Unknown input observers for singular systems designed by eigenstructure assignment

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Unknown input observers for singular systems

designed byeigenstructure assignment

Sheng-Fuu Lin*, An-Ping Wang

Department of Electrical and Control Engineering, National Chiao Tung University, 1001 Ta Hsueh Road, Hsinchu, Taiwan 30050, R.O.C.

Accepted 12 February2003

Abstract

In this paper, anyunknown input observers with orders between minimum and full orders can be established for singular systems by eigenstructure assignment method. The complete and parametric solutions for the observer matrices and for the generalized eigenvectors are obtained. The completeness and parametric forms of the solutions and the ﬂexibilityin selecting the observer order allow the designer to choose a suitable observer according to the control purposes; hence, the solutions are quite suitable for advanced applications.

1. Introduction

Over the past two decades, manyresearchers have paid attention to the problem of estimating the states of a regular dynamic system subjected to both known and

unknown inputs[1–12]. Theyhave developed the reduced and/or full order unknown

input observers bydifferent approaches. This problem is of considerable importance,

because plant disturbances mayexist and some of the inputs to the system are

inaccessible. Recently, some researchers have further investigated this problem for

the singular systems[13–18]. The state responses of the singular systems may contain

differential terms of the input. It is verysensitive even if the input is changed slightly. Hence, the unknown input observer problem is especiallymeaningful for singular systems.

*Corresponding author.

E-mail address:sﬂin@cc.nctu.edu.tw (S.-F. Lin).

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In this paper, a new set of solutions of the unknown input observers are established for the singular systems by eigenstructure assignment method. The eigenstructure assignment method has been shown to be useful in the feedback

design for regular dynamic systems[19–23], and for singular systems[24–26], since

the eigenvalues and the corresponding eigenvectors can be assigned simultaneously. Furthermore, the solutions obtained byeigenstructure assignment are often represented in parametric forms. Hence, more system properties such as robust, sensitivity, minimum gain, etc., can be achieved by choosing the free parameters

from optimizing certain objective functions[27–31].

In this paper, the solutions of the unknown input observer with order which is between minimum and full orders are developed. The complete and parametric solutions for the observer matrices and for generalized eigenvector are obtained. The completeness and parametric forms of the solutions and the ﬂexibilityin selecting the observer order allow the designer to choose a suitable observer according to the control purposes; hence, the solutions are quite suitable for advanced applications. In most eigenstructure assignment approaches which are applied to feedback design, it is assumed that the system is controllable, so the generalized eigenvectors with any uncontrollable eigenvalue have not been discussed. The transmission zeros in unknown input observer design playthe same roles as the uncontrollable eigenvalues in feedback design. If the system contains a transmission zero, it must be an eigenvalue of anypossible unknown input observers. In this paper, we also develop the solutions of the generalized eigenvectors with the transmission zeros to complete the problem.

The organization of the paper is as follows. In Section 2, an unknown input observer is introduced. Some preliminaryresults are presented in Section 3. In Section 4, the problem is formulated and in Section 5 the solutions for the unknown input observers are established. The illustrative examples are provided in Section 6 and Section 7 concludes the paper.

2. Observer design

Consider the following singular system

E’x ¼ Ax þ Bu þ Dv;

y ¼ Cx þ Fu; ð1Þ

where xARn _{is the state variable, uAR}p _{is the input variable, vAR}m_{is the unknown}

disturbance, yARq_{is the measurable output, EAR}nn_{is a singular matrix and A, B,}

D, C, F are matrices with appropriate dimensions.

In this paper, the main purpose is to design an rth order observer of the following type

’z ¼ Nz þ Ly þ Gu;

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using onlythe information of the input u and measurable output y and without any knowledge of the unknown disturbance v, where r is an integer in the range

n qprpn, #x is the estimation of x; zARr is the state variable of the observer, and

N; L; G; P; Q; Z are matrices with appropriate dimensions.

From Eqs. (1) and (2), the following relations can be easilyderived for any

T ARrn_:

ð’z TE ’xÞ ¼ Nðz TExÞ þ ðNTE þ LC TAÞx TDv þ ðG TB þ LFÞu;

ð#x xÞ ¼ Pðz TExÞ þ ðQC I þ PTEÞx þ ðZ þ QFÞu; ð3Þ

If the following relations hold:

NTE TA þ LC ¼ 0; ð4Þ

TD ¼ 0; ð5Þ

G TB þ TF ¼ 0; ð6Þ

PTE þ QC ¼ I ; ð7Þ

Z þ QF ¼ 0; ð8Þ

we can obtain that

ð’z TE ’xÞ ¼ Nðz TExÞ;

ð#x xÞ ¼ Pðz TExÞ:

If N is a stable matrix, it follows that #x-x: The behaviors of x and #x are

demonstrated by u, v, x(0) and z(0). Therefore, if one of relations (4)–(8) does not

satisfyfor anyT ARrn; there must exist some u, v, x(0) and z(0) which make ð#x xÞ

not converge to zero. Hence, Eqs. (4)–(8) are the sufﬁcient and necessarycondition for the existence of the proposed observer and the problem of the rth order unknown input observer design is to ﬁnd matrices N; L; G; P; Q; Z and T satisfying Eqs. (4)–(8).

