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The effects of redundant control inputs in optimal control DUAN ZhiSheng† , HUANG Lin & YANG Ying State Key Lab for Turbulence and Complex Systems and Department of Mechanics and Aerospace Engineering, College of Engineering, Peking University, Beijing 100871, China

For a stabilizable system, the extension of the control inputs has no use for stabilizability, but it is important for optimal control. In this paper, a necessary and sufficient condition is presented to strictly decrease the quadratic optimal performance index after control input extensions. A similar result is also provided for H 2 optimal control problem. These results show an essential difference between single-input and multi-input control systems. Several examples are taken to illustrate related problems. redundant control, optimal control, Riccati equation, Hamilton matrix

1 Introduction and problem formulation Wonham[1] proved that a multi-input linear system can be transformed equivalently into a singleinput system by introducing a state feedback and finding an auxiliary vector in the column space of the input matrix. But for a long time, the essential differences between single-input linear systems and multi-input linear systems have not received much attention. It is known that if a single-input linear system is controllable, increasing the columns of the control input matrix does not affect the stabilizability, controllability and poles assignment of the system under state-feedback. But what is the case when other system performances are considered? What is the role of the increased columns of the input matrix when they are linearly dependent with the original columns? Clearly, these problems are fundamentally important in modern

control theory. Research on such problems would be of great importance to understanding the redundant control inputs, especially for modern aircraft and robot systems, where redundant actuators are widely used to improve system reliability and performance[2−7] . Modern aircrafts, especially warcrafts, often use redundant effectors, such as canard or thrust vectoring, to improve maneuverability and reliability. New generation aircrafts are even innovated aerodynamic layout, such as the Lockheed-Martin Innovative Control Effector (LM-ICE) aircraft[8] . Since inappropriate control allocation among different types of actuators may reduce the performance of the aircraft, the problem of control allocation has become an important research subject in modern aircraft control. Theoretically speaking, the dynamics of any system can be described

Received January 20, 2009; accepted August 7, 2009 doi: 10.1007/s11432-009-0189-3 † Corresponding author (email: [email protected]) Supported by the National Natural Science Foundation of China (Grant Nos. 90916003, 60674093, 60874011), and the Key Projects of Educational Ministry of China (Grant No. 107110)

Citation: Duan Z S, Huang L, Yang Y. The effects of redundant control inputs in optimal control. Sci China Ser F-Inf Sci, 2009, 52(11): 1973–1981, doi: 10.1007/s11432-009-0189-3

by the following nonlinear differential equation:

V0 A + AT V0 + C T C − V0 B0 R0−1 B0T V0 = 0,

x˙ = f (t, x, u), where x stands for the system states and u the system inputs. The function f (t, x, u) can describe the cases of input variation with system states, control redundancy and control effector failure. Since it is difficult to analyze the equation in such a general form, an effective way is to start from simple systems and then extend the results to complicated ones. For a linear time-invariant system, the redundancy of the control inputs is equivalent to the increase of the columns of the input matrix. Since the linear quadratic optimal control problem can be reduced to the solvability of an algebraic Riccati equation, in this paper, we consider the problem of redundant control for linear time-invariant systems with quadratic performance index. A necessary and sufficient condition of strict decreasing of the quadratic performance index is established when the columns of the input matrix are increased. Both theoretical analysis and numerical simulation results show that the system performance can be improved by increasing the number of control inputs. The effect of redundant control inputs on the H2 norm and the controller gain is also discussed. Considering theoretical study and engineering applications simultaneously[9−13] , there are many problems worth to be studied in the future. Given a linear system ( x˙ = Ax + B0 u0 , A ∈ Rn×n , B0 ∈ Rn×r0 , (1) y = Cx, C ∈ Rm×n with quadratic performance index Z +∞ J0 = (y T (τ )y(τ ) + uT 0 (τ )R0 u0 (τ ))dτ.

