lecture robust zhou

7/30/2019 Lecture Robust Zhou

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Lecture Note on Robust Control

Jyh-Ching JuangDepartment of Electrical Engineering

National Cheng Kung UniversityTainan, TAIWAN

[email protected]

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Motivation of Robust ControlJ.C. Juang

For a physical process, uncertainties may appear as

Parameter variation

Unmodeled or neglected dynamics

Change of operating condition

Environmental factors

External disturbances and noises

Others

Robust control, in short, is to ensure that the performance and stability of feedbacksystems in the presence of uncertainties.

Objectives of the course

Recognize and respect the effects of uncertainties in dynamic systems

Understand how to analyze the performance and robustness of the dynamicsystems in the presence of uncertainties

Learn to use computer aided tools for control system analysis and design

Design of robust controllers to accommodate uncertainties

Apply robust control techniques to engineering problems

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Course RequirementsJ.C. Juang

Prerequisite:

Control engineering

Linear system theory

Multivariable systems

Working knowledge of Matlab

References:

Kemin Zhou with John Doyle, Essentials of Robust Control, Prentice Hall,1998.

G. E. Dullerud and F. Paganini, A Course in Robust Control Theory, Springer-Verlag, 2000.

Doyle, Francis, and Tannenbaum, Feedback Control Theory, Maxwell MacMil-lan, 1992.

Boyd, El Ghaoui, Feron, and Balakrishnan, Linear Matrix Inequalities inSystem and Control Theory, SIAM, 1994.

Zhou, Doyle, and Glover, Robust and Optimal Control, Prentice Hall, 1996.

Maciejowski, Multivariable Feedback Design, Addison-Wiley, 1989.

Skogestad and Postlethwaite, Multivariable Feedback Control, John Wiley& Sons, 1996.

Dahleh and Diaz-Bobillo, Control of Uncertain Systems, Prentice Hall, 1995.

Green and Limebeer, Linear Robust Control, Prentice Hall, 1995.

Many others

Grading:

50% : Homework50% : Term project: paper study and design report

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Course OutlineJ.C. Juang

1. Introduction

2. Preliminaries

(a) Linear algebra and matrix theory

(b) Function space and signal

(c) Linear system theory

(d) Measure of systems (different norms)

(e) Equations for analysis and design

3. Internal Stability

(a) Feedback structure and well-posedness

(b) Coprime factorization

(c) Linear fractional map

(d) Stability and stabilizing controllers

4. Performance and Robustness

(a) Feedback design tradeoffs

(b) Bodes relation

(c) Uncertainty model

(d) Small gain theorem

(e) Robust stability

(f) Robust performance

(g) Structured singular values

5. Controller Design

(a) H2state feedback

(b) Hstate feedback

(c) Observer-based controller

(d) Output injection and separation principle

(e) Output feedback H2controller design

(f) Loop transfer recovery

(g) Output feedback Hcontroller design

(h) synthesis

6. Miscellaneous Topics

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Review of Linear Algebra and Matrix Theory

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Linear SubspacesJ.C. Juang

Let R and C be real and complex scalar field, respectively.

The set of all linear combinations of x1, x2, , xk V is a subspace called the spanof x1, x2, , xk, denoted by

span{x1, x2, , xk} = {x|x = 1x1 + 2x2 + + kxk; i F}

A set of vectors x1, x2, , xk Ware said to be linearly dependent over F if thereexists 1, 2, , k F not all zero such that 1x1 + 2x2 + + kxk = 0; otherwisethey are linearly independent.

Let Wbe a subspace of a vector space V, then a set of vectors {x1, x2, , xk} Wis said to be a basis for W if x1, x2, , xk are linearly independent and W =span{x1, x2, , xk}.

The dimension of a vector subspace Wequals the number of basis vectors. A set of vectors {x1, x2, , xk} in Ware mutually orthogonal if xi xj = 0 for all i = j,

where xi is the complex conjugate transpose of xi.

A collection of subspaces W1, W2, . . ., Wk of V is mutually orthogonal if xy = 0whenever x Wi and y Wj for i = j.

The vectors are orthonormal if xi xj = ij =

1 when i = j0 otherwise

, the Kronecker delta.

Let Wbe a subspace of V. The set of all vectors in Vthat are orthogonal to everyvector in W is the orthogonal complement of Wand is denoted by W.

W

= {y V : y

x = 0, x W} Each vector x in Vcan be expressed uniquely in the form x = xW + xW for xW W

and xW W. A set of vectors {u1, u2, . . . , uk} is said to be an orthonormal basis for a k-dimensional

subspace Wif the vectors form a basis and are orthonormal. Suppose that the di-mension ofVis n, it is then possible to find a set of orthonormal basis {uk+1, . . . , un}such that

W = span{uk+1, . . . , un} Let M

Rmn be a linear transformation from Rn to Rm. M can be viewed as an

m n matrix. The kernel or null space of a linear transformation M is defined as

kerM =N(M) = {x|x Rn : Mx = 0}

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The image or range of M is

imgM = {y|y Rm : y = Mx for some x Rn}

The rank of a matrix M is defined as the dimension of imgM.

An m n matrix M is said to a have full row rank if m n and rank(M) = m. Ithas a full column rank if m

n and rank(M) = n.

Let M be an m n real, full rank matrix with m > n, the orthogonal complementof M is a matrix M of dimension m (m n) such that M M is a square,nonsingular matrix with the following property: MTM = 0.

The following properties hold: (kerM) = imgMT and (imgM) = kerMT.

Homework: Given M =

1 4 20 1 1

. Find kerM, imgM, kerMT, and imgMT.

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Eigenvalues and EigenvectorsJ.C. Juang

The scalar is an eigenvalue of the square matrix M Cnn ifdet(I M) = 0

There are n eigenvalues, which are denoted by i(M), i = 1, 2, , n. The spectrum of M, denoted by spec(M), is the collection of all eigenvalues of M,

i.e., spec(M) = {1, 2, , n}. The spectral radius is defined as

(M) = maxi

|i(M)|

where i is an eigenvalue of M.

If M is a Hermitian matrix, i.e., M = M, then all eigenvalues of M are real. Inthis case, max(M) is used to represent the maximal eigenvalue of M and min isused to represent the minimal eigenvalue of M.

If i spec(M), then any nonzero vector xi Cn that satisfiesMxi = ixi

is a (right) eigenvector of M. Likewise, any nonzero vector wi Cn that satisfieswi M = iw

i

is a left eigenvector of M.

If M is Hermitian, then there exists a unitary matrix U, i.e., UU = I and a realdiagonal such that

M = UU

In this case, the columns of U are the (right) eigenvectors of M.

Homework: Given M =

1 4 24 0 1

2 1 1

. What is (M)? Find U and .

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Matrix Inversion and Pseudo-InverseJ.C. Juang

For a square matrix M, its inverse, denoted by M1 is the matrix such that MM1 =M1M = I.

Let M be a square matrix partitioned as follows

M = M11 M12M21 M22 Suppose that M11 is nonsingular, then M can be decomposed as

M =

I 0

M21M111 I

M11 0

0 S11

I M111 M120 I

where S11 = M22 M21M111 M12 is the Schur complement of M11 in M.

If M (as partitioned above) is nonsingular, then

M1 =

M111 + M

111 M12S

111 M21M

111 M111 M12S111

S111 M21M

111 S

111

If both M11 and M22 are invertible,

(M11 M12M122 M21)1 = M111 + M111 M12(M22 M21M111 M12)1M21M111 The pseudo-inverse (also called Moore-Penrose inverse) of a matrix M is the matrix

M+ that satisfies the following conditions:

1. MM+M = M

2. M+MM+ = M+

3. (MM+) = MM+

4. (M+M) = M+M

Consider the linear matrix equation with unknown X,

AXB = C

The equation is solvable if and only if

AA+CB+B = C

All solutions can be characterized as

X = A+CB+ A+AYBB+ + Yfor some Y.

In the case of no solutions, the best approximation is

Xappr = A+CB+

Homework: Given M =

M11 M12M21 M22

=

1 4 24 0 1

3 1 1

. Use the formula above to

determine M1.

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Invariant SubspacesJ.C. Juang

Let M be a square matrix. A subspace Cn is invariant for the transformationM, or M-invariant, if Mx for every x .

is invariant for M means that the image of under M is contained in : M .

Examples of M-invariant subspaces: eigenspace of M, kerM, and imgM. If is a nontrivial subspace and is M-invariant, then there exist x and such

that Mx = x.

Let 1, . . ., k be eigenvalues of M (not necessarily distinct), and let xi be the cor-responding (generalized) eigenvectors. Then = span{x1, . . . , xk} is an M-invariantsubspace provided that all the lower-rank generalized eigenvectors are included.

More specifically, let 1 = 2 = = l be eigenvalues of M, and let x1, x2, . . ., xl becorresponding eigenvector and the generalized eigenvectors obtained through thefollowing equations:

(M 1I)x1 = 0(M 1I)x2 = x1...

(M 1I)xl = xl1Then a subspace with xj for some j l is an M-invariant subspace if andonly if all lower-rank eigenvectors and generalized eigenvectors of xj are in , i.e.,xi , i j.

An M-invariant subspace Cn is called a stable invariant subspace if all theeigenvalues of M constrained to have negative real parts.

Example

M

x1 x2 x3 x4

=

x1 x2 x3 x4

1 11

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with {1} < 0 (the real part of 1 is less than zero), {3} < 0, and {4} > 0.M-invariant subspace? stable invariant subspace?

span{x1}span{x2}span{x3}span{x4}

span{x1, x2}span{x1, x2, x3}span{x1, x2, x4}

span{x2, x3}

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Vector Norm and Matrix NormJ.C. Juang

Consider a linear space V over the field F. A real-valued function defined on allelements x from Vis called a norm, written , if it satisfies the following axioms:

1. (Nonnegativity) x 0 for all x Vand x = 0 if and only if x = 0.2. (Homogeneity) x = || x for all x Vand F.3. (Triangle inequality) x + y x + y for all x and y in V.

A linear space together with a norm defined on it becomes a normed linear space.

Vector norm: Let x =

x1x2...

xn

be a vector in Cn. The followings are norms on Cn

1. Vector p-norm (for 1 p < ):

xp = ( ni=1

|xi|p)1/p

2. Vector 1-norm: x1 =n

i=1 |xi|3. Vector 2-norm or Euclidean norm: x2 =

xx =

ni=1 |xi|2

4. Vector -norm: x = max1in |xi| If p is a vector norm on Cn and M Cnn. The matrix norm induced by the

vector p-norm p is

M

p = sup

xCn,x=0

Mxpxp

Matrix Norm: Let M be a matrix in Cmn. M =

m11 m12 m1nm21 m22 m2n

.... . .

mm1 mmn

1. Matrix 1-norm (column sum): M1 = maxjm

i=1 |mij|2. Matrix 2-norm: M = M2 =

max(MM).

