Bài giảng Control system design - Chapter VI: The stability of linear feedback systems - Nguyễn Công Phương

Nguyễn Công Phương CONTROL SYSTEM DESIGN The Stability of Linear Feedback Systems Contents I. Introduction II. Mathematical Models of Systems III. State Variable Models IV. Feedback Control System Characteristics V. The Performance of Feedback Control Systems VI. The Stability of Linear Feedback Systems VII. The Root Locus Method VIII.Frequency Response Methods IX. Stability in the Frequency Domain X. The Design of Feedback Control Systems XI. The Design of State Variable Feedback Systems XII. Robust Control Systems XIII.Digital Control Systems sites.google.com/site/ncpdhbkhn 2 The Stability of Linear Feedback Systems 1. The Concept of Stability 2. The Routh – Hurwitz Stability Criterion 3. The Stability of State Variable Systems 4. System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 3 The Concept of Stability (1) • Stability is of the utmost importance. • A close – loop feedback system that is unstable is of little value. • A stable system is a dynamic system with a bounded response to a bounded input. sites.google.com/site/ncpdhbkhn 4 The Concept of Stability (2) cycles/cycle-styles/city-bike sites.google.com/site/ncpdhbkhn 5 The Concept of Stability (4) 1 MNMNA B s C Y( s ) i  k k  y ( t )  1  A eit  D e   k t sin( t   )  2 2 2 i  k k k si1 s  i k  1s2k s  (  k   k ) i  1 k  1 j 1 1 10 1 0 0 0 0 -1 -1 -1 -10 0 5 10 0 5 10 0 5 10 0 5 10 1 1 10 1 0 0 0 0 -1 -1 -1 -10 0 5 10 0 5 10 0 5 10 0 5 10 1 1 2 10  0.5 0.5 1 5 0 0 0 0 0 5 10 0 5 10 0 5 10 0 5 10 sites.google.com/site/ncpdhbkhn 6 The Stability of Linear Feedback Systems 1. The Concept of Stability 2. The Routh – Hurwitz Stability Criterion 3. The Stability of State Variable Systems 4. System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 7 The Routh – Hurwitz Stability Criterion (1) n n1 n  2 qsasas( )n  n1  as n  2  ...  asa 1  0  0 n s an an2 an4  1 an a n2 an1 a n  2 a n a n  3 bn1   , n1 aa a a s an1 an3 an5  n1n1 n  3 n  1 sn2 b b b  n1 n3 n5 1 a a b  n n4 , n3 n3 s cn1 cn3 cn5  an1 an1 a n  5      1 a a 0 c  n1 n  3 , s hn1 n1 bn1 bn1 b n  3 The Routh – Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array sites.google.com/site/ncpdhbkhn 8 The Routh – Hurwitz Stability Criterion (2) n n1 n  2 qsasas( )n  n1  as n  2  ...  asa 1  0  0 st n 1. No element in the 1 column is s an an2 an4  zero. st sn1 a a a  2. There is a zero in the 1 column, n1 n3 n5 but some other elements of the row st sn2 b b b  containing the zero in the 1 n1 n3 n5 column are nonzero. n3 st s cn1 cn3 cn5  3. There is a zero in the 1 column, and the other elements of the row      containing the zero are also zero. 0 4. Repeated roots of the characteristic s hn1 equation on the jω – axis. The Routh – Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array sites.google.com/site/ncpdhbkhn 9 The Routh – Hurwitz Stability Ex. 1 Criterion (3) 2 sn a a a  q() s a2 s  a 1 s  a 0 n n2 n4 2 2 n1 s a2 a0 s a2 a0 s an1 an3 an5  1 1 n2 s a1 0 s a1 0 s bn1 bn3 bn5  0 0 n3 s b1 0 s a0 0 s cn1 cn3 cn5  1 a a 1 a a      b  n n2 b 2 0  a n1 1 0 0 an1 an1 a n  3 a1 a1 0 s hn1 The Routh – Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array The system is stable if a2, a1 & a0 are all positive or all negative sites.google.com/site/ncpdhbkhn 10 The Routh – Hurwitz Stability Ex. 2 Criterion (4) 3 2 n q( s ) s  s  2 s  50 s an an2 an4  3 3 n1 s 1 2 s 1 2 s an1 an3 an5  2 2 n2 s 1 50 s 1 50 s bn1 bn3 bn5  1 1 n3 s b1 b0 s 48 0 s cn1 cn3 cn5  0 0      s c1 c0 s 50 0 0 s hn1 1 1 2 1 1 0 1 1 50 1 1 0 b   48, b   0, c   50, c   0 1 1 1 50 0 1 1 0 1 48 48 0 0 48 48 0 The Routh – Hurwitz criterion states that the number of roots of q(s) with positive real parts is equal to the number of changes in sign of the first column of the Routh array sites.google.com/site/ncpdhbkhn 11 The Routh – Hurwitz Stability Ex. 3 Criterion (5) 3 