From Eq. (5), it follow that rankTpn rankD and from Eq. (7), it should be

satisﬁed that rankTXrankTEXnrankC for the existence of P and Q; Hence, it

follows that nrankDXnrankC, i.e. rankDprankC. If the condition is not held,

the observer does not exist. From this fact, we assume in the whole paper that C is of full row rank, D is of full column rank and qXm without loss of generality.

3. Some preliminary results

In this section, the possible eigenvalues of the matrix N are discussed.

Deﬁnition 1. A complex number liis a transmission zero of the quartet (E, A, D, C) if

and onlyif

rank ðA liEÞ D

C 0

" #

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Deﬁnition 2. If there exist a set of linear independent vectors vk ij wk ij " # ; j ¼ 1; y; %yi; k ¼ 1; y;%rij; satisfying that ðA liEÞ D C 0 " # vk ij wk ij " # ¼ E 0 0 0 " # vk1 ij wk1 ij " # ; v0_ij¼ 0; ð9Þ then we saythat vk

ij is a right generalized transmission vector of order k with the

transmission zero li. Furthermore, if rank

ðA liEÞ D

C 0

" #

¼ n þ m %yi

and anynonzero linear combination of v%ri1 i1 0 " # ; y; v %ri %yi i %yi 0 2 4 3 5 is not in the column space of

ðA liEÞ D

C 0

" #

;

then we saythat vk_ij; j ¼ 1; y; %yi; k ¼ 1; y;%rij; form a complete set of right

transmission vectors with the transmission zero li.

The following lemma discusses the role of transmission zero in an unknown input observer design.

Lemma 3. Assume that observer (2) exists. The eigenvalues of N must contain the transmission zeros of the quartet (E, A, D, C) counting multiplicity and the rest of the eigenvalues can be assigned arbitrarily.

Proof. Let F ACns _{be a matrix whose columns are all generalized transmission}

vectors and LTZ be a Jordan form matrix with all transmission zeros as its

eigenvalues. Since CF=0, TEF is of full column rank; otherwise, from Eq. (7), the

observer does not exist. We can, therefore, ﬁnd a matrix QACnðrsÞ such that

F Q

½ and TE F½ Q are of full column rank. From right-multiplying by F½ Q

and left-multiplying by TE F½ Q1 to (4), we have

LTZ LTZ N1 A1þ L1C2

0 N2 A2þ L2C2

" #

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where

N TEF½ TEQ ¼ TEF½ TEQ LTZ N1

0 N2 " # ; ð11Þ TA F½ Q ¼ TEF½ TEQ LTZ A1 0 A2 " # ; ð12Þ

C F½ Q ¼ 0½ C2 and L ¼ TEF½ TEQ

L1

L2

" #

ð13Þ and the last three Eqs. (11)–(13) can be derived from Eqs. (4), (5) and (9). It can be

seen that if the matrices pair (A2,C2) is unobservable, then its unobservable

eigenvalue is a transmission zero of the original system. So, if LTZ contains all

transmission zeros, (A2, C2) is an observable pair. Hence, from Eqs. (10) and (11), the

result follows. &

The following notations are given for gathering the chains of generalized transmission vectors with the same lengths to the same group, and then the lengths of generalized transmission vectors in different group would have different lengths.

For a transmission zero li, assume that the elements of the set f%ri1; y;%ri %yig are

arranged as follows: _%ri1p%ri2p?p%ri %yi: Denote fi as the number of distinct

elements in the set f_%ri1; y;%ri %yig: And the notations si1; si2; y; sifi; satisfying

silosi2o?osifi; represent all distinct elements of the set f%ri1; y;%ri %yig: Assume

that there are Zilelements with value sil; l ¼ 1; y; fi; within the set f%ri1; y;%ri %yig: Let

