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(i) A precise relation Let λ C. It is shown in [16] that if (A, B) is controllable, between eigenstructure and topology is characterized (Sec- then there∈ exists tion III). (ii) A method for imposing topological constraints on eigenstructure assignment is presented (Section III.B). N1(λ) (iii) More general cases where the initial inter-agent topol- N(λ) := C(n+m)×m (2) N2(λ) ∈ ogy is arbitrary and/or there exist non-stabilizable agents are addressed (Section IV). (iv) The problem of achieving with linearly independent columns such that cooperative circular motion is solved. (v) All the proofs are provided. N1(λ) We note that [22] also proposed an eigenstructure assign- λI A B =0. (3) − N2(λ) ment method and applied it to the multi-agent consensus Thus the columns of N(λ) form a basis of Ker[λI A B]; problem. Their approach is bottom-up: first a communication − topology is imposed among the agents, then local control Ker denotes kernel. Also we will use Im to denote image. strategies are designed based on eigenstructure assignment Lemma 1: ([16]) Consider the system (1) and suppose that respecting the topology, and finally the correctness of the (A, B) is controllable and KerB = 0. Let λ1,...,λn be { } proposed strategies is verified at the global level. By con- a set of distinct complex numbers, and v1,...,vn a set of { } trast, our approach is top-down: no topology is imposed linearly independent vectors in Cn. Then there is a unique a priori, and topology is a result of control synthesis. F such that for every i [1,n], (A + BF )vi = λivi if and ∈ Moreover, we characterize the relation between topology and only if eigenstructure, and design special topologies by selecting ( i [1,n])v ImN (λ ) (4) special eigenstructures. In addition, the problems addressed ∀ ∈ i ∈ 1 i in the paper are distinct, namely scalable/rigid formation where N1( ) is in (2). and circular motion on the plane, which involve complex Lemma· 1 provides a necessary and sufficient condition eigenvalues and eigenvectors. of eigenstructure assignment. When the condition holds and The rest of the paper is organized as follows. In Section II thus F exists for assigning distinct complex eigenvalues λi we review the basics of eigenstructure assignment and formu- and the corresponding eigenvectors vi (i [1,n]), F may be late the multi-agent formation control problem. In Section III constructed by the following procedure [16].∈ we solve the problem by eigenstructure assignment, and (i) For each λ compute a basis of Ker[λ I A B]. Stack discuss the relations between eigenvalues/eigenvectors and i i − topologies. In Section IV we study the more general cases the basis vectors to form N(λi) in (2); partition N(λi) where the initial inter-agent topology is arbitrary and/or there properly to get N1(λi) and N2(λi). (ii) Find w = N (λ )k , where k C is such that exist non-stabilizable agents. In Section V we present a i − 2 i i i ∈ hierarchical synthesis procedure to reduce computation time, N1(λi)ki = vi (the condition KerB = 0 in Lemma 1 and in Section VI extend the method to achieve rigid forma- ensures that N1(λi) has independent columns; thus ki may tion and circular motion. Simulation examples are given in be uniquely determined). Section VII and our conclusions stated in Section VIII. (iii) Compute F by −1 II.PRELIMINARIESANDPROBLEM F = [w1 wn][v1 vn] . (5) ··· ··· FORMULATION Note that the entries of F may include complex numbers A. Preliminaries on Eigenstructure Assignment in general. If λ ,...,λ is a self-conjugate set of distinct { 1 n} First we review the basics of eigenstructure assign- complex numbers, and vi =v ¯j wherever λi = λ¯j (¯ denotes ment [16]. Consider a linear time-invariant finite-dimensional complex conjugate), then all entries of F are real numbers.· system modeled by The procedure (i)-(iii) of computing F has complexity O(n3), inasmuch as the calculations involved are solving x˙ = Ax + Bu (1) systems of linear equations, matrix inverse and multiplication where x Cn is the state vector, u Cm the input vector, (e.g. [24]). and A ∈Rn×n, B Rn×m. ∈ We note that the above eigenstructure assignment result Suppose∈ we modify∈ (1) by state feedback u = F x. It may be extended to the case of repeated eigenvalues with is well-known (e.g. [23]) that F may be chosen to assign generalized eigenvectors. For details refer to [17] or Ap- any (self-conjugate) set of closed-loop eigenvalues for x˙ = pendix. (A + BF )x if and only if (A, B) is controllable. Unless m =1 (single input), however, F is not uniquely determined B. Problem Formulation by a set of closed-loop eigenvalues. Indeed, with state Consider a heterogeneous multi-agent system where each feedback F one has additional freedom to assign certain sets agent is modeled by a first-order ODE: of closed-loop eigenvectors. Simultaneously assigning both eigenvalues and eigenvectors is referred to as eigenstructure assignment. x˙ i = aixi + biui, i =1, . . . , n. (6) Here xi C is the state variable, ui C the control variable, with the corresponding eigenvector f, and other eigenvalues a R and∈ b (= 0) R are constant∈ parameters. Thus each have negative real parts, then for every initial condition x(0), i ∈ i 6 ∈ agent is a point mass moving on the complex plane, with lim →∞ x(t)= cf for some c C. t ∈ possibly stable (ai < 0), semistable (ai = 0) or unstable Proof: The solution of the closed-loop system (8) (a > 0) dynamics. The requirement b = 0 is to ensure is x(t) = e(A+BF )tx(0). Since A + BF has a simple i i 6 stabilizability/controllability of (ai,bi); thus each agent is eigenvalue 0, with the corresponding eigenvector f, and other stabilizable/controllable. Note that represented by (6), the eigenvalues have negative real parts, it follows from the agents are independent (i.e. uncoupled) and no inter-agent standard linear systems analysis that x(t) (w⊤x(0))f as topology is imposed at this stage. t . Here w Cn is the left-eigenvector→ of A+BF with → ∞ ∈ In vector-matrix form, the system of n independent agents respect to the eigenvalue 0. Therefore limt→∞ x(t) = cf, is where c := w⊤x(0). In view of Proposition 1, if the specified eigenvalues and x˙ = Ax + Bu (7) the corresponding eigenvectors may be assigned by state where x := [x x ]⊤ Cn, u := [u u ]⊤ Cn, feedback u = F x, then Problem 1 is solved. To this end, 1 ··· n ∈ 1 ··· n ∈ A := diag(a1,...,an) and B := diag(b1,...,bn); here we resort to eigenstructure assignment. diag( ) denotes a diagonal matrix with the specified diagonal III. MAIN RESULTS entries.