Technical report from Automatic Control at Linköpings universitet
Generalized Riccati Equations for
H
∞Synthesis
Anders Helmersson
Division of Automatic Control
E-mail: anders.helmersson@liu.se
20th May 2005
Report no.: LiTH-ISY-R-3085
Address:
Department of Electrical Engineering Linköpings universitet
SE-581 83 Linköping, Sweden
WWW: http://www.control.isy.liu.se
AUTOMATIC CONTROL REGLERTEKNIK LINKÖPINGS UNIVERSITET
Technical reports from the Automatic Control group in Linköping are available from http://www.control.isy.liu.se/publications.
Abstract
Conditions for the existence of controllers for linear time-varying (LTV) systems is given. The closed loop performance is specified in terms of energy gain, which also includes terminal constraints. The conditions can be formulated either as linear matrix inequalities (LMIs) or as solutions to Riccati differental equations with algebraic constraints.
Generalized Riccati Equations
for H
∞
Synthesis
Anders Helmersson
Linköpings universitet
SE-581 83 Linköping, Sweden
email: anders.helmersson@liu.se
2015-05-20
Abstract
Conditions for the existence of controllers for linear time-varying (LTV) systems is given. The closed loop performance is specified in terms of energy gain, which also includes terminal constraints. The conditions can be formulated either as linear matrix inequalities (LMIs) or as solutions to Riccati differental equations with algebraic con-straints.
1
Introduction
Two of the most used methods for synthesis of H∞ controller are based on
Riccati equations and linear matrix inequalities (LMI), respectively.
In the standard Riccati approach [3], two Riccati equations are solved to produce the two Lyapunov matrices, X and Y . If a set of rank conditions on the system matrices hold, the Riccati equations produce the extremal solutions, which is tested against an eigenvalue condition, maxiλi(XY ) < γ2,
where γ is the bound of the H∞gain. A bisection technique can then be used
to find the lowest possible gain, γ.
In the LMI approach, two convex conditions are used, FX(γ, X) ≺ 0 and
FY(γ, Y ) ≺ 0, together with a connecting LMI,
X I
I Y
0, or, equiva-lently, miniλi(XY ) ≥ 1. The rank conditions on the system matrices are
Some of the rank conditions in the Riccati approach are needed for as-suring that the extremal solution exists. If this is the case, then the LMI ap-proach can be twisted to find the extremal solutions of FX and FY separately
for a given γ, and then to test the connecting LMI. If the rank conditions do not hold, the extremal solution may not exist, since X and Y may be unbounded. When this occurs, we can structure the LMI problem using the Kronecker Canonical Form [5] into unbounded and bounded blocks of X and Y .
In this paper, we exploit these ideas in order to tackle the H∞ problems
for continuous-time, linear time-varying (LTV) systems, both using Riccati and LMI methods. The methods can be used for solving the LTI and LTV synthesis problem in a closely related fashion, using Riccati differential equa-tions with algebraic constraints.
The paper is organized as follows. We start with the definition of the linear time-varying problem and its performance bound in section 2. We also elaborate on the synthesis conditions. In section 3 we employ the structure of the synthesis conditions in order to obtain extremal solutions. The algorithm is applied on an example and finally the conclusions are given.
2
Continuous time
We start by assuming that the continuous-time, linear system, G, is governed by the following differential equation:
(
˙x(t) = A(t)x(t) + B(t)w(t) z(t) = C(t)x(t) + D(t)w(t), which we also denote by G := A(t) B(t)
C(t) D(t)
. The H∞ gain from w to z is
defined by: J = xT(T )PTx(T ) + kγ−1/2zk22− kγ 1/2 wk22− xT(0)P0x(0), where kγ−1/2zk2 2 = RT 0 γ −1zT(t)z(t)dt and kγ1/2wk2 2 = RT 0 γw T(t)w(t)dt. The
term xT(T )PTx(T ) is the final cost caused by a non-zero final state. The term
xT(0)P
0x(0) is the initial energy in the system available at time t = 0. The
gain, γ > 0, can be generalized to a time-varying function in this formulation. We often drop the explicit time argument (t) in the sequel.
