Technical report from Automatic Control at Linköpings universitet

Employing Kronecker Canonical Form forH∞ Synthesis Problems

Anders HelmerssonDivision of Automatic ControlE-mail: andersh@isy.liu.se

23rd March 2011

Report no.: LiTH-ISY-R-3007Submitted to IEEE Transactions of Automatic Control, 2010-10-20

Address:Department of Electrical EngineeringLinköpings universitetSE-581 83 Linköping, Sweden

WWW: http://www.control.isy.liu.se

AUTOMATIC CONTROLREGLERTEKNIK

LINKÖPINGS UNIVERSITET

Technical reports from the Automatic Control group in Linköping are available fromhttp://www.control.isy.liu.se/publications.

AbstractThe Kronecker canonical form (KCF) can be employed when solving H∞synthesis problem. The KCF structure reveals variables that can be elimi-nated in the semidefinite program that defines the controller. The structurecan also be used to remove states in the controller without sacrificing per-formance.In order to find the KCF structure, we can transform the relevant matricesto a Guptri (Generalized upper triangular) form using orthogonal trans-formations. Thus we can avoid finding the KCF structure explicitly, whichis a badly conditioned problem.

Keywords: KCF, GUPTRI, LMI, semidefinite programming, controllersynthesis, model reduction

Employing Kronecker Canonical Form for H∞Synthesis Problems

Anders Helmersson

2011-03-23

Abstract

The Kronecker canonical form (KCF) can be employed when solvingH∞ synthesis problem. The KCF structure reveals variables that can beeliminated in the semidefinite program that defines the controller. Thestructure can also be used to remove states in the controller without sac-rificing performance.

In order to find the KCF structure, we can transform the relevant ma-trices to a Guptri (Generalized upper triangular) form using orthogonaltransformations. Thus we can avoid finding the KCF structure explicitly,which is a badly conditioned problem.

1 IntroductionIn the control community, there was a shift in the 1980s from the more tradi-tional Linear Quadratic Gaussian (LQG) control towards H∞ control. Zames’article from 1981 [15] was an important influence for the development of H∞synthesis algorithms, initially based on Riccati equations [8, 3]. Soon after, Lin-ear Matrix Inequalities (LMIs) were used to solve the same synthesis problem[4, 9].

In the original algorithms for solving theH∞ synthesis problem using Riccatiequation, there exist a conditions on the rank on some of the state-space ma-trices, in particual the D12 and D21 matrices. The LMI approach, on the otherhand, these conditions are not present, and a controller can be found withoutmodifying the problem as needed in the Riccati approach.

In the general case the controller will have the same number of states as thesystem it is designed to control. This includes any states added to the plant inorder to specify the performance criteria of the closed loop system. As we willsee in this paper, the rank deficiencies in D12 and D21 can actually be employedto reduce the order of the controller even without sacrificing performance.

This paper discusses the structure of the LMI formulation, which can bedescribed using Kronecker Canonical Forms (KCF) [7]. The KCF structure isused to partition the variables in the LMI program into two parts: unboundedand bounded variables. In order to find the minimum H∞ gain, γ, of theclosed-loop system, the unbounded variables can be removed, while the boundedvariables must be kept. This can be used to reduced the complexity of the LMIproblem and it can also improve the numerical properties. In addition, the KCF

1

structure can be used to reduce the order of the controller, without reducing itsperformance.

Finding the KCF of a system is a badly conditioned problem. However, wedo not necessarily need the explicit KCF structure. Instead we can use nu-merically well-behaved algorithms like Guptri (Generalized Upper Triangular)decomposition [1, 2].

When unbounded elements are present in the LMI problem, this also indi-cates that the optimal controller may have very high gains (increasing to infinitywhen γ approaches its minimum value). Also, such a controller may be fragilein the sense that it becomes sensitive to parameter variations [10]. Due to theseaspects, we could even say that such a synthesis problem is not properly definedsince some properties are missing in the specification.

The paper is organized as follows. We start by making a short recapitulationof the Real Bounded Lemma and the LMI approach to solve H∞ synthesisproblems. We also discuss how rank deficiencies in D21 can be used to reducethe order of the controller. We then give an overview of the Kronecker CanonicalForm and the Guptri form.

Before presenting the main result we use the KCF structure on an examplefrom the COMPleib’s benchmark suite. We also discuss how the controller canbe recovered when infinite variables are present in the LMIs. Finally, we presentthe results from the COMPleib benchmarks followed by conclusions.

1.1 NotationsFor symmetric or Hermitian matricesX ≺ (�) 0 denotes negative (semi-)definitematrices. In sums, ()T denotes the transpose of the previous term. X∗ de-notes the complex conjugate transpose of a complex matrix and X† denotes thepseudo-inverse of X.

2 LMI formulationWe start by taking a quick review of the LMI formulation used to measure theH∞ performance of a system. We then proceed to the LMI synthesis techniqueused for finding an optimal H∞ controller. Finally, we discuss how rank defi-cencies in the D12 and D21 matrices can be used for reducing the order of thecontroller.

2.1 The Bounded Real LemmaConsider the Bounded Real Lemma (BRL) [13, 5], which can be used to deter-mine the H∞ gain, γ, of a continuous-time system, G(s) = C(sI −A)−1B +D, PA+ATP PB CT

BTP −γI DT

C D −γI

≺ 0, (1)

where P � 0.Using Schur complement the BRL (1) can also be written as[

0 00 −γI

]+ γ−1

[CT

DT

] [C D

]+

[I0

]P[A B

]+

( )T≺ 0,

2

where ()T denotes the transpose of the previous term. This LMI can haveunbounded solutions in P , if there are strictly stable, uncontrollable modes. Tosee this, let x belong to the null space of

[A− λI B

]∗, and let P = P0+αxx∗,where P0 � 0 is any solution to (1). If Reλ ≤ 0 then P is also a solution forany α ≥ 0, since[

I0

]xx∗

[A B

]+

( )∗= 2(Reλ)

[xx∗ 00 0

]� 0.

In addition we must assure that A is strictly Hurwitz (Reλ < 0); otherwise (1)is not feasible.