· Consider modifying (7) by a state feedback u = F x and thus the closed-loop system is In this section, we solve Problem 1, the formation control problem of multi-agent systems, by the method of eigen- x˙ = (A + BF )x. (8) structure assignment. The following is our first main result. Straightforward calculation shows that the diagonal entries Theorem 1: Consider the multi-agent system (7) and let f be a desired formation configuration. Then there always of A+BF are ai +biFii, and the off-diagonal entries biFij . exists a state feedback control u = F x that solves Problem 1, Since bi = 0, the off-diagonal entries (A + BF )ij = 0 if and only6 if F = 0 (i = j). 6 i.e. ij 6 6 In view of the structure of A + BF , we can define a ( x(0) Cn)( c C) lim x(t)= cf. corresponding directed graph = ( , ) as follows: the ∀ ∈ ∃ ∈ t→∞ node set := 1,...,n withG nodeVi E standing for agent i (orV state{x ); the}edge set ∈ V with edge Proof: By Proposition 1, the multi-agent system (7) i E ⊆V×V (j, i) if and only if F ’s off-diagonal entry Fij = 0 . with u = F x achieves a formation configuration f n ∈ Since ∈F V = 0 implies that x uses x in its state update,6 C if the closed-loop matrix A + BF has the following ij 6 i j we say for this case that agent communicates its state eigenstructure: (i) its eigenvalues λ1, λ2,...,λn satisfy j { } xj to agent i, or j is a neighbor of i. The graph is 0= λ1 < λ2 λn , and ( i [2,n])Re(λi) < 0 (9) therefore called a communication network among agents,G | | ≤ · · · ≤ | | ∀ ∈ whose topology is decided by the off-diagonal entries of F . (ii) the corresponding eigenvectors Thus the communication topology is not imposed a priori, v , v ,...,v are linearly independent, and v = f but emerges as the result of applying the state feedback 1 2 n 1 { } (10) control u = F x. Now we define the formation control problem of the multi- So we must verify that the above eigenstructure is assignable agent system (7). by state feedback u = F x for the multi-agent system (7). Problem 1: Consider the multi-agent system (7) and spec- Note that except for (λ1, v1) which is fixed, we have freedom n ify a vector f C (f = 0). Design a state feedback to choose (λi, vi), i [2,n]. For simplicity we let λi be all ∈ control u = F x∈such that6 for every initial condition x(0), distinct, and thus Lemma 1 can be applied. lim →∞ x(t)= cf for some constant c C. In (7), we have A = diag(a1,...,an), B = t ∈ In Problem 1, the specified vector f represents a de- diag(b1,...,bn), and bi =0 for all i [1,n]. Thus it is easily 6 ∈ sired formation configuration in the plane. By formation checked that the pair (A, B) is controllable and KerB =0. configuration we mean that the geometric information of To show that there exists F such that (A + BF )vi = λivi, the formation remains when scaling and rotational effects for each i [1,n] (with λi, vi specified above), it suffices to ∈ are discarded. Indeed, by writing the constant c C in verify the condition (4) of Lemma 1. jθ ∈ the polar coordinate form (i.e. c = ρe , j = √ 1), the First, for λ1 =0, we find a basis for final formation cf is the configuration f scaled by− ρ and Ker[λ I A B]= Ker[ A B] rotated by θ. The constant c is unknown a priori and in 1 − − general depends on the initial condition x(0). Note also that and derive N1(λ1) = B and N2(λ1) = A (N1( ),N2( ) in Problem 1 includes the consensus problem as a special case (2)). Thus ImN (λ )= Cn, and hence v = f ·ImN (·λ ), 1 1 1 ∈ 1 1 when f = 1 := [1 1]⊤. i.e. the condition (4) of Lemma 1 holds. To solve Problem··· 1, we note the following fact. Next, let i [2,n]; we find a basis for Ker[λ I A B] ∈ i − Proposition 1: Consider the multi-agent system (7) and and derive N1(λi) = B and N2(λi) = A λiI. So again state feedback u = F x. If A+BF has a simple eigenvalue 0, ImN (λ )= Cn, and v ImN (λ ), i.e. the− condition (4) of 1 i i ∈ 1 i control gain). As an example, for (ii) in Example 1 assign the second largest eigenvalue λ2 = 2 (originally 1), and 1 − − change v4 = [0 0 2 1]. This results in a new feedback matrix − 4 1 2 1 − ′ 1 22 1 F = 2 −2− 0 0 2 − 0 1 2 3 − which has zero entries at the same locations as F2. Thus with ′ the same topology, F2 achieves faster convergence speed. Fig. 1. Example 1 As we have seen in Example 1, the feedback matrix F ’s off-diagonal entries, which determine the topology of , are dependent on the choice of eigenvalues as well Lemma 1 holds. Therefore, we conclude that there always asG eigenvectors. Namely different sets of eigenvalues and exists a state feedback u = F x such that the multi-agent eigenvectors result in different inter-agent communication system (7) achieves the formation configuration f. topologies. Our next result characterizes a precise relation
In the proof we considered distinct eigenvalues λi, i between the eigenvalues/eigenvectors and the topologies. [1,n]; hence the control gain matrix F may be computed∈ Theorem 2: Consider the multi-agent system (7) and f by (5). The computed F in turn gives rise to the agents’ a desired formation configuration. Let the eigenstructure λi communication graph . The following is an illustrative and vi (i = 1, ..., n) be as in (9) and (10), and denote the G −1 ∗ example. rows of [v1 vn] by vi . Then the communication graph = ( , ) ···of the (closed-loop) multi-agent system is such Example 1: Consider the multi-agent system (7) of 4 Gthat V E single integrators (that is, ai =0 and bi =1, i =1, ..., 4). ⊤ (i) Square formation with f = [1 j 1 j] (i1, j1),..., (iK , jK ) / (K 1) (j = √ 1). Let the desired closed-loop eigenvalues− − be ∈E ≥ − λ1 =0, λ2 = 1, λ3 = 2, λ4 = 3 and the corresponding if and only if the vector − − − ⊤ eigenvectors be v1 = f, v2 = [1 1 0 0] , v3 = ⊤ ⊤ ⊤ [λ2 λn] [0 1 1 0] , v4 = [0 0 1 0] . By (5) one computes the ··· control gain matrix F1, which determines the corresponding is orthogonal to the subspace spanned by the K vectors: communication graph 1 (see Fig. 1). G ∗ ∗ ⊤ ∗ ∗ ⊤ Observe that F contains complex entries, which may be [v2 i1 v vn i1 v ] , , [v2 iK v vn iK v ] . 