V_ilk¼ vk_iðZ i0þZi1þ?þZiðl1Þþ1Þ v k iðZi0þZi1þ?þZiðl1Þþ2Þ ? v k iðZi0þZi1þ?þZiðl1ÞþZilÞ h i ; W_ilk¼ wk iðZi0þZi1þ?þZiðl1Þþ1Þ w k iðZi0þZi1þ?þZiðl1Þþ2Þ ? wk iðZi0þZi1þ?þZiðl1ÞþZilÞ h i ; k ¼ 1; y; sil; l ¼ 1; y; fi and Zi0¼ 0;

where the column vectors of V_ilk are the right generalized transmission vectors of

grade k, in all chains with length sil. Then by(9), we have

ðA liEÞ D C 0 " # V_ilk W_ilk " # ¼ E 0 0 0 " # V_ilk1 W_ilk1 " # ; k ¼ 1; y; sil; Vil0¼ 0: ð14Þ A series of matrices V1 il; Vil2; y; V sil

il ; group those chains of right generalized

transmission vectors with the same length sl

i; l ¼ 1; y; fi: Let Uil¼ Vi1si1 ? V sil il

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which gathers all last generalized transmission vectors in those chains with length

less than or equal to sil. And it can be seen that

Uil¼ Uiðl1Þ Vilsil

:

4. Problem formulation

The keyproblem in this chapter is now stated as follows:

Unknown input observer design by eigenstructure assignment: Given a symmetric set

of complex numbers fli; y; lpg which contain the transmission zeros, and a set of

positive integers r_i1; y; r_iy_i; i ¼ 1; y; p; representing the multiplicities and satisfying

thatPp_i¼1Pyi

j¼1rij¼ r; we want to ﬁnd the parametric solutions of the matrices N, L,

G, P, Q, Z and T over the ﬁeld of real number satisfying (4)–(7) where the matrix N

has eigenvalues fli; y; lpg and left generalized eigenvectors h1ij; y; h

rij

ij ; j ¼

1; yyi; i ¼ 1; y; p satisfying the relation:

hk_ijN ¼ lihkijþ hk1ij ; for i ¼ 1; y; p; j ¼ 1; y; yi;

k ¼ 1; y; r_ij; and h0_ij¼ 0: ð15Þ

Since hk

ij; i ¼ 1; y; p; j ¼ 1; y; yi; k ¼ 1; y; rij are linearlyindependent, Eqs. (4)

and (5) are equivalent to the following conditions:

hk_ijTðA liI Þ þ hkijLC ¼ h k1 ij T ; h k ijTD ¼ 0; i ¼ 1; y; p; j ¼ 1; y; yi; k ¼ 1; y; rij: ð16Þ Let *tk

ij¼ hkijT and *lkij¼ hkijL; then Eq. (16) is equivalent to

*tk ij *lkij h i ðA liEÞ D C 0 " # ¼ *tk1 ij E 0 h i : ð17Þ Deﬁne H as H ¼ H1 ^ Hp 2 6 4 3 7 5; where Hi¼ Hi1 ^ Hiyi 2 6 4 3 7 5; and Hij¼ h1_ij ^ hrij ij 2 6 6 4 3 7 7 5 ð18Þ

and the matrices *T and *L are deﬁned in the same way.

If the solutions of *t1 ij; y;*t rij ij and *l1ij; y; *l rij ij ; j ¼ 1; y; yi corresponding to

eigenva-lues li; i ¼ 1; y; p have been found from Eq. (17), then bychoosing a nonsingular

H, the solutions of T, L, N and G can be obtained as follows:

T ¼ H1*T; L ¼ H1*L; N ¼ H1LH and G ¼ TB; ð19Þ

where L is a Jordan form matrix in lower case with eigenvalues fl1; y; lmg and

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Furthermore, if the matrix *T or T satisﬁes the following relation: rank TE C " # ¼ rank *TE C " # ¼ n;

we can obtain a pseudo-inverse of TE

C as follows: M ¼ TE C " #H TE C " # 0 @ 1 A 1 TE C " #H ;

where XHmeans the Hermitian adjoint of the matrix X. The solutions of P and Q are

as follows: P Q ½ ¼ M þ K Irþq TE C " # M ! ;

where KARnðrþqÞ_{represents free parameters. Then the matrix Z can be obtained by}

Z ¼ QF :

From the above discussion, the following proposition, similar to that given by

Moore[19] and Klein and Moore [20], characterizing all possible solutions of the

observer is given.

Lemma 4. Let fl1; y; lmg be a symmetric set of complex numbers and let

fr_i1; y; r_iy_i; i ¼ 1; y; mg be a set of positive integers satisfying that Pm_i¼1Pyi

j¼1rij¼

r: There exist N, G, P, Q, T of real number and h1

ij; y; h

rij

ij; j ¼ 1; yyi; i ¼ 1; y; p,

satisfying (4)–(7) and (15) if and only if (a) The matrix H is nonsingular.