2.2 SynthesisWe will now consider the system, G, which takes the inputs w and u and pro-duces the outputs z and y:

G :=

A B1 B2

C1 D11 D12

C2 D21 D22

to be controlled by

K :=

[A B

C D

],

which connects to y and u of G, see Figure 1. The closed loop system, G fromw to z, is given by

G :=

[A BC D

]=

A 0 B1

0 0 0C1 0 D11

+

0 B2

I 00 D12

[ A B

C D

] [0 I 0C2 0 D21

].

(2)

K �

G-

- -w z

u y

Figure 1: The synthesis problem consists of the design task of finding a con-troller K to the system G such that the H∞ norm of the closed-loop system,from w to z, is minimized.

The Bounded Real Lemma (1) applied to G(s) = C(sI−A)−1B+D, togetherwith the Elimination Lemma [4, 9] can be used to derive the necessary andsufficient conditions for the existence of a controller with a closed-loop H∞

3

norm less than γ. The conditions are

FX(γ,X) :=

[NX 00 I

]T XA+ATX XB1 CT1BT1 X −γI DT

11

C1 D11 −γI

[ NX 00 I

]≺ 0,

(3a)

FY (γ, Y ) :=

[NY 00 I

]T AY + Y AT Y CT1 B1

C1Y −γI D11

BT1 DT11 −γI

[ NY 00 I

]≺ 0,

(3b)

where NX and NY designate any bases of the null spaces of[C2 D21

]and[

BT2 DT12

], respectively. In addition X and Y are connected by[

X II Y

]� 0 (4)

and rank(XY − I) ≤ r, where r is the order of the controller.When X and Y are found, we proceed by finding a P � 0 such that

P =

[X ∗∗ ∗

]=

[Y ∗∗ ∗

]−1, (5)

which together with the Bounded Real Lemma of the closed loop system givesan LMI condition for the controller.

2.3 Rank deficiencies in D21

In order to illustrate how the structure of the synthesis problem can be used toreduce the order of the controller, we assume that D21 in the augmented plantis rank deficient. Rank deficencies in the D12 matrix can be handled similarly.

Without loss of generality, we can assume that[C2 D21

]has full rank,

which means that the number of outputs to the controller,m2 = rank[C2 D21

].

If this is not the case we can reduce the number of outputs to the controller.When D21 ∈ Rm2×p1 has full row rank, then we can choose

NX =

[NX1

NX2

]=

[I

−D†21C2

],

which means that NX1 ∈ Rn×n has full rank. When D21 looses row rank thenalso NX1 will do so, which in turn means that some elements in X becomefree and will not affect FX(γ,X) in (3a). Note that D21 looses row rank whenp1 < m2 (more control outputs than disturbance inputs) or when D21 is zero orhas linearly dependent rows.

Suppose that NX1 looses rank. We assume that NX2 =

[0∗

], possibly after

a similarity transformation. In this case X can be partitioned into

X =

[X11 X12

X21 X22

],

where X11 does not affect FX(γ,X).

4

Using Schur complement, (4) can be rewritten as X−Y −1 � 0 and the rankconstraint becomes rank(X−Y −1) ≤ r. Note that (4) always implies that bothX � 0 and Y � 0 so Y = Y −1 is well defined. Then (4) is equivalent to[

X11 − Y11 X12 − Y12X21 − Y21 X22 − Y22

]� 0.

When X22 − Y22 is nonsingular, this condition is equivalent to

X11 − Y11 −(X12 − Y12

) (X22 − Y22

)−1 (X21 − Y21

)� 0,

X22 − Y22 � 0.

Otherwise (when X22 − Y22 is singular) we need to replace the inverse by thepseudo inverse. In this case the null space of X22− Y22 belongs to the null spaceof X12 − Y12.

We now assume that X11 can be chosen freely without affecting FX(γ,X).Let

X11 = Y11 +(X12 − Y12

) (X22 − Y22

)† (X21 − Y21

),

which implies that rank(X − Y −1

)= rank

(X22 − Y22

)≤ rankNX1.

This means that when D21 looses row rank, we can reduce the number ofstates in the controller accordingly (without affecting the performance, γ). Forinstance if D21 = 0 we can remove m2 states from the controller, from n ton−m2 states. In the state feedback case, when m2 = n, the controller becomesstatic (state feedback).

This illustrates that the structure of the H∞ problem sometimes can beemployed for both reducing the number of variables in the LMI problem and alsofor reducing the number of states in the optimal controller. We will generalizethis in the sequel.

3 Kronecker Canonical Form (KCF)We will now take a more general view of the structure of the H∞ synthesisproblem.

3.1 The Structure of the Synthesis LMIsUsing Schur complement we can rewrite (3a) as

FX(γ,X) := N TX

([XA+ATX XB1

BT1 X −γI

]+ γ−1

[CT1DT

11

] [C1 D11

])NX ≺ 0

(6)

This matrix inequality is on the form

Q(γ) + UXV T + V XUT ≺ 0, (7)

withUT =

[A B1

]NX and V T =

[I 0

]NX

The LMI (3b) has the same structure.

5

In the discrete-time case the matrix inequalities will be of the type

Q(γ) + UXUT − V XV T ≺ 0. (8)

We can treat the discrete-time case using similar techniques as the for continuous-time case, but we will not pursue that track here.

3.2 The KCF structureWe will now consider (7) and (4) where we assume that Y is given.

Q(γ) + UXV T + V XUT ≺ 0,

X � Y −1 � 0.

In order to solve this type of LMIs we can employ the Kronecker CanonicalForm (KCF). The KCF is a generalization of the Jordan form and describes thestructure of the generalized eigenvalues and eigenvectors of the pencil U − λV .By applying nonsingular P and Q, we can partition the pencil into a block-diagonal structure:

P−1(U − λV )Q = diag(Ui − λVi),

where each block Ui−λVi must be on one of the following four forms [7]: Jj(α),Nj , Lj , and LTj . The Jordan j-by-j block is defined by

Jj(µ) :=

µ− λ 1

. . . . . .. . . 1

µ− λ

and

Nj :=

1 −λ

. . . . . .. . . −λ

1

∈ Rj×j

corresponds to Jj(∞). The remaining two blocks, Lj ∈ Rj×(j+1) and LTj ∈R(j+1)×j , are rectangular blocks defined by

Lj :=

−λ 1. . . . . .

