Mixed Uncertainties
A. Helmersson
Department of Electrical Engineering Linkoping University
S-581 83 Linkoping, Sweden tel: +46 13 281622 fax: +46 13 282622 email:
andersh@isy.liu.seSeptember 26, 1994 1995 ECC
Abstract
This paper presents the solution to the mixed synthesis problem, and how to design gain scheduling controllers with linear fractional transfor- mations (LFTs). The system is assumed to have a parameter dependences described by LFTs The goal of the controller is, with real-time knowledge of the parameters, to provide disturbance and error attenuation. The paper treats both complex and real parametric dependencies.
Keywords:
Gain sceduling, linear fractional transforms, structured singular values, parametric dependent systems, synthesis.
1 Introduction
During the last couples of year synthesis methods for gain scheduling using linear fractional transformations (LFTs) and structured singular values have been developed, see e.g 10, 9, 7]. The idea behind this approach is to let the controller have access to some of the uncertainties or perturbations of the system to be controlled. The design leads to two linear matrix inequalities (LMIs), which are coupled by a third LMI and non-convex rank conditions.
This is analogous with the two-Riccati algorithm for solving
H1synthesis.
In the referred papers 10, 9, 7], the shared uncertainties have been assumed
to be complex, possibly repeated scalar blocks. In this paper the mixed
design
problem (that is with both real and complex uncertainties) is elaborated. The
synthesis algorithm has the same structure as in the pure complex case with two
LMIs coupled by a third LMI and rank conditions. The di erence to the pure
M
-
Figure 1: System with uncertainties.
complex case is that blocks corresponding to real uncertainties in the connecting LMI disappear only the rank conditions remain.
The structure of the problem is treated in a unied and general way cover- ing both discrete and continuous systems, which are equivalent using bilinear transformation between the
z-domain and the
s-domain 2].
1.1 Notations
X
denotes the complex conjugate transpose of
XX >( ) 0 a hermitian (
X=
X
) positive denite (semidenite) matrix
X;= (
X)
;1 Xyis the Moore- Penrose pseudo inverse of
Xdiag
X1X2] a block-diagonal matrix composed of
X1and
X2rank
Xdenotes the rank of the matrix
Xherm
X=
12(
X+
X)
S
(
::) denotes the Redhe er star product
(
X) the maximal singular value of
X
.
2 -analysis and LFTs
This section gives a short review on structured singular values and linear frac- tional transformations (LFTs), see also e.g. 2].
2.1 Denitions
We will here use a similar notations and denitions as is used in 12, 13]. The denition of
depends upon the underlying block structure
Nof the uncer- tainties , which could be either real or complex, see Figure 1. For notational convenience we assume that all uncertainty blocks are square. This can be done without loss of generality by adding dummy inputs or outputs.
Given a matrix
M 2Cn
n and three non-negative integers
fr ,
fc and
fC , with
f=
fr +
fc +
fC
nthe block structure is an
f-tuple of positive integers
N
=
n1:::nf
rnf
r+1:::nf
r+f
cnf
r+f
c+1::: nf
r+f
c+f
C] (1)
where
Pf i
=1ni =
nfor dimensional compatibility. The set of allowable per-
turbations is dened by a set of block diagonal matrices
X 2 Cn
n dened
by
X
=
f= diag
r
1In
1:::rn
frIn
frc
1
I
n
fr+1:::cf
cIn
fr+fcC
1:::Cn
C] :
ri
2Rci
2CCi
2Cn
fr+fc+in
fr+fc+ig:(2) Assuming the uncertainty structure
N, the structured singular value
of a matrix
M2Cn
n is dened by
=
min
2X
f
() : det(
I;M) = 0
g
;1
(3)
and if no
2Xsatises det(
I;M) = 0 then
(
M) = 0.
2.2 Upper and Lower Bounds
Generally the structured singular value cannot be exactly computed, and instead we have to resort to upper and lower bounds, which are usually sucient for most practical applications. A tutorial review of the complex structured singular value is given in 8].
An upper bound can be determined using convex methods, either involving minimization of singular values with respect to a scaling matrix or by solving a linear matrix inequality (LMI) problem. The upper bound is conservative in the general case, but can be improved by branch and bound schemes.
A lower bound can be found by maximizing the real eigenvalue of a scaled matrix. This bound is nonconservative in the sense that if the true global maximum is found it is equal to
. However, since the problem is not convex, we cannot guarantee that we nd the global maximum.
