Proceedings of the 26th Chinese Control Conference July 26-31, 2007, Zhangjiajie, Hunan, China
A Proof of the Discrete-Time KYP Lemma Using Semidefinite Programming Duality* Cheng Yiping School of Electronic and Information Engineering Beijing Jiaotong University, Beijing 100044, P.R.China E-mail:
[email protected] Abstract: Balakrishnan and Vandenberghe have given an elegant proof of the KYP lemma based on their theorems of alternatives in semidefinite programming. Based also on these theorems, this paper gives a proof of a generalized version of the discrete-time KYP lemma. In addition, we point out that in the nonstrict case of Balakrishnan and Vandenberghe’s (continuous-time) KYP formulation the hypothesis that M22 be positive semidefinite can be dropped. Key Words: Discrete-time KYP lemma, Linear matrix inequalities, Semidefinite programming, Control education
1 INTRODUCTION
algebraic part, uses only linear algebra. We will formulate the lemma in Section 2 and give the proof in Section 3. The secondary purpose of this paper is to point out that the BV proof can still be further refined. We will show that in the nonstrict case, the hypothesis that M22 be positive semidefinite can be dropped from the BV formulation. This issue is quite subtle and will be discussed in Section 4.
The Kalman-Yakubovich-Popov (KYP) Lemma (see e.g. [1, 2]) has been recognized as one of the cornerstones of control theory. It establishes the equivalence between the feasibility of a state-space LMI and a frequency inequality. From this lemma many important results, such as the positive real lemma and the bounded real lemma, can be conveniently derived. There is a discrete-time analog of the KYP lemma, namely the discrete-time KYP lemma, which plays the same role for discrete-time systems as the KYP lemma plays for continuous-time systems. There are many variants both of the KYP lemma and of the discrete-time KYP lemma. However, the ones proposed by Rantzer in [2] are generally recognized as standard formulations, and [2] also contained the first elementary proof of the KYP lemma. Recently, Balakrishnan and Vandenberghe gave in [3, 4] a generalized formulation and a new proof of the KYP lemma, which, for ease of reference, will here be called the BV formulation and the BV proof, respectively. Both proofs are short, elementary, and elegant, requiring only some basic facts from convex analysis and linear algebra. However, it is our perception that the BV proof is even more elegant than Rantzer’s, because
2
GENERALIZED FORMULATION OF THE DISCRETE-TIME KYP LEMMA
We use standard notations: R and C denote the set of real numbers and the set of complex numbers, respectively. tr(·) denotes trace of a matrix, and (·)∗ denotes conjugate transpose. If A is a matrix, then A > 0, A > 0, and A = 0 mean A is positive definite, positive semidefinite, and a zero matrix, respectively; A 0 means A > 0 and A 6= 0. H n denotes the set of hermitian n × n complex matrices. The following is a generalized formulation of the discretetime KYP lemma. Theorem 1 Let A ∈ Cn×n , B ∈ Cn×m , and M ∈ H n+m . Then the following two statements are equivalent. 1. There exists P ∈ H n such that · ∗ ¸ A P A − P A∗ P B M− >0 B∗P A B∗P B
• The BV proof is more general as it allows A to have imaginary eigenvalues. • The theorems of alternatives, which were developed in [3, 4] to be the basis of the BV proof, are fundamental results of the semidefinite programming duality theory, and can be used to prove many other interesting results on linear time-invariant systems. This feature is lacking in Rantzer’s proof.
· 2. For all ω ∈ R, u ∈ Cn , v ∈ Cm , if (ejω I − A)u = Bv, then
£
u∗
v∗
¤
M
u ·v
¸ 6= 0 and ¸ u >0 v
If (A, B) is controllable, then equivalence also holds between the two statements with all the occurrences of > replaced by >.
The main purpose of this paper, therefore, is to give a proof of the discrete-time KYP lemma (which [3, 4] did not touch on), also using the theorems of alternatives in [3, 4] as basic tools. The proof presented here has a similar structure to that of the BV proof. It is divided into two parts. One part, which we call the duality part, uses theorems of alternatives and some linear algebra. The other part, which we call the linear
In the sequel we shall refer to the first statement of Theorem 1 as the state-space statement, and the second statement as the frequency statement. Note that the above formulation is more general than the standard version in [2] in that it imposes no restriction on the location of the eigenvalues of A.
* This work was supported by the Chinese National Natural Science Foundation under Grant 60404010.
156
3 PROOF OF THE DISCRETE-TIME KYP LEMMA
Theorem ALT2 Let V , H, A, A0 be as in Theorem ALT1. Then at most one of the following statements is true.
