CONTROL OF BLOCK CIRCULANT SYSTEMSbroucke/Webpapers/SBDSIAM13.pdf · 1. Introduction. Most complex...

CONTROL OF BLOCK CIRCULANT SYSTEMS∗

ADAM C. SNIDERMAN† , MIREILLE E. BROUCKE‡ , AND GABRIELE M. T. D’ELEUTERIO§

Abstract. Many physical and engineering systems have a notable structure which must be adhered to by any control lawdesigned for that system. Despite this, most methods for control synthesis provide general, unstructured controllers. This paperaddresses structured control synthesis for structured systems with a particular structure: a topological ring. We show thatseveral classic control synthesis problems for a ring system are amenable to a structured control synthesis respecting the ringtopology. Moreover, the conditions for solvability of all these structured control problems are the same as for general controlsystems, meaning that the control aspects of the problem are almost completely decoupled from the structural aspects. Ourfindings suggest that structured systems naturally admit structured controllers.

1. Introduction. Most complex engineering systems consist of subsystems interacting in a predefinedmanner. This structure often takes the form of highly recognizable repeated patterns, and these patternsmanifest themselves algebraically in the system model. For linear control systems, such patterns arise inthe system matrices of the state space model. If a pattern is recognized in a control system, then a naturalquestion is whether the pattern can be preserved in control synthesis. Suprisingly, this question has not beenwidely studied from a synthesis perspective — existing synthesis methods give only unstructured feedbacklaws. While a general feedback may solve the mathematical control problem, it may not be implementablegiven the structure of the physical system. Structured control synthesis for structured systems remains anopen research area.

In this work we focus on one specific system structure: a topological ring. A ring system is comprised of anumber of identical subsystems whose interactions occur in a closed loop. The system can be visualized as agraph with a single cycle, where each node of the graph represents a subsystem; moreover, the first subsystemtreats the second in the same way that the second subsystem treats the third, and so on. Mathematicallythe ring structure manifests itself as a pattern in the system matrices of the state space model: every systemmatrix is block circulant. This paper will show that several classic control synthesis problems are amenableto a structured synthesis realizing block circulant feedbacks. The proposed framework provides a templatefor future studies of more general structures.

Block circulant dynamical systems and block circulant matrices have been the focus in several applicationareas, including: cross-directional control of paper machines [19, 27]; control of partial differential equations[6]; finding the natural modes of a diamatic dome [18]; control of a crystal growth furnace [1]; reconstructionof a 3D image [20]; and analysis of cellular chemical reactions [29]; among many other areas.

Exploiting structure among subsystems in order to simplify analysis of an overall system is not a newconcept. It has particularly been studied within the context of symmetries. Among the earliest work in thecontrol literature is [13, 30, 31]. Symmetries are studied for coupled subsystems in [22, 23]. The behaviouralapproach was used to study systems with symmetries in [9, 32]. Assuming identical interactions between allpairs of subsystems, [28] developed decentralized stabilizing controllers. H∞ control design for symmetricallyinterconnected systems was presented in [17], and [8] presented computational techniques for H∞ controlof block circulant systems. Recently, [2, 3] studied stabilization and optimal distributed stabilization ofspatially invariant systems. System identification of circulant systems was studied in [24]. Controllabilityon the more general class of bilinear Toeplitz systems was explored by [26]. An in-depth study of control ofidentical dynamically-coupled systems was given in [25] using an LMI approach.

∗A significantly condensed version of this paper will be submitted to IEEE Conference on Decision and Control, 2013.†Institute for Aerospace Studies, University of Toronto. Email: [email protected]. Supported by the

Natural Sciences and Engineering Research Council of Canada (NSERC).‡Dept. of Electrical and Computer Engineering, University of Toronto. Email: [email protected]. Supported

by NSERC.§Institute for Aerospace Studies, University of Toronto. Email: [email protected]. Supported by the

NSERC.

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The approach of this paper extends and differs from the aforementioned works in several significant ways.We follow classical developments by providing explicit constructive procedures for finding time-invariantcontrollers. While we only consider systems with identical subsystems, we do not require them to interactwith each other identically; instead, we invoke the much less restrictive constraint that these interactionsare invariant under cyclic shift of indices. We make no reference to distributed control — all we seek isa controller that has block circulant structure. We do not end our analysis with system stabilization butstudy a number of classic multivariable control problems. While block circulant systems admit the symmetryalgebra given by the group ring of a cyclic group [16], we do not impose structure from a Lie group on oursystems. Finally, in contrast to most prior work, we examine structured control systems through the lensof linear geometric control. We reexamine the results of [33], taking system structure into account: given ablock circulant system, can the necessary and sufficient conditions of [33] recover control laws that preservethe block circulant structure?

We are not the first to study structured systems through the lens of linear geometric control. Our resultsare inspired primarily by two previous works. First, [6] studied controllability and observability of blockcirculant systems. Using the simultaneous block diagonalization property of block circulant matrices viathe Fourier matrix, the full system can be decomposed into modal subsystems; by maintaining certainconditions on those subsystems when designing control laws, the full system remains block circulant. Inthis way, control problems for block circulant systems can be reduced to a collection of control problemson smaller modal systems with easily satisfied constraints. Most other researchers (e.g. [3, 8, 17, 25]) havefollowed suit, focusing solely on this decomposition in analyzing block circulant systems. We will also use thisFourier decomposition at times when it is convenient to do so. However, rather than verifying our matrices’structure through block diagonalization and the resulting Fourier decomposition, we instead exploit thecommutative properties of block circulant matrices. This allows us to forgo the Fourier decomposition infavour of the standard ones (controllable decomposition, observable decomposition, etc.) used in geometriccontrol [4, 33].

Second, [15] formulated the geometric approach for patterned systems by encoding the structure in a basematrix, of which all system matrices are a polynomial. While circulant matrices are patterned, block circulantmatrices are not, so a generalization of their encoding method is needed. A contribution of this work is torecognize that a suitable encoding of the block circulant pattern is via commuting relationships of blockcirculant matrices rather than by block diagonalization [6] or by polynomial functions of a base matrix [15].The idea of exploiting commuting relationships has also been explored in [16]. Finally, using the device ofblock circulant subspaces, we bridge the algebraic domain (of system matrices) and the geometric domain (ofthe system’s state space).

The paper is organized as follows. Section 2 contains background with the primary intention to set notation.Sections 3 and 4 present relevant results on commuting and block circulant matrices. Section 5 introducesblock circulant subspaces. With this background, we then turn our attention to control. Sections 6 and 7focus on system properties. Sections 8-11 study a number of multivariable controller synthesis problems.These include the stabilization problem, output stabilization problem, disturbance decoupling problem,measurement feedback problem, and restricted regulator problem. Concluding remarks in Section 12 assessthe significance of the findings.

2. Background. The Kronecker product of two matrices A ∈ Cn×m and B ∈ Cp×q, denoted by A⊗B,is the Cmp×nq matrix obtained by replacing the (ij)th element of A by aijB, for all i, j. Useful propertiesof the Kronecker product are that (A⊗B)(C ⊗D) = AC ⊗BD and (A⊗B)∗ = A∗ ⊗B∗, where ∗ denotesthe complex conjugate transpose. Let σ(A) denote the spectrum of a square matrix A.

2.1. Fourier Matrix. Define ω = e2πi/r where i =√−1 and let Ωr = diag(1, ω, ω2, . . . , ωr−1), whose

diagonal elements are the r roots of unity. Let Ir denote the r×r identity matrix and Πr the r×r fundamental

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permutation matrix. Let Fr denote the r × r Fourier matrix given by

F ∗r =1√r

1 1 1 · · · 11 ω ω2 · · · ωr−1

1 ω2 ω4 · · · ω2(r−1)

......

......

1 ωr−1 ω2(r−1) · · · ω(r−1)(r−1)

.Note that Fr = F ᵀ

r and FrF∗r = Ir, i.e. it is unitary. One can verify the following diagonalization formula

for Πr:

Πr = F ∗r ΩrFr . (2.1)

2.2. Invariant Subspaces. We assume that the reader is familiar with the tools of linear geometriccontrol theory [4, 33]. We make use of two projection maps which are standard in linear system theory:the insertion and the natural projection. Let X be an n-dimensional vector space and let V,W ⊂ X betwo subspaces such that X = V ⊕W and dim(V) = k. The insertion map S : V → X maps x ∈ V to thecorresponding element x ∈ X ; that is, Sx := x. In coordinates, it maps the k× 1 coordinate vector of x in abasis for V to the corresponding n× 1 coordinate vector for x in a basis for X . The natural projection on Valong W, denoted Q : X → V, maps x ∈ X to its component in V; that is, given the unique representationx = v + w with v ∈ V, w ∈ W, Qx = v. Note that QS = IV , where IV is the identity map on V.

Let A : X → X be a linear map and V ⊂ X a subspace. Suppose V is A-invariant; that is, AV ⊂ V. Therestriction of A to V, denoted AV , is the unique solution of AS = SAV . It has the action of A on V andis not defined off V. Let W ⊂ X be any complementary subspace to V, i.e. V ⊕W = X . In a basis for X

adapted to V, A has the matrix representation

[A1 ∗0 A2

], where A1 is a matrix representation of AV . Let

S1, Q1 and S2, Q2 denote the insertions and natural projections on V and W, respectively. The coordinate

transformation T = [S1 S2] where T−1 =

[Q1

Q2

]yields T−1AT in the form above with A1 = Q1AS1 and

A2 = Q2AS2. We deduce the following relationships:

AS1 = S1A1 (2.2)

Q2A = A2Q2 . (2.3)

If further, W is also A-invariant, then V is said to decompose X relative to A. Then A has the matrix

representation

[A1 00 A2

]where A2 is a matrix representation of AW . With a decomposition of this form,

all four projection maps will satisfy commutative relationships with A:

ASi = SiAi (2.4)

QiA = AiQi . (2.5)

2.3. Modal Decomposition. Let A : X → X be a linear map and let ψ(s) be its minimal polynomial.Suppose ψ is factored as ψ(s) = ψ1(s) · · ·ψp(s), where the ψi are pairwise coprime. Define Xi(A) :=Kerψi(A). Because the ψi are coprime, X = X1(A) ⊕ · · · ⊕ Xp(A), and AXi(A) ⊂ Xi(A). In the sequel wewill consider the case when X is decomposed as X = Xm(A)⊕X c(A) where Xm(A) is a modal subspace andX c(A) is its unique complement. We assume that corresponding to this decomposition is a disjoint partitionof C = Cm ∪ Cc with Cm ∩ Cc = ∅. The minimal polynomial of A then factors as ψ(s) = ψm(s)ψc(s) withXm(A) = Kerψm(A) and X c(A) = Kerψc(A).

Lemma 2.1 ([33]). Let A : X → X be a linear map and let V,W ⊂ X be A-invariant subspaces such thatV ⊕W = X . Let Q : X → V be the natural projection on V along W. Then the following hold:

(i) QXm(A) = Vm(AV).(ii) Xm(A) ⊂ V if and only if σ(AW) ⊂ Cc.

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3. Commuting Matrices. Our framework for control of block circulant systems is built up fromproperties of commuting matrices. Here we review some results on commuting matrices from Chapter VIIIof [11].

Definition 3.1. Let F be a field. We say A ∈ CF(U, V ) if A, U , and V all take entries in F and

UA = AV . (3.1)

We assume the field of real numbers when not otherwise specified, i.e. C(U, V ) := CR(U, V ). If A ∈ CF(U,U),we write A ∈ CF(U), for short.

Lemma 3.2. The following hold:

(i) If A,B ∈ CF(U, V ), then A+B ∈ CF(U, V ).(ii) If A ∈ CF(U, V ) and B ∈ CF(V,W ), then AB ∈ CF(U,W ).

(iii) If A ∈ CF(U, V ), then Aᵀ ∈ CF(V ᵀ, Uᵀ).(iv) If A ∈ CF(U, V ), then A−1 ∈ CF(V,U) if A is nonsingular.

In the results to follow, let ΛU := Γ−1U UΓU and ΛV := Γ−1V V ΓV be the Jordan forms of U and V respectively.

Lemma 3.3 ([11]). Let A and ΛA satisfy ΛA := Γ−1U AΓV . Then A ∈ CF(U, V ) if and only if ΛA ∈CC(ΛU ,ΛV ).

Theorem 3.4 ([11]). Let A and ΛA satisfy ΛA := Γ−1U AΓV . Partition Γ−1U AΓV according to the Jordanblocks of ΛU and ΛV :

Γ−1U AΓV =

Z11 · · · Z1r2...

...Zr11 · · · Zr1r2

.Suppose the ith Jordan block of U corresponds to eigenvalue δi for i = 1, ..., r1, and the jth Jordan block of Vcorresponds to eigenvalue λj for j = 1, ..., r2. Then for all i = 1, . . . , r1 and j = 1, . . . , r2, Zij = 0 if δi 6= λj,and Zij is an upper triangular Toeplitz matrix if δi = λj.