We will here focus on the computation of the upper bound, which we here denote
, in order to distinguish it from the true
function. The upper bound
can be computed as a convex optimization problem. For complex uncertainties it is determined by
(
M) = inf D
2D
(
DMD;1) (4)
where
Dis the set of block diagonal Hermitian matrices that commute with
X, that is
D
=
f0
<D=
D2Cn
n :
D=
D82Xg:(5) This problem is equivalent to an LMI problem
(
M) = inf >
P
2D0f
:
MPM<2Pg:(6) Real uncertainties can be included in the LMI problem for computing the upper bound (see e.g. 3, 12, 13]):
(
M) = inf >
P G
2D2G0f
:
MPM+
j(
GM;MG)
<2Pg(7)
where
G
=
fG:
G=
G2Cn
n
G=
G82Xg:(8) Every
G2Gis block diagonal with zero blocks for complex uncertainties. If we let
G=
f0
gin (7) we recover the complex upper bound (6).
We can reformulated (7) as a positive real property
(
M) = inf >
W
2W0
: herm
;(
I+
1M
)
W(
I;1M
)
>0
(9) where herm
X=
12(
X+
X) and
W=
fW=
P+
jG:
P 2DG2Gg. Note that herm
W >0 always and that
W=
Wfor complex uncertainties. Another equivalent reformulation of (7) is
(
M) = inf >
P G
2D2G0f
:
;1DMD;1;jG
(
I+
G2)
;12 <1
g:(10)
2.3 Linear Fractional Transformations (LFTs)
Suppose
Mis a complex matrix partitioned as
M
=
M
11 M
12
M
21 M
22
2C
(
p
1+p
2)(m
1+m
2)(11) and let u
2Cm
1p
1and l
2Cm
2p
2. The upper and lower linear fractional transformations (LFTs) are dened by
F
u (
Mu ) =
M22+
M21u (
I;M11u )
;1M12(12) and
F
l (
Ml ) =
M11+
M12l (
I;M22l )
;1M21(13) respectively. Clearly, the existence of the LFTs depends on the invertibility of
I;M
11
u and
I;M22l respectively.
The Redhe er star product 11] is a generalization of the LFTs. Assume that
Qis partitioned similarly to
M. Then the star product is dened by
S
(
QM) =
F
l (
QM11)
Q12(
I;M11Q22)
;1M12M
21
(
I;Q22M11)
;1Q21 Fu (
MQ22)
:
(14) A block diagram illustrating the star product is given in Figure 2.
Note that the denition above is dependent on the partitioning of the ma- trices
Qand
M. The LFTs can be dened by the
Snotation, as
F
u (
Mu ) =
S( u
M) and
F
l (
Ml ) =
S(
Ml )
:The star product is associative, that is
S
(
AS(
BC)) =
S(
S(
AB)
C)
:M
!
!
!
!
!
a a a a a
,
S
(
QM)
Figure 2: The Redhe er star product.
3 Synthesis
In this section we will treat the
synthesis problem when uncertainties in the controller may be shared by the controller.
3.1 Structure of shared uncertainties
Formally we treat this by letting the controller
Khave access to a copy of the system's uncertainties. Not all blocks in the uncertainty matrix are accessible and we denote the accessible uncertainties by
L
R
() = diag
1Ir
12Ir
2:::f
1Ir
f1] (15) where
f1=
fr +
fc . We only allow blocks that are scalar repeated blocks of the type
i
Ir
iwhere
could be either real or complex. The structure of the controller accessible uncertainties are given by
R
=
r1:::rf
rrf
r+1:::rf
r+f
c0
:::0] (16) Note that all blocks corresponding to full complex blocks are not allowed and, consequently,
ri = 0 for those.
3.2 The General Synthesis Problem
The general synthesis problem can be depicted as in Figure 3. Find a controller
K
, possibly with a given order, to the system
M(or the augmented system ~
M), such that
(
S(
MK)) =
(
S( ~
MK)) is minimized or below a specied value.
The diagrams below show three equivalent representations of the
-synthesis problem. In the left diagram
Mand
Khave separated uncertainty blocks, and
LR() respectively. In the middle diagram the uncertainties are explicitly shared. Here ~ = diag
LR()]. In the right diagram through connections between
Kand have been included in ~
M.