3. Matrix -norm (row sum): M = maxi

nj=1 |mij|

4. Frobenius norm: MF = (mi=1nj=1 |mij |2) 12 = trace(MM) Homework: Let x =

13

2

and M =

4 0 24 0 1

3 1 1

. Determine x1, x2, x,

M1, M2, M, and MF.

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Singular Value DecompositionJ.C. Juang

The singular values of a matrix M are defined as

i(M) =

i(MM)

Here, i(M)s are real and nonnegative.

The maximal singular value, max, can be shown to be

max(M) = M = maxx=0

Mx2x2

When M is invertible, the maximal singular value of M1 is related to the minimalsingular value of M, min(M), by

max(M1) =

1

min(M)

The rank of M is the same as the number of nonzero singular values of M.

The matrix M and its complex conjugate transpose M have the same singularvalues, i.e., i(M) = i(M

).

Let M Cmn, there exists unitary matrices U = u1 u2 um Cmm andV =

v1 v2 vn

Cnn such thatM = UV

where

= 1 00 0 with 1 =

1 0 00 2 0..

.

. . .00 0 r

with 1 2 r 0 where r = min{m, n}. Singular Value Decomposition. The matrix admits the decomposition

M =r

i=1

iuivi =

u1 u2 ur

1 0 00 2 0...

. . . 00 0 r

v1 v2 vr

kerM = span{

vr+1,

, vn}

and (kerM) = span{

v1,

, vr}

.

imgM = span{u1, , ur} and (imgM) = span{ur+1, , um}. The Frobenius norm of M equals MF =

21 +

22 + + 2r .

The condition number of a matrix M is defined as (M) = max(M) max(M1) pro-

vided that M1 exists.

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Semidefinite MatricesJ.C. Juang

A square Hermitian matrix M = M is positive definite (semidefinite), denoted byM > 0 ( 0), if xMx > 0 ( 0) for all x = 0.

All eigenvalues of a positive definite matrix are positive.

Let M = M11 M12

M21 M22

be a Hermitian matrix. Then M > 0 if and only if M11 > 0

and S11 = M22 M21M111 M12 > 0. Let M 0. There exists a unitary matrix U and nonnegative diagonal such that

M = UU.

The square root of a positive semidefinite matrix M, denoted by M1/2, satisfiesM1/2 M1/2 = M and M1/2 0. It can be shown that M1/2 = U1/2U.

For a positive definite matrix M, its inverse M1 exists and is positive definite.

For two positive definite matrices M1 and M2, we have M1 + M2 > 0 when > 0and 0.

Homework: Let M =

5 33 2

and assume the 2 1 vector x has a 2-norm that is

less than or equal to 1. In the 2-dimensional plane, draw the possible values ofMx.

Homework: Let M =

5 3 13 2 2

1 2 x

. Find the minal value of x such that M is positive

definite.

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Matrix CalculusJ.C. Juang

Let x = [xi] be an n-vector and f(x) be a scalar function on x, then the firstderivative gradient vector is defined as

f(x)x

= [gi] =

f(x)x1

f(x)x2

...f(x)xn

and the second derivative Hessian matrix is

2f(x)

x2= [Hij] =

2f(x)

x21

2f(x)x1 xn

......

2f(x)xn x1

2f(x)x2n

Let X = [Xij] be an mn matrix and f(X) be a scalar function on X. The derivativeof f(X) with respect to X is defined as

Xf(X) =

Xijf(X)

Formula for derivatives (assuming that x and y are vectors and A, B, and X arereal matrices)

x(yTAx) = ATy

x (xTAx) = (A + AT)x

(traceX)

X= I

(traceAXB)

X= ATBT

(traceAXTB)

X= BA

(traceXTAX)

X= (A + AT)X

(traceAXBX)

X = AT

XT

BT

+ BT

XT

AT

(log det X)

X= (XT)1

(det X)

X= (det X)(XT)1

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Functional Space and Signal

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Function SpacesJ.C. Juang

Let V be a vector space over C and be a norm defined on V. Then V is anormed space.

The space Lp(, ) (for 1 p < ) consists of all Lebesgue measurable functionsw(t) defined on the interval (

,

) such that

wp := (

|w(t)|pdt)1/p <

The space L(, ) consists of all Lebesgue measurable functions w(t) with abounded

w := ess suptR

|w(t)|

Let Vbe a vector space over C. An inner product on Vis a complex value function, : V V C

such that for any u,v,w

Vand ,

C

1. u,v + w = u, v + u, w2. u, v = v, u3. u, u > 0 if u = 0

A vector space Vwith an inner product is an inner product space. A Hilbert space is a complete inner product space with the norm induced by its

inner product.

The space L2 = L2(, ) is a Hilbert space of matrix-valued functions on R, withinner product

u, v :=

trace(u

(t)v(t)) dt

The space L2+ = L2[0, ) is a subspace ofL2(, ) with functions zero for t < 0. The space L2 = L2(, 0] is a subspace ofL2(, ) with functions zero for t > 0. Let Vbe a Hilbert space and U Va subset. Then the orthogonal complement ofU, denoted by U is defined as

U = {u : u, v = 0, v U, u V}The orthogonal complement is a Hilbert space.

Let

W=

L2+

L2, then

W =

L2.

Let U and W be subspaces of a vector space V. V is said to be the direct sumof U and W, written V= U W, if UW= {0} and every element v V can beexpressed as v = u + w with u Uand w W. IfVis an inner product space and Uand Ware orthogonal, then Vis said to be the orthogonal direct sum ofUand W.

The space L2 is the orthogonal direct sum of L2+ and L2.

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Power and Spectral SignalsJ.C. Juang

Let w(t) be a function of time. The autocorrelation matrix is

Rww() := limT

1

2T T

T

w(t + )w(t) dt

if the limit exists and is finite for all .

The Fourier transform of the autocorrelation matrix is the spectral density, de-noted as Sww(j)

Sww(j) :=

Rww()ejd

The autocorrelation can be obtained from the spectral density by performing aninverse Fourier transform

Rww() =1

2

Sww(j)ej d

A signal is a power signal if the autocorrelation matrix Rww() exists for all andthe power spectral density function Sww(j) exists.

The power of w(t) is defined as

wrms :=

limT

1

2T

TT

w(t)2 dt =

trace[Rww(0)]

In terms of the spectral density function,

w2rms = 12

trace[Sww(j)] d

Note that rms is not a norm, since a finite-duration signal has a zero rms value.

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Signal QuantizationJ.C. Juang

For a signal w(t) mapping from [0, ) to R (or Rm), its measures can be defined as1. -norm (peak value)

w = supt0

|w(t)|or

w = supt0

max1im

|wi(t)| = max1im

wi

2. 2-norm (total energy)

w2 =

0

w2(t) dt

or

w2 =

0

m

i=1w2i (t) dt =

m

i=1wi22 =

0

wT(t)w(t) dt

3. 1-norm (resource consumption)

w1 =0

|w(t)| dt

or

w1 =0

mi=1

|wi(t)| dt =mi=1

wi1

4. p-norm

w

p = (

0 |w(t)

|p dt)1/p

5. rms (root mean square) value (average power)

wrms =

limT

1

T

T0

w2(t) dt

or

wrms =

limT

1

T

T0

wT(t)w(t) dt

If

w

2 0, and final state xf, there exists a piecewisecontinuous input w() such that x(t) = xf.

2. The controllability matrix B AB A2B An1B has full row rank, i.e.,< A|imgB >:= ni=1 img(Ai1B) = Rn.

3. The matrix

Wc(t) =

t0

eABBTeAT d

is positive definite for any t > 0.

4. PBH test: The matrix

A I B has full row rank for all in C.5. Let and x be any eigenvalue and any corresponding left eigenvector of A, i.e.,

xA = x, then xB = 0.6. The eigenvalues of A + BF can be freely assigned by a suitable choice of F.

The pair (A, B) is stabilizable if and only if

1. The matrix

A I B has full row rank for all in C with {} 0.2. For all and x such that xA = x and {} 0, we have xB = 0.3. There exists a matrix F such that A + BF is Hurwitz.

The pair (C, A) is observable if and only if

1. For any t > 0, the initial state xo can be determined from the time history ofthe input w(t) and the output z(t) in the interval [0, t].

2. The observability matrix

CCA

CA2

...CAn1

has full column rank, i.e.,n

i=1 ker(CAi1) =

0.

3. The matrix

Wo(t) =

t0

eATCTCeA d

is positive definite for any t > 0.

4. The matrix A IC has full column rank for all in C.5. The eigenvalues of A + HC can be freely assigned by a suitable choice of H.

6. For all and x such that Ax = x, we have Cx = 0.7. The pair (AT, CT) is controllable.

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The pair (C, A) is detectable if and only if

1. The matrix

A I

C

has full-column rank for all {} 0.

2. There exists a matrix H such that A + HC is Hurwitz.

3. For all and x such that Ax = x and {} 0, we have Cx = 0. Kalman canonical decomposition: through a nonsingular coordinate transforma-

tion T, the realization admits

T AT1 T BCT1 D

=

Aco 0 A13 0 BcoA21 Aco A23 A24 Bco

0 0 Aco 0 00 0 A43 Aco 0

Cco 0 Cco 0 D

In this case, the transfer function

M(s) = D + C(sI A)1B = D + Cco(sI Aco)1Bco The state space realization (A,B,C,D) is minimal if and only if(A, B) is controllable

and (C, A) is observable.

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State Space AlgebraJ.C. Juang

Let

M1(s) :

x1 = A1x1 + B1w1z1 = C1x1 + D1w1

and M2(s) :

x2 = A2x2 + B1w2z2 = C2x2 + D2w2

In terms of the compact representation, M1(s) s= A1 B1C1 D1 and M2(s) s= A2 B2C2 D2 . Parallel connection of M1(s) and M2(s).

w

-w2 M2(s)

-w1 M1(s)

z2

z1

6?c -z

The new state is x =

x1x2

, input w = w1 = w2, and output z = z1 + z2.

Thus

x =

A1 00 A2

x +

B1B2

w and z =

C1 C2

x + (D1 + D2)w

or M1(s) + M2(s)s

=

A1 0 B10 A2 B2

C1 C2 D1 + D2

.