(2)

0

The following lemma shows a basic result in quadratic performance index optimal control problem, which is also called linear quadratic regulator problem by some authors[12] . Lemma 1[9,13] Suppose that (A, B0 ) is stabilizable, (A, C) is observable and R0 = R0T > 0. Then, the state feedback such that the index (2) achieves minimum is u0 = −R0−1 B0T V0 x, where V0 is the unique positive definite solution of the Ric1974

cati equation (3)

with the minimum performance index J0 = xT 0 V0 x0 , where x0 is the initial condition of (1). Quadratic performance index optimal control is closely related to the Riccati equation (3) which is generally solved by Hamilton matrix method. The corresponding Hamilton matrix of eq. (3) is ! A −B0 R0−1 B0T H0 = . −C T C −AT Increasing the number of the control inputs of (1), one gets ( x˙ = Ax + Bu, (4) y = Cx, T T n×r1 where B = (B0 , B1 ), u = (uT 0 , u1 ) , B1 ∈ R and the quadratic performance index of system (4) is Z +∞ J= (y T (τ )y(τ ) + uT (τ )Ru(τ ))dτ, (5) 0

where R is an extension of R0 , a compatible positive definite matrix with the form of ! R0 R01 (6) R= = RT > 0. T R01 R1 In the following we will discuss the relations between the performance indexes of systems (1) and (4).

2 The effects of the redundant control inputs in quadratic performance indexes In order to discuss the problem in a simple way, the following lemma is introduced to simplify the extension matrix R. Lemma 2. For an extension of R0 as in (6), if B1 − B0 R0−1 R01 6= 0, without loss of generality R can be supposed to be a diagonal extension ! R0 0 R= = RT > 0. (7) 0 R1 Proof. For any extended matrix R (6), and matrix B, the corresponding Riccati equation becomes V A + AT V + C T C − V BR−1 B T V = 0.

DUAN Z S et al. Sci China Ser F-Inf Sci | Nov. 2009 | vol. 52 | no. 11 | 1973-1981

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Taking T = ( −RTI1R−1 I02 ) and making a congruence 01 0 transformation on R, eq. (8) becomes V A + AT V + C T C − V BT T (T RT T )−1 T B T V = 0. Obviously, T RT T = ( R00 R1 −RT 0R−1 R01 ). On the 01 0 other hand, BT T = (B0 , B1 − B0 R0−1 R01 ). Based on this transformation, if B1 − B0 R0−1 R01 6= 0, any extension of R0 as in (6) can be supposed to be a diagonal extension as in (7) by changing B1 correspondingly. Remark 1. Clearly, if R(B1 ) 6⊂ R(B0 ), where R(·) denotes the column space of the corresponding matrix, then B1 − B0 R0−1 R01 6= 0 for any R; if R(B1 ) ⊂ R(B0 ), it is possible that B1 − B0 R0−1 R01 = 0. In this case, one can choose R01 such that B1 − B0 R0−1 R01 6= 0. So, the assumption on the diagonal characterization of R is not so restrictive. By Lemma 2, suppose that the matrix R has the diagonal structural characteristic (7) in Riccati equation (8). Then, eq. (8) becomes T

T

VA+A V +C C −V −V

B1 R1−1 B1T V

B0 R0−1 B0T V

= 0.

(9)

The Hamilton matrix corresponding to Riccati equation (9) is ! A −B0 R0−1 B0T − B1 R1−1 B1T H= . −C TC −AT To obtain the condition for the decrease of quadratic performance index, the following lemma is needed for the positive definiteness of the solution to Riccati equation. [12]

Suppose that (A, B0 ) is staLemma 3 . bilizable and (A, C) has no unobservable modes on the imaginary axis. Then, Riccati equation (3) has a positive semi-definite solution V0 , and A − B0 R0−1 B0T V0 is stable. In addition, V0 > 0 if and only if (A, C) has no stable unobservable modes. Combining the above discussions with Riccati equation theory, one can get the following result on the decrease of the solution to Riccati equation with the increasing columns of the control input matrix. Theorem 1. Suppose that (A, B0 ) is stabilizable and (A, C) is observable. Then, for Ric-

cati equations (3) and (9), V0 − V is positive semidefinite. Furthermore, Ker(V0 − V ) 6= {0} if and only if Hamilton matrices H0 and H have a common stable eigenvalue and corresponding common eigenvector simultaneously. Proof.