Introduce a positive definite, time-varying, Lyapunov function, V (x) = xT(t)P (t)x(t), where P (t) 0 and P (T ) = P
T. We further assume that
˙
Then J = xT(T )PTx(T ) + γ−1kzk22− γkwk 2 2− x T(0)P 0x(0) = xT(0) (P (0) − P0) x(0) + Z T 0 ˙V + γ−1 zTz − γwTw | {z } ≤0 dt ≤ xT(0) (P (0) − P 0) x(0).
Here we should assume that P (0) P0, since otherwise the cost would not
have any upper bound if x(0) → ∞.
The assumption (1) can be rewritten as x w T P A + ATP + ˙P P B BTP −γI + γ−1 CT DT C D x w ≤ 0. since ˙V = xTP (Ax + Bw) + (Ax + Bw)TP x + xTP x. This should hold for˙
all possible x and w, from which we can infer that P A + ATP + ˙P P B BTP −γI + γ−1 CT DT C D 0. This can be rewritten using Schur complement as
P A + ATP + ˙P P B CT BTP −γI DT C D −γI 0. (2)
This is the Bounded Real Lemma (BRL) for a linear time-varying (LTV) system. If ¯σ(D) < γ, we can rewrite this as
P A + ATP + ˙P + P B CT γI −DT −D γI −1 BTP C 0, from which we can integrate P in backward time (if we assume equality instead of ). Introduce the Hamiltonian
H = A 0 0 −AT + B 0 0 −CT γI −DT −D γI −1 0 BT C 0 . The Riccati differential equation
P −I H I P + ˙P = 0
can be solved using the associated Hamiltonian differential equation ˙ ξ1 ˙ ξ2 = H ξ1 ξ2 .
If ξ1 is nonsingular then P = ξ2ξ1−1, since
˙ P = ˙ξ2ξ1−1− ξ2ξ1−1ξ˙1ξ−11 = −ξ2ξ1−1 I ˙ ξ1 ˙ ξ2 ξ−11 = −P I H ξ1 ξ2 ξ−11 = −P I H I P .
We can verify the performance of the system by letting ξ(T ) =
I PT
, then integrating it backwards, and checking the non-singularity of ξ1(t) and
γI −DT
−D γI
at each instant, 0 ≤ t ≤ T , and, finally, verifying that ξ2(0)ξ1−1(0) P0.
Alternatively, we can use the differential algebraic equation A B 0 C D −γI 0 −AT −CT γI −BT −DT ξ1 ξ2 ξ3 ξ4 = ˙ ξ1 0 ˙ ξ3 0 , which gives P = ξ3ξ1−1.
2.1
The Synthesis Problem
We will now consider the system, G, which takes the inputs w and u and produces the outputs z and y:
G := A B1 B2 C1 D11 D12 C2 D21 D22 to be controlled by K := " ˆ A Bˆ ˆ C Dˆ # ,
which connects to y and u of G, see Figure 1. The closed loop system, G from w to z, is given by G := A B C D = A 0 B1 0 0 0 C1 0 D11 + 0 B2 I 0 0 D12 ˆ A Bˆ ˆ C Dˆ 0 I 0 C2 0 D21 . (3) K G -- -w z u y
Figure 1: The synthesis problem consists of the design task of finding a controller K to the system G such that the H∞ norm of the closed-loop
system, G, from w to z, is minimized.