Here we use a somewhat more general structure of ~ than used so far. With this formulation it is possible to catch most of the relevant synthesis problems in the
setting. The generalized structure of ~ is nothing more than a reordering of rows and columns, so that both
Mand
Kmay share the same uncertainties.
For instance, if includes frequency, either
sor
z, we can handle continuous-
time and discrete-time problems respectively.
L
R
()
K
- -
M
-
,
K M
!
!
!
!
!
a a a a a
~
- -
,
K -
~
M
~
-
Figure 3: Equivalent representations of a system
Mcontrolled by
Kthat have access to a reduced copy
LR() of the system's uncertainties . The augmented uncertainty block ~ = diag
LR()].
3.3 An Ane Problem
We assume that
Mis partitioned as:
M
=
M
11 M
12
M
21 M
22
2C
(
n
+m
)(n
+p
):(17) Without restriction we may assume that
M22= 0 (
Mis strictly proper). If
M
22
6
= 0, we replace
Mwith
M
=
S(
MNM ) =
M
11 M
12
M
21
0
(18) with
N
M =
0
II ;M
22
(19) Thus, if we can nd a controller
Kto the modied system
M, then
K=
S
(
NM
K) will solve the original problem. Since the star product inverse of
NM
exists for all
M, the modied problem is equivalent to the original one as long as
I;M22Kis not singular.
Due to the simpler structure of
M, the star product can be rewritten as the following matrix expression, which is ane in
K:
S
(
MK) =
M11+
M12KM 21=
Q+
UKV:(20)
3.4 Mathematical Preliminaries
In this section we give two important and related lemmas to be used for solv- ing
synthesis problems. The lemmas, theorems and proofs are made on the assumption that the matrices are complex. The real case can be obtained by replacing
Cwith
Rwithout a ecting the validity of the theory.
We rst consider an ane optimization problem 1].
Lemma 3.1 Assume that
Q2Cn
n and that both
U 2Cn
p and
V 2Cm
n have full column rank and row rank respectively. Suppose
U? 2Cn
(n
;p
)and
V
? 2 C
(
n
;m
)n are chosen such that
U U?,
V
V
?
are both invertible, and that
UU?= 0 and
VV?= 0. Then
K
2Cinf
p m(
Q+
UKV)
<(21) if and only if
V
?
;
Q
Q; 2
I
V
?
<
0 (22)
and
U
?
;
; 2
I
U
?
<
0
:(23)
Proof: See 1].
2A related lemma 4, 5], goes as follows.
Lemma 3.2 Given a hermitian matrix
Q 2 Cn
n and
U,
V,
U?and
V?as above. Then
Q
+
UKV+ (
UKV)
<0 (24) is solvable for
K2Cp
m if and only if
U
? QU
?
<
0
V
? QV
?
<
0
:(25)
Proof: Necessity of (25) is clear: for instance,
U?U= 0 implies
U?QU?<0 when pre- and post-multiplying (24) by
U?and
U?respectively. For details on
the suciency part, see 4, 5].
2In 5] the set of
Ksolving (24) is parametrized.
3.5 Complex -synthesis
If we restrict the problem to only have complex uncertainties (real uncertainties can be conservatively treated us complex ones), the following lemma from 10]
applies.
Lemma 3.3 Let
Q,
U,
V,
U?and
V?be given as above. Then
K
2CD inf
2Dp m ;D(
Q+
UKV)
D;1<(26) if and only if there exists a
P 2Dsuch that
V
?
;
Q
PQ; 2
P
V
?
<
0 (27)
and
U
?
;
QP
;1
Q
; 2
P
;1
U
?
<
0
:(28)
Proof: The necessity of (27) follows by rewriting the inequality in (26) as (
Q+
UKV)
P(
Q+
UKV)
;2P<0
and pre and post-multiplying by
V?and
V?respectively. The necessity of (28)
is obtained analogously. For the suciency part refer to 10].
23.6 Mixed -synthesis
If both real and complex uncertainties are involved, we have a mixed
problem.
The following lemma then applies.