Subtracting (9) from (3), one gets

(V0 − V )A + AT (V0 − V ) − V B0 R0−1 B0T V0 − (V0 − V )B0 R0−1 B0T (V0 − V ) − V0 B0 R0−1 B0T V + 2V B0 R0−1 B0T V + V B1 R1−1 B1T V = 0.

(10)

Rewriting eq. (10) as (V0 − V )(A − B0 R0−1 B0T V ) + (A − B0 R0−1 B0T V )T (V0 − V ) − (V0 − V )B0 R0−1 B0T (V0 − V ) + V B1 R1−1 B1T V = 0,

(11)

or (V0 − V )(A − B0 R0−1 B0T V0 ) + (A − B0 R0−1 B0T V0 )T (V0 − V ) + (V0 − V )B0 R0−1 B0T (V0 − V ) + V B1 R1−1 B1T V = 0. By Riccati equation (3) and Lemma 3, A − B0 R0−1 B0T V0 is stable. Then, V0 − V is positive semi-definite according to the above equation. In the following, the condition for V0 > V will be discussed. If Ker(V0 − V ) 6= {0}, then by Lemma 3 and eq. (11), there exist x ∈ Ker(V0 − V ) and one eigenvalue λ of A − B0 R0−1 B0T V such that (A − B0 R0−1 B0T V )x = λx, B1T V x = 0. This eigenvalue is an unobservable mode of (A − B0 R0−1 B0T V, B1T V ). Simultaneously, λ and x are an eigenvalue of A − B0 R0−1 B0T V − B1 R1−1 B1T V and its corresponding eigenvector, i.e., (A − B0 R0−1 B0T V − B1 R1−1 B1T V )x = λx. In addition, by V0 x = V x, λ and x are also an eigenvalue of A − B0 R0−1 B0T V0 and the corresponding eigenvector, i.e., (A − B0 R0−1 B0T V0 )x = λx. Obviously, λ is a stable eigenvalue. By Riccati equation theory[12] , λ is also a common eigenvalue of Hamilton matrices H0 and H. The following discussion shows that H0 and H also have a common eigenvector. Combining Riccati equation with


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Hamilton matrix, ! I H0 = V0 ! I H = V

one gets ! I (A − B0 R0−1 B0T V0 ), V0 ! I (A − B0 R0−1 B0T V V

H (12)

(13)

−B1 R1−1 B1T V ).

Based on the discussions on eigenvalue and eigenvector, multiplying the right-hand side of the above two equations by x leads to ! ! I I H0 x=λ x, V0 V0 and H

I V

!

x=λ

I V

!

H0

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x y

!

=λ

x y

!

y

!

=λ

x y

!

.

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According to Hamilton matrix theory, under the assumptions of the theorem, X1 and Y1 are both nonsingular. Hence, the above equations can be rewritten as ! ! I I H0 = X1 Λ1 X1−1 , −1 −1 X2 X1 X2 X1 and ! ! I I H = Y1 Λ2 Y1−1 . −1 −1 Y2 Y1 Y2 Y1 Comparing the above equations with eqs. (12) and (13), one gets X1 Λ1 X1−1 = A − B0 R0−1 B0T V0 ,

x.

