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    Linear Multivariable Theory

    Linear Multivariable Control Theory

    D Viswanath

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    Linear Multivariable Theory

    Outline

    1 Introduction

    2 Geometric Control Preliminaries

    3 Results and Discussion

    4 Conclusions and Future work

    5 Publications

    6 References

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    Introduction

    Overview

    Geometric term is used since the setting is linear state space and the math-ematics used is chiefly linear algebra in abstract (geometric) style.

    The concepts of controllability and observability are the geometric propertiesof distinguished state subspaces.

    The geometric approach does not directly look for a feedback law (say,u = F x)as a solution to the synthesis problem. Instead it first characterizes solvabilityas a verifiable property of some state space, say, . ThenFis calculated from.

    Thus an intractable nonlinear problem in F is converted to a straightforward

    quasilinear one in F.

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    Geometric Control Preliminaries

    Linear Spaces

    A linear (vector) space consists of an additive group, of elements calledvectors,together with an underlying field ofscalars.

    The field F of real numbers R or complex numbers C is considered.

    Linear spaces are denoted by capital scripts X, Y, ...and their elements (vec-tors) by lower case Roman letters x, y,...; and the field elements (scalars) bylower case Roman or Greek letters.

    Properties of vector addition and multiplication of vectors by scalars apply, i.e.,ifx1, x2 X and c1, c2 Fthenc1x1 X, c1(x1+x2) =c1x1+c1x2, (c1+c2)x1 =c1x1+c2x1,(c1c2)x1 =c1(c2x1).

    Ifk is a positive integer, k denotes the set of integers {1, 2, 3,...,k}.

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    Geometric Control Preliminaries

    Linear Spaces

    Let x1, x2,...,xk X where X is defined over F. Their spanis the set of all

    linear combinations of the xi, with coefficients inF

    .X isfinite dimensionalif there exists a (finite) kand a set{xi, i k; xi X}.

    IfX = 0, the least k for which this happens is the dimension ofX writtend(X). When X = 0, d(X) := 0. Ifk=d(X)= 0, a spanning set {xi, i k} is

    a basis for X.A set of vectors {xi X, i m} is (linearly) independent (over F) if for allsets of scalars {ci F, i m}, the relation

    m

    i=1

    cixi = 0 (1)

    impliesci= 0for alli m. If thexi are independent, and ifx Span{xi, im}, then the representation x= c1x1+c2x2+...+cmxm is unique.

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    Geometric Control Preliminaries

    Linear SubspaceA linear subspace S of the linear space X is a non-empty subset ofX which isa linear space under the operations of vector addition and scalar multiplication.

    S X and for all x1, x2 S and c1, c2 F, we have c1x1+c2x2 S.

    The notation S X means that S is a subset ofX.

    Ifxi X; (i k), then Span{xi, i k} is a subspace ofX.

    Geometrically, a subspace may be pictured as a hyperplane passing throughthe origin ofX.

    Thus the vector 0 S for every subspace S X.

    0 d(S) X, with d(S) = 0(resp.d(X)) if and only ifS = 0.(resp.X)

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    Geometric Control Preliminaries

    Linear Subspace

    IfR,S X, subspaces R+S X and RS X can be defined according

    to

    R+S := {r+s: r R, s S} (2)

    RS := {x: x R & x S} (3)R+S is the span ofR and S and may be much larger than the set-theoreticunion.R

    S is generally not a subspace.

    As the zero subspace0 R and0 S, it is always true that 0 RS =.

    In other words, two subspaces ofX are never disjoint in the set-theoretic

    sense.

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    Geometric Control Preliminaries

    Linear Subspace-Lattice

    The family of all subspaces is partially ordered by subspace inclusion ().

    Under the operations+and , this forms alattice, namelyR +S is the small-est subspace containing both R and S, while R

    S is the largest subspace

    contained in both R and S.A lattice diagram may picture inclusion relations among subspaces in whichthenodesrepresent subspaces and a rising branch from R to S means R S.

    Thus shown below is the lattice diagram for arbitrary R and S X:-

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    Geometric Control Preliminaries

    Linear Subspace-Lattice

    Let R,S,T X and suppose R S.

    Then

    R

    (S+ T) = RS + R

    T (4)

    = S + RT (5)

    Equation5is called the modular distributive rule. The lattice in which thisrule holds i called modular.

    If no inclusion, i.e., R,S,T X, is postulated, then

    R

    (S+ T) = RS + R

    T (6)

    It is also true that

    S

    (R + T) = SR + S

    T (7)

    and by symmetry

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    Linear Subspace - Linear Independence

    Two subspaces R,S X arelinearly independentifR

    S = 0.

    A family ofk subspaces is independentif

    Ri

    (R1+...+ Ri1+ Ri+1+...+ Rk) = 0 (9)

    for all i k.