Series connection of two systems

-w2 M2(s) -z2 = w1 M1(s) -

z1

The connection gives M1(s)M2(s)s

=

A1 B1C1 D1

A2 B2C2 D2

=

A1 B1C2 B1D20 A2 B2

C1 D1C2 D1D2

A2 0 B2B1C2 A1 B1D2D1C2 C1 D1D2

. Inverse system M11 (s)

s=

A1 B1D11 C1 B1D11

C1D11 D11

provided that D1 is invertible.

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Can be verified that M1(s)M11 (s) = I.

w1 = D11 z1 D11 C1x1

x1 = A1x1 + B1(D11 z1 D11 C1x1) = (A B1D11 C1)x1 + B1D11 z1

Transpose or dual system MT1 (s)s

=

AT1 C

T1

BT1 DT1

.

Conjugate system M1 (s) = MT1 (s) s=

AT1 CT1BT1 D

T1

. Thus, M1 (j) = M

1 (j).

Feedback connection of M1(s) and M2(s).

-c6

-w1w M1(s) -z = z1

w2M2(s)z2

The connection gives z = z1 = w2 and w1 = w + z2.

Thus,

z = C1x1 + D1(w + z2)

z2 = C2x2 + D2z

z = (I D1D2)1(C1x1 + D1C2x2 + D1w)z2 = (I D2D1)1(D2C1x1 + C2x2 + D2D1w)x1 = A1x1 + B1w + B1(I D2D1)1(D2C1x1 + C2x2 + D2D1w)x2 = A2x2 + B2(I D1D2)1(C1x1 + D1C2x2 + D1w)

(I M1(s)M2(s))1M1(s)s

=

A1 + B1(I D2D1)1D2C1 B1(I D2D1)1C2 B1 + B1(I D2D1)1D2D1B2(I D1D2)1C1 A2 + B2(I D1D2)1D1C2 B2(I D1D2)1D1

(I D1D2)1C1 (I D1D2)1C2 (I D1D2)1D1

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System Poles and ZerosJ.C. Juang

Let M(s) = D + C(sI A)1B s=

A BC D

. The eigenvalues of A are the poles of

M(s).

The system matrix of M(s) is defined as

Q(s) =

A sI B

C D

which is a polynomial matrix.

The normal rank of Q(s), denoted normal rank, is the maximally possible rank ofQ(s) for at least one s C.

A complex number o C is called an invariant zero of the system realization if itsatisfies

rank A oI BC D < normal rank A sI BC D The invariant zeros are not changed by constant state feedback, constant output

injection, or similarity transformation.

Suppose that

A sI B

C D

has full-column normal rank. Then o C is an invari-

ant zero if and only if there exist 0 = x Cn and w Cm such thatA oI B

C D

xw

= 0

Moreover, if w = 0, then o is also a nonobservable mode. When the system is square, i.e., equal number of input and output, then the

invariant zeros can be computed by solving a generalized eigenvalue problem.A BC D

xw

=

I 00 0

xw

for some generalized eigenvalue and generalized eigenvector

xw

.

Suppose that A sI BC D has full-row normal rank. Then o C is an invariantzero if and only if there exist 0 = y Cn and v Cp such that

y v A oI B

C D

= 0

Moreover, if v = 0, then o is also a noncontrollable mode.

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The system M(s) has full-column normal rank if and only if

A sI B

C D

has

full-column normal rank.

Note that A sI B

C D

=

I 0

C(A sI)1 I

A sI B0 M(s)

and

normal rank

A sI B

C D

= n + normal rank M(s)

Let M(s) be a p m transfer matrix and let (A,B,C,D) be a minimal realization. Ifo is a zero of M(s) that is distinct from the poles, then there exists an input andinitial state such that the output of the system z(t) is zero for all t.

Let xo and wo be such that

A oI B

C D xowo = 0

i.e.,Axo + Bwo = oxo and Cxo + Dwo = 0

Consider the input w(t) = woeot and initial state x(0) = xo.

The output is

z(t) = CeAtxo +

t0

CeA(t)Bw() d + Dw(t)

= CeAtxo + CeAt

t0

e(oIA)(oI A)xo d + Dwoeot

= CeAtxo + CeAt e(oIA)|t0xo + Dwoeot= 0

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Measure of System and Fundamental Equations

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H2 SpaceJ.C. Juang

Let Sbe an open set in C and let f(s) be a complex-valued function defined on S.Then f(s) is said to be analytic at a point zo in S if it is differentiable at zo andalso at each point in some neighborhood of zo.

If f(s) is analytic at zo then f has continuous derivatives of all orders at zo. A function f(s) is said to be analytic in S if it has a derivative or is analytic at

each point of S. A matrix-valued function is analytic in Sif every element of the matrix is analytic

in S. All real rational stable transfer function matrices are analytic in the right-half

plane.

Maximum modulus theorem. Iff(s) is defined and continuous on a closed-boundedset

Sand analytic on the interior of

S, then

|f(s)

|cannot attain the maximum in

the interior of Sunless f(s) is a constant. |f(s)| can only achieve its maximum on the boundary of S, i.e.,

maxsS

|f(s)| = maxsS

|f(s)|

where Sdenotes the boundary of S. The space L2 = L2(jR) is a Hilbert space of matrix-valued functions on jR and

consists of all complex matrix functions M such that

trace[M(j)M(j)] d 0. The corresponding norm is defined as

M22 := sup>0

{ 12

trace[M( +j)M( +j)] d}

It can be shown that

M22 =1

2

trace[M(j)M(j)] d

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The real rational subspace of H2, which consists of all strictly proper and realrational stable transfer function matrices is denoted by RH2.

The space H2 is the orthogonal complement of H2 in L2. IfM is a strictly proper, stable, real rational transfer function matrix, then M H2

and M H2 . Parsevals Relation: Isometric isomorphism between the

L2 spaces in the time

domain and the L2 spaces in the frequency domain.L2(, ) L2(jR)

L2[0, ) H2L2(, 0] H2

If m(t) L2(, ) and its bilateral Laplace transform is M(s) L2(jR), thenM2 = m2

Define an orthogonal projection P+ :

L2(

,

)

L2[0,

) such that for any func-

tion m(t) L2(, )P+m(t) =

m(t) for t 0

0 otherwise

On the other hand, the operator P from L2(, ) to L2(, 0] is defined as

Pm(t) =

0 for t > 0

m(t) for t 0

Relationships among function spaces

-

-

-

? ?

6 6

Inverse transform

Laplace transform

Inverse transform

Laplace transform

Inverse transform

Laplace transform

L2(, 0]

L2(, )

L2[0, )

H2

L2(jR)

H2

P+

P

P+

P

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H SpaceJ.C. Juang

The space L(jR) is a Banach space of matrix-valued functions that are essentiallybounded on jR, with norm

M

:= ess sup

R

max[M(j)]

The rational subspace ofL denoted by RL consists of all proper and real rationaltransfer function matrices with no poles on the imaginary axis.

The space H is a subspace of L with functions that are analytic and boundedin the open right-half plane. The H norm is defined as

M := sup{s}>0

max[M(s)] = supR

max[M(j)]

The real rational subspace of H is denoted by RH, which consists of all properand real rational stable transfer function matrices.

The space H is a subspace of L with functions that are analytic and boundedin the open left-half plane. The H norm is defined as

M := sup{s}


31/141

Lyapunov EquationJ.C. Juang

Given A Rnn and Q = QT Rnn, the equation on X Rnn

ATX+ XA + Q = 0

is called a Lyapunov equation. Define the map

: Rnn Rnn, (X) = ATX + XAThen the Lyapunov equation has a solution X if and only if Q img.

The solution is unique if and only if is injective.

The Lyapunov equation has a unique solution if and only if A has no two eigenvaluessum to zero.

Assume that A is stable, then

1. X = 0 eATtQeAt dt.2. X > 0 if Q > 0 and X 0 if Q 0.3. If Q 0, then (Q, A) is observable if and only if X > 0.

Suppose A, Q, and X satisfy the Lyapunov equation, then

1. {i(A)} 0 if X > 0 and Q 0.2. A is stable if X > 0 and Q > 0.

3. A is stable if (Q, A) is detectable, Q 0, and X 0. Proof:

(a) Let there be x = 0 such that Ax = x with {} 0.(b) Form x(ATX+XA+Q)x = 0 which can be reduced to 2{} xXx+xQx =

0.

(c) Thus xQx = 0

(d) Leads to Qx = 0

(e) Implies that

A I

Q

x = 0, which contradicts the detectability assump-

tion.

Homework: Let A =

2 1 10 2 51 4 4

and Q =

1 0 00 1 00 0 1

. Determine X such that

ATX+ XA + Q = 0.

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Controllability and Observability GramiansJ.C. Juang

M(s)- -w z

6xo

Consider the stable transfer function from M(s) with z = M(s)w. Assume thefollowing minimal realization

M(s) = C(sI A)1BThat is,

x = Ax + Bwz = Cx

Suppose that there is no input from 0 to , then the output energy generated bythe initial state can be computed as follows:

[

0

zT(t)z(t) dt]1

2

= {xT0 [0

eATtCTCeAt dt]xo} 12

= [xTo Woxo]1

2

where Wo = 0 eATtCTCeAt dt is the observability gramian of the system M(s). Indeed, Wo is the positive definite solution of the Lyapunov equation

ATWo + WoA + CTC = 0

When A is stable, the system is observable if and only if Wo > 0.

On the other hand, the input from to 0 that drives the state to xo satisfies

xo =

0

eAtBw(t) dt

If we minimize the input energy subject to the reachability condition, the optimalinput can be found to be

w(t) = BTeATtW1c xo

where

Wc =

0

eAtBBTeATt dt =

0

eAtBBTeATt dt

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The matrix Wc is the controllability gramian of the system M(s). It satisfies theLyapunov equation

AWc + WcAT + BBT = 0

When A is stable, the system is controllable if and only if Wc is positive definite.

Note that the input energy (square) needed is

[0

wT(t)w(t) dt]1

2

= {xTo [0

eAtBBTeATt dt]xo}12

= [xTo W1c xo]

1

2

In summary, the observability gramian Wo determines the total energy in thesystem output starting from a given initial state (with no input) and the control-lability gramian Wc determines which points in the state space that can be reachedusing an input with total energy one.

Both Wo and Wc depend on the realization. Through a similarity transformation,the realization (T AT1, T B, CT1) gives the gramians as T WcT

T and TTWoT1,

respectively.

The eigenvalues of WcWo are invariant under similarity transformation, thus, asystem property.