(b) For each iAf1; y; pg there exists i0Af1; y; pg such that l_i¼ ðl_i0Þ; y_i¼ y_i0; r_ij¼

ri0_j; j ¼ 1; y; y_i; and hk_ij¼ ðhk_ijÞ; i ¼ 1; y; p; j ¼ 1; y; y_i; k ¼ 1; y; r_ij; where

ðhk

ijÞ

_{means the component-wise conjugate of the vector h}k

ij:

(c) For each iAf1; y; mg, there exists a set of vectors f*tk

ij; *lkij; i ¼ 1; y; p; j ¼

1; y; yi; k ¼ 1; y; rijg satisfying Eq. (17) and

rank *TE

C

" #

¼ n: ð20Þ

Proof. The sufﬁciencyhas been stated above. Assume that the solutions exist. Since the matrix H represents the left generalized eigenvectors, it is nonsingular. If Eq. (20)

is not satisﬁed, P and Q do not exist. Hence, the necessityfollows. &

In the above deriving process, it can be seen that the matrix H can be arbitrary chosen to satisfyconditions (a) and (b) in Lemma 4. Hence, the existing condition of

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satisfying Eqs. (17) and (20) exists. The following lemma discusses the possible orders of the observer.

Lemma 5. If the observer exists, its order can be chosen within n qprpn:

Proof. If an observer of order r exists, a matrix *TARrn_{satisfying Eqs. (17) and (20)}

can be found. If follows from Eq. (20) that rank *TEXn q: Hence, we have rXn q:

This means that the order of the observers must be at least nq. If an observer of

order nq is required, a new matrix *T0_A_RðnqÞn_{satisfying Eq. (20) can be obtained}

bycarefullydeleting r(nq) rows of *T and an observer of order nq can be

obtained bythis new matrix *T0_{from the process discussed above. If an observer of}

order n is required, a new matrix *T00ARnnsatisfying (20) can be obtained by adding

nr additional solutions of Eq. (17) to the row vectors of *T and an observer of order

n can be obtained from this new matrix *T00: Other observers of order within

n qprpn can be obtained in a similar way. &

5. Main results

From Lemma 4, the problem now is to ﬁnd the solutions of Eq. (17). In this

section, we shall establish the complete parametric solutions of *t1

ij; y;*t rij ij and *l1 ij; y; *l rij

ij with an eigenvalue lifrom Eq. (17). The properties of Eq. (17) depend on

whether the eigenvalue liis a transmission zero or not; hence, we shall ﬁrst establish

the parametric solutions of Eq. (17) for eigenvalues which are not transmission zeros and then the parametric solutions of Eq. (17) for eigenvalues which are transmission zeros.

5.1. The solutions for eigenvalues which are not transmission zeros

If liis not a transmission zero, then

rank ðA liEÞ D

C 0

" #

¼ n þ m:

We can obtain a nonsingular transformation as follows:

U_i11 U_i12 U_i21 U_i22 " # ðA liEÞ D C 0 " # C1_i C2_i " # ¼ IðnþmÞðnþmÞ 0 " # ; ð21Þ where U_i11ACðnþmÞn; U_i12ACðnþmÞq; U_i21ACðqmÞn; U_i22ACðqmÞq; C1_iACnðnþmÞ; C2_iA CmðnþmÞ_{and I}

ðnþmÞðnþmÞis an ðn þ mÞ ðn þ mÞ identitymatrix. We can obtain the

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Theorem 6. If li is not a transmission zero, the complete solution of Eq. (17) are as follows: *tk ij *l k ij h i ¼h*tk1_ij EC1_i f_ijk1i U 11 i U 12 i U_i21 U_i22 " # ; k ¼ 1; y; r_ij; *t0_ij¼ 0; ð22Þ

where f_ijk1 are vectors representing the free parameters.

Proof. Necessity: Introduce the following variable transformation: *tk ij *lkij h i ¼ ak ij fikk1 h i U11 i Ui12 U21 i Ui22 " # ; k ¼ 1; y; r_ij: ð23Þ

From Eqs. (17) and (21), we have *tk1 ij E 0 h i C1 i C2_i " # ¼ ak ij fijk h i IðnþmÞðnþmÞ 0 " # ; k ¼ 1; y; r_ij; hence ak_ij¼*tk1_ij EC1_i; k ¼ 1; y; r_ij:

Byapplying this relation into Eq. (23), Eq. (22) is obtained. Sufficiency: Using Eqs. (22) and (21), we have

*tk ij *lkij h i ðA liEÞ D C 0 " # ¼ *tk1 ij EC1i *zkij h i U11 i Ui12 U_i21 U_i22 " # ðA liEÞ D C 0 " # ¼ *tk1 ij EC 1 i *zkij h i IðnþmÞðnþmÞ 0 " # C1_i C2_i " #1 ¼ *tk1 ij E 0 h i C1 i C2_i " # C1_i C2_i " #1 ¼h*tk1_ij E 0i:

Therefore the vectors given byEq. (22) satisfyEq. (17). &

5.2. The solutions for eigenvalues which are transmission zeros

If liis a transmission zero, then

rank ðA liEÞ D

C 0

" #

¼ rion þ m:

We can obtain the following nonsingular transformation:

U11 i Ui12 U_i21 U_i22 " # ðA liEÞ D C 0 " # C1 i C1i C2_i C2_i " # ¼ Iri 0 0 0 " # ; ð24Þ

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where U_i11ACrin; U_i12ACriq; U_i21ACðnþqriÞn; U_i22ACðnþqriÞq; C1_iACnri; C1_iA CnðnþmriÞ_{; C}2

iACmri; C2iACmðnþmriÞ; and Iri is an ri ri identitymatrix. We can

obtain the following lemma.

