inner outer and spectral factorizations

Inner-Outer and Spectral Factorizations

SOLO HERMELIN

Updated: 27.05.11

1

Table of Content

SOLO Inner-Outer and Spectral Factorizations

2

Linear Quadratic Regulator (LQR) Problem


3

Given the Linear Time-Invariant System 00 xxuBxAx

and the Quadratic Cost Function:

TT

T

TTc PPRRdt

u

x

RS

SPuxJ

&02

1

0

The Optimal Regulator that Minimizes the Cost Function Jc is given by the following procedure:Define the Hamiltonian of the Optimization Problem:

uBxAu

x

RS

SPuxuxH T

T

TT

2

1,,

The Euler-Lagrange Equations are:

TTTT

T

T

T

BxSRuBxSuRu

H

uSxPAx

H

td

d

uBxAH

td

xd

10



4

TTTT

T

T

T

BxSRuBxSuRu

H

uSxPAx

H

td

d

uBxAH

td

xd

10

B

A

s

In

Kc

xu

or:

xA

x

SRBASRSP

BRBSRBAxHTTT

TT

11

11

TTT

TT

HSRBASRSP

BRBSRBAA

11

11

:

We want to find X (t) such that: , then txtXt

XBSRKxKxXBSRu TTcc

TT 11 :



5

xA

x

SRBASRSP

BRBSRBAxHTTT

TT

11

11

We want to find X (t) such that: , then txtXt

txtXtxtXt

xXBRBxSRBAXxXxXSRBAxSRSP TTTTT 1111

txxSRSPXBRBXSRBAXXSRBAX TTTTT 01111

TTTTT SRSPXBRBXSRBAXXSRBAX 1111 We obtain the following Differential Riccati Equation for :X

0

0

1

11

11

1111

PXBSRXBSAXXA

X

IAIX

X

I

SRBASRSP

BRBSRBAIX

SRSPXBRBXSRBAXXSRBA

TTTTTT

HTTT

TT

TTTTT

If a Steady-state Solution exists for t→∞ then and:0X

Continuous Algebraic

RiccatiEquation(CARE)



B

A

s

In

Kc

xu

Theorem 1 A Unique Stabilizing( A+BKc=[A-BR-1(ST+BTX)] is Stable) Solution of CARE, X=XT, is obtained iff:a.(A,B) is Stabilizableb.Re[λi(AH)]≠0 for all i=1,2,…,2nThe Unique Stabilizing Solution is denoted X=Ric [AH] Proof

a. It is clear that the condition that (A,B) is Stabilizable is a necessary condition to stabilize A+BKc.

b. Rewrite

IX

I

XBSRBA

BRBXBSRBA

IX

I

SRBASRSP

BRBSRBAA

TTT

TTT

TTT

TT

H

0

0

0:

1

11

11

11

If λi is an eigenvalue of AH, - λi is also; hence AH has n eigenvalues λi s.t. Re[λi] ≤ 0 and n eigenvalues λj s.t. Re[λj] ≥ 0. If AH has no eigenvalue of jω axis, then we can find a solution X s.t. all the eigenvalues of A-BR-1(ST+BTX)=A+BKc are the stable eigenvalues of AH.

011

1 PXBSRXBSAXXA TTTTTT



B

A

s

In

Kc

xuTheorem 1 A Unique Stabilizing( A+BKc=[A-BR-1(ST+BTX)] is Stable) Solution of CARE is obtained iff:a.(A,B) is Stabilizableb.Re[λi(AH)]≠0 for all i=1,2,…,2nThe Unique Stabilizing Solution is denoted X=Ric [AH]

Proof (continue -1)c. Assume that there are two Stabilizing Solutions X1=X1

T and X2 =X2T that satisfy the

CARE 0

0

21

222

11

111

PXBSRXBSAXXA

PXBSRXBSAXXA

TTTTTT

TTTTTT

Subtracting those two equations we obtain:

