Lecture introduction to control systems chapter 6 design of continuous control systems (dr huynh thai hoang)

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/ November 2012 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Chapter DESIGN OF CONTINUOUS CONTROL SYSTEMS November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Content        Introduction Effect of controllers on system performance Control systems design using the root locus method Control systems design in the frequency domain Design g of PID controllers Control systems design in state-space Design of state estimators November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Introduction November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Introduction to design process  Design is a process of adding/configuring hardware as well as software in a system so that the new system satisfies the d i d specifications desired ifi ti November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Series compensator  The controller is connected in series with the plant plant R(s) + GC(s) G(s) Y(s)  Controllers: phase lead, phase lag, lead-lag compensator, P, PD PI, PD, PI PID,… PID  Design method: root locus, locus frequency response November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ State feedback control  All the states of the system are fed back to calculate the control rule r(t) +  u(t) x (t )  Ax (t )  Bu (t ) x(t) C y(t) K  State feedback controller: u (t )  r (t )  Kx (t ) K  k1 k2  kn   Design method: pole placement, placement LQR,… LQR November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Effects of controller on system performance November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Effects of the addition of poles  The addition of a pole (in the left left-half half ss-plane) plane) to the open openloop transfer function has the effect of pushing the root locus to the right, tending to lower the system’s relative stability and t slow to l d down th settling the ttli off the th response Im s Re s November 2012 Im s Re s © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Im s Re s Effects of the addition of zeros  The addition of a zero (in the left left-half half ss-plane) plane) to the open openloop transfer function has the effect of pulling the root locus to the left, tending to make the system more stable and to speed up the th settling ttli off the th response November 2012 Im s Im s Im s Re s Re s Re s © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 10 Pole placement method If the system is controllable, controllable then it is possible to determine the feedback gain K so that the closed-loop system has the poles at any location  Step 1: Write the characteristic equation of the closed-loop system ((1)) det[ sI  A  BK ]   Step 2: Write the desired characteristic equation: n  ( s  pi )  (2) i 1 th desired d i d poles l pi , (i  1, n) are the  Step 3: Balance the coefficients of the equations (1) and (2), we can find the state feedback gain K November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 89 Pole placement method – Example  Problem: Given a system described by the state state-state state equation:  x (t )  Ax(t )  Bu (t )   y (t )  Cx(t ) 0 0  0     B  A 0     1    3  C  0 1 Determine the state feedback controller u (t )  r (t )  Kx(t ) so that the closed-loop system has complex poles with   0,6;n  10 and the third pole at 20 November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 90 Pole placement method – Example (cont’)  Solution  The characteristic equation of the closed-loop system: det[ sI  A  BK ]   0   1 0     det d t  s 0   0   3k1        0 1    3 1        k2   k3      s  (3  3k  k3 ) s  (7  3k1  10k  21k3 ) s  (4  10k1  12k3 )  (1)  The desired characteristic equation: ( s  20)( s  2 n s   n2 )   s  32 s  340 s  2000  November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ (2) 91 Pole placement method – Example (cont’)  B l Balance th coefficients the ffi i t off the th equations ti (1) and d (2), (2) we have: h 3  3k  k3  32  7  3k1  10k  21k3  340 4  10k  12k  2000   Solve the above set of equations, we have: k1  220,578  k  3,839 k  17,482   Conclusion: November 2012 K  220,578 3,839 17,482  © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 92 D Design i off state t t estimators ti t November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 93 The concept of state estimation  To be able to implement state feedback control system, system it is required to measure all the states of the system  However, However in some application, application we can only measure the output, but cannot measure the states of the system problem is to estimate the states of the system y from the  The p output measurement  State estimator (or state observer) November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 94 Observability    x (t )  Ax(t )  Bu (t ) Consider a system:   y (t )  Cx(t ) The system y is complete p state observable if g given the control law u(t) and the output signal y(t) in a finite time interval t0  t tf , it is possible to determine the initial states x(t0) Qualitatively the system is state observable if all state variable Qualitatively, x(t) influences the output y(t) y(t) Signal flow graph of an incomplete state observable system November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 95 Observability condition   x (t )  Ax(t )  Bu (t )   y (t )  Cx(t ) It is necessaryy to estimate the state xˆ (t ) from mathematical model of the system and the input-output data S t System  C   CA    O   CA       CAn 1   Observability matrix:  The necessary and sufficient condition for the observability is: rank (O )  n November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 96 Observability – Example   x (t )  Ax(t )  Bu (t ) Consider the system   y (t )  Cx(t ) 1 0 1  C  1 3 where: A B     3  2 Evaluate the observability of the system  Solution: Observability matrix:  Because C  O   CA  3 1 O       det(O )  10  rank (O )   The system is observable November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 97 State estimator r(t)  u(t) x (t )  Ax (t )  Bu (t ) x(t) C + L B ++ +  y(t)  xˆ (t ) C yˆ (t ) A K  State estimator: where: November 2012  xˆ (t )  Axˆ (t )  Bu (t )  L( y (t )  yˆ (t ))   yˆ (t )  Cxˆ (t ) T L  [l1 l2  ln ] © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 98 Design of state estimators   Requirements:  The state estimator must be stable, estimation error should approach to zero  Dynamic response of the state estimator should be fast enough in comparison with the dynamic response of the control loop It is required to chose L satisfying:  All the roots of the equation det( sI  A  LC )  locates i the in th h half-left lf l ft s-plane l  The roots of the equation det( sI  A  LC )  are further from the imaginary axis than the roots of the equation det( sI  A  BK )   Depending on the design of L, we have different state estimator:  Luenberger state observer  Kalman filter November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 99 Procedure for designing the Luenberger state observer  St 1: Step Write W it the th characteristic h t i ti equation ti off the th state t t observer b det[ sI  A  LC ]   (1) Step 1: Write the desired characteristic equation: n  ( s  pi )  (2) i 1 pi , (i  1, n) are the desired poles of the state estimator  Step 3: Balance the coefficients of the characteristic equations (1) and (2), we can find the gain L November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 100 Design of state estimators – Example  Problem: Given a system described by the state equation:  x (t )  Ax(t )  Bu(t )  C (t )  y (t )  Cx 0 0   A 0      3  0  B  3   1 C  1 0 Assuming that the states of the system cannot be directly measured Design the Luenberger state estimator so that the poles of the state estimator lying p y g at 20,, 20 and 50 November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 101 Design of state estimators – Example (cont’)  Solution  The characteristic equation of the Luenberger state estimator: det[ sI  A  LC ]    1 0     l1          d t  s 0    det   l2 1 0    0 1    3 l      3     s  (l1  3) s  (3l1  l2  7) s  (7l1  5l2  l3  4)   (1) The desired characteristic equation: ( s  20) ( s  50)   s  90s  2400s  20000  November 2012 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ (2) 102 Design of state estimators – Example (cont’)  B l Balancing i the th coefficients ffi i t off the th equ (1) and d (2) leads l d to: t l1   90  3l1  l2   2400 7l  3l  l   20000   Solve the above set of equations, we have: l1  87  l2  2132 l  12991 3  Conclusion November 2012 L  87 2132 12991 T © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 103 ... Control systems design using the root locus method Control systems design in the frequency domain Design g of PID controllers Control systems design in state-space Design of state estimators November.. .Chapter DESIGN OF CONTINUOUS CONTROL SYSTEMS November 2012 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Content        Introduction Effect of controllers on system performance Control systems. .. 0, 16 T2  Calculate the pole of the lag compensator: 1    ((0, 16) T2 T2 16  November 2012  0.01 T2 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 48 Example of designing a lead lag compensator