MIMO Stability Margins

Stability Margins of
Multivariable Feedback Systems

OUTLINE
• Introduction
• Loop Gain Margins
- Time Domain Method
- Frequency Domain Method
• Loop Phase Margins
• Conclusions

2011年2月8日星期二 2

OUTLINE
• Introduction
• Conclusions


HISTORICAL REMARKS
• Uncertain systems
• Robust control: dominant approach
• Margin based approach


SISO STABILITY MARGINS
The unity output feedback configuration.
G(s)is the process

C (s)is the compensator.

SISO stability margins
- Gain margin: GM  1/ G( j ) .
p

- Phase margin: PM  180  G( jc ).


MIMO MOTIVATION EXAMPLE
Consider 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 


The 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 )  0

The 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)

The 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.

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.


• 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.



• A similar problem in concept is phase margin but different for
computation
• Our goal is to define meaningful margins and compute them


OUTLINE
• Introduction
• Conclusions


PROBLEM FORMULATION
Consider an m m square plant with dimensional state-space
realization:
x(t )  Ax (t )  Bu (t ),

y (t )  Cx(t ),

x , y  m B and C are real constant matrices with appropriate
n

dimensions. 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 

PROBLEM FORMULATION
k1i , k2 jand k3l are scalars to be determined, I1i, I 2 jand I 3l are
identity 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 3l

Then,
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 1

where I vi  diag 0, ,0, I vi ,0, ,0  mm
, v  1, 2,3, i  1, 2, , rv .


PROBLEM FORMULATION
Problem 1. For a given plant G(s) under the controller u(t ), find the
ranges of scalars k1i , k2 j and k3l , i  1, , r1 , j  1, , r2 , l  1, , r3
such that the closed-loop system is stable for all allowable k1i , k2 j and
k3l in these ranges.

Remark
The 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
1
K ( s)  diag k   diag k   sdiag k  corresponds to the specially
1i 2i 3i
s
chosen r1  r2  r3  m (or 　m1i  m2 j  m3l  1).
.

PROBLEM FORMULATION
For an m m square plant G(s) under the decentralized proportional
control 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 the
system. Graphically, such a stabilizing range is the largest line segment
of k1  k2   km available in the stabilizing region for
ki , i  1, 2, , m.


PROBLEM FORMULATION
Definition
If K (s)  diag k1 , k1 , , km  with probably different gains for
different 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.
 
Remark
The 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 respective
ranges.

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  .

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 .


PROPOSED APPROACH
Definition
A 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 k3l
i  1, 2, , r1 , j  1, 2, , r2 , l  1, 2, , r3 , such that the closed-loop
system is admissible for all allowable k1i , k2 j and k3l in these ranges.


PROPOSED APPROACH
0
Let k0
vi , v  1, 2,3, i  1, 2, be such that ( E, A ) with
, rv cl

Acl  A  i 1 k10i A1i  i 1 k2i A2i  i 1 k3i A3i , is admissible.
0 r 1 r 0 2 r 3 0

Set 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  2r0
vertices 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
 

PROPOSED APPROACH
Lemma
The pair  E, A( )  is robustly admissible if and only if there exist
parameter-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 by
symmetry.
Proof: The proof is parallel to that for standard systems in Geromel et
al. (1998) and Peaucelle et al. (2000).


SPECIAL CASES
Proposition
The pair  E, Acl is robustly admissible if and only if there exist
matrices Pj , F j , H j and X jl with X jj  X T , l  j, j, l  1, 2, , r,
jj

such 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


SPECIAL CASES
In 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 ),
1

where 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 


SPECIAL CASES
Proposition
The polytope A is robustly stable if there exist matrices P  0 , and
F 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.

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
0

Step 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
0

Step 3. that LMIs are feasible for kvi  kvi  (k min  k max ) / 2. Obtain the
0 0

   0 , 0 , , 0 , 0 

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 0

Step 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
0
Step 6. If  vi  0
* low upp

such that LMIs are still feasible for i  1, 2, , m.
0
Step 7. Calculate the range of kvi by kvi  k vi   vi , vi 

low upp


Illustrative Example
Consider 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.