Remark 3.5. We state some of our notational conventions. Let σd(V ) denote the set of distinct eigenvaluesof V . Suppose σd(V ) = λ1, . . . , λr and σd(U) ⊂ σd(V ). Our convention is that the indexing of eigenvaluesin σd(U) follows that of σd(V ); for example, σd(V ) = λ1, λ2, λ3 and σd(U) = λ1, λ3. Suppose λi hasalgebraic multiplicity ki and mi in σ(U) and σ(V ), respectively. Note that some ki’s may be zero. We usethe following preferred ordering of eigenvalues of V and U :

σ(V ) =λ1, . . . , λ1︸︷︷︸m1 times

, . . . , λr, . . . , λr︸︷︷︸mr times

σ(U) =λ1, . . . , λ1︸︷︷︸k1 times

, . . . , λr, . . . , λr︸︷︷︸kr times

.

We also partition ΓU and ΓV as

ΓU =[U1 · · · Ur

], ΓV =

[V1 · · · Vr

](3.2)

where each Ui has ki columns and each Vi has mi columns (note again some Ui may have “zero width”). Thegeneralized eigenvectors are chosen so that if λi = λj, then Ui = U j. If V is real, there exists a permutation`1, . . . , `r of 1, . . . , r such that λì = λi for each i = 1, . . . , r. C

If A ∈ C(U, V ) we would like to know what complex conjugate pattern arises in the blocks Zij of ΛA :=Γ−1U AΓV . First we need a minor technical result about Jordan form.

Lemma 3.6. Let ΓU be partitioned as (3.2) and partition Γ−1U as col(U ′1, . . . , U′r), where the number of rows

of U ′i is equal to the number of columns of Ui. Suppose there exists a permutation `1, . . . , `r of 1, . . . , rsuch that Ui = U ì for each i. Then, U ′i = U

′ì for each i.

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Proof. Since Γ−1U ΓU = I, U ′iUi = I and U ′iUj = 0 for i 6= j. Also U ′ìUì = I = I, and so U ′iUi =

U′ìU ì . By assumption, Ui = U ì giving U ′iUi = U

′ìUi. Then we have U ′iΓU =

[0 · · · U ′iUi · · · 0

]=[

0 · · · U′ìUi · · · 0

]= U

′ìΓU . Since ΓU is full rank, it follows that U ′i = U

′ì .

Lemma 3.7. Let σd(V ) = λ1, . . . , λr and suppose σd(U) ⊂ σd(V ). Suppose λi has algebraic multiplicityki in U and mi in V (some ki’s may be zero). Let A ∈ C(U, V ) and define ΛA = Γ−1U AΓV . Then ΛA =diag(A1, . . . , Ar) with Ai ∈ Cki×mi and Ai = Aì for each i = 1, . . . , r.

Proof. Using the ordering of Remark 3.5, Theorem 3.4 gives that Γ−1U AΓV = diag(A1, . . . , Ar), a block diago-nal matrix. Partition ΓU =

[U1 · · · Ur

]and ΓV =

[V1 · · · Vr

], as in (3.2). Let Γ−1U = col(U ′1, . . . , U

′r).

Then

ΛA :=

A1

. . .

Ar

=

U′1AV1

. . .

U ′rAVr

.By Lemma 3.6, U ′i = U

′ì and by Remark 3.5, Vi = V ì . Since A is real, we conclude Ai = Aì for each

i = 1, . . . , r.

The next result tells us when a block diagonal matrix can be converted back to a real matrix.

Lemma 3.8. Let σd(V ) = λ1, . . . , λr and suppose σd(U) ⊂ σd(V ). Suppose λi has algebraic multiplicityki in U and mi in V (some ki’s may be zero). Let ΛA := diag(A1, . . . , Ar) with Ai ∈ Cki×mi , and defineA := ΓUΛAΓ−1V . If for all i, j ∈ 1, . . . , r, Ai = Aj whenever λi = λj, then A is real.

Proof. Partition ΓU =[U1 · · · Ur

]and ΓV =

[V1 · · · Vr

]according to (3.2) such that Ui has ki

columns and Vi has mi columns. Following Remark 3.5 there exists a permutation `1, . . . , `r of 1, . . . , rsuch that Ui = U ì and Vi = V ì for every i = 1, . . . , r. By assumption Ai = Aì for every i = 1, . . . , r. Let

V −1 = col(V ′1 , . . . , V′r ). By Lemma 3.6, V ′i = V

′ì . Then we have

A := ΓUΛAΓ−1V =

|U1

|· · ·

|Ur|

A1

. . .

Ar

— V ′1 —

...— V ′r —

= U1A1V

′1 + · · ·+ UrArV

′r .

If ì = i, then Ui, V′i , and Ai are real. If ì = j 6= i, then UiAiV ′i = UjAjV

′j and so UiAiV

′i +UjAjV

′j is real.

It follows that A is real, as desired.

Next we give sufficient conditions, (A1)-(A2) below, for a matrix A to satisfy UA = AV . These conditionswill be seen to be reasonable for our control results. Conditions (A3)-(A5) are simply to fix notation andfollow Remark 3.5.

Assumption 3.9. We assume that the pair (U, V ) satisfies:

(A1) U and V are diagonalizable.(A2) σd(U) ⊂ σd(V ).(A3) σd(V ) = λ1, . . . , λr where λi has algebraic multiplicity 0 ≤ ki ≤ n in σ(U) and m in σ(V ) for

some n,m > 0.(A4) The eigenvalues of U and V are ordered according to the preferred ordering of Remark 3.5.(A5) There exists a permutation `1, . . . , `r of 1, . . . , r such that λì = λi for each i = 1, . . . , r.

Lemma 3.10. Suppose Assumption 3.9 holds. Let ΛA := diag(A1, . . . , Ar).

(i) If Ai ∈ Cki×m and A := ΓUΛAΓ−1V , then A ∈ CC(U, V ).(ii) If Ai ∈ Cm×ki and A := ΓV ΛAΓ−1U , then A ∈ CC(V,U).

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Proof. We only prove (i); the proof of (ii) is similar. We have

ΛUΛA =

λ1Ik1×k1. . .

λrIkr×kr

A1

. . .

Ar

=

λ1A1

. . .

λrAr

=

A1

. . .

Ar

λ1Im×m

. . .

λrIm×m

= ΛAΛV .

By Lemma 3.3, UA = AV .

We conclude this section with an examination of the eigenvalues of A ∈ C(U). The main result is that theeigenvalues of A follow the same complex conjugate pattern as those of U .

Definition 3.11. Let σd(U) = λ1, . . . , λr be the distinct eigenvalues of U . A spectrum L is called U -patterned if it has the same cardinality as σ(U), and it can be ordered and partitioned as L = L1 ] · · · ]Lr

such that Li = L j whenever λi = λj.

Lemma 3.12. Let σd(U) = λ1, . . . , λr and let A ∈ C(U). Then σ(A) is U -patterned.

Proof. Let ΛA := Γ−1U AΓV . By Lemma 3.7, ΛA = diag(A1, . . . , Ar) and Ai = Aj whenever λi = λj . DefineLi = σ(Ai). Thus, σ(A) = L1 ] · · · ] Lr and Li = Lj whenever λi = λj , and so σ(A) is U -patterned.

4. Block Circulant Matrices. Let A1, . . . , Ar ∈ Cn×m. A block circulant matrix is an rn×rm matrixof the form

A =

A1 A2 · · · ArAr A1 · · · Ar−1...

......

A2 A3 · · · A1

.Every block circulant matrix A can be represented as

A =

r∑i=1

(Πi−1r ⊗Ai) . (4.1)

A well-known result from [7] is that a matrix A ∈ Crn×rn is block circulant if and only if it commutes withΠr ⊗ In. The result easily generalizes to nonsquare matrices.

Lemma 4.1. A ∈ Crn×rm is block circulant if and only if A ∈ CC(Πr ⊗ In,Πr ⊗ Im).

The vector space of rn × rm block circulant matrices will henceforth be denoted as CC(Πr ⊗ In,Πr ⊗ Im).Consider (Fr ⊗ In)∗ and note that (Fr ⊗ In)∗(Fr ⊗ In) = Irn. An important fact is the following blockdiagonalization formula for Πr ⊗ In:

Πr ⊗ In = (Fr ⊗ In)∗(Ωr ⊗ In)(Fr ⊗ In) . (4.2)

Block circulant matrices are well-known to be block diagonalizable.

Theorem 4.1 ([7]). A ∈ CC(Πr ⊗ In,Πr ⊗ Im) if and only if it is of the form

A = (Fr ⊗ In)∗ diag(M1,M2, . . . ,Mr)(Fr ⊗ Im) (4.3)

where M1, . . . ,Mr ∈ Cn×m.

Using the fact that (Fr ⊗ In)∗(Fr ⊗ In) = Irn, (4.3) gives the following block diagonal matrix:

ΛA := diag(M1,M2, . . . ,Mr) = (Fr ⊗ In)A(Fr ⊗ Im)∗ . (4.4)

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By placing conditions on the Mis, we obtain a characterization of C(Πr⊗ In,Πr⊗ Im), the class of real blockcirculant matrices.

Theorem 4.2. A ∈ C(Πr ⊗ In,Πr ⊗ Im) if and only if it is of the form

A = (Fr ⊗ In)∗ diag(M1,M2, . . . ,Mr)(Fr ⊗ Im) (4.5)

where M1, . . . ,Mr ∈ Cn×m satisfy:

(i) M1 ∈ Rn×m.(ii) Mr−i+2 = M i , i = 2, . . . , r.

Next we characterize the spectra of real square block circulant matrices.

Lemma 4.2 ([6]). If A ∈ C(Πr ⊗ In), then the spectrum of A satisfies:

(i) It is symmetric with respect to the real axis.(ii) If r is odd, then there are no more than n real eigenvalues of odd multiplicity.

(iii) If r is even, then there are no more than 2n real eigenvalues of odd multiplicity.

Definition 4.3. A spectrum L satisfying (i)-(iii) of Lemma 4.2 is called a block circulant spectrum.

Lemma 4.4. A spectrum is block circulant if and only if it is (Πr ⊗ In)-patterned.

Remark 4.5. While the class of block circulant spectra is somewhat more restrictive than the class ofsymmetric spectra, it is always possible to find stable block circulant spectra. This makes block circulantspectra sufficiently versatile for most pole placement-related design problems of control theory. C

Lemma 4.6. Let L be a block circulant spectrum. There exists A ∈ C(Πr ⊗ In) such that σ(A) = L .

Proof. We present a proof sketch of how to construct a matrix A ∈ C(Πr ⊗ In) with σ(A) = L . (Somedetails are left out due to space restrictions.) To that end, we write L in a special order so that suitablematrices Mi can be constructed according to Theorem 4.2. By Definition 4.3, L is symmetric with respectto the real axis and contains rn eigenvalues, where r and n satisfy conditions (ii)-(iii) of Lemma 4.2. Wewill split L into r spectra, L1, . . . ,Lr, each containing n eigenvalues. We have two cases: either r is oddor even.

First, suppose r is odd. The number of distinct real eigenvalues of odd multiplicity will have the same parityas n. Place one of each into L1; an even number of “spots” are still available in L1. It can be shown thatan even number of eigenvalues remain, all of which come in complex conjugate pairs (some of which maybe real eigenvalues with even multiplicity). Thus, L1 can be filled by complex conjugate pairs. ConstructL2, . . . ,Lr from complex conjugate pairs, ensuring that Li = L r−i+2 for each i = 2, . . . , r (where thecomplex conjugate of the set is taken element-wise).

Second, suppose r is even and let ` = r2 + 1. There will be an even number of distinct real eigenvalues of odd

multiplicity. Place one of each into either L1 or L`, ensuring that the parity of the number of eigenvaluesplaced in each set is equal to the parity of n. An even number of “spots” are still available in each of L1 andL`. It can be shown that an even number of eigenvalues remain, all of which come in complex conjugate pairs(some of which may be real eigenvalues with even multiplicity). Thus, L1 and L` can be filled by complexconjugate pairs. Construct the remaining Lis from complex conjugate pairs, ensuring that Li = L r−i+2.

Now, for i = 1, . . . , r, define Mi to be the companion matrix with eigenvalues Li. The Mis satisfy theconditions of Theorem 4.2, so defining A = (Fr ⊗ In)∗ diag(M1, . . . ,Mr)(Fr ⊗ In) gives A ∈ C(Πr ⊗ In) withσ(A) = L .

5. Block Circulant Subspaces. In the previous two sections we presented algebraic tools used in ourframework for control of block circulant systems. In this section we introduce the main geometric constructthat will allow us to link the algebraic properties of block circulant and commuting matrices with the theory

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of linear geometric control [4, 33].

Definition 5.1. We say that V ⊂ X is a block circulant subspace if it is (Πr ⊗ In)-invariant. That is,(Πr ⊗ In)V ⊂ V.

Lemma 5.2. Let V,W ⊂ X be block circulant subspaces. Then V + W and V ∩ W are block circulantsubspaces.