This section is organized as follows. Subsections 3.1 and 3.2 introduce some basic concepts, and subsection 3.3 states three theorems of alternatives. In Subsection 3.4 (the duality part) we establish the equivalence between the state-space statement and an intermediate statement, then in Subsection 3.5 (the linear algebraic part) we establish the equivalence between the intermediate statement and the frequency statement. So these two parts constitute a full proof of Theorem 1.
1. There exists X ∈ V such that A(X) + A0 0. 2. There exists Z ∈ H such that Z > 0, Aadj (Z) = 0, and hA0 , ZiH 6 0. Moreover, if A0 = A(X0 ) for some X0 ∈ V , or if there exists no X ∈ V with A(X) 0, then exactly one of the statements is true. Theorem ALT3 Let V , H, A, A0 be as in Theorem ALT1. Then at most one of the following statements is true.
3.1 The Hilbert Spaces of Hermitian Matrices It is clear that H n constitutes an n2 -dimensional linear space over R. It is also possible to define the standard innerproduct associated with H n by hA, BiH n = tr(A∗ B) = tr(AB)
1. There exists X ∈ V such that A(X) + A0 > 0. 2. There exists Z ∈ H such that Z 0, Aadj (Z) = 0, and hA0 , ZiH < 0.
(1)
Moreover, if A0 = A(X0 ) for some X0 ∈ V , or if there exists no X ∈ V with A(X) 0, then exactly one of the statements is true.
We see that the inner-product always takes value in R, and H n is complete. Thus H n constitutes a finite-dimensional Hilbert space over R.
3.4 Duality Part of the Proof Theorem ALT1 will be used to establish Theorem 1 in the strict case, and Theorem ALT3 will be used in the nonstrict case. We notice that Theorem ALT3 generally gives a pair of weak alternatives, i.e., two statements at most one of which is true; they become strong alternatives when a certain condition holds. We shall use Theorem ALT2 to show that such a condition holds when the controllability condition specified in Theorem 1 holds. The following lemma is a well-known result on discrete Lyapunov equations [5].
3.2 The DKYP Linear Mappings and Their Adjoint Recall that given two Hilbert spaces V and W and a linear mapping A : V → W , the adjoint mapping of A, denoted by Aadj , is a linear mapping from W to V such that ∀x ∈ V , y ∈ W ,
hA(x), yiW = hx, Aadj (y)iV
The existence of an adjoint mapping is unique. Now we have a special class of so-called DKYP linear mappings which play a key role in the discrete-time KYP lemma. To each matrix-pair (A, B) ∈ Cn×n × Cn×m , there corresponds a DKYP linear mapping D : H n → H n+m defined by · ∗ ¸ A P A − P A∗ P B D(P ) = (2) B∗P A B∗P B
Lemma 1 Let A ∈ Cn×n and B ∈ Cn×m . If A has all its eigenvalues inside the unit disk and (A, B) is controllable, then there exists X ∈ H n with X > 0 and AXA∗ − X + BB ∗ = 0
It is easily verified that Dadj : H n+m → H n is given by ∗ Dadj (Z) =AZ11 A∗ − Z11 + BZ12 A∗ ∗ ∗ + AZ12 B + BZ22 B · ¸ Z11 Z12 where Z = . ∗ Z12 Z22
We further have the following lemma. Lemma 2 Given A ∈ Cn×n , B ∈ Cn×m , and the linear mapping D defined by (2). If (A, B) is controllable, then there exists no X ∈ H n such that D(X) 0.
(3)
Proof Since (A, B) is controllable, there exists K ∈ Cm×n such that A + BK has all its eigenvalues inside the unit disk. Obviously (A+BK, B) is also controllable, thus by Lemma 1, there exists W ∈ H n with W > 0 and
3.3 The Theorems of Alternatives The theorems of alternatives established in [3, 4] are very deep results on semidefinite programming duality, and their validity is ultimately based on the existence of a separating hyperplane between two disjoint convex sets. The following are the three theorems of alternatives that are directly relevant to our purpose.
(A + BK)W (A + BK)∗ − W + BB ∗ = 0 Let
· Z=
Theorem ALT1 Let V be a finite-dimensional Hilbert space over R and let H be one H n . Let A : V → H be a linear mapping and A0 ∈ H. Then exactly one of the following statements is true.
