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ = ) ( ) ( ) ( 2 1 t x t x t M x ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎣ − − − − − − 0 1 0 2 0 1 0 1 0 0 0 a a a a a a a a n n n K K ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎣1 0 ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ ) (t xn M ⎦ ⎣ 0 0 0 0 ⎥ ⎤ ⎢ ⎡ = − 0 0 0 1 b b b m m C 20 September 2011 88 ⎥ ⎦ ⎢ ⎣ 0 0 0 0 0 K K a a a C © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 89. State State- -space equations in controllable canonical form (cont’) space equations in controllable canonical form (cont’) Write the controllable canonical state equations of the following system: Write the controllable canonical state equations of the following system: ) ( 3 ) ( ) ( 4 ) ( 5 ) ( ) ( 2 t u t u t y t y t y t y + = + + + & & & & & & & & S l i Solution: ⎩ ⎨ ⎧ = + = ) ( ) ( ) ( ) ( ) ( t t y t r t t Cx B Ax x ⎤ ⎡ ⎥ ⎤ ⎢ ⎡ 0 1 0 0 1 0 ⎤ ⎡0 where: ⎩ ) ( ) ( y ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ − − − = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ − − − = 5 . 0 5 . 2 2 1 0 0 0 1 0 1 0 0 0 1 0 1 2 3 a a a A ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 1 0 0 B ⎦ ⎣ ⎥ ⎦ ⎢ ⎣ 0 0 0 a a a ⎦ ⎣ [ ] 5 . 0 0 5 . 1 0 1 2 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = b b b C 20 September 2011 89 [ ] 0 0 0 ⎥ ⎦ ⎢ ⎣ a a a © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 90. Method for establishing state equations from block diagrams Method for establishing state equations from block diagrams Example Example Example Example Establish the state equations describing the system below: R(s) + − Y(s) ) 3 )( 1 ( 10 + + s s s Define the state variables as in the block diagram: R(s) Y(s) 10 1 1 X1(s) X2(s) X3(s) ( ) + − Y(s) ) 3 ( 10 + s ) 1 ( 1 + s s 1 X1(s) 2(s) X3(s) 20 September 2011 90 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 91. Method for establishing state equations from block diagrams Method for establishing state equations from block diagrams Example (cont’) Example (cont’) Example (cont ) Example (cont ) From the block diagram, we have: 10 ) ( 3 10 ) ( 2 1 s X s s X + = • ) ( 10 ) ( 3 ) ( 2 1 1 s X s X s sX = + ⇒ ) ( 10 ) ( 3 ) ( 2 1 1 t x t x t x + − = ⇒ & (1) ) ( 1 1 ) ( 3 2 s X s X = • ) ( ) ( ) ( 3 2 2 s X s X s sX = + ⇒ ) ( 1 ) ( 3 2 s + ) ( ) ( ) ( 3 2 2 ) ( ) ( ) ( 3 2 2 t x t x t x + − = ⇒ & (2) ( ) ) ( ) ( 1 ) ( 3 s Y s R s s X − = • ) ( ) ( ) ( 1 3 s X s R s sX − = ⇒ ) ( ) ( ) ( & (3) 20 September 2011 91 ) ( ) ( ) ( 1 3 t r t x t x + − = ⇒ & (3) © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 92. Method for establishing state equations from block diagrams Method for establishing state equations from block diagrams Example (cont’) Example (cont’) Example (cont ) Example (cont ) Combining (1), (2), and (3) leads to the state equations: ) ( 0 0 ) ( ) ( 1 1 0 0 10 3 ) ( ) ( 2 1 2 1 t r t x t x t x t x ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ + ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ − − = ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎡ & & { 1 ) ( ) ( 0 0 1 ) ( ) ( 3 3 t t x t t x B x A x ⎥ ⎦ ⎢ ⎣ ⎥ ⎦ ⎢ ⎣ ⎥ ⎦ ⎢ ⎣− ⎥ ⎦ ⎢ ⎣ 3 2 1 4 4 3 4 4 2 1 3 2 1 & & ⎤ ⎡ ) (t x Output equation: [ ] ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = = ) ( ) ( ) ( 0 0 1 ) ( ) ( 3 2 1 1 t x t x t x t x t y 4 3 4 2 1 C 20 September 2011 92 ⎥ ⎦ ⎢ ⎣ ) ( 3 t x C © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 93. State equation to transfer function State equation to transfer function Gi d ib d b h i Given a system described by the state equations: ⎩ ⎨ ⎧ + = ) ( ) ( ) ( ) ( ) ( t t t u t t C B Ax x & ⎩ ⎨ = ) ( ) ( t t y Cx Then the transfer function of the system is: ( ) B A I C 1 ) ( ) ( ) ( − − = = s s U s Y s G 20 September 2011 93 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 94. State equation to transfer function State equation to transfer function – – Example Example Calculate the transfer function of the system described by the state Calculate the transfer function of the system described by the state equation: ⎨ ⎧ + = ) ( ) ( ) ( t u t t B Ax x & ⎩ ⎨ = ) ( ) ( t t y Cx ⎤ ⎡ where ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − − = 3 2 1 0 A ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 1 3 B [ ] 0 1 = C Solution: The transfer function of the system is: ( ) B A I C 1 ) ( ) ( ) ( − − = = s s U s Y s G 20 September 2011 94 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 95. Calculate transfer functions from state equations Calculate transfer functions from state equations Example (cont’) Example (cont’) Example (cont ) Example (cont ) ( ) ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = − 3 2 1 3 2 1 0 1 0 0 1 s s s s A I ⎦ ⎣ + ⎦ ⎣ − − ⎦ ⎣ 3 2 3 2 1 0 s ( ) ⎥ ⎤ ⎢ ⎡ + = ⎥ ⎤ ⎢ ⎡ − = − − − s s s 1 3 1 1 1 1 A I ( ) ⎥ ⎦ ⎢ ⎣ − − − + ⎥ ⎦ ⎢ ⎣ + s s s s 2 ) 1 .( 2 ) 3 ( 3 2 ( ) [ ] [ ] 1 3 1 1 3 0 1 1 1 + = ⎥ ⎤ ⎢ ⎡ + = − − s s s A I C( ) [ ] [ ] 1 3 2 3 2 0 1 2 3 2 2 + + + = ⎥ ⎦ ⎢ ⎣ − + + = s s s s s s s A I C ( ) [ ] 1 ) 3 ( 3 3 1 3 1 1 + + = ⎥ ⎤ ⎢ ⎡ + = − − s s s B A I C( ) [ ] 2 3 1 1 3 2 3 2 2 + + = ⎥ ⎦ ⎢ ⎣ + + + = − s s s s s s B A I C 10 3 ) ( + s s G ⇒ 20 September 2011 95 2 3 ) ( 2 + + = s s s G ⇒ © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 96. Solution to state equations Solution to state equations Solution to the state equation ? ) ( ) ( ) ( t u t t B Ax x + = & ∫ − Φ + Φ = + t d u t t t ) ( ) ( ) 0 ( ) ( ) ( τ τ τ B x x ∫ − Φ + Φ = d u t t t 0 ) ( ) ( ) 0 ( ) ( ) ( τ τ τ B x x )] ( [ ) ( 1 s t Φ = Φ − L where transient matrix )] ( [ ) ( s t Φ = Φ L 1 ) ( ) ( − − = Φ A I s s where transient matrix System response? ) ( ) ( t t y Cx = 20 September 2011 96 Example: © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 97. Relationship between the mathematical models Relationship between the mathematical models Diff. equation L L -1 Define x Transfer function State equation ( ) B A I C 1 ) ( − − = s s G 20 September 2011 97 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 98. Linearized