Lecture "Fundamentals of control systems - Chapter 8: Analysis of discrete control systems" presentation of content: Stability conditions for discrete systems, extension of Routh - Hurwitz criteria, jury criterion, root locus,... Invite you to reference.
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/ www4 hcmut edu vn/ hthoang/ December 2013 © H T Hoang - www4.hcmut.edu.vn/~hthoang/ Chapter ANALYSIS OF DISCRETE CONTROL SYSTEMS December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Content Stability conditions for discrete systems Extension of Routh-Hurwitz criteria Jury J criterion it i Root locus Steady St d state t t error Performance of discrete systems December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Stability conditions for discrete systems December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Stability conditions for discrete systems A system is defined to be BIBO stable if every bounded input to the system results in a bounded output I s Im Stable Res I z Im Re s | z | Re z z eTs The region of stability for a contin o s system continuous s stem is the left-half s-plane December 2013 Stable The region of stability for a di discrete t system t iis th the interior of the unit circle © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Characteristic equation of discrete systems Discrete systems described by block diagram: R(s) + GC(z) ZOH G(s) T Y(s) H(s) Characteristic equation: GC ( z )GH ( z ) Discrete systems described by the state equation x( k 1) Ad x( k ) Bd r ( k ) y ( k ) Cd x( k ) Characteristic equation: det( zI Ad ) December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ Methods for analysis the stability of discrete systems Algebraic stability criteria The extension of the Routh-Hurwitz criteria Jury’s J ’ stability t bilit criterion it i The root locus method December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ The extension of the RouthRouth-Hurwitz criteria December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ The extension of the Routh Routh Hurwitz criteria Characteristic C a acte st c equat equation o o of d discrete sc ete syste systems: s a0 z n a1 z n 1 an Im z Im w Region R i off stability Re z Region of stability 1 w z 1 w Re w The extension of the Routh-Hurwitz criteria: transform zw,, and then apply pp y the Routh – Hurwitz criteria to the characteristic equation of the variable w December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ The extension of the Routh Routh Hurwitz criteria – Example Analyze the stability of the following system: R(s) + T 0.5 ZOH G(s) Y(s) H(s) 3e s Gi Given that: th t G( s) s3 H ( s) s 1 Solution: Sol tion The characteristic equation of the system: GH ( z ) December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 10 Performance of discrete system – Example (cont’) G( z) Gcl ( z ) G( z) G( z) 0.042 z 0.036 ( z 0.819)( z 0.741) 0.042 z 0.036 ( z 0.819)( z 0.741) 0.042 z 0.036 1 ( z 0.819)( z 0.741) 0.042 z 0.036 Gcl ( z ) z 1.518 z 0.643 December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 48 Performance of discrete system – Example (cont’) The Th time ti response off the th system t to t step t input i t Y ( z ) Gk ( z ) R( z ) 0.042 z 0.036 R( z ) z 1.518 z 0.643 Gk ( z ) 0.042 z 0.036 z 1.518 z 0.643 0.042 z 1 0.036 z 2 R( z ) 1 2 1.518 z 0.643z (1 