Lemma 3.4 Let
Q,
U,
V,
U?and
V?be given as above. Then
K
2CD G inf
2D2Gp m
;
1
D(
Q+
UKV)
D;1;jG(
I+
G2)
;12 <1
(29) if and only if there exists an
W 2Wwhere
W=
fW=
P+
jG:
P 2DG2Gg, such that
herm
;V?(
I+
1Q)
W(
I ;1Q
)
V?>0 (30) and
herm
;U?(
I+
1Q
)
W;1(
I;1Q
)
U?>0 (31) where herm(
X) =
12(
X+
X).
Proof: Let ^
Q=
1Q
, ~
Q= (
DQD^
;1 ;jG)(
I+
G2)
;12, ~
U=
DU, ~
V=
VD
;1
(
I+
G2)
;12and ^
K=
1K. Then
Q
~ + ~
UK^
V~
<1 (32) if and only if
~
U
?
( ~
QQ~
;I)~
U?=
U?D;1(
DQD^
;1;jG)(
I+
G2)
;1(
D;Q^
D+
jG)
;I D;U?=
U?QP^ U
Q^
+
j( ^
QGU
;GU
Q^
)
;PU
U
?
= herm
U?( ^
Q+
I)(
PU
;jGU )( ^
Q;I)
U? <0
(33) with
PU =
D;1(
I+
G2)
;1D;and
GU =
D;1(
I+
G2)
;1GD;. Analogously we obtain
~
V
?
( ~
QQ~
;I)~
V?=
V?D;1(
D;Q^
D+
jG)(
DQD^
;1;jG)
;(
I+
G2)
D;V?=
V?Q^
PV
Q^ +
j(
GV
Q^
;Q^
GV )
;PV
V
?
= herm
V?( ^
Q+
I)(
PV +
jGV )( ^
Q;I)
V? <0
(34) with
PV =
DDand
GU =
DGD. Next, by introducing
X=
PV +
jGV
2Wand observing that
W;1=
PU
;jGU , we can conclude the proof.
2Remark 3.1 Lemma 3.4 follows by letting
W=
W=
P.
3.7 Shared uncertainties
We will now consider the problem with shared uncertainties. We augment the system ~
Mwith through connections between the uncertainty block ~ and
K. We assume that
M22= 0.
~
M
=
2
4
~
M
11 M
~
12~
M
21
0
3
5
=
2
6
6
6
4 M
11
0
M120
0 0 0
Ir
M
21
0 0 0
0
Ir 0 0
3
7
7
7
5
=
2
6
6
6
4
Q
0
U0 0 0 0
Ir
V
0 0 0
0
Ir 0 0
3
7
7
7
5
(35)
:Thus, ~
U= diag
M12Ir ] and ~
V= diag
M21Ir ]. Then, ~
U?= diag
U?0] and
~
V
?
= diag
V?0]. Applying Lemma 3.4 we obtain
herm
;V?(
I+
1Q
)
S(
I;1Q
)
V?>0 (36) and
herm
;U?(
I+
1Q
)
R(
I;1Q
)
U?>0 (37) where
S,
R2Cn
n are the upper left block of
Wand
W;1respectively.
W
=
S
(38) and
W
;1
=
R
:
(39)
The matrices
Rand
Sare mutually constrained by these two equations. Next, we will show that this constraint can be reformulated as a rank condition on
I;R S
and, in the case of complex uncertainties, an additional LMI.
Before going much further we need to specify the structure of ~,
Wand
W;1more explicitly. The uncertainty matrix ~ is assumed to be a block diagonal matrix such that
= diag
LR()]
= diag
1In
1:::f
1In
f1f
1+1:::f
1Ir
1:::f
1Ir
f1] (40) where
f1=
fr +
fc is the number of scalar blocks and
fis the total number of blocks. Shared blocks are restricted to the type
i
Iwhere
i is either a real or a complex scalar.
We assume that
Whave the following block structure.
W
=
2
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
4 W
111
0
0
W1210
0 0
W1120 0
W1220 ... ... ... ... ... ... ... ...
0 0
W11f 0 0
W12f
W
211
0
0
W2210
0 0
W2120 0
W2220 ... ... ... ... ... ... ... ...
0 0 f 0 0 f
3
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
5
(41)
For notational convenience, all
W12i ,
W21i and
W22i are included even if they are not shared and are in such case empty matrices. The structure of
W;1will be identical to the structure of
W.
We now denote
W
i =
W
11
i
W12i
W
21
i
W22i
:
Specically, for uncertainties that are not shared we have
Wi =
W11i .
