[PDF] Lecture 13 Linear quadratic Lyapunov theory

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EE363Winter 2008-09

Lecture 13

Linear quadratic Lyapunov theory

•the Lyapunov equation

•Lyapunov stability conditions

•the Lyapunov operator and integral

•evaluating quadratic integrals

•analysis of ARE

•discrete-time results

•linearization theorem

13-1

The Lyapunov equation

theLyapunov equationis A

TP+PA+Q= 0

whereA, P, Q?Rn×n, andP, Qare symmetric interpretation: for linear systemx=Ax, ifV(z) =zTPz, then

V(z) = (Az)TPz+zTP(Az) =-zTQz

i.e., ifzTPzis the (generalized)energy, thenzTQzis the associated (generalized)dissipation linear-quadratic Lyapunov theory:lineardynamics,quadraticLyapunov function

Linear quadratic Lyapunov theory13-2

we consider systemx=Ax, withλ1,...,λnthe eigenvalues ofA ifP >0, then •the sublevel sets are ellipsoids (and bounded)

•V(z) =zTPz= 0?z= 0

boundedness condition:ifP >0,Q≥0then

•all trajectories ofx=Axare bounded

block of size one)

Linear quadratic Lyapunov theory13-3

Stability condition

ifP >0,Q >0then the systemx=Axis (globally asymptotically) stable,i.e.,?λi<0 to see this, note that

λmax(P)zTPz=-αV(z)

whereα=λmin(Q)/λmax(P)>0

Linear quadratic Lyapunov theory13-4

An extension based on observability

(Lasalle"s theorem for linear dynamics, quadratic function) ifP >0,Q≥0, and(Q,A)observable, then the systemx=Axis (globally asymptotically) stable now suppose thatv?= 0,Av=λv,?λ= 0 thenA¯v=¯λ¯v=-λ¯v, so

Q1/2v???2=v?Qv=-v??ATP+PA?v=λv?Pv-λv?Pv= 0

which impliesQ1/2v= 0, soQv= 0, contradicting observability (by PBH) interpretation: observability condition means no trajectory canstay in the "zero dissipation" set{z|zTQz= 0}

Linear quadratic Lyapunov theory13-5

An instability condition

ifQ≥0andP?≥0, thenAis not stable to see this, note that sinceP?≥0, there is awwithV(w)<0; trajectory starting atwdoes not converge to zero invariant

Linear quadratic Lyapunov theory13-6

The Lyapunov operator

theLyapunov operatoris given by

L(P) =ATP+PA

special case of Sylvester operator Lis nonsingular if and only ifAand-Ashare no common eigenvalues, i.e.,Adoes not have pair of eigenvalues which are negatives of each other •ifAis stable, Lyapunov operator is nonsingular •ifAhas imaginary (nonzero,iω-axis) eigenvalue, then Lyapunov operator is singular thus ifAis stable, for anyQthere is exactly one solutionPof Lyapunov equationATP+PA+Q= 0

Linear quadratic Lyapunov theory13-7

Solving the Lyapunov equation

A

TP+PA+Q= 0

we are givenAandQand want to findP if Lyapunov equation is solved as a set ofn(n+ 1)/2equations in n(n+ 1)/2variables, cost isO(n6)operations fast methods, that exploit the special structure of the linear equations, can solve Lyapunov equation with costO(n3) based on first reducingAto Schur or upper Hessenberg form

Linear quadratic Lyapunov theory13-8

The Lyapunov integral

ifAis stable there is an explicit formula for solution of Lyapunov equation: P=? 0 etATQetAdt to see this, we note that A

TP+PA=?

0?

ATetATQetA+etATQetAA?

dt 0? d dtetATQetA? dt =etATQetA???∞ 0 =-Q

Linear quadratic Lyapunov theory13-9

Interpretation as cost-to-go

ifAis stable, andPis (unique) solution ofATP+PA+Q= 0, then

V(z) =zTPz

=zT??∞ 0 etATQetAdt? z 0 x(t)TQx(t)dt wherex=Ax,x(0) =z thusV(z)is cost-to-go from pointz(with no input) and integral quadratic cost function with matrixQ

Linear quadratic Lyapunov theory13-10

ifAis stable andQ >0, then for eacht,etATQetA>0, so P=? 0 etATQetAdt >0 meaning: ifAis stable, •we can chooseanypositive definite quadratic formzTQzas the dissipation,i.e.,-V=zTQz •then solve a set of linear equations to find the (unique) quadratic form

V(z) =zTPz

•Vwill be positive definite, so it is a Lyapunov function that provesAis stable in particular:a linear system is stable if and only if there is a quadratic

Lyapunov function that proves it

Linear quadratic Lyapunov theory13-11

generalization:ifAstable,Q≥0, and(Q,A)observable, thenP >0 to see this, the Lyapunov integral showsP≥0 ifPz= 0, then

0 =zTPz=zT??∞

0 etATQetAdt? z=? 0???

