1
|
- Algebraic Riccati Equation
- Solving ARE
- Bounded Real Lemma
- Standard ARE
|
2
|
- Algebraic Riccati Equation:
- A*X+XA+XRX+Q=0, R=R*, Q=Q*
- An ARE may have many solutions. We are only interested in symmetric
solutions. In particular, we are interested in the symmetric solution
such that A+RX is stable. This solution is called stabilizing solution.
- Consider the associated Hamiltonian matrix:
- Then
- so H and -H* are similar. Thus l is an eigenvalue iff -
l* is.
- Thus we conclude: eig(H) ¹ jw Û H has n
eigenvalues in Re s<0 and n eigenvalues in Re s>0.
|
3
|
- Let X_(H) be the n-dimensional spectral subspace corresponding to
eigenvalues in Re s <0:
- where X1, X2
Î Cn´n
(X1 and X2 can be chosen to be real
matrices.) If X1
is nonsingular, define
- X:=Ric(H)= X2 X1-1: dom(Ric) Ì R2n´2n
® R2n´2n
- where dom(Ric) consists of all H matrices such that
- H has no eigenvalues on the imaginary axis
- are
complementary (or X1 is nonsingular.)
- Then X is a solution of the ARE. (see next theorem)
- >>[X1,X2] = ric_schr(H), X = X2/X1
|
4
|
- Theorem 12.1: Suppose H Î dom(Ric) and X=Ric(H). Then
- (i) X is real symmetric
- (ii) X satisfies the algebraic Riccati equation
- A*X+XA+XRX+Q=0
- (iii) A+RX is stable.
- Proof: (i) Let . We show X1*X2 is
symmetric.
- Note that there exists a stable matrix H_ in Rn´n
such that
- Pre-multiply this equation by
- to get
|
5
|
- Since JH is symmetric Þ :
- (-X1*X2+X2*X1) H_=H_*(-X1*X2+X2*X1
)*= -H_*(-X1*X2+X2*X1 )
- This is Lyapunov equation. Since H_ is stable, the unique solution is
- -X1*X2+X2*X1 =0.
- i.e., X1*X2 is symmetric. Þ X=(X1-1)*(X1*X2)X1-1
is symmetric.
- (ii) Start with the equation
- and post-multiply by X1-1 to get
- now pre-multiply by [X -I]:
- This is precisely the Riccati equation.
- (iii)
- Thus A + RX is stable because H_ is.
|
6
|
- Theorem 12.2: Suppose eig(H) ¹ jw and R is semi-definite (³ 0 or £0). Then H Î dom(Ric) Û (A, R)
is stabilizable.
- Proof. (Ü ) Note
that
- We need to show that X1 is nonsingular, i.e., Ker X1=0.
- Claim: Ker X1 is H_-invariant.
- Let x Î Ker X1
and note that X2*X1 is symmetric and
- AX1+RX2=X1H_
.
- Pre-multiply by x* X2* , post multiply by x to get
- x* X2* RX2 x=0 Þ RX2
x=0 Þ X1H_ x=0
- i.e., H_ x Î Ker
X1.
- Suppose Ker X1¹0. Then H_ |Ker X1
has an eigenvalue, l, and a corresponding eigenvector, x:
- H_ x= l x, Rel <0, 0¹ xÎ Ker X1
|
7
|
- Note that
- -QX1-A*X2=X2 H_
- Post-multiply the above equation by x:
- (A*+lI)X2 x=0
- Recall that RX2x = 0, we have
- x*X2*[A+ l*I R]=0.
- (A,R) stabilizable Þ X2 x=0 Þ
- has full column rank, which is a
contradiction.
- (Þ ) H Î dom(Ric) Þ A+RX is stable Þ (A, R) is
stabilizable.
|
8
|
- Corollary 12.3: Let g>0. G(s)=C(sI-A)-1B+D Î RH¥ and
- where R=g2I-D*D.
Then the following conditions are equivalent:
- (i) ||G||¥ < g .
- (ii) ||D||< g and H has no eigenvalues on the
imaginary axis.
- (iii) ||D||< g and H Î dom(Ric).
- (iv) ||D||< g
, H Î dom(Ric)
and Ric(H)³0
(Ric(H)>0 if (C,A) is observable).
|
9
|
- (v) ||D||< g and
there exists an X³0 such that
- X(A+BR-1D*C)+(A+BR-1D*C)*X+XBR-1B*X+C*(I+DR-1D*)C=0
- and A+BR-1D*C+BR-1B*X has no eigenvalues on the imaginary
axis.
- (vi) ||D||< g and
there exists an X > 0 such that
- X(A+BR-1D*C)+(A+BR-1D*C)*X+XBR-1B*X+C*(I+DR-1D*)C<0
- (vii) there exists and X > 0 such that
- Proof: We have already known: (i) Û (ii).