models of nonlinear systems Linearized models of nonlinear systems 20 September 2011 98 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 99. N li t d t ti f th iti i i l d Nonlinear systems Nonlinear systems Nonlinear systems do not satisfy the superposition principle and cannot be described by a linear differential equation. Most of the practical systems are nonlinear: Most of the practical systems are nonlinear: Fluid system (Ex: liquid tank,…) Thermal system (Ex: furnace,…) Mechanical system (Ex: robot arm,….) Electro-magnetic system (TD: motor,…) Hybrid system ,… y y , 20 September 2011 99 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 100. Mathematical model of nonlinear systems: Mathematical model of nonlinear systems: Input output relationship of a continuous nonlinear system can be Input – output relationship of a continuous nonlinear system can be expressed in the form of a nonlinear differential equations. ⎞ ⎛ − ) ( ) ( ) ( ) ( ) ( 1 t d t d t d t d t d m n n ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − ) ( , ) ( , , ) ( ), ( , ) ( , , ) ( ) ( 1 1 t u dt t du dt t u d t y dt t dy dt t y d g dt t y d m m n n n n L L where: u(t): input signal, y(t): output signal, g(.): nonlinear function g(.): nonlinear function 20 September 2011 100 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 101. Nonlinear system Nonlinear system – – Example 1 Example 1 f th di h l a: cross area of the dischage valve. A: cross area of the tank g: gravity acceleration ( ) u(t) qin k: constant CD: discharge constant y(t) qout Balance equation: ) ( ) ( ) ( t q t q t y A out in − = & ) ( ) ( t ku t qi = where: ) ( ) ( t ku t qin ) ( 2 ) ( t gy aC t q D out = where: ⇒ (first order li ) ( ) ) ( 2 ) ( 1 ) ( t gy aC t ku t y D − = & 20 September 2011 101 ⇒ nonlinear system ) ( ) ) ( 2 ) ( ) ( t gy aC t ku A t y D = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 102. Nonlinear system Nonlinear system – – Example 2 Example 2 J: moment inertia of the robot arm J: moment inertia of the robot arm M: mass of the robot arm m: object mass l l h f b l l: length of robot arm lC : distance from center of gravity to rotary axis B: friction constant m u θ g: gravitational acceleration u(t): input torque θ(t): robot arm angle θ(t): robot arm angle According to Newton’s Law ) ( cos ) ( ) ( ) ( ) ( 2 t u g Ml ml t B t ml J C = + + + + θ θ θ & & & C ⇒ ) ( ) ( 1 cos ) ( ) ( ) ( ) ( ) ( 2 2 2 t u ml J g ml J Ml ml t ml J B t C + + + + − + − = θ θ θ & & & 20 September 2011 102 (second order nonlinear system) © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 103. Nonlinear system Nonlinear system – – Example 3 Example 3 δ: steering angle ψ: ship angle k: constant τi: constant ψ(t) δ(t) Moving direction ψ( ) The differential equation describing the steering dynamic of a ship: ( ) ( ) ) ( ) ( ) ( ) ( 1 ) ( 1 1 ) ( 3 3 t t k t t t t δ δ τ ψ ψ ψ ψ + ⎟ ⎟ ⎞ ⎜ ⎜ ⎛ + + ⎟ ⎟ ⎞ ⎜ ⎜ ⎛ − ⎟ ⎟ ⎞ ⎜ ⎜ ⎛ + − = & & & & & & & & (third order nonlinear system) ( ) ( ) ) ( ) ( ) ( ) ( ) ( ) ( 