Case 1: P control
• Consider a common gain controller K (s)  kI 2 , to stabilize the
plant 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 the
stabilizing range of k . Thus, the stabilizing range of k is obtained as
k  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 independent
gains of ki is ki [0.5522,1.5513],i  1, 2.
0
• Suppose that k1 is more important than k2 and choose k1  0.5522
and 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

• Let   0.5 and relax  as  *  . The sequence of range
shifting is as follows: firstly find the lower bound of k1 , secondly the
upper bound of k2 , thirdly the upper bound of k1 and finally the lower
bound of k2 .
• Fix the stabilizing range of k1 as k1 [1.6556,1.2]and compute
the 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] and
the 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].

30

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]

Remark
When the stabilizing range for one loop is the same, the range for other
loop is much more conservative by Ho’s method than ours.


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 systems
uncertainties as complex valued.
• Stabilizing ranges are generally not symmetric with respect to the
nominal value, while the allowable perturbations in the analysis is
always symmetric.

Case 2: PI control
1
Let K (s)  diag k1 , k2   diag k3 , k4 .
s
• Choose k1  1, k2  5, k3  1 and k40  5, the stabilizing ranges
0 0 0

are 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
0

ranges are calculated as
k1  [1.9482, 0.7263], k2  [0.1479,100]
k3  [0.1, 0.05], k4  [0.1,1.1035].

33

Case 3: PD control
Let 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].

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].

34

Case 4: PID control

Let K ( s)   k1  k2 / s  k3 s  I 2


k1 [1.5344,100], k2  [5.4603,100],k3  [5.2821,100].

35

OUTLINE
• Introduction
• Conclusions


The Proposed Approach
Under the nominal stabilization (   I m ), the closed-loop system can
be 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 vectors
z  C such that ( I  G) z  0 .
m

v z y
+

 G(s)
_

Figure 1: Diagram of a MIMO control system.

37

Consider a diagonal real matrix . Let v  v1 , v2 , , vm  , and
T

z   z1 , z2 , , zm  .
T

It follows from Figure 1 that v  Gz and z  v, which implies
zi  ki vi or ki  zi / vi .

Basic Idea
We try to find the solution z through some optimization technique,
and then obtain v and  .

38

Take the following quadratic function as the cost function for ease of
optimization:
z*G* z
Because v z  k v v  ki vi is a real number, and the unit vectorz
*
i i
*
i i i
2

meets 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, otherwise

39

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 m

Convert to an equivalent real
constrained optimization problem by
decomplexification.


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 m
40

• To solve the above constrained optimization problem, we use the
Lagrange multiplier:
   
m
f ( )  Z G Z  1 Z Z  1   i 1 Z G H W Z
T T T T T i

i 1

where   [ 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 

41

• 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 ).
42

• If J is singular, then a Moore-Penrose inverse is used. To see
whether the iteration routine achieves the maximum or minimum, the
eigenvalues 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 minimum
case and all non-positive for the maximum case.

• To find, say, the maximum from the minimum, a new search is
carried out with the initial search direction set as opposite to the
eigenvector of H corresponding to the largest positive eigenvalue.
43


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)


We 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 1
also 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 )  ,
2i

and  ( P( j)) to be an eigenvalue of P( j ).
45

The set ( P)  z Pz : z z  1, z  is commonly called the numerical
* * m

range of (Ballantine, 1987). Since P( j)is a Hermitian matrix, the
numerical range of P( j) is the segment of the real axis bounded by
the 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 the
numerical range of P( j)contain the origin and there will exist z
satisfying (5) at the underlying frequency  .

46

• We first find all the real roots l such that det( P( j ))  0 . Hence, by
l

calculating all  ( P( j)) for one   ( ,  ) , we know their sign
l l 1

distribution and can then determine whether or not (5) may have a
solution. If the answer is yes, we assign (l , l 1 )  .

However, the calculation by numerical range method to meet (5) may
not necessarily meet the latter and so-calculated  is overestimated.

47

• Denote  by   (1, 1 ) (r , r ) in the ascending order of
the 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 denote
it by kmax and enlarge and enclose this region by a hypercube defined
by ki [kmax , kmax ].

48

Note that for any frequency    , the gain solutions not inside the
*
n

hypercube will not affect the stability boundaries will not reduce the
previously determined region.

Otherwise, the stability region may be reduced by the new boundaries
in the hypercube.

The key issue is then to check if the gain solutions of the remaining
frequency intervals (n1, n1 ), , lies inside the hypercube.