Lemma 7. U_i21EUilis of full column rank for l ¼ 1; y; fi.

Proof. If U_i21EUilis not of full column rank, then we can ﬁnd a nonzero vector f such

that U_i21EUilf ¼ 0: Since the matrix

Iri 0

C1

i C1i

C2_i C2_i

" #1

is of full row rank, we can ﬁnd a vector %v_%w such that

Iri 0 C1 i C 1 i C2_i C2_i " #1 %v %w " # ¼ U11 i Ui12 EUilf 0 " # :

Then we can obtain that

A liE D C 0 " # %v %w " # ¼ EUilf 0 " # : So, EUilf 0 h i

is in the column space of

A liE D

C 0

" #

:

It is a contradiction to the deﬁnition of Uil. &

From Lemma 7, for each Uil; l ¼ 1; y; fi; we can obtain the following

nonsingular transformation: S1 il S2 il " # U_i21EUil¼ IZi1þ?þZil 0 " # ; ð25Þ where S1_ilARðZi1þ?þZilÞðnþqriÞ_{; S}2

ilARðnþqriZi1?ZilÞðnþqriÞ and the matrix S1 il S2 il h i is nonsingular. To ﬁnd S1

iland S2ilfor each lAf1; y; fig; we onlyneed to calculate S1ifi

and S2 ifi: Let S1 ifi S2 ifi " # ¼ si1 ^ sifi siðfiþ1Þ 2 6 6 6 4 3 7 7 7 5;

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where silARZilðnþqriÞ_{; l ¼ 1; y; f} i and siðfiþ1ÞAR ðnþqriZi1?ZifiÞðnþqriÞ_{: Then S}1 il and S2 il can be selected as S_il1¼ si1 ^ Sil 2 6 4 3 7 5 and S1i2¼ siðlþ1Þ ^ siðfiþ1Þ 2 6 4 3 7 5:

The following theorem can now be given.

Theorem 8. Assume that liis a transmission zero.

(a) If r_ijpsi1; the complete solutions of Eq. (17) can be represented as follows:

*tk ij *lkij h i ¼ *tk1 ij EC 1 i fijk1 h i U11 i Ui12 U_i21 U_i22 " # *t0 ij¼ 0; k ¼ 1; y; rij; ð26Þ where fk1

ij ; k ¼ 1; y; rijare vectors representing the free parameters.

(b) If si1orijpsifi there exists b such that siborijpsiðbþ1Þ, and if rij> sifi; then

b=fi. Under the above conditions, the complete solutions of Eq. (17) can be

represented as follows: *tk ij *lkij h i ¼h*tk1_ij EC1_i f_ijk1i I U 11 i EUibS1ib 0 S2 ib " # U_i11 U_i12 U_i21 U_i22 " # ; *t0 ij¼ 0; k ¼ 1; y; ðrij sibÞ; *tk ij *lkij h i ¼ *tk1 ij EC1i fijk1 h i I U11 i EUilS1il 0 S2 il " # U_i11 U_i12 U_i21 U_i22 " # ; k ¼ ðr_ij siðlþ1ÞÞ þ 1; y; ðrij silÞ; l ¼ ðb 1Þ; ?; 1; *tk ij *lkij h i ¼ *tk1 ij EU1i fijk1 h i U11 i Ui12 U_i21 U_i22 " # ; k ¼ ðr_ij si1Þ þ 1; y; rij; ð27Þ where fk1

ij ; k ¼ 1; y; rij are vectors representing the free parameters.

Proof. Here, we will show that Eq. (26) is equivalent to Eqs. (17) and (27) is equivalent to Eq. (17).

