Lecture Notes 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/ Chapter MATHEMATICAL MODELS OF CONTINUOUS CONTROL SYSTEMS 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Content The concept of mathematical model Transfer function Block diagram algebra Signal flow graph State space equation Linearized models of nonlinear systems Nonlinear state equation Linearized state equation 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ The concept of mathematical models 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Question If you design a control system, system what you need to know about the plant or the process to be control? What are the advantages of mathematical models? 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 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 ddifferent e e t type of o control co t o systems syste s ⇒ Mathematics at e at cs The relationship between input and output of a LTI system of can be described byy linear constant coefficient equations: q u(t) Linear TimeI Invariant i t System S t y(t) d n y (t ) d n −1 y (t ) dy (t ) d mu (t ) d m −1u (t ) du (t ) L a0 + a + + a + a y ( t ) = b + b + + b + bmu (t ) L n −1 n m −1 n n −1 m m −1 dt dt dt dt dt dt n: system order, for proper systems: n≥m ai, bi: parameter of the system 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 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 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Example: Car suspension d y (t ) dy (t ) M +B + Ky (t ) = f (t ) dt dt M: equivalent mass B friction constant, K spring stiffness f(t): external force: input y(t): travel of the car body: output 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Example: Elevator d y (t ) dyy (t ) ML +B + M T g = Kτ (t ) + M B g dt dt ML Cabin & load MB Counterbalance ML: mass off cabin bi andd load, l d MB: counterbalance b l B friction constant, K gear box constant τ((t): ) drivingg moment of the motor: input p y(t): position of the cabin: output 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 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 ) + + + + an y (t ) = L a0 a a n −1 n n −1 dt dt dt d mu (t ) d m −1u (t ) du (t ) b0 + b1 + L + bm −1 + bmu (t ) m m −1 dt dt dt 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 analysis and design of control systems easier: transfer function state space equation 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 10 State space model of nonlinear system – Example State Differential equation: l u m θ&&(t ) = − θ B &(t ) − (ml + MlC ) g cosθ + θ u (t ) 2 ( J + ml ) ( J + ml ) ( J + ml ) Define the state variable: State equation: q where ⎧ x1 (t ) = θ (t ) ⎨ & ⎩ x2 (t ) = θ (t ) ⎧ x& (t ) = f ( x (t ), u (t )) ⎨ ) u (t )) ⎩ y (t ) = h( x (t ), ⎡ x2 (t ) ⎤ ⎥ B f ( x , u ) = ⎢ ( ml + MlC ) g x2 (t ) + u (t ) ⎥ cos x1 (t ) − ⎢− 2 ( J + ml ) ( J + ml ) ⎣ ( J + ml ) ⎦ h( x (t ), u (t )) = x1 (t ) 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 106 Equilibrium points of a nonlinear system Consider a nonlinear system described by the diff equation: ⎧ x& (t ) = f ( x (t ), u (t )) ⎨ ⎩ y (t ) = h( x (t ), u (t )) The state x is called the equilibrium point of the nonlinear system if the system is at the state x and the control signal is fixed at u then