(iii) Þ (ii) is obvious. To show that (ii) Þ (iii), we need to show that (A+BR-1D*C,
BR-1B*) is stabilizable (Theorem 12.2). In fact, we will show
that A+BR-1D*C is
stable for all those g such that ||G||¥ < g .
|
10
|
- Note that we can write
- A+BR-1D*C= A+B(g2I-D*D)-1D*C= A+B(I-DD1)-1DC1
- with D=D*/g, D1 =D/g, and C1 =C/g. Then ||D||<1 and
- ||C1(sI-A)-1B+D1||¥ = g--1 ||G||¥ <1.
- Hence by small gain theorem, A+B(I-DD1)-1DC1 is stable for all D with ||D||<1. Thus A+BR-1D*C
is stable for all g- such that ||G||¥ <g-.
- (iii) Þ (iv) follows from the fact that the
ARE
- X(A+BR-1D*C)+(A+BR-1D*C)*X+XBR-1B*X+C*(I+DR-1D*)C=0
- can be regarded as a Lyapunov equation with
- A1 := A+BR-1D*C, Q := XBR-1B*X+C*(I+DR-1D*)C
- Hence X ³ 0 since
A1 is stable and Q³0.
- (v) Þ (i): Assume D = 0 for simplicity. Then
there is an X³0
- and A+BB*X/g2 has no jw-axis eigenvalue.
|
11
|
- Hence
- has no zeros on the imaginary
axis since
- has no poles on the imaginary axis. Next, note that
- -X(jwI-A)-(jwI-A)*X+XBB*X/g2I +C*C=0
- Multiply B*{(jwI-A)*}-1
on the left and (jwI-A)-1B on the right of the above
equation to get
- -B*{(jwI-A)*}-1XB-B*X(jwI-A)-1B+B*{(jwI-A)*}-1C*C
(jwI-A)-1B
- + B*{(jwI-A)*}-1XBB*X(jwI-A)-1B/g2 =0
- Completing square, we have
- G*(jw)G(jw)= g2I-W*(jw)W(jw)
|
12
|
- Since W(s) has no jw-axis zeros, we conclude that ||G||¥ <
g .
- (vi) Û (vii): follows from Schur complement.
- (vi) Þ (i): by following the similar
procedure as above.
- (i) Þ (vi): let
- Then there exists an e>0 such that ||Ge||¥ <
g . Now
(vi) follows by applying part (v) to Ge .
|
13
|
- Theorem 12.4: Suppose H has the form
- Then H Î dom(Ric)
iff (A,B) is stabilizable and (C,A) has no unobservable modes on the
imaginary axis. Furthermore, X=Ric(H)³0. And X > 0 if and only if (C,A) has no
stable unobservable modes.
- Proof: Only need to show that, assuming (A,B) is stabilizable, H has no jw eigenvalues iff (C,A) has no
unobservable modes on the imaginary axis.
- Suppose that jw
is an eigenvalue and
is a corresponding eigenvector. Then
- Ax-BB*z=jw x, -C*Cx-A*z=jw z
- Re-arrange: (A- jwI)x=BB*z, -(A-jwI)*z=C*Cx
|
14
|
- Thus <z, (A- jwI)x>=<z,BB*z>=||B*z||2
- - <x, (A- jwI)*z>=<x,C*Cx>=||Cx||2
- so <x,(A-jwI)*z>
is real and
- -||Cx||2=< (A-jwI)x,z>=<z,(A-jwI)x>*=||B*z||2
- Therefore B*z = 0 and Cx = 0. So
- (A-jwI)x=0, (A-jwI)*z=0
- Combine the last four equations to get
- The stabilizability of (A,B) gives z=0. Now it is clear that jw is an eigenvalue of H
if jw is an
unobservable mode of (C,A).
- (A-BB*X)*X+X(A-BB*X)+XBB*X+C*C=0
- X³0 since A-BB*X
is stable.
|
15
|
- Corollary 12.5: Suppose (A,B) is stabilizable and (C,A) is detectable.
Then
- A*X+XA-XBB*X+C*C=0
- has a unique positive semidefinite solution. Moreover, it is
stabilizing.
- Corollary 12.7: Suppose D has full column rank and denote R = D*D>0.
Let H have the form
- Then H Î dom(Ric)
iff (A,B) is stabilizable and has full column
rank for all w. Furthermore, X=Ric(H)³0 if H Î dom(Ric) and Ker(X) = 0 if and only if (D^*C, A-BR-1D*C)
has no stable unobservable modes.
- Proof: This is because has full
column rank for all w
- Û ((I-DR-1D*)C,
A-BR-1D*C) has no observable modes on jw-axis.
|