Y1 Λ2 Y1−1 = A − B0 R0−1 B0T V

(15)

−B1 R1−1 B1T V,

By V0 x = V x, ( VI )x is a common eigenvector of H0 and H corresponding to the eigenvalue λ. On the other hand, suppose that H0 and H have a common stable eigenvalue λ and the corresponding common eigenvector. According to the blocks of Hamilton matrix, suppose that this eigenvector is ( xy ). Based on the Hamilton matrix theory[12] , under the assumptions of the theorem, H0 and H have no eigenvalues on the imaginary axis, and all their eigenvalues are symmetric about the imaginary axis. Furthermore, assume that the Jordan standard form corresponding to stable eigenvalues of H0 and H are Λ1 and Λ2 , respectively, and the matrices composed of eigenvectors and generalized eigenvectors with stable eigenvalues of H0 and H 1 ) and ( YY12 ), respectively. Without loss of are ( X X2 generality, suppose that their first columns are the same, i.e., the common eigenvector of H0 and H, ( xy ). Then, ! ! X1 X1 H0 = Λ1 , X2 X2 ! ! Y1 Y1 H = Λ2 , Y2 Y2 and

x

,

where V0 = X2 X1−1 and V = Y2 Y1−1 are solutions to Riccati equations (3) and (9). Clearly, the first column of X1 and Y1 , x, is the common eigenvector of A−B0 R0−1 B0T V0 and A−B0 R0−1 B0T V −B1 R1−1 B1T V corresponding to the eigenvalue λ, and the first column of X2 and Y2 , y, satisfies V0 x = V x = y. Furthermore, by eq. (14) B1 R1−1 B1T y = 0, i.e., B1T y = 0. So B1T V x = 0. Combining the result with eq. (15), one observes that λ is a stable unobservable mode of (A−B0 R0−1 B0T V, B1T V ), whose the corresponding eigenvector x satisfies (V0 −V )x = 0. This leads to the conclusion. Remark 2. Let T0 = {x | xT V0 x = 1, V0 > 0, x ∈ Rn } and T = {x | xT V x = 1, V > 0, x ∈ Rn }, which denote two elliptical spheres in ndimensional space. If V0 > V , T strictly includes T0 . If V0 6> V , but V0 − V > 0, it implies that T0 and T are tangent at least at one point; that is, there is an x such that (V0 − V )x = 0. By the proof of Theorem 1, this characteristic determines one common eigenvector of H0 and H. This together with the discussions on the eigenvalue leads to the theorem. Corollary 1. Suppose that (A, B0 ) is stabilizable and (A, C) is observable. Hamilton matrices H0 and H have a common stable eigenvalue and common corresponding eigenvector if and only if H0 has a stable eigenvalue and its corresponding


eigenvector ( xy ) satisfies B1T y = 0. Furthermore, by V0 x = y, one has B1T V0 x = 0. Corollary 2. Suppose that (A, B0 ) is stabilizable and (A, C) is observable, and the extended control input matrix satisfies B1 = B0 . Then, Hamilton matrices H0 and H have a common stable eigenvalue and corresponding common eigenvector if and only if H0 and A have a common stable eigenvalue, and the corresponding eigenvector x of A satisfies B0T V0 x = 0. Remark 3. If Hamilton matrices H0 and H have a common stable eigenvalue, they do not necessarily have a common eigenvector. Theorem 1 requires that they have a common stable eigenvalue and eigenvector simultaneously. Example 1. Consider systems (1) and (4) with matrix data     −1 0 0 0     A =  0 0 1  , B1 = B0 =  0  , −2 1 3

1

C = ( 1 1 0 ), R1 = R0 = 1.

Since B1 = B0 , and −1 is an uncontrollable mode of (A, B0 ) and (A, [B0 , B1 ]), −1 is a common eigenvalue of H0 and H. However, in this case, A does not satisfy the condition of Corollary 2, so H0 and H do not have a common eigenvector. And correspondingly, the solutions to Riccati equations (3) and (9) satisfy V0 > V . Change B1 into B1 = (0, 1, 2)T . Then −1 is still an uncontrollable mode of (A, B0 ) and (A, [B0 , B1 ]). So, −1 is a common eigenvalue of H0 and H. And the corresponding eigenvector of H0 satisfies the condition of Corollary 1. In this case, V0 6> V , but V0 − V > 0. Combining Theorem 1 with Lemma 1, one gets the following result on the decrease of quadratic performance index of system (4).