the system will stay at state x forever If ( x , u ) is equilibrium point of the nonlinear system then: f ( x (t ), u (t )) x = x ,u =u = The equilibrium point is also called the stationary point of the nonlinear system 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 107 Equilibrium point of nonlinear system – Example Consider a nonlinear system described by the state equation: ⎡ x&1 (t ) ⎤ ⎡ x1 (t ).x2 (t ) + u ⎤ ⎢ x& (t )⎥ = ⎢ x (t ) + x (t ) ⎥ ⎣ ⎦ ⎣ ⎦ Find the equilibrium point when u (t ) = u = Solution: The equilibrium point(s) are the solution to the equation: f ( x (t ), u (t )) x = x ,u =u = ⇔ ⎧ x1.x2 + = ⎨ ⎩ x1 + x2 = ⇔ ⎧ x1 = ⎪ ⎨ ⎪⎩ x2 = − 20 September 2011 or ⎧ x1 = − ⎪ ⎨ ⎪⎩ x2 = + © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 108 Equilibrium point of nonlinear system – Example Consider a nonlinear system described by the state equation: 2 ⎡ x&1 ⎤ ⎡ + x2 + x3 + u ⎤ ⎢ x& ⎥ = ⎢ x3 + sin( x1 − x3 )⎥ ⎥ ⎢⎣ x&3 ⎥⎦ ⎢ x u + ⎣ ⎦ y = x1 Find the equilibrium point when u (t ) = u = 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 109 Linearized model of a nonlinear system around an equilibrium point Consider a nonlinear system described by the differential equation: ⎧ x& (t ) = f ( x (t ), u (t )) ⎨ ) u (t )) ⎩ y (t ) = h( x (t ), (1) Expanding p g Taylor y series for f( f(x,u)) and h(x,u) ( ) around the equilibrium q point ( x , u ) , we can approximate the nonlinear system (1) by the following linearized state equation: ⎧ x~& (t ) = Ax~(t ) + Bu~(t ) ⎨~ ~(t ) + Du~(t ) y ( t ) = C x ⎩ where: x~ (t ) = x (t ) − x u~ (t ) = u (t ) − u ~ y (t ) = y (t ) − y 20 September 2011 (2) ( y = h( x , u )) © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 110 Linearized model of a nonlinear system around an equilibrium point Th matrix The i off the h linearized li i d state equation i are calculated l l d as follow: f ll ⎡ ∂f1 ⎢ ∂x ⎢ ⎢ ∂f A = ⎢ ∂x1 ⎢ M ⎢ ∂f n ⎢ ⎣ ∂x1 ∂f1 ∂x2 ∂f ∂x2 M ∂f n ∂x2 ⎡ ∂h C=⎢ ⎣ ∂x1 ∂h ∂x2 20 September 2011 ∂f1 ⎤ ∂xn ⎥ ⎥ ∂f ⎥ L ∂xn ⎥ O M ⎥ ∂f n ⎥ K ⎥ ∂xn ⎦ ( x,u ) ⎡ ∂f1 ⎤ ⎢ ∂u ⎥ ⎢ ∂f ⎥ ⎢ B = ∂u ⎥ ⎢ ⎥ ⎢ M ⎥ ⎢ ∂f n ⎥ ⎣⎢ ∂u ⎥⎦ ( x,u ) ∂h ⎤ K ⎥ ∂xn ⎦ ( x,u ) ⎡ ∂h ⎤ D=⎢ ⎥ ⎣ ∂u ⎦ ( x,u ) L © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 111 Linearized state state space model – Example The parameter of the tank: a = 1cm , A = 100cm qin u(t) y(t) Nonlinear state equation: qout k = 150cm3 / sec V , CD = 0.8 g = 981cm / sec2 ⎧ x& (t ) = f ( x (t ), u(t )) ⎨ ) u(t )) ⎩ y (t ) = h( x (t ), where aCD g gx1 (t ) k f ( x, u ) = − + u (t ) = −0.3544 x1 (t ) + 0.9465u (t ) A A h( x (t ), u (t )) = x1 (t ) 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 112 Linearized state state space model – Example (cont’) Linearize the system around y = 20cm: The equilibrium point: x1 = 20 f ( x , u ) = −0.3544 x1 + 1.5u = 20 September 2011 ⇒ u = 0.9465 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 113 Linearized state state space model – Example (cont’) The matrix matri of the linearized lineari ed state-space state space model: A= aC g ∂f1 =− D ∂x1 ( x,u ) A x1 ∂h C= =1 ∂x1 ( x,u ) = −0.0396 ( x,u ) ∂f1 k = = 1.5 B= ∂u ( x,u ) A ( x,u ) D= ∂h =0 ∂u ( x,u ) The linearized lineari ed state equation eq ation describing the system s stem around aro nd the equilibrium point y=20cm is: ⎧~ x& (t ) = −0.0396 ~ x (t ) + 1.5u~aC (t ) gx1 (t ) k D ⎨~ ~ (t ) f ( x , u ) = − + u (t ) y ( t ) = x ⎩ A A h( x (t ), u (t )) = x1 (t ) 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 114 Linearized state state space model – Example The parameters of the robot: l u l = 0.5m, lC = 0.2m, m = 0.1kg m M = 0.5kg , J = 0.02kg.m θ B = 0.005, g = 9.81m / sec2 ⎧ x& (t ) = f ( x (t ), u (t )) Nonlinear state equation : ⎨ ⎩ y (t ) = h( x (t ), u (t )) where: ⎡ x2 (t ) ⎤ ⎥ B f ( x , u ) = ⎢ ( ml + MlC ) g x2 (t ) + u (t ) ⎥ cos x1 (t ) − ⎢− 2 ( J + ml ) ( J + ml ) ⎣ ( J + ml ) ⎦ h( x (t ), u (t )) = x1 (t ) 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 115 Linearized state state space model – Example (cont’) Linearize the system around the equilibrium point y = π/6 (rad): Finding the equilibrium point: x1 = π / ⎡ x2 ⎤ ⎥=0 ⇒ B f ( x , u ) = ⎢ ( ml + MlC ) g cos x1 − x2 + u⎥ ⎢− 2 ( J + ml ) ( J + ml ) ⎦ ⎣ ( J + ml ) ⎧ x2 = ⎨ ⎩u = 1.2744 Then the equilibrium point is: ⎡ x1 ⎤ ⎡π / 6⎤ x=⎢ ⎥=⎢ ⎥ x ⎦ ⎣ 2⎦ ⎣ u = 1.2744 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 116 Linearized state state space model – Example (cont’) The system s stem matrix matri around aro nd the equilibrium eq ilibri m point: ⎡ a11 a12 ⎤ A=⎢ ⎥ ⎣a21 a22 ⎦ a11 = ∂f1 =0 ∂x1 ( x,u ) a12 = a21 = ∂f (ml + MlC ) = sin x1 (t ) ∂x1 ( x,u ) ( J + ml ) ( x,u ) a22 = ∂f ∂x2 =− ( x,u ) ∂f1 ∂x2 =1 ( x,u ) B ( J + ml ) ( x,u ) ⎤ ⎡ x2 (t ) ⎥ B f ( x , u ) = ⎢ ( ml + MlC ) g − cos x ( t ) − x ( t ) + u ( t ) ⎢ ⎥ 2 2 ( ) J ml J + ml ( J + ml ) ( ) + ⎣ ⎦ 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 117 Linearized state state space model – Example (cont’) The input inp t matrix matri around aro nd the equilibrium eq ilibri m point: ⎡ b1 ⎤ B=⎢ ⎥ ⎣b2 ⎦ b1 = ∂f1 =0 ∂u ( x,u ) b2 = ∂f ∂u = ( x,u ) J + ml ⎡ x2 (t ) ⎤ ⎥ B f ( x , u ) = ⎢ ( ml + MlC ) g − cos x ( t ) − x ( t ) + u ( t ) ⎢ ⎥ 2 2 + J + ml ( J + ml ) ( ) ( ) J ml ⎣ ⎦ 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 118 Linearized state state space model – Example (cont’) The output matrix around the equilibrium point: C = [c1 c2 ] D = d1 ∂h c1 = =1 ∂x1 ( x,u ) d1 = c2 = ∂h ∂x2 =0 ( x,u ) ∂h =0 ∂u ( x,u ) ⎧ x~& (t ) = A~ x (t ) + Bu~ (t ) Then the linearized state equation is: ⎨ ~ ~ y ( t ) = C x (t ) + Du~ (t ) ⎩ ⎤ ⎡0 A=⎢ ⎥ a a ⎣ 21 22 ⎦ ⎡0⎤ B=⎢ ⎥ ⎣b2 ⎦ C = [1 0] D=0 h( x , u ) = x1 (t ) 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 119 Regulating nonlinear system around 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 system around the equilibrium point r(t) + e(t) Linear control − u(t) Nonlinear system y(t) ON-OFF Mode select 20 September 2011 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 120 ... RCs 20 Transfer function of some type of controllers (cont’) Passive compensators C Phase lead compensator: R1 R2 KC = R1 + R2 R2 R2 R1C T= R1 + R2 Phase lag compensator : R2 R1 G ( s) = KC 20 .. .Chapter MATHEMATICAL MODELS OF CONTINUOUS CONTROL SYSTEMS 20 September 20 11 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Content The concept of mathematical model Transfer... Derivative controller (PID) ( ) KI G(s) = K P + + KDs s R1C1 + R2C2 KI = − KP = − R1C2 R1C2 K D = − R2C1 20 September 20 11 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ 23 Transfer function of DC motors