W KW
W K∗ I + KW K ∗
¸
then it is readily verified that Z ∈ H n+m , Z > 0 (by Schur complement formula), and ∗ AZ11 A∗ − Z11 + BZ12 A∗ + AZ12 B ∗ + BZ22 B ∗ = 0
1. There exists X ∈ V such that A(X) + A0 > 0. 2. There exists Z ∈ H such that Z 0, Aadj (Z) = 0, and hA0 , ZiH 6 0.
This amounts to saying that there exists Z ∈ H n+m such that Z > 0 and Dadj (Z) = 0. 157
Now in Theorem ALT2 we take V = H n , H = H n+m , A = −D, and A0 = 0. Because A(0) = A0 , Theorem ALT2 now provides “strong alternatives”, i.e. exactly one of the statements is true. Since the second statement is true, the first statement must be false. Therefore there exists no X ∈ H n such that A(X) 0, i.e., D(X) 0.
2. There exist ω ∈ R, u ∈ Cn , v ∈ Cm , such that · ¸ u 6= 0 v £
u∗
Then we are ready to give the main result of this subsection. Proposition 1 Let A, B, M be as in Theorem 1. Then the following two statements are equivalent. 1. There exists no P ∈ H n such that · ∗ ¸ A P A − P A∗ P B M− >0 B∗P A B∗P B
(10) (11)
The equivalence also holds between statement 1 with the 6 in (8) replaced by is equivalent to statement 2 with the 6 in (5) replaced by tr(M Z) · ¸ £ ¤ U = tr( U ∗ V ∗ M ) V· ¸ £ ¤ U = tr(Q∗ U ∗ V ∗ M Q) V ¸ · r £ ¤ P U = (qi∗ U ∗ V ∗ M qi ) V i=1 where qi is the i-th column of Q. Hence there must be k ∈ {1, · · · , r} such that
Lemma 3 Let F and G be complex matrices of the same size. Then F F ∗ = GG∗ if and only if there exists a unitary matrix R (R unitary means R square and RR∗ = I) such that F = GR.
qk∗
The main result of this subsection is as follows.
V
∗
¤
· M
U V
¸ qk 6 0
(14)
¸ U Now let u = U qk and v = V qk . The fact that has V full column rank and qk 6= 0 implies (9). Furthermore, it follows from (12),(13) that Au + Bv = U Rqk = σk U qk = σk u
(6)
Because R is unitary, σk must be on the unit circle, i.e. there is ω ∈ R such that σk = ejω , hence (10) holds. And finally, (14) implies (11).
∗ AZ11 A∗ +BZ12 A∗ +AZ12 B ∗ +BZ22 B ∗ = Z11 (7)
tr(M Z) 6 0
U
∗
·
Proposition 2 Let A, B, M be as in Theorem 1. Then the following two statements are equivalent. 1. There exists Z ∈ H n+m such that · ¸ Z11 Z12 Z= 0 ∗ Z12 Z22
£
(8) 158
· 2 ⇒ 1: Let Z =
u v
¸
£
¤ v ∗ . Then (9) implies (6).
u∗
in · the¸frequency statement for the strict case, the condition u 6= 0 is now changed to u 6= 0. Third, in the nonstrict v case the positive semidefiniteness of M22 totally disappears. Fourth, in ·the frequency statement for the nonstrict case, the ¸ u condition 6= 0 now disappears. v The first difference is made only to formulate an unconditional equivalence, it actually offers hardly anything more. The second difference actually changes nothing and is made only to simplify the writing. The third difference is the one we wish to highlight. The fourth difference is easy to understand, as we are working with > rather than > in the nonstrict case. As a part of the proof of Theorem 2, we now paraphrase the main result of the duality part of the BV proof as follows.
And = = = =
∗ AZ11 A∗ + BZ12 A∗ + AZ12 B ∗ + BZ22 B ∗ Auu∗ A∗ + Bvu∗ A∗ + Auv ∗ B ∗ + Bvv ∗ B ∗ (Au + Bv)(Au + Bv)∗ ejω u(ejω u)∗ uu∗ = Z11
by (10)
Finally, from (11) we have tr(M Z) =
£
u
∗
v
∗
¤
· M
u v
¸ 60
We are now in a position to obtain Theorem 1 as an easy consequence of Propositions 1 and 2.
Proposition 3 Let A, B, M be as in Theorem 2. Then the following two statements are equivalent.
4 REFINEMENT OF THE LINEAR ALGEBRAIC PART OF THE BV PROOF
1. There exists no P ∈ H n such that ¸ · ∗ A P + PA PB >0 M− B∗P 0
In this section we turn to the (continuous-time) KYP lemma. In the BV formulation of the KYP lemma, the positive (semi)definiteness of M22 is used as a hypothesis. Rantzer’s formulation, however, does not have such a hypothesis but instead it extends the domain of the frequency inequality to include ω = ∞. It turns out that in the strict case, explicit specification of the definiteness property of M22 , or the inclusion of ω = ∞ in the domain of the frequency inequality, is essential. But in the nonstrict case it is not, as we shall demonstrate. To facilitate further discussion, we now give our refined BV formulation of the KYP lemma.