−λ 1

.The null space of Lj is given by

[1 λ · · · λj

]T for any value of λ.We illustrate the KCF structure on the following example:

U =

[1 2 34 5 7

], V =

[7 8 910 11 13

],

6

where we can use

P =

[6 16 0

]and Q =

−1 −5 −25 1 3−3 3 −1

,to obtain

P−1(U − λV )Q =

[−λ 1 0

0 0 1− λ

]= L1 ⊕ J1(1).

Here we use the notation A ⊕ B = diag[A,B]. Thus, this eigenvalue problemhas a finite eigenvalue in 1 and an undefined eigenvalue (which can take anyvalue).

3.3 The Generalized Upper Triangular (Guptri) FormIn order to solve these types of generalized eigenvalue problems we employalgorithms like Guptri (Generalized upper triangular) form [1, 2]. Unitarymatrices, P and Q, are used here, which makes the algorithm numerically well-conditioned. For the previous example we can use

P =

[−0.7071 −0.7071−0.7071 0.7071

]and

Q =

0.1690 0.8407 0.5145−0.8452 −0.1449 0.51450.5071 −0.5218 0.6860

to obtain

PT (U − λV )Q =

[0 1.4349 −9.21640 0 4.1231

]− λ

[1.4343 −0.0410 −23.7685

0 0 4.1231

].

The KCF structure can be derived from the staircase block structure of theGuptri form. Due to numerical tolerances in the computations and the finiterepresentation, the elements below the staircase may not be exactly equal tozero but they may differ with some given tolerance. This also means that theGuptri form is not unique in the general case: for a given tolerance we cansometimes find several forms. As the tolerance is increased more forms maybecome acceptable.

The basic algorithm for computing the Guptri form is based on the singularvalue decomposition (SVD). For the example, we first apply the SVD on U tofind obtain a Q1 such that the first column of UQ1 becomes zero. Next, find aP1 such that PT1 V Q1 becomes zero in its left, bottom element. We have thenfound the first step and can proceed to the next step of the staircase structure.When ready we have separated the Lj and Jj(0) blocks.

The LTj andNj blocks can be extracted using the same algorithm but appliedon V T and UT . Finally, the remaining blocks, which are Jj(µ) blocks, areobtained by using a standard eigenvalue decomposition. The order of extractionof blocks can be tailored by applying the algorithm on (U, V ), (UT , V T ), (V,U),or (V T , UT ).

7

4 The ROC7 ExampleIn order to illustrate how the LMI structure can be used for solvingH∞ synthesisproblems, we have taken the COMPleib’s benchmarks as examples [11]. We startby considering the ROC7 benchmark. Here we have removed the augmentedstate before applying the H∞ synthesis. This means that the reduced ROC7plant has four states, two outputs to the controller, y, and one control input, u.

A B1 B2

C1 D11 D12

C2 D21 D22

=

0 1 0 0 0 0−1 0 0 0 1 −0.20 0 0 1.02 0 0

0.2 0 0 0 −0.2 10.1 0 0 0 0 00 0 0.1 0 0 00 0 0 0 0 0.21 0 0 0 0 00 0 1 0 0 0

The KCF structure of FX(γ,X) is L1 ⊕N2, which can be written as:

FX(γ,X) := Q(γ) +

0 1 0 00 0 0 10 0 0 0

X 1 0 0 0

0 0 1 00 0 0 1

T +

T

= Q(γ) +

2x12 x23 + x14 x24x23 + x14 2x34 x44

x24 x44 0

≺ 0 (9)

where

X =

x11 x12 x13 x14x12 x22 x23 x24x13 x23 x33 x34x14 x24 x34 x44

� Y −1. (10)

Note that FX(γ,X) only has five linearly independent variables: x12, x23 +x14, x24, x34 and x44.

Here we can modify x12 to decrease without bounds and at the same timeincreasing x11 and x22 in order to satisfy (10). This means that x12 will dominatethe first row and column in (9). After removing the first row and column weget [

q22(γ) q23(γ)q23(γ) q33(γ)

]+

[2x34 x44x44 0

]≺ 0. (11)

In addition (10) can be reduced to[x33 x34x34 x44

]� Y22, (12)

by removing the two first rows and columns. We can go one step further sinceboth x44 and x34 can grow without bounds. If there exists a solution to (11)then the lower right element, q33(γ), must be negative, say −ε < 0. For instance,let [

x33 x34x34 x44

]= ε

[5t3 2t2

2t2 t

]� ε

6

[t3 00 t

]� Y22

8

by choosing t large enough. This allows the 3-4 block to dominate both in Xand FX .

Using this argument we can get rid of X in (9) altogether. What remains isjust a lower bound on γ given by q33(γ) < 0, which here is 0. In this way wehave reduced the number of variables from 10 + 10 + 1 = 21 to 10 + 1 = 11.

Next, looking at the structure of the FY (γ, Y ) ≺ 0, which is LT2 ⊕ LT2 , or

FY (γ, Y ) := QY (γ) +

0 0 0 01 0 0 00 1 0 00 0 0 00 0 1 00 0 0 1

Y

1 0 0 00 1 0 00 0 0 00 0 1 00 0 0 10 0 0 0

T

+

T

≺ 0.

We first look at the upper left 3× 3 block, which must be negative definite, q11 q12 q13q12 q22 q23q13 q23 q33

+

0 y11 y12y11 2y12 y22y12 y22 0

≺ 0.

This block bounds all elements in the upper left block of Y . To see this we firstconsider y12 in the upper right corner. This element is bounded by the first andlast elements on the diagonal as |q13 + y12|2 < q11q33. Using the same argumenty12 in turn bounds y11 and y22. Similarly, the lower right block puts bounds ony33, y34 and y44. Finally, the elements in the off-diagonal blocks are boundedby the diagonal blocks. Thus all elements in Y are bounded.

Solving FY (γ, Y ) ≺ 0 with Y � 0 yields γ = 1.12033. Using the full set ofLMIs gives a slightly higher value, γ = 1.12037. Elements in X become large(∼ 105) and the solver also gives warnings about problems when finding thesolution.

When recovering a controller that achieves the optimal gain we can observethat the controller has very high gains:[

A B

C D

]≈

2858.981 982.834 −41590.878 8451.580−8656.141 −2970.635 125899.336 −25584.559

1649.430 566.424 −23991.035 4875.370

.Even worse, the controller is fragile, since small changes in its parameters maycause large increase in γ. When the parameters are rounded to three decimalsas above we obtain γ = 1.1233; when rounded to one decimal the gain morethan doubles, γ = 2.3921, for the closed loop system.