Proof. We have (Πr ⊗ In)(V +W) = (Πr ⊗ In)V + (Πr ⊗ In)W ⊂ V +W. Similarly, (Πr ⊗ In)(V ∩W) ⊂(Πr ⊗ In)V ∩ (Πr ⊗ In)W ⊂ V ∩W.

Consider A ∈ C(Πr⊗ In) and let V be an A-invariant subspace. Is V a block circulant subspace? The answeris not generally. Fortunately it is possible to identify several A-invariant subspaces, useful in a control theorycontext, that are also block circulant subspaces.

Lemma 5.3. Let A ∈ C(Πr ⊗ In), B ∈ C(Πr ⊗ In,Πr ⊗ Im), and let ρ(s) be a polynomial. Then, Im B,KerB, Im ρ(A), and Ker ρ(A) are block circulant subspaces.

Proof. First, let x ∈ Im B. There exists ξ ∈ X such that x = Bξ. Thus, (Πr⊗In)x = (Πr⊗In)Bξ = B(Πr⊗Im)ξ ∈ Im B. Second, let x ∈ KerB. Then, B(Πr ⊗ In)x = (Πr ⊗ Im)Bx = 0, and so (Πr ⊗ In)x ∈ KerB.Next, by Lemma 3.2, ρ(A) ∈ C(Πr ⊗ In). Then by the previous argument with B = ρ(A), Im ρ(A) andKer ρ(A) are block circulant subspaces, as desired.

Given a block circulant matrix A ∈ C(Πr ⊗ In) and a block circulant subspace V ⊂ X , the image andpreimage of V under A are themselves block circulant subspaces.

Lemma 5.4. Let A ∈ C(Πr ⊗ In) and let V ⊂ X be a block circulant subspace. Then AV, the image of Vunder A, and A−1V = x ∈ X | Ax ∈ V, the preimage, are both block circulant subspaces.

Proof. First, (Πr⊗In)(AV) = A(Πr⊗In)V ⊂ AV, so AV is a block circulant subspace. Second, let x ∈ A−1V.Since Ax ∈ V and (Πr ⊗ In)V ⊂ V, we have A(Πr ⊗ In)x = (Πr ⊗ In)Ax ∈ V. Thus, (Πr ⊗ In)x ∈ A−1V, asdesired.

Another useful property is that a block circulant subspace has a block circulant complement.

Lemma 5.5. Let V ⊂ X be a block circulant subspace. Then V⊥ is a block circulant subspace.

Proof. Let w ∈ V⊥. For any v ∈ V, 〈v, w〉 = 0, where 〈·, ·〉 denotes the inner product on X . Consider(Πr ⊗ In)w. Since (Πr ⊗ In)r−1v ∈ V, we have 〈v, (Πr ⊗ In)w〉 = 〈(Πr ⊗ In)ᵀv, w〉 = 〈(Πr ⊗ In)r−1v, w〉 = 0.Hence, (Πr ⊗ In)w ∈ V⊥ and V⊥ is a block circulant subspace.

The significance of Lemma 5.5 is that any block circulant subspace V decomposes X relative to Πr⊗In. Thismeans that the restrictions (Πr ⊗ In)V and (Πr ⊗ In)W are both defined (where W is any block circulantcomplement of V). This important property will allow us to preserve structure in restrictions of blockcirculant maps; certain commuting relationships inherited from the original block circulant structure willhold for such maps. The main result is the following block circulant version of the standard representationtheorem for linear maps with respect to invariant subspaces.

Theorem 5.1 (Block Circulant Representation Theorem). Let A ∈ C(Πr ⊗ In) and let V ⊂ X be a blockcirculant subspace such that dim(V) = k < rn and AV ⊂ V. Let W ⊂ X be a block circulant complement ofV. Then A has a matrix representation

[A1 ∗0 A2

], (5.1)

where A1 is a matrix representation of AV . Moreover, A1 ∈ C((Πr ⊗ In)V) and A2 ∈ C((Πr ⊗ In)W).

Proof. Let v1, . . . , vk, vk+1, . . . , vrn be a preferred basis of X such that V = spanv1, . . . , vk and W =spanvk+1, . . . , vrn. This determines projection maps S1 : V → X , S2 :W → X , Q1 : X → V, Q2 : X → W,

8

and the coordinate transformation T = [ S1 S2 ]. We obtain the standard representation

T−1AT =

[Q1AS1 ∗

0 Q2AS2

].

Then A1 := Q1AS1 = Q1S1AV = AV , and A2 := Q2AS2. Thus, we obtain (5.1). Now we show AV ∈C((Πr ⊗ In)V). We have

(Πr ⊗ In)VAV = (Πr ⊗ In)VQ1AS1 = Q1(Πr ⊗ In)AS1

= Q1A(Πr ⊗ In)S1 = Q1AS1(Πr ⊗ In)V = AV(Πr ⊗ In)V .

The argument for A2 ∈ C((Πr ⊗ In)W) is analogous.

Theorem 5.1 takes a full block circulant matrix and pushes it down to a restriction in C((Πr ⊗ In)V).Conversely, given a matrix in C((Πr ⊗ In)V) we would like to be able to lift it back up to a real blockcirculant matrix. This is doable with the correct choice of complementary subspace.

Lemma 5.6 (Lifting Lemma). Let V ⊂ X be a block circulant subspace and W ⊂ X a block circulantcomplement. Let S1 : V → X and Q1 : X → V be the insertion and natural projection on V.

(i) If A1 ∈ C((Πr ⊗ In)V) then S1A1Q1 ∈ C(Πr ⊗ In).(ii) If B1 ∈ C((Πr ⊗ In)V ,Πr ⊗ Im) then S1B1 ∈ C(Πr ⊗ In,Πr ⊗ Im).

(iii) If K1 ∈ C(Πr ⊗ Im, (Πr ⊗ In)V) then K1Q1 ∈ C(Πr ⊗ Im,Πr ⊗ In).

Proof. Using (2.2), we have

(S1A1Q1)(Πr ⊗ In) = S1A1(Πr ⊗ In)VQ1 = S1(Πr ⊗ In)VA1Q1 = (Πr ⊗ In)(S1A1Q1)

(S1B1)(Πr ⊗ Im) = S1(Πr ⊗ In)VB1 = (Πr ⊗ In)(S1B1)

(K1Q1)(Πr ⊗ In) = K1(Πr ⊗ In)VQ1 = (Πr ⊗ Im)(K1Q1) .

The ability to transition between a full space and a subspace without sacrificing commuting relationships,shown by Theorem 5.1 and Lemma 5.6, is what allows us to recover feedbacks that preserve a control system’sblock circulant structure using the usual decompositions of linear geometric control. We conclude this sectionby identifying pairs (U, V ) useful in block circulant control design, that satisfy Assumption 3.9.

Lemma 5.7. Let V ⊂ X be a block circulant subspace (including possibly V = X ), and suppose (U, V ) is thepair (U, V ) = ((Πr ⊗ In)V ,Πr ⊗ Im). Then Assumption 3.9 holds.

We have laid the foundations in the areas of commuting matrices, block circulant matrices, block circulantsubspaces, and restrictions and lifts of block circulant maps. We now begin our study of block circulantcontrol systems.

6. Controllability. Consider the linear time-invariant system given by

x(t) = Ax(t) +Bu(t) (6.1a)

y(t) = Cx(t), (6.1b)

where x(t) ∈ Rrn is the vector of states, u(t) ∈ Rrm is the vector of inputs, and y(t) ∈ Rrp is the vector ofmeasurements. We denote the state space, input space and measurement space by X , U and Y, respectively.Assume a real system with matrices A ∈ Rrn×rn, B ∈ Rrn×rm and C ∈ Rrp×rn. We refer to such asystem in shorthand by the triple (C,A,B) or simply by the pair (A,B) or the pair (C,A), when thethird transformation is not applicable. If further A ∈ C(Πr ⊗ In), B ∈ C(Πr ⊗ In,Πr ⊗ Im), and C ∈C(Πr ⊗ Ip,Πr ⊗ In), then (C,A,B) is called a block circulant system. The open loop poles of the system arethe eigenvalues of A; for block circulant systems the open loop poles of the system form a block circulantspectrum.

9

Let B := Im B. The controllable subspace 〈A|B〉 of the pair (A,B) is given by 〈A|B〉 = B+AB+· · ·+Arn−1B.Our focus on block circulant subspaces has been driven by the following observation.

Lemma 6.1. The controllable subspace is a block circulant subspace.

Proof. By Lemma 3.2, Ai−1B is a block circulant matrix and by Lemma 5.3, Im (Ai−1B) is a block circulantsubspace for all i ∈ N. Using Lemma 5.2, we conclude 〈A|B〉 is a block circulant subspace.

It is well known that the spectrum of A+BK can be arbitrarily assigned to any symmetric set of poles bychoice of K : X → U if and only if (A,B) is controllable. For a block circulant system, the question arisesof what possible poles can be achieved by choice of block circulant state feedback. We address this questionin the more general setting when A ∈ C(U) and B ∈ C(U, V ) for some linear maps U and V .

Theorem 6.1 (Pole Placement). Let V ⊂ X be a block circulant subspace and suppose (U, V ) = ((Πr ⊗In)V ,Πr ⊗ Im). Let A ∈ C(U) and B ∈ C(U, V ). The pair (A,B) is controllable if and only if for everyU -patterned spectrum L , there exists a map K : V → U with K ∈ C(V,U) such that σ(A+BK) = L .

Proof. (Only If) From Assumption 3.9, σd(V ) = λ1, . . . , λr and there is a permutation `1, . . . , `r of1, . . . , r such that λi = λì for i = 1, . . . , r. By Lemma 3.7, ΛA := Γ−1U AΓU = diag(A1, . . . , Ar) andΛB := Γ−1U BΓV = diag(B1, . . . , Br), where Ai = Aì and Bi = Bì i = 1, . . . , r. Since (A,B) is controllable,(ΛA,ΛB) is controllable and, in turn, each pair (Ai, Bi) is controllable. Then, for any spectrum Li, thereexists Ki such that σ(Ai + BiKi) = Li. By Lemma 3.12 and the index convention of (A3), L can alsobe partitioned as L = L1 ] · · · ] Lr where Li = L ì , i = 1, . . . , r. Thus, the Kis can be chosen suchthat Ki = Kì , i = 1, . . . , r. Let ΛK = diag(K1, . . . ,Kr) and define K := ΓV ΛKΓ−1U . By Lemma 3.10(ii),K ∈ CC(V,U). By Lemma 3.8, K is real. We conclude that K ∈ C(V,U).

(If) The argument is a minor variation of the proof of Theorem 2.1 in [33], so it is omitted.

Remark 6.2. The proof of Theorem 6.1 provides an explicit method to construct a block circulant feedbackfor pole placement in any block circulant spectrum. As in most prior work on block circulant systems, weaccomplish this by block diagonalization. First, transform the system into Fourier coordinates, i.e. puteach matrix in Jordan form. This decomposes the system into r subsystems with the property that the firstsubsystem is real, and the ith and (r − i + 2)th subsystems are complex conjugates for each i = 2, . . . , r.Second, place the poles in each subsystem independently via any general unstructured feedback, ensuringthat the feedback matrix used for the first subsystem is real, and the feedback matrices used for the ith and(r− i+ 2)th subsystems are complex conjugates for each i = 2, . . . , r. Third, transform the system back intostandard coordinates. The resulting closed-loop system remains block circulant as does the overall feedbacklaw.

6.1. Controllable Decomposition. Suppose we have a block circulant system that is not fully con-trollable, i.e. 〈A|B〉 6= X . Then there is a basis in which the controllable and uncontrollable parts of thesystem are displayed transparently. This is the content of our first decomposition theorem.

Theorem 6.2 (First Decomposition Theorem). Let C := 〈A|B〉 and suppose dim(C) = k < rn. Then thereexists a coordinate transformation T : X → X such that (A, B) := (T−1AT, T−1B) has the form

A =

[A1 ∗0 A2

], B =

[B1

0

], (6.2)

where A1 = AC ∈ C((Πr⊗ In)C), A2 ∈ C((Πr⊗ In)C⊥), and B1 ∈ C((Πr⊗ In)C ,Πr⊗ Im). Moreover, the pair(A1, B1) is controllable.

Proof. By Lemma 6.1 and Lemma 5.5, C and C⊥ are block circulant subspaces. We choose projection mapsS1, S2, Q1, and Q2 according to Theorem 5.1 and apply the coordinate transformation (ξ1, ξ2) = T−1xwhere T =

[S1 S2

]. The forms of A, A1, and A2 follow from Theorem 5.1. Now consider

B = T−1B =

[Q1

Q2

]B =:

[B1

B2

]10

Since B := Im B ⊂ C, B2 = Q2B = 0 and S1Q1B = B. Thus, B = S1B1. Using (2.5) applied to (Πr ⊗ In),we get

(Πr ⊗ In)CB1 = Q1(Πr ⊗ In)B = Q1B(Πr ⊗ Im) = B1(Πr ⊗ Im) .

