3. STUDY OF ROBUSTNESS PERFORMANCE OF SDRE+ISMC
4.2. Design of Active Reliable Control Law …
4.3.1. Design of Fault Diagnosis and Detection (FDD)
−ρ(x, t) [D ˜BH(x, t)]Ts
||[D ˜BH(x, t)]Ts|| + ∆dm
)
≤ −||[D ˜BH(x, t)]Ts|| · ρ(x, t) + ||[D ˜BH(x, t)]Ts|| · ||∆dm||
≤ ||[D ˜BH(x, t)]Ts|| · [−ρ(x, t) + ρm(x, t)]
< 0.
To sum up with an important theorem,
Theorem 4.1 : Suppose that System (4.1) experiences actuator faults at the control channels in F with estimated value ˆuF and error ∆uF given by FDD mechanism (4.5).
Then the origin of System (4.1) under Assumptions 4.1 - 4.4 and the control law given by (4.13)-(4.14) is globally asymptotically stable (GAS).
4.3 Application to Satellite Attitude Control
In this section, we use the same satellite attitude control model as in Section 3.3.1.
In the following, we first detail the design of fault detection and diagnosis (FDD) and compare the simulating results using different control methods.
4.3.1 Design of Fault Diagnosis and Detection (FDD)
In this section, we investigate the deign of FDD observer mentioned in Section 3.3.1 for the satellite attitude control. The main idea of this design is to decouple the control input so that the fault associated with each channel can be diagnosed and distinguished from the healthy ones. And the following system dynamics, same as (3.1), is considered.
( ˙x1 = x2
˙x2 = ˜f(x, t) + ˜B(x, t)u + ˜d. (4.15)
Fig. 4.1. FDD diagram
Fig. 4.1 shows the relation between FDD and system. Since the three Euler rates can be expressed in terms of angular velocity vector, which is available through accelerometer and gyroscope [34], in this section, we assume that all of the state variables are available for measurement and that ˜B(x, t) in (4.15) is a constant matrix. We adopt the observer and residual signals ri from [32] as (4.16) and (4.17) below:
ξi = finew(z) + ui+ liu4+ ki· (zi+3− ξi), i = 1, 2, 3 (4.16) and
ri = zi+3− ξi, i = 1, 2, 3 (4.17) where ki > 0. It was shown in [32] that any single actuator fault can be detected and diagnosed at an exponential rate depending on ki.
When the residual signals are larger than a selected threshold, the alarm will be set to be on.
4.3.2 Simulation Results
In this section, we still use MATLAB software to simulate the satellite attitude control under SDRE and ISMC approach. For both control approaches, we check whether the sys-tem with disturbances can be stabilized and compare their performances (e.g. quadratic performance index and convergence time).
The following Table 4.1 shows the simulating parameters in this chapter: (Note that for SDRE approach, the procedure of factorizing f = A(x, t)x is described in Appendix)
Table 4.1. Simulation parameters.
Ix 2000 N · m · s2
Iy 400 N · m · s2
Iz 2000 N · m · s2
ω0 1.0312 × 10−3 rad/s
d˜ (0.01 sin(t), 0.01 cos(2t), 0.01 sin(3t))T A(x, t) see Appendix 3A in Chapter 3
D I3
Q I6
R I4
u0 SDRE approach
ρ(x, t), ρm(x, t) ||∆dm||∞+ 0.5
x0 (0.7, 0.07, −1.5, −0.3, −1.3, 0.2)T
k1 10
k2 10
k3 10
Furthermore, to alleviate chattering, we modify the control law (4.13) into:
u =
B˜H+(x, t)[ ˜B(x, t)k0− ˜BF(x, t)ˆuF] − ρ(x, t)||[D ˜[D ˜BBHH(x,t)](x,t)]TTss|| if ||[D ˜BH(x, t)]Ts|| ≥ ǫ B˜H+(x, t)[ ˜B(x, t)k0− ˜BF(x, t)ˆuF] − ρ(x, t)[D ˜BH(x,t)]ǫ Ts if ||[D ˜BH(x, t)]Ts|| < ǫ(4.18) where we choose ǫ = 0.02. We simulate the faulty situation by that u2 fails at time 1 and alarm signals as soon as |ri| ≥ 0.01.
The simulation results are shown in Figs. 4.2-4.6, and the summary of comparison of performance are shown in Table 4.2.