However, if we relax the performance by fixing γ to say 1.13, we obtain acontroller with much less gain and higher robustness for parameter variations:[

A B

C D

]≈

−1.3 −12.0 −4.7 −1.31.8 −15.7 −12.6 −0.1−1.3 −18.2 −7.7 −1.8

.The conclusion of this elaborated example is that depending on the KCF struc-ture it is sometimes possible to remove elements from X or Y when solvingthe LMIs during the synthesis step. For KCF blocks like Lj , Jj(0) and Nj thecorresponding elements become unbounded and can be removed. When these

9

kinds of KCF blocks are present, the order of the controller can also be reducedwithout sacrificing performance.

However, when the corresponding controller is recovered it may become frag-ile (sensitive to parameter variations) and may have unrealistically high gainswhen the γ approaches its optimal value. We can even argue that the synthesisproblem is not well formulated since there are no bounds on the controller gainsor on its sensitivity to parameter variations.

5 KCF PartitioningWe will now take a more general look at how the various types of KCF blocksaffect the solution of the LMIs. The matrices U and V are on a block diagonalKCF structure and the Q and X = X∗ matrices have a corresponding full blockstructure. Note the use of complex conjugate transpose ()∗, since the eigenvaluesand eigenvectors may be complex.

FX(γ,X) = Q(γ) + UXV ∗ + V XU∗ =[Q11 Q12

Q∗12 Q22

]+

[U1X11V

∗1 U1X12V

∗2

U2X∗12V

∗1 U2X22V

∗2

]+

( )∗=[

Q11 + U1X11V∗1 + V1X11U

∗1 Q12 + U1X12V

∗2 + V1X12U

∗2

Q∗12 + U2X∗12V

∗1 + V2X

∗12U

∗1 Q22 + U2X22V

∗2 + V2X22U

∗2

]5.1 Lj blocksWe first consider the case when (U1, V1) =

([0 I

],[I 0

]), which corre-

sponds to Lj . The 1-1-block is dominated by

U1X11V∗1 +V1X11U

∗1 =

x12 + x21 x13 + x22 . . . x1n + x2(n−1)x22 + x31 x23 + x32 . . . x2n + x3(n−1)

......

. . ....

xn1 + x(n−1)2 xn2 + x(n−1)3 . . . xn(n−1) + x(n−1)n

.(13)

To see this we first let x21 to go to −∞ and at the same time letting x11 approach∞ in order to satisfy X � Y −1. Then the first row and column will dominate.We continue with the second row and column in the same way until the finalrow and column. Thus only

Q22(γ) + U2X22V∗2 + V2X22U

∗2 ≺ 0 (14)

remains.

5.2 Jj and Nj blocksWe next consider the case when (U1, V1) = (µI + T, I), which corresponds toJj(µ), where

T = U1 − µV1 =

0 1

. . . . . .. . . 1

0

.

10

Then

FX(γ,X) =

[Q11 + 2(Reµ)X11 + TX11 +X11T

∗ Q12 +X12(U2 + µV2)∗ + TX12V∗2

Q∗12 + (U2 + µV2)X∗12 + V2X∗12T∗ Q22 + U2X22V

∗2 + V2X22U

∗2

].

If Reµ < 0 then X11 = X∗11 dominates in the first j rows and columns. Theremaining condition in FX(γ,X) ≺ 0 becomes (14).

Next, assume that Reµ = 0 by letting µ = jω where ω ∈ R. Then

TX11 +X11T∗ =

x12 + x21 . . . x1n + x2(n−1) x2n

.... . .

......

xn1 + x(n−1)2 . . . xn(n−1) + x(n−1)n xnnxn2 . . . xnn 0

.Here we can identify the framed block as (13), which according to previousdiscussion, can be made arbitrarily negative definite. Thus it will dominate andonly the last row and column remain. Consequently, the first j − 1 rows andcolumns can be removed and the term TX12V

∗2 disappears altogether, since the

last row of T is zero.We can now use the Elimination lemma [4, 9] when solving for the last row

of X12. Let N = diag[N1,N2] be defined as a matrix spanning the null space of(U − jωV )∗ =

[(U1 − jωV1)∗ (U2 − jωV2)∗

]. The two equivalent conditions

of the elimination lemma become N ∗Q(γ)N ≺ 0 and (14), where we have usedthe fact that

N ∗2 (U2X22V∗2 + V2X22U

∗2 )N2

= N ∗2 (jωV2X22V∗2 − jωV2X22V

∗2 )N2 = 0,

since (U2 − jωV2)∗N2 = 0. Note that the condition N ∗Q(γ)N ≺ 0 does notinvolve X and can be replaced by a lower bound on γ.

When Reµ > 0, the 1-1-block becomes

Q11 + 2(Reµ)X11 + TX11 +X11T∗ ≺ 0,

or

Q11 + (2 Reµ)︸ ︷︷ ︸>0

x11 . . . x1(n−1) x1n...

. . ....

...x(n−1)1 . . . x(n−1)(n−1) x(n−1)nxn1 . . . xn(n−1) xnn

+

x12 + x21 . . . x1n + x2(n−1) x2n

.... . .

......

xn1 + x(n−1)2 . . . xn(n−1) + x(n−1)n xnnxn2 . . . xnn 0

≺ 0.

Looking at the rightmost, bottom element, we conclude that xnn is bounded byQ11 and X � Y −1 � 0.

Next, looking at the second last element along the diagonal, the elementsx(n−1)(n−1) and xn(n−1) are bounded, since |xn(n−1)|2 < xnnx(n−1)(n−1) (X � 0)and q(n−1)(n−1) + (2 Reµ)x(n−1)(n−1) < 2|xn(n−1|. We can now proceed bytaking x(n−2)(n−2) and so on until x11. We conlude that all elements in X11

when Reµ > 0 are bounded.

11

5.3 LTj blocks

We next look at the last type of KCF block: LTj for which U1 =

[0I

]and

V1 =

[I0

]:

Q11 +

0 x11 . . . x1(n−1) x1nx11 x12 + x21 . . . x1n + x2(n−1) x2n...

.... . .