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Balanced Realization and Hankel Singular ValuesJ.C. Juang

Let Wc and Wo be the controllability gramian and observability gramian of thesystem (A,B,C), respectively, i.e.,


andATWo + WoA + CTC = 0

The Hankel singular values are defined as the square roots of the eigenvalues ofWcWo, which are independent of the particular realization.

Let T be the matrix such that T WcWoT1 = =

21 00 22

. . .2n

.

The Hankel singular values of the system are 1, 2, , n (in descending order). The matrix T diagonalizes the controllability and observability gramians. Indeed,

the new realization (T AT1, T B, CT1) admits T WcTT = TTWoT

1 =

1 00 2

. . .n

A realization (A, B, C) is balanced if its controllability and observability gramians

are the same.

The maximal gain from the past input to the future output is defined as the Hankelnorm.

-

6

t

x(t)

sxo

M(s)h = supu(.)L2(,0)

zL2(0,)uL2(,0)

= supxo

(xTo Woxo)1

2

(xTo W1c xo)

1

2

= 1

2max(WcWo)

= 1

The Hankel norm is thus the maximal Hankel singular value.

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Quantification of SystemsJ.C. Juang

The norm of a system is typically defined as the induced norm between its outputand input.

M(s)- -w z

Two classes of approaches

1. Size of the output due to a particular signal or a class of signal2. Relative size of the output and input

System 2-norm (SISO case).

Let m(t) be the impulse response of the system M(s). Its 2-norm is defined as

M2 = (

|m(t)|2 dt) 12

The system 2-norm can be interpreted as the 2-norm of the output due toimpulse input.

By Parseval theorem,

M2 = ( 12

|M(j)|2 d) 12

If the input signal w and output signal z are stochastic in nature. Let Szz()and Sww() be the spectral density of the output and input, respectively. Then

Szz() = Sww() |M(j)|2

Note that

zrms = (1

2 S

zz() d)1

2 = (1

2 |M(j)|

2

Sww() d)1

2

The system 2-norm can then be interpreted as the rms value of the outputsubject to unity spectral density white noise.

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System 2-norm (MIMO case). The system 2-norm is defined as

M2 = ( 12

trace[M(j)M(j)] d)1

2

= (trace

0

m(t)mT(t) dt)1

2

= (1

2

i i(M(j))2 d)1

2

where m(t) is the impulse response.

Let ei be the i-th standard basis vector ofRm. Apply the impulsive input (t)ei to

the system to obtain the impulse response zi(t) = m(t)ei. Then

M22 =mi=1

zi22

The system 2-norm can be interpreted as

1. The rms response due to white noise: M2 = zrms|w=white noise2. The 2-norm response due to impulse input M2 = z2|w=impulse

The system norm can be regarded as the peak value in the Bode magnitude(singular value) plot.

M = sup{s}>0

max(M(s))

= sup

max(M(j)) (When M(s) is stable)

= supwrms=0

Mwrmswrms

= supw2=0

Mw2w2

= supw2=1

Mw2

System 1-norm: peak to peak ratio

M1 = sup

Mww

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State-Space Computation of 2-NormJ.C. Juang

Consider the state space realization of a stable transfer function M(s)

x = Ax + Bw

z = Cx + Dw

In 2-norm computation, A is assumed stable and D is zero.

Recall the impulse response of M(s), m(t), is

m(t) = CeAtB

The 2-norm, according to the definition, satisfies

M22 = trace0

mT(t)m(t) dt

= trace [B

T

(

0 e

ATt

C

T

Ce

At

dt)B]= trace BTWoB

Thus, the 2-norm of M(s) can be computed by solving a Lyapunov function for Wo

ATWo + WoA + CTC = 0

and taking the square root of the trace of BTWoB.

M2 =

trace BTWoB

Similarly, let Wc be the controllability gramian


then,

M2 =

trace CWcCT

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H Norm ComputationJ.C. Juang

The H norm of M(s) can be computed through a search of the maximal singularvalue of M(j) for all .

M(s)

= ess sup

max(M(s))

Let M(s)s

=

A BC D

. Then

M(s) < if and only if max(D) < and H has no eigenvalues on the imaginary axis where

H =

A 0

CTC AT

+

B

CTD

(2I DTD)1 DTC BT Computation of the H norm requires iterations on .

M(s) < if and only if

(s) = 2I M(s)M(s) > 0if and only if (j) is nonsingular for all if and only if 1(s) has no imaginaryaxis pole. But

1(s)s

=

H

B(2I DTD)1

CTD(2I DTD)1

(2I DTD)1DTC (2I

DTD)1BT (2I D

TD)1

Homework: For the system M(s)

s=

2 1 1 11 1 1 2

4 2 1 01 0 1 0

, determine

1. The controllability and observability gramians of the system.

2. The Hankel singular value.

3. The Balanced realization of the system.

4. The H2 and H norms of M(s).

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Linear Matrix InequalityJ.C. Juang

Linear matrix inequality (LMI)

F(x) = F0 +m

i=1xiFi < 0

where x Rm is the variable and the symmetric matrices Fi = FTi Rnn fori = 1, , m are given.

The inequality < must be interpreted as the negative definite.

The LMI is a convex constraint on x, i.e., the set {x|F(x) < 0} is convex. That is, ifx1 and x2 satisfy F(x1) < 0 and F(x2) < 0, then F(x1 + x2) < 0 for positive scalars and such that + = 1.

Multiple LMIs can be rewritten as a single LMI in view of the concatenation

F1(x) < 0, and F2(x) < 0

F1(x) 0

0 F2(x) < 0 Schur complement technique: Assume that Q(x) = QT(x), R(x) = RT(x), and S(x)

depend affinely on x, thenQ(x) S(x)

ST(x) R(x)

< 0 R(x) < 0 and Q(x) S(x)R1(x)ST(x) < 0

That is, nonlinear inequalities of Q(x) S(x)R1(x)ST(x) < 0 can be represented asan LMI.

Let Z(x) be a matrix on x. The constraint Z(x) < 1 is equivalent to I >Z(x)ZT(x) and can be represented as the LMI

I Z(x)ZT(x) I

> 0

Let c(x) be a vector and P(x) be a symmetric matrix, the constraints P(x) > 0and cT(x)P1(x)c(x) < 1 can be represented as the LMI

P(x) c(x)cT(x) 1

> 0

The constrainttrace ST(x)P1(x)S(x) < 1, P(x) > 0

where P(x) = PT(x) and S(x) depend affinely on x can be restated as

trace Q < 1, ST(x)P1(x)S(x) < Q, P (x) > 0

and hence

trace Q < 1,

Q ST(x)

S(x) P(x)

> 0

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Orthogonal complement of a matrix. Let P Rnm be of rank m < n, the orthogonalcomplement of P is a matrix P Rn(nm) such that

PTP = 0

and

P P

is invertible. Finslers lemma. Let P Rnm and R Rnn where rank(P) = m < n. Suppose P

is the orthogonal complement of P then

P PT + R < 0

for some real if and only ifPTRP < 0

To see the above, note that

P PT + R < 0

PTPT

(P PT + R) P P < 0

PTP PTP + PTRP PTRP

PTRP PTRP

< 0

PTRP < 0 and(PTP PTP) + PTRP PTRP(PTRP)1PTRP < 0

PTRP < 0

Generalized Projection lemma. Given P Rnm (rank m < n), Q Rnl (rankl < n), and R = RT

Rnn, there exists K

Rml such that

R + P KQT + QKTPT < 0

if and only ifPTRP < 0 and QTRQ < 0

where P and Q be the orthogonal complements of P and Q, respectively.

Homework: Let P =

2 11 1

3 0

, Q =

1 42 3

3 2

, and R =

4 1 21 8 3

2 3 2

. Find a

matrix K such that R + P KQT + QKTPT < 0.

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Stability and Norm Computation: LMI ApproachJ.C. Juang

Lyapunov Stability. The systemx = Ax

is stable if and only if all the eigenvalues of A are in the left half plane if and only

if there exists an X > 0

such thatAX+ XAT < 0

H Norm Computation. The H norm of the system

A BC D

is less than if

and only if 2I > DTD and there exists an X > 0 such that

ATX+ XA + CTC+ (XB + CTD)(2I DTD)1(BTX + DTC) = 0if and only if 2I > DTD and there exists an X > 0 such that

ATX+ XA + CTC+ (XB + CTD)(2I DTD)1(BTX + DTC) < 0if and only if

ATX+ XA + CTC XB + CTDBTX+ DTC DTD 2I

< 0

if and only if there exists an X > 0 such that

ATX+ XA XB C T

BTX I DTC D I

< 0

H2 Norm Computation. The H2 norm of the system M(s) s=

A BC 0

can be de-

termined from the gramian. Indeed, we have M(s)22 = trace BTWoB where Wosatisfies

ATWo + WoA + CTC < 0

Thus, to find the 2-norm, one can solve positive definite X and Y such that

ATX+ XA + CTC < 0

and Y BTXXB X > 0The 2-norm is determined from trace Y.

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Stability

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Well-PosednessJ.C. Juang

Well-posedness means

Given any initial state and input signals, the state and output are uniquelydetermined.

All transfer functions between different nodes exist and are proper.

In order to investigate the normal behavior of the system and to avoid possiblepathologies, we are in need of this well-posedness condition.

Consider the feedback system in which both Pyu and K are linear, time-invariant,real rational, and proper.

K(s)

Pyu(s)i -6

- i -?-

yu

y

v1

v2

The transfer function from external input

v1v2

to

uy

admits

I KPyu I

uy

=

v1v2

or

(I KPyu)u = v1 + Kv2(I PyuK)y = Pyuv1 + v2

Thus,uy

=

I K

Pyu I1

v1v2

=

(I KPyu)1 0

0 (I PyuK)1

I KPyu I

v1v2

The feedback system is well-posed if and only if the inverse of(I Pyu(s)K(s)) exists

and is proper.

Note that

(I K(s)Pyu(s))1

= I+ K(s) (I Pyu(s)K(s))1

Pyu(s) The feedback system is well-posed if and only if (I Pyu()K()) is invertible or

I K()Pyu() I

is invertible.

In particular, the feedback system is well-posed if either K() or Pyu() is zero.

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Internal StabilityJ.C. Juang

Concepts of Stability

(i) Equilibrium State: An equilibrium state is asymptotically stable (in the senseof Lyapunov) if the state trajectory returns to the original equilibrium point

under perturbations.(ii) Input-Output Behavior: A system is stable if bounded input results in bounded

output.