Hence, B1 ∈ C((Πr ⊗ In)C ,Πr ⊗ Im). Finally, the pair (A1, B1) is controllable if we can show S1〈A1|B1〉 = Cwhere B1 = Im B1. Using (2.2), we have

S1〈A1|B1〉 = S1B1 + S1A1B1 + · · ·+ S1Arn−11 B1

= S1B1 +AS1B1 + · · ·+Arn−1S1B1= B +AB + · · ·+Arn−1B = 〈A|B〉 .

Example 6.3.

Consider the block circulant system (A,B), where A ∈ C(Π4 ⊗ I2) and B ∈ C(Π4 ⊗ I1), given by

x =

A︷︸︸︷

−1 5 3 1 −4 −3 3 10 −7 0 −1 0 −3 0 −13 1 −1 5 3 1 −4 −30 −1 0 −7 0 −1 0 −3−4 −3 3 1 −1 5 3 1

0 −3 0 −1 0 −7 0 −13 1 −4 −3 3 1 −1 50 −1 0 −3 0 −1 0 −7

x+

B︷︸︸︷

1 0 1 01 0 −1 00 1 0 10 1 0 −11 0 1 0−1 0 1 0

0 1 0 10 −1 0 1

u (6.3)

The controllable subspace of (A,B) is given by 〈A|B〉 = spanc1, c2, c3, c4, c5, c6 where c1 := e1, c2 := e3,c3 := e5, c4 := e7, c5 := (0, 1, 0, 0, 0,−1, 0, 0), and c6 := (0, 0, 0, 1, 0, 0, 0,−1). Following Lemma 6.1, 〈A|B〉is a block circulant subspace. A basis for 〈A|B〉⊥ is given by r1, r2, where r1 := (0, 1, 0, 0, 0, 1, 0, 0) and r2 :=(0, 0, 0, 1, 0, 0, 0, 1) and, by Lemma 5.5, 〈A|B〉⊥ is also a block circulant subspace. Thus, c1, c2, c3, c4, c5, c6, r1, r2is a preferred basis for X adapted to 〈A|B〉. We will perform a controllable decomposition of the system usingthis basis. Define the insertion maps

S1 =[c1 c2 c3 c4 c5 c6

]S2 =

[r1 r2

]This gives the coordinate transformation T =

[S1 S2

]. The natural projection maps Q1 and Q2 are then

uniquely defined by T−1 = col(Q1, Q2), where Q1 is the first six rows and Q2 is the final two rows. Thecontrollable decomposition of (A,B) is then given by

T−1AT =

−1 3 −9 3 8 0 2 23 −1 3 −9 0 8 2 2−9 3 −1 3 −8 0 2 2

3 −9 3 −1 0 −8 2 20 0 0 0 −4 0 0 00 0 0 0 0 −4 0 0

−10 −2−2 −10

, T−1B =

1 0 1 00 1 0 11 0 1 00 1 0 11 0 1 00 1 0 1

(6.4)

Denote the top-left subsystem by (A1, B1) and the bottom-right subsystem by (A2, B2). First, we will showthat A2 ∈ C((Π4 ⊗ I2)〈A|B〉⊥). We have

(Π4 ⊗ I2)〈A|B〉⊥ := Q2(Π4 ⊗ I2)S2 =

[0 11 0

]11

By direct calculation,

(Π4 ⊗ I2)〈A|B〉⊥A2 =

[−2 −10−10 −2

]= A2(Π4 ⊗ I2)〈A|B〉⊥

and so A2 ∈ C((Π4 ⊗ I2)〈A|B〉⊥) by definition. Following similar arguments, it can be shown that A1 ∈C((Π4 ⊗ I2)〈A|B〉), B1 ∈ C((Π4 ⊗ I2)〈A|B〉,Π4 ⊗ I1) and B2 ∈ C((Π4 ⊗ I2)〈A|B〉⊥ ,Π4 ⊗ I1). Finally, it can beverified that 〈A1|Im B1〉 = R6 and so the pair (A1, B1) is controllable, as predicted by Theorem 6.2. C

6.2. Stabilizability. A system, or equivalently the pair (A,B), is stabilizable if there exists K : X → Usuch that σ(A+BK) ⊂ C−. A system is stabilizable if and only if X+(A) ⊂ C where C := 〈A|B〉 [33]. For ablock circulant system the question arises of whether the system can be stabilized by a block circulant statefeedback. We present here a more general result to be used in later synthesis problems.

Theorem 6.3 (Stabilizability). Let V ⊂ X be a block circulant subspace. Suppose A ∈ C(Πr ⊗ In) andAV ⊂ V. Let A1 ∈ C((Πr ⊗ In)V) be the restriction of A to V and let B1 ∈ C((Πr ⊗ In)V ,Πr ⊗ Im). Thereexists a state feedback K1 : V → U , K1 ∈ C(Πr ⊗ Im, (Πr ⊗ In)V), such that σ(A1 +B1K1) ⊂ C− if and onlyif

X+(A1) ⊂ C1

where C1 is the controllable subspace of the pair (A1, B1).

Proof. (If) Suppose X+(A1) ⊂ C1 and let dim C1 = k. First we show that C1 = Im (B1) + Im (A1B1) +· · ·+ Im (Ak−11 B1) is a block circulant subspace. To that end, we will show that (Πr ⊗ In)SIm (Ai−11 B1) ⊂SIm (Ai−11 B1) for i = 1, . . . , k, where S : V → X is the insertion map. Using Lemmas 5.3 and 5.6, we have

(Πr ⊗ In)SIm (Ai−11 B1) = (Πr ⊗ In)Im (S(Ai−11 B1))

= (Πr ⊗ In)Im (Ai−1(SB1)) ⊂ Im (Ai−1SB1) = SIm (Ai−11 B1) .

Thus, C1 is a block circulant subspace. Now suppose X+(A1) ⊂ C1. First, suppose dim C1 = dimV. BecauseC1 = V is a block circulant subspace, Lemma 5.7 is verified with (U, V ) = ((Πr ⊗ In)V ,Πr ⊗ Im). Then wecan apply Theorem 6.1 to obtain K1 ∈ C(Πr ⊗ Im, (Πr ⊗ In)V) satisfying σ(A1 +B1K1) ⊂ C−.

Second, suppose dim C1 < dimV. We will decompose the system (A1, B1) into its controllable and uncon-trollable parts. By Lemma 5.5, C⊥1 is a block circulant subspace, and by Lemma 5.2, so is W := V ∩ C⊥1 .Because C1 ⊂ V, by the modular distributive rule of subspaces [33, (0.3.1)]

V = (V ∩ C1)⊕ (V ∩ C⊥1 ) = C1 ⊕W .

Let S11, Q11 and S12, Q12 be the insertions and natural projections on C1 and W respectively, and letT :=

[S11 S12

]. Define transformed coordinates (ξ1, ξ2) := T−1x (where T−1 = col(Q11, Q12)). The

transformed system is [ξ1ξ2

]=

[A11 ∗0 A12

] [ξ1ξ2

]+

[B11

B12

]u (6.5)

where A1i := Q1iA1S1i and B1i := Q1iB1 for i = 1, 2. Observe that (Πr ⊗ In)C1 = Q11(Πr ⊗ In)VS11. As inthe First Decomposition Theorem 6.2, we can show that A11 ∈ C((Πr⊗In)C1), B11 ∈ C((Πr⊗In)C1 ,Πr⊗Im),and the pair (A11, B11) is controllable. Because C1 ⊂ V is block circulant, Lemma 5.7 is verified with (U, V ) =((Πr⊗In)C1 ,Πr⊗Im). Then we can apply Theorem 6.1 to obtainK11 : C1 → U , K11 ∈ C(Πr⊗Im, (Πr⊗In)C1),such that σ(A11 +B11K11) ⊂ C−. Let K1 := K11Q1. Using (2.5),

(Πr ⊗ Im)K1 = K11(Πr ⊗ In)C1Q11 = K11Q11(Πr ⊗ In)V = K1(Πr ⊗ In)V .

Hence, K1 ∈ C(Πr⊗Im, (Πr⊗In)V). Now we have u = K1x = K11ξ1. Substituting into (6.5), the closed-loopsystem satisfies

σ(A1 +B1K1) = σ(A11 +B11K11) ] σ(A12) .

12

By assumption, X+(A1) ⊂ C1, so by Lemma 2.1(ii), σ(A12) ⊂ C−. Thus, σ(A1 +B1K1) ⊂ C−, as desired.

(Only If) The solvability condition is identical to that for general stabilizability. Since it is necessary forthe existence of a general feedback, it is also necessary for the existence of a special feedback.

Example 6.4. Recall the system (A,B) from (6.3) of Example 6.3. The eigenvalues of A are given by theblock circulant spectrum σ(A) = −16,−12,−8,−4,−4,−4, 8, 8, and so the unstable subspace correspondsto the eigenvalue 8 with algebraic multiplicity 2. It can be verified that the minimal polynomial of A is givenby ψ(s) = (s+16)(s+12)(s+8)(s+4)(s−8), and so the unstable subspace of A is given by X+(A) := Ker(A−8I) = a1, a2 where a1 = (1, 0, 0, 0,−1, 0, 0, 0) and a2 = (0, 0, 1, 0, 0, 0,−1, 0). Considering the controllablesubspace 〈A|B〉 = spanc1, · · · , c6 (where the cis were defined in Example 6.3), it can immediately be seenthat a1 = c1 − c3 and a2 = c2 − c4. Thus, X+(A) ⊂ 〈A|B〉 and the system is stabilizable by Theorem 6.3.We will now explicitly find a stabilizing feedback.

From Theorem 6.4, the system spilts into subsystems (A1, B1) and (A2, B2) with σ(A1) = −16,−4,−4,−4, 8, 8and σ(A2) = −12,−8. Thus, all instabilities of the system are contained in (A1, B1), which was shown inExample 6.3 to be controllable. By Theorem 6.1, the closed-loop poles of (A1, B1) can then be placed in anyarbitrary (Π4 ⊗ I2)〈A|B〉-patterned spectrum by a feedback K1 ∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)〈A|B〉). Since our goal isstabilization, we will place them in the left half-plane. In particular, the feedback

K1 =1

8

−25 0 25 0 −32 0

0 −25 0 25 0 −3225 0 −25 0 32 00 25 0 −25 0 32

∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)〈A|B〉)

gives σ(A1 +B1K1) = −16,−4,−2,−2,−2,−2 ⊂ C−, and so A1 +B1K1 is stable. By Lemma 5.6, liftingK1 to the full space X (using the natural projection map Q1 defined in Example 6.3) gives

K = K1Q1 =1

8

−25 −16 0 0 25 16 0 0

0 0 −25 −16 0 0 25 1625 16 0 0 −25 −16 0 00 0 25 16 0 0 −25 −16

∈ C(Π4 ⊗ I1,Π4 ⊗ I2)

and the poles of the closed-loop system are given by σ(A + BK) = −16,−12,−8,−4,−2,−2,−2,−2,confirming that the system has been stabilized. C

7. Observability. Consider again the block circulant system given in (6.1). The unobservable subspaceN of the pair (C,A) is given by N =

⋂rni=1 Ker(CAi−1).

Lemma 7.1. The unobservable subspace is a block circulant subspace.

Proof. By Lemma 3.2, CAi−1 is a block circulant matrix and by Lemma 5.3, Ker(CAi−1) is a block circulantsubspace for all i ∈ N. Using Lemma 5.2, we conclude N is a block circulant subspace.

Using duality and Theorem 6.1, we have the following result about observability of block circulant systems.

Theorem 7.1. Let V ⊂ X be a block circulant subspace and suppose Assumption 3.9 holds for (U, V ) =((Πr ⊗ In)V ,Πr ⊗ Ip). Let A ∈ C(U) and C ∈ C(V,U). The pair (C,A) is observable if and only if for everyU -patterned spectrum L , there exists a map K : Y → V with K ∈ C(U, V ) such that σ(A+KC) = L .

7.1. Second Decomposition Theorem. Suppose we have a block circulant system that is not fullyobservable, i.e. N 6= 0. There is a basis in which the unobservable and observable parts of the system aredisplayed transparently. This is the content of our second decomposition theorem.

Theorem 7.2 (Second Decomposition Theorem). Suppose dim(N ) = k 6= 0. There exists a coordinatetransformation T : X → X such that (C, A) := (CT, T−1AT ) has the form

A =

[A1 ∗0 A2

], C =

[0 C2

], (7.1)

13

where A1 = AN ∈ C((Πr ⊗ In)N ), A2 ∈ C((Πr ⊗ In)N⊥), and C2 ∈ C(Πr ⊗ Ip, (Πr ⊗ In)N⊥). Moreover, thepair (C2, A2) is observable.

Proof. By Lemma 7.1 and Lemma 5.5, N and N⊥ are block circulant subspaces. We choose projectionmaps S1, S2, Q1, and Q2 according to Theorem 5.1 and apply the coordinate transformation (ξ1, ξ2) = T−1xwhere T =

[S1 S2


C = CT =[CS1 CS2

]=:[C1 C2

].