......

x(n−1)1 x(n−1)2 + xn1 . . . x(n−1)n + xn(n−1) xnnxn1 xn2 . . . xnn 0

≺ 0

together with X11 � 0.Let kI � Q11 be an upper bound of Q11 and use Rex ≤ |x|. Considering

the elements on the diagonal and sub and super-diagonals we get

x211 < k (k + 2|x12|)|x12|2 < x11x22

x222 < (k + 2|x12|)(k + 2|x23|). . .

|x(n−1)n|2 < x(n−1)(n−1)xnn

x2nn < k(k + 2|x(n−1)n|

).

Replacing the xi(i+1) terms with their upper bounds gives

x211 < k (k + 2√x11x22)

x222 < (k + 2√x11x22) (k + 2

√x22x33)

. . .

x2(n−1)(n−1) <(k + 2

√x(n−2)(n−2)x(n−1)(n−1)

) (k + 2

√x(n−1)(n−1)xnn

)x2nn < k

(k + 2

√x(n−1)(n−1)xnn

).

Next, choose a κ such that κ√xiix(i+1)(i+1) ≥ k holds for all i = 1, . . . , n− 1:

x411 < k2(κ+ 2)2x11x22

x422 < (κ+ 2)4x11x222x33

. . .

x4(n−1)(n−1) < (κ+ 2)4x(n−2)(n−2)x2(n−1)(n−1)xnn

x4nn < k2(κ+ 2)2x(n−1)(n−1)xnn.

Further we can show that

x311 < k2(κ+ 2)2x22

x522 < k2(κ+ 2)4×22−2x333

. . .

x2n−1(n−1)(n−1) < k2(κ+ 2)4(n−1)2−2x2n−3nn

x3nn < k2(κ+ 2)2x(n−1)(n−1).

12

Table 1: Classification of KCF structure.Lj unboundedJj(µ),Reµ < 0 unboundedNj ∼ Jj(∞) mixed, elimination lemmaJj(µ),Reµ = 0 mixed, elimination lemmaJj(µ),Reµ > 0 boundedLTj bounded

Using the two last inequalities, we get xnn < k(κ + 2)n−1 and x(n−1)(n−1) <

k(κ+2)3n−5, which also means that all xii < k(κ+2)(2i−1)(n+1−i)−i are bounded(using κ =

√2 − 1 can be used for all n ≥ 2). Further, all off-diagonal xij ele-

ments are bounded by the xii elements. Thus, all elements in X11 are bounded.

5.4 SummaryWe summarize this section by stating the following lemma.

Lemma 1. Consider the problem of minimizing γ with respect to Q(γ)+UXV T+V XUT < 0 and X > 0. Partition (U, V ) into KCF blocks and partition Q andX accordingly.

(i) Blocks corresponding to Lj and Jj(µ) where Reµ < 0, can be truncated.

(ii) Blocks of the type Nj and Jj(µ) where Reµ = 0, can be truncated andreplaced by a additional condition N ∗Q(γ)N < 0, where N designate anybasis of the null space of (U − µV )∗.

(iii) The remaining blocks in X corresponding to LTj and Jj(µ) where Reµ > 0are bounded.

For some examples in our test suite, both the FX and FY inequalities canbe removed and replaced by a lower bound on γ. These are AC1, AC12, DLR1,UWV, PAS and ROC5.

When solving the LMIs (FX and FY ), we start by finding the KCF structureusing the Guptri algorithm. For finding the optimal γ we sort the blocks inthe following order: Lj , Jj(0), Nj , Jj(µ), LTj . The Jj(µ) blocks are sorted sothat the eigenvectors corresponding to Reµ ≤ 0 come first. After the optimalγ is obtained, it can be relaxed to a slightly higher value and the controller canbe computed. The blocks are then sorted in the order Lj , Jj(µ), Jj(0), Nj , LTj .The Jj(µ) blocks are sorted so that the eigenvectors corresponding to Reµ ≤ −εcome first. These, together with the Lj blocks, are truncated in the LMI whilethe others are kept. The truncated blocks will correspond to uncontrollable orunobservable modes in the closed-loop system. The parameter ε should selectedto get sufficient closed-loop decay rate.

6 Recovering X, Y and the controller, KAfter finding X and Y we need to compute the controller, K. This can be donein two steps: first, find a P that satisfies (5) and, secondly, compute a controllersuch that (1) holds for the closed-loop system.

13

When X and Y are bounded, we can perform a similarity transformation ofthe system so that X = Y = Σ � I becomes diagonal. Then we can use

P =

[Σ ΓΓ Σ

]=

[Σ −Γ−Γ Σ

]−1,

where Γ2 = Σ2− I � 0. When Σ = diag[Σr, In−r] contains ones in its diagonal,the corresponding states can be removed in the controller by using

P =

Σr 0 Γr0 In−r 0

Γr 0 Σr

.If there are unbounded elements in X and Y then this scheme needs to

be modified accordingly. The unbounded elements correspond to uncontrol-lable and unobservable modes in the closed-loop system. The closed-loop statespace can be divided into four subspaces corresponding to uncontrollable andunobservable modes, uncontrollable modes, unobservable modes, and both con-trollable and observable modes:

P =

∞ 0 0 0 0 ∞ 0 0 00 ∞ 0 0 0 0 ∞ 0 00 0 I 0 0 0 0 I 00 0 0 Σ 0 0 0 0 Γ0 0 0 0 I 0 0 0 0∞ 0 0 0 0 ∞ 0 0 00 ∞ 0 0 0 0 ∞ 0 00 0 I 0 0 0 0 I 00 0 0 Γ 0 0 0 0 Σ

=

∞ 0 0 0 0 −∞ 0 0 00 I 0 0 0 0 −I 0 00 0 ∞ 0 0 0 0 −∞ 00 0 0 Σ 0 0 0 0 Γ0 0 0 0 I 0 0 0 0−∞ 0 0 0 0 ∞ 0 0 0

0 −I 0 0 0 0 I 0 00 0 −∞ 0 0 0 0 ∞ 00 0 0 −Γ 0 0 0 0 Σ

−1

.

There are still some questions that remain: how do we recover the removedvariables in an efficient way and can we even find a reduced order controller inthe same process? There even may be no need to recover the complete X andY , and we can possibly find the controller parameters anyway.