1 2 j1 ··· n j1 ··· 2 jK ··· n jK viewed as control gains for the real and imaginary axes, Proof: For each λ , i [1,n], we derive from (3) that respectively, or scaling and rotating gains on the complex i ∈ plane. Also note that 1 has a spanning tree with node 4 (λiI A)N1(λi)+ BN2(λi)=0 the root, and the computedG feedback control u = F x can be − implemented by the four agents individually. Choose N1(λi) = B and N2(λi) = (λiI A) to satisfy ⊤ −1− − −1 (ii) Consensus with f = [1 1 1 1] . Let the desired the above equation. Then ki = N1 (λi)vi = B vi and −1 eigenvalues be λ1 = 0, λ2 = 1, λ3 = 3, λ4 = 4 and wi = N2(λi)ki = (λiI A)B vi. By (5) we have − − − ⊤ − − the corresponding eigenvectors be v1 = f, v2 = [1 1 0 1] , −1 ⊤ ⊤ F = [w1 wn][v1 vn] v3 = [1 0 0 1] , v4 = [0 0 1 1] . Again by (5) one ··· ··· − −1 −1 −1 computes the control gain matrix F2 and the corresponding = [(λ1I A)B v1 (λnI A)B vn][v1 vn] −1 − ··· − ··· −1 graph 2 (see Fig. 1). = B ( A[v1 vn] + [λ1v1 λnvn])[v1 vn] NoteG that in this case is real and strongly connected. − ··· ··· ··· F2 2 = B−1A + B−1[v v ]diag(λ ,...,λ )[v∗ v∗ ]⊤. But unlike the usual consensus algorithmG (e.g. [2]), F − 1 ··· n 1 n 1 ··· n − 2 is not a graph Laplacian matrix for the entries (2, 1) and Thus the closed-loop matrix (3, 1) are positive. Thus our eigenstructure assignment based ∗ ∗ ⊤ approach may generate a larger class of consensus algorithms A + BF = [v1 vn]diag(λ1,...,λn)[v1 vn] . (11) ··· ··· with negative weights. The (i, j)-entry of A + BF is We remark that in our approach, the convergence speed (A + BF ) = λ v v∗ + λ v v∗ + + λ v v∗ to the desired formation configuration is assignable. This is ij 1 1i 1j 2 2i 2j ··· n ni nj because the convergence speed is dominated by the eigen- = λ v v∗ + + λ v v∗ (λ = 0) 2 2i 2j ··· n ni nj 1 value λ2, with the second largest real part, of the closed- ∗ ∗ ⊤ = [λ2 λn][v2iv2j vnivnj ] . loop system x˙ = (A + BF )x; and in our approach λ2 ··· ··· is freely assignable. The smaller the Re(λ2) is, the faster Therefore (A + BF )i1j1 = = (A + BF )iK jK =0, i.e. in the convergence to formation occurs (at the cost of higher the communication graph (···i , j ),..., (i , j ) / , if and 1 1 K K ∈ E only if the vector [λ λ ]⊤ is orthogonal to each of the we derive 2 ··· n following K vectors: 0 0 0 λ2f2 ··· ∗ ∗ ⊤ ∗ ∗ ⊤ λ2 0 [v v v v ] , , [v v v v ] . f1 2 i1 2 j1 n i1 n j1 2 iK 2 jK n iK n jK − ··· ··· ··· ··· A + BF = ...... ⊤ . . . . Namely [λ2 λn] is orthogonal to the subspace spanned ··· λnfn by these K vectors. 0 λn − f1 ··· Once the desired eigenvalues and eigenvectors are chosen, Therefore the corresponding graph is a star topology with Theorem 2 provides a necessary and sufficient condition node 1 the root. G to check the interconnection topology among the agents, 2) Cyclic Topology: A directed graph = ( , ) is a without actually computing the feedback matrix F . On cyclic topology if = (1, 2), (2, 3), ..., (nG 1,nV), (En, 1) . the other hand, the problem of choosing an appropriate Consider the followingE eigenstructure.{ − } eigenstructure to match a given topology is more difficult, eigenvalues: λ , λ ,...,λ = 0,ω 1,...,ωn−1 1 inasmuch as there are many free variables to be determined in { 1 2 n} { − − } f1 f1 f1 the eigenvalues and eigenvectors. While we shall investigate ··· n−1 f2 f2ω f2ω ··· the general problem of eigenstructure design for imposing f f ω2 f ω2(n−1) eigenvectors: [v1 v2 vn] = 3 3 ··· 3 . . . . particular topologies in our future work, in the next subsec- ··· . . . . (independent) f f ωn−1 f ω(n−1)(n−1) tion, nevertheless, we show that choosing certain appropriate n n ··· n eigenstructures results in certain special (sparse) topologies. (13) With these topologies the synthesized control u = F x may where ω := e2πj/n (j = √ 1). be implemented in a distributed fashion. − Proposition 3: Consider the multi-agent system (7). If the A. Special Topologies eigenstructure (13) is used in the synthesis of feedback We show how to derive the following three types of special control u = F x, then Problem 1 is solved and the resulting topologies by choosing appropriate eigenstructures. is a cyclic topology. G 1) Star Topology: A directed graph = ( , ) is a star Proof: Following the same lines as the proof of topology if there is a single root node,G sayV nodeE 1, and Proposition 2, and using the eigenstructure in (13) and = (1,i) i [2,n] . Thus all the other nodes receive 1 1 1 f1 f2 fn E { | ∈ } ··· n−1 information from, and only from, the root node 1. In terms 1 ω¯ ω¯ f1 f2 fn of the total number of edges, a star topology is one of the 2 ··· 2(n−1) −1 1 1 ω¯ ω¯ sparsest topologies, with the least number (n 1) of edges, [v1 v2 vn] = f1 f2 fn − ··· n . . ··· . that contain a spanning tree. Now consider the following . . . . . . eigenstructure. 1 ω¯(n−1) ω¯ (n−1)(n−1) f1 f2 fn eigenvalues: λ =0, λ ,...,λ distinct ··· 1 2 n we derive and Re(λ ),..., Re(λ ) < 0 2 n 1 f1 0 0 − f2 ··· f1 0 0 0 1 f2 0 ··· f3 f2 1 0 − ··· ··· . . . . . eigenvectors: [v1 v2 vn] = . . . . (12) A + BF = ...... . ··· . . .. . (independent) . . fn−1 fn 0 1 0 0 0 . fn ··· f Proposition 2: Consider the multi-agent system (7). If the n 0 0 1 f1 ··· − eigenstructure (12) is used in the synthesis of feedback By inspection, we conclude that the corresponding graph control u = F x, then Problem 1 is solved and the resulting is a cyclic topology. G graph is a star topology. 3) Line Topology: Proof:G First, it follows from the proof of Theorem 1 A directed graph = ( , ) is a (directed) line topology if there is a singleG rootV node,E say that the eigenstructure (12) can be assigned to the closed- node 1, and = (1, 2), (2, 3), ..., (n 1,n) . A line loop matrix A + BF . Then by Proposition 1, Problem 1 is E { − } solved. topology is also one of the sparsest topologies containing a spanning tree. Now consider the following eigenstructure. Now proceed analogously to the proof of Theorem 2 and derive A + BF as in (11). Substituting into (11) the eigenvalues: λ =0, λ = = λ = 1 1 2 ··· n − eigenvalues and eigenvectors in (12), as well as f 0 0 1 ··· 1 0 0 f2 0 f2 f1 . . ··· −. f ··· . . . 2 1 0 eigenvectors: [v1 v2 vn] = . . . −1 − f1 ··· ··· [v1 v2 vn] = . . . . (independent) . . ··· . . . . fn−1 0 . fn−1 . . . . − fn fn fn fn 0 1 − ··· − − f1 ··· (14) Proposition 4: Consider the multi-agent system (7). If the By (16) we set the new feedback matrix eigenstructure (14) is used in the synthesis of feedback control u = F x, then Problem 1 is solved and the resulting F1 is a line topology. . . G Note that in (14) we have repeated eigenvalues (λ ) 2 Fˆ := Fˆ Fˆ 0 Fˆ Fˆ and the corresponding generalized eigenvectors. As a result, i1 ··· i(j−1) i(j+1) ··· in . Lemma 1 and (5) for computing the control gain matrix . F cannot be applied to this case. Instead, we resort to F n the generalized method of [17], and provide a proof of (17) Proposition 4 in Appendix. ˆ B. Topology Constrained Eigenstructure Assignment namely F is the same as the originally computed F except for the ith row replaced by Fˆ⊤ computed in (16) and Fˆ = We end this section by presenting an alternative approach i ij 0.1 to imposing topological constraints on eigenstructure assign- While the new