We denote the results:
• SDRE : the system without disturbance (nominal system) under SDRE approach only
• SDREr : the disturbed and actuator-failed system under SDRE approach only
• SDRE+ISMCr : the disturbed and actuator-failed system using SDRE-ISMC com-bined approaches
In addition, in Table 4.2, we also compare the performances under the Sliding Mode Control (SMC, see Section 4.2.2 in [42]), LQR (see Section 4.2.3 in [42]), and LQR-ISMC combined approach (see Section 4.2.1 in [42]), respectively.
• SMCr : the disturbed and actuator-failed system under nonlinear SMC approach only
• LQR : the system without disturbance (nominal system) under nonlinear LQR ap-proach only
• LQRr : the disturbed system and actuator-failed under nonlinear LQR approach only
• LQR+ISMCr : the disturbed and actuator-failed system using LQR-ISMC combined approaches
It is observed from Fig. 4.2 that the stabilization performance is, as expected, achieved for the SDRE and the SDRE+ISMC designs. Besides, the state trajectories of the ISMC and those for nominal design (SDRE) are found almost identical, which agrees with the theoretical conclusion. From Fig. 4.6, the sliding variables of the SDRE+ISMC design are seen to keep at zero all the time. It implies that the system states remain on the sliding manifold for all t, which also agrees with the main results. In Fig. 4.4, the actuator fault is successfully detected by both designs, since the magnitude of the second residual signal exceeds the threshold near tSDRE+ISMC ≈ 1.04 and tSDRE ≈ 1.067, respectively. This can also be seen from the alarm signals given in Fig. 4.5 where alarm2 denotes the fault of the second actuator. After the fault is successfully detected, the associated active reliable controllers are activated and the magnitude of the residual signals soon decreases, as shown in Fig. 4.4. The persistent oscillation of the residual signal comes from the effect of the disturbance ˜d, which also contributes to the oscillating control inputs (u1, u3, u4) of SDREr and SDRE+ISMCr in Fig. 4.3. It is also noted from Figs. 4.3 that SDRE+ISMC design is observed to require larger control efforts than SDRE design due to the additional control u1 in (4.13) and (4.14). Finally, since the SDRE+ISMC design of this example adopts the SDRE scheme for the nominal healthy subsystem, its performances are close to those of SDRE except for the requirement of extra control component to compensate for the uncertainties.
Table 4.2 shows the comparison of performance, including energy consumptionR uTu,
quadratic performance index R(xTx + uTu), required maximum control magnitude ||u||∞, and convergence time (when the magnitude of state is less than 0.01 at first time). For nominal system, LQR [42] approach has better performance than SDRE for all consid-ered performance indexes. For the system with disturbances, LQR+ISMCr approach also has better performance than SDRE+ISMCr. Moreover, both SDRE+ISMCr and LQR+ISMCr consumes more control energy than the corresponding nominal control law SDRE and LQR [42], respectively. This is because the additional part, u1 in (4.13) and (4.14), is required in the ISMC design. Last but not least, we see that SMCr [42] and SDREr succeeds to stabilize, but LQRr [42] fails. To sum up, we conclude that in this study SDRE control law possesses certain robustness but not reliable.
0 5 10 15 20
Fig. 4.2. Time history of the six state variables.
0 5 10 15 20
Fig. 4.3. Time history of the four control inputs.
0 2 4 6 8 10 12 14 16 18 20
Fig. 4.4. Time history of the three residual signals.
0 2 4 6 8 10 12 14 16 18 20
Fig. 4.5. Time history of the three alarm signals.
0 2 4 6 8 10 12 14 16 18 20
Fig. 4.6. Time history of the three sliding variables (SDRE+ISMCr).
Table 4.2. Comparison of performance.
Performance Index Controller
||x(t)||t→∞ <10−3 R uTu R xTx R(xTx + uTu) ||u||∞ Convergence time
LQR+ISMCr 1.6149 4.4142 6.0291 2.1232 8.844
SDRE+ISMCr Yes
4.0208 4.7619 8.7827 2.3917 11.61
SMCr Yes 1.8763 6.0829 7.9592 2.2829 7.094
LQRr No X X X X X
SDREr Yes 3.5434 5.066 8.6094 2.3917 16.982
LQR 1.5576 4.4156 5.9732 2.1232 8.799
SDRE Yes
2.7756 4.7638 7.5395 2.3917 11.598
CHAPTER FIVE
ON FACTORIZATION OF THE DRIFT TERM IN SDRE SCHEME
5.1 Problem Statement
Although the SDRE algorithm fully captures the nonlinearities of the system, bringing the nonlinear system to a (non-unique) linear structure having state-dependent coefficient (SDC) matrices, and minimizing a nonlinear performance index having a quadratic-like structure, it has some drawbacks. First, it is known that the conditions“[A(x, t), B(x, t)]
is stabilizable” and“[A(x, t), C(x)] is observable” are required for the existence of a unique positive definite solution P (x) in Eq. (2.3) [24]; however, these symbolic checking con-ditions are in general not easy to implement, especially when the system dynamics is complicated. Next, there is no guideline provided for the factorization f(x, t) = A(x, t)x.