2. There exists Z ∈ H n+m such that · ¸ Z11 Z12 Z= 0 ∗ Z12 Z22 ∗ Z11 A∗ + AZ11 + BZ12 + Z12 B ∗ = 0 tr(M Z) 6 0
(15)
(16)
If all uncontrollable modes of (A, B) are nondefective and correspond to imaginary eigenvalues, then statement 1 with the > in (15) replaced by > is equivalent to statement 2 with the 6 in (16) replaced by 0 B∗P 0
Proposition 4 Let A, B, M be as in Theorem 2. Then the following three statements are equivalent.
2. M22 > 0 and for all ω ∈ R, u ∈ Cn , v ∈ Cm·, if u ¸6= 0 £ ¤ u and (jωI − A)u = Bv, then u∗ v ∗ M > v 0.
1. There exists Z ∈ H n+m such that · ¸ Z11 Z12 Z= 0 ∗ Z12 Z22 ∗ Z11 A∗ + AZ11 + BZ12 + Z12 B ∗ = 0 tr(M Z) < 0
If all uncontrollable modes of (A, B) are nondefective and correspond to imaginary eigenvalues, then the following two statements are equivalent.
(17) (18) (19)
2. M22 0, or there exist ω ∈ R, u ∈ Cn , v ∈ Cm , such that
1. There exists P ∈ H n such that · ∗ ¸ A P + PA PB M− >0 B∗P 0
£
2. For all ω ∈ R, u ∈ C·n , v ¸∈ Cm , if (jωI − A)u = Bv, £ ¤ u then u∗ v ∗ M >0 v
u∗
u 6= 0 (jωI − A)u = Bv · ¸ ¤ u v∗ M < 0 v
(20) (21) (22)
3. There exist ω ∈ R, u ∈ Cn , v ∈ Cm , such that (21), (22) hold and · ¸ u 6= 0 (23) v
Let us note four differences between the above formulation and the original BV formulation. First, in the strict case the positive definiteness of M22 is now not a hypothesis but instead placed into the frequency statement. Second, 159
Furthermore, equivalence also holds between statement 1 with the < in (19) replaced by 6, and statement 2 with M22 0 replaced by M22 ≯ 0 and the < in (22) replaced by 6.
It is obvious that for sufficiently large ω ∈ R, Ãs ! −δ −δ −1 k(jω − A) Bvk2 < min , 3ρ1 6ρ2 kvk2
Remark Unlike before, the primary case in Proposition 4 now corresponds to the nonstrict case in Theorem 2. This is because we wish to highlight the nonstrict case. It is also important to note that we have only two equivalent statements in the secondary, strict, case.
Then we can choose such a ω, and set u = (jω − A)−1 Bv. It is clear ω, u, v satisfy (21)∼(23). For the secondary case, the equivalence between the modified statement 1 and the modified statement 2 can be shown by similar arguments to the preceding 1 ⇒ 2 and 2 ⇒ 1 arguments.
Proof We consider first the primary case. 1 ⇒ 2: Use the argument used in [3, Proposition 11 and Lemma 1]. m 2 ⇒ 1: If M22 0, then there · exists 0¸6= v ∈ C such 0 0 that v ∗ M22 v < 0. Let Z = , then Z satisfies 0 vv ∗ (17∼19). Otherwise, i.e., there are ω ∈· R, ¸u ∈ Cn , v ∈ ¤ u £ ∗ u v∗ . Cm such that (20)∼(22) hold. Let Z = v Then (20) implies (17). And = = = =
∗ + Z12 B ∗ Z11 A∗ + AZ11 + BZ12 ∗ ∗ ∗ ∗ uu A + Auu + Bvu + uv ∗ B ∗ (Au + Bv)u∗ + u(Au + Bv)∗ jωuu∗ − jωuu∗ 0
Finally, from (22) we have £ tr(M Z) = u∗
v∗
¤
· M
u v
We then obtain Theorem 2 from Propositions 3 and 4. In particular we have justified the disappearance of the positive semidefiniteness of M22 in the nonstrict case of Theorem 2. In fact, from the proof of Proposition 4 we can find that in the nonstrict case the positive semidefiniteness of M22 is automatically implied by both the state-space and the frequency statements.
5
We have presented a proof of the discrete-time KYP lemma based on Balakrishnan and Vandenberghe’s semidefinite programming duality theory, and proposed a refinement on Balakrishnan and Vandenberghe’s formulation and proof of the KYP lemma. These two pieces of work are well suited for control educational purposes.
by (21)
¸
REFERENCES