3 2 1 2 1 2 1 t t t t t t δ δ τ τ τ ψ ψ τ τ ψ τ τ ψ + ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ + + ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ ⎟ ⎟ ⎠ ⎜ ⎜ ⎝ + 20 September 2011 103 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 104. Describing nonlinear system by state equations Describing nonlinear system by state equations A continuous nonlinear system can be described by the state equation: ⎧ = )) ( ) ( ( ) ( t u t t x f x & ⎩ ⎨ ⎧ = = )) ( ), ( ( ) ( )) ( ), ( ( ) ( t u t h t y t u t t x x f x where: u(t): input, y(t): output, x(t): state vector, x(t) = [x1(t), x2(t),…,xn(t)]T f(.), h(.): nonlinear functions 20 September 2011 104 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 105. State State- -space model of nonlinear system space model of nonlinear system – – Example 1 Example 1 Differential equation: Differential equation: ( ) u(t) qin ( ) ) ( 2 ) ( 1 ) ( t gy aC t ku A t y D − = & Define the state variable: ) ( ) ( y(t) qout A ) ( ) ( 1 t y t x = S i ⎨ ⎧ = )) ( ), ( ( ) ( t u t t x f x & State equation: ⎩ ⎨ ⎧ = )) ( ), ( ( ) ( )) ( ), ( ( ) ( t u t h t y x f ) ( 2 k t gx aC ) ( ) ( 2 ) , ( 1 t u A k A t gx aC u D + − = x f ) ( )) ( ), ( ( 1 t x t u t h = x trong ñoù: 20 September 2011 105 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 106. State State- -space model of nonlinear system space model of nonlinear system – – Example 2 Example 2 Differential equation: Differential equation: m l ) ( ) ( 1 cos ) ( ) ( ) ( ) ( ) ( 2 2 2 t u ml J g ml J Ml ml t ml J B t C + + + + − + − = θ θ θ & & & Define the state variable: ⎩ ⎨ ⎧ = = ) ( ) ( ) ( ) ( 2 1 t t x t t x θ θ & u θ ⎩ = ) ( ) ( 2 t t x θ State equation: ⎩ ⎨ ⎧ = )) ( ) ( ( ) ( )) ( ), ( ( ) ( h t u t t x f x & q ⎩ ⎨ = )) ( ), ( ( ) ( t u t h t y x ⎥ ⎤ ⎢ ⎡ ) ( 2 t x where ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ + + + − + + − = ) ( ) ( 1 ) ( ) ( ) ( cos ) ( ) ( ) , ( 2 2 2 1 2 t u ml J t x ml J B t x ml J g Ml ml u C x f 20 September 2011 106 ) ( )) ( ), ( ( 1 t x t u t h = x © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 107. Equilibrium points of a nonlinear system Equilibrium points of a nonlinear system Consider a nonlinear system described by the diff. equation: ⎩ ⎨ ⎧ = = )) ( ), ( ( ) ( )) ( ), ( ( ) ( t u t h t y t u t t x x f x & Consider a nonlinear system described by the diff. equation: ⎩ )) ( ), ( ( ) ( t u t h t y x The state is called the equilibrium point of the nonlinear system if the system is at the state and the control signal is fixed at then x x u If is equilibrium point of the nonlinear system then: ) ( u x the system is at the state and the control signal is fixed at then the system will stay at state forever. x u x If is equilibrium point of the nonlinear system then: ) , ( u x 0 )) ( ), ( ( , = = = u u t u t x x x f The equilibrium point is also called the stationary point of the nonlinear system. 