Q1/2etAz???2dt

so we concludeQ1/2etAz= 0for allt≥0 this implies thatQz= 0,QAz= 0, . . . ,QAn-1z= 0, contradicting (Q,A)observable

Linear quadratic Lyapunov theory13-12

Monotonicity of Lyapunov operator inverse

supposeATPi+PiA+Qi= 0,i= 1,2 ifQ1≥Q2, then for allt,etATQ1etA≥etATQ2etA ifAis stable, we have P 1=? 0 etATQ1etAdt≥? 0 etATQ2etAdt=P2 in other words: ifAis stable then Q

1≥Q2=? L-1(Q1)≥ L-1(Q2)

i.e., inverse Lyapunov operator is monotonic, or preserves matrix inequality, whenAis stable (question: isLmonotonic?)

Linear quadratic Lyapunov theory13-13

Evaluating quadratic integrals

supposex=Axis stable, and define J=? 0 x(t)TQx(t)dt to findJ, we solve Lyapunov equationATP+PA+Q= 0forP then,J=x(0)TPx(0) in other words: we can evaluate quadratic integral exactly, by solving a set of linear equations, without even computing a matrix exponential

Linear quadratic Lyapunov theory13-14

Controllability and observability Grammians

forAstable, the controllability Grammian of(A,B)is defined as W c=? 0 etABBTetATdt and the observability Grammian of(C,A)is W o=? 0 etATCTCetAdt these Grammians can be computed by solving the Lyapunov equations AW c+WcAT+BBT= 0, ATWo+WoA+CTC= 0 we always haveWc≥0,Wo≥0; W c>0if and only if(A,B)is controllable, and W o>0if and only if(C,A)is observable

Linear quadratic Lyapunov theory13-15

Evaluating a state feedback gain

consider x=Ax+Bu, y=Cx, u=Kx, x(0) =x0 with closed-loop systemx= (A+BK)xstable to evaluate the quadratic integral performance measures J u=? 0 u(t)Tu(t)dt, Jy=? 0 y(t)Ty(t)dt we solve Lyapunov equations (A+BK)TPu+Pu(A+BK) +KTK= 0 (A+BK)TPy+Py(A+BK) +CTC= 0 then we haveJu=xT

0Pux0,Jy=xT

0Pyx0

Linear quadratic Lyapunov theory13-16

Lyapunov analysis of ARE

write ARE (withQ≥0,R >0) A

TP+PA+Q-PBR-1BTP= 0

as (A+BK)TP+P(A+BK) + (Q+KTRK) = 0 withK=-R-1BTP we conclude: ifA+BKstable, thenP≥0(sinceQ+KTRK≥0) i.e., any stabilizing solution of ARE is PSD if also(Q,A)is observable, then we concludeP >0 to see this, we need to show that(Q+KTRK,A+BK)is observable if not, there isv?= 0s.t. (A+BK)v=λv,(Q+KTRK)v= 0

Linear quadratic Lyapunov theory13-17

which implies v ?(Q+KTRK)v=v?Qv+v?KTRKv=?Q1/2v?2+?R1/2Kv?2= 0 soQv= 0,Kv= 0 (A+BK)v=Av=λv, Qv= 0 which contradicts(Q,A)observable the same argument shows that ifP >0and(Q,A)is observable, then

A+BKis stable

Linear quadratic Lyapunov theory13-18

Monotonic norm convergence

suppose thatA+AT<0,i.e., (symmetric part of)Ais negative definite can express asATP+PA+Q= 0, withP=I,Q >0 meaning:xTPx=?x?2decreases along every nonzero trajectory,i.e.,

• ?x(t)?is alwaysdecreasing monotonicallyto0

•x(t)is always moving towards origin

this impliesAis stable, but the converse is false: for a stable system, we need not haveA+AT<0 (for a stable system withA+AT?<0,?x(t)?converges to zero, but not monotonically)

Linear quadratic Lyapunov theory13-19

for a stable system we can always change coordinates so we havemonotonic norm convergenceletPdenote the solution ofATP+PA+I= 0

takeT=P-1/2 in new coordinatesAbecomes˜A=T-1AT,

A+˜AT=P1/2AP-1/2+P-1/2ATP1/2

=P-1/2?PA+ATP?P-1/2 =-P-1<0 in new coordinates, convergence isobviousbecause?x(t)?is always decreasing

Linear quadratic Lyapunov theory13-20

Discrete-time results

all linear quadratic Lyapunov results have discrete-time counterparts thediscrete-timeLyapunov equation is A

TPA-P+Q= 0

meaning:ifxt+1=AxtandV(z) =zTPz, thenΔV(z) =-zTQz •ifP >0andQ >0, thenAis (discrete-time) stable (i.e.,|λi|<1) •ifP >0andQ≥0, then all trajectories are bounded •ifP >0,Q≥0, and(Q,A)observable, thenAis stable

•ifP?>0andQ≥0, thenAis not stable

Linear quadratic Lyapunov theory13-21

Discrete-time Lyapunov operator

the discrete-time Lyapunov operator is given byL(P) =ATPA-P Lis nonsingular if and only if, for alli, j,λiλj?= 1 (roughly speaking, if and only ifAandA-1share no eigenvalues) ifAis stable, thenLis nonsingular; in fact

P=∞?

t=0(AT)tQAt is the unique solution of Lyapunov equationATPA-P+Q= 0 the discrete-time Lyapunov equation can be solved quickly (i.e.,O(n3)) and can be used to evaluate infinite sums of quadratic functions, etc.