To avoid these difficulty, in this study, we consider the following approach instead.
Problem A: At any nonzero state x and time t, f := f(x, t) is a constant vector, while B := B(x, t) and C := C(x) are constant matrices. Find a matrix A := A(x, t) ∈ IRn×n pointwise such that Ax = f, (A, B) is stabilizable and (A, C) is observable.
To demonstrate the benefits of the alternative approach, we give an example below which shows the traditional SDRE scheme does not work when a specific factorization of f = Ax is adopted, but the alternative approach do work.
Example: Let f = (x1+ x21x32, x21x22)T, B = (0, 1)T and C = I2. Suppose that a specific factorization for f = Ax is given as A := 1 x21x22
0 x21x2
!
. Clearly, (A, C) is observable, but (A, B) is not stabilizable when x1 = 0 or x2 = 0. Thus, the SDRE, given by (2.3), might fail to have a positive definite solution P (x) when x1 = 0 or x2 = 0, which will result in the SDRE scheme failing to operate. However, since Q(x) = CT(x)C(x) = I2, Problem
A is solvable for this case (see Corollary 5.2).
It is also worth noting that Problem A is always solvable if the SDRE problem for some specific factorization can be continuously operated. We first consider the Problem A at a specific nonzero state, as described in Problem B below:
Problem B: Given two constant vectors x, f ∈ IRn, and two constant matrices B ∈ IRn×m and C ∈ IRp×n with x 6= 0, rank(B) ≥ 1 and rank(C) ≥ 1, when does there exist a matrix A ∈ IRn×n pointwise such that Ax = f, (A, B) is stabilizable and (A, C) is observable?
Note that, Problem A (and B) are always solvable for the case of n = 1. Therefore, in the following we only consider the case of n > 1. To answer Problem B, we denote (IRn)∗ = {xT|x ∈ IRn}, which is known to be the dual space of IRn [22]. Suppose that p1, · · · , pk ∈ IRn and qT1, · · · , qTl ∈ (IRn)∗. We denote {p1, · · · , pk}⊥ = {qT ∈ (IRn)∗ | qTpi = 0 for 1 ≤ i ≤ k} and {qT1, · · · , qTl }⊥ = {p ∈ IRn|qTi p = 0 for 1 ≤ i ≤ l}.
In addition, we denote B⊥ := {qT ∈ (IRn)∗|qTB = 0} and C⊥:= {p ∈ IRn|Cp = 0}.
5.2 Solvability Condition
We assume that the matrix A is diagonalizable in the form of
A = MDM−1 (5.1)
where D =diag[λ1, · · · , λn] ∈ IRn×n, M = [p1, · · · , pn] ∈ IRn×n, M−1 = [q1, · · · , qn]T ∈ IRn×n, and λ1, · · · , λn are distinct. Clearly, λ1, · · · , λn are the eigenvalues of A, pi and qTi are the right and the left eigenvectors of A associated with eigenvalues λi, respectively.
We have the following lemma:
Lemma 5.1 Let A be factorized in the form of (5.1). Then
(i) Ax = f ⇐⇒ λiqTi x = qTi f for all i ⇐⇒ qTi (λix − f) = 0 for all i.
(ii) (A, C) is observable if and only if pi 6∈ C⊥ for all i = 1, · · · , n.
(iii) (A, B) is controllable if and only if qTi 6∈ B⊥ for all i = 1, · · · , n.
(iv) (A, B) is stabilizable if and only if qTi 6∈ B⊥ whenever λi ≥ 0.
Proof: (i) The result follows from writing Ax = f in the form of DM−1x = M−1f and then comparing both sides componentwise.
(ii) It is known from the PBH test [8] that the pair (A, C) is observable if and only if rank
C λiI − A
!!
= n for all i = 1, · · · , n, i.e., C λiI − A
!
p 6= 0 for any p 6= 0.
It is clear that (λiI − A)p = 0 if and only if (λi, p) is an eigenpair of A or p = 0. It follows that (A, C) is observable if and only if Cpi 6= 0 for all i, that is, pi 6∈ C⊥ for all i = 1, · · · , n.
(iii) It is known that (A, B) is controllable if and only if (AT, BT) is observable [8].