The Bounded Real Lemma (2) applied to G(s) = D + C(sI − A)−1B, to-gether with the Elimination Lemma [4, 6] can be used to derive the necessary and sufficient conditions for the existence of a controller with a closed-loop H∞ norm less than γ. Let P =
X N NT L = Y M MT ∗ −1 in XA + ATX + ˙X ATN + ˙N XB 1 C1T NTA + ˙NT L˙ NTB 1 0 BT 1X B1TN −γI D11T C1 0 D11 −γI + N XB2 L NTB 2 0 0 0 D12 ˆ A Bˆ ˆ C Dˆ 0 I 0 0 C2 0 D21 0 + T ≺ 0.
The conditions are FX := NX 0 0 I T XA + ATX + ˙X XB1 C1T BT 1X −γI D11T C1 D11 −γI NX 0 0 I ≺ 0, (4a) FY := NY 0 0 I T AY + Y AT − ˙Y Y CT 1 B1 C1Y −γI D11 BT 1 DT11 −γI NY 0 0 I ≺ 0, (4b)
where NX and NY designate any bases of the null spaces of C2 D21 and
BT
2 DT12 , respectively. In addition, X and Y are connected by
P = X ∗ ∗ ∗ = Y ∗ ∗ ∗ −1 0 ∈ R(n+r)×(n+r), (5)
which is equivalent to the condition
X I
I Y
0, (6)
and, the non-convex constraint, rank(XY −I) ≤ r, where r is the order of the controller. Note that (6) is in turn equivalent to Y 0 and X Y−1 0. Also, P0 X(0) Y−1(0) and X(T ) Y−1(T ) PT shall apply.
In order to find a feasible solution we can proceed as follows:
(i) Let Y (T ) = PT−1 and integrate (4b) in backward time down to t = 0. (ii) Let X(0) = P0 and integrate (4a) in forward time up to t = T .
(iii) Check that (6) holds in the interval t ∈ [0, T ].
If any of these steps fails, we need to adjust one or several of the control objectives defined by γ, P0, and PT.
In the steps (i) and (ii) we should find the maximizing solution. Note that the finite maximizing solution does not exist for all problems, since it may be unbounded. This depends on the structure of the LMIs in (4) as we will elaborate on in the sequel.
3
Structured algorithms
3.1
The KCF approach
According to [5] the LTI H∞ synthesis problem in continuous time can be
Lemma 1 ([5]). Consider the problem of minimizing γ with respect to X Y−1 0 subject to Q(γ) + U XVT + V XUT ≺ 0. Partition (U − λV ) into
KCF blocks and partition Q and X accordingly.
(i) Blocks corresponding to Jj(µ) where Re µ < 0 and Lj can be truncated.
(ii) Blocks of the type Jj(µ) where Re µ = 0 and Nj can be truncated and
replaced by an additional condition Nµ∗Q(γ)Nµ≺ 0, where Nµdesignate
any basis of the null space of (U − µV )∗.
(iii) The remaining blocks in X, corresponding to Jj(µ) where Re µ > 0 and
LT
j, are bounded.
We assume that, after a congruence transformation, U − λV is block diagonal and where each block is Jj(µ), Nj, Lj or LTj. The Jordan blocks, Jj
and Nj, are square, while Lj and LTj are rectangular.
Jj(µ) := µ − λ 1 . .. ... . .. 1 µ − λ ∈ Rj×j, Nj := 1 −λ . .. ... . .. −λ 1 ∈ Rj×j, and Lj := −λ 1 . .. ... −λ 1 ∈ R j×(j+1).
3.2
LTV synthesis
We start with (4a), which we rewrite using Schur complement as NT X XA + ATX + ˙X XB 1 BT 1X −γI + γ−1 CT 1 DT11 C1 D11 NX = Q(γ) + U XVT + V XUT + V ˙XVT ≺ 0, (7) where UT = A B N X, VT = I 0 NX and Q(γ) = NXT 0 0 0 −γI + γ−1 CT 1 DT 11 C1 D11 NX.
Here NX spans the null space of C2 D21 .
We will now structure (7) in a similar way as in the KCF approach. Step A) Let us assume that V = V1 0 where V1 has full column rank,
possibly after a