    An independent set of vectors cannot include the zero vector. But any inde-pendent family of subspaces remains independent if one or more zero subspacesare adjoined.

    The following statement are equivalent:-(a) The family {Ri, i k} is independent.

    (b)k

    i=1(Ri

    j=iRj) = 0.(c)k

    i=2(Rii1

    j=1Rj) = 0.(d) Every vectorx R1+...+Rkhas auniquerepresentationx= r1+...+rkwith ri Ri.

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    Linear Subspace - Linear Independence

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    Geometric Control Preliminaries

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    Maps and Matrices

    Usually CL(X,Y) can be written as C : X Y.

    Let xi, i n be a basis for X and yj, j p be a basis for Y.IfC : X Y is a map, then

    Cxi =c1iy1+c2iy2+...+cpiyp, i n (13)

    for uniquely determined elements cji

    F.

    It can be seen that ifx X, then Cx is completely determined by the Cxi;linearity does the rest.

    The array

    Mat C=

    c11 . . . c1n...

    .

    ..cp1 . . . cpn

    (14)is the matrix ofC relative to the given basis pair.

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    Maps and Matrices

    Fundamentally, Mat Ccan be considered as a function p n F.

    The symbol Fpn denotes the class of all p n matrices with elements in F.

    It is turned into a linear space over F, of dimension pn, by the usualoperations of matrix addition and multiplication of matrices by scalars.

    Let C : X Ybe a map.

    X is the domain ofCand Y is the co-domain; the array size ofMat C is

    thus d(X) d(Y).The kernel (ornull space) ofCis the sub-space

    Ker C :={x: x X & Cx= 0} X (15)

    while the imageorrange ofC is the subspace

    Im C := {y:y Y & x X, y=C x (16)

    = {Cx: x X} Y (17)

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    Linear Invariance

    Let A: X X and let S X have the property AS S.

    S is said to be A invariant.

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    Disturbance Decoupling Problem(DDP)

    Consider the system

    x(t) = Ax(t) +Bu(t) +Sq(t); t 0 (18)

    z(t) = Dx(t); t 0 (19)

    The term q(t) represents a disturbance which is assumed not to be directly

    measurable by the controller.The DDP problem is to find (if possible) state feedback F such that q(.) hasno influence on the controlled output z(.).

    Assume that q(.) belongs to a rich function class Qsince the specific featuresof the disturbance are not known.

    Setting Q := R with the continuous R -valued functions on [0, inf), weassume that S: Q X.

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    Disturbance DecouplingLet u(t) =F x(t) in equation18with state feedback F introduced.

    The system represented by equations 18 and 19 are said to be disturbancedecoupledrelative to the pair q(.), z(.), if for each initial state x(0) X, theoutput z(t), t 0, is the same for every q(.) Q.

    Thus disturbance decoupling means that the forced response

    z(t) =D

    t0

    e(ts)(A+BF)Sq(s)ds= 0 (20)

    for all q(.) Q and t 0.

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    Let K:=Ker D and S :=Im S.

    Then from equation20the following result is obtained.

    Lemma 1. The system 18 and 19 is disturbance decoupled if and only ifA+BF|S K.

    In algebraic terms the realization of disturbance decoupling amounts to thefollowing.

    Disturbance Decoupling Problem(DDP). Given

    A: X X, B: U X,S X, and K X,

    find (if possible) F : X U such that

    A+BF|S K (21)

    The subspace on the left on eqn.21is (A + BF)-invariant and if eqn.21is true,it belongs to K.

    Intuitively, DDP will be solvable if and only if the largest subspace havingthese properties contains S.

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    (A, B)-Invariant SubspacesLet A: X X and B: U X.

    A subspace V X isA, B-invariantif there exists a map F : X Usuch that

    (A+BF)V V (22)

    The class of A, B invariant spaces ofX are denoted by T(A, B;X), orsimply T(X) where A and B are fixed.

    It can be observed that any A invariantsubspace is automatically A, Binvariantwith F = 0.

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    Let V be A, B invariant and state feedback F chosen so as to satisfy

    Equation22.The disturbance-free closed loop system

    x= Ax+Bu; u= F x (23)

    then has the property that ifx(0) x0 V,

    x(t) =et(A+BF)x0 V (24)

    for all t, i.e., V is invariantunder the motion.

    Thus V has the property that ifx(0) V then there exists a control u(t)(t

    0) such that x(t) V for all t 0.In other words, the state x() can be held in V by suitable choice ofu().

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    An explicit test to determine whether a given subspace is A, B invariantisgiven by the following result.

    Lemma 4.2LetV X andB =Im B. Then V T(A, B;X) if and only if

    AV V + B (25)

    Proof. Suppose V T(A, B;X) and let v V. By equation 23, (A+BF)v=w for some w V, or

    Av=

    w

    BF vV

    +B

    (26)Conversely, suppose equation25is true. Let v1, v2,...,v be a basis for V.