K(s)

Pyu(s)j -6

- j -?-

yu

y

v1

v2

The system is internally stable if the output

uy

is bounded under bounded

excitation

v1v2

.

Since

uy

= I KPyu I

1 v1v2

= (I KPyu)1 (I KPyu)1K

(I PyuK)1Pyu (I PyuK)1 v1

v2

The system is internally stable if all four transfer function matrices are stablerational transfer function matrices, i.e.,

(I KPyu)1 (I KPyu)1K(I PyuK)1Pyu (I PyuK)1

RH

Internal stability guarantees that all signals in a system are bounded provided thatthe injected signals are bounded.

Internal stability cannot be concluded even if three of the four transfer functionmatrices are stable.

For example,

Pyu =s 1s + 1

and K = 1s 1

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46/141

Concepts of Coprime FactorizationJ.C. Juang

Two elements are coprime if their greatest common divisor (g.c.d.) is a unit.

For example, 8 and 11 are coprime in the set of integer because their g.c.d. is 1.

If a and b are coprime, then there exist (coprime) integers x and y such thatax + by = 1.

In the previous example, 8 7 + 11 (5) = 1. A rational number is the ratio of two integers and indeed any rational number can

be canonically written as a ratio of two coprime numbers.

Two polynomials are coprime if they do not share common zeros.

Two polynomials n(s) and d(s) are coprime if and only if there exists polynomialsu(s) and v(s) such that

n(s)u(s) + d(s)v(s) = 1

For example, for the coprime polynomials n(s) = 2(s 1) and d(s) = s2 s + 1, thereexist u(s) = 12s and v(s) = 1 such that n(s)u(s) + d(s)v(s) = 1.

A rational transfer function can be represented as a fraction of two coprime poly-nomials. For example, the transfer function

2(s 1)s2 s + 1

is a fraction of the polynomials 2(s 1) and s2 s + 1. The Euclids algorithm can be used to find the great common divider and hence

to check coprimeness. A rational transfer function can also be written as a ratio of two stable transfer

functions.2(s 1)

s2 s + 1 =

2(s 1)s2 + s + 1

s2 s + 1s2 + s + 1

1 Benefits:

Stability can be treated in an algebraic manner

State space computations can be carried out

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Coprime FactorizationJ.C. Juang

Let N(s) and D(s) be in RH. N(s) and D(s) are right coprime over RH if thereexist U(s) and V(s), both in RH, such that

U(s)N(s) + V(s)D(s) = I

They are left coprime if there exist U(s) and V(s) in RH such thatN(s)U(s) + D(s)V(s) = I

The above equations are called Bezout identities.

When N(s) and D(s) are right coprime, the concatenated matrix

N(s)D(s)

has a left

inverse, namely

U(s) V(s)

, in RH. Likewise, N(s) and D(s) are left coprimeif and only if

N(s) D(s)

has a right inverse.

G(s) is said to admit a right coprime factorization if G(s) = Nr(s)D1

r (s) for someright coprime Nr(s) and Dr(s).

G(s) = D1l (s)Nl(s) is a left coprime factorization if Nl(s) and Dl(s) are left coprime.

Coprime Factorization Theorem. For every real rational transfer function matrixG(s), there exist right coprime factors Nr(s), Dr(s) and left coprime factors Nl(s),Dl(s) such that

G(s) = Nr(s)D1r (s) = D

1l (s)Nl(s)

Proof: (by construction)

Assume that G(s) admits the (minimal) state space realization

G(s) = D + C(sI A)1B s=

A BC D

Let F and H be stabilizing state feedback gain and output injection gain,respectively. That is, both A + BF and A + HC are stable.

Then, the coprime factors admit the following state space realizations

Nr(s)Dr(s) s= A + BF B

C+ DF DF I

Nl(s) Dl(s) s= A + HC (B + HD) HC D I

Checks:

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Nr(s), Nl(s), Dr(s), and Dl(s) are in RH. This is true because A + BF andA + HC are stable matrices.

G(s) = Nr(s)D1r (s) = D

1l (s)Nl(s).

Nr(s)D1r (s)

s=

A + BF BC+ DF D

A + BF B

F I

1

= A + BF BC+ DF D A BF I =

A + BF BF B0 A B

C+ DF DF D

similarity transformation

=

A + BF 0 00 A B

C+ DF C D

removal of uncontrollable modes

=

A BC D

= G(s)

The coprimeness.

Nr(s)Dr(s)

admits a left inverse in RH. Let

Ur(s) Vr(s)

s=

A + HC H (B + HD)

F 0 I

Then

Ur(s) Vr(s) Nr(s)

Dr(s)

s= A + HC H (B + HD)

F 0 I

A + BF BC+ DF DF I

=

A + HC HC BF B0 A + BF B

F F I

=

A + HC 0 00 A + BF B

F 0 I

= I

Let G = NrD1r where Nr and Dr be right coprime, then G is stable if and only ifD1r is stable.

If G is stable, then we can take Nr = Nl = G, Dr = I, Dl = I, Ur = 0, Ul = 0, Dr = Iand Dl = I.

48


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Double Coprime FactorizationJ.C. Juang

A transfer function matrix Pyu(s) is said to have a double coprime factorizationif there exist a right coprime factorization Pyu(s) = Nr(s)D

1r (s) and a left coprime

factorization Pyu(s) = D1l (s)Nl(s) such that

Vr(s) Ur(s)Nl(s) Dl(s)

Dr(s) Ul(s)Nr(s) Vl(s) = I 00 I Here, Nr, Dr, Nl, Dl, Ur, Ul, Vr, and Vl are all in RH.

The double coprime equations

UrNr + VrDr = I

DlVl + NlUl = I

DlNr + NlDr = 0UrVl + VrUl = 0

Suppose that Pyu(s) is a proper real rational transfer function matrix with thestabilizable and detectable realization

Pyu(s)s

=

A B0C0 D00

and let F and H be such that A + B0F and A + HC0 are both stable. Then aparticular state space realization of the double coprime factors is

Dr(s) Ul(s)

Nr(s) Vl(s)

s=

A + B0F B0 HF I 0

(C0 + D00F) D00 I

Vr(s) Ur(s)Nl(s) Dl(s)

s

=

A + HC0 B0 + HD00 HF I 0

C0 D00 I

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Stability and Coprime FactorizationJ.C. Juang

K(s)

Pyu(s)

j-

6

-

j-?-

yu

y

v1

v2

Internal stability requires thatI K

Pyu I1

=

(I KPyu)1 (I KPyu)1K

(I PyuK)1Pyu (I PyuK)1

RH

A matrix transfer function E(s) RH is called unimodular if its inverse exist andE1(s) RH.

Assume that Pyu(s) admits the following right and left coprime factorizations

Pyu(s) = Nr(s)D1r (s) = D

1l (s)Nl(s)

and assume that K(s) admits the following coprime factorizationsK(s) = V1r (s)Ur(s) = Ul(s)V1l (s)

Then the following statements are equivalent

1. K(s) stabilizes Pyu(s).

2. Ur(s)Nr(s) + Vr(s)Dr(s) is unimodular.

3. Dl(s)Vl(s) + Nl(s)Ul(s) is unimodular.

4.

Dr(s) Ul(s)

Nr(s) Vl(s)

is unimodular.

5. Vr(s) Ur(s)Nl(s) Dl(s)

is unimodular. Proof of 1 and 2.

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Assume Pyu = NrD1r and K = V1r Ur. Let E = Ur(s)Nr(s) + Vr(s)Dr(s). Then

(I KPyu)1 = (I+ V1r UrNrD1r )1 = Dr(UrNr + VrDr)1Vr= DrE

1Vr

(I KPyu)1K = DrE1Ur(I PyuK)1Pyu = NrE1Vr

(I

PyuK)

1 = I + Pyu(I

KPyu)

1K = I

NrE

1Ur

Clearly, when E is unimodular, the system is internally stable.

On the other hand, internal stability implies thatNrDr

E1

Ur Vr

RH Since Nr and Dr are right coprime, there exist Ur and Vr in RH such that

Ur Vr Nr

Dr

= I. Likewise, there exist Nr and Dr such that

Ur Vr

NrDr

=

I.

Hence, Ur Vr

NrDr

E1

Ur Vr

NrDr

= E1 RH

That is, E is invertible in RH. Proof of 1 and 4.

Note thatDr(s) Ul(s)

Nr(s) Vl(s)

=

I UlV

1l

NrD1r I

Dr 00 Vl

=

I K

Pyu I

Dr 00 Vl

The term is invertible provided that the system is well-posed.

Thus, I K

Pyu I1

=

Dr 00 Vl

X

Dr(s) Ul(s)

Nr(s) Vl(s)1

Y1

However, Xand Y are right coprime in the sense that there exist W =

Vr Ur

Nl Dl

and Z =

0 Ur

Nl 0

both in RH such that

W X+ ZY = Vr Ur

Nl Dl Dr 0

0 Vl + 0 Ur

Nl 0 Dr Ul

Nr Vl = I 0

0 I Thus,

I K(s)

Pyu(s) I1

is stable if and only if

Dr(s) Ul(s)

Nr(s) Vl(s)1

is stable if

and only if

Dr(s) Ul(s)

Nr(s) Vl(s)

is unimodular in RH.

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54/141

Standard Form for Feedback DesignJ.C. Juang

Canonical form for feedback systems design and analysis problem.

P(s)

K(s)

-

- -w z

uy

In the above figure, w stands for the exogenous input signal which could be thecommand signal, the disturbance, or the sensor noise depending on the applica-tion. u is the control signal generated by the controller to satisfy certain designrequirements. y is the measured output, the signal used for controller synthesis.Finally, z is the output signal to be controlled which represents the response ofthe system, the tracking error, or the actuation signal.

zy

=

Pzw(s) Pzu(s)Pyw(s) Pyu(s)

wu

The matrix transfer function from

wu

to

zy

, P(s) =

Pzw(s) Pzu(s)Pyw(s) Pyu(s)

, is de-

rived based on the plant dynamics, the design requirements, the interconnection,and the weightings.

u = K(s)y

The objective of the regulator design problem is to synthesize the controller K(s)

such that the design requirements are satisfied. Design requirements may includeinternal stability, achievement of certain input-output relationships, robustness,among others.