Necessity: A variable transformation is adopted as follows: *tk ij *lkij h i ¼ ak ij ckij h i U11 i Ui12 U_i21 U_i22 " # ; k ¼ 1; y; r_ij; ð28Þ

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where ak

ijAC1r and ckijAC1ðmþnrÞ: It can be seen that Eq. (17) is equivalent to the

following relations:

ak_ij¼*tk1_ij EC1_i; ð29Þ

0 ¼ *tk1_ij EC1_i; k ¼ 1; y; r_ij: ð30Þ

Observe that the column space of C1_i is the same as the column space of V1

il; l ¼

1; y; f_i: Hence, Eq. (30) is equivalent to

*tk1

ij EVil1¼ 0; l ¼ 1; ?; fi; k ¼ 1; y; rij:

From Eqs. (14) and (17), it can be further shown that *tk1 ij EVil1¼ *tðksilÞ ij EV sil il if k > sil; *t0 ijEVilk 0 if kpsil: ( ð31Þ

We discuss the problem in the following two cases:

(a) It can be seen from Eq. (31) that if r_ijpsi1; Eq. (30) would be satisﬁed for all

possible *tk1_ij : Hence, Eq. (17) is equivalent to Eq. (29). Let f_ijk1 ¼ ck

ij; k ¼

1; y; rij: Byapplying Eqs. (29) to (28), Eq. (26) is obtained.

(b) If si1orijpsifi; there exists a number b such that siborijpsiðbþ1Þ; and if rij>

sifi; then b ¼ fi: Under the above conditions, from Eq. (31), Eq. (30) is

equivalent to the following relations: *tk1

ij EUib¼ 0; where k ¼ 2; y; ðrij sbiÞ þ 1;

*tk1

ij EUiðb1Þ¼ 0; where k ¼ ðrij sibÞ þ 2; y; ðrij siðb1ÞÞ þ 1;

^ *tk1

ij EUi1¼ 0; where k ¼ ðrij si2Þ þ 2; y; ðrij si1Þ þ 1:

From Eq. (28), we have

ðak1

ij Ui11þ ck1ik Ui21ÞEUib¼ 0; k ¼ 2; y; ðrij sibÞ þ 1;

ðak1_ij U_i11þ ck1_ij U_i21ÞEUil¼ 0; k ¼ ðrij siðlþ1ÞÞ þ 2; y; ðrij silÞ þ 1;

l ¼ 1; ?; ðb 1Þ:

From Eqs. (25) and (29), it can be seen that the above equations are equi-valent to ck_ij¼ *tk1_ij EC_i1U_i11EU1bS1ibþ f k1 ij S 2 ib; k ¼ 1; y; ðrij sibÞ; ck ij¼ *tk1ij Ec 1 iUi11EU1lS1ilþ fijk1Sil2; k ¼ ðrij siðlþ1ÞÞ þ 1; y; ðrij silÞ; l ¼ 1; ?; ðb 1Þ; ð32Þ

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where fk

ij are column vectors with appropriate dimension and represent the free

parameters. For k ¼ ðr_ij silÞ; y; rij; ckij is free; hence, deﬁne

ck_ij¼ fk1

ij ; k ¼ ðrij si1Þ þ 1; y; rij: ð33Þ

Byapplying Eqs. (29), (32) and Eqs. (33) to (28), Eq. (27) is obtained. Sufficiency: Deﬁne the variable transformation (28). Since the matrix

U_i11 U_i12

U_i21 U_i22

" #

is nonsingular, Eq. (28) is a nonsingular transformation. It can also be seen that Eq. (31) is a equivalent relation; Hence, from Eqs. (26) or (27), we can derive Eq. (17)

byreversing the procedure in the necessity. &

5.3. The realness of the solutions

The realness of the solutions is guarantee bythe following lemma:

Lemma 9. Under the conditions proposed by Lemma 4, if the free parameters in

Theorems 6 and 8 are chosen as fk

ij ¼ ðfik0_jÞwhen li¼ ðli0Þ, then the solutions N, L, G,

P, Q and W are real.

Proof. Observe that hk_ij; the row vector of H, is the left generalized eigenvector of N.

Let V ¼ H1_{; and v}k

ij be the corresponding column vector of V, representing the

right generalized eigenvector of V. Then, under the conditions (2) of Lemma 4,

we have vk

ij¼ ðvki0_jÞ when li¼ ðli0Þ: Then, we have ðNÞ¼ ðH1LHÞn¼

Pp i¼1 Pyi j¼1 Prij k¼1ðliÞðvkijÞ _ðhk ijÞ _¼Pp i¼1 Pyi j¼1 Prij k¼1livkijhkij¼ H1LH ¼ N: Hence,

the matrix N is real.

If fk

ij ¼ ðfik0_jÞ; when li¼ ðli0Þ; then from Lemma*and*, we have *tk

ij¼ ð*tki0_jÞ; and *lk ij¼ ð*lki0_jÞ when li¼ ðli0Þ: Then ðT Þ¼ ðH1 *TÞ¼ Pp i¼1 Pyi j¼1 Prij k¼1ðliÞðvkijÞ ð*tk ijÞ _¼Pp i¼1 Pyi j¼1 Prij k¼1livkij*tkij¼ H1*T ¼ T and ðLÞ _{¼ ðH}1*LÞ_¼Pp i¼1 Pyi j¼1 Prij k¼1ðliÞðvkijÞ _ð*lk ijÞ _¼Pp i¼1 Pyi j¼1 Prij

k¼1livkij*lkij¼ H1*L ¼ L: Hence, the

ma-trices T and L are real.