feedback matrix Fˆ satisfies the imposed ment. topological constraint, the eigenstructure of A + BFˆ is Suppose that we have computed by (5) a feedback matrix generally different from that of A + BF . Therefore, we F to achieve a desired formation, i.e. the closed-loop matrix must verify if the new eigenvalues/eigenvectors still satisfy (A+BF )’s eigenvalues λ , ..., λ and eigenvectors v , ..., v 1 n 1 n (9) and (10), i.e. achieve formation control. This verification satisfy (9) and (10). Now assume that a topological constraint need not always be successful, but in case it does turn is imposed such that agent i cannot receive information from out successful, we are guaranteed to achieve the desired agent j (for reasons such as cost or physical impossibility). formation with a feedback matrix satisfying the imposed But unfortunately, in the computed F the (i, j)-entry F = ij 6 topological constraint. We illustrate the above method by the 0. Thus our goal is to derive a new feedback matrix Fˆ, by following example. suitably modifying F , such that Fˆij = 0. In doing so, the Example 2: Consider the multi-agent system (7) of 5 sin- new closed-loop matrix A+BFˆ will generally have different gle integrators (that is, a =0 and b =1, i =1, ..., 5), and eigenvalues λˆ , ..., λˆ and eigenvectors vˆ , ..., vˆ . Hence we i i 1 n 1 n the desired formation is simply consensus ( ⊤). must check if these new eigenvalues and eigenvectors still f = [1 1 1 1 1] Choose the following eigenvalues and eigenvectors satisfy (9) and (10). Our approach proceeds as follows, which is inspired by eigenvalues: λ , λ , λ , λ , λ = 0, 1, 2, 3, 4 the “constrained feedback” method in [18]. Writing V := { 1 2 3 4 5} { − − − − } [v1 vn] and J := diag(λ1, ..., λn), we have from (11): 1000 0 ··· 1100 1 (A + BF )V = VJ eigenvectors: [v v v v v ]= 1 0 1 1 1 1 2 3 4 5 − 1001 0 BF V = VJ AV ⇒ − 1010 1 F V = B−1(VJ AV ). ⇒ − −1 and compute by (5) the feedback matrix F to achieve Let ψ := B (VJ AV ) and denote by ψi the ith row − consensus: of ψ, also Fi the ith row of F . Then the above equation is rewritten in terms of Kronecker product and row stacking as 000 0 0 follows: 2.5 1 1.5 1.5 1.5 ⊤ ⊤ ⊤ − − − V 0 F1 ψ1 F = 2 0 30 1 . . . − .. . = . . 3 0 0 3 0 . . . − 3 0 1 1 3 0 V ⊤ F ⊤ ψ⊤ n n − − To constrain the (i, j)-entry of F to be zero, we focus on Suppose that the topological constraint is that agent 2 ⊤ ⊤ ⊤ ⊤ the equation V Fi = ψi . Deleting Fij from Fi as well cannot receive information from agent 4, i.e. the (2, 4)-entry ⊤ as the jth column of V , we obtain the reduced equation: of F (F24 = 1.5) must be set to be zero. For this we first − ⊤ ⊤ ⊤ ˆ ⊤ ˆ⊤ ⊤ derive equation V Fi = ψi as follows: V Fi = ψi (15) where Vˆ ⊤ Cn×(n−1) is the matrix V ⊤ with jth col- 11 1 11 F21 0 ∈ umn deleted and Fˆ⊤ Cn−1 the vector F ⊤ with jth 01 0 00 F22 1 i ∈ i − element deleted. Now view the entries of ˆ⊤ as the un- 00 1 01 F23 = 0 . Fi knowns (i.e. (15) contains n equations with n 1 unknowns 00 1 10 F24 0 − ˆ ˆ ˆ ˆ ). Using the pseudoinverse of 0 1 1 0 1 F25 4 Fi1, ..., Fi(j−1) Fi(j+1), ..., Fin − − Vˆ ⊤, denoted by (Vˆ ⊤)†, we derive 1 ˆ⊤ ˆ ⊤ † ⊤ The above derivation may be readily extended to deal with more than Fi = (V ) ψi . (16) one topological constraint. ⊤ ⊤ Deleting F24 from Fi and the 4th column of V , we obtain A. Arbitrary Inter-Agent Connections the reduced equation (15): First we consider the case where the agents have arbitrary initial interconnection, while keeping the assumption that 11 1 1 ˆ 0 F21 they are individually stabilizable. That is, we consider the 01 0 0 1 Fˆ22 following multi-agent system 00 1 1 = −0 . Fˆ 00 1 0 23 0 ˆ x˙ = Ax + Bu (18) 0 1 1 1 F25 4 − − where x Cn, u Cn, A Rn×n and B = ∈ ∈ ∈ Solve this equation for the unknowns Fˆ , Fˆ , Fˆ , Fˆ , we diag(b1,...,bn) (bi =0). The matrix A is now an arbitrary 21 22 23 25 6 compute by (16): real matrix, modeling an arbitrary (initial) communication topology among the agents. ˆ F21 1.8571 It turns out, despite the general A matrix, that the same ˆ ⊤ F22 1.4286 conclusion as Theorem 1 holds. Fˆ = = − . 2 Fˆ 0.8571 Theorem 3: Consider the multi-agent system (18) and let 23 ˆ 1.2857 f be a desired formation configuration. Then there always F25 − exists a state feedback control u = F x that achieves Finally set as (17) the new feedback matrix formation control, i.e. 0 0 000 ( x(0) Cn)( c C) lim x(t)= cf. t→∞ 1.8571 1.4286 0.8571 0 1.2857 ∀ ∈ ∃ ∈ Fˆ = 2− 0 3 0− 1 . − Proof: The proof proceeds similarly to that of The- 3 0 0 3 0 − orem 1. First, for the diagonal matrix in (18), we have 3 0 1 1 3 B − − (regardless of A) that the pair (A, B) is controllable and Observe that (only) the second row of Fˆ has entries all KerB =0. different from that of the original F , and the (2, 4)-entry It is left to verify if the (distinct) eigenvalues λ1, ..., λn and eigenvectors as in (9) and (10) satisfy the Fˆ24 = 0. Moreover the eigenstructure of the new closed- v1, ..., vn ˆ ˆ condition (4) of Lemma 1. Let i [1,n]. Since loop matrix A + BF (= F ) is ∈ ˆ ˆ ˆ ˆ ˆ B eigenvalues: λ1, λ2, λ3, λ4, λ5 = 0, 1.4286, 2, 3, 4 λiI A B −1 =0 { } { − − − − } − B (A λiI)B eigenvectors: [ˆv1 vˆ2 vˆ3 vˆ4 vˆ5] − we find a basis for Ker[λ I A B] and derive 10 0 0 0 i − 1 2.2361 1.0477 0.8047 1.1352 N1(λi)= B = 1 0 −1.3969− 1.4753− 1.3623 . − −1 10 0 1.4753 0 N2(λi)= B (A λiI)B. − 1 0 1.3969 0 1.3623 n − − Thus ImN1(λi) = C , and vi ImN1(λi), i.e. the condi- ∈ Hence these new eigenvalues/eigenvectors still satisfy (9) tion (4) of Lemma 1 holds. Therefore, there always exists and (10), and therefore consensus is achieved despite of the a state feedback u = F x such that the multi-agent system (18) achieves the formation configuration . imposed topological constraint. (In fact, since λˆ2 < λ2, we f have faster convergence with the new Fˆ.) In the proof of Theorem 1, we derived N1(λi) = B and N (λ ) = A λ I. Since A was diagonal and diagonal 2 i − i IV. GENERAL MULTI-AGENT SYSTEMS matrices commute, there held B So far we have considered the multi-agent system in λiI A B =0. − A λiI (7), where the agents are uncoupled and each is (self-) − stabilizable (the matrices A, B are diagonal and B’s diagonal For a general A as in Theorem 3, we have found instead −1 entries nonzero). For (7) we have shown in Theorem 1 that a N1(λi) = B and N2(λi) = B (A λiI)B that deal state feedback control, based on eigenstructure assignment, with arbitrary A without depending on− the commutativity always exists to drive the agents to a desired formation. of matrices. More generally, however, the agents may be initially Theorem 3 asserts that, as long as the