Stability Margins ofMultivariable Feedback Systems
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method   - Frequency Domain Method• Loop Phase Margins   - Time Do...
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method  - Frequency Domain Method• Loop Phase Margins  - Time Doma...
HISTORICAL REMARKS• Uncertain systems• Robust control: dominant approach• Margin based approach2011年2月8日星期二               ...
SISO STABILITY MARGINSThe unity output feedback configuration.G(s)is the processC (s)is the compensator.SISO stability mar...
MIMO MOTIVATION EXAMPLEConsider a 2 × 2 system                               1        2                                ...
MIMO MOTIVATION EXAMPLEThe characteristic equation of the closed-loop system is            Pc (s)  P (s) PK (s) det  I ...
MIMO MOTIVATION EXAMPLEThe stabilizing range is drawn as the share region.                                      • Stabiliz...
MIMO RESEARCH STATUS• Doyle (1982) developed the u-analysis, and treats system  uncertainties as complex valued.  Thus, it...
MIMO RESEARCH STATUS• Gershgorin band method is also used to calculate the gain margin of  multivariable system (Hong, Yan...
MIMO RESEARCH STATUS• A similar problem in concept is phase margin but different for  computation• Our goal is to define m...
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method  - Frequency Domain Method• Loop Phase Margins  - Time Doma...
PROBLEM FORMULATIONConsider an m m square plant with dimensional state-spacerealization:                       x(t )  Ax...
PROBLEM FORMULATIONk1i , k2 jand k3l are scalars to be determined, I1i, I 2 jand I 3l areidentity matrices with dimensions...
PROBLEM FORMULATIONProblem 1. For a given plant G(s) under the controller u(t ), find theranges of scalars k1i , k2 j and ...
PROBLEM FORMULATIONFor an m m square plant G(s) under the decentralized proportionalcontrol K (s)  kI m, suppose that th...
PROBLEM FORMULATIONDefinitionIf K (s)  diag k1 , k1 , , km  with probably different gains fordifferent loops, suppose t...
PROPOSED APPROACH                                                     t By introducing augmented state x  [ x (t ), 0 yT...
PROPOSED APPROACH The closed-loop system is             Ex (t )  ( A  BKC ) x (t )                               r1     ...
PROPOSED APPROACHDefinitionA descriptor system is called admissible if the system, or say, the pair E, Acl  is regular, ...
PROPOSED APPROACH                                                          0Let k0     vi   , v  1, 2,3, i  1, 2,   be s...
PROPOSED APPROACHLemmaThe pair  E, A( )  is robustly admissible if and only if there existparameter-dependent matrices ...
SPECIAL CASESPropositionThe pair  E, Acl is robustly admissible if and only if there existmatrices Pj , F j , H j and X ...
SPECIAL CASESIn this special case,   K2  0   and   K3  0 . Then,                              x(t )  Ax(t )  Bu (t ), ...
SPECIAL CASESPropositionThe polytope A is robustly stable if there exist matrices P  0 , andF H        and X with X  X T...
ALGORITHM        Algorithm 1.Step 1. Find a common gain controller, K (s), to stabilize the plant, G(s). If          K (s)...
ALGORITHM         mutually independent stabilizing range of         kvi   as                             kvi  kvi  0 ,...
Illustrative ExampleConsider a process (Bryant and Yeung, 1996)                               s 1           4          ...
Illustrative ExampleCase 1: P control• Consider a common gain controller K (s)  kI 2 , to stabilize theplant G(s) Since G...
Illustrative Example• Let   0.5 and relax  as  *  . The sequence of rangeshifting is as follows: firstly find the l...
Illustrative Example                          k1                            k2  Ho’s result     k1 [1.6556,1.2]         ...
Illustrative Example                          k1                            k2  Our result      k1 [1.6556,1.2]         ...
Illustrative ExampleCase 2: PI control                             1Let K (s)  diag k1 , k2   diag k3 , k4 .        ...
Illustrative ExampleCase 3: PD controlLet K (s)   k1  k2 s  I 2• Choose k 0  0 , the stabilizing ranges are calculate...
Illustrative ExampleCase 4: PID controlLet   K ( s)   k1  k2 / s  k3 s  I 2• Choose k 0  5 , the stabilizing ranges ...
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method  - Frequency Domain Method• Loop Phase Margins  - Time Doma...
The Proposed ApproachUnder the nominal stabilization (   I m ), the closed-loop system canbe destabilized if and only if...
The Proposed ApproachConsider a diagonal real matrix . Let   v  v1 , v2 ,   , vm  , and                                ...
The Proposed ApproachTake the following quadratic function as the cost function for ease ofoptimization:                  ...
The Proposed Approach                    min  z *G * z     or   max  z *G * z  ,                          z* z  1   ...
The Proposed Approach• To solve the above constrained optimization problem, we use theLagrange multiplier:                ...
The Proposed Approach• A numerical solution to is sought by the Newton-Raphson algorithm:                       n1  n ...
The Proposed Approach• If J is singular, then a Moore-Penrose inverse is used. To seewhether the iteration routine achieve...
The Proposed Approach        Algorithm 2.Step 1. Run the Newton-Raphson iteration. If the iteration is convergent,        ...
The Proposed ApproachWe look for the relevant frequency range , such that the solution to(4) may exist while there is no ...
The Proposed ApproachThe set ( P)  z Pz : z z  1, z  is commonly called the numerical                   *     *     ...
The Proposed Approach• We first find all the real roots l such that det( P( j ))  0 . Hence, by                        ...
The Proposed Approach• Denote  by   (1, 1 ) (r , r ) in the ascending order ofthe frequency. Suppose that we have c...
The Proposed ApproachNote that for any frequency    , the gain solutions not inside the                              * ...
The Proposed ApproachIt is well known that the spectral radius forms a lower bound on anycompatible matrix norm (Morari an...
The Proposed Approach        Algorithm 3.Step 1. Calculate all the real roots l of det( P( j))  0. Divide the entire   ...
The Proposed ApproachStep 3. Run Algorithm 2 for (r , r ]to get gain solution of (4) and plot them.        If a closed r...
Illustrative ExampleConsider a time delay process                                  s 1                0.5s  1        ...
Illustrative Example• Running Step 3 of Algorithm 3 for the first two frequency intervals,[0,2.652] and (2.876,6.918], yie...
Illustrative Example• The region including   I 2 found above is thus the stabilizing oneand is marked in yellow in Figur...
Illustrative Example                                           1Figure 2: The curves of   G( j ) 1 and          .   Figur...
Illustrative Example• Comparison with LMI method                                k1                          k2   Our metho...