The structure of
Wi is given by the type of uncertainty it refers to. We will now look specically at two kinds of uncertainties that may be shared: complex and real. They are both of the type i =
i
I, where
i is either complex or real. For complex uncertainties we have
Wi =
Wi
>0 for real uncertainties
W
i could be any square matrix such that herm
Wi
>0.
We will now assume that
Rand
Sare diagonal block matrices with a struc- ture identical to
W11. Depending on the structure of , and consequently of
W, the matrices
Rand
Sare related to each other. The next two sections treats the two cases of shared uncertainties: complex and real. For notational convenience we drop the sux
iand consider each block at a time.
3.7.1 Complex uncertainties
If we are dealing with complex uncertainties we restrict
Wto be hermitian and positive denite
W=
W=
P >0. Then we have the following lemma from
7]. Lemma 3.5 Suppose that
R=
R 2Cn
n ,
S=
S 2Cn
n , with
R >0 and
S >
0. Let
rbe a positive integer. Then there exists matrices
N 2Cn
r and
L2C
r
r such that
L=
Land
P
=
S N
N
L
>
0 and
P;1=
R
if and only if
R I
n
I
n
S
0 and rank
;S;R;1r:Proof: (
)) Using Schur complements,
R=
S;1+
S;1N(
L;NS;1N)
;1NS;1. Inverting, using the matrix inversion lemma, gives that
R;1=
S;NL;1N. Hence
S;R;1=
NL;1N0, and indeed, rank(
S;R;1) = rank(
NL;1N)
r
. (
() By assumption, there is a matrix
N 2Cn
r such that
S;R;1=
NN.
Dening
L=
Ir completes the construction.
23.7.2 Real uncertainties
For real uncertainties we have a more general set of matrices such that herm
W >0. Note that herm
W >0 implies that
W;1exists and that herm(
W;1)
>0.
The following lemma then applies.
Lemma 3.6 Suppose that
R 2 Cn
n ,
S 2 Cn
n , with herm
R >0 and herm
S>0. Let
rbe a positive integer. Then there exists matrices
N 2Cn
r ,
M 2C
r
n and
L2Cr
r such that
Lis nonsingular and
W
=
S N
M L
herm
W >0 and
W;1=
R
if and only if
rank
;S;R;1r:Proof: (
)) Using Schur complements,
R=
S;1+
S;1N(
L;MS;1N)
;1MS;1. Inverting, using the matrix inversion lemma, gives that
R;1=
S;NL;1M. Hence,
S;R;1=
NL;1M, and indeed, rank(
S;R;1) = rank(
NLM)
r. (
() Dene
t= rank(
S;R;1). Find any
M2Ct
n of full row rank, such that
S;R
;1
=
KMt , with
K2Cn
t having full column rank. Choose
Nt as a linear function of
Lt by letting
Nt = (
S;R;1)
Mt
yLt where
Mt
y=
Mt
(
Mt
Mt
)
;1 2C
n
t is the Moore-Penrose inverse of
Mt . By dening
M=
M
t
0
,
N=
N
t 0
and
L= diag
Lt
Ir
;t ], we obtain
W
=
S N
M L
=
2
4
S
0 0
M
t 0 0 0 0
Ir
;t
3
5
+
2
4
(
S;R;1)
Mt
yI
t
0
3
5
L
t
0
It 0
:Then, herm
Wi
>0 is equivalent to
S
+
S Mt
M
t 0
+
(
S;R;1)
Mt
yI
t
L
t
0
It
+ ]
>0
(42) where ]
denotes the complex conjugate transpose of the previous term. Using lemma 3.2 with
U?=
In (
R;1;S)
Mt
yand
V?=
In 0
, this is in turn equivalent to
U
?
S
+
S Mt
M
t 0
U
?
=
S+
S+ (
Mt
yMt )
(
R;1;S)
+ (
R;1;S)
Mt
yMt
=
R;1+
KMt +
R;+ (
KMt )
;(
KMt
Mt
yMt )
;KMt
Mt
yMt
=
R;1+
R;= 2herm(
R;1)
>0
and
V
?
S
+
S Mt
M
t 0
V
?
=
S+
S= 2herm
S>0
:Thus (42) is satised implying that
Lis nonsingular, which concludes the proof.
2
3.8 Summary
It has been shown that the matrix inequality
S R;1or equivalently
S I
I R