Since N ⊂ Ker(C), C1 = CS1 = 0 and CS2Q2 = C. Also

(Πr ⊗ Ip)C2 = (Πr ⊗ Ip)CS2 = C(Πr ⊗ In)S2 = C2(Πr ⊗ In)N⊥ .

Hence, C2 ∈ C(Πr ⊗ Ip, (Πr ⊗ In)N⊥). Finally, we verify that (C2, A2) is observable. We must show

N2 =⋂rn−ki=1 Ker

(C2A

i−12

)= 0. Let ξ2 ∈ N2. Using the Cayley-Hamilton theorem, C2A

i−12 ξ2 = 0 for all

i ∈ N. There exists x ∈ X such that ξ2 = Q2x. Using (2.3)

CAi−1x = C2Q2Ai−1x = C2A

i−12 Q2x = C2A

i−12 ξ2 = 0 , i ∈ N .

Thus, x ∈ N . However, N = KerQ2, so ξ2 = Q2x = 0. Thus, N2 = 0, as desired.

Example 7.2. Consider the block circulant system (C,A,B), where A and B are as in (6.3) (of Example 6.3)and C ∈ C(Π4 ⊗ I1,Π4 ⊗ I2) is defined in the measurement equation

y =

C︷︸︸︷7 5 −1 1 −5 −3 −1 1−1 1 7 5 −1 1 −5 −3−5 −3 −1 1 7 5 −1 1−1 1 −5 −3 −1 1 7 5

x (7.2)

The unobservable subspace of the pair (C,A) is given by N = spann1, n2, n3, n4 where n1 = (2,−3, 0, 0,−2, 3, 0, 0),n2 = (0, 0, 2,−3, 0, 0,−2, 3), n3 = (1, 0, 1, 0, 1, 0, 1, 0), and n4 = (0, 1, 0,−1, 0, 1, 0,−1). Following Lemma 7.1,N is a block circulant subspace. A basis for N⊥ is given by r1, r2, r3, r4, where r1 = (7, 5,−1, 1,−5,−3,−1, 1),r2 = (−1, 1, 7, 5,−1, 1,−5,−3), r3 = (−5,−3,−1, 1, 7, 5,−1, 1), and r4 = (−1, 1,−5,−3,−1, 1, 7, 5). Also,N⊥ is a block circulant subspace by Lemma 5.5. Thus, n1, n2, n3, n4, r1, r2, r3, r4 is a preferred basis forX adapted to N . We will perform the observable decomposition using this basis. Define the insertion maps

S1 =[n1 n2 n3 n4

]S2 =

[r1 r2 r3 r4


[S1 S2


uniquely defined by T−1 = col(Q1, Q2), where Q1 is the upper four rows and Q2 is the lower four rows. Theobservable decomposition of (C,A,B) is then given by

T−1AT =

−4 16 0 −16 0−4 0 16 0 −16

−4 4 4 4 4−8 0 0 0 0

−3 1 −11 11 −3 1 −11

−11 1 −3 11 −11 1 −3

, T−1B =

1

104

−24 0 24 00 −24 0 24

52 52 52 520 0 0 0

17 −13 9 −13−13 17 −13 9

9 −13 9 17−13 9 −13 17

,

CT =1

16

7 0 −6 00 7 0 −6−6 0 7 0

0 −6 0 7

(7.3)

14

Denote the top-left subsystem by (C1, A1, B1) and the bottom-right subsystem by (C2, A2, B2). It can beverified that A1 ∈ C((Π4 ⊗ I2)N ), A2 ∈ C((Π4 ⊗ I2)N⊥), B1 ∈ C((Π4 ⊗ I2)N ,Π4 ⊗ I1), B2 ∈ C((Π4 ⊗I2)N⊥ ,Π4 ⊗ I1), C1 ∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)N ), and C2 ∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)N⊥). As well, C2 is full rank,

so the pair (C2, A2) is observable, i.e. N2 :=⋂4i=1 Ker(C2A

i−12 ) = 0. C

7.2. Detectability. Detectability is the dual concept of stabilizability. We say (C,A) is detectable ifthere exists K such that σ(A+KC) ⊂ C−. Thus, (C,A) is detectable if and only if (Aᵀ, Cᵀ) is stabilizable.From Theorem 6.3 we immediately obtain the following.

Theorem 7.3 (Detectability). Let V ⊂ X be a block circulant subspace. Suppose A ∈ C(Πr ⊗ In) andAV ⊂ V. Let A1 ∈ C((Πr ⊗ In)V) be the restriction of A to V and let C1 ∈ C(Πr ⊗ Ip, (Πr ⊗ In)V). Thereexists a state feedback K1 : Y → V, K1 ∈ C((Πr ⊗ In)V ,Πr ⊗ Ip), such that σ(A1 +K1C1) ⊂ C− if and onlyif N1 ⊂ X−(A1), where N1 is the unobservable subspace of (C1, A1).

Example 7.3. Recall the system (C,A) from (6.3)-(7.2) of Examples 6.3 and 7.2 and the unstable subspaceX+(A) derived in Example 6.4. It can be verified that X+(A) ∩ N = 0, and so (C,A) is detectable. Fromthe observable decomposition (7.3), the spectrum of A splits into σ(A1) = −4,−4,−4,−8 and σ(A2) =−16,−12, 8, 8. Defining the state variables of each subsystem by ξ1 = Q1x and ξ2 = Q2x (where Q1

and Q2 are defined as in Example 7.2), all unstable modes of the system are contained in ξ2. Also, fromExample 7.2, the subsystem (C2, A2) is observable. Thus, all instabilities in the system can be observed. Inthis specific example, C2 is invertible, and so the unstable modes can be recovered by ξ2 = C−12 y. C

8. Output Stabilization. Consider the linear time-invariant system

x(t) = Ax(t) +Bu(t)

z(t) = Dx(t),

where x(t) ∈ Rrn, u(t) ∈ Rrm, and z(t) ∈ Rrq. The Output Stabilization Problem (OSP) is to find a statefeedback u(t) = Kx(t) such that z(t)→ 0 as t→∞. The problem can be restated in more geometric termsas finding a state feedback K : X → U that makes the unstable subspace unobservable at the output z(t).Equivalently,

X+(A+BK) ⊂ KerD.

The solution to the OSP requires the notion of controlled invariant subspaces. A subspace V ⊂ X is said tobe controlled invariant if there exists a map F : X → U such that (A+BF )V ⊂ V. In this case F is calleda friend of V. Let I (X ) denote the set of all controlled invariant subspaces in X . Similarly, for any V ⊂ X ,let I (V) denote the set of all controlled invariant subspaces in V. It is well-known that the OSP is solvableif and only if X+(A) ⊂ C+V?, where C = 〈A|B〉 and V? := sup I (KerD), the supremal controlled-invariantsubspace contained in KerD. Consider now the OSP for block circulant systems.

Problem 8.1 (Block Circulant Output Stabilization Problem). Given a block circulant triple (D,A,B),find a block circulant state feedback K : X → U such that

X+(A+BK) ⊂ KerD .

In contrast with the standard OSP, in the block circulant OSP it is necessary to be able to construct blockcirculant friends for block circulant controlled invariant subspaces.

Lemma 8.1. Let (A,B) be a block circulant pair with A ∈ C(Πr ⊗ In) and B ∈ C(Πr ⊗ In,Πr ⊗ Im). IfV ⊂ X is a block circulant subspace and V ∈ I (X ), then there exists F ∈ C(Πr ⊗ Im,Πr ⊗ In) such that(A+BF )V ⊂ V.

Proof. We construct a candidate F ∈ C(Πr⊗ Im,Πr⊗ In) such that V is (A+BF )-invariant. Let k = dimV,and let v1, . . . , vk, vk+1, . . . , vrn be a basis for X adapted to V. By Lemma 4.2 of [33], AV ⊂ V +B. Thus,

15

for each i = 1, . . . , k, there exists wi ∈ V and ui ∈ U such that Avi = wi −Bui. This gives

A[v1 · · · vk

]=[w1 · · · wk

]−B

[u1 · · · uk

](8.1)

A[v1 · · · vrn

]=[w1 · · · wk ∗ · · · ∗

]−B

[u1 · · · uk 0 · · · 0

].

Block diagonalizing A and B gives

(Fr ⊗ In)∗ΛA(Fr ⊗ In)[v1 · · · vrn

]=[

w1 · · · wk ∗ · · · ∗]− (Fr ⊗ In)∗ΛB(Fr ⊗ Im)

[u1 · · · uk 0 · · · 0

].

Define [e1,1 · · · er,n

]:= (Fr ⊗ In)

[v1 · · · vrn

] [v1 · · · vrn

]−1(Fr ⊗ In)∗[

w1,1 · · · wr,n]

:= (Fr ⊗ In)[w1 · · · wk ∗ · · · ∗

] [v1 · · · vrn

]−1(Fr ⊗ In)∗[

u1,1 · · · ur,n]

:= (Fr ⊗ Im)[u1 · · · uk 0 · · · 0

] [v1 · · · vrn

]−1(Fr ⊗ In)∗ ,

where the order of indexing for e, w, and u is e1,1, . . . , e1,n, e2,1, . . . , e2,n, . . . , er,1, . . . , er,n, etc. (Noticethat

[e1,1 · · · er,n

]= Irn.) With these definitions, we have

ΛA[e1,1 · · · er,n

]=[w1,1 · · · wr,n

]− ΛB

[u1,1 · · · ur,n

]. (8.2)

Thus, finding w1, . . . , wk and u1, . . . , uk that solve (8.1) is equivalent to finding w1,1, . . . , wr,n and u1,1, . . . , ur,nthat solve (8.2). We will choose the ui,js to make F block circulant.

Let ΛA := diag(A1, . . . , Ar) and ΛB := diag(B1, . . . , Br). For each i = 1, . . . , r and j = 1, . . . , n, partitionei,j = col(e1i,j , . . . , e

ri,j), where each eì,j is an n × 1 column vector. Similarly partition the wi,js and ui,js.

We have the equations

Aèì,j = wì,j −Bùì,j , i = 1, . . . , r, j = 1, . . . , n, ` = 1, . . . , r .

Since ei,j is the (n(i− 1) + j)th standard basis vector of X = Rrn, eì,j = 0 whenever ` 6= i. We can thus set

wì,j = 0 and uì,j = 0 for each ` 6= i. We are left with the equations

Aieii,j = wii,j −Biuii,j , i = 1, . . . , r, j = 1, . . . , n .

For each i = 1, . . . , r, define i′ = (r−i+2)mod(r). By Theorem 4.2, Ai = Ai′ and Bi = Bi′ . Also, by definition

eii,j = ei′

i′,j for each j = 1, . . . , n. Therefore, we can choose the wii,j ’s and uii,j ’s such that wii,j = wi′

i′,j and

uii,j = ui′

i′,j .

Now define Fi :=[uii,1 · · · uii,n

]for i = 1, . . . , r and ΛF := diag(F1, . . . , Fr). Then Fi = F (r−i+2)mod(r)

for each i = 1, . . . , r. Defining F := (Fr ⊗ Im)∗ΛF (Fr ⊗ In), it follows from Theorem 4.2 that F ∈C(Πr ⊗ Im,Πr ⊗ In). Further, observe that ΛF =

[u1,1 · · · ur,n

]. Thus,

F = (Fr ⊗ Im)∗[u1,1 · · · ur,n

](Fr ⊗ In)

=[u1 · · · uk 0 · · · 0

] [v1 · · · vrn

]−1.

For each i = 1, . . . , k, we have

(A+BF )vi = Avi +B[u1 · · · uk 0 · · · 0

] [v1 · · · vrn

]−1vi

= Avi +B[u1 · · · uk 0 · · · 0

]ei

= Avi +Bui

= wi ∈ V .

16

Hence V is (A+BF )-invariant.

The next result verifies that the supremal controlled invariant subspace contained in a block circulant sub-space is itself a block circulant subspace.

Lemma 8.2. Let (D,A,B) be a block circulant triple with A ∈ C(Πr ⊗ In), B ∈ C(Πr ⊗ In,Πr ⊗ Im), andD ∈ C(Πr ⊗ Iq,Πr ⊗ In). Then V? = sup I (KerD) is a block circulant subspace.

Proof. Consider the recursive algorithm in Theorem 4.3 of [33]:

V0 = Ker(D)

Vi = Ker(D) ∩A−1(B + Vi−1) .(8.3)

The sequence is nonincreasing and has a lower bound of 0, so it must have a fixed point. Using Lemmas 5.2and 5.4, each Vi is a block circulant subspace; hence, so is the fixed point. By [33, Theorem 4.3], the fixedpoint is V?.

Given a block circulant friend F of a block circulant subspace V, there is a basis in which A + BF has auseful matrix representation. This is the content of our third decomposition theorem.