Since (λi, qi), i = 1, · · · , n, are eigenpairs of AT, we have from the proof of (ii) that (AT, BT) is observable if and only if BTqi 6= 0, i.e., qTi B 6= 0, for all i. Thus, (A, B) is controllable if and only if qTi 6∈ B⊥ for all i.
(iv) (A, B) is stabilizable if and only if rank([λiI − A ... B]) = n for those i in which λi ≥ 0 [8]. This is equivalent to qTi B 6= 0 whenever λi ≥ 0, that is, qTi 6∈ B⊥ whenever λi ≥ 0.
We also need the following three results:
Lemma 5.2 Let V be a k dimensional vector subspace of (IRn)∗, k < n, and {qT1, · · · , qTk} are linearly independent (LI) vectors with qT1 6∈ V. Then there exists qTk+1 ∈ V such that {qT1, · · · , qTk+1} are LI.
Proof: Suppose that such qTk+1 does not exist. Then V ⊂ span{qT1, · · · , qTk}. Since both V and span{qT1, · · · , qTk} have dimension k, we have V =span{qT1, · · · , qTk}, and thus qT1 ∈ V, a contradiction. This completes the proof.
Lemma 5.3 Let V be a k−1 dimensional vector subspace of (IRk)∗ and {vT1, · · · , vkT} be a basis of (IRk)∗with vTi 6∈ V for all i. Besides, let Wi := span{vT1, · · · , viT−1, vTi+1, · · · , vTk}.
Then V 6⊂ ∪ki=1Wi. As a result, there exists a nonzero vT ∈ V such that vT =Pki=1αivTi and αi 6= 0 for all i = 1, · · · , k.
Proof: Note that, for all i = 1, · · · , k, Wi is a vector space of dimension k − 1 and V 6= Wi; Otherwise, vTj ∈ V for all j 6= i, which contradicts to the assumption vjT 6∈ V for all j. Since ∪ki=1Wi is not a vector space, we thus have V 6⊂ ∪ki=1Wi. This fact together with {vT1, · · · , vTk} being a basis implies there exists a nonzero vT ∈ V such that vT =Pki=1αiviT with αi 6= 0 for all i; Otherwise, each v ∈ V will belong Wi for some i, which contradicts V 6⊂ ∪ki=1Wi.
Lemma 5.4 Let {qT1, · · · , qTn−1, cT} are LI and qTn := αccT +Pnj=1−1αjqTj, αc 6= 0 and αj 6= 0 for all j = 1, · · · , n − 1. Then
(i) {qT1, · · · , qTn} are LI.
(ii) For any i ∈ {1, · · · , n}, the n vectors {qT1, · · · , qTi−1, qTi+1, · · · , qTn, cT} are LI.
(iii) For any i ∈ {1, · · · , n}, {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥6⊂ (cT)⊥.
Proof: (i) Suppose that Pni=1kiqTi = 0T. Inserting the expression of qTn into the equation yields Pni=1−1(ki+ knαi)qTi + knαccT = 0T. Since {qT1, · · · , qTn−1, cT} are LI, we have knαc = 0 and ki+ knαi = 0 for all i = 1, · · · , n − 1. Since αc 6= 0, we have kn = 0 and ki = 0 for i = 1, · · · , n − 1. This proves the linear independency of {qT1, · · · , qTn}.
(ii) Suppose that Pj6=ikjqTj + kccT = 0T. Inserting qTn into the equation, we have
Pn−1
j6=i(kj+ knαj)qTj + knαiqTi + (knαc+ kc)cT = 0T. Since {qT1, · · · , qTn−1, cT} are LI and αi 6= 0, we have from the coefficient of qTi that kn = 0, and thus kc = 0 and kj = 0 for all j 6= i and j ≤ n − 1. Thus, {qT1, · · · , qTi−1, qTi+1, · · · , qTn, cT} are LI.
(iii) Suppose, on the contrary, that {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ ⊂ (cT)⊥. Then any nonzero vector p ∈ {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ has the property cTp = 0 and qTjp = 0 for all j 6= i. Since {qT1, · · · , qTi−1, qTi+1, · · · , qTn, cT} is a basis for (IRn)∗, it follows that p must be a zero vector, which contradicts the fact that {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ is a vector space of dimension 1. This proves that {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ 6⊂ (cT)⊥.
In the following, we denote IR− the set of negative real numbers. A necessary and sufficient condition for Problem B is now stated as Theorem 5.1 below:
Theorem 5.1 Problem B is unsolvable if and only if {x, f} are linearly dependent (LD) and Cx = 0.