Illustrative ExampleThe Nyquist curve of (6) for two cases, (k1  2.0233, k2  3.2188)from the LMI method, and, (k1  2...
Illustrative ExampleNyquist curve of (6) for (k1  1.1502, k2  5.0708)from by LMI method,and (k1  1.1502, k2  3.6852) f...
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method  - Frequency Domain Method• Loop Phase Margins  - Time Doma...
Problem FormulationProblem 2. For an        m  m square system, G( s) , under the decentralizedphase perturbation,       ...
Problem Formulation• Time domain method                  x(t )  Ax(t )  Bu (t ),• System:                          y ...
Problem FormulationTheorem 1. For given scalars Lk  0, k  1, 2, , m, the closed-loopsystem is asymptotically stable if t...
Problem Formulationk  1, 2,   , m; i  1, 2,        , m  1; j  i  1, i  2,   , m, such that the followingLMIs hold:  ...
Problem Formulationwhere               P  P 0 0  0,               A   A  BI1C      BI 2C   BI mC ,             ...
Problem FormulationProof. Choose the cadidate Lyapunov-Krasovskii functional to be                                        ...
Problem Formulation To find the range of delays, let Lk = Lk + ¢Lk ; k = 1; 2; ¢ ¢ ¢ ; m, where                           ...
Problem Formulation          Algorithm 4.Step 1.   Choose the initial Lk , k  1, 2, , m, such that LMIs in Theorem 3 are ...
Problem FormulationStep 5. Let   Li  ( Li  Li ) / 2 and di  ( Li  Li ) / 2 . If i  m,              ˆ                 ...
Problem Formulation                                                       ˆStep 3. Else, let mid  (min  max) / 2, rlow ...
Problem FormulationStep 2. If new LMIs are feasible, let   rupp = max, then go to Step 6.Step 3. Else, let                ...
Problem FormulationProposition 2. (Bar-on and Jonckheere 1998). There exists a unitary¢in the feedback path which destabil...
Problem Formulation               G : x1 = Ax1 + Bu; y1 = Cx1 ;                    _             GH : x2 = ¡AT x2 + C T y1...
Problem Formulation        Algorithm 5.Step 1. Calculate the purely imaginary eigenvalues, !i; i = 1; 2; ¢ ¢ ,¢ of        ...
Illustrative Example    Example (Bar-on and Jonckheere ,1998)       2                               3           2         ...
Illustrative ExampleExample (cont’d)2011年2月8日星期二                              76
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method  - Frequency Domain Method• Loop Phase Margins  - Time Doma...
The Proposed Approach• Frequency domain method¢(s) = diagfejÁi g; i = 1; 2; ¢ ¢ ¢ ; m:     det(I + G(j!)¢) = 0 ( 9z 2 Cm; ...
The Proposed ApproachLagrange multiplier                           m                                              X  F (·)...
The Proposed Approach           @f (·)    @ 2 F (·)J[f(·)] =          =           =2           @·         @·2             ...
The Proposed Approach          Algorithm 6.Step 1.   Determine Ð = f! j 0 · ¾(G(j!)) · 1 · ¾(G(j!))gfrom                  ...
Illustrative ExampleExample (Bar-on and Jonckheere, 1998)                    ^                    Ð = (0:643; 1:613)      ...
Illustrative Example                                    Loop Phase Margin       Method                                    ...
OUTLINE• Introduction• Loop Gain Margins  - Time Domain Method  - Frequency Domain Method• Loop Phase Margins  - Time Doma...
Conclusions1.  The Loop gain and phase margins of MIMO feedback systems has   been defined2. The algorithms presented for ...
2011年2月8日星期二   86
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MIMO Stability Margins

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MIMO Stability Margins

  1. 1. Stability Margins ofMultivariable Feedback Systems
  2. 2. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 2
  3. 3. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 3
  4. 4. HISTORICAL REMARKS• Uncertain systems• Robust control: dominant approach• Margin based approach2011年2月8日星期二 4
  5. 5. SISO STABILITY MARGINSThe unity output feedback configuration.G(s)is the processC (s)is the compensator.SISO stability margins- Gain margin: GM  1/ G( j ) . p- Phase margin: PM  180  G( jc ).2011年2月8日星期二 5
  6. 6. MIMO MOTIVATION EXAMPLEConsider a 2 × 2 system  1 2   s 1 s  1 G ( s)     3 4   s 1  s  1 under a decentralized proportional control K (s)  diag k1 , k2 2011年2月8日星期二 6
  7. 7. MIMO MOTIVATION EXAMPLEThe characteristic equation of the closed-loop system is Pc (s)  P (s) PK (s) det  I  G(s)K (s) G  s  (k1  4k2  2)s  (k1  4k2  1  2k1k2 )  0The closed-loop system is stable if and only if k1  4k2  2  0,  k1  4k2  1  2k1k2  0.or.,  1 1 k1   k1  ,  4 2  k1  1 k2  , if , k1  2,  2(k1  2)  k 1  k2  1 , if , k1  2.   2(k1  2)2011年2月8日星期二 7
  8. 8. MIMO MOTIVATION EXAMPLEThe stabilizing range is drawn as the share region. • Stabilizing range of k depends 2 on the value of k and vice versa. 1 (e.g. k  1 yields k  3/ 4 ) 1 2 • More useful to prescribe a range for k when finding stabilizing 1 range for k .(e.g. k [1, 2] 2 1 yeilds k 2 [3/ 4, ). • Graphically a rectangle. (e.g. ABCD) • Not Unique.2011年2月8日星期二 8