0

0

11

211

2

21

211

11

21211

2121

XBRBXXBRBX

XBRBXXBRBXSRBXXXXBRSAXXXXATT

TTTTT

or: 011

212121 XBSRBAXXXXXBSRBA TTTTT



B

A

s

In

Kc

xuTheorem 1 A Unique Stabilizing( A+BKc=[A-BR-1(ST+BTX)] is Stable) Solution of CARE is obtained iff:a.(A,B) is Stabilizableb.Re[λi(AH)]≠0 for all i=1,2,…,2nThe Unique Stabilizing Solution is denoted X=Ric [AH]

Proof (continue -2)c. Assume that there are two Stabilizing Solutions X1=X1

T and X2 =X2T that satisfy the

CARE

This is a Sylvester Matrix Equation and since both X1 and X2 are Stabilizing Solutions we must have:

011

212121 XBSRBAXXXXXBSRBA TTTTT

jiXBSRBAXBSRBA

njXBSRBA

niXBSRBATT

jTT

iTTj

TTi

,0ReRe,,10Re

,,10Re2

11

1

21

11

Therefore the solution of the Sylvester Matrix Equation is Unique, and by substitution we can see that X1 = X2 is the Unique Solution.

q.e.d.



Proof (1)

(1) The Matrix YTZ is Symmetric.(2) If Y-1 exists, the X=ZY-1 is the Solution of CARE such that the Matrix ((A-BR-1ST)-BR-1BT X) has the Eigenvalues λ1,λ2,…,λn. X is Symmetric.(3) If (A,B) Stabilizable and Re[λi(AH)]≠0 for all i=1,2,…,2n (Theorem 1) Y-1 exists.

Theorem 2 Let the columns of the Matrix be the Eigenvectors

of AH corresponding to the Stable Eigenvalues λ1,λ2,…,λn, and J the corresponding Jordan Matrix. Then

nxnnxn RZYRZ

Y

,2

Define TTT SRSPQBRBWSRBAE 111 :,:,:

TTTT

TT

HEQ

WE

SRBASRSP

BRBSRBAA

11

11

:

JZZEQY

JYZWYEJ

Z

Y

Z

Y

EQ

WEJ

Z

Y

Z

YA

TTHWe have

ZYJZWZZEYZYJWZEYJYZWYE TTTTTTTTTTT

JZYQYYZEYJZYZEYQYYJZZEQYY TTTTTTTTTT



Proof (1) (continue – 1)

Since –YTQTY-ZTWZ is symmetric so is (YTZ)J+JT(YTZ) or

ZYJJZYZWZQYYZWZZYJJZYQYYJZYQYYZEY

ZYJZWZZEY TTTTTTTTTT

TTTT

TTTTT

TTTTTTTTTTT ZYJJZYZYJJZYZYJJZY

Because J represents the stable Eigenvalues Re[λi(J)+λj(J)]≠0 for all I,j=1,…,n, the Unique Solution of the Sylvester Matrix Equation is

0TTTTTTT ZYZYJJZYZY Sylvester Matrix Equation

SymmetricisZYZYZYZYZY TTTTTTT 0




nxnnxn RZYRZ

Y

,2



Proof (2)

JZZEQY

JYZWYET

111 YJYYZWEYJYZWYE1111111 YJZYZWYZEYZYJYYZWEYZ

111 YJZYZEQYJZZEQY TT

01111 QYZWYZEYZYZET

Define TTT SRSPQBRBWSRBAE 111 :,:,:

Therefore X=ZY-1 is the Unique Stabilizing Solution of the CARE

01111 TTTTT SRSPXBRBXSRBAXXSRBA q.e.d.




nxnnxn RZYRZ

Y

,2

Assume Y-1 exists:



Proof (2) (continue – 1)

q.e.d.