agents are in- interconnected (owing to physical coupling or existence of dividually stabilizable, formation control is achievable by communication channels), and/or some agents might not be eigenstructure assignment regardless of how the agents are capable of stabilizing themselves (though they can receive initially interconnected. The final topology, on the other information from others). It is thus of interest to inquire, hand, is in general determined by the initial connections based on the eigenstructure assignment approach, what con- ‘plus’ additional ones resulted from the chosen eigenval- clusions we can draw for formation control in these more ues/eigenvectors (as has been discussed in Section III). It general cases. may also be possible, however, that the initial connections are ‘decoupled’ by the corresponding entries of the synthesized where x Cn, u Cm, A Rn×n and feedback matrix. This is illustrated by the following example. ∈ ∈ ∈ b1 Consider again Example 1(i), but change A from the zero .. matrix to the following . bm Rn×m B = (m 1 0 0 j Proof: First observe from (19) that KerB =0, since B’s − 1 2 0 2 j columns are linearly independent. Now let i [1,n]. Since A + BF = − − − ∈ 1 1 3 1 2j (A, B) is controllable (condition (i)), there exist N1(λi) and − − − 000 0 N2(λi) such that (3) holds. Setting N2(λi) = I, we derive from (3) the following matrix equation which is the same as the feedback matrix F1 (as well as (A λiI) N1(λi)= B. (20) the closed-loop matrix) in Example 1(i). Thus despite the − Since ImB Im(A λ I) (condition (ii)), this equation has initial coupling, the final topology turns out to be the same ⊆ − i as that of Example 1(i). In particular, in the final topology a solution N1(λi) (which is determined by A, B, λi). Finally, since v ImN (λ ) (condition (iii)), the condition (4) of agents1 and 2, 4 and 3 are uncoupled – their initial couplings i ∈ 1 i are ‘canceled’ by the corresponding entries of the feedback Lemma 1 is satisfied. Therefore the desired eigenvalues and eigenvectors satisfying (9) and (10) may be assigned by a matrix F . state feedback control u = F x, i.e. formation control is achieved. Theorem 4 provides sufficient conditions to ensure solv- B. Existence of Non-Stabilizable Agents ability of the formation control problem for multi-agent Continuing to consider arbitrary initial topology (i.e. gen- systems with non-stabilizable agents. In the following we eral A), we further assume that some agents cannot stabilize illustrate this result by working out a concrete example, themselves (i.e. the corresponding diagonal entries of B where A represents a directed line topology and there is only in (18) are zero). Equivalently, the non-stabilizable agents one agent that is stabilizable (i.e. B is simply a vector). have no control inputs. In this case, achieving a desired Example 3: Consider the multi-agent system x˙ = Ax + formation is possible only if those non-stabilizable agents Bu with may take advantage of information received from others (via a1 0 0 b1 connections specified by A). This is a problem of global aˆ2 a2 0 formation stabilization with locally unstabilizable agents, A = . . , B = . .. .. . which has rarely been studied in the literature. We aim 0a ˆ a 0 to provide an answer using our top-down eigenstructure n n assignment based approach. where a1, ..., an, aˆ2, ..., aˆn,b1 are nonzero. Namely A repre- Without loss of generality, assume that only the first m sents a directed line topology with agent 1 the root, and B (< n) agents are stabilizable. Thus the multi-agent system means that only agent 1 is stabilizable. Thus this is a single- we consider in this subsection is input multi-agent system – by controlling only the root of a directed line. First, it is verified that (A, B) is controllable, i.e. condi- x˙ = Ax + Bu (19) tion (i) of Theorem 4 is satisfied. To ensure condition (ii), ImB Im(A λ I), it suffices to choose each desired Now for the configuration f and x,u,A,B in (7), write ⊆ − i eigenvalue λi (i [1,n]) such that λi = aj for j [1,n] (i.e. in accordance with the partition (possibly with reordering) every eigenvalue∈ is distinct from the nonzero6 diagonal∈ entries g y w of A). At the same time, these eigenvalues must satisfy (9). 1 1 1 . . . Having condition (ii) hold, equation (20) has a solution f = . , x = . ,u = . , N (λ ). Let us solve (20) gl yl wl 1 i A1 B1 a1 λi 0 0 b1 − A = .. ,B = .. aˆ2 a2 λi 0 . . − . . N1(λi)= . A B .. .. . l l nk nk×nk 0a ˆn an λi 0 where gk,yk, wk C and Ak,Bk C , k [1,l]. − Thus for each group∈ k, the dynamics∈ is ∈ and obtain an explicit solution y˙k = Akyk + Bkwk. (22) b1 a1−λi aˆ2b1 For later use, also write gk1,yk1, wk1 (resp. Ak1,Bk1) (a −λ )(a −λ ) − 1 i 2 i for the first component of gk, yk, wk (resp. (1,1)-entry N1(λi)= . . . ⊤ ⊤ . of Ak,Bk), and g0 := [g11 gl1] , y0 := [y11 yl1] , ··· ⊤ ··· ··· ( 1)n−1 aˆ2 aˆnb1 w0 := [w11 wl1] , A0 := diag(A11, , Al1), B0 := (a1−λi)(a2−λi)···(an−λi) ··· ··· − diag(B11, ,Bl1). Hence to ensure condition (iii) of Theorem 4, we must The vector··· g (k [1,l]) is the local formation configu- k ∈ choose each desired eigenvector vi (i [1,n]) such that ration for group k, while g0 is the formation configuration v ImN (λ ) and (10) is satisfied. In particular,∈ for i =1 for the set of the first component agent from each group. We i ∈ 1 i we have λ = 0 and v = f; thus v ImN (λ ) means assume that these configurations are all nonzero, i.e. g =0 i 1 1 ∈ 1 1 k that the formation vector f must be such that for k [1,l] and g = 0. Now we present the hierarchical6 ∈ 0 6 ⊤ synthesis procedure. f = c b1 aˆ2b1 ( 1)n−1 aˆ2···aˆnb1 (21) (i) For each group k [1,l] and its dynamics (22), a1 − a1a2 ··· − a1a2···an ∈ h i compute Fk by (5) such that Ak + BkFk has a simple where c C (c = 0). This characterizes the set of eigenvalue 0 with the corresponding eigenvector gk, and all achievable∈ formation6 configurations for the single-input other eigenvalues have negative real parts; moreover the multi-agent system under consideration. topology defined by Fk has a unique root node yk1 (e.g. We conclude that, by controlling only one agent, indeed star or line by the method given in Section III.A). the root agent of a directed line topology, it is not possible (ii) Treat y k [1,l] (the group leaders) as a higher- { k1| ∈ } to achieve arbitrary formation configurations but those deter- level group, with the dynamics mined by the nonzeros entries of the matrices A and B in y˙ = A y + B w . (23) the specific manner as given in (21). 