Even if the LMIs don’t bound some elements in X and Y this does notnecessarily mean the they should approach infinity. Often, we can find boundedsolutions and in that case X and Y can be fully reconstructed and K can besearched for. This is also needed when Jj(µ) where Reµ = 0 are involved sinceP in (1) cannot be unbounded for these elements, since this would imply anuncontrollable mode on the imaginary axis.

On the other hand, we can always let the unbounded elements in X andY approach infinity. Then they will correspond to unbounded KCF blocks (Ljand Jj(µ) where Reµ < 0) in FX(γ,X) or FY (γ, Y ).

If the generalized eigenvalue, Reλ ≤ 0, corresponds to an unbounded elementin FX(γ,X) < 0, then there exists a vector xλ such that

xTλ[A− λI B1

]NX = 0. (15)

Suppose there exists an uncontrollable mode in the closed-loop system (2).Then, there exist x 6= 0 and λ such that

xT[A− λI B

]= 0.

Note that the eigenvalues corresponding to the uncontrollable modes must bedefined since there cannot exist LTi blocks due to the λI-term.

14

Expanding this we get

xT[A− λI B

]=

[x1x2

]T ([A− λI 0 B1

0 −λI 0

]+

[0 B2

I 0

] [A B

C D

] [0 I 0C2 0 D21

])=[xT1 (A− λI) −xT2 λI xT1 B1

]+[xT2 xT1 B2

] [ A B

C D

] [0 I 0C2 0 D21

]= 0,

This has a solution, K(λ) = C(λI − A)−1B + D, if and only if

xT1[A− λI B1

]NX = 0

N TZ x

T1

[A− λI B1

]= 0

where NZ designates any base of the null space of[BT2 x1x2

].

Note that NZ only exists if both BT2 x1 and x2 are zero; otherwise thesecond condition disappears. When NZ exists the second condition becomesxT1[A− λI B1

]= 0, which makes the first condition redundant. This case

corresponds to an uncontrollable mode in λ for the original system described bythe pair (A,

[B1 B2

]). If we assume that the original system is controllable,

NZ disappears and only the first condition remains, which is equivalent to (15).What happens if Reλ = 0? In this case we can choose the corresponding

element in P arbitrarily large, which drives λ towards the imaginary axis. How-ever, we must choose a finite P in order to obtain a stable closed loop system.Thus, we need to treat the case when Reµ = 0 specially, compared to Jj(µ)blocks with Reµ < 0.

In some cases the performance bounds obtained using the reduced LMIs aretheoretical lower bounds. For instance for AC1 the lower bound is zero, butthis would require a controller with infinite gain. In order to obtain a practical,finite-gain controller, we must settle for some γ > 0, which gives reasonablegains in the controller. This can also be interpreted as if the problem is notwell formulated and more constraints should be added to limit the gain of thecontroller.

6.1 Rank ReductionThe KCF structure reveals if we can reduce the order of the controller by choos-ing the free elements in X or Y appropriately. In general, we can reduce therank of XY − I by one for each block of the type Lj . In addition we can furtherreduce the rank by one if there exists a block of the type Jj(0) or Nj . If thereexist several such blocks they can only be combined if the eigenvalues are equal.For instance, for the KCF structure defined by L1 ⊕ L2 ⊕ J1(0)⊕N2 the rankcan be reduced by 3 (from 8 to 5); the J1(0) block cannot be combined withthe N2 block, but it can be combined with both L1 and L2. Another exampleis DLR1, in which FX has the structure J1(0) ⊕ N1 ⊕ 8J1(−). The controllercan be reduced by 1 from 10 to 9 according to the rule above.

For some block structures we can reduce the rank even more. In many casesthe possibility to reduce the rank depends on the values of the fixed elements (in

15

Table 2: Jet engine (JE) and reactor (REA) modelsCase n X Y r γJE1 30 5N1, 25LT1 5LT6 25 3.8500 (3.8702)JE2 21 3N1, 18LT1 15LT1 , 3L

T2 18 42.348 (42.508)

JE3 24 24J1(−) 6LT4 24 2.8833 (2.8860)REA1 4 3N1, L

T1 2LT2 1 0.86172

REA2 4 2N1, 2LT1 2LT2 2 1.1341

REA3 12 3N1, 9LT1 10LT1 , L

T2 9 74.251

X−Y −1 or Y −X−1). The conditions are in general non-convex and algebraic.For instance, if x1, x2 and x3 are free variables and a, b and c are fixed in x1 a b− x2

a b+ x2 cb− x2 c x3

� 0,

we can obtain rank 1, either when b2 ≥ ac 6= 0, when a = c = 0, or when b > 0and ac = 0. This example corresponds to an L2 block.

7 ExamplesIn order to illustrate the method using KCF, we have taken the COMPleib’sbenchmarks as examples [11]. In COMPleib the ROC examples (all but ROC3)have been augmented with internal states (1, 2 or 3), which corresponds to theminimum order of a stabilizing controller. These extra states have been removedbefore the controller synthesis. If they are kept addition KCF blocks of the typeL0 appear, one for each added state.

In the Tables 2-10, the order of the original system and the optimal controllerare denoted by n and r, respectively. The H∞ norm of the closed loop systemis given by γ. The existence of an optimal controller has been verified for mostsystems, but not all (unverified systems are marked by * in the γ-column). Forsome systems the gain can be made arbitrarily small, denoted by → 0, but as γapproaches zero the controller gain is likely to increase to infinity. If a γ-valueis given within paranthesis, this gives the best performance obtained for theclosed loop system using the synthesized controller.

The LMIs have been solved in Matlab using the Yalmip interpreter [12]together with SeDuMi 1.3 [14] and LMILAB [6] solvers.

8 ConclusionsIn many control synthesis problems, the structure of the plant augmented withweighting functions can be employed to reduce the complexity of the LMI prob-lem and to reduce the number of states in the controller. There are still somedetails to be elaborated, such that recovering the removed variables in the LMIsin an efficient way.

