Lecture Control system design: The stability of linear feedback systems - Nguyễn Công Phương

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Lecture Control system design: The stability of linear feedback systems - Nguyễn Công Phương

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Lecture Control system design: The stability of linear feedback systems include all of the following content: The concept of stability, the Routh – Hurwitz stability criterion, the stability of state variable systems, system stability using control design software.

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 The Stability of Linear Feedback Systems The Concept of Stability The Routh – Hurwitz Stability Criterion The Stability of State Variable Systems System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 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 The Concept of Stability (2) http://www.ctc.org.uk/cyclists-library/bikes-and-othercycles/cycle-styles/city-bike sites.google.com/site/ncpdhbkhn The Concept of Stability (4) Y ( s)   s M  i 1 Ai  s  i Bk s  Ck  y (t )   2 k 1 s  2 k s  ( k  k ) N  M  Ae i  i t i 1  N D e k  k t sin(k t   k ) k 1 j 1 0 -1 -1 10 1 0 -1 -1 10 10 0 -1 10 10 10 10 -10 1 10 0.5 0.5 0 10 0 10 10 10 -1 10 -10 sites.google.com/site/ncpdhbkhn 10  10 The Stability of Linear Feedback Systems The Concept of Stability The Routh – Hurwitz Stability Criterion The Stability of State Variable Systems System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn The Routh – Hurwitz Stability Criterion (1) q( s )  an s n  an 1s n 1  an 2 s n 2   a1s  a0  sn an an  an   s n 1 an 1 an  an   s n 2 bn 1 bn 3 bn 5  s n 3 cn 1 cn 3 cn 5       s hn 1 1 an an 2 an 1an 2  an an 3 bn 1   , an 1 an 1 an 3 an 1 bn 3  1 an an 4 , an 1 an 1 an 5 cn 1  1 an 1 an 3 , bn 1 bn 1 bn 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 The Routh – Hurwitz Stability Criterion (2) q( s )  an s n  an 1s n 1  an 2 s n 2   a1s  a0  sn an an  an   s n 1 an 1 an  an   s n 2 bn 1 bn 3 bn 5  s n 3 cn 1 cn 3 cn 5       s0 hn 1 No element in the 1st column is zero There is a zero in the 1st column, but some other elements of the row containing the zero in the 1st column are nonzero There is a zero in the 1st column, and the other elements of the row containing the zero are also zero Repeated roots of the characteristic 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 The Routh – Hurwitz Stability Criterion (3) Ex q( s )  a2 s  a1s  a0 sn an an  an   s2 a2 a0 s2 a2 a0 s n 1 an 1 an  an   s1 a1 s1 a1 s n 2 bn 1 bn 3 bn 5  s0 b1 s0 a0 s n 3 cn 1 cn 3 cn 5       s0 hn 1 1 an an 2 bn 1  an 1 an 1 an 3 1 a2  b1  a1 a1 a0  a0 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 Criterion (4) Ex q( s )  s  s  s  50 sn an an  an   s3 s3 s n 1 an 1 an  an   s2 50 s2 50 s n 2 bn 1 bn 3 bn 5  s1 b1 b0 s1 48 s n 3 cn 1 cn 3 cn 5  s0 c1 c0 s0 50      s0 hn 1 b1  1 1  0,  48, b0  1 1 50 c1  1 50 1  50, c0  0 48 48 48 48 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 Criterion (5) Ex q( s )  a3s  a2 s  a1s  a0 s3 a3 a1 s2 a2 a0 s1 b1 s0 c1 a3  a    b1  c1  b1  a3  a     a2 a1  a0a3 0  a2  a0  a2 a1  a0a3 , c1  a0 a2 a3  a    a2 a1  a0a3  a0  q( s )  s  s  6s  10, a2 a1  a0a3    10   sites.google.com/site/ncpdhbkhn 12 Ex The Routh – Hurwitz Stability Criterion (6) q( s )  s  s  s  s  11s  10 s5 11 s5 11 s4 10 s4 10 s3 s3  s2 c1 10 s2 c1 10 s1 d1 0 s1 d1 0 s0 10 s0 10 c1  4  12 12 6c  10   , d1    10   c1 Two sign changes  two roots