Redraw the previous figure as

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Closed-Loop MapJ.C. Juang

The system is governed by

z = Pzww + Pzuu

y = Pyww + Pyuu

andu = Ky

Linear Fractional Map. When the fictitious signals are zero, the transfer function

from the input w to the output z is a linear fractional map. Let P =

Pzw PzuPyw Pyu

,

the linear fractional transform is

Fl(P, K) = Pzw + Pzu K(I Pyu K)1Pyw = Pzw + Pzu(I K Pyu)1K Pyw Scattering Matrix Representation. Let S be the matrix that governs the map from

yu to wz , that is,

wz

= S

yu

=

Swy SwuSzy Szu

yu

Then,

w = P1yw Pyu u + P1yw yand

z = (Pzu Pzw P1yw Pyu)u + PzwP1yw yThus,

S = Swy SwuSzy Szu = P1yw

P1yw Pyu

Pzw P1yw Pzu PzwP1yw Pyu It can be shown that

w = Swyy + Swuu = (Swy + Swu K)y

andz = Szyy + Szuu = (Szy + Szu K)y

Thus,z = (Szy + Szu K)(Swy + Swu K)

1w

This is a scattering matrix representation of the closed-loop map.

State Space Representation. Assume that the transfer function matrix P(s) admitsthe state space realization:

x = Ax + B1w + B0u

z = C1x + D11w + D10u

y = C0x + D01w + D00u

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64/141

All Stabilizing ControllersJ.C. Juang

All stabilizing controllers can be represented in terms of a linear fractional map

K(s) = [Vr(s) Q(s)Nl(s)]1[Ur(s) + Q(s)Dl(s)]=

[Ul(s) + Dr(s)Q(s)][Vl(s)

Nr(s)Q(s)]

1

and thusK(s) = Fl(J(s), Q(s))

for Q(s) in RH and

J(s) =

Ul(s)V1l (s) V1r (s)V1l (s) V

1l (s)Nr(s)

or

V1r (s)Ur(s) V1r (s)V1l (s) Nl(s)V

1r (s)

To see this,

K(s)

= (Ul + DrQ)(Vl NrQ)1= Ul(Vl NrQ)1 DrQ(Vl NrQ)1= UlV1l UlV1l NrQ(Vl NrQ)1 DrQ(Vl NrQ)1= UlV1l (Dr + UlV1l Nr)Q(Vl NrQ)1= UlV1l V1r Q(I V1l NrQ)1V1l= Fl(J(s), Q(s))

The matrix J(s) in the linear fractional controller parametrization is

J(s)s

=A + B0F + HC0 + HD00F H B0 + HD00

F 0 IC0 + D00F I D00

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Block Diagram InterpretationJ.C. Juang

Controller design problem

K(s)

P(s)

- -

-

w z

u y

After controller parametrization

Q(s)

J(s)

P(s)- --

-

w z

u y

The closed-loop map after stabilization

Q(s)

T(s)- --

w z

or

- T01(s) - Q(s) - T10(s) -b -zw- T11(s)

?

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Feedback Control and Performance Limitations

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The sensitivity function also governs the response from the command r to thetracking error e. Thus for good tracking performance, i.e., small tracking error,S must be small as well.

Let Lo(s) be the nominal loop transfer function, then the map from the com-mand r to the output y is characterized by the transfer function To(s) where

To(s) =Lo(s)

1 + Lo(s)Suppose that due to neglected dynamics, perturbations, and so forth, theactual loop transfer function becomes L(s), the resulting closed-loop transferfunction becomes

T(s) =L(s)

1 + L(s)

The sensitivity function S(s) of the closed-loop map T(s) with respect to thevariation of the loop transfer function L is

S(s) =TToT

LLo

L

=1

1 + Lo(s)Thus, S(s) must be maintained small in order to reduce the effect of plantuncertainty on the closed-loop map.

But, how small is small? Recall that in the open-loop situation, the transferfunction from d to y is 1, thus, it is generally required that |S(s)| to be smallerthan 1 at all ss.

Let D C be an open set and let f(s) be a complex valued function definedon D. Then f(s) is said to be analytic at a point zo in D if its is differentiableat zo and also at each point in some neighborhood of zo. In other words, thefunction has a power series representation around zo.

A function f(s) is said to be analytic in D if it is analytic at each point of D. The maximum modulus theorem states that if a complex valued function f

of complex variable is analytic inside and on the boundary of some domainD, then the maximum modulus (magnitude) of the function f occurs on theboundary of the domain D.

In other words, since S(s) is required to be stable, i.e., it is analytic in the closedright half plane (including at infinity), the maximum of |S(s)| is achieved onthe imaginary axis. It is thus equivalent to evaluating |S(j)|.

6a-r - e u ?d

- K(s) - G(s) -a -yc G(s) -a -?d

yo

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-

6

u Re L(j)

Im L(j)

1

Note that the number of poles of 1 + L in RHP is the same as the number ofunstable open loop poles.

Assume that both G(s) and K(s) are stable, then closed-loop stability meansthat the Nyquist plot is not allowed to circle around 1.

The critical issue for the encirclement is the phase 180o with unity gain. Let L be a perturbation that creates marginally instability, i.e.,

Lo(j) + L = 1Then, to have robust stability, L must be large. This is to require that

L

Lo= 1 + Lo

Lo

be large. In other words,

T(s) =Lo(s)

1 + Lo(s)

must be small in magnitude at the critical frequencies. The function T(s) is

the complementary sensitivity function. Another Limitation: The sensitivity and complementary sensitivity functionssum to one.

S(s) + T(s) = 1 , sS(s) and T(s) cannot be made small at the same time. There is a conflictbetween achieving good sensitivity and good stability margin.

In summary, feedback can be used to

Shape system response

Attenuate disturbance

Accommodate uncertainty Adapt toward variation

and yet it is subject to fundamental limitations on the achievable performance ofa feedback system. In particular, the tradeoffs between sensitivity function andcomplementary sensitivity function must be accounted for. A small sensitivityfunction implies that

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Small sensitivity toward variations of GK

Good disturbance rejection

Small tracking error

On the other hand, a small complementary sensitivity function means

Large stability margin

Good noise rejection Robust against variation of sensor dynamics

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Multivariable Feedback SystemJ.C. Juang

Consider the feedback design problem in the figure

r -f - K(s) -u f -upd1

G(s) -f -?

yp yo

d2

f n6

??

The input loop transfer function (matrix) or the transfer function at the inputnode Lin(s) is defined as

Lin(s) = K(s)G(s)

Likewise, the output loop transfer function is

Lout(s) = G(s)K(s)

The input sensitivity function is defined as the transfer function matrix from theinput disturbance d1 to the plant input signal up

Sin(s) = [I+ Lin(s)]1 = [I+ K(s)G(s)]1

andup = Sind1

The output sensitivity function is defined as the transfer function matrix from the

output disturbance d2 to the output signal yo

Sout(s) = [I+ Lout(s)]1 = [I + G(s)K(s)]1

andyo = Soutd2

The matrix I+ Lin(s) is called input return difference matrix and I+ Lout(s) is theoutput return difference matrix.

The input and output complementary sensitivity functions are defined, respec-tively, as

Tin(s) = I Sin(s) = Lin(s)[I+ Lin(s)]1 = K(s)G(s)[I+ K(s)G(s)]1

andTout(s) = I Sout(s) = Lout(s)[I + Lout(s)]1 = G(s)K(s)[I + G(s)K(s)]1

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For a nonminimum phase system, the constraint becomes more strigent. Indeed,consider

L(s) =s + zs + z

Lmp(s)

where Lmp(s) is stable and minimum phase and z > 0. Then

L(jo) = Lmp(jo) + jo + zjo + z

Since jo+zjo+z

< 0, an extra phase is introduced. In particular, when o = z,

jo+zjo+z

= 90 and when o = z/2, jo+zjo+z = 53.13

A rule of thumb is to have the crossover frquency

c < z/2

Let L(s) be the open-loop transfer function with at least two more poles than zerosand let pis be the open right-half plane poles of L(s). Then, the Bodes sensitivity

integral states that 0

ln |S(j)| d = mi=1

{pi} 0

If the sensitivity is kept low in certain frequency range, there will exist somefrequency range in which |S(j)| is large. This is the so-called water bed effect.

Suppose that L(s) has a single real right-half plane zero z, then0

ln |S(j)| 2zz2 + 2

d = lnmi=1

|pi + zpi z|

The function 2zz2+2

acts as a weighting on the sensitivity function and further limitsthe design tradeoffs.

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82/141

Robustness Analysis MotivationJ.C. Juang

Let Go(s) =s1

s(s2)and K(s) be a stabilizing controller.

Suppose that the open loop plant is perturbed as G(s) = s1+2s(s21) for some 1 and

2.

6c-

- K(s) -c - 1s - 1s - 1 -c -

1

26c

6?

- 2

6

w1 z1 z2 w2

G(s)

Go(s)

The problem of robust stability is to assess how much tolerance on the perturba-tions 1 and 2 the system can endure before going unstable.

The robustness analysis problem can be formulated as the following standard formwhere the perturbation is cast into an uncertainty block and the known dynamicis represented in another block.

1

s21

1+KGo 1

s2K

1+KGo1

s(s2)1

1+KGo 1

s(s2)K

1+KGo

1 00 2

-

-

w1

w2

z2

z1

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Plant Uncertainty and RobustnessJ.C. Juang

Consider the system in the figure where Lo(s) is the nominal loop transfer func-tion and (s) is a multiplicative perturbation. The actual loop transfer functionbecomes L(s) = (I + (s))Lo(s).

6e-

- Lo(s) - I + (s) -

L(s)

The system can be redrawn as the standard form

Lo(I+ Lo)1

-

w z

If the nominal system is closed-loop stable, (s) is open-loop stable, and

max((j)) max[Lo(j)(I + Lo(j))1] < 1for all , then the perturbed system is closed-loop stable.

Too see this, recall that the characteristic equation is det[I + (I + (s))Lo(s)] = 0.Since (I + Lo) is invertible,

det[I+(I+(s))Lo(s)] = det[(I+Lo(s))+(s)Lo(s)] = det[I+Lo(s)]det[I+(s)Lo(s)[I+Lo(s)]1]

In order for det[I + (s)Lo(s)[I + Lo(s)]1

] = 0, it is necessary and sufficient thatI + (s)Lo(s)[I+ Lo(s)]

1 is nonsingular. In other words,

max[Lo(s)(I+ Lo(s))1] max((s)) < 1

will ensure the nonsingularity of I + (I + (s))Lo(s) and hence the stability of theclosed-loop system.

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86/141

Singular Value Plots and UncertaintiesJ.C. Juang

Complementary sensitivity function at the output node.