0 0 0 0 0 Compute F Cl×l by (5) such that A + B F has a 0 ∈ 0 0 0 V. HIERARCHICAL EIGENSTRUCTURE ASSIGNMENT simple eigenvalue 0 with the corresponding eigenvector g0, and other eigenvalues have negative real parts. In the previous sections, we have shown that a control gain (iii) Set the control gain matrix F := F low + F high, matrix F can always be computed (as long as every agent is where stabilizable) such that the multi-agent formation Problem 1 is solved. Computing such F by (5) (see the eigenstructure F1 3 low . assignment procedure in Section II) has complexity O(n ), F := .. where n is the number of agents. Consequently the com- Fl putation cost becomes expensive as the number of agents increases. and F high is partitioned according to F low, with each block To address this issue of centralized computation, we (i, j), i, j [1,l] propose in this section a hierarchical synthesis procedure. ∈ 1 0 0 We shall show that the control gain matrix F computed by ··· 0 0 0 this hierarchical procedure again solves Problem 1, which high ··· (F )ij = (F0)ij . . . . moreover significantly improves computational efficiency · . . .. . (empirical evidence provided in Section VII). 0 0 0 ··· For clarity of presentation, let us return to consider the (F ) 0 0 0 ij ··· multi-agent system (7), and Problem 1 with the desired 0 0 0 formation configuration f Cn (f =0). Partition the agents = . . ···. . . ∈ 6 . . .. . into l ( 1) pairwise disjoint groups. Let group k ( [1,l]) have n≥( 1) agents; n may be different and Σl n∈ = n. 0 0 0 k ≥ k k=1 k ··· The computational complexity of Step (i) is O(ˆn3), where VI. RIGID FORMATION AND CIRCULAR MOTION nˆ := max n , ..., n ; and Step (ii) is O(l3). Let n˜ := { 1 l} In this section we show that our method of eigenstructure max n,ˆ l . Then the complexity of the entire hierarchical assignment may be easily extended to address problems of { } 3 synthesis procedure is O(˜n ). With proper group partition, rigid formation and circular motion. this hierarchical procedure can significantly reduce com- putation time, as demonstrated by an empirical study in A. Rigid Formation Section VII. First, we extend our method to study the problem of Note that in Step (i) of the above procedure, requiring achieving a rigid formation, one that has translational and the topology defined by each Fk to have a unique root, i.e. rotational freedom but fixed size. a single leader, is for simplicity of presentation. It can be Problem 2: Consider the multi-agent system (7) and spec- extended to the case of multiple leaders, and then in Step (ii) ify f Cn (f =0) and d> 0. Design a control u such that ∈ 6 jθ treat all the leaders at the higher-level. On the other hand, the for every initial condition x(0), limt→∞ x(t) = c1 + dfe number of leaders should be kept small such that the high- for some c C and θ [0, 2π). level control synthesis in Step (ii) can be done efficiently. In Problem∈ 2, the goal∈ of the multi-agent system (7) is to The correctness of the hierarchical synthesis procedure is achieve a rigid formation df, with translational freedom in asserted in the following. c, rotational freedom in θ, and fixed size d. Theorem 5: Consider the multi-agent system (7) and let f We now present the rigid-formation synthesis procedure. be a desired formation configuration. Then the state feedback (i) Compute F by (5) such that A+BF has two eigenval- control u = F x synthesized by the hierarchical synthesis ues 0 with the corresponding (non-generalized) eigenvectors procedure solves Problem 1, i.e. 1 and f, and other eigenvalues have negative real parts;2 3 ( x(0) Cn)( c C) lim x(t)= cf. moreover the topology defined by F is 2-rooted with exactly ∀ ∈ ∃ ∈ t→∞ 2 roots (say nodes 1 and 2). This topology may be achieved by assigning appropriate eigenstructures, e.g. Proof: For each k [1,l] let y′ := [y y ]⊤ and ∈ k k2 ··· knk g′ := [g g ]⊤ Cnk −1. Thus y′ and g′ are y ,g eigenvalues: λ1 = λ2 =0, λ3,...,λn distinct k k2 ··· knk ∈ k k k k with the first element removed. By Step (i) of the hierarchical and Re(λ3),..., Re(λn) < 0 synthesis procedure, since yk1 is the unique root node, we 1 f1 0 0 can write y˙ = (A + B F )y as follows: ··· k k k k k 1 f2 0 0 ··· y˙k1 0 0 yk1 eigenvectors: [v v v v ] = 1 f3 1 0 | 1 2 3 ··· n . . . ··· . (independent) ...... ′ = ′ . . . . . . y˙k Hk Gk yk 1 fn 0 1 ··· (25) Then by the eigenstructure of Ak +BkFk, all the eigenvalues (ii) Let f1,f2 be the first two components of f, and set of Gk have negative real parts and 2 2 2 x˙ 1 (x2 − x1)(||x2 − x1|| − d |f2 − f1| ) gk1 = 2 2 2 =: r(x1,x2). x˙ 2 (x1 − x2)(||x1 − x2|| − d |f1 − f2| ) H G =0. (24) k k g′ k (iii) Set the control r(x , x ) Reorder x = [y⊤ y⊤]⊤ to get xˆ := [y⊤ y′ ⊤ y′ ⊤]⊤. u := F x + B−1 1 2 . (26) 1 l 0 1 n 0 Then there is a permutation··· matrix that similarly··· transforms the control gain matrix F in Step (iii) to Fˆ, and xˆ˙ = Fˆxˆ is The idea of the above synthesis procedure is to first use eigenstructure assignment to achieve a desired formation y˙0 A0 + B0F0 0 y0 ′ ′ configuration with two leaders, and then control the size of y˙1 H1 G1 y1 . = . . . . the formation by stabilizing the distance between the two . .. .. . leaders to the prescribed d. The latter is inspired by [13]. y˙′ H G y′ Our result is the following. l l l l Proposition 5: Consider the multi-agent system (7) and It then follows from the eigenstructure of A0 + B0F0 ˆ let f Cn, d > 0. Then the control u in (26) synthesized assigned in Step (ii) and (24) above that the matrix F has ∈ a simple eigenvalue 0 with the corresponding eigenvector by the rigid-formation synthesis procedure solves Problem 2 ′ ′ ˆ ⊤ ⊤ ⊤ ⊤ for all initial conditions x(0) with x1(0) = x2(0). f := [g0 g1 gl ] , and other eigenvalues have negative 6 real parts. Hence··· (cf. Proposition 1), 2For repeated eigenvalues with non-generalized eigenvectors, the eigen- ( xˆ(0) Cn)( cˆ C) lim xˆ(t)=ˆcf.ˆ structure assignment result Lemma 1 and the computation of control gain ∀ ∈ ∃ ∈ t→∞ matrix F in (5) remain the same as for the case of distinct eigenvalues. 3 ˆ A 2-rooted topology is one where there exist 2 nodes from which every Since xˆ (resp. f) is just a reordering of x (resp. f), the other node v can be reached by a directed path after removing an arbitrary conclusion follows and the proof is complete. node other than v [13]. Proof: First, by a similar argument to that in the proof of Theorem 1, we can show that the desired eigen- values/eigenvectors (two eigenvalues at 0 with eigenvectors 1 1 and f; all other eigenvalues with negative real parts) may always be assigned for the multi-agent system (7). As a result 0.5 (cf. Proposition 1), ( x(0) Cn)( c,c′ C) lim x(t)= c1 + c′f. 0 ∀ ∈ ∃ ∈ t→∞ Moreover, choosing the eigenstructure in (25) and following position(Im) -0.5 Agent 1 similarly to the proof of Proposition 2, we can show that Agent 2 Agent 3 the resulting topology defined by F is 2-rooted with nodes -1 Agent 4 1 and 2 the only two roots. Agent 5 With the 2-rooted topology and the design in Step (ii), -1 0 1 it follows from [13, Theorem 4.4] that for all x(0) with position(Re) x (0) = x (0), we have c′ = dfejθ for some θ [0, 2π). 