(4), if one increases B0 repeatedly every time, after n times of extensions, there are n + 1 B0 in B = (B0 , B0 , . . . , B0 ). Suppose that R is an identity matrix with compatible dimension. Denote by Vn the solution to the Riccati equation corresponding to the nth extension system. If the Hamilton matrices do not have common stable eigenvalues and eigenvectors for every control input matrix extension, then B0T Vn B0 → 0 when n → ∞. Correspondingly, if the initial state x0 of the system belongs to the column space of B0 , the quadratic performance index approaches zero. Proof. By Theorem 1, one has V0 > V1 > V2 > · · · > Vn > · · · > 0. If Vn → 0, the conclusion holds. If Vn 6→ 0, there exists an infimum V (V 6= 0) being positive semi-definite such that V0 > V1 > V2 > · · · > Vn > · · · > V. For any arbitrarily small ǫ > 0, there exists a number N such that VN − V < ǫI. There are N + 1 B0 in the control input matrix B after N times of extensions, and then BB T = (N + 1)B0 B0T . So, the corresponding Riccati equation is VN A + AT VN + C T C − (N + 1)VN B0 B0T VN = 0. Similarly, after 2N times of extensions, the Riccati equation is V2N A+AT V2N +C T C −(2N +1)V2N B0 B0T V2N = 0. Similar to the proof of Theorem 1, computing the difference of the above two Riccati equations gives (VN − V2N )A + AT (VN − V2N ) − (N + 1)VN B0 B0T VN + (2N + 1)V2N B0 B0T V2N = 0.

Corollary 3. For systems (1) and (4), the quadratic performance indexes satisfy J 6 J0 . Besides, if the corresponding Hamilton matrices do not have common eigenvalues and eigenvectors, then J < J0 for any initial conditions.

Obviously, VN − V2N → 0, VN → V and V2N → V , when N → ∞. Then, one gets N V B0 B0T V = 0 when N → ∞ from the above equation, which leads to V B0 B0T V = 0. So, V B0 = 0 by positive semi-definiteness of V . Consequently, B0T V B0 = 0. Therefore, if the initial state x0 ∈ R(B0 ), one has x0 V x0 = 0 which means that the quadratic performance index is zero by Lemma 1.

Corollary 4. During the control input matrix extension process from system (1) to system

Example 2. Reconsider Example 1 with A and B0 as above. Then the solution to Riccati


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equation (3) is 

2.1677

 V0 =  0.1530

0.1530 −2.8153 2.2590

−2.8153 2.4142



 2.4142  . 6.7187

As discussed in Corollary 4, repeatedly increasing B0 in B and taking R as an identity matrix with compatible dimension, and extending B0 100 times, we have the solution to the corresponding Riccati equation (9)   0.3593 0.4109 0.0366   V =  0.4109 0.5288 0.1099  ; 0.0366 0.1099 0.0850

and extending B0 1000 times, we have the solution to the corresponding Riccati equation (9)   0.2117 0.2357 0.0221   V =  0.2357 0.2696 0.0326  . 0.0221 0.0326 0.0116

From this example, it follows that B0T V B0 → 0 with the extension of B0 .

3 The effects of redundant control inputs in H 2 optimal control It is well known that the robustness of the controller from quadratic performance optimization is generally bad. In this section, we consider the H2 optimal control problem. Consider system  n×n ,   x˙ = Ax + Bw w + B0 u0 , A ∈ R (16) z = C1 x + D12 u0 ,   y = C2 x + D21 w,

where w ∈ Rnw is arbitrary exogenous input, including plant disturbances, measurement noises, etc., u0 ∈ Rr0 is the control input, z ∈ Rnz is the regulated output and y ∈ Rny is the measured output. The following additional assumptions are generally made for standard H2 control problem: (i) (A, B0 ) is stabilizable and (A, C2 ) is detectable; T T (ii) D12 D12 = Ir0 and R2 = D21 D21 = Iny ; A−jwI B0 (iii) ( C1 D12 ) has full column rank for all w ∈ R; 1978

Bw (iv) ( A−jwI ) has full row rank for all w ∈ R. C2 D21 For simplicity, in this paper we assume that the T orthogonal condition D12 C1 = 0 holds. With these assumptions, one can get the following result on H2 optimal control problem[12] .