On the other hand, it can also be argued that if the LMIs are associatedwith unbounded elements in X or Y (that is if Lj and some Jj blocks are

16

Table 3: 2D heat flow problemsCase n X Y r γ

HF2D10 5 3N1, 2LT1 5LT1 2 79536

HF2D11 5 3N1, 2LT1 5LT1 2 76443

HF2D12 5 4N1, LT1 5LT1 1 1037666

HF2D13 5 4N1, LT1 5LT1 1 101548

HF2D14 5 4N1, LT1 5LT1 1 526354

HF2D15 5 4N1, LT1 5LT1 1 173042

HF2D16 5 4N1, LT1 5LT1 1 444144

HF2D17 5 4N1, LT1 5LT1 1 300236

HF2D18 5 2N1, 3LT1 5LT1 3 93.153

HF2D_CD4 7 2N1, 5LT1 7LT1 5 24.458

HF2D_CD5 7 2N1, 5LT1 7LT1 5 25.771

HF2D_CD6 7 2N1, 5LT1 7LT1 5 15.922

HF2D_IS5 5 4N1, LT1 5LT1 1 1131511

HF2D_IS6 5 4N1, LT1 5LT1 1 132263

HF2D_IS7 5 2N1, 3LT1 5LT1 3 36.550

HF2D_IS8 5 2N1, 3LT1 5LT1 3 91.297

Table 4: Decentralized interconnected systems (DIS) and Euler-Bernoulli beams(EB)

Case n X Y r γDIS1 8 L0, N1, 2L2 LT1 , 2L

T2 , L

T3 4 4.1593

DIS2 3 2N1, LT1 LT3 1 0.94741

DIS3 6 4N1, 2LT1 LT1 , L

T5 2 1.0424

DIS4 6 6N1 2LT3 0 0.73147DIS5 4 LT4 LT4 3 666.27EB1 10 LT10 LT10 9 3.1041EB2 10 LT10 LT10 9 1.7676EB3 10 LT10 LT10 9 1.7976EB4 20 LT20 LT20 19 1.7973EB5 40 LT40 LT40 39 1.7973

Table 5: Helicopter problemsCase n X Y r γHE1 4 N1, L

T3 J1(−), 3J1(+) 3 0.073615

HE2 4 2N1, 2LT1 2LT2 2 2.4181

HE3 8 2L0, 3L1 4LT1 , 2LT2 3 0.79897

HE4 8 6N1, 2LT1 8LT1 , 2L

T2 2 22.838

HE5 8 LT8 6J1(−), 2J1(+) 7 1.7650HE6 20 18J1(−), 2J1(+) 4LT1 , 8L

T2 19 2.3876

HE7 20 J1(−), 2LT6 , LT7 4LT1 , 8L

T2 19 2.6130

17

Table 6: Aircraft models

Case n X Y r γAC1 5 2N1, N2, J1(−) L4 2 → 0AC2 5 2N1, N2, J1(−) J1(0), 2LT2 2 0.11149AC3 5 4N1, L

T1 LT1 , 2L

T2 1 2.9674

AC4 4 3J1(−), J1(+) LT4 3 0.55729AC5 4 2N1, 2L

T1 2LT2 2 657.06

AC6 7 4N1, 3LT1 3LT1 , 2L

T2 3 3.4273

AC7 9 N1, LT0 , L

T8 7J1(−), 2J1(+) 8 0.037862 (0.037865)

AC8 9 LT0 , LT1 , L

T2 , 2L

T3 LT9 8 1.6165

AC9 10 2LT1 , LT2 , 2L

T3 L3, L4, N1 7 0.99998

AC10 55 N1, N2, LT51 2LT18, L

T19 53 3.2324 *

AC11 5 4N1, LT1 LT1 , 2L

T2 1 2.8079

AC12 4 L1, 2N1 2L1 1 → 0 (0.13)AC13 28 4N1, 24LT1 22LT1 , 3L

T2 24 156.23

AC14 40 39J1(−), J1(+) 2LT2 , LT3 , 2L

T6 , 3L

T7 39 100.00

AC15 4 3N1, LT1 4LT1 1 14.863

AC16 4 4N1 4LT1 0 14.856AC17 4 2N1, 2L

T1 2LT1 , L

T2 2 6.6124

AC18 10 2N1, LT8 2LT3 , L

T4 8 5.3839 (5.3991)

Table 7: Second order problemsCase n X Y r γCM1 20 L19 LT1 , L

T19 19 0.81650

CM2 60 L59 LT1 , LT59 59 0.81650

CM1_IS 20 L19 LT1 , LT19 19 0.81650

CM2_IS 60 L59 LT1 , LT59 59 0.81650 (0.81817)

TMD 6 3L0, J2(0), J1(−) 2J2(0), LT2 2 2.1207FS 5 3N1, 2L

T1 3LT1 , L

T2 2 77919

DLR1 10 J1(0), N1, 8J1(−) 10J1(−) 9 0.061930DLR2 40 L37, N2 40LT1 38 82.679 *DLR3 40 L38, N1 40LT1 38 82.679 *LAH 48 48J1(−) LT0 , L

T48 47 5.3727× 10−5

18

Table 8: Problems for various physical systems

Case n X Y r γTG1 10 2N1, 8L

T1 6LT1 , 2L

T2 8 3.4652

AGS 12 2N1, 10LT1 8LT1 , 2LT2 10 8.1732

WEC1 10 4N1, 6LT1 4LT1 , 3L

T2 6 3.6363

WEC2 10 4N1, 6LT1 4LT1 , 3L

T2 6 3.5981

WEC3 10 4N1, 6LT1 4LT1 , 3L

T2 6 3.7685

BDT1 11 L0, L4, N5 LT1 , 2LT5 8 0.26621

MFP 4 2N1, 2LT1 LT4 2 4.1865

UWV 8 N2, N6 L7 6 → 0IH 21 10N1, 11LT1 21J1(−) 11 → 0

CSE1 20 9L1, N1, J1(−) J1(−), LT1 , 9LT2 10 0.019871

CSE2 60 29L1, N1, J1(−) J1(−), LT1 , 29LT2 30 0.019996 (0.02445)PAS 5 L0, N1, N2, J1(−) N4, J1(−) 2 → 0 (0.000765)TF1 7 2L1, L2 LT3 , L

T4 4 0.24636

TF2 7 L0, L1, J1(0), 3J1(−) LT3 , LT4 4 5200

TF3 7 L0, L3, 2J1(−) LT3 , LT4 5 0.24636

PSM 7 L0, 2N3 LT1 , 2LT3 4 0.92024

Table 9: Academic test problems

Case n X Y r γNN1 3 2N1, L

T1 LT1 , L

T2 1 13.130

NN2 2 N1, LT1 LT2 1 1.7644

NN3 4 N2, 2J1(+) N2, J1(−), J1(+) 3 7.9019 (7.97)NN4 4 3N1, L

T1 LT1 , L

T4 1 1.2862

NN5 7 2N1, 5LT1 5LT1 , L

T2 5 237.87

NN6 9 4N1, 5LT1 7LT1 , L

T2 5 126.58

NN7 9 4N1, LT5 LT1 , L

T8 5 33.039 (33.046)