Linear quadratic Lyapunov theory13-22

Converse theorems

supposext+1=Axtis stable,ATPA-P+Q= 0

•ifQ >0thenP >0

•ifQ≥0and(Q,A)observable, thenP >0

in particular, a discrete-time linear system is stable if and onlyif there is a quadratic Lyapunov function that proves it

Linear quadratic Lyapunov theory13-23

Monotonic norm convergence

supposeATPA-P+Q= 0, withP=IandQ >0 this meansATA < I,i.e.,?A?<1 meaning:?xt?decreases on every nonzero trajectory; indeed, when?A?<1, •system is calledcontractivesince norm is reduced at each step the converse is false: system can be stable without?A?<1

Linear quadratic Lyapunov theory13-24

now supposeAis stable, and letPsatisfyATPA-P+I= 0 takeT=P-1/2 in new coordinatesAbecomes˜A=T-1AT, so

AT˜A=P-1/2ATPAP-1/2

=P-1/2(P-I)P-1/2 =I-P-1< I i.e.,?˜A?<1 so for a stable system, we can change coordinates so the system is contractive

Linear quadratic Lyapunov theory13-25

Lyapunov"s linearization theorem

we consider nonlinear time-invariant systemx=f(x), wheref:Rn→Rn supposexeis an equilibrium point,i.e.,f(xe) = 0, and let

A=Df(xe)?Rn×n

the linearized system, nearxe, isδx=Aδx linearization theorem: •if the linearized system is stable,i.e.,?λi(A)<0fori= 1,...,n, then the nonlinear system is locally asymptotically stable •if for somei,?λi(A)>0, then the nonlinear system is not locally asymptotically stable

Linear quadratic Lyapunov theory13-26

stability of the linearized system determines the local stability of the nonlinear system,exceptwhen all eigenvalues are in the closed left halfplane, and at least one is on the imaginary axis examples likex=x3(which is not LAS) andx=-x3(which is LAS) show the theorem cannot, in general, be tightened examples: eigenvalues ofDf(xe)conclusion aboutx=f(x)-3,-0.1±4i,-0.2±iLAS nearxe -3,-0.1±4i,0.2±inot LAS nearxe -3,-0.1±4i,±ino conclusion

Linear quadratic Lyapunov theory13-27

Proof of linearization theorem

let"s assumexe= 0, and express the nonlinear differential equation as x=Ax+g(x) suppose thatAis stable, and letPbe unique solution of Lyapunov equation A

TP+PA+I= 0

the Lyapunov functionV(z) =zTPzproves stability of the linearized system; we"ll use it to prove local asymptotic stability of the nonlinear system

Linear quadratic Lyapunov theory13-28

V(z) = 2zTP(Az+g(z))

=zT(ATP+PA)z+ 2zTPg(z) =-zTz+ 2zTPg(z) =-?z?2(1-2K?P??z?)

2λmax(P)zTPz=-1

2?P?zTPz

Linear quadratic Lyapunov theory13-29

finally, using

λmin(P)zTPz

we have

2?P?V(z)

and we"re done comments: •proof actually constructs an ellipsoid inside basin of attractionof x e= 0, and a bound on exponential rate of convergence •choice ofQ=Iwas arbitrary; can get better estimates using otherQs, better bounds ong, tighter bounding arguments . . .

Linear quadratic Lyapunov theory13-30

Integral quadratic performance

considerx=f(x),x(0) =x0 we are interested in the integral quadratic performance measure

J(x0) =?

0 x(t)TQx(t)dt for any fixedx0we can find this (approximately) by simulation and numerical integration (we"ll assume the integral exists; we do not requireQ≥0)

Linear quadratic Lyapunov theory13-31

Lyapunov bounds on integral quadratic performance

suppose there is a functionV:Rn→Rsuch that

•V(z)≥0for allz

upper bound on the integral quadratic cost conclude that trajectories are bounded)

Linear quadratic Lyapunov theory13-32

to show this, we note that

V(x(T))-V(x(0)) =?

T T 0 x(t)TQx(t)dt and so T 0 since this holds for arbitraryT, we conclude 0

Linear quadratic Lyapunov theory13-33

Integral quadratic performance for linear systems

for a stable linear system, withQ≥0, the Lyapunov bound is sharp,i.e., there exists aVsuch that

•V(z)≥0for allz

and for whichV(x0) =J(x0)for allx0 (takeV(z) =zTPz, wherePis solution ofATP+PA+Q= 0)

Linear quadratic Lyapunov theory13-34

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