Theorem 8.1 (Third Decomposition Theorem). Let V ⊂ X be a block circulant subspace with V ∈ I (X )and let W ⊂ X be a block circulant complement of V. There exists a state and feedback transformation (T, F )with T : X → X and F ∈ C(Πr ⊗ Im,Πr ⊗ In) such that (A, B) := (T−1(A+BF )T, T−1B) has the form

A =

[A1 ∗0 A2

], B =

[B1

B2

], (8.4)

where A1 = (A + BF )V ∈ C((Πr ⊗ In)V), A2 ∈ C((Πr ⊗ In)W), B1 ∈ C((Πr ⊗ In)V ,Πr ⊗ Im), and B2 ∈C((Πr ⊗ In)W ,Πr ⊗ Im).

Proof. Since V ∈ I (X ), by Lemma 8.1 there exists F ∈ C(Πr ⊗ Im,Πr ⊗ In) such that (A + BF )V ⊂ V.Let u = Fx+ v to obtain the new system x = (A+BF )x+Bv. By Lemma 3.2, A+BF ∈ C(Πr ⊗ In). Wechoose projection maps S1, S2, Q1, andQ2 according to Theorem 5.1 and apply the coordinate transformation(ξ1, ξ2) = T−1x where T =

[S1 S2


B = T−1B =

[Q1

Q2

]B =:

[B1

B2

].

Using (2.5) we have

(Πr ⊗ In)VB1 = (Πr ⊗ In)VQ1B = Q1(Πr ⊗ In)B = B1(Πr ⊗ Im) .

Hence B1 ∈ C((Πr ⊗ In)V ,Πr ⊗ Im). Similarly, B2 ∈ C((Πr ⊗ In)W ,Πr ⊗ Im).

Example 8.3. Consider the block circulant system (D,A,B), where A and B are as in (6.3) (of Example6.3), and D ∈ C(Π4 ⊗ I1,Π4 ⊗ I2) is defined in the output equation

z =

D︷︸︸︷−5 0 −1 0 1 0 −1 0−1 0 −5 0 −1 0 1 0

1 0 −1 0 −5 0 −1 0−1 0 1 0 −1 0 −5 0

x (8.5)

Following the algorithm in the proof of Lemma 8.2, the supremal controlled invariant subspace contained inKerD of the system can be shown to be V? := sup I (KerD) = spanv1, v2, where v1 = (0, 1, 0, 0, 0, 1, 0, 0)and v2 = (0, 0, 0, 1, 0, 0, 0, 1); it can also be verified that V? is a block circulant subspace. Following Lemma8.1, a block circulant feedback F for which V? is (A+BF )-invariant is

F =1

2

0 −1 0 −1 0 −1 0 −10 −1 0 −1 0 −1 0 −10 −1 0 −1 0 −1 0 −10 −1 0 −1 0 −1 0 −1

∈ C(Π4 ⊗ I1,Π4 ⊗ I2) . (8.6)

17

Defining the feedback transformation u = Fx+ v, we then have the system x = (A+BF )x+Bv.

A basis for (V?)⊥ is given by r1, r2, r3, r4, r5, r6, where r1 = e1, r2 = e3, r3 = e5, r4 = e7, r5 =(0, 1, 0, 0, 0,−1, 0, 0), and r6 = (0, 0, 0, 1, 0, 0, 0,−1). By Lemma 5.5, (V?)⊥ is a block circulant subspace.Thus, v1, v2, r1, r2, r3, r4, r5, r6 is a preferred basis for X adapted to V?. We will perform the third decom-position using this basis. Define the insertion maps

S1 =[v1 v2

]S2 =

[r1 r2 r3 r4 r5 r6


[S1 S2


uniquely defined by T−1 = col(Q1, Q2), where Q1 is the first two rows and Q2 is the remaining six rows. Thetriple (D,A+BF,B) can then be decomposed as

T−1(A+BF )T =

−10 −2−2 −10

−1 3 −9 3 8 03 −1 3 −9 0 8−9 3 −1 3 −8 0

3 −9 3 −1 0 −80 0 0 0 −4 00 0 0 0 0 −4

, T−1B =

0 0 0 00 0 0 01 0 1 00 1 0 11 0 1 00 1 0 11 0 1 00 1 0 1

DT =

5 −1 1 −1 0 0−1 5 −1 1 0 0

1 −1 5 −1 0 0−1 1 −1 5 0 0

(8.7)

Denote the top-left subsystem by (D1, A1, B1) and the bottom-right subsystem by (D2, A2, B2). It can be

verified that A1 ∈ C((Π4 ⊗ I2)V?), A2 ∈ C((Π4 ⊗ I2)(V?)⊥), B1 ∈ C((Π4 ⊗ I2)V? ,Π4 ⊗ I1), and B2 ∈C((Π4 ⊗ I2)(V?)⊥ ,Π4 ⊗ I1). As well, we have D1 = 0, verifying that the modes of A1 will not show up in theoutput (as guaranteed in the construction of V? ⊂ KerD). C

Using the previous results on controlled invariant subspaces, we proceed to the solution of the block circulantOSP.

Theorem 8.2. The block circulant OSP is solvable if and only if

X+(A) ⊂ C + V? . (8.8)

Proof. (If) Suppose (8.8) holds. By Lemmas 5.5 and 8.2 there exists a block circulant complement Wof V?. Since V? ∈ I (KerD), by the Third Decomposition Theorem 8.1 there exists a state and feedbacktransformation (T, F ) with ξ = T−1x, u = Fx+ v, and F ∈ C(Πr ⊗ Im,Πr ⊗ In) such that[

ξ1ξ2

]=

[A1 ∗0 A2

] [ξ1ξ2

]+

[B1

B2

]v

where A1 ∈ C((Πr⊗In)V?), A2 ∈ C((Πr⊗In)W), B1 ∈ C((Πr⊗In)V? ,Πr⊗Im), and B2 ∈ C((Πr⊗In)W ,Πr⊗Im). Let Q2 : X → W be the natural projection on W. Since KerQ2 = V? ⊂ KerD, there exists D2 suchthat D = D2Q2. Since ξ2 = Q2x, we have

z = Dx = D2ξ2 .

It is then clear that a sufficient condition for solvability of the block circulant OSP is that (A2, B2) isstabilizable using a feedback K ′2 ∈ C(Πr ⊗ Im, (Πr ⊗ In)W). By Lemma 5.7, Assumption 3.9 holds with(U, V ) = ((Πr ⊗ In)W ,Πr ⊗ Im). Then by Theorem 6.3, (A2, B2) is stabilizable if and only if

W+(A2) ⊂ 〈A2 | B2〉18

where B2 = Im (B2). Using Lemma 2.1(i), (2.3), and Q2V? = 0, this is equivalent to

Q2X+(A) ⊂ Q2〈A|B〉 = Q2(〈A|B〉+ V?) .

This is achieved since X+(A) ⊂ 〈A|B〉 + V? by (8.8). Finally, let K ′ := K ′2Q2. By Lemma 5.6, K ′ ∈C(Πr ⊗ Im,Πr ⊗ In). The overall feedback is K := F + K ′, and by Lemma 3.2, K ∈ C(Πr ⊗ Im,Πr ⊗ In),as desired.

(Only If) The condition is exactly the necessary condition for the general OSP. Since it is necessary for theexistence of a general state feedback, it is also necessary for the existence of a block circulant state feedback.

Example 8.4. Consider the system (D,A,B) from (6.3)-(8.5) of Examples 6.3 and 8.3. Also recall theunstable subspace X+(A) from Example 6.4, the controllable subspace 〈A|B〉 from Example 6.3, and thesupremal controlled invariant subspace V? from Example 8.3. It was already shown (in Example 6.4) thatX+(A) ⊂ 〈A|B〉, and so we immediately have that X+(A) ⊂ V? + 〈A|B〉. By Theorem 8.2, the system isoutput stabilizable. We will now explicitly find an output-stabilizing feedback.

In Example 8.3 we found a friend F of V? (given in (8.6)) and utilized the feedback transformation u =Fx + v, resulting in a decomposition of the system (D,A + BF,B) for which only the modes of the second

subsystem (A2, B2) show up in the output. Therefore, to output-stabilize (D,A,B), it is sufficient to stabilize

(A2, B2). From (8.7), it can easily be verified that (A2, B2) is controllable, so its closed-loop poles can beplaced arbitrarily (by Theorem 6.1). For stabilization, we choose to place them in C−. In particular, thestate feedback

K ′2 =1

2

−25 0 25 0 −32 0

0 −25 0 25 0 −3225 0 −25 0 32 00 25 0 −25 0 32

∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)V)

gives σ(A2 +B2K′2)=−16,−23.2,−23.2,−4.8,−4.8,−4. From Lemma 5.6, K ′2 can be lifted back to the full

space X (using the natural projection map Q2 defined in Example 8.3), giving

K ′ = K ′2Q2 =1

2

−25 −16 0 0 25 16 0 0

0 0 −25 −16 0 0 25 1625 16 0 0 −25 −16 0 00 0 25 16 0 0 −25 −16

∈ C(Π4 ⊗ I1,Π4 ⊗ I2)

With K ′, we have output-stabilized the feedback-transformed system (D,A + BF,B). The overall outputstabilizing feedback for the original system (D,A,B) is then given by

K = F +K ′ =1

2

−25 −17 0 −1 25 15 0 −1

0 −1 −25 −17 0 −1 25 1525 15 0 −1 −25 −17 0 −10 −1 25 15 0 −1 −25 −17

∈ C(Π4 ⊗ I1,Π4 ⊗ I2) .

C

9. Disturbance Decoupling. Consider the linear time-invariant system

x(t) = Ax(t) +Bu(t) + Ew(t) (9.1)

z(t) = Dx(t) (9.2)

where x(t) ∈ Rrn, u(t) ∈ Rrm, z(t) ∈ Rrq, and w(t) ∈ Rrs. The signal w(t) is a disturbance which is assumednot to be directly measurable by the controller. We would like to find a state feedback u = Kx so that thecontrolled output z(t) is not affected by any disturbance w(t). The disturbance is assumed to belong to some

19

sufficiently rich class of signals, reflecting our lack of knowledge about the characteristics of this signal. Wesay the closed-loop system is disturbance decoupled if for each initial condition x(0) ∈ X , the output z(t) isthe same for every w(t). Let E = Im E. Turning to block circulant systems, a geometric statement of theproblem is as follows.

Problem 9.1 (Block Circulant Disturbance Decoupling Problem (DDP)). Given the block circulant system(D,A,B,E) with A ∈ C(Πr ⊗ In), B ∈ C(Πr ⊗ In,Πr ⊗ Im), D ∈ C(Πr ⊗ Iq,Πr ⊗ In), and E ∈ C(Πr ⊗In,Πr ⊗ Is), find a block circulant state feedback u = Kx such that

〈A+BK | E〉 ⊂ KerD . (9.3)

Theorem 9.1. The block circulant DDP is solvable if and only if

E ⊂ V? (9.4)

where V? = sup I (KerD).

Proof. (If) Since V? ∈ I (KerD), by Lemma 8.1 there existsK ∈ C(Πr⊗Im,Πr⊗In) such that (A+BK)V? ⊂V?. Then using (9.4)

〈A+BK | E〉 ⊂ 〈A+BK | V?〉 = V? ⊂ KerD .

(Only If) If u = Kx solves the DDP then the subspace V := 〈A + BK | E〉 belongs to I (KerD) soE ⊂ V ⊂ V?.

Example 9.1. Recall the triple (D,A,B) from (6.3)-(8.5) of Examples 6.3 and 8.3, and consider the blockcirculant disturbance matrix given by

E =

0 0 0 01 0 1 00 0 0 00 1 0 10 0 0 01 0 1 00 0 0 00 1 0 1

∈ C(Π4 ⊗ I2,Π4 ⊗ I1) .

We now have the system

x = Ax+Bu+ Ew

z = Dx(9.5)

where the signal w(t) represents a disturbance that is not directly measurable by the controller. This distur-bance can affect the system state x via vectors in E := Im E = span(0, 1, 0, 0, 0, 1, 0, 0), (0, 0, 0, 1, 0, 0, 0, 1).It can immediately be seen that E = V? (where V? := sup I (KerD), as given in Example 8.3), and sothe system can be disturbance decoupled by block circulant state feedback by Theorem 9.1. In particular, thefriend F ∈ C(Π4⊗ I1,Π4⊗ I2) found in Example 8.3 satisfies 〈A+BF |E〉 ⊂ V?, and so the disturbance doesnot show up in the output of the closed-loop system with control law u = Fx. C

10. Measurement Feedback. It is well known that one can stabilize a controllable, observable triple(C,A,B) by using dynamic observers and pole placement. Here we study stabilization of block circulantsystems (C,A,B) that are stabilizable and detectable. We study two cases: either the observable states areavailable for feedback, or only the measurements are available. The latter is often called the static outputfeedback problem or static measurement feedback problem.