Lemma 4. There exists a unique optimal controller such that the minimum H2 norm of the closed-loop system is q γ = trace(LT XL) + trace(C1 Y C1T ), where X is the solution to the Riccati equation XA + AT X + C1T C1 − XB0 B0T X = 0,

(17)

Y is the solution to the Riccati equation T T Y (A − Bw D21 C2 )T + (A − Bw D21 C2 )Y T + Bw (I − D21 D21 )Bw − Y C2T C2T Y = 0,(18)

and the optimal dynamic output feedback controller is ( x˙ k = Ak xk − Ly, (19) u = F xk , T and Ak = A+ F = −B0TX, L = −Y C2T − Bw D21 B0 F + LC2 . Increasing the control inputs of (16), we get a new system    x˙ = Ax + Bw w + Bu, (20) z = C1e x + D12e u,   y = C2 x + D21 w,

where u and B are given as in (4), and C1e = ( C01 ), D12e = ( D012 Ir0 ). 1 Suppose that the Hamilton matrix corresponding to Riccati equation (17) is ! A −B0 B0T H0 = , −C1T C1 −AT and the corresponding new Hamilton matrix related to system (20) is ! A −BB T . H= −C1T C1 −AT By Theorem 1 and Lemma 4, we can get the following result. Theorem 2. Suppose that (A, C1 ) is observable and the above assumptions hold. If H0 and H have no common stable eigenvalues and corresponding eigenvectors simultaneously, then there exist optimal controllers such that the minimum


H2 norm of the closed-loop system corresponding to (20) is strictly smaller than the one corresponding to (16). Remark 4. Note that the two Riccati equations in Lemma 4 are separated. Then one can get F and L independently. Then by Theorem 1 and Lemma 4, if the solutions to the corresponding Riccati equation (17) strictly decrease for new systems with increased control inputs, the corresponding H2 norms will strictly decrease. For simplicity, this paper assumes the orthogonal condiT C1 = 0. When this condition does not tion D12 hold, one can get a similar result through tedious discussions on observability conditions.

4 Further topics for research Although a necessary and sufficient condition is given in Theorem 1 for the strict decrease of the solution to the Riccati equation corresponding to the extended control system (4), which shows the possibility of the strict decrease for the quadratic performance index of the system with extended control inputs, there are many problems left in the future studies. First, we are confronted with the controller gain problem. In this paper, let the largest singular value of the matrix denote the controller gain. The controller gain is important for the implementation of practical actuators. The larger the gain is, the faster the actuator is saturated. From a practical viewpoint, the controller gain should be small. In the quadratic performance index, it includes a term uT Ru. Intuitively, if the quadratic performance index decreases, the controller gain should be smaller. However, the controller is given by u = −R−1 B T V x. The extension of the control input matrix possibly makes the controller gain larger. On the other hand, the decrease of matrix V can make the controller gain smaller. All these factors make it hard to solve the problem. In addition, how can we extend the control input matrix? And will the controller gain decreases faster? Examples show that the controller gain may either increase or decrease. Example 3.

Consider system (1) with matri-

ces 

  0     A =  0 0 1  , B0 =  0  , −2 1 3 1 C = 1 1 0 , R0 = 1. −1 0 0



Here the optimal controller for (2) is K = (2.8153, −2.4142, −6.7187). The controller gain is 7.6743. Taking the extended matrix B and weighted matrix R as   ! 0 0 1 0   Be1 =  0 0  , R = 0 1 1 1 leads to the controller K=

1.3060 −1.3660 −3.4016 1.3060 −1.3660 −3.4016

!

whose gain is 5.5032. On the other hand, taking the extended matrix B and weighted matrix R as   ! 0 0 1 0   Be2 =  0 1  , R = 0 1 1 0 leads to the controller K=

2.9535 −1.6950 −6.0850 1.5116 −1.2317 −1.6950

!

whose gain is 7.2170. In Example 3, change A into   −1 0 0   A =  0 0 1 , −2 1 −3

Example 4.

and keep the other matrices the same as above. For this matrix A, the optimal controller for (2) is K = (0.8586, −2.4142, −0.7187) whose gain is 2.6612. For the extended matrix Be1 , the optimal controller is ! 0.3879 −1.3660 −0.4016 K= 0.3879 −1.3660 −0.4016 whose gain is 2.0870.