1 6 2 ∈ Fig. 2. Scalable regular pentagon formation (x: initial positions, ◦: steady- An illustrative example of achieving rigid formations is state positions) provided in Section VII below. B. Circular Motion system (7) can be made to achieve circular motion while We apply the eigenstructure assignment approach to solve keeping a rigid formation with some specified size d> 0. a cooperative circular motion problem, in which the agents Circular motion may be applied to the task of target all circle around the same center while keeping a desired encircling, which is illustrated by an example in the next formation configuration. This cooperative task may find section. useful applications in target tracking and encircling (e.g. [25], [26]). VII. SIMULATIONS Problem 3: Consider the multi-agent system (7) and spec- We illustrate the eigenstructure assignment based approach ify f Cn (f = 0) and b R (b = 0). Design a state by several simulation examples. For all the examples, we feedback∈ control 6u = F x such∈ that for every6 initial condition consider the multi-agent system (7) with 5 heterogeneous x(0), lim x(t) = c1 + c′febjt for some c,c′ C and t→∞ agents, where j = √ 1. ∈ − In Problem 3, the goal is that the agents of (7) all A = diag(1.6, 4.7, 3.0, 0.7, 4.2) circle around the same center at rate , while keeping the − − c b B = diag(0.2, 1.5, 0.5, 3.3, 3.7). formation configuration f scaled by c′ . − − − | | Our result is the following. Thus the first 3 agents are unstable while the latter 2 are Proposition 6: Consider the multi-agent system (7) and stable; all agents are stabilizable. let f Cn, b R. Then there always exists a state feedback First, to achieve a scalable (regular) pentagon formation, ∈ ∈ control u = F x that solves Problem 3. assign the following eigenstructure: Proof: By a similar argument to that in the proof of Theorem 1, we can show that for (7) there always exists F eigenvalues: λ1,...,λ5 = 0, 1, 2, 3, 4 { } {2πj×−1 − − − } such that (A + BF ) has the following eigenstructure: e 5 10 00 2πj×2 − eigenvalues: λ =0, λ = bj, λ ,...,λ distinct e 5 1 0 00 1 2 3 n 2πj×3 eigenvectors: [v v ]= 5 . and Re(λ ),..., Re(λ ) < 0 1 5 e 2 1 1 0 3 n ··· 2πj×4 − − 1 e 5 2 0 1 0 eigenvectors: v1 = , v2 = f, v1, v2, ..., vn independent 2πj×5 − − { } e 5 2 0 2 1 Then (cf. Proposition 1), − − By (5) we compute the control gain matrix ( x(0) Cn)( c,c′ C) lim x(t)= c1 + c′febjt. t→∞ 10.5 1.8164j 2.5 1.8164j 0 0 0 ∀ ∈ ∃ ∈ − − − 0.3333+ 0.2422j 3.4667+ 0.2422j 0 0 0 That is, Problem 3 is solved. F = 0.4721 2.3511j −0.4721 2.3511j 10 2 0 . − − − − − 0.0442 0.4403j 1.1679 0.4403j 0 0.697 0 The key point to achieving circular motion is to assign − − − 0.3979+ 0.4391j 0.1427+ 0.4391j 0 0.5405 0.0541 one, and only one, pure imaginary eigenvalue bj, associated − − − with the formation vector f. The circular motion is counter- Simulating the closed-loop system with initial condition clockwise if b> 0, and clockwise if b< 0. One may easily x(0) = [1+j 1 0.5j 1 j 1+j]⊤, the result is displayed speed up or slow down the circular motion by specifying the in Fig. 2. Observe− that a regular− pentagon is formed, and the value b . topology determined by F contains a spanning tree. Also| note| that, by a similar synthesis procedure to that for Next, to achieve a rigid pentagon formation, we follow the rigid formation in the previous subsection, the multi-agent method presented in Section VI.A: first assign the following agent1 agent2 2.5 10 agent3 agent4 agent5 2 5 1.5 1 0 target 0.5 position(Im) position(Im) -5 0 agent1 agent2 agent3 -0.5 agent4 -10 agent5 -1 0 5 10 15 20 25 -2 -1 0 1 2 position(Re) position(Re) Fig. 3. Rigid regular pentagon formation, with size d = 5, 10, 15 (x: Fig. 4. Target encircling by circular motion (x: initial positions, ◦: steady- initial positions, ◦: steady-state positions) state positions) Moreover, choose the following eigenstructure eigenstructure: eigenvalues: λ ,...,λ = 0, j, 1, 2, 3, 4 { 1 6} { − − − − } 2πj×1 1 e 5 0 00 0 2πj×2 eigenvalues: λ1,...,λ5 = 0, 0, 1, 2, 3 1 e 5 1 00 0 { } { 2πj−×1 − − } 2πj×3 1 e 5 0 0 0 1 e 5 0 10 0 eigenvectors: × 2πj×2 [v1 v6]= 2πj 4 . 1 e 5 0 0 0 ··· 1 e 5 0 01 0 2πj×3 2πj×5 eigenvectors: [v v ]= 5 1 e 5 0 00 1 1 5 1 e 1 0 0 ··· 2πj×4 5 1 0 0000 1 e 0 1 0 2πj×5 1 e 5 0 0 1 Thus the desired formation is (again) a regular pentagon and the target x6 is at the center of this pentagon. Moreover, by the eigenvectors the resulting topology will contain a span- and by (5) compute the control gain matrix ning tree with x6 (the target) being the root. Corresponding to the formation vector is the eigenvalue j; hence the agents will perform circular motion at rate 1. Since the center will 8 0 000 − 0 3.1333 0 0 0 not move (x6 at the center is the root), this makes the first F = 0.618+ 1.9021j 2.618− 1.9021j 80 0 . − − 5 agents encircle the target x6. 0.303+0.9326j 0.303 0.9326j 0 0.3939 0 − − − 1.0614+ 0.7711j 0.2506 0.7711j 0 0 0.3243 By (5) we compute the control gain matrix − − − 8+5j 000 0 5j − − 0.428+0.84j 3.800 0 1.0947 0.84j 4−.4116 0.7331j −0 10 0 0 8.4116+− 0.7331j F = − − Thus the topology determined by is 2-rooted with nodes 0.5574+ 0.7795j 0 00.697 0 1.4664 0.7795j F − − 0.5911+ 0.9447j 0 0 0 0.0541 0.49 0.9447j − − − − 1 and 2 the only two roots. Then for different sizes (d = 0 0000 0 ), we obtain by (26) the control . Simulating the 5, 10, 15 u to assign the above eigenstructure. Indeed, the corresponding closed-loop system with the same initial condition x(0) as topology has a spanning tree whose root is x and x˙ =0. above, the result is displayed in Fig. 3, where pentagons with 6 6 Simulating the closed-loop system with the initial condition specified sizes are formed. x(0) = [1 0.5j 2+2j 2+j 1+j j 1+j]⊤, the Further, we consider the task of target encircling and solve result is displayed− − in Fig. 4.