From (1) YTZ is Symmetric, so

YZZYZY TTTT

TTTTTTTT ZYXZXYZYZYYYZZY 11

1 YZX

TTTT XYZZYX 1 X is Symmetric




nxnnxn RZYRZ

Y

,2

References

SOLO

13

S. Hermelin, “Robustness and Sensitivity Design of Linear Time-Invariant Systems”,PhD Thesis, Stanford University, 1986

G.Stein, J.Doyle, B.Francis, “Advances in Multivariable Control”, ONR/Honeywell Workshop, 1984

Inner-Outer and Spectral Factorizations

T. Kailath, A.H. Sayed, B. Hassibi, “Linear Estimation”. Prentice Hall, 2000

14

SOLO

TechnionIsraeli Institute of Technology

1964 – 1968 BSc EE1968 – 1971 MSc EE

Israeli Air Force1970 – 1974

RAFAELIsraeli Armament Development Authority

1974 – 2013

Stanford University1983 – 1986 PhD AA

http://upload.wikimedia.org/wikipedia/en/a/ae/RAFAEL_logo.png

Sylvester Matrix Equation : Anxn Xnxm+Xnxm Bmxm=Cnxm

SOLO

15

Consider the Sylvester Matrix Equationwhere Anxn, Bmxm, Cnxm are given matrices. Then there exists a Unique Solution Xnxm if and only if

Matrices

James Joseph Sylvester(1814 – 1887)

nxmmxmnxmnxmnxn CBXXA

njiAA nxnjnxni ,,1,0ReRe

0

dteCeX tBnxm

tAnxm

mxmnxn

Note : If B=AT : Lyapunov EquationThe Necessary and Sufficient Condition for the existence of a Unique Solution is

nxnT

nxnnxnnxnnxn CAXXA

In particular if λi(A)=- λj(A) = j ω a Unique Solution does not exist.

Aleksandr Mikhailovich Lyapunov

1857 - 1918

mjniBA mxmjnxni ,,1,,,10

the Unique Solution is given byIf mjniBA mxmjnxni ,,1,,,10ReRe

http://en.wikipedia.org/wiki/File:James_Joseph_Sylvester.jpg

http://en.wikipedia.org/wiki/File:Alexander_Ljapunow_jung.jpg


SOLO Matrices

Proof

Let rewrite

0

21

22221

11211

21

22221

11211

21

22221

11211

21

22221

11211

21

22221

11211

nmnn

m

m

mmmm

m

m

nmnn

m

m

nmnn

m

m

nnnn

n

n

nxmmxmnxmnxmnxn

ccc

ccc

ccc

bbb

bbb

bbb

xxx

xxx

xxx

xxx

xxx

xxx

aaa

aaa

aaa

CBXXA

nmnn

m

m

nnnn

n

n

nxmnxn

xxx

xxx

xxx

aaa

aaa

aaa

XA

21

22221

11211

21

22221

11211

nxmnxnc

mnxn

nxn

nxn

Xvec

m

nm

m

n

n

nxn

nxn

nxn

nxmcnxnm XAvec

cA

cA

cA

c

x

x

c

x

x

c

x

x

A

A

A

XvecAI

nxmc

2

1

1

2

2

12

1

1

11

00

00

00

nxmc Xvec

xmn

m

nm

m

n

n

mcnxmc

c

x

x

c

x

x

c

x

x

cccvecXvec

1

1

2

2

12

1

1

11

21 :

Define the VectorizationOperator vec:


SOLO Matrices

Proof (continue – 1)