6b-

K

- G -b-

? -

Robustness Test

max(GK(I + GK)1) max() < 1

Uncertainty:

Output (sensor) error

Neglected HF dynamics

Changing number of RHP zeros

Multiplicative uncertainty: G

=(I+ )G

Performance:

Sensor noise attenuation

Output response to output com-

mand

Control function at the input node.

6b-K

- G -b

-

? -

Robustness Test

max(K(I + GK)1) max() < 1

Uncertainty:

Additive plant error

Uncertain RHP zeros

Additive uncertainty: G

= G +

Performance:

Input response to output com-mand

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Complementary sensitivity function at the input node.

6b-

K

G --b-

?-

Robustness Test

max(KG(I + KG)1) max() < 1

Uncertainty:

Input (actuator) error

Neglected HF dynamics

Changing number of RHP zeros

Multiplicative uncertainty: G

=

G(I+ )

Performance:

Input response to input com-mand

Sensitivity function at the output node.

6b-

K

- G -b?

-

Robustness Test

max((I + GK)1) max() < 1

Uncertainty:

LF plant parameter errors

Changing number of RHP poles

G

= (I + )1G

Performance:

Output sensitivity

Output error to output com-mands and disturbances

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Robustness MarginJ.C. Juang

Gain margin problem

Real perturbation kg

Test:

|1 + kgL(j)

|= 0

At the phase crossover frequency p, the imaginary part of L(jp) is zero, i.e.,L(jp) = Lr; thus, kg = L(jp)1 = 1/|Lr|.

Phase margin problem

Unity gain, complex perturbation ej

Test: |1 + ejL(j)| = 0 At the gain crossover frequency g, |L(jg)| = 1, i.e., L(jg) = ejL(jg); thus,

= + L(jg).

Robustness margin problem

Complex perturbation kr

Test: |1 + krL(j)| = 0 At the worst-case frequency o, L(jo) = Lr +jLi; thus, |kr| = (

L2r + L

2i )1.

-

6

Re

Im

q1

L(j)

-

6

Re

Im

q1

L(j)

-

6

Re

Im

q1

L(j)

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Small Gain TheoremJ.C. Juang

M(s)

-

Suppose that M(s) RH, then the system is well-posed and internally stable forall (s) RH if and only if

M(s) (s) < 1 Critical stability occurs when

det(I M(j)) = 0for a frequency and an uncertainty .

Since M(s) is stable, the condition for robust stability can be stated as

det(I M(j)) = 0for all and . This is guaranteed if

max(M(j)(j)) < 1 Note that

sup

max(M) sup

[max(M)max()]

sup

max(M) sup

max()

M On the other hand, we can construct the worst-case perturbation as follows. As-

sume that M =

M is achieved at the frequency o and let M(jo) = UMV

(singular value decomposition). Then, by assigning = 1M V U, one has

M(s) (s) = M 1M

= 1

anddet(I M(jo)) = det(I 1M UMVV U) = 0

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The small gain theorem is conservative.

However, when the perturbation is unstructured, complex, and independent, thesmall gain theorem gives a necessary and sufficient condition for robust stability.

Assume that A is stable and the perturbed system has a state transition matrixA + BEC for some known B and C (structure) and unknown E.

The perturbed system can be arranged as a feedback system

B (sI A)1 C

E

- - -

Thus, a sufficient condition for robust stability is that, according to the small gaintheorem,

max(E) 0, the measure ismax

i(M) real;i(M)>0i(M)

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8. For all Q Q and D D,(M) = (MQ) = (QM) (D 12 MD 12 )

9.

maxQQ

(QM) maxB

(M) = (M) infDD

max(D1

2 MD1

2 )

10. max(D1

2 MD1

2 ) is convex in ln(D1

2 ).11. Linear matrix inequality

max(D1

2MD1

2 ) <

max(D 12MD 12D 12MD12 ) < 2 D12MD 12D 12MD 12 2I < 0 MDM 2D < 0

12. Computation (lower bound)

maxQQ (QM) = (M)

Q

M Q-

- Nonconvex, local maximum

Power method

Projection method

13. Computation (Upper bound)

(M) infDD

max(D1

2 MD1

2 )

D1

2 D1

2

D1

2 M D1

2- -

-

Convex optimization

Frobenius

Perron eigenvalue

Linear matrix inequality The bound is tight if 2s + f 3.

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Main Loop TheoremJ.C. Juang

For all 1 1 with max(1) 1, the perturbed closed-loop system is well-posedand stable if and only if

sup(s)0

1(M22) <

1

M =

M11 M12M21 M22

-

- -

For all 1 1 with max(1) 1, the perturbed closed-loop system is well-posed,stable, and

sup(s)0

2Fl(M, 1) <

if and only ifsup(s)0

(M) <

=

1 00 2

M =

M11 M12M21 M22

-

-

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Kharitonov ApproachJ.C. Juang

Stability can be checked by the roots of characteristic equation det(I M) = 0. Let the characteristic polynomial be

p(s, q) =

ni=0

qisi = qnsn + qn1s(n1) + + q1s + q0

where qis are in the interval

q Q or qi qi q+i i

The polynomial is called an interval polynomial p(s, q) =n

i=0 [qi , q

+i ] s

i.

Kharitonov Theorem: The interval polynomial is robustly stable, i.e., all the rootsare in the LHP for all qis in the interval, if and only if its four invariant degreeKharitonov polynomials

k1(s) = q0 + q

1 s + q

+s s

2 + q+3 s3 + q4 s

4 + q5 s5 + q+6 s

6 + k2(s) = q

+0 + q

+1 s + q

s s

2 + q3 s3 + q+4 s

4 + q+5 s5 + q6 s

6 + k3(s) = q

+0 + q

1 s + q

s s

2 + q+3 s3 + q+4 s

4 + q5 s5 + q6 s

6 + k4(s) = q

0 + q

+1 s + q

+s s

2 + q3 s3 + q4 s

4 + q+5 s5 + q+6 s

6 + are stable.

Proof: Fix a frequency o and consider the range p(jo, Q) = {p(jo, q) : q Q}.p(jo, Q) = q0 + q1(jo) + q2(jo)

2 + q3(jo)3 + q4(jo)

4 +

= q0 q22

o + q4

4

o q66

o + +j(q1o q33

o + q5

5

o )Note that (Re and Im stand for realpart and imaginary parts, respec-tively)

minqQ

Rep(jo, q) = Re k1(jo)

maxqQ

Rep(jo, q) = Re k2(jo)

minqQ

Imp(jo, q) =

Im k3(jo), o 0Im k4(jo), o < 0

maxqQ

Imp(jo, q) = Im k4(jo), o 0

Im k3(jo), o < 0-

6

Re

Im

k1 k3

k4 k2

o 0

Thus, the range of p(jo, q) in the complex is bounded by four vertices k1(jo),k2(jo), k3(jo), and k4(jo). When this is true for every frequency, the intervalpolynomial is stable.

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Controller Design

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State Feedback DesignJ.C. Juang

Assume that the state is accessible for feedback.

In terms of the standard design formulation:

x = Ax + B0u + B

1w

y = x

z = C1x + D10u + D11w

Typically, the feedback control is assumed to be a state feedback

u = F x

for some feedback gain F.

The closed-loop map becomes

x = (A + B0F)x + B1wz = (C1 + D10F)x + D11w

Issues in the feedback design

1. Stability

2. Robustness

3. Performance

4. Boundedness of z

5. Boundedness of the map from w to z

Stability The closed-loop (A+B0F) is stable if there exists a positive definite matrixX such that

X(A + B0F) + (A + B0F)TX < 0

Performance The H2-norm of the map from w to z is the square root of traceBT1 XB1where X is the observability gramian

X(A + B0F) + (A + B0F)TX+ (C1 + D10F)

T(C1 + D10F) = 0

Robustness The H-norm of the map from w to z is bounded above by if thereexists a positive definite matrix X such that

X(A + B0F) + (A + B0F)TX+ (C1 + D10F)T(C1 + D10F)

+[XB1 + (C1 + D10F)TD11](

2I DT11D11)1[BT1 X + DT11(C1 + D10F)] = 0

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Quadratic StabilizationJ.C. Juang

Key Concept: Construct a quadratic Lyapunov function to govern the stability(and robustness) of the closed-loop system.

Assume that the Lyapunov function is

V(x) = xTP x

for some positive definite matrix P.

The time derivative of V(x) is

V(x) = xTP(A + B0F)x + xT(A + B0F)

TP x

= xT[P A + ATP + P B0F + FTBT0 P]x

A necessary and sufficient condition for stability is that

P A + ATP + P B0F + FTBT

0P < 0

It suffices to have

P A + ATP + P B0F + FTBT0 P + F

TRF < 0

for some positive definite R.

The latter can be manipulated to become

(F + R1BT0 P)TR(F + R1BT0 P) + P A + A

TP P B0R1BT0 P < 0

By assigning

F = R1BT0 Pwith P satisfying

P > 0 and P A + ATP P B0R1BT0 P < 0a stabilizing state feedback gain is constructed.

Equivalently (by treating P = X1), a stabilizing state feedback gain can be de-signed by solving an X such that

X > 0

and

AX+ XAT B0R1BT0 < 0This is the linear matrix inequality (LMI) approach.

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Matched and Unmatched UncertaintiesJ.C. Juang

Consider the systemx = (A + A)x + (B0 + B)u

for some uncertainties A and B with

A = B0EA and B = B0EB

for some unknown EA and EB.

Uncertainties of the above form, i.e., the uncertainties are in the range space ofB0, are said to satisfy the matching condition.

Assume that

1. The matrix A is asymptotically stable

2. The uncertainty Ea is bounded: Ea eA3. The uncertainty Eb is bounded:

Eb

eB < 1

Again, consider the Lyapunov function

V(x) = xTP x

and the Lyapunov equationATP + P A + Q = 0

for some positive definite Q and positive definite P.

The state feedback gain F is assumed to be

u = F x = r1BT0 P xHere, r is a positive scalar to be determined.

The derivative of V(x) is

V(x) = xTP x + xTPx

= xTQx r1xTP B0(2I+ EB + ETB)BT0 P x + xT(P B0EA + ETABT0 P)x

Note thatQ min(Q)I

P B0(2I+ EB + ETB)B

T0 P 2(1 eB)P B0BT0 P

and P B0Ea + ETa B

T0 P P B0BT0 P + ETa Ea P B0BT0 P + e2AI

Thus, the derivative of the Lyapunov function

V(x) min(Q)xTx 2(1 eB)r1xTP B0BT0 P x + xTP B0BT0 P x + e2AxTx= xT[e2A min(Q)]x + xTP B0[1 2r1(1 eB)]BT0 P x

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By selecting Q and r such that

min(Q) e2A and r 2(1 eB)the system is stable for all permissible A and B.