Proof: We divide the proof into the following four cases:
Case 1: ({x, f} are LI and C[x, f] 6= 0)
Note that, C[x, f] 6= 0 implies that there exists a nonzero row vector cT of C with cT 6∈ {x, f}⊥. Choose λ1, · · · , λn ∈ IR−such that the n vectors {λix − f | i = 1, · · · , n} are distinct and cT(λix−f) 6= 0 for all i = 1, · · · , n. If n > 2, since dim((λix−f)⊥) = n−1 for all i, we may easily choose qTi ∈ (λix − f)⊥, 1 ≤ i ≤ n − 2, satisfying qT1 6∈ (λn−1x − f)⊥,
qT1(λnx − f) > 0, qTi(λnx − f) ≥ 0 for i = 2, · · · , n − 2 and {qT1, · · · , qTn−2, cT} are LI.
Since qT1 6∈ (λn−1x − f)⊥ and dim((λn−1x − f)⊥) = n − 1, it follows from Lemma 5.2 that there exists a qTn−1 ∈ (λn−1x − f)⊥ such that {qT1, · · · , qTn−1, cT} are LI. We also select qTn−1 satisfying qTn−1(λnx − f) ≥ 0. Define qTn = αcT +Pni=1−1qTi, α = −[Pni=1−1qTi (λnx − f)]/[cT(λnx−f)]. Clearly, α 6= 0 since cT(λnx−f) 6= 0, qT1(λnx−f) > 0 and qTi (λnx−f) ≥ 0 for 2 ≤ i ≤ n − 1. Moreover, it is easy to check that qTn(λnx − f) = 0. Thus, from (i) of Lemma 5.1, we have Ax = f. From the structure of qTn, the fact pi ∈ {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ and Lemma 5.4, we have pi 6∈ (cT)⊥ for all i. This together with C⊥⊂ (cT)⊥ and (ii) of Lemma 5.1 implies that (A, C) is observable. Finally, (A, B) is stabilizable since λi ∈ IR− for all i. Thus, Problem B is solvable. If n = 2, the proof of this case can also be easily derived if we choose qT1 ∈ (λ1x − f)⊥ and qT1(λ2x − f) > 0.
Case 2: ({x, f} are LI and C[x, f] = 0)
This case implies that each nonzero row vector cT of C satisfies cT ∈ {x, f}⊥. Similar to that of Case 1, we choose λ1, · · · , λn ∈ IR−such that the n vectors {λix − f | i = 1, · · · , n}
are distinct. Suppose that n > 2. Since (λix − f)⊥∩ c⊥ is a vector space of dimension n − 2 for all i, we may choose qTi ∈ {(λix − f)⊥∩ c⊥}\(λnx − f)⊥ for 1 ≤ i ≤ n − 2 and qT1 6∈ (λn−1x − f)⊥ such that {qT1, · · · , qTn−2} are LI. Since W := (λn−1x − f)⊥ ∩ c⊥ has dimension n − 2, W ⊂ V := c⊥ and qT1 6∈ W, we have from Lemma 5.2 that there exists a vector qTn−1 ∈ W\(λnx − f)⊥ such that {qT1, · · · , qTn−1} are LI. Under these settings, {qT1, · · · , qTn−1, cT} are also LI since qTi c = 0 for all 1 ≤ i ≤ n − 1. Now, from Lemma 5.3, there exists a vT ∈ (λnx − f)⊥ such that vT = Pni=1−1αiqTi and αi 6= 0 for all i.
Since both cT and vT belong to (λnx − f)⊥, we have qTn := cT + vT ∈ (λnx − f)⊥. Thus, from (i) of Lemma 5.1, we have Ax = f. Besides, from the structure of qTn, the fact pi ∈ {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ and Lemma 5.4, we have pi 6∈ (cT)⊥ for all i.
This together with C⊥ ⊂ (cT)⊥ and (ii) of Lemma 5.1 implies that (A, C) is observable.
Finally, (A, B) is stabilizable since λi ∈ IR− for all i. Thus, Problem B is solvable. The case for n = 2 can be similarly proved if we choose qT1 ∈ (λ1x − f)⊥\(λ2x − f)⊥.