with positive real part  the system is unstable sites.google.com/site/ncpdhbkhn 13 Ex The Routh – Hurwitz Stability Criterion (7) q( s )  s  s  s  s  K s4 1 K s4 1 K s3 1 s3 1 s2 K s2  K s1 c1 0 s1 c1 0 s1 K s1 K c1   K K 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 Ex The Routh – Hurwitz Stability Criterion (8) 3 q( s )  s  s  s  24 s  3s  63  ( s  s  s  21)( s  3) s5 s5  s  s  24 s  3s  63 s  s4 24 63 s3 20 60 s  s  24 s  3s  63 s2 21 63 s4 s 0 s  s  s 21  3s s5  3s s  21s  3s  63  3s s3  U ( s )  21s  63  21( s  3) 21s  63 21s  63 sites.google.com/site/ncpdhbkhn 15 Ex The Routh – Hurwitz Stability Criterion (9) 3 q( s )  s  s  s  24 s  3s  63  ( s  s  s  21)( s  3) s5 s4 24 63 s s s 20 21 60 63 0 0 s3 1 s2 21 s1 20 s0 21 Two sign changes  two roots with positive real part  the system is unstable sites.google.com/site/ncpdhbkhn 16 The Routh – Hurwitz Stability Criterion (10) Ex K (s  a) G( s)  s  s( s  2)( s  3) T ( s)  R( s ) ( ) s( s  2)( s  3) G( s) K (s  a)   G ( s ) s  6s  11s  ( K  6) s  Ka  q( s )  s  6s  11s  ( K  6) s  Ka s4 11 Ka s3 K 6 s2 b3 Ka s1 c3 0 s1 Ka b3  K (s  a) s 1  60  K  0   b3 ( K  6)  Ka 0  b   Ka   0  K  60   (60  K )( K  6) a   36 K b ( K  6)  Ka 60  K , c3  b3 sites.google.com/site/ncpdhbkhn 17 Y ( s) The Stability of Linear Feedback Systems The Concept of Stability The Routh – Hurwitz Stability Criterion The Stability of State Variable Systems System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 18 Ex The Stability of State Variable Systems (1) 3  x1  3x1  x2   x2  x2  Kx1  Ku K U ( s) 1/ s L1  s 1 , L2  3s 1 , L3   Ks 2  1 N L n n 1   Ln Lm  n ,m nontouching  X ( s) 1 X1 ( s) 1/ s X2 Ln Lm Lp  n ,m , p nontouching   ( L1  L2  L3 )  ( L1L2 )   2s 1  ( K  3) s 2  q( s )  s  s  ( K  3)  K 3 0 K  sites.google.com/site/ncpdhbkhn 19 Ex The Stability of State Variable Systems (2)  a   dx     dt     0 1 0 u  0 x         u2  0 0 0   0   a    det( I  A )  det            0         0           0          [  (   )  (   )]  q( )   [  (   )  (   )]          sites.google.com/site/ncpdhbkhn 20 The Stability of Linear Feedback Systems The Concept of Stability The Routh – Hurwitz Stability Criterion The Stability of State Variable Systems System Stability Using Control Design Software sites.google.com/site/ncpdhbkhn 21 Ex System Stability Using Control Design Software (1) R( s ) ( ) sites.google.com/site/ncpdhbkhn 10( s  2) s 1 s( s  2)( s  3) 22 Y ( s) Ex System Stability Using Control Design Software (2) q( s )  s  s  3s  K sites.google.com/site/ncpdhbkhn 23 Ex System Stability Using Control Design Software (3) Given a characteristic equation q(s) = s4 + 8s3 + 17s2 + (K + 10)s + aK = Find a & K such that the system is stable s4 17 aK s3 K  10 s2 b3 b1 s1 c3 0 s1 d3 126  K b3   [( K  10)   17]  8 b1   (1   8aK )  aK c3   (126  K )( K  10) [8b1  b3 ( K  10)]   8aK b3 126  K 0  b3     (126  K )( K  10)  8aK  c3    d    aK   sites.google.com/site/ncpdhbkhn d3   (b3   b1c3 )  b1  aK c3 24 ... s) The Stability of Linear Feedback Systems The Concept of Stability The Routh – Hurwitz Stability Criterion The Stability of State Variable Systems System Stability Using Control Design Software... 20 The Stability of Linear Feedback Systems The Concept of Stability The Routh – Hurwitz Stability Criterion The Stability of State Variable Systems System Stability Using Control Design Software... 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

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