1.518 z 1 0.643 z 2 )Y ( z ) (0.042 z 1 0.036 z 2 ) R ( z ) y (k ) 1.518 y (k 1) 0.643 y (k 2) 0.042r (k 1) 0.036r (k 2) y (k ) 1.518 y (k 1) 0.643 y (k 2) 0.042r (k 1) 0.036r (k 2) December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 49 Performance of discrete system – Example (cont’) U it step Unit t input: i t r ( k ) 1, kk Initial condition: y ( 1) y ( 2) Substitute the initial condition to the recursive equation off y(k), (k) we have: h y ( k ) 0; 0.0420; 0.1418; 0.2662; 0.3909; 0.5003; 0.5860; 0.6459; 0.6817;0.6975; 0.6985; 0.6898; 0.6760; 0.6606; 0.6461; 0.6341; 0.6251; 0.6191; c(k ) 1.518c(k 1) 0.643c(k 2) 0.042r (k 1) 0.036r (k 2) December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 50 Performance of discrete system – Example (cont’) Step Response 0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0 0.5 1.5 2.5 Time (sec) December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 51 Performance of discrete system – Example (cont’) Transient performances: The steady state response: yss lim li (1 z 1 )Y ( z ) z 1 lim(1 z 1 )Gk ( z ) R( z ) 0.042 z 0.036 Gk ( z ) z 1.518 z 0.643 R( z ) z 1 z 1 0.042 z 0.036 lim(1 z ) 1 z 1 z 1.518 z 0.643 z 1 y ss 0.624 The maximum value: ymax 0.6985 Percentage of overshoot: ymax yss 0.6985 0.624 POT 100% 100% 11.94% yss 0.624 December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 52 Performance of discrete system – Example (cont’) Settling time (5% criterion): First, we need to find ks satisfying: 1 yss y (k ) 1 yss , k ks cxl 0.624 5% 0.05 0.593 y (k ) 0.655, k ks From the time response calculated before ks 14 ts ksT 14 0.1 ts 1.4 sec St d c(state Steady kt) t error: 0; 0.0420; 0.1418; 0.2662; 0.3909; 0.5003; 0.5860 0.6459negative ; 0.6817;feedback, 0.6975; 0.6985 0.6898; Since the system is;unity we ;have: 6606 ; 0.6461; 0.6341e; 0.6251 ; 0.6191; ess rss 0.6760 yss ; 10. 0.624 376 ss December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 53 Performance of discrete system – Example (cont’) Note: It is possible to calculate POT and ts based on the dominant poles The poles of the closed-loop system are the roots of the equation: z 1.518 z 0.643 z1*, 0.7590 j 0.2587 0.80190.3285 ln r ln 0.8019 0.5579 2 2 (ln r ) (ln 0.8019) 0.3285 1 2 n (ln r ) (ln 0.8019) 0.3285 0.3958 T POT exp t qñ .100% exp 0.5579 3.14 .100% 12.11% 5579 3 1.36 sec n 0.5579 0.3958 December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 54 Performance of discrete system – Example r(t) + e(t) T with T = 0.1 e(kT) ZOH eR(t) y(t) G (s ) 2( s 5) G( s) ( s 2)( s 3) Formulate F l t the th state t t equations ti d describing ibi the th system t Calculate the response of the system to unit step input (assuming the initial conditions are zeros) using the state equation formulated above Calculate POT, POT settling time, time steady state error December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 55 Performance of discrete system – Example (cont’) Solution: S l ti Formulate the state equation: 2( s 5) s 10 Y ( s) G( s) E R ( s ) ( s 2)( s 3) s 5s The state equation of the continuous plant: x1 (t ) x1 (t ) 0 x (t ) 5 x (t ) 1eR (t ) B A x1 (t ) y (t ) 10 2 x2 (t ) C December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 56 Performance of discrete system – Example (cont’) The transient matrix: 1 1 1 0 s -1 ( s ) sI A s 5 s 5 s5 s 1 ( s 