  9. 9. MIMO RESEARCH STATUS• Doyle (1982) developed the u-analysis, and treats system uncertainties as complex valued. Thus, it is inevitable to bring the conservativeness because the gain change in each loop is of real number.• Baron and Jonckheere (1991) viewed the gain margin for a multivariable system as the minimal complex matrix perturbation. Such a definition allows the gain perturbation to be a full matrix, not necessarily to be diagonal.2011年2月8日星期二 9
  10. 10. MIMO RESEARCH STATUS• Gershgorin band method is also used to calculate the gain margin of multivariable system (Hong, Yang, 2005; Ho, Lee and Gan. 1997). It gives conservative results because it requires the diagonal dominance of the system, which brings some limitations to its applications.• Li and Lee (1993) showed that H norm of a sensitivity function  matrix for a stable multivariable closed-loop system is related to some common gain and phase margins for all the loops.2011年2月8日星期二 10
  11. 11. MIMO RESEARCH STATUS• A similar problem in concept is phase margin but different for computation• Our goal is to define meaningful margins and compute them2011年2月8日星期二 11
  12. 12. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 12
  13. 13. PROBLEM FORMULATIONConsider an m m square plant with dimensional state-spacerealization: x(t )  Ax (t )  Bu (t ),  y (t )  Cx(t ),x , y  m B and C are real constant matrices with appropriate ndimensions. The PID control K2 K ( s)  K1   K3 s s   diag k11I11 , , k1r1 I1r1  1   diag k21I 21 , s , k2 r2 I 2 r2    sdiag k31I31 , , k3r3 I3r3 2011年2月8日星期二 13
  14. 14. PROBLEM FORMULATIONk1i , k2 jand k3l are scalars to be determined, I1i, I 2 jand I 3l areidentity matrices with dimensions m , m and 1i 2j m3l, respectively, and  m   m   m  m. r1 r2 r3 i 1 1i j 1 2j l 1 3lThen, t u (t )   K y(t )  K  y ( )d  K y (t ) 1 2 0 3 r1 r2 r3 :  k1i I1i y (t )   k2i I 2i  y ( )d   k3i I 3i y (t ) t 0 i 1 i 1 i 1where I vi  diag 0, ,0, I vi ,0, ,0  mm , v  1, 2,3, i  1, 2, , rv .2011年2月8日星期二 14
  15. 15. PROBLEM FORMULATIONProblem 1. For a given plant G(s) under the controller u(t ), find theranges of scalars k1i , k2 j and k3l , i  1, , r1 , j  1, , r2 , l  1, , r3such that the closed-loop system is stable for all allowable k1i , k2 j andk3l in these ranges.RemarkThe controller of the form K (s)   k  ks  k s  I corresponds to the 1 2  3 m  specially chosen r1  r2  r3  1; the controller of the form 1K ( s)  diag k   diag k   sdiag k  corresponds to the specially 1i 2i 3i schosen r1  r2  r3  m (or  m1i  m2 j  m3l  1). .2011年2月8日星期二 15
  16. 16. PROBLEM FORMULATIONFor an m m square plant G(s) under the decentralized proportionalcontrol K (s)  kI m, suppose that the solution to Problem 1. is k  k , k  .  This stabilizing range is called as the common gain margin of thesystem. Graphically, such a stabilizing range is the largest line segmentof k1  k2   km available in the stabilizing region forki , i  1, 2, , m.2011年2月8日星期二 16
  17. 17. PROBLEM FORMULATIONDefinitionIf K (s)  diag k1 , k1 , , km  with probably different gains fordifferent loops, suppose that the solution to Problem 1. is ki  k i , k i  , i  1, 2,   , m.Then, k i , k i  is called the gain margin for the i-th loop, subject to  other loops’ gain margins within k j , k j  , j  1, 2, , m, j  i.  RemarkThe closed-loop remains stable even when the gain for the i-th loop, ki varies between ki and ki, provided that other loop gains, k j j  1, 2, , m, j  i. are (arbitrary but) also within their respectiveranges.2011年2月8日星期二 17
  18. 18. PROPOSED APPROACH t By introducing augmented state x  [ x (t ), 0 yT ( )d , yT (t )]T and T t output y(t )  [ y (t ), 0 yT ( )d , yT (t )]T, system with PID control is T transformed into a descriptor control system Ex (t )  Ax (t )  Bu (t ), y (t )  Cx (t ), u (t )   K y (t ), In 0 0  A 0 0  B C E 0  Im 0 , A   C 0 0  , B   0  , C         Im   0  0 0  CA 0  I m    CB      In  K   K1 K2 K3  . 2011年2月8日星期二 18
  19. 19. PROPOSED APPROACH The closed-loop system is Ex (t )  ( A  BKC ) x (t ) r1 r2 r3  ( A   k1i A1i   k2 j A2 j   k3l A3l ) x (t ) i 1 j 1 l 1 : Acl x (t ),  BI 1i C 0 0 0 BI 2 j C 0  0 0 BI 3l C       A1i   0 0 0  , A2 J  0 0 0  , A3l  0 0 0  CBI 1i C 0 0 0 CBI 2 j C 0  0 0 CBI 3l C        for i  1, 2, , r1 , j  1, 2, , r2 and l  1, 2, , r3 . 2011年2月8日星期二 19
  20. 20. PROPOSED APPROACHDefinitionA descriptor system is called admissible if the system, or say, the pair E, Acl  is regular, impulse-free and stable.So far, Problem 1. has been converted to the following problem.Problem 2. Find the ranges of scalars k1i , k2 j and k3li  1, 2, , r1 , j  1, 2, , r2 , l  1, 2, , r3 , such that the closed-loopsystem is admissible for all allowable k1i , k2 j and k3l in these ranges.2011年2月8日星期二 20
  21. 21. PROPOSED APPROACH 0Let k0 vi , v  1, 2,3, i  1, 2, be such that ( E, A ) with , rv clAcl  A  i 1 k10i A1i  i 1 k2i A2i  i 1 k3i A3i , is admissible. 0 r 1 r 0 2 r 3 0Set k vi  kvi  kvi and suppose k vi   vi , vi . Then 0  low upp  Acl  A  A  i11 k10i A1i  i2 1 k2i A2i  i31 k3i A3i , 0 r 0r r 0 cl which is equivalently recast as a matrix polytope with r  2r0vertices denoted by A j ( )  ( n2m)( n2m) ,  r r  Acl   A( ) : A( )    j A j (  );   j  1;  j  0; j  1, 2 ,r  j 1 j 1  r0  r1  r2  r3 and    1 , 1 , ,  r0 ,  r0  . low upp low upp   2011年2月8日星期二 21