20 September 2011 107 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 108. Equilibrium point of nonlinear system Equilibrium point of nonlinear system – – Example 1 Example 1 Consider a nonlinear system described by the state equation: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + + = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ) ( 2 ) ( ) ( ). ( ) ( ) ( 2 1 2 1 2 1 t x t x u t x t x t x t x & & Consider a nonlinear system described by the state equation: ⎦ ⎣ ⎦ ⎣ ) ( ) ( ) ( 2 1 2 Find the equilibrium point when 1 ) ( = = u t u Solution: 0 )) ( ), ( ( , = = = u u t u t x x x f The equilibrium point(s) are the solution to the equation: , u u x x ⎩ ⎨ ⎧ = + = + 0 2 0 1 . 2 1 2 1 x x x x ⇔ ⎩ 2 1 ⎪ ⎪ ⎨ ⎧ = 2 2 1 x ⎪ ⎪ ⎨ ⎧ − = 2 2 1 x ⇔ or 20 September 2011 108 ⎪ ⎩ ⎨ − = 2 2 x ⎪ ⎩ ⎨ + = 2 2 x © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 109. Equilibrium point of nonlinear system Equilibrium point of nonlinear system – – Example 2 Example 2 Consider a nonlinear system described by the state equation: ⎥ ⎤ ⎢ ⎡ + + + ⎤ ⎡ u x x x 2 3 2 2 1 1 & ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ + − + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ u x x x x x x x 2 3 3 1 3 3 2 3 2 1 ) sin( & & Find the equilibrium point when 0 ) ( = = u t u 1 x y = 20 September 2011 109 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 110. Linearized model of a nonlinear system around an equilibrium point Linearized model of a nonlinear system around an equilibrium point Consider a nonlinear system described by the differential equation: ⎩ ⎨ ⎧ = = )) ( ) ( ( ) ( )) ( ), ( ( ) ( t u t h t y t u t t x x f x & Consider a nonlinear system described by the differential equation: (1) ⎩ = )) ( ), ( ( ) ( t u t h t y x Expanding Taylor series for f(x,u) and h(x,u) around the equilibrium p g y f( ) ( ) q point , we can approximate the nonlinear system (1) by the following linearized state equation: ⎧ ) , ( u x ⎩ ⎨ ⎧ + = + = ) ( ~ ) ( ~ ) ( ~ ) ( ~ ) ( ~ ) ( ~ t u t t y t u t t D x C B x A x & (2) where: u t u t u t t − = − = ) ( ) ( ~ ) ( ) ( ~ x x x 20 September 2011 110 y t y t y − = ) ( ) ( ~ ) ( ) ( )) , ( ( u h y x = © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 111. Linearized model of a nonlinear system around an equilibrium point Linearized model of a nonlinear system around an equilibrium point Th i f h li i d i l l d f ll 1 1 1 f f f ⎥ ⎤ ⎢ ⎡ ∂ ∂ ∂ 1 f ⎤ ⎡ ∂ The matrix of the linearized state equation are calculated as follow: 2 2 2 1 2 1 1 1 n x f x f x f x f x f x f A ⎥ ⎥ ⎥ ⎥ ⎢ ⎢ ⎢ ⎢ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ = L L 2 1 u f u f B ⎥ ⎥ ⎥ ⎥ ⎤ ⎢ ⎢ ⎢ ⎢ ⎡ ∂ ∂ ∂ = 2 1 n n n n x f x f x f x x x A ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎣ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ K M O M M ) ( u n u f u , x ⎥ ⎥ ⎥ ⎥ ⎦ ⎢ ⎢ ⎢ ⎢ ⎣ ∂ ∂ ∂ M ) ( 2 1 u n x x x , x ⎦ ⎣ ∂ ∂ ∂ ⎤ ⎡ h⎤ ⎡∂ ) ( 2 1 u n x h x h x h , x C ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∂ ∂ ∂ ∂ ∂ ∂ = K ) ( u u h , x D ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ∂ ∂ = 20 September 2011 111 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 112. Linearized state Linearized state- -space model space model – – Example 1 Example 1 The parameter of the tank: u(t) qin The parameter of the tank: 3 2 2 8 0 / 150 100 , 1 C V k cm A cm a = = y(t) u(t) qout 2 3 sec / 981 8 . 