It is always possible to stabilize bounded, matched uncertainties provided that thestate is accessible.

Consider the systemx = (A + A)x + B0u

whereA =

i

iEiFi

where both Ei and Fi are known (structure of the uncertainty) and i is unknown.A bound on i, however, is available: |i| i.

The uncertainty is unmatched in the sense that the uncertainty is not necessarilyin the range space of B0.

A robustly stabilizing state feedback gain F can be constructed as

F = R1BT0 Pwhere P is the positive definite solution to

ATP + P A + Q 2P B0R1BT0 P + P(i

iEiETi )P +

i

iFTi Fi = 0

for some positive definite Q and R.

To verify the above claim, check the Lyapunov function

V(x) = xTP x

and its time derivative

V(x) = xTQx

i

ixT(ETi P Fi)T(ETi P Fi)x

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Algebraic Riccati EquationJ.C. Juang

Let A, Q, and R be real n n matrices with Q and R symmetric. Define the 2n2nmatrix

H :=

A RQ

AT A matrix of this form is called a Hamiltonian matrix.

The Hamiltonian matrix is related to the following algebraic Riccati equation

ATX+ XA + XRX + Q = 0

The spectrum of H, spec(H), is symmetric with respect to the imaginary axis. Toprove this, introduce the 2n 2n matrix

J :=

0 II 0

having the property J2 = I. Then,J1HJ = JHJ = HT

so H and HT are similar. Thus, spec(H) spec(HT) spec(H)

Assume H has no eigenvalues on the imaginary axis. Then it must have n eigen-values in {s} < 0 and n in {s} > 0. Thus the two spectral subspaces, (H) and+(H), have dimension n. We will be interested in the stable subspace (H) andits associated properties. Partitioning the subspace,

(H) = img

X1X2

where X1 and X2 are in R

nn. If X1 is nonsingular, i.e., if the two subspaces

(H), img

0I

are complementary, we can define

X = X2X11

to get

(H) = img

X1X2

= img

IX

X1 = img

IX

Notice that X is uniquely determined by H, we can thus have

X = Ric(H)

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Ric is a function R2n2n Rnn which maps H to X where

(H) = img

IX

The domain of Ric, denoted by domRic, consists of Hamiltonian matrices H with

two properties:

1. H has no eigenvalues on the imaginary axis,

2. (H) and img

0I

are complementary.

Suppose H domRic and X = Ric(H). Then1. X is symmetric

2. X satisfies the equation

ATX+ XA + XRX Q = 03. A + RX is stable

Proof of 1.

Let X1 and X2 be as above, i.e., (H) = img

X1X2

. It is claimed that

XT1 X2 is symmetric

To prove this, note that there exists a stable matrix H in Rnn such that

H

X1X2

=

X1X2

H

Premultiply this equation by

X1X2 T

J

to get X1X2

TJH

X1X2

=

X1X2

TJ

X1X2

H

Now, JH is symmetric; hence so is the left hand side of the above equation. Henceso is the right hand side:

(XT1 X2 + XT2 X1)H = HT(XT1 X2 + XT2 X1)T = HT(XT1 X2 + XT2 X1)But this is a Lyapunov equation. Since H is stable, the unique solution is

XT1 X2 + XT2 X1 = 0We have

XX1 = X2

Premultiply this by XT1 , we have XT1 XX1 is symmetric. Hence, X is symmetric

since X1 is nonsingular.

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Bounded Real LemmaJ.C. Juang

Suppose that M(s) = D+C(sIA)1B with A asymptotically stable. Then M(s) < if and only if

1. D <

2. There exists X = XT

satisfying the Riccati equationATX+ XA + CTC+ (XB + CTD)(2I DTD)1(BTX+ DTC) = 0

such that A + BR1(DTC+ BTX) is asymptotically stable.

Furthermore, when such an X exists, X 0. Proof

1. It amounts to performing a spectral decomposition, i.e., finding a stable N(s)such that

2I M(s)M(s) = N(s)N(s)2. Recall that

2I M(s)M(s) s= A 0 BCTC AT CTD

DTC BT 2I DTD

Through a similarity transformation of

I 0

X I

s=

A 0 BCTC XA ATX AT XB + CTD

DTC+ BTX BT 2I DTD

3. Let N(s) s= A B

Cx Dx

. Then

N(s)N(s)s

=

AT CTxB DTx

A B

Cx Dx

s

=

A 0 BCTx Cx AT CTx Dx

DTx Cx BT DTx Dx

4. Comparing the coefficients

DTx Dx = 2I DTD

DTx Cx = BTX DTC

C

T

x Cx = CT

C XA AT

X5. This leads to the Riccati equation.

A dual version is to find a positive semidefinite matrix Y such that

AY + Y AT + BBT + (Y CT + BDT)(2I DDT)1(CY + DBT) = 0

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Riccati Equation in Control DesignJ.C. Juang

Theorem: Suppose H =

A BBT

CTC AT

with (A, B) stabilizable and (C, A) de-

tectable. Then H domRic and Ric(H) 0. If (C, A) is observable, Ric(H) > 0.

The corresponding algebraic Riccati equation isATX+ XA XBBTX + CTC = 0

Proof:

1. H is in the domain of Riccati.

(a) H has no eigenvalues on the imaginary axis.

Let j and

xz

be an eigenpair. Then

(A

j)x = BBTz and

(A

j)z = CTCx

or z, (A j)x = BTz2 and x, (A j)z = Cx2. Thus, Cx2 =z, (A j)x = BTz2. This implies that BTz = 0 and Cx = 0.

The assumptions on stabilizability and detectability are violated since

z

A jI B = 0 and A jIC

x = 0. Hence, there are no eigenval-

ues on the j axis.

(b) The two subspaces (H) and img

0I

are complementary. Recall that

(H) = imgX1X2 and H

X1X2 =

X1X2 H. The two subspaces are com-plementary is equivalent to that X1 is nonsingular, i.e., kerX1 = 0.

The subspace kerX1 is H-invariant. To see this, note that

AX1 BBTX2 = X1HLet x kerX1, then

xX2 (AX1 BBTX2)x = xX2X1Hxor (by recalling that X1X2 is symmetric) xX2BBTX2x = xX1X2Hx, orxX2BBTX2x = 0, or BTX2x = 0. This implies that

X1Hx = 0

That is, Hx is in the kernel of X1.

The fact that X1 is nonsingular is proved by contradiction. SupposekerX1 = 0, then H|kerX1 has an eigenvalue and eigenvector x such that

Hx = x {} < 0, 0 = x kerX1

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Note that CTCX1 ATX2 = X2H. Postmultiplying by x gives(AT + )X2x = 0

Since BTX2x = 0, we have

xX2

A + B

= 0

Stabilizability implies that X2x = 0. But if X

1x = 0 and X

2x = 0, then

x = 0, a contradiction.

2. To prove that X 0. Set X = Ric(H). ThenATX + XA XBBTX+ CTC = 0

or(A BBTX)TX+ X(A BBTX) + (XBBTX + CTC) = 0

Note that A BBTX is stable, we have

X =

0

e(ABBTX)Tt(XBBTX+ CTC)e(ABB

TX)t dt

Since XBBTX+ CTC is positive semidefinite, so is X.

3. The statement that X > 0 when (C, A) is observable is proved by contradiction.

Suppose that there exists a stable unobservable mode with {} < 0 suchthat

A I

C

x = 0 for a nonzero vector x.

From the Riccati equation

2({}) xXx xXBBTXx = 0 Thus, xXx = 0. That is X is singular.

Conversely, if there exists a nonzero vector x kerX, then the Riccatiequation implies that Cx = 0 and XAx = 0.

The space kerX is A-invariant. There then exists a such that x = Ax =(A BBTX)x and Cx = 0 with {} < 0. That is, there exists a stable,unobservable mode.

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Linear Quadratic RegulatorJ.C. Juang

Consider the linear system

x = Ax + B0u , x(0) = xo

and the quadratic cost functional to be minimized0

xTQx + uTRudt

Theorem: Assume that

1. The pair (A, B0) is stabilizable.

2. The matrix Q is symmetric and positive semidefinite.

3. The pair (C, A) is detectable where C satisfies Q = CTC, i.e., C is a Choleskyfactor of Q.

4. The matrix R is symmetric and positive definite.

Then, the optimal, stabilizing control exists and is of the state feedback form

u = F x

where the state feedback gain F is

F = R1BT0 Xwith X satisfying the algebraic Riccati equation

X = Ric

A B0R1BT0

CTC AT

orATX+ XA + Q XB0R1BT0 X = 0

Q governs the rate of convergence of the state and R controls the extent of controlactivity.

Stabilizable = the integral is finite; Observable = guarantee x(t) approacheszero as t goes to infinity; R positive definite = no impulsive controls.

The assumptions imply that

A B0R1BT0

CTC AT

dom Ric and X 0.

Define

z = C

0

x +

0R

1

2

u = C1x + D10u

Then, the cost function becomes0

xTQx + uTRudt =

0

zTz dt = z22

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Approach 1

Define the Hamiltonian

H(x,u,p) = 12

(xTQx + uTRu) +pT(Ax + B0u)

where p is the costate.

Optimality implies that

Hu

= 0 = Ru + BT0 p

Hp

= x = Ax + B0u

Hx

= p = Qx + ATp

Thus, the optimal controlu = R1BT0 p

and xp = A B0R1BT0Q AT xp

Assume that the state and costate are related

p = Xx

for some X. Then,

u = R1BT0 Xxx = (A B0R1BT0 X)x

Xx = (Q ATX)xThe last equation then leads to the algebraic Riccati equation

ATX+ XA + Q XB0R1BT0 X = 0 Approach 2

The closed-loop system is

x = (A + B0F)x , x(0) = xo

z = (C1 + D10F)x

which can also be rewritten as

x = (A + B0F)x + xo(t) , x(0) = 0

z = (C1 + D10F)x

Let X be the observability gramian of the closed-loop system, that is,

(A + B0F)TX+ X(A + B0F) + (C1 + D10F)

T(C1 + D10F) = 0

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Consider the Lyapunov function V(x) = xTXx and its derivative

V(x) = xTXx + xTXx

= (Ax + B0u)TXx + xTX(Ax + B0u)

= [(A + B0F)x + B0(u F x)]TXx + xTX[(A + B0F)x + B0(u F x)]= xT[(A + B0F)X+ X(A + B0F)]x + 2B0(u

lecture robust zhou

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