  22. 22. PROPOSED APPROACHLemmaThe pair  E, A( )  is robustly admissible if and only if there existparameter-dependent matrices P( ), F ( ) and H ( ) such that P( )T E  EP( )  0,  F ( ) A( )  A( )T F ( )T    0  P( )  F ( )  H ( ) A( )  H ( )  H ( )  T T T T  Here and in the sequel, an ellipsis  denotes a block induced bysymmetry.Proof: The proof is parallel to that for standard systems in Geromel etal. (1998) and Peaucelle et al. (2000).2011年2月8日星期二 22
  23. 23. SPECIAL CASESPropositionThe pair  E, Acl is robustly admissible if and only if there existmatrices Pj , F j , H j and X jl with X jj  X T , l  j, j, l  1, 2, , r, jjsuch that PjT E  EPj  0, (1)  jl  lj  X jl  X T , j  1, 2, , r , l  j , jl (2)  X 11    X X 22   (3)  21       X r1  X r2 X rr    Fj Al (  )  Al (  )T Fj T    jl   .where  Pj  Fj  H j Al (  )  T T T H j  H j  T 2011年2月8日星期二 23
  24. 24. SPECIAL CASESIn this special case, K2  0 and K3  0 . Then, x(t )  Ax(t )  Bu (t ), y (t )  Cx(t ), u (t )   K1 y (t ).or, rewritten as x(t )   A  ir1 k1i A1i ,  x(t ) : Acl x(t ), 1where A1i  BI 1iC, i  1, 2, , r1.Thus  r r  Acl   A( ) : A( )   j Aj (  ); j  1; j  0; j  1, 2, , r .  j 1 j 1 2011年2月8日星期二 24
  25. 25. SPECIAL CASESPropositionThe polytope A is robustly stable if there exist matrices P  0 , andF H and X with X  X T , l  j, j, l  1, 2, , r, such that cl j j j jl jj jj  jl  lj  X jl  X T , j  1, 2, jl , r , l  j,  X 11    X X 22    21       X r1  X r2 X rr    Fj Al (  )  Al (  )T Fj T  where  jl   .  Pj  Fj  H j Al (  ) H j  H j  T T T T  Similar steps were carried out for PD and PI Control.2011年2月8日星期二 25
  26. 26. ALGORITHM Algorithm 1.Step 1. Find a common gain controller, K (s), to stabilize the plant, G(s). If K (s)G(s) is stable, take k 0  0 otherwise, use any standard ; technique (Cao et al., 1998; Zheng et al., 2002; Lin et al., 2004) to find the scalar k 0. Let kvi  k 0. If G(s) can not be stabilized by 0 any common gain controller, find a controller in the general form, i.e., Let k  k  k 0 or k vi  kvi  kvi . Calculate the stabilizing ranges of 0Step 2. by formula of Barmish (1994) for P/PI control or Lee et al. (1997) for PD/PID control. Reset kvi , v  1, 2,3, i  1, 2, , rv . and find the maximum 0  0 such 0Step 3. that LMIs are feasible for kvi  kvi  (k min  k max ) / 2. Obtain the 0 0    0 , 0 , , 0 , 0  2011年2月8日星期二 26
  27. 27. ALGORITHM mutually independent stabilizing range of kvi as kvi  kvi  0 , kkvi  0  .  0 0 Step 4. Arrange kvi in decreasing order of their importance and choose G(s) 0 0 initial values k vi  kvi  0or k vi  kvi  0   LMIs are still 0 0   0 . Thus,     01 ,  v1 , ,  vm ,  vm  , where  vi  kvi  0  k vi 0 0 0 feasible for  v 0 0    and  vi  kvi  0  k vi . 0 0 0Step 5. Relax  as  *   ,  (0,1) with   0.5 by default. If  *  0 vi (or    0) , find vi  0 (or  vi  0) such that LMIs are *  vi low upp feasible for i  1, 2, , m. (or    0) , find vi    vi ) (or  vi    vi ) * 0  vi 0Step 6. If  vi  0 * low upp such that LMIs are still feasible for i  1, 2, , m. 0Step 7. Calculate the range of kvi by kvi  k vi   vi , vi   low upp  2011年2月8日星期二 27
  28. 28. Illustrative ExampleConsider a process (Bryant and Yeung, 1996)  s 1 4   ( s  1)( s  3) s 3 G ( s)     1 3    s2 s  2 Its state-space minimum realization is  1.255 0.2006 0.1523  0.1458 0.03811A   1.452  0.465 0.8761  ,  B   0.113 0.06613 ,    0.5496 4.108   4.28    0.4033 0.7228     0.3773 7.907 4.831C .  5.273 8.879 3.06 Suppose the largest available range of parameters is 100.2011年2月8日星期二 28
  29. 29. Illustrative ExampleCase 1: P control• Consider a common gain controller K (s)  kI 2 , to stabilize theplant G(s) Since G(s) is stable, take k 0  0 . .• Let k  k  k 0 . By Barmish’s formula, k [0.5522,1.5513] is thestabilizing range of k . Thus, the stabilizing range of k is obtained ask  k 0  k [0.5522,1.5513]• Reset k10  k2  k10  (0.5522  1.5513) / 2  0.4995 and calculate 00  1.0518. Then, the stabilizing range with mutually independentgains of ki is ki [0.5522,1.5513],i  1, 2. 0• Suppose that k1 is more important than k2 and choose k1  0.5522and k 2  1.5513 as initial values. Then, LMIs are still feasible for 0  [0, 2.1035, 2.1035,0].2011年2月8日星期二 29
  30. 30. Illustrative Example• Let   0.5 and relax  as  *  . The sequence of rangeshifting is as follows: firstly find the lower bound of k1 , secondly theupper bound of k2 , thirdly the upper bound of k1 and finally the lowerbound of k2 .• Fix the stabilizing range of k1 as k1 [1.6556,1.2]and computethe stabilizing range of k2 as k1 [1.6556,1.2], k2 [0.45225,100].• If the stabilizing range of k2 is fixed to k2 [0.3529, 4.0967] andthe stabilizing range of k1 is calculated as  1low , 1upp    3.2436,1.94905 ,  which yields the stabilizing proportional controller gain ranges as k1 [3.7958,1.39685], k2 [0.3529, 4.0967].2011年2月8日星期二 30