Case 3: ({x, f} are LD and Cx 6= 0)
Let cT be a nonzero row vector of C such that cTx 6= 0, and b be a nonzero column vector of B. We choose n − 1 distinct real numbers λ1, · · · , λn−1 ∈ IR−, and n − 1 LI row
vectors qT1, · · · , qTn−1 ∈ x⊥. This implies that {qT1, · · · , qTn−1, cT} are LI since cT 6∈ x⊥. If {x, b} are LD (i.e., x⊥ = b⊥), we choose qTn := αcT +Pni=1−1qTi , α 6= 0. It follows that qTn 6∈ b⊥, and thus qTn 6∈ B⊥ since B⊥ ⊂ b⊥. On the other hand, if {x, b} are LI, the above-mentioned qT1, · · · , qTn−1 may be chosen from x⊥ satisfying qT1b > 0 and qTi b ≥ 0 for all i = 2, · · · , n − 1. It follows thatPni=1−1qTi b > 0, and therefore there exists a nonzero constant α such that (αcT +Pni=1−1qTi )b = αcTb +Pni=1−1qTi b 6= 0 no matter cTb is zero or not. Here, we also choose qTn := αcT +Pni=1−1qTi as before. Clearly, qTn 6∈ B⊥ since qTn 6∈ b⊥ and B⊥ ⊂ b⊥. Finally, we choose λn such that qTn(λnx − f) = 0. From these discussions, we have qTi (λix − f) = 0 for all i = 1, · · · , n, which implies from (i) of Lemma 5.1 that Ax = f. Besides, due to the special structure of qTn and (ii) of Lemma 5.4, we have {qT1, · · · , qTi−1, qTi+1, · · · , qTn, cT} are LI for any i = 1, · · · , n. This fact together with (iii) of Lemma 5.4 and pi ∈ {qT1, · · · , qTi−1, qTi+1, · · · , qTn}⊥ leads to cpi 6= 0 for all i, and thus Cpi 6= 0 for all i. That is, by (ii) of Lemma 5.1, (A, C) is observable. Since λ1, · · · , λn−1 ∈ IR− and qTn 6∈ B⊥, (A, B) is stabilizable by (iv) of Lemma 5.1. Thus, Problem B is solvable.
Case 4: ({x, f} are LD and Cx = 0)
Since {x, f} are LD, we have f = λx for some constant λ. Suppose that there exists A such that Ax = f. Then Ax = f = λx, and thus (λ, x) is an eigenpair of A. This fact together with the condition Cx = 0 results in C
λI − A
!
x = 0, which implies that (A, C) is unobservable and Problem B is unsolvable.
Summarizing the above three cases gives the result.
From Theorem 5.1, we have the next two trivial results:
Corollary 5.1 Problem B is solvable if and only if Cx 6= 0 or {x, f} are LI.
Corollary 5.2 Problem A is always solvable if any one of the following two conditions holds:
(i) Q(x) is a nonsingular matrix for all x 6= 0.
(ii) Q(x) = Q is a constant matrix and rank(Q) = n.
5.3 Implementation
We look forward to implement by usage of orthogonal matrices because the condition number of them equals 1 [40]. In particular, we choose the Householder matrix [20], denoted as H ∈ IRn×n, which is orthogonal and possesses following properties:
1. H = I − 2vvT, where I is the (n × n) identity matrix, v ∈ IRn is an unit vector.
2. -1 is one eigenvalue of H with corresponding eigenvector, v. Moreover, 1 of multi-plicity (n-1) are the other eigenvalues with corresponding eigen-space being perpen-dicular to v.
3. HT = H, det(H) = −1, and tr(H) = n − 2.
4. Let H′ = H2H1, where Hi = I − 2viviT, ||vi|| = 1, i = 1, 2, then H’ is an orthog-onal matrix. Moreover, if v1 and v2 are LI, then the subspace perpendicular to span(v1, v2) is the eigen-space of H′ corresponding to eigenvalue 1 of multiplicity (n − 2).
To implement, we need the following lemmas:
Lemma 5.5 Assume that A ∈ IRn×n is orthogonal, diagonalizable, and has four distinct eigenvectors: p1 = pR+ ipI, p2 = pR − ipI, p3 = qR+ iqI, and p4 = qR− iqI, with corresponding distinct eigenvalues λ1 = αp + iβp, λ2 = αp − iβp, λ3 = αq + iβq, and λ4 = αq+ iβq, where βp 6= 0 and βq 6= 0. Then
(i) A can be represented as A = Ψ S 0 0 T
!
Ψ−1, where S = αp βp
−βp αp
!
∈ IR2×2, T =diag[λ3, λ4, · · · , λn] ∈ IR(n−2 )×(n−2 ), and Ψ =
pR...pI...p3...p4...···...pn. (ii) pR⊥pI and ||pR|| = ||pI|| = √12.
(iii) (pR, pI, qR, qI) is an orthogonal set.