0 , . sec / 150 cm g C V cm k D = = = Nonlinear state equation: ⎩ ⎨ ⎧ = = )) ( ) ( ( ) ( )) ( ), ( ( ) ( t u t h t y t u t t x x f x & ⎩ = )) ( ), ( ( ) ( t u t h t y x ) ( 9465 0 ) ( 3544 0 ) ( ) ( 2 ) ( 1 k t gx aCD f where ) ( 9465 . 0 ) ( 3544 . 0 ) ( ) ( ) , ( 1 1 t u t x t u A k A g u D + − = + − = x f ) ( )) ( ), ( ( 1 t x t u t h = x 20 September 2011 112 ) ( )) ( ), ( ( 1 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 113. Linearized state Linearized state- -space model space model – – Example 1 (cont’) Example 1 (cont’) Linearize the system around y = 20cm: Linearize the system around y = 20cm: The equilibrium point: 20 20 1 = x 0 5 . 1 3544 . 0 ) , ( 1 = + − = u x u x f 9465 . 0 = u ⇒ 20 September 2011 113 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 114. Linearized state Linearized state- -space model space model – – Example 1 (cont’) Example 1 (cont’) The matri of the lineari ed state space model: 0396 . 0 2 2 1 − = − = ∂ ∂ = D x A g aC x f A 5 . 1 1 = = ∂ ∂ = A k u f B The matrix of the linearized state-space model: 2 ) ( 1 ) ( 1 ∂ u u x A x , x , x ) ( ) ( ∂ u u A u , x , x 1 = ∂ = h C 0 = ∂ ∂ = h D ) ( 1 ∂ u x , x ) ( ∂ u u , x The lineari ed state eq ation describing the s stem aro nd the The linearized state equation describing the system around the equilibrium point y=20cm is: ⎧ + = ) ( ~ 5 1 ) ( ~ 0396 0 ) ( ~ t u t t x x & ) ( ) ( 2 ) , ( 1 t u A k A t gx aC u D + − = x f ⎩ ⎨ ⎧ = + − = ) ( ~ ) ( ~ ) ( 5 . 1 ) ( 0396 . 0 ) ( t t y t u t t x x x 20 September 2011 114 ) ( )) ( ), ( ( 1 t x t u t h = x © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 115. Linearized state Linearized state- -space model space model – – Example 2 Example 2 The parameters of the robot: The parameters of the robot: 2 C 02 0 5 0 1 . 0 , 2 . 0 , 5 . 0 m kg J kg M kg m m l m l = = = m l 2 sec / 81 . 9 , 005 . 0 . 02 . 0 , 5 . 0 m g B m kg J kg M = = = = u θ Nonlinear state equation : ⎩ ⎨ ⎧ = = )) ( ), ( ( ) ( )) ( ), ( ( ) ( t u t h t y t u t t x x f x & ⎩ )) ( ), ( ( ) ( t u t h t y x ⎥ ⎤ ⎢ ⎡ ) ( 2 t x where: ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ + + + − + + − = ) ( ) ( 1 ) ( ) ( ) ( cos ) ( ) ( ) , ( 2 2 2 1 2 t u ml J t x ml J B t x ml J g Ml ml u C x f 20 September 2011 115 ) ( )) ( ), ( ( 1 t x t u t h = x © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 116. Linearized state Linearized state- -space model space model – – Example 2 (cont’) Example 2 (cont’) Linearize the system around the equilibrium point y = π/6 (rad): Linearize the system around the equilibrium point y = π/6 (rad): Finding the equilibrium point: 6 / 1 π = x 0 1 ) ( ) ( 2 ⎥ ⎤ ⎢ ⎡ + B g Ml ml x u x f ⇒ ⎨ ⎧ = 0 2 x 0 ) ( 1 ) ( cos ) ( ) ( ) , ( 2 2 2 1 2 = ⎥ ⎥ ⎦ ⎢ ⎢ ⎣ + + + − + + − = u ml J x ml J B x ml J g Ml ml u C x f ⇒ ⎩ ⎨ = 2744 . 