  31. 31. Illustrative Example k1 k2 Ho’s result k1 [1.6556,1.2] k2 [0.3529, 4.0967] Our result k1 [1.6556,1.2] k2 [0.45225,100] k1 [3.7958,1.39685] k2 [0.3529, 4.0967]RemarkWhen the stabilizing range for one loop is the same, the range for otherloop is much more conservative by Ho’s method than ours.2011年2月8日星期二 31
  32. 32. Illustrative Example k1 k2 Our result k1 [1.6556,1.2] k2 [0.45225,100]  -analysis k1 [0.4516, 0.0040] k2 [0.8710,98.6768] k1 [1.2494, 0.79638] k2 [13.8371,85.7107]Remark --Analysis gives conservative results.• PID parameters are real, while the -analysis treats systemsuncertainties as complex valued.• Stabilizing ranges are generally not symmetric with respect to thenominal value, while the allowable perturbations in the analysis isalways symmetric. 2011年2月8日星期二 32
  33. 33. Illustrative ExampleCase 2: PI control 1Let K (s)  diag k1 , k2   diag k3 , k4 . s• Choose k1  1, k2  5, k3  1 and k40  5, the stabilizing ranges 0 0 0are calculated as k1  [15.4516,1.0542], k2 [4.9458,100] k3  [1.0545, 0.0001], k4  [4.9451,5.0550]• Choose k10  1, k20  1, k30  0.1 and k4  0.1 , the stabilizing 0ranges are calculated as k1  [1.9482, 0.7263], k2  [0.1479,100] k3  [0.1, 0.05], k4  [0.1,1.1035].2011年2月8日星期二 33
  34. 34. Illustrative ExampleCase 3: PD controlLet K (s)   k1  k2 s  I 2• Choose k 0  0 , the stabilizing ranges are calculated as k1 [0.5522,0.6578], k2 [0.6577, 2.0822].• Choose k 0  5 , the stabilizing ranges are calculated as k1 [1.5515,100], k2 [4.2389,100].• Choose k 0  5, the stabilizing ranges are calculated as k1 [100, 0.5683], k2 [100, 0.2361].2011年2月8日星期二 34
  35. 35. Illustrative ExampleCase 4: PID controlLet K ( s)   k1  k2 / s  k3 s  I 2• Choose k 0  5 , the stabilizing ranges are calculated as k1 [1.5344,100], k2  [5.4603,100],k3  [5.2821,100].2011年2月8日星期二 35
  36. 36. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 36
  37. 37. The Proposed ApproachUnder the nominal stabilization (   I m ), the closed-loop system canbe destabilized if and only if there is a gain perturbation  such that det( I  G( j))  0,which is equivalent to exist some non-zero unit vectorsz  C such that ( I  G) z  0 . m v z y +  G(s) _ Figure 1: Diagram of a MIMO control system.2011年2月8日星期二 37
  38. 38. The Proposed ApproachConsider a diagonal real matrix . Let v  v1 , v2 , , vm  , and T  z   z1 , z2 , , zm  . TIt follows from Figure 1 that v  Gz and z  v, which implies zi  ki vi or ki  zi / vi .Basic IdeaWe try to find the solution z through some optimization technique,and then obtain v and  .2011年2月8日星期二 38
  39. 39. The Proposed ApproachTake the following quadratic function as the cost function for ease ofoptimization: z*G* zBecause v z  k v v  ki vi is a real number, and the unit vectorz * i i * i i i 2meets z* z  1 . Then, the constraint conditions are given as  z* z  1     Im   z*G* H i z   0, i=1, ,m.where H i   hpq    is given by 1, p  q  i h pq  0, otherwise2011年2月8日星期二 39
  40. 40. The Proposed Approach min  z *G * z  or max  z *G * z  ,  z* z  1  s.t.    Im   z *G * H i z   0, i=1, .m. z  C mConvert to an equivalent realconstrained optimization problem bydecomplexification.  min Z G Z T T  or  max Z G Z T T   Z T Z  1, (4)  s.t.  T T i  Z G H W Z  0,  i=1,2, ,m. Z  R 2 m2011年2月8日星期二 40
  41. 41. The Proposed Approach• To solve the above constrained optimization problem, we use theLagrange multiplier:     m f ( )  Z G Z  1 Z Z  1   i 1 Z G H W Z T T T T T i i 1where   [ Z , 1, 2 , , m1 ]T . The stationery condition for optimality is T  T m  (G  G ) Z  21 Z   i 1[(G H W )T  G H W ]Z  T i T i   i 1   T Z Z 1  q( ) : f ( ) /      0.  T T Z G H WZ 1       T T m   Z G H WZ 2011年2月8日星期二 41
  42. 42. The Proposed Approach• A numerical solution to is sought by the Newton-Raphson algorithm: n1  n  J 1[q(n )]q(n ) q(n )  2 f (n )where J [q( n )]   n   n2   ( G T  G )  2 I  T 1 T m   1 2m [(G H W )T  [(G H W )T   m 2Z   i 1[(G H W )  G H W ] T i T T i T 1 T m  G H W ]Z G H W ]Z   i 1   2Z T 0 0 0   T   Z [(G H W )T  G H W ]  T 1 T 1 0 0 0      T T m T m   Z [(G H W )T  G H W ] 0 0 0 is the Jacobian matrix of q( n ).2011年2月8日星期二 42
  43. 43. The Proposed Approach• If J is singular, then a Moore-Penrose inverse is used. To seewhether the iteration routine achieves the maximum or minimum, theeigenvalues of the Hessian matrix T H  (G  G)  21 I 2 m m i 1   T i   i 1  G H W   G T T i H W  are calculated and they should be all non-negative for the minimumcase and all non-positive for the maximum case.• To find, say, the maximum from the minimum, a new search iscarried out with the initial search direction set as opposite to theeigenvector of H corresponding to the largest positive eigenvalue.2011年2月8日星期二 43
  44. 44. The Proposed Approach Algorithm 2.Step 1. Run the Newton-Raphson iteration. If the iteration is convergent, obtain z , v and the gains.Step 2. Calculate the eigenvalues of H and decide if the solution is for the minimum (or maximum) case. Set the new initial vector 0 as opposite to the eigenvector of H corresponding to the largest positive (or smallest negative) eigenvalue.Step 3. Go to Step 1 once more for the maximization (or minimization) 2011年2月8日星期二 44