Proof: (i) Since A is diagonalizable, let A = Pni=1λipiqTi = P2i=1λipiqTi +Pni=3λipiqTi , where pi and qi are right and left eigenvectors of A associated with A’s eigenvalues λi, respectively, for i = 1, 2, · · · , n. Consider the first summation only,
P2 w1 and w2 are complex LI vectors. In order to yield two real-valued LI vectors from w1
and w2, let r1 = 12(w1+ w2) = pRαp− pIβp and r2 = −i2 (w1− w2) = pRβp+ pIαp. Hence for i = 1, 2, are both Householder matrices. Let α ± iβ denote two eigenvalues of H (else being 1) with corresponding eigenvectors, p = pR+ ipI and ¯p. Moreover, we choose LI unit vectors u3, u4 ∈ {u1, u2}⊥, then
(ii)From the definition of trace of a matrix, we have tr(H2H1) = n−2+2α. In addition, tr(H2H1) = trh(I − 2u2uT2)(I − 2u1uT1)i = trhI − 2u1uT1 − 2u2uT2 + 4(uT2u1)(u2uT1)i = n − 4 + 4 cos2θ, where cos θ = uT1u2 = tr(uT1u2) = tr(uT2u1). Combining these two results yields α = −1 + 2 cos2θ. Moreover, since det(H2H1) = 1, we have α2 + β2 = 1 ⇒ β =
√1 − α2.
(iii)By (i), span(u1, u2) =span(pR, pI). Given that u3 ∈ {u1, u2}⊥, then u3 ∈ {pR, pI}⊥ ⇒ pHu3 = ¯pHu3 = 0. Hence H3Hp = (I − 2u3uT3)Hp = (I − 2u3uT3)(α + iβ)p = (α + iβ)p. Similarly, it is true for ¯p. By similar procedure, it is true for H4H3H.
From Lemma 5.6, we have the following two results:
Corollary 5.3 Recall that C is defined in Problem A in Section 5.1.
1. span(u1, u2) 6⊂ C⊥ ⇔ span(pR, pI) 6⊂ C⊥ ⇔ CT(p) = CT(pR + ipI) 6= 0 ⇔ CT(¯p) = CT(pR− ipI) 6= 0.
2. We categorize θ into the the following:
(a) θ ∈ (0,π4) ∪ (3π4, π) : α > 0.
(b) θ ∈ (π4,3π4 ) : α < 0.
(c) θ = π4,3π4 : (α, β) = (0, 1).
(d) θ = π2 : (α, β) = (−1, 0).
By using above lemmas, we obtain the next important result described by Theorem 5.2 as below:
Theorem 5.2 We consider that rank(C) =rank(B) = 1( other cases can be discussed similarly) and denote cT and b as one row vector of C and one column vector of B, respectively. Under the following two cases of assumptions:
(i) If n is odd, i.e., n = 3, 5, 7, · · ·, we assume that {x, f} are LI, and {cT, f} are LI.
(ii) If n is even, i.e., n = 4, 6, 8, · · ·, we assume that {x, f, cT} are LI. Moreover, if fTx ≤ 0, we also assume that {x, f, b} are LI.
Then we can factorize A, which solves Problem A, into products of Householder matrice, therefore the condition number of A equals 1 [40].
Proof: (i)Let K = ||x||||f ||. If fTx ≤ 0, then let u1 = ||Kx+f||Kx+f ∈ IRn; else, let u1 = ||Kx−f||Kx−f ∈ IRn. Hence H1x = f, where H1 = I − 2u1u1T. Then choose an unit vector u2 ∈ IRn such that u2 ∈ f⊥ , and u2 6∈ C⊥. Since u2 ∈ f⊥, we have H2H1x = f, where H2 = I − 2u2u2T. If fTx ≤ 0, then let u1 = ||Kx+f||Kx+f ∈ IRn. By Lemma 5.6, we know that the two eigenvalues (6= 1) of H2H1 will have positive real parts; else, we require that |uT2u1| = ξ(2,1) < √12, hence the eigenvalues(6= 1) of H2H1 will have negative real parts, else if n = 3, we also require that b 6∈ span(u1, u2).
When n = 3, we require that cT 6∈ span(u1, u2). Then we can construct A. If fTx ≤ 0, then let A = −H2H1, hence all eigenvalues of A will have negative real parts, which implies that A is stabilizable. On the other hand, since u2 6∈ C⊥, by Lemma 5.6 and Corollary 5.3, we have the two right eigenvectors(6= −1) of A not perpendicular to cT. Moreover, since cT 6∈ span(u1, u2), the right eigenvector corresponding to eigenvalue -1 will not be perpendicular to cT, which means A is observable; else, we let A = H2H1. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of real part (1) is not perpendicular to b. On the other hand, A is observable by same reason as the other case of fTx ≤ 0.