NN8 3 2N1, LT1 LT3 1 2.3573

NN9 5 N1, 2J1(−), 2J1(+) L0, N1, 2LT1 , J1(+) 3 13.639 (13.641)

NN10 8 L1, 2L2, 3LT0 N1, L2, L3, L

T0 5 → 0

NN11 16 3N1, 13J1(−) N1, N3, 10J1(−), J1(+) 13 0.011994 (0.01760)NN12 6 2N1, 4L

T1 2LT1 , 3L

T2 4 6.2830

NN13 6 LT6 LT6 5 10.184NN14 6 LT6 LT6 5 9.4314NN15 3 N1, L1 LT1 , L

T2 1 0.097741

NN16 8 4N1, 4LT1 2J1(+), 6J1(±j) 4 0.95556

NN17 3 N1, J1(−), J1(+) N1, J1(−), J1(+) 2 2.6364

19

Table 10: Reduced order controller models

Case n X Y r γROC1 8+1 LT8 LT8 7 1.1267ROC2 9+1 2LT4 , N1 8J1(−), J1(+) 8 0.041030 (0.041032)ROC3 11 4N1, 7L

T1 4LT1 , 2L

T2 , L

T3 7 42.647

ROC4 8+1 3J1(−), 2J1(+), N3, LT0 7J1(−), N1, L

T0 7 7.6111 (12.56)

ROC5 6+1 L0, 2N2, N1 2N2, 2J1(−) 2 → 0 (1× 10−6)ROC6 3+2 LT1 , L

T2 LT1 , L

T2 2 21.528

ROC7 4+1 L1, N2 2LT2 2 1.1203ROC8 6+3 6J1(±j) 6LT1 5 3.4870ROC9 4+2 4J1(±j) 4LT1 3 2.2361ROC10 5+1 2L1, N1, L

T0 N2, J1(−), J1(+), LT1 2 0.071440 (0.07188)

present), this would indicate that the synthesis problem is badly formulatedsince constraints limiting the controller gain could be missing. If so, the KCFstructure will readily provide a tool for diagnosing this kind of problems.

References[1] J. Demmel and B. Kågström. The generalized Schur decomposition of an

arbitrary pencil A − λB: Robust software with error bounds and applica-tions. Part I: Theory and algorithms. ACM Transactions of MathematicalSoftware, 19(2):160–174, 1993.

[2] J. Demmel and B. Kågström. The generalized Schur decomposition of anarbitrary pencil A − λB: Robust software with error bounds and applica-tions. Part II: Software and applications. ACM Transactions of Mathemat-ical Software, 19(2):175–201, 1993.

[3] J. C. Doyle, K. Glover, P. Khargonekar, and B. A. Francis. State-spacesolutions to the standard H2 and H∞ control problems. IEEE Transactionson Automatic Control, 34(8):831–847, August 1989.

[4] P. Gahinet and P. Apkarian. A linear matrix inequality approach to H∞control. International Journal of Robust and Nonlinear Control, 1:656–661,October 1992.

[5] P. Gahinet and P. Apkarian. An LMI-based parametrization of all H∞controllers with applications. In IEEE Proceedings of the 32nd Conferenceon Decision and Control, volume 1, pages 656–661, San Antonio, Texas,December 1993.

[6] P. Gahinet and A. Nemirovsikii. LMI-Lab – A Package for Manipulatingand Solving LMIs, version 2.0, 1993.

[7] F. R. Gantmacher. The Theory of Matrices, volume II. New York: ChelseaPublishing Company, 1959, 1971.

20

[8] K. Glover and J. C. Doyle. State-space formulae for all stabilizing con-trollers that satisfy an H∞ norm bound and relations to risk sensitivity.Systems and Control Letters, 11:167–172, 1988.

[9] T. Iwasaki and R. E. Skelton. All controllers for the general H∞ controlproblem: LMI existence conditions and state space formulas. Automatica,30:1307–1317, August 1994.

[10] L. H. Keel and S. P. Bhattacharyya. Robust, fragile or optimal. IEEETransactions on Automatic Control, 42(8):1098–1105, August 1997.

[11] F. Leibfritz. Compleib: Constraint matrix-optimization problem library– a collection of test examples for nonlinear semidefinite programs, con-trol system design and related problems. Technical report, Department ofMathematics, University of Trier, Trier, Germany, 2004.

[12] J. Löfberg. YALMIP : A toolbox for modeling and optimization in MAT-LAB. In Proceedings of the CACSD Conference, Taipei, Taiwan, 2004.

[13] C. Scherer. The Riccati Inequality and State-Space H∞-Optimal Control.PhD thesis, Universitat Wurtzburg, Wurtzburg, Germany, 1990.

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21

Avdelning, InstitutionDivision, Department

Division of Automatic ControlDepartment of Electrical Engineering

DatumDate

2011-03-23

SpråkLanguage

2 Svenska/Swedish

2 Engelska/English

2

�

RapporttypReport category

2 Licentiatavhandling

2 Examensarbete

2 C-uppsats

2 D-uppsats

2 Övrig rapport

2

�

URL för elektronisk version

http://www.control.isy.liu.se

ISBN

—

ISRN

—

Serietitel och serienummerTitle of series, numbering

ISSN

1400-3902

LiTH-ISY-R-3007

TitelTitle

Employing Kronecker Canonical Form for H∞ Synthesis Problems

FörfattareAuthor

Anders Helmersson

SammanfattningAbstract

The Kronecker canonical form (KCF) can be employed when solving H∞ synthesis problem.The KCF structure reveals variables that can be eliminated in the semidefinite program thatdefines the controller. The structure can also be used to remove states in the controllerwithout sacrificing performance.

In order to find the KCF structure, we can transform the relevant matrices to a Guptri(Generalized upper triangular) form using orthogonal transformations. Thus we can avoidfinding the KCF structure explicitly, which is a badly conditioned problem.

NyckelordKeywords KCF, GUPTRI, LMI, semidefinite programming, controller synthesis, model reduction