  45. 45. The Proposed ApproachWe look for the relevant frequency range , such that the solution to(4) may exist while there is no solution in its complement set for which(4) will not be performed. m• We know that v z  k v v  k i vi v z   k i vi 2 * i i * i i i is real so that * 2 is i 1also real, which leads to v* z  z*v *  G ( j )  G ( j )  * Im(v z )  * z  z  0 . (5) 2i  2i Let G( j )  G( j )* P( j )  , 2iand  ( P( j)) to be an eigenvalue of P( j ).2011年2月8日星期二 45
  46. 46. The Proposed ApproachThe set ( P)  z Pz : z z  1, z  is commonly called the numerical * * mrange of (Ballantine, 1987). Since P( j)is a Hermitian matrix, thenumerical range of P( j) is the segment of the real axis bounded bythe smallest and largest eigenvalue of P( j)(Horn and Johnson, 1991)• As a result, if the eigenvalues of P( j)spread across zero, that is,there are opposite-sign eigenvalues or zero eigenvalue, then thenumerical range of P( j)contain the origin and there will exist zsatisfying (5) at the underlying frequency  .2011年2月8日星期二 46
  47. 47. The Proposed Approach• We first find all the real roots l such that det( P( j ))  0 . Hence, by lcalculating all  ( P( j)) for one   ( ,  ) , we know their sign l l 1distribution and can then determine whether or not (5) may have asolution. If the answer is yes, we assign (l , l 1 )  .However, the calculation by numerical range method to meet (5) maynot necessarily meet the latter and so-calculated  is overestimated.2011年2月8日星期二 47
  48. 48. The Proposed Approach• Denote  by   (1, 1 ) (r , r ) in the ascending order ofthe frequency. Suppose that we have computed the gain solutions for (1 , 1 ) (n , n ) and formed a closed stability region including   Im .• Find the maximum ki on the boundaries of this region and denoteit by kmax and enlarge and enclose this region by a hypercube definedby ki [kmax , kmax ].2011年2月8日星期二 48
  49. 49. The Proposed ApproachNote that for any frequency    , the gain solutions not inside the * nhypercube will not affect the stability boundaries will not reduce thepreviously determined region.Otherwise, the stability region may be reduced by the new boundariesin the hypercube.The key issue is then to check if the gain solutions of the remainingfrequency intervals (n1, n1 ), , lies inside the hypercube.2011年2月8日星期二 49
  50. 50. The Proposed ApproachIt is well known that the spectral radius forms a lower bound on anycompatible matrix norm (Morari and Zafiriou, 1989 ):  (G)  max i (G)  G 1  G 1  1• It follows that 1   (G* )  max i (G* )  G 1 *  max( k1* , , km ) G 1  kmax G 1 * 1Hence, for any frequency   ( ,  ), where r  n , if 1  k G( j ) , * r r max * 1the gain solution * would lie inside the hypercube, and thecorresponding frequency interval ( ,  ) will be taken into account r rfurther. Otherwise, will be removed in .2011年2月8日星期二 50
  51. 51. The Proposed Approach Algorithm 3.Step 1. Calculate all the real roots l of det( P( j))  0. Divide the entire frequency interval of zero to infinity into [0, 1 ] (1, 2 ] (l , l 1 ] ;Step 2. Take one   (l , l 1 ] and calculate  ( P( j))for l  1,2, . If max ( P( j))min ( P( j))  0 , then (l , l 1 ]   ; Arrange   ( ,  ] 1 1 ( ,  ] r r which is in the ascending order of the frequency. Set r  1 . 2011年2月8日星期二 51
  52. 52. The Proposed ApproachStep 3. Run Algorithm 2 for (r , r ]to get gain solution of (4) and plot them. If a closed region including   I mis formed, let n  r and find the maximum ki on the boundaries of this region, and denote it as kmax . Otherwise, re-do this step with r  r  1; 1Step 4. Remove the frequency interval ( ,  ] from , if G( j )  k r r 1 for all   ( ,  ] and r  n. max r rStep 5. Run Algorithm 2 for all the remaining (r , r ]in , r  n, if any, and plot them. Determine the refined stabilizing region and loop gain margins from it. 2011年2月8日星期二 52
  53. 53. Illustrative ExampleConsider a time delay process  s 1 0.5s  1   ( s  1)( s  3) 0.5s 3  2.5s 2  s  3  G(s)     0.3 0.3s 0.5s  1 0.7 s   s2e  s 2  2s  5 e  • Calculate the real roots of det( P( j))  0 to get 1  2.652,2  2.876,   6.918,   11.135,   15.757. Check the 3 4 5condition of  ( P( j))min  ( P( j))max  0 for any one in each interval,and find   [0, 2.652] (2.876,6.918] (11.135,15.757] .For   0in [0,2.652], the gain solutions are given by  z1  k1    z1 / 3  z2 / 3  k2  z22011年2月8日星期二   0.3z1 / 2  z2 / 5 53
  54. 54. Illustrative Example• Running Step 3 of Algorithm 3 for the first two frequency intervals,[0,2.652] and (2.876,6.918], yields a closed region including   I 2 .The maximum value of ki on the boundaries of this region iskmax  6.659. m•Plot G( j)  max  g ( j) with respect to  and the horizontal 1 j ij i 1straight line in Figure 2. One sees that G( j) 1 marked by blue line is 1always below the red line k , where   8.2. max•So, (11.135,15.757] is all removed from .2011年2月8日星期二 54
  55. 55. Illustrative Example• The region including   I 2 found above is thus the stabilizing oneand is marked in yellow in Figure 3.• Fix the range of k1 as k1 [2.0233,1.1502] . The maximumrectangle is then determined in the stable region and the correspondingrange of k2 is k2 [3.84,3.6852] , which yields the stabilizingproportional controller gain ranges as k1  [2.0233,1.1502] k 2  [3.84,3.6852]2011年2月8日星期二 55