20

10.1. Stabilization by Observable States. We are interested in finding a feedback u = K2ξ2, whereξ2 corresponds to observable states, such that x(t)→ 0 as t→∞. Let N denote the unobservable subspaceof (C,A), and let S2 : N⊥ → X and Q2 : X → N⊥ denote the insertion and natural projection on N⊥.Then ξ2 = Q2x. Defining A2 := Q2AS2 and C2 := Q2C, (C2, A2) is the observable pair obtained fromTheorem 7.2. Also, set B2 := Q2B. Since (C2, A2) is observable we can define an observer

˙ξ2 = A2ξ2 +B2u+ L2(y − y) , y = C2ξ2

such that σ(A2 − L2C2) ⊂ C− and e2(t) := ξ2(t) − ξ2(t) → 0 as t → ∞. Now set u = K2ξ2 and defineK := K2Q2. Noting that u = K2(ξ2 − e2) = Kx−K2e2 we obtain

x = (A+BK)x−BK2e2 .

Thus, stability is achieved using u = K2ξ2 if and only if

• e(A+BK)t → 0 as t→∞.•∫ t0

e(A+BK)(t−τ)BK2e(A2−L2C2)τdτ → 0 as t→∞.

The first constraint is equivalent to σ(A+BK) ⊂ C−. If the first constraint is true, so is the second one, sowe only concentrate on the first requirement. Thus, we assume that ξ2 = Q2x is directly measurable, andthe observer can be introduced later in the design. Now we want to find u = Kx where K = K2Q2. Thelatter equation has a solution for K2 so long as N ⊂ KerK. Thus, we arrive at the following problem.

Problem 10.1 (Block Circulant Stabilization by Observable State Feedback Problem (SOSFP)). Given ablock circulant triple (C,A,B), find a block circulant state feedback u = Kx such that

N ⊂ KerK (10.1)

σ(A+BK) ⊂ C− . (10.2)

Theorem 10.1. The block circulant SOSFP is solvable if and only if (A,B) is stabilizable and (C,A) isdetectable. That is,

X+(A) ⊂ 〈A|B〉 , (10.3)

X+(A) ∩N = 0 . (10.4)

Proof. (If) Suppose (10.3)-(10.4) hold. By the Second Decomposition Theorem 7.2 we have[ξ1ξ2

]=

[A1 ∗0 A2

] [ξ1ξ2

]+

[B1

B2

]u , y = C2ξ2 , (10.5)

where A1 ∈ C((Πr⊗In)N ), A2 ∈ C((Πr⊗In)N⊥), B2 ∈ C((Πr⊗In)N⊥ ,Πr⊗Im), C2 ∈ C(Πr⊗Ip, (Πr⊗In)N⊥),and C2Q2 = C. Using any u = K2ξ2 and setting K = K2Q2, the requirement (10.1) is satisfied since

KerQ2 = N ⊂ KerK .

Then using (10.5)

σ(A+BK) = σ(A1) ] σ(A2 +B2K2) . (10.6)

By (10.4) and Lemma 2.1, σ(A1) ⊂ C−. By (10.3), Lemma 2.1(i), and (2.3),

(N⊥)+(A2) = Q2X+(A) ⊂ Q2〈A|B〉 = 〈A2|B2〉

where B2 = Im B2. By Theorem 6.3, (A2, B2) is stabilizable and there exists K2 ∈ C(Πr ⊗ Im, (Πr ⊗ In)N⊥)such that σ(A2 + B2K2) ⊂ C−. Let K := K2Q2. By Lemma 5.6, K ∈ C(Πr ⊗ Im,Πr ⊗ In). Finally, using(10.6), we get σ(A+BK) ⊂ C−, as desired.

(Only If) Conditions (10.3)-(10.4) are exactly the necessary conditions for the general SOSFP. Since theyare necessary for the existence of a general state feedback, they are also necessary for the existence of a blockcirculant state feedback.

21

10.2. Stabilization by Measurement Feedback. We next study the more general problem of findinga static measurement feedback u = K ′y such that x(t) → 0 as t → ∞. Following our geometric approach,we would like to find an A-invariant subspace L ⊂ X and a state feedback u = Kx such that

KerC ⊂ L ⊂ KerK .

The requirement L ⊂ KerK gives the interpretation to L as a “masking subspace” that characterizes whatstate information cannot be used in state feedback, or equivalently what state information is masked out byK. Intuitively, the larger the dimension of L, the less state information that can appear in the feedback.The requirement KerC ⊂ L imposes that only the measurements y can be used in the feedback. Now ifKerC were A-invariant, then the best choice for L would be L = KerC = N , and the control design forSOSFP would apply. Generally, KerC is not A-invariant, and the next best choice is the smallest A-invariantsubspace containing KerC, namely 〈A | KerC〉. Thus we arrive at the following problem for block circulantsystems.

Problem 10.2 (Block Circulant Stabilization by Measurement Feedback Problem (SMFP)). Given a blockcirculant triple (C,A,B), find a block circulant state feedback u = Kx such that

KerC ⊂ KerK (10.7)

σ(A+BK) ⊂ C− . (10.8)

Lemma 10.1. The subspace 〈A | KerC〉 is a block circulant subspace.

Proof. By Lemma 3.2, Ai is a block circulant matrix for all i ∈ N. By Lemma 5.3, KerC is a block circulantsubspace and by Lemma 5.4, Ai KerC is a block circulant subspace. Using Lemma 5.2, we conclude that〈A | KerC〉 is a block circulant subspace.

We have the following sufficient condition for solving the block circulant SMFP.

Theorem 10.2. The block circulant SMFP is solvable if

X+(A) ⊂ 〈A|B〉 , (10.9)

X+(A) ∩ 〈A | KerC〉 = 0 . (10.10)

Proof. Let L := 〈A | KerC〉. By Lemma 10.1, L is a block circulant subspace. Let W be a blockcirculant complement of L, which exists by Lemma 5.5. Let S1, Q1 and S2, Q2 denote the insertion andnatural projection on V and W, respectively. Using Theorem 7.2 with N and N⊥ replaced by L and W,respectively, and applying the coordinate transformation (ξ1, ξ2) = T−1x where T =

[S1 S2

], we obtain

the transformed system [ξ1ξ2

]=

[A1 ∗0 A2

] [ξ1ξ2

]+

[B1

B2

]u , (10.11)

where A1 ∈ C((Πr⊗ In)L), A2 ∈ C((Πr⊗ In)W), B2 ∈ C((Πr⊗ In)W ,Πr⊗ Im), C2 ∈ C(Πr⊗ Ip, (Πr⊗ In)W),and C2Q2 = C. Using any u = K2ξ2 and setting K := K2Q2, the requirement (10.7) is satisfied since

KerC ⊂ KerQ2 = L ⊂ KerK .

Then using (10.11)

σ(A+BK) = σ(A1) ] σ(A2 +B2K2) . (10.12)

By (10.10) and Lemma 2.1(ii), σ(A1) ⊂ C−. By (10.9), Lemma 2.1(i), and (2.3),

W+(A2) = Q2X+(A) ⊂ Q2〈A|B〉 = 〈A2|B2〉22

where B2 = Im B2. By Theorem 6.3, (A2, B2) is stabilizable and there exists K2 ∈ C(Πr ⊗ Im, (Πr ⊗ In)W)such that σ(A2 + B2K2) ⊂ C−. Let K := K2Q2. By Lemma 5.6, K ∈ C(Πr ⊗ Im,Πr ⊗ In). Finally, using(10.12), we get σ(A+BK) ⊂ C−, as desired.

Example 10.2. Consider the system (C,A,B) as defined in (6.3)-(7.2) of Examples 6.3 and 7.2. As shownin Examples 6.4 and 7.3, the system is stabilizable and detectable, and so it is stabilizable by observable statefeedback by Theorem 10.1. Let L := 〈A|KerC〉. It can be verified that in this example, L = N = KerC(where N is given in Example 7.2). Therefore, X+(A) ∩ L = X+(A) ∩ N = 0, and the system is alsostabilizable by block circulant measurement feedback by Theorem 10.2. In fact, since L = N , the same statefeedback u = Kx can be used for each. Here, we will find such a feedback. Later on in the control design, Kcan be used to construct a block circulant observer to solve SOSFP and a static block circulant measurementfeedback u = K ′y to solve SMFP.

Recall the observable decomposition (7.3) from Example 7.2, where the subsystem (C2, A2, B2) is observable.We will show that this subsystem is stabilizable, and then we will stabilize it through a controllable decom-position on the pair (A2, B2). In dealing with this subsystem, we will work in the coordinates of ξ2 := Q2x,where Q2 : X → N⊥ is the natural projection map defined in Example 7.2.

Let C2 := 〈A2|B2〉 represent the controllable subspace of this subsystem. A basis for C2 is given by b1, b2, b3,where b1 = (1, 0,−1, 0), b2 = (0, 1, 0 − 1), and b3 = (1,−1, 1,−1). Similarly to the first part of the proof ofTheorem 6.3, it can be verified that C2 is a block circulant subspace. It can also be shown that the unstablesubspace of (A2, B2) is given by the block circulant subspace (N⊥)+(A2) = spanb1, b2 ⊂ C2. Thus, thesubsystem is stabilizable.

By Lemma 5.5, C⊥2 is a block circulant subspace; by Lemma 5.2, so is W := C⊥2 ∩ N⊥. Using the modulardistributive rule of subspaces [33, (0.3.1)], also N⊥ = C2 ⊕W. Taking W = spanw where w = (1, 1, 1, 1),a basis for N⊥ is then given by b1, b2, b3, w.

Now, define the insertion maps S21 : C2 → N⊥ and S22 :W → N⊥ by

S21 =[b1 b2 b3

]S22 =

[w]

Let T2 =[S21 S22

]. The corresponding natural projection maps Q21 and Q22 are then uniquely defined

by T−12 = col(Q21, Q22), where Q21 is the first three rows and Q22 is the last row. (A2, B2) can then befurther decomposed (by controllable decomposition as in Theorem 6.2) as

T−12 A2T2 =

8

8−16

−12

, T−12 B2 =

4 0 −4 00 4 0 −4

13 −13 13 −13

(10.13)

Denote the top-left subsystem by (A21, B21) and the bottom-right subsystem by (A22, B22). It can be verifiedthat A21 ∈ C((Π4 ⊗ I2)C2), A22 ∈ C((Π4 ⊗ I2)W), B21 ∈ C((Π4 ⊗ I2)C2 ,Π4 ⊗ I1), and B22 ∈ C((Π4 ⊗I2)W ,Π4 ⊗ I1). Clearly, σ(A1) = 8, 8,−16 and σ(A2) = −12, and so all instabilities are contained inthe 21-subsystem. It can also be verified that (A21, B21) is controllable, and so the closed-loop poles can beplaced arbitrarily. For stabilization, we will place them in C−. In particular, the feedback matrix

K21 = 208

−1 0 0

0 −1 01 0 00 1 0

∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)C2)

gives σ(A21 + B21K21) = −8,−8,−16. K21 can be lifted back to N⊥ by K2 := K21Q21, then back to the

23

full space X by K := K2Q2. Thus, we get the block circulant state feedback matrix

K = (K21Q21)Q2 =

−6 −4 0 0 6 4 0 0

0 0 −6 −4 0 0 6 46 4 0 0 −6 −4 0 00 0 6 4 0 0 −6 −4

∈ C(Π4 ⊗ I1,Π4 ⊗ I2)

The poles of the closed-loop system are given by σ(A+BK) = −16,−12,−8,−8,−8,−4,−4,−4, verifyingthat the system has been stabilized. Also, it can be shown that KerK ⊂ KerC, confirming that the feedbackonly uses states which can be seen in the measurement. C

11. Output Stabilization by Measurement Feedback. Consider the linear time-invariant system

x(t) = Ax(t) +Bu(t) (11.1)

y(t) = Cx(t) (11.2)

z(t) = Dx(t) (11.3)

where x(t) ∈ Rrn, u(t) ∈ Rrm, y(t) ∈ Rrp, and z(t) ∈ Rrq. The output stabilization by measurement feedbackproblem is to find a measurement feedback u = Ky such that z(t)→ 0 as t→∞. If (C,A,B) is controllableand observable, then the problem can be solved by observer-based feedback as in Section 10.1. In the moregeneral case, ideas originating from the solutions of OSP and SMFP must be used.

Recall from SMFP that the set of states which cannot be used in a measurement feedback can be characterizedby some invariant subspace L. If L is given, then the state feedback u = Kx is enforced to satisfy L ⊂ KerK.If L is not given, then one would choose it to be as small as possible and, ideally, one would choose it to bethe unobservable subspace of (C,A), as in Theorem 10.1. However, this choice may not always be feasible,given the constraint that z(t) → 0. If a valid L can be found, then the problem can be converted to themore tractable Restricted Regulator Problem (RRP) [33], stated here for block circulant systems.