Let rewrite

0

21

22221

11211

21

22221

11211

21

22221

11211

21

22221

11211

21

22221

11211

nmnn

m

m

mmmm

m

m

nmnn

m

m

nmnn

m

m

nnnn

n

n

nxmmxmnxmnxmnxn

ccc

ccc

ccc

bbb

bbb

bbb

xxx

xxx

xxx

xxx

xxx

xxx

aaa

aaa

aaa

CBXXA

mxmnxmc

mmmmm

mm

mm

Xvec

m

nm

m

n

n

nmmnmnm

nmnm

nmnn

nxmcnT

mxm BXvec

cbcbcb

cbcbcb

cbcbcb

c

x

x

c

x

x

c

x

x

IbIbIb

IbIbIb

IbIbIb

XvecIB

nxmc

2211

2222112

1221111

1

2

2

12

1

1

11

21

22212

12111

mmmm

m

m

nmnn

m

m

mxmnxm

bbb

bbb

bbb

xxx

xxx

xxx

BX

21

22221

11211

21

22221

11211


SOLO Matrices


We have

0 nxmcmxmnxmcnxmnxncnxmmxmnxmnxmnxnc CvecBXvecXAvecCBXXAvec

nxmcnxmcnT

mxmnxnm CvecXvecIBAI or

Let use the Jordan decomposition for Anxn and Bmxm11 & TTT BBB

TmxmAAAnxn SJSBSJSA

n

TTT

m

TT

TTT

I

AABBBAAA

I

BB

nBBBAAAmnT

mxmnxnm

SSSJSSJSSS

ISJSSJSIIBAI1111

11

111

111

1111

ATB

T

nTBATB

TT

ATBAm

TT

SS

AB

IJSS

ABB

SSJI

AABAB

DBCADCBA

BABASSSJSSJSSS

1 ABnBAmABnT

mxmnxnm SSIJJISSIBAIT

TT

This equation has a Unique Solution iff is Nonsingular. nTmxmnxnm IBAI


SOLO Matrices


where

1 ABnBAmABnT

mxmnxnm SSIJJISSIBAIT

TT

lJ

J

J

J

00

00

00

2

1

i

i

i

i

i

xkki iiJ

0000

1000

0000

0010

0001

Ji are Upper-Triangular Matrices

nBAm IJJI T Therefore is also Upper Triangular with diagonal elements λi[A] + λj[B](i=1,…,n, j=1,…,m), that are the Eigenvalues of and therefore to the Similar Matrix . This Matrix is Nonsingular iff

nBAm IJJI T

nTmxmnxnm IBAI

mjniBA mxmjnxni ,,1,,,10


SOLO Matrices

Since 0lim,0lim,,1,,,10ReRe

tB

t

tA

tmxmjnxni eemjniBA


Let rewrite

nxm

mSnnxm

nxm

m

nxnnxm

mxmnnxmnxmmxmnxmnxmnxn X

IAIX

X

I

AC

BIXCBXXA

00

nxmnxmtB

nxmtA

nxm CPeCetP mxmnxn 0:Define

By differentiation

mxmnxmnxmnxnmxmtB

nxmtAtB

nxmtA

nxnnxm BPPABeCeeCeAtd

tPdmxmnxnmxmnxn

Let Integrate the Differential Equation

mxmnxmnxmnxn

nxmnxmnxm BdtPdtPAdt

td

tPdPP

000

0

We have and where

0

dteCeX tBnxm

tAnxm

mxmnxn

0nxmP

mxmnxmnxmnxnmxmnxmnxmnxnnxm BXXABdtPdtPAC

00

Matrix Differential Riccati Equation


tDtXtCtXtBtXtXtAtX

where all matrices are nxn and A (t),B (t), C(t), D(t) are integrable over an interval t0 ≤ t ≤ tf.

This Matrix Differential Equation can be decompose in the following 2 Linear Differential Equations:

The Nonlinear Matrix Differential Riccati Equation is given by

tZtAtYtDtZ

tZtCtYtBtY

If in the interval t0 ≤ t ≤ tf the Matrix Y(t) is nonsingular, then the solution of the Riccati Differential Equation is 10

1 ttttYtZtX

To verify we have tYtd

tYdtYtY

td

dtY

td

dtZtY

td

tZdtX

td

d 11111 &

tXtCtXtBtXtXtAtD

tYtZtCtYtBtYtZtYtZtAtYtDtXtd

d

111

Matrix Differential Riccati Equation


tDtXtCtXtBtXtXtAtX

If yhe matrices are nxn and A ,B , C, D are constant, then

This Matrix Differential Equation can be decompose in the following 2 Linear Differential Equations:

The Nonlinear Matrix Differential Riccati Equation is given by

tZAtYDtZ

tZCtYBtY

If in the interval t0 ≤ t ≤ tf the Matrix Y(t) is nonsingular, then the solution of the Riccati Differential Equation is 10

1 ttttYtZtX

Define

AD

CBG

tZ

tYtW :&:

to obtain which have the solution tGWtW 00 tWetW ttG

inner outer and spectral factorizations

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