  56. 56. Illustrative Example 1Figure 2: The curves of G( j ) 1 and . Figure 3: Stabilizing region of (k1 , k2 ) kmax 2011年2月8日星期二 56
  57. 57. Illustrative Example• Comparison with LMI method k1 k2 Our method k1  [2.0233,1.1502] k2 [3.84,3.6852] LMI method k1  [2.0233,1.1502] k2 [3.2188,5.0708]For comparison, the loop 2 has the equivalent transfer function asfollows:  kg g   g 22  1 12 21 k 2 g2   (6)  1  k1 g11  2011年2月8日星期二 57
  58. 58. Illustrative ExampleThe Nyquist curve of (6) for two cases, (k1  2.0233, k2  3.2188)from the LMI method, and, (k1  2.0233, k2  3.84) from theproposed method.2011年2月8日星期二 58
  59. 59. Illustrative ExampleNyquist curve of (6) for (k1  1.1502, k2  5.0708)from by LMI method,and (k1  1.1502, k2  3.6852) from the proposed method.2011年2月8日星期二 59
  60. 60. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 60
  61. 61. Problem FormulationProblem 2. For an m  m square system, G( s) , under the decentralizedphase perturbation,  K  diag e j1 , , e jm  , find the ranges,  ,  , i i   i  i   , i  1, , m , such that the closed-loop system is stablewhen i  i , i  for all i, but marginally stable when i  i or i   ifor some i .Definition. The solution to Problem 2, i  i , i  , is called the phasemargin of the i-th loop of G(s) under other loops phases  j  i , i  , j  i, i  1, , m. If i   j   and i   j   , then  ,   is called thecommon phase margin.2011年2月8日星期二 61
  62. 62. Problem Formulation• Time domain method  x(t )  Ax(t )  Bu (t ),• System:   y (t )  Cx(t ), x  n , y  m – Controller:  K  diag e L1s , e L2s ,  , e Lm s ,  y1 (t  L1 )   e1 Cx(t  L1 )  T  y (t  L )   T  e2 Cx(t  L2 )  m u (t )    2 2     I k Cx(t  Lk ),     k 1    T   ym (t  Lm )  emCx(t  Lm )    – Closed-loop system: m x(t )  Ax(t )   BI k Cx (t  Lk ). k 12011年2月8日星期二 62
  63. 63. Problem FormulationTheorem 1. For given scalars Lk  0, k  1, 2, , m, the closed-loopsystem is asymptotically stable if there exists P  PT  0, Qk  QkT  0,Wk  WkT  0, Zij  Zij  0, T  X 00 ) (k X 01 ) (k X 0k )  ( m Mk0   (k ) k  M   X 10 X 11 ) (k X 1(m)  Xk   0, M k   k1  ,      (k ) (k     X m1 X mk2) X mm)  (    M km  Y00ij ) Y01ij ) ( ( Y0(m )  ij  N 0ij )  (  (ij ) ij   (ij )  Y10 Y11ij ) ( Y1(m )  N Yij    0, Nij   1  ,      (ij ) ( ij )   (ij )  Ym1 Ym 2 Ymm   Nm  ( ij )    2011年2月8日星期二 63
  64. 64. Problem Formulationk  1, 2, , m; i  1, 2, , m  1; j  i  1, i  2, , m, such that the followingLMIs hold: ,   P A  A P  Q  A  A   M k Gk  Gk K k  Lk X k  m T T T T T k 1 m 1   m   Nij H ij  H ij Nij  L j  Li Yij  0 T T i 1 j i 1 X Mk  k   k T   0, k  1, , m,  M k Wk   Yij Nij  ij   T   0, i  1, , m  1; j  i  1, , m,  Nij Zij 2011年2月8日星期二 64
  65. 65. Problem Formulationwhere P  P 0 0  0, A   A  BI1C  BI 2C   BI mC , m  Q  diag  Qk  Q1  Q2   Qm ,  k 1  m 1 m m     L j  Li Z ij   LkWk , i 1 j i 1 k 1   Gk   I 0 0  I 0 0,    k 1 mk    H ij  0 0 I 0 0  I 0 0.     i j i 1 m j 2011年2月8日星期二 65
  66. 66. Problem FormulationProof. Choose the cadidate Lyapunov-Krasovskii functional to be m V  xt  : xk (t ) Px(t )    t T xT ( s )Qk x( s ) s t  Lk k 1 m   0 t k 1   Lk t  xT ( s )Wk x( s ) s ,   m1 m  Li t    sgn( L j  Li )   xT ( s ) Z ij x( s ) s ,    L j t  i 1 j i 1and show V ( xt )  0 . 2011年2月8日星期二 66
  67. 67. Problem Formulation To find the range of delays, let Lk = Lk + ¢Lk ; k = 1; 2; ¢ ¢ ¢ ; m, where ^ j¢Lk j · dkwith dk ¸ 0are given large scalars. The first LMI in Theorem 1 becomes Xn m o^ ¹ T ¹ ¹T ¹ ¹ ¹T^ ¹© = P A + A P + Q + A ¦A + T T ^ Mk Gk + Gk Mk + (Lk + dk )Xk k=1 m¡1 X X n m ³¯ ¯ ´ o T T ¯^ ^ ¯ + Nij Hij + Hij Nij + ¯Lj ¡ Li ¯ + dj + di Yij < 0; i=1 j=i+1 where m¡1 X X ³¯ m ¯^ ¯ ¯ ´ m X ^ ¦= ^ ¯Lj ¡ Li ¯ + dj + di Zij + ^ (Lk + dk )Wk : i=1 j=i+1 k=1 2011年2月8日星期二 67
  68. 68. Problem Formulation Algorithm 4.Step 1. Choose the initial Lk , k  1, 2, , m, such that LMIs in Theorem 3 are feasible and set L  L , k  1, 2, , m. k kStep 2. For Lk , find a maximum d  0 such that new LMIs are feasible when dk  d . Let dk  d , k  1, 2, , m and i  1.Step 3. For fixed dk  d , k  i, find a maximum di  d such that new LMIs are feasible.Step 4. If d  L , let Li  Li  di , then go to Procedure A. Else, let i i Li  0 , then go to Procedure B. 2011年2月8日星期二 68
  69. 69. Problem FormulationStep 5. Let Li  ( Li  Li ) / 2 and di  ( Li  Li ) / 2 . If i  m, ˆ let i  i  1 go to Step 3.Step 6. The ranges of Lk [ Lk , Lk ], k  1, 2, , m, are those for guaranteeing the stability of closed-loop system. Procedure A.Step 1. Let rlow = Li ¡ di; rupp = Li + di; min = 0; max = rlow; and ^ ^ ^ Li = rupp=2; di = rupp=2:Step 2. If new LMIs are feasible, let rlow = 0, then go to Step 6. 2011年2月8日星期二 69
  70. 70. Problem Formulation ˆStep 3. Else, let mid  (min  max) / 2, rlow  mid , Li  (rupp  rlow ) / 2 and di = (rupp ¡ rlow)=2:Step 4. If new LMIs are feasible, let max = mid, else, let min = mid:Step 5. Step 5. If jmax ¡ minj < ,² let rlow = max Else, go to Step 3. .Step 6. Step 6. Let Li = rlow then return to Step 5 of Algorithm 2. , Procedure B. ^ ^Step 1. Let rlow = 0; rupp = Li + di; min = rupp; max = ±; Li = max=2 and di = max=2: 2011年2月8日星期二 70