− Observe− that the− target stays put it by circular motion introduced in Section VI.B. Suppose at its initial position 1+ j, while other agents circle around that there is a static target, say x6 with x˙ 6 = u6 (u6 is it. constantly zero), and the goal is make the same 5 agents as Finally, we present an empirical study on the computation above circle around x6. For this we treat the target x6 as time of synthesizing feedback matrix F . In particular, we part of the multi-agent system; hence the augmented A′ and compare the centralized synthesis by (5) and the hierarchical B′ are synthesis in Section V; the result is listed in Table I for differ- ent numbers of agents.4 Here for the hierarchical synthesis, we partition the agents in such a way that the number of A′ = diag(1.6, 4.7, 3.0, 0.7, 4.2, 0) − − 4Computation is done by Matlab R2014b on a laptop with Intel(R) B′ = diag(0.2, 1.5, 0.5, 3.3, 3.7, 1). − − − Core(TM) i7-4510U [email protected] 2.60GHz and 8.00GB memory. TABLE I as to deal with agents with higher-order (heterogeneous, COMPARISON OF COMPUTATION TIME (UNIT: SECONDS) possibly non-stabilizable) dynamics. agent # centralized method by (5) hierarchical method in Sec. V PPENDIX 100 0.398 0.027 IX. A 500 57.308 0.179 We provide the proof of Proposition 4. For this we 900 552.8419 0.394 first briefly review the eigenstructure assignment method in 1000 1068.729 0.525 [17] for dealing with repeated eigenvalues and generalized eigenvectors. Lemma 2: ([17]) Consider the system (1) and sup- groups and the number of agents in each group are ‘balanced’ pose that (A, B) is controllable and KerB = 0. Let (to make n˜ small): e.g. 100 agents are partitioned into 10 d1,...,dk (k n) be a set of positive integers satisfying groups of 10 agents each; 500 agents are partitioned into 16 { k } ≤ d = n, λ ,...,λ be a set of complex numbers, groups of 23 agents each plus 6 groups of 22 each. Observe i=1 i 1 k and v , ..., v { ,...,v ,} ..., v a set of linearly inde- that the hierarchical synthesis is significantly more efficient P 11 1d1 k1 kdk pendent{ vectors in Cn. Then there} is a feedback matrix F than the centralized one, and the efficiency increases as the such that for every i [1, k], number of agents increases. In particular, for 1000 agents ∈ only 0.525 seconds needed, the hierarchical approach might (λ I A BF )v =0 i − − i1 well be sufficient for many practical purposes. (λ I A BF )v = v , j =2,...,d i − − ij i(j−1) i VIII. CONCLUDING REMARKS if and only if m We have proposed a top-down, eigenstructure assignment ( wij C ) ∃ ∈ based approach to synthesize state feedback control for vi1(λi) λiI A B =0 solving multi-agent formation problems. The relation be- − − wi1(λi) tween the eigenstructures used in control synthesis and the vij (λi) resulting topologies among agents has been characterized, λiI A B = vi(j−1), j =2,...,di. − − wij (λi) and special topologies have been designed by choosing (27) appropriate eigenstructures. More general cases where the Lemma 2 provides a necessary and sufficient condition initial inter-agent coupling is arbitrary and/or there exist for assigning repeated eigenvalues λi (i [1, k]) with non-stabilizable agents have been studied, and a hierarchical ∈ eigenvector vi1 and generalized eigenvectors vi2, ..., vidi synthesis procedure has been presented that improves com- (corresponding to a Jordan block of the closed-loop matrix putational efficiency. Further, the approach has been extended A + BF ). When the condition holds and thus F exists, F to achieve rigid formation and circular motion. may be constructed by the following procedure [17]. In our view, the proposed top-down approach to multi- agent formation control is complimentary to the existing (i) Compute the following (maximal-rank) matrices (mainstream) bottom-up approach (rather than opposed to). N1i S1i Ni = , Si = Indeed, the bottom-up approach, if successful, can produce N2i S2i scalable control strategies effective for possibly time-varying satisfying topologies, nonlinear agent dynamics, and robustness issues like communication failures, which are the cases very diffi- N1i S1i [λiI A B] =0, [λiI A B] = I. cult to be dealt with by the top-down approach. On the other − N2i − S2i − hand, bottom-up design is generally challenging, requiring (S needs to be computed only when d > 1.) significant insight into the problem at hand and possibly i i (ii) From the following vector chain many trial-and-errors in the design process; by contrast, top- down design is straightforward and can be automated by vi1 = N1ipi1 algorithms. Hence we suggest the following. When a control v = N p S v i2 1i i2 − 1i i1 researcher or engineer faces a distributed control design . problem for achieving some new cooperative tasks, one can . start with a linear time-invariant version of the problem and vidi = N1ipidi S1ivi(d −1) try the top-down approach to derive a solution. With the ideas − i and insights gained from such a solution, one may then try find the vectors pi1, ..., pidi . Then generate a new vector the bottom-up design possibly for time-varying and nonlinear chain as follows: cases. w = N p In future work, we aim to apply the top-down, eigen- i1 − 2i i1 w = N p + S v structure assignment based approach to solve more complex i2 − 2i i2 2i i1 cooperative control problems of multi-agent systems. In . . particular, our immediate goals are to achieve formations in w = N p + S v . three dimensions with obstacle avoidance abilities, as well idi − 2i idi 2i i(di−1) (iii) Compute F satisfying F vij = wij for all i eigenstructure can be assigned to the closed-loop matrix [1, k], j [1, d ]. (If no solution exists, alter one or more∈ A + BF . Then by Proposition 1, Problem 1 is solved. ∈ i of the vectors pij in Step (ii).) Finally we compute the feedback matrix F . Let W = Now we are ready to prove Proposition 4. [w11 w21 w2(n−1)]. Since Proof of Proposition 4: Consider the multi-agent sys- ··· 1 0 0 0 tem (7) and assign the following eigenstructure: f1 ··· 0 0 1 1 ··· fn−1 − fn eigenvalues: λ1 =0, λ2 = = λn = 1 . . ··· − −1 . . f 0 0 V = . . 1 ··· f2 0 f2 . ··· − 1 . . . . 0 . eigenvectors: [v v v ]= . . . f2 11 21 2(n−1) 1 1 ··· . . 0 f − 0 . f − f1 f2 n 1 − n 1 − fn fn fn − ··· − we obtain by F = W V −1 that First, for λ1 =0 (with d1 =1), in Step (i) we only need −a1 0 0 to compute N1 and obtain N11 = B, N21 = A. Then in b1 ··· Step (ii), from f2 −(1+a2) b2f1 b2 F = . . . .. .. v11 = N11p11 fn −(1+an) −1 0 b f − b we have p11 = B v11. Hence n n 1 n Therefore the closed-loop matrix is −1 a1f1 anfn ⊤ w11 = N21p11 = AB v11 = [ ] . − − − b1 ···− bn 0 0 0 f2 ··· It is verified that w11, together with v11, λ1, satisfies the first 1 f1 − equation of (27). A + BF = . . .. .. Second, for λ2 = 1 (with d2 = n 1), in Step (i) we derive N = B, −N = (A + I), S− = B, S = 0 fn 1 12 22 12 22 fn−1 (A + I) B−1. 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