Since T is real. It follows that P, Q, G and W are real. &

6. Example

Example 1. Consider system (1) with the following matrices[18]

E ¼ 1 0 0 0 0 1 1 0 0 2 6 4 3 7 5; A ¼ 5 0 0 0 1 0 0 0 1 2 6 4 3 7 5; D ¼ 1 0 0 2 6 4 3 7 5; C ¼ 1₀ 0₁ 0₀ " # :

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This system contain no transmission zero, so eigenvalues of the observer can be arbitrarilyassigned. We illustrate the design of the proposed method for unknown input observer from minimum order to full order.

(a) First(minimum)-order observer: Let the eigenvalue of the observer be l, then T ¼ h1f₁₁0 0 1 l; L ¼ h1f₁₁0hl2 ₁i_{; P} _Q_¼ 0 1 0 0 0 1 h f0 11 l 0 2 6 6 4 3 7 7 5: Let a ¼ h1_f0

11; then the parametric solutions of the observers are

’z ¼ lz þ al2 a y; #x ¼ 0 0 1 a 2 6 4 3 7 5z þ 1 0 0 0 l 0 2 6 4 3 7 5y: ðbÞ

(b) Second-order observer: Let the eigenvalues of the observers be two complex conjugate numbers 1+i and 1i, then

*t1¼ f110 0 1 1 i ;*t2¼ f210 0 1 1 þ i ; *l1¼ f110 2i 1 ; *l2¼ f210 2i 1 :

For considering the realness of the observers, choose the free parameters to be

complex conjugate as f0 11 ¼ %f210 ¼ 1 þ i; H ¼ 1 1 þ i 1 1 i " # and K ¼ 1 0 1 0 1 0 2 1 0 1 0 0 2 6 4 3 7 5;

then the observer can be obtained as

’z ¼ 2 2 1 0 " # z þ 4 0 2 1 " # y; #x ¼ 1 0 1 0 0 1 2 6 4 3 7 5z þ 1 0 2 1 0 0 2 6 4 3 7 5y:

For this system, the minimum order of the observer is 1. However, if the convergence of the error vector is required to be sinusoidal, then its order must be at least 2 for the realness of the observer. Thus, if onlythe method for minimum order observer design is obtained, the observer with sinusoidal convergence cannot be obtained; while bythe present method, this requirement can be easilyachieved as shown in the example.

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(c) Third(full)-order observer: Let the eigenvalues of the observers be 1+i,1i, and 2 then *t1¼ f110 0 1 2 ; *t2¼ f210 0 1 1 i ; *t3 ¼ f310 0 1 1 þ i ; *l1¼ f110 4 1 ; *l2¼ f210 2i 1 ; *l3¼ f310 2i 1 :

For considering the realness of the observers, the free parameters are chosen

to be complex conjugate. Then arbitrarilythe parameters are chosen as f0

11 ¼ f210 ¼ f0 31¼ 1; H 0 1 0 1 1 þ i 1 i 1 1 i 1 þ i 2 6 4 3 7 5 and K ¼ 2 0 1 1 0 0 1 1 1 1 1 1 1 1 0 2 6 4 3 7 5; then the observer can be obtained as

’z ¼ 0 2 2 0 2 0 1 0 2 2 6 4 3 7 5z þ 2 1 4 1 6 1 2 6 4 3 7 5y; #x ¼ 0 1 0 0 1 1 1 1 1 2 6 4 3 7 5z þ 1 0 1 1 1 0 2 6 4 3 7 5y:

In this example, the unknown input observers for the three possible orders are designed in a uniﬁed way. In practice, a suitable order and free parameters can be chosen bypractical consideration and speciﬁc control requirement.

Example 2. Consider system (1) with the following matrices

E ¼ 0 0 1 1 1 1 0 0 1 2 6 4 3 7 5; A ¼ 1 0 2 3 2 3 0 0 1 2 6 4 3 7 5; C ¼ 1 0 1 0 0 1 " # ; and D ¼ 2 1 1 2 6 4 3 7 5:

This system has one transmission zero 2 with %y1¼ 1 and %r11 ¼ 1; so it is an

eigenvalue of the observer.