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For the extended matrix Be2 , the optimal controller is ! 0.0125 −0.3050 −0.0850 K= −0.1708 −1.2317 −0.3050 whose gain is 1.3178. The above two examples show that the controller gain deceases with the control input matrix extensions. Clearly, the controller gain decreases faster in Example 3 for the extended matrix Be1 , but it decreases faster in Example 4 for the extended matrix Be2 . Change A again into   −1 0 0   A= 0 0 1 ,

Example 5.

−2 −1 −3

and keep the other matrices the same as in Example 3. For this new matrix A, the optimal controller for (2) is K = (0.0177, −0.4142, −0.1350) whose gain is 0.4360. For the extended matrix Be1 , the optimal controller is ! −0.0056 −0.3660 −0.1174 K= −0.0056 −0.3660 −0.1174 whose gain is 0.5437. For the extended matrix Be2 , the optimal controller is ! −0.0384 −0.1865 −0.0559 K= −0.3286 −0.7695 −0.1865 whose gain is 0.8790. Comparing Example 5 with Examples 3 and 4, one can see that the controller gain increases with the control input matrix extensions in this example. It is interesting and important to discover a rule for the increase or decrease of controller gains with control input matrix extensions. Second, let us consider the changes of the L∞ norms of the transfer functions. Although one can see that the optimal performance index decreases continually with the control input matrix extensions from Corollary 4 and Example 2, it is hard to be implemented in practical systems because of the increase of the complexity and cost of controllers. In addition, as the practical system is generally 1980

nonlinear, system (4) can be viewed as a linearized system of the real nonlinear system. Along with the extension of the control input matrix, the error of the linearization will be larger and larger. During the extending process of the control input matrices, faster increase of the error should be avoided. To this end, the analysis of the norms of the linearized systems can give some implications. For example, the transfer function of system (1) is G0 (s) = C(sI − A)−1 B0 , whose L∞ norm[12] is supposed to be γ0 . After N times of extensions in system (4) with repeatedly increasing B0 in B, the transfer function is G(s) = C(sI − A)−1 B satisfying G(s)GTp (−s) = (N + 1)G0 (s)GT 0 (−s) whose L∞ norm is γ0 (N + 1). The increase of L∞ norms shows the increase of input and output errors to some degree. Then, how can we extend the matrix B and will the increase of the norms of transfer functions be made slower? The properties of the transfer functions after control input matrix extensions are worth further studying. Third, we should deal with the problems of finite time optimization and nonlinear optimal control. In many practical problems, finite time optimizations are often encountered. In such problems, time-varying Riccati equations are involved in quadratic performance optimization. Can a similar result to Theorem 1 be obtained in finite time problems? Further, how can we solve the nonlinear optimal control problems? The nonlinear optimal control problem is much harder to solve than the quadratic performance problem in linear systems. To find a useful algorithm is a right choice.

5 Conclusions In this paper, by using Riccati equation theory, a necessary and sufficient condition is presented for the strict decrease of solutions to Riccati equations corresponding to the systems with control input matrix extensions, which leads to a decrease inquadratic performance indexes for any initial states. A similar result is presented for H2 optimal control problems. Other related problems are also illustrated through taking some examples. The result shows that a repeated extension of the control input matrix can also result in a decrease in the


performance index. This paper gives a theoretical result on the effects of the redundant control inputs in quadratic performance optimal control problems

of linear systems. Many problems related to practical applications will be further studied in the future.

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