  71. 71. Problem FormulationStep 2. If new LMIs are feasible, let rupp = max, then go to Step 6.Step 3. Else, let ^ mid = (min + max)=2; rlow = mid; Li = (rupp + rlow)=2Step 4. and di = (rupp ¡ rlow)=2:Step 5. If new LMIs are feasible, let max = mid , else, let min = mid:Step 6. If jmax ¡ minj < ² , let rlow = max . Else, go to Step 3. Let Li = rlow, then return to Step 5 of Algorithm 2. 2011年2月8日星期二 71
  72. 72. Problem FormulationProposition 2. (Bar-on and Jonckheere 1998). There exists a unitary¢in the feedback path which destabilizes the system, G(s), if and only ifthere exists an ! such that ¾(G(j!)) ¸ 1 and 0 · ¾(G(j!)) · 1 . ¹Let Ð = f! j 0 · ¾(G(j!)) · 1 · ¾(G(j!))g and ^ ^ !g = minf! j ! 2 Ðg:The closed-loop system remains stable for all Ák 2 (!g Lk ; !g Lk ) := (Ák ; Ák ):Solution of Ð ^ q ¾(G(j!)) = GH (s)G(s); G(s) = C(sI ¡ A)¡1 B; GH (s) = GT (¡s) = [C(¡sI ¡ A)¡1 B]T = ¡B T (sI + AT )¡1 C T :2011年2月8日星期二 72
  73. 73. Problem Formulation G : x1 = Ax1 + Bu; y1 = Cx1 ; _ GH : x2 = ¡AT x2 + C T y1 ; y2 = ¡B T x2 : _ · ¸ · ¸· ¸ · ¸ x_ A 0 x1 B GH G : 1 = + u; x2 _ C T C ¡AT x2 0 | {z } |{z} ~ A ~ B · ¸ £ ¤ x1 y = 0 ¡B T : | {z } x2 £ ¤ h~ C i H ~ ~ det I ¡ G (s)G(s) = det I ¡ C(sI ¡ A) B¡1 ~ h i ~~ ~ = det I ¡ B C(sI ¡ A) ¡1 h i. ~ ~~ ~ = det sI ¡ (A + B C) det(sI ¡ A):2011年2月8日星期二 73
  74. 74. Problem Formulation Algorithm 5.Step 1. Calculate the purely imaginary eigenvalues, !i; i = 1; 2; ¢ ¢ ,¢ of the matrix A + BC. ~ ~~Step 2. Choose any ! 2 (!i; !i+1); ! > 0, and calculate ¾(G(j!)) and ¹ ¾(G(j!)). If ¾(G(j!)) ¸ 1 and ¾(G(j!)) · 1 , then ¹ [ ^ Ð = (!i ; !i+1):Step 3. Let !g = minf! j ! 2 Ðg then the stabilizing range of Ák ^, is calculated as Ák 2 (!g Lk ; !g Lk ) := (Ák ; Ák ): Remarks: Conservativeness comes from both the LMI conditions and the determination of crossover frequency . 2011年2月8日星期二 74
  75. 75. Illustrative Example Example (Bar-on and Jonckheere ,1998) 2 3 2 3 0 1 0 0 0 0 · ¸ 6 ¡3 0:75 1 0:25 7 6 7 6 A=4 7; B = 6 0 1 7 ; C= 1 0 0 0 : 0 0 0 1 5 4 0 0 5 0 0 1 0 4 1 ¡4 ¡1 0:25 0G(s) = C(sI ¡ A)¡1 B · 2 ¸ 1 0:0625s + 0:25 s +s+4 = 4 + 1:75s3 + 7:5s2 + 4s + 8 0:25s2 + 0:1875s + 0:75 : s s+4 L0 L0 L1 L2 ^ Ð !g Á1 Á2 1 2 0 0 [0, 0,1979] [0, 0.1967] [0. 0.1272] [0, 0.1265] [0.643, 1.613] 0.643 0.1 0 [0, 0.2920] [0, 0.1914] [0, 0.1878] [0, 0.1231] 2011年2月8日星期二 75
  76. 76. Illustrative ExampleExample (cont’d)2011年2月8日星期二 76
  77. 77. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 77
  78. 78. The Proposed Approach• Frequency domain method¢(s) = diagfejÁi g; i = 1; 2; ¢ ¢ ¢ ; m: det(I + G(j!)¢) = 0 ( 9z 2 Cm; s.t. z = ¢v = ¡¢Gz: )Solution of z (constrained optimization) object: z¤ (G¤ + G)z ½ ¤ z z = 1; s.t. z¤ (Hk ¡ G¤ Hk G)z = 0; k = 1; 2; ¢ ¢ ¢ ; m:2011年2月8日星期二 78
  79. 79. The Proposed ApproachLagrange multiplier m X F (·) = z¤(G¤ + G)z + ¸1 (z¤ z ¡ 1) + ¸k+1 z¤(Hk ¡ G¤ Hk G)z; k=1with · = [z1 ; ¢ ¢ ¢ ; zm; ¸1 ; ¸2 ; ¢ ¢ ¢ ; ¸m+1 ]T : 2 m 3 X ¤ 6(G + G)z + ¸1 z + ¸k+1 (Hk ¡ G¤ Hk G)z7 6 7 6 k=1 7 @F (·) 6 6 ¤ z z¡1 7 f (·) = =6 7; @· ¤ ¤ z (H1 ¡ G H1 G)z 7 6 7 6 . . 7 4 . 5 z¤ (Hm ¡ G¤ Hm G)zNewton-Raphson: ·n+1 = ·n ¡ J ¡1[f(·n)]f(·n):2011年2月8日星期二 79
  80. 80. The Proposed Approach @f (·) @ 2 F (·)J[f(·)] = = =2 @· @·2 3 (G¤ + G) + ¸1 Im6 X m 76 + z (H1 ¡ G¤ H1 G)z ¢ ¢ ¢ (Hm ¡ G¤ Hm G)z76 ¸k+1 (Hk ¡ G¤ Hk G) 76 k=1 76 76 2z¤ 0 0 ¢¢¢ 0 76 76 2z¤ (H1 ¡ G¤ H1 G) 0 0 ¢¢¢ 0 76 76 . . . . . . .. . . 74 . . . . . 5 2z¤ (Hm ¡ G¤ Hm G) 0 0 ¢¢¢ 0 is the Jacobian matrix of f(·) ½ 1; i = j = k; Hk = [hi;j ] 2 Rm£m; and hi;j = 0; otherswise. 2011年2月8日星期二 80
  81. 81. The Proposed Approach Algorithm 6.Step 1. Determine Ð = f! j 0 · ¾(G(j!)) · 1 · ¾(G(j!))gfrom ^ Proposition 2 and Algorithm 3.Step 2. Construct the framework of constrained optimization.Step 3. For each ! 2 Ð, solve the optimization problem with Lagrange ^ multiplier and find z with Newton-Raphson method.Step 4. Stabilizing boundary of phase perturbation is given by Ái = argfzi=vig; i = 1; 2; ¢ ¢ ¢ ; m:2011年2月8日星期二 81
  82. 82. Illustrative ExampleExample (Bar-on and Jonckheere, 1998) ^ Ð = (0:643; 1:613) Ð = (0:764; 0:884) [ (1:533; 1:572)2011年2月8日星期二 82
  83. 83. Illustrative Example Loop Phase Margin Method Common Phase Margin ������₁ ������₂ Frequency domian (−������, ������) (−0.2423, 0.2423) (−0.3108, 0.3108) Bar-on and Jonckheere N.A. N.A. (−0.2701, 0.2701) Time domain [0, 0.1878] [0, 0.1231] [0, 0.1265]2011年2月8日星期二 83
  84. 84. OUTLINE• Introduction• Loop Gain Margins - Time Domain Method - Frequency Domain Method• Loop Phase Margins - Time Domain Method - Frequency Domain Method• Conclusions2011年2月8日星期二 84
  85. 85. Conclusions1. The Loop gain and phase margins of MIMO feedback systems has been defined2. The algorithms presented for computations of such margins in both time and frequency domains3. The computations involved are substantial, and further improvement is under consideration4. Use of these margins for controller tuning is under progress2011年2月8日星期二 85
  86. 86. 2011年2月8日星期二 86

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