2)( s 3) ( s 2)( s 3) s s ( s 5) s ( s 2)( s 3) ( s 2)( s 3) 1 1 L L s s 3 s s 3 1 (t ) L [( s)] L 1 L 1 s s s s (3e 2t 2e 3t ) (e 2t e 3t ) (t ) 2t 3t 2t 3t (6e 6e ) (2e 3e ) December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 57 Performance of discrete system – Example (cont’) The state equation of the x[(k 1)T ] Ad x(kT ) Bd eR (kT ) discrete open-loop system: y (kT ) Cd x(kT ) (3e 2T 2e 3T ) 0.9746 0.0779 (e 2T e 3T ) Ad (T ) 0.4675 0.5850 2T 3T 2T 3T ( 6 ) ( ) e e e e T 2 3 (e 2 e 3 ) 0 (3e 2e ) d Bd ( ) Bd 2 3 2 3 0 (6e 6e ) (2e 3e ) 1 T T 2 3 0.1 e (e 2 e 3 ) ( e 0.0042 ) d 2 3 0779 ( ) e e 0 (e e ) T Cd C 10 2 December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 58 Performance of discrete system – Example (cont’) The state equation of the discrete closed-loop system: x[(k 1)T ] Ad Bd Cd x(kT ) Bd r (kT ) y (kT ) Cd x(kT ) with 0.9326 0.0695 0.9746 0.0779 0.0042 Ad Bd Cd 10 2 1.2465 0.4292 0.4675 0.5850 0.0779 x1 (k 1) 0.9326 0.0695 x1 (k ) 0.0042 x (k 1) 1.2465 0.4292 x (k ) 0.0779 r (kT ) x1 (k ) y (k ) 10 2. x ( k ) December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 59 Performance of discrete system – Example (cont’) Time response of the system: From the closed-loop state equations, we have: x1 (k 1) 0.9326 x1 (k ) 0.0695 x2 (k ) 0.0042r ( k ) x2 (k 1) 1.2465 x1 (k ) 0.4292 x2 (k ) 0.0779r (t ) With initial condition x1(1)=x2(1)=0, unit step input, we can calculate the solution to the state equation: x1 (k ) 10 3 0; 4.2; 13.5; 24.2; 34.2; 42.6; 49.1; 54.0; 57.4; 59.7; 61.2; 62.0; 62.5; 62.7; 62.8; 62.8; 62.7; 62.7; 62.6; 62.6 x2 (k ) 10 3 0; 77.9; 106.1; 106.6; 93.5; 75.4; 57.2; 41.2; 28.3; 18.5; 11 4; 6.5; 11.4; 5; 3.4; 4; 1.4; 4; 0.3; 3; -0.3; 3; -0.5; 5; -0.5; 5; -0.5; 5; -0.4 The closed-loop system response: y (k ) 10 x1 (k ) x2 (k ) y (k ) 0; 0.198;; 0.348;; 0.455;; 0.529;; 0.577;; 0.606;; 0.622;; 0.631;; 0.634; ; 0.635; 0.634; 0.632; 0.630; 0.629; 0.627; 0.627; 0.626; 0.625; 0.625 December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 60 Performance of discrete system – Example (cont’) Step Response espo se 0.7 0.6 Amplitude 0.5 0.4 0.3 0.2 0.1 0 0.5 1.5 2.5 Time (sec) December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 61 Performance of discrete system – Example (cont’) Performances of the system: Percentage of overshoot: ymax 0.635 63 yss 0.625 POT ymax yss 100% 1.6% yss The settling time: 1 0.05 y y ( k ) 1 0.05 y , k ks According to the response of the system: 0.594 y (k ) 0.656, k ks ts ksT 0.6 sec c(k ) 0; 0.198; 0.348; 0.455; 0.529; 0.577; 0.606; 0.622; 0.631; 0.634; S d state error: ess rss yss 0.625 0.375 Steady 0.635; 0.634; 0.632; 0.630; 0.629; 0.627; 0.627; 0.626; 0.625; 0.625 December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 62 ... 21 .83 K cr 21 .83 December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ 29 The root locus of discrete systems – Example (cont’) Im z 5742+j0 81 87 0.5742+j0 .81 87 +j 2.506 506 2 3 2 1 0 .85 7... 1.1 485 z z 0.5742 j 0 .81 87 Then the intersection of the RL with the unit circle are: z 0.5742 j 0 .81 87 December 2013 © H T Hồng - www4.hcmut.edu.vn/~hthoang/ 27 The root locus of. . .Chapter ANALYSIS OF DISCRETE CONTROL SYSTEMS December 2013 © H T Hoàng - www4.hcmut.edu.vn/~hthoang/ Content Stability conditions for discrete systems Extension of Routh-Hurwitz criteria