Problem 11.1 (Block Circulant Restricted Regulator Problem (RRP)). Given a block circulant subspace Lsuch that AL ⊂ L, find K ∈ C(Πr ⊗ Im,Πr ⊗ In) such that

L ⊂ KerK , (11.4)

X+(A+BK) ⊂ KerD . (11.5)

Theorem 11.1. The block circulant RRP is solvable if and only if there exists a block circulant subspaceV ∈ I (KerD) such that

A(L ∩ V) ⊂ L ∩ V (11.6)

X+(A) ∩ L ⊂ L ∩ V (11.7)

X+(A) ⊂ 〈A|B〉+ V . (11.8)

Proof. (If) Suppose (11.6)-(11.8) hold for some block circulant subspace V ∈ I (KerD). By Lemma 5.2,L + V and L ∩ V are block circulant subspaces. By Lemma 5.5, they have block circulant complements Rand W, respectively; that is

X = (L+ V)⊕RX = (L ∩ V)⊕W .

Also, L := L∩W and V := V ∩W are block circulant subspaces by Lemma 5.2. By the modular distributiverule of subspaces [33, (0.3.1)]

L = (L ∩ V)⊕ L

V = (L ∩ V)⊕ V .

24

Then we obtain L+V = (L∩V)⊕L⊕V. We conclude the space X splits into four block circulant subspaces:

X = (L ∩ V)⊕ L ⊕ V ⊕ R . (11.9)

Let Si, Qi, i = 1, . . . , 4, be the insertion and natural projection maps of these four spaces in the order of(11.9).

By Lemma 8.1, V has a friend F ∈ C(Πr ⊗ Im,Πr ⊗ In) such that (A + BF )V ⊂ V. We will modify F toanother friend F ′ ∈ C(Πr ⊗ Im,Πr ⊗ In) such that

F ′L = 0 , (A+BF ′)V ⊂ V . (11.10)

To that end, define F ′ := FS3Q3. First, since L ∩ V = 0, Q3L = 0 and so F ′L = 0. Second, we show F ′ isa friend of V. Since S3Q3V = V, we have F ′S3 = FS3. Then using F ′L = 0 and (11.6),

(A+BF ′)V = (A+BF ′)[(L ∩ V)⊕ V

]= A(L ∩ V) + (A+BF ′)S3Q3V

= A(L ∩ V) + (AS3 +BFS3)Q3V = A(L ∩ V) + (A+BF )V ⊂ (L ∩ V) + V = V .

Third, we show that F ′ ∈ C(Πr ⊗ Im,Πr ⊗ In). Using (2.4)-(2.5), we have

(Πr ⊗ Im)F ′ = (Πr ⊗ Im)FS3Q3 = F (Πr ⊗ In)S3Q3

= FS3(Πr ⊗ In)VQ3 = FS3Q3(Πr ⊗ In) = F ′(Πr ⊗ In) .

From (11.10), (A+BF ′)L = AL. Thus,

(A+BF ′)(L+ V) ⊂ AL+ (A+BF ′)V ⊂ L+ V . (11.11)

We conclude that L, V, L ∩ V, and L+ V are all (A+BF ′)-invariant.

Now define T =[S1 S2 S3 S4

]. Using the state and feedback transformation (T, F ′) with ξ = T−1x

and u = F ′x+ v, the transformed system has the form

ξ =

A1 ∗ ∗ ∗0 A2 0 ∗0 0 A3 ∗0 0 0 A4

ξ +

B1

B2

B3

B4

vz =

[D1 D2 D3 D4

]ξ

(11.12)

where the zeros arise from the (A + BF ′)-invariance of L, V, L ∩ V, and L + V. By Theorem 5.1, A4 ∈C((Πr ⊗ In)R). By similar arguments to those in Theorem 5.1, B4 ∈ C((Πr ⊗ In)R,Πr ⊗ Im) and D4 ∈C(Πr ⊗ Iq, (Πr ⊗ In)R). Since V ⊂ KerD, D1 = DS1 = 0 and D3 = DS3 = 0. Thus,

z = D2ξ2 +D4ξ4 . (11.13)

It is clear from (11.12) and (11.13) that stabilizing z(t) is equivalent to stabilizing ξ2(t) and ξ4(t). By (11.7),X+(A)∩L ⊂ L∩V, which implies L+(AL) = L+((A+BF ′)L) ⊂ L∩V (using (11.10)), so by Lemma 2.1(ii),σ(A2) ⊂ C−. Thus, we are only concerned with σ(A4).

By Lemma 6.2 of [33] and (11.8) X+(A + BF ′) + 〈A|B〉 = X+(A) + 〈A|B〉 ⊂ 〈A|B〉 + V. Combined withLemma 2.1 of [33] that 〈A+BF ′|B〉 = 〈A|B〉 we get

X+(A+BF ′) ⊂ 〈A|B〉+ V = 〈A+BF ′|B〉+ V . (11.14)

Using Lemma 2.1(i), (11.14), Q4V = 0, and (2.5),

R+(A4) = Q4X+(A+BF ′) ⊂ Q4〈A+BF ′|B〉+Q4V = 〈A4|B4〉 ,25

where B4 := Im (B4). By Theorem 6.3 there exists K4 ∈ C(Πr ⊗ Im, (Πr ⊗ In)R) such that σ(A4 +B4K4) ⊂C−. Define K = K4Q4. By Lemma 5.6, K ∈ C(Πr ⊗ Im,Πr ⊗ In). The overall feedback solving the problemis (F ′ +K) ∈ C(Πr ⊗ Im,Πr ⊗ In) and we have (11.5).

Finally we verify (11.4). First L ⊂ Ker(F ′) by (11.10). Also, L ⊂ L+V = Ker(Q4) ⊂ Ker(K4Q4) = Ker(K).Thus, L ⊂ Ker(F ′) ∩Ker(K) ⊂ Ker(F ′ +K), giving (11.4).

(Only If) Suppose there exists K ∈ C(Πr⊗Im,Πr⊗In) for which (11.4)-(11.5) hold. Let V = X+(A+BK).Since V is a modal subspace, AV ⊂ V. From (11.5), also V ⊂ KerD. Thus, V ∈ I (KerD). Further, byLemma (5.3), V is a block circulant subspace.

Since L ⊂ KerK (from (11.4)), it follows that (A+BK)L = AL ⊂ L, and so (A+BK)L is well-defined; infact, (A+ BK)L = AL. Using Lemma 6.1 of [33], we get X+(A) ∩ L = X+(A+ BK) ∩ L = V ∩ L. Hence,condition (11.7) is satisfied. Using Lemma 6.2 of [33], we get that

X+(A) ⊂ 〈A|B〉+ X+(A) = 〈A|B〉+ X+(A+BK) = 〈A|B〉+ V .

Hence, condition (11.8) also holds. Lastly, since V and L are both A-invariant, condition (11.6) must alsohold.

Example 11.1. Recall the system (D,C,A,B) from (6.3)-(8.5) of Examples 6.3, 7.2, and 8.3, and takeV = sup I (KerD) (as given in “V?” of Example 8.3). Also L = N := spann1, n2, n3, n4, where the niswere defined in Example 7.2.

First, we will check the conditions for the solvability of RRP (as given in (11.6)-(11.8)). We have L ∩ V =spann4, which can be shown to be A-invariant, satisfying condition (i). Since the system is detectable (asshown in Example 7.3) and since L = N , we have X+(A) ∩ L = 0 ⊂ L ∩ V, and so condition (ii) is alsosatisfied. Since the system is output stabilizable (as shown in Example 8.4), condition (iii) is satisfied aswell. Thus, by Theorem 11.1, the block circulant RRP is solvable.

Starting from L ∩ V, a basis for L can be completed by n1, n2, n3 and a basis for V can be completed by

v where v = (0, 1, 0, 1, 0, 1, 0, 1). Define L = spann1, n2, n3 and V = spanv. Using the decompositionof (11.9), R := (L+ V)⊥ = spanr1, r2, r3, where r1 = (1, 0,−1, 0, 1, 0,−1, 0), r2 = (3, 2, 0, 0,−3,−2, 0, 0),and r3 = (0, 0, 3, 2, 0, 0,−3,−2). We now have a preferred basis for X given by n4, n1, n2, n3, v, r1, r2, r3,which is adapted to the decomposition X = (L ∩ V) ⊕ L ⊕ V ⊕ R. Define the insertion maps S1 =

[n4],

S2 =[n1 n2 n3

], S3 =

[v], and S4 =

[r1 r2 r3

]. Let T =

[S1 S2 S3 S4

]. The natural projection

maps Q1, Q2, Q3, and Q4 are then uniquely defined by T−1 = col(Q1, Q2, Q3, Q4), where Q1 is the first row,Q2 is the following three rows, Q3 is the fifth row, and Q4 is the final three rows.

In (8.6) of Example 8.3, we found a friend F ∈ C(Π4 ⊗ I1,Π4 ⊗ I2) of V, i.e. (A + BF )V ⊂ V. To satisfyconstraint (11.10), we instead use the friend F ′ = FS3Q3. (In this example, F ′ = F , so there is no need toverify that it is still a block circulant friend of V.) Using the feedback transformation u = F ′x+ v, we then

have the system x = Ax+Bv where A = A+BF ′. Based on the above decomposition of X , this system canbe decomposed into four subsystems:

T−1AT =

−8 0 0 0 0 0 0 0−4 0 0 8 0

−4 0 0 0 8−4 0 0 0 0

−12 0 0 0−16

88

, T−1B =

1

26

0 0 0 0−6 0 6 0

0 −6 0 613 13 13 130 0 0 0

13 −13 13 −134 0 −4 00 4 0 −4

DT =

8 0 4 8 12 00 8 4 −8 0 12−8 0 4 8 −12 0

0 −8 4 −8 0 12

26

It can be verified that all submatrices satisfy the desired commuting properties with the various restrictionsof Π4⊗ I2. Denote the bottom-right subsystem by (D4, A4, B4); it can clearly be seen (since T−1AT is upper

triangular) that all unstable modes of the system come from A4. It can also be verified that R+(A4) ⊂〈A4|B4〉, and so the fourth subsystem is stabilizable. In particular, the feedback matrix

K ′4 = 52

0 −1 00 0 −10 1 00 0 1

∈ C(Π4 ⊗ I1, (Π4 ⊗ I2)R)

sets the closed-loop eigenvalues of the fourth subsystem to σ(A4+B4K4) = −16,−8,−8. Using Lemma 5.6,K ′4 can be lifted to the block circulant matrix

K ′ = K ′4Q4 =

−6 −4 0 0 6 4 0 0

0 0 −6 −4 0 0 6 46 4 0 0 −6 −4 0 00 0 6 4 0 0 −6 −4

∈ C(Π4 ⊗ I1,Π4 ⊗ I2)

With K ′, we have stabilized the output of the feedback-transformed system (D,A + BF ′, B). The overallfeedback solving the block circulant RRP for the original system (D,A,B) is then given by

K = F ′ +K ′ =1

2

−12 −9 0 −1 12 7 0 −1

0 −1 −12 −9 0 −1 12 712 7 0 −1 −12 −9 0 −10 −1 12 7 0 −1 −12 −9

∈ C(Π4 ⊗ I1,Π4 ⊗ I2)

C

12. Concluding Remarks. In this paper we showed that control systems possessing a certain structure— that of subsystems connected in a ring — can be controlled in a way that respects and maintains thatstructure. The ring structure is algebraically encoded in commuting properties of the system’s block circulantmatrices. The results strongly suggest that such commuting properties are an integral consideration incarrying out the control design. To bridge the algebraic and geometric domains, we introduced block circulantsubspaces. This device allows to make a connection between matrix structure and subspace structure throughU -commuting matrices and U -invariant subspaces.

The control design tools we have used are standard and familiar to any researcher versed in geometric controltheory. Whereas up to this point, the control of block circulant systems has been studied primarily throughthe Fourier decomposition, we have found the use of commuting properties to provide a more direct andintuitive approach to control design. For example, the decompositions used in designing control laws can bedone in the precisely same way here as for general linear systems. The only control designs that resort toblock diagonalization are block circulant pole placement and construction of block circulant friends. As aresult, the conditions for solvability of each of the design problems are almost completely decoupled from theblock circulant structure. Indeed, a significant outcome of this work is that block circulant control of blockcirculant systems works exactly the same way as unstructured control of general linear systems; we did nothave to modify any linear geometric control conditions in order to guarantee the possibility of recovering ablock circulant control law. In other words, we have established not only that a block circulant system canbe controlled by block circulant feedback, but that it can be done in the “usual” way.

Our work suggests that structured systems naturally admit structured controllers. We have provided amechanism by which to provide such a controller in a number of control design problems, and the evidencestrongly suggests that other control problems should be amenable to study in the same way.

This paper has focused on ring systems; we are also interested in looking at systems whose subsystemsinteract in other manners, such as in chains and trees [14]. We suspect that certain other structures can also

27

be encoded through commuting properties of the system matrices, as is the case for patterned systems corre-sponding to one-dimensional subsystems [15]. Confirming these suspicions will require further investigationbut, once a pattern in the matrices is found, our hope is that it will be natural to construct a feedback thatfollows that pattern and respects the system structure.

Acknowledgements: The numerical examples of the paper were generated using the Matlab toolboxprovided by Giovanni Marro.

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