(a) First(minimum)-order observer: The eigenvalue should be 2 and it follows that *t1 ij *l1ij h i ¼ f_ij0 1 1 1 0 0 1 2 0 1 1 " # :

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Let f0 11 ¼ g 1 g2and T ¼ 1 hðg1 g2Þ 1 hðg1þ 2g2Þ 1 hf1 ; L ¼ 1 hg 2 1 hg 2 ; P½ Q ¼ 0 1 1 h g1 2g2 1 1 g1 2g2 g2 0 0 1 2 6 6 6 4 3 7 7 7 5:

Let a1¼ h1g1 and a2¼ h1g2; then the parametric solutions of the observers are

’z ¼ 2z þ a½ 2 a2y; #x ¼ 0 1 a1 2a2 0 2 6 6 6 4 3 7 7 7 5z þ 1 1 1 a2 a1 2a2 0 1 2 6 6 6 4 3 7 7 7 5y:

(b) Third(full)-order observer: Here, we assigned the eigenvalues of the observer to

be the transmission zero l1=2 with y1=1 and r11=3 to illustrate the assignment

of generalized eigenvectors with an eigenvalue which is a transmission zero. Bya simple computation, it follows that

*tk 1j *lk1j h i ¼h*tk1_1j f_1jk1i 1 0 2 1 2 0 0 0 1 0 1 0 2 1 2 1 0 2 1 1 2 6 6 6 4 3 7 7 7 5; k ¼ 1; 2; *t3 1j *l 3 1j h i ¼h*t2_1j f_1j2i 1 2 0 1 0 0 0 0 1 0 1 2 0 1 0 1 1 1 0 0 1 2 0 1 1 2 6 6 6 6 6 6 4 3 7 7 7 7 7 7 5 : Choose f0 11¼ 1; f111 ¼ 2; f112 ¼ 1½ 1; K ¼ 1 0 1 0 0 0 0 1 0 1 0 1 0 0 1 2 6 4 3 7 5 and H ¼ 1 0 1 1 1 0 1 0 0 2 6 4 3 7 5:

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The obtained observer is ’z ¼ 1 1 0 0 3 1 1 1 2 2 6 4 3 7 5z þ 0 1 1 1 1 2 2 6 4 3 7 5y; #x ¼ 1 0 1 0 0 1 1 1 0 2 6 4 3 7 5z þ 1 0 1 1 0 0 2 6 4 3 7 5y:

(c) Third(full)-order observer: Let the eigenvalues of the observer be l1 ¼ 2 with

r₁¼ 1 and l2 ¼ 1 with r2 ¼ 2 to illustrate the assignment of the generalized

eigenvectors with an eigenvalue which is not a transmission zero. The solutions of

the eigenvalues l2=1 are

*tk 2j *lk2j h i ¼ *tk1 2j f2jk1 h i 0 0 1 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 1 2 6 6 6 4 3 7 7 7 5; k ¼ 1; 2: Choose f0 11¼ 1 1 ; f0 21¼ 1; f211 ¼ 1; H ¼ 1 0 0 1 1 1 1 0 1 2 6 4 3 7 5 and K ¼ 1 0 1 0 1 0 0 1 0 1 1 1 1 0 0 2 6 4 3 7 5:

The obtained observer is

’z ¼ 1 0 0 1 1 1 2 1 3 2 6 4 3 7 5z þ 0 1 0 0 0 2 2 6 4 3 7 5y; #x ¼ 1 0 1 0 1 1 1 0 1 2 6 4 3 7 5z þ 1 1 0 0 0 1 2 6 4 3 7 5y:

Example 3. Consider system (1) with the following matrices

E ¼ 1 0 0 0 0 1 0 0 0 2 6 4 3 7 5; A ¼ 5 0 0 0 1 0 0 0 1 2 6 4 3 7 5; D ¼ 1 0 1 2 6 4 3 7 5; C ¼ 1₀ 0₁ 0₀ " # :

This system has been studied by Kawaji and Sawada[17] for the design of a

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observer be l, then it follows that T ¼ h1f₁₁0 l 1 l; L ¼ h1f₁₁0 lð5 þ lÞ 1; P Q ¼ 0 1 0 0 0 1 h f0 11 l 0 2 6 6 4 3 7 7 5: Let a ¼ h1f0

11; then the parametric solutions of the observers are

’z ¼ lz þ alð5 þ lÞ a y; #x ¼ 0 0 1 a 2 6 4 3 7 5z þ 1 0 0 1 l 0 2 6 4 3 7 5y: ð34Þ

In Eq. (34), there is one more free parameter a than the results given byKawaji and

Sawada[14]where onlythe solutions when a=1 are obtained. It can be seen that a is

not a trivial parameter and it provides a more degree of freedom to assign a suitable observer according to the control requirement.

7. Conclusions

In this paper, the complete parametric solutions of the unknown input observers with orders between minimum and full orders for singular systems have been established. The complete and parametric solutions of the observer matrices and of the generalized eigenvector are obtained. In the illustrative examples, it can be seen that the solutions obtained bythe proposed method have more free parameters than the previous results. Furthermore, in the proposed method, the order of the observer can be chosen from the range between nq and n, according to the control purposes. The completeness and parametric form of the solutions and ﬂexibilityin selecting the observer order make them more suitable for advanced applications.

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