1 u Then the equilibrium point is: ⎥ ⎤ ⎢ ⎡ = ⎥ ⎤ ⎢ ⎡ = 6 / 1 π x x ⎥ ⎦ ⎢ ⎣ ⎥ ⎦ ⎢ ⎣ 0 2 x x 2744 . 1 = u 20 September 2011 116 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 117. Linearized state Linearized state- -space model space model – – Example 2 (cont’) Example 2 (cont’) The s stem matri aro nd the eq ilibri m point: The system matrix around the equilibrium point: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 12 11 a a a a A 0 ) ( 1 1 11 = ∂ ∂ = u x f a , x 1 ) ( 2 1 12 = ∂ ∂ = x f a ⎦ ⎣ 22 21 a a ) ( u , x ) ( 1 2 ) ( 1 2 21 ) ( sin ) ( ) ( C t x ml J Ml ml x f a + + = ∂ ∂ = ) ( 2 u , x ) ( ) ( 1 ) ( u u ml J , x , x 2 2 22 ) ( ml J B x f a + − = ∂ ∂ = ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ + = ) ( 1 ) ( ) ( ) ( ) ( ) ( 2 B g Ml ml t x u C x f ) ( ) ( 2 ) ( u u ml J x , x , x + ∂ 20 September 2011 117 ⎥ ⎦ ⎢ ⎣ + + + − + + − ) ( ) ( 1 ) ( ) ( ) ( cos ) ( ) ( ) , ( 2 2 2 1 2 t u ml J t x ml J B t x ml J g Ml ml u C x f © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 118. Linearized state Linearized state- -space model space model – – Example 2 (cont’) Example 2 (cont’) The inp t matri aro nd the eq ilibri m point: The input matrix around the equilibrium point: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ = 1 b b B 0 1 ∂f b ⎦ ⎣ 2 b 0 ) ( 1 1 = ∂ = u u b , x 1 f ∂ 2 ) ( 2 2 1 ml J u f b u + = ∂ ∂ = , x ⎥ ⎥ ⎤ ⎢ ⎢ ⎡ + = ) ( 1 ) ( ) ( ) ( ) ( ) ( 2 B g Ml ml t x u C x f 20 September 2011 118 ⎥ ⎦ ⎢ ⎣ + + + − + + − ) ( ) ( 1 ) ( ) ( ) ( cos ) ( ) ( ) , ( 2 2 2 1 2 t u ml J t x ml J B t x ml J g Ml ml u C x f © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 119. Linearized state Linearized state- -space model space model – – Example 2 (cont’) Example 2 (cont’) The output matrix around the equilibrium point: The output matrix around the equilibrium point: 1 1 1 = ∂ ∂ = x h c [ ] 2 1 c c = C 0 ) ( 2 2 = ∂ ∂ = x h c ) ( 1 ∂ u x , x ) ( 2 u , x 1 d = D 0 ) ( 1 = ∂ ∂ = u h d ) ( ∂ u u , x Then the linearized state equation is: ⎩ ⎨ ⎧ + = ) ( ~ ) ( ~ ) ( ~ ) ( ~ ) ( ~ ) ( ~ t t t t u t t D C B x A x & ⎩ + = ) ( ) ( ) ( t u t t y D x C ⎥ ⎤ ⎢ ⎡ = 1 0 A ⎥ ⎤ ⎢ ⎡ = 0 B [ ] 0 1 = C 0 = D ⎥ ⎦ ⎢ ⎣ = 22 21 a a A ⎥ ⎦ ⎢ ⎣ = 2 b B [ ] 0 1 C 0 = D 20 September 2011 119 ) ( ) , ( 1 t x u h = x © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/
- 120. Regulating nonlinear system around equilibrium point Regulating nonlinear system around equilibrium point Drive the nonlinear system to the neighbor of the equilibrium point Drive the nonlinear system to the neighbor of the equilibrium point (the simplest way is to use an ON-OFF controller) Around the equilibrium point, use a linear controller to maintain the Around the equilibrium point, use a linear controller to maintain the system around the equilibrium point. Linear r(t) Nonlinear system + − y(t) Linear control u(t) e(t) system ON-OFF Mode select 20 September 2011 120 © H. T. Hoang - www4.hcmut.edu.vn/~hthoang/