2 q() s a3 s  a 2 s  a 1 s  a 0 3 s a3 a1 2 s a2 a0 a2 a 1 a 0 a 3 b1, c 1  a 0 1 a s b1 0 2 0 s c1 0 a3  0 a3  0 a3  0 a  0   2  a2  0  a2  0   a a a a   b  0 2 1 0 3  0 a a a a  0  1  a  2 1 0 3   2  c1  0 a0  0 a0  0 3 2 q() s s  2 s  610, s  a2 a 1  a 0 a 3  261012     sites.google.com/site/ncpdhbkhn 12 The Routh – Hurwitz Stability Ex. 4 Criterion (6) q( s ) s5  2 s 4  2 s 3  4 s 2  11 s  10 s5 1 2 11 s5 1 2 11 s4 2 4 10 s4 2 4 10 s3 0 6 0 s3  6 0 2 2 s c1 10 0 s c1 10 0 1 1 s d1 0 0 s d1 0 0 s0 10 s0 10 4 12 12 6c1  10  c1 4  , d 1   6  10   c1 Two sign changes  two roots with positive real part  the system is unstable sites.google.com/site/ncpdhbkhn 13 The Routh – Hurwitz Stability Ex. 5 Criterion (7) q() s s4  s 3  s 2  s  K s4 1 1 K s4 1 1 K s3 1 1 0 s3 1 1 0 s2 0 K 0 s2  K 0 1 1 s c1 0 0 s c1 0 0 s1 K s1 K   KK c  1  1   One sign change  one root with positive real part  the system is unstable for all values of K sites.google.com/site/ncpdhbkhn 14 The Routh – Hurwitz Stability Ex. 6 Criterion (8) q( s ) s5  s 4  4 s 3  24 s 2  3 s  63 (s3  s 2  s  21)( s 3  3) s5 1 4 3 s5 s 4 4 s 3  24 s 2  3 s  63 s 2  3 s4 1 24 63 s5 3 s 3 s3 s2 s 21 3 s 20 60 0 s4 s 3 24 s 2  3 s  63 2 4 2 s 21 63 0 s 3 s 3 2 1 s21 s  3 s  63 s 0 0 0 s3  3 s U( s )  21 s2  63  21( s 2  3) 21s2  63 21s2  63 0 sites.google.com/site/ncpdhbkhn 15 The Routh – Hurwitz Stability Ex. 6 Criterion (9) q( s ) s5  s 4  4 s 3  24 s 2  3 s  63 (s3  s 2  s  21)( s 3  3) s5 1 4 3 s4 1 24 63 3 3 s 1 1 s 20 60 0 2 2 s 1 21 s 21 63 0 1 s1 0 0 0 s 20 0 0 s 21 0 Two sign changes  two roots with positive real part  the system is unstable sites.google.com/site/ncpdhbkhn 16 The Routh – Hurwitz Stability Ex. 7 Criterion (10) K( s a ) 1 R() s K() s a 1 Y() s G() s  s( s 2)( s  3) s1 s ( s  2)( s  3) () s 1 G()() s K s a T() s   1 G ( s ) s46 s 3  11 s 2  ( K  6) s  Ka 60  K q( s )  s4  6 s 3  11 s 2  ( K  6) s  Ka  0  6  4 b( K 6)  6 Ka s 1 11 Ka  3  0 b3 3  s 6 K  6 0 Ka  0  2  s b3 Ka 0 1 0K  60 s c3 0 0    (60KK )(  6) s1 a  Ka  36K 60  K b3( K 6)  6 Ka b3, c 3  6 b3 sites.google.com/site/ncpdhbkhn 17 The Stability of Linear Feedback Systems 1. The Concept of Stability 2. The Routh – Hurwitz Stability Criterion 3. The Stability of State Variable Systems 4. System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 18 The Stability of State Variable Ex. 1 Systems (1) 1 3 x1 3 x 1  x 2 1 K 1  U() s X1() s x2 x 2  Kx 1  Ku 1/ s 1/ s  X2 () s X 2 1 1  1  2 L1 s, L 2   3 s , L 3   Ks N  1 LLLLLLn   n m   n m p  ... n1 n , m n , m , p nontouching nontouching 1  2 1  (LLLLL1  2  3 )  ( 1 2 ) 1  2s  ( K  3) s q( s )  s2  2 s  ( K  3) KK 3  0   3 sites.google.com/site/ncpdhbkhn 19 The Stability of State Variable Ex. 2 Systems (2) a  0   1 0  dx u     0 x   0 1  1       dt u2    0   0 0  0 0  a   0      0   det(IA )det0   0       0            0   0 0     0       [ 2  (    )   (    2 )] q( )   [ 2  (    )   (    2 )]    0   2    0 sites.google.com/site/ncpdhbkhn 20 The Stability of Linear Feedback Systems 1. The Concept of Stability 2. The Routh – Hurwitz Stability Criterion 3. The Stability of State Variable Systems 4. System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 21 System Stability Using Control Ex. 1 Design Software (1) R() s 10(s  2) 1 Y() s s( s 2)( s  3) () s 1 sites.google.com/site/ncpdhbkhn 22 System Stability Using Control Ex. 2 Design Software (2) q( s ) s3  2 s 2  3 s  K sites.google.com/site/ncpdhbkhn 23 System Stability Using Control Ex. 3 Design Software (3) Given a characteristic equation q(s) = s4 + 8s3 + 17s2 + (K + 10)s + aK = 0. Find a & K such that the system is stable. 4 1 126  K s 1 17 aK b [( K  10)  8  17]  3 8 8 s3 8 K 10 0 1 2 b1  (1  0  8 aK )  aK s b3 b1 0 8 1 s c3 0 0 1 (126KK )(  10) c3 [8 b 1  b 3 ( K  10)]   8 aK 1 b3 8 s d3 126  K 1  0 d3 ( b 3  0  b 1 c 3 )  b 1  aK  8 c3 b3  0   (126KK )(  10) c0    8 aK  0 3 8 d  0   3 aK  0  sites.google.com/site/ncpdhbkhn 24