When n 6= 3, we need an iteration. Iterate this step over i = 3, 5, 7, · · · , n − 2.
Choose unit vectors ui and ui+1, which form a sub-basis of (u1, u2, · · · , ui−1, f)⊥ and cTui 6= 0. Therefore we have Hi+1· · · H2H1x = f, and by Lemma 5.6, that the eigenvec-tors of Hi−1· · · H2H1 corresponding to span(u1, u2, · · · , ui−1) will still be eigenvectors of Hi+1· · · H2H1. If i = n − 2, we also require that cT 6∈ span(u1, u2, · · · , un−1). Moreover, if fTx ≤ 0, then we require that |uTi+1ui| = ξ(i+1,i) > √12 and ξ(i+1,i) 6= ξ(i,i−1) 6= · · · 6= ξ(2,1); else, we require that |uTi+1ui| = ξ(i+1,i) < √12 and ξ(i+1,i) 6= ξ(i,i−1) 6= · · · 6= ξ(2,1), else if i = n − 2, we also require that b 6∈ span(u1, u2, · · · , un−1). As all iterations are done, we still need the final step to fully construct A. If fTx ≤ 0, then A = −Hn−1· · · H2H1, where Hj = I − 2ujuTj, ∀j = 1, 2, · · · , n − 1. By Lemma 5.6, we have all eigenvalues of A having negative real parts, which implies that A is stabilizable. On the other hand, since ui 6∈ c⊥,
by Lemma 5.6 and Corollary 5.3, we have the n-1 eigenvectors of A corresponding to eigenvalues(6= −1) not perpendicular to cT. Moreover, since cT 6∈ span(u1, u2, · · · , un−1), the right eigenvector of A corresponding to eigenvalue -1 will not be perpendicular to cT, which means A is observable; else, we let A = Hn−1· · · H2H1. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of positive real part, which equals 1 and this left eigenvector is perpendicular to span(u1, u2, · · · , un−1), is not per-pendicular to b. On the other hand, A is observable by same reason as the other case of fTx ≤ 0.
(ii)Let k and u1 as given in (i). Then choose an unit vector u2 ∈ IRn such that u2 ∈ f⊥ and u2 6∈ c⊥. Since u2 ∈ f⊥, we have H2H1x = f, where H2 = I − 2u2u2T. If fTx ≤ 0, we require that |uT2u1| = ξ(2,1) > √12, and by Lemma 5.6, the two eigenvalues (6= 1) of H2H1 will have positive real parts; else, we require that |uT2u1| = ξ(2,1) < √12, hence the eigenvalues(6= 1) of H2H1 have negative real parts.
When n = 4, we choose an unit vector u3 ∈ {u1, u2, f}⊥ and require that cT 6∈
span(u1, u2, f), which implies cTu3 6= 0. Finally, we can construct A. If fTx ≤ 0, then A = −H2H1, hence all eigenvalues of A have negative real parts, which implies that A is stabilizable. On the other hand, since u2 6∈ c⊥and cTu3 6= 0, by Lemma 5.6 and Corollary 5.3, we know that the three right eigenvectors of A corresponding to eigenvalues (6= −1) is not perpendicular to cT. Moreover, since the right eigenvector of A corresponding to eigenvalue -1 is perpendicular to span{u1, u2, u3} and {x, f, c} are LI, we obtain that {u1, c} are LI and therefore this right eigenvector is not perpendicular to c, which means A is observable; else, we let A = H2H1. By similar derivation of c, we conclude that the right eigenvector corresponding to eigenvalue 1 is not perpendicular to b. Therefore A is stabilizable since the left eigenvector corresponding to the only eigenvalue of real parts (equals 1) is not perpendicular to b. On the other hand, A is observable by same reason as the other case of fTx ≤ 0.
When n 6= 4, we need an iteration. Iterate this step over i = 3, 5, · · · , n − 3.
Choose unit vectors ui and ui+1, which form a sub-basis of (u1, u2, · · · , ui−1, f)⊥ and cTui 6= 0. Therefore we have Hi+1· · · H2H1x = f, and by Lemma 5.6, that the eigen-vectors of Hi−1· · · H2H1 corresponding to span(u1, u2, · · · , ui−1) are still eigenvectors of
Hi+1· · · H2H1. If i = n − 3, we also require that cT 6∈ span(u1, u2, · · · , un−2, f). Moreover,
Hi+1· · · H2H1. If i = n − 3, we also require that cT 6∈ span(u1, u2, · · · , un−2, f). Moreover,