    By Eqn.25there exists wi V and ui U (i ) such that

    Avi =wi Bui, i (27)

    Defining F0 : V UbyF0vi =ui, i (28)

    and let Fbe any extension ofF0 to X.

    Then (A+BF)vi =wi V, i.e.,(A+BF)V V, so that V T(A, B;X).

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    IfV T(A, B;X)then the class of mapsF : X U, such that(A+BF)V Vis written as F(A, B;V), or simply F(V).

    The notation F F(V) is read F is a friend ofV.

    From the proof of Lemma 4.2, if F F(V) then F F(V) if and only if(F F)V B1V.

    In particular, (F F)|V = 0 ifB is monic and BV = 0.

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    Proposition 4.1Let (A, B) be controllable and V T(A, B;X), with d(V) =v.

    IfF0F(A, B;V)and

    is a symmetric set ofn v complex numbers, thereexists F : X U, such that

    F|V =F0|V (29)

    and(A+BF) =[(A+BF)|V]

    (30)

    Proof

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    Lemma 4.3

    The class of subspacesT(A, B;X) is closed under the operation of subspaceaddition.

    Proof. From Lemma 4.2, ifV1,V2 T(X), then

    A(V1+ V2) = AV1+AV2 (31)

    V1+ V2+ B (32)

    HenceV1+ V2 T(X).

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    IfB is a family of subspaces ofX, the largestorsupremalelement V ofBis defined as that member ofB (when it exists) which contains every memberofB.

    Thus, V B and ifV Bthen V V.

    V

    is unique.We write

    V =sup {V : V B} (33)

    In simple terms, this can be written as V =sup B.

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    Lemma 4.4

    LetB be a nonempty class of subspaces ofX, closed under addition. Then Bcontains a supremal elementV.

    Proof. Since X is finite-dimensional there is an element V B of greatest

    dimension.IfV B, V+V B and sod(V) d(V+V) d(V); i.e., V = V+V

    hence V V and so V is supremal.

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    Let K X be arbitrary, and let T(A, B;K) denote the subclass of (A, B)invariantsubspaces contained in Ksuch that

    T(A, B;K) :={V : V T(A, B;X) & V K (34)

    With A and B fixed, T(A, B;K) can be written as T(K).

    Trivially,0 K, so T(K)=.Since Kis a subspaceLemma 4.3 implies that T(K) is closed under addition.

    Then, Lemma 4.4guarantees the existence of the supremal element

    V :=sup T(K) (35)

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    Theorem 4.1

    Let A : X X and B : U X. Every subspaceK X contains a uniquesupremal(A, B)-invariant subspaceT(K).

    Thus when K=Ker D and z=Dx, a choice of feedback control F F(V)

    where V = sup T(Ker D) renders the system maximally unobservable fromz.

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    Theorem 4.2

    DDP is solvable if and only ifV S (36)

    whereV :=sup T(A, B;K) (37)

    Proof. (If) From Lemma 4.2, F F(V), i.e., (A+BF)V V.

    Using equation36,

    A+BF|S A+BF|V= V K (38)

    (Only if) IfFsolves DDP, the subspace

    V :=A+BF|S (39)

    clearly belongs to T(K), and therefore

    V V S (40)

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    Algorithm for V: Theorem 4.3

    An algorithm by which V can be computed efficiently in a finite number of

    steps can help in determining V without the need for checking condition36ofTheorem 4.2.

    Theorem 4.3. LetA: X X, B: U X, andK X. Define the sequenceV according to

    V0

    = KV = K

    A1(B+V1); = 1, 2, .. (41)

    Then V V1 and for somekd(K),

    V

    =sup T(A, B;K) (42)

    for all k.

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    Principle of Solution to DDP

    IfV T(X) and q() = 0, then by suitable choice of u() the system statex() can always be held in V if it starts there.

    Ifq= 0but ImS V, then the contribution to x(t)by the disturbance Sq(t)(i.e., the first-order-effect ofq() on x()) is also localized to V.

    Under these conditions, the integrated contribution to x() by q() can be

    controlled to remain in V.This contribution is unobservable at z just when V Ker D, and so it isenough to work with V.

    In actual operation with a control of form u = F x+Gv, where F F(V)and v() is a new external input, the system state will generally not remain in

    V

    ; however linearity ensures that the contribution to x()from q() is held inV, which is all that is required to decouple q() from z().

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    References

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    References

    [1] Morgan Jr., The synthesis of linear multivariable systems by state variablefeedback, Proc. 1964 JACC, Stanford, California, pp. 468-472., 1964.

    [2] Wonham, W.M., Linear Multivariable Control - A Geometric Approach,Springer-Verlag, New York, 1979.

    [3] Isidori, A., Nonlinear Control Systems., Springer-Verlag, London, 3rd Ed., 1995.

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