49

It is well known that the spectral radius forms a lower bound on any
compatible 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
*
1

Hence, for any frequency   ( ,  ), where r  n , if 1  k G( j ) ,
*
r r max
*
1

the gain solution * would lie inside the hypercube, and the
corresponding frequency interval ( ,  ) will be taken into account
r r

further. Otherwise, will be removed in .

50


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 .



Step 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;
1
Step 4. Remove the frequency interval ( ,  ] from , if G( j )  k
r r 1

for all   ( ,  ] and r  n.
max
r r

Step 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.


Consider 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 5

condition 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  z2
2011年2月8日星期二 
 0.3z1 / 2  z2 / 5 53

• 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 is
kmax  6.659.

m

•Plot G( j)  max  g ( j) with respect to  and the horizontal
1 j
ij
i 1
straight line in Figure 2. One sees that G( j) 1 marked by blue line is
1
always below the red line k , where   8.2.
max

•So, (11.135,15.757] is all removed from .

54


• The region including   I 2 found above is thus the stabilizing one
and is marked in yellow in Figure 3.

• Fix the range of k1 as k1 [2.0233,1.1502] . The maximum
rectangle is then determined in the stable region and the corresponding
range of k2 is k2 [3.84,3.6852] , which yields the stabilizing
proportional controller gain ranges as
k1  [2.0233,1.1502] k 2  [3.84,3.6852]

55


1
Figure 2: The curves of G( j ) 1 and . Figure 3: Stabilizing region of (k1 , k2 )
kmax

56

• 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 as
follows:
 kg g 
 g 22  1 12 21 k 2
g2   (6)
 1  k1 g11 


57

The 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 the
proposed method.

58

Nyquist 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.

59

OUTLINE
• Introduction
• Conclusions


Problem Formulation
Problem 2. For an m  m square system, G( s) , under the decentralized
phase 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 stable
when i  i , i  for all i, but marginally stable when i  i or i   i
for some i .

Definition. The solution to Problem 2, i  i , i  , is called the phase
margin 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 the
common phase margin.

61

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 1


Problem Formulation
Theorem 1. For given scalars Lk  0, k  1, 2, , m, the closed-loop
system 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 )
   


Problem Formulation
k  1, 2, , m; i  1, 2, , m  1; j  i  1, i  2, , m, such that the following
LMIs 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 


Problem Formulation
where
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 


Problem Formulation
Proof. 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 1

and show V ( xt )  0 .



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


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 k

Step 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.


Problem Formulation

Step 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.


Problem Formulation

ˆ
Step 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:


Problem Formulation

Step 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)=2
Step 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.


Problem Formulation
Proposition 2. (Bar-on and Jonckheere 1998). There exists a unitary¢
in the feedback path which destabilizes the system, G(s), if and only if
there exists an ! such that ¾(G(j!)) ¸ 1 and 0 · ¾(G(j!)) · 1 .
¹

Let 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 :


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):


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 .


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 0

G(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]


Example (cont’d)


OUTLINE
• Introduction
• Conclusions


• 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:


Lagrange multiplier m
X
F (·) = z¤(G¤ + G)z + ¸1 (z¤ z ¡ 1) + ¸k+1 z¤(Hk ¡ G¤ Hk G)z;
k=1

with · = [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)z

Newton-Raphson: ·n+1 = ·n ¡ J ¡1[f(·n)]f(·n):


@f (·) @ 2 F (·)
J[f(·)] = = =
2 @· @·2 3
(G¤ + G) + ¸1 Im
6 X m 7
6 + z (H1 ¡ G¤ H1 G)z ¢ ¢ ¢ (Hm ¡ G¤ Hm G)z7
6 ¸k+1 (Hk ¡ G¤ Hk G) 7
6 k=1 7
6 7
6 2z¤ 0 0 ¢¢¢ 0 7
6 7
6 2z¤ (H1 ¡ G¤ H1 G) 0 0 ¢¢¢ 0 7
6 7
6 .
. .
. .
. .. .
. 7
4 . . . . . 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.



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:


Example (Bar-on and Jonckheere, 1998)

^
Ð = (0:643; 1:613)
Ð = (0:764; 0:884) [ (1:533; 1:572)



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]

OUTLINE
• Introduction
• Conclusions


Conclusions
1. The Loop gain and phase margins of MIMO feedback systems has
been defined
2. The algorithms presented for computations of such margins in
both time and frequency domains
3. The computations involved are substantial, and further
improvement is under consideration
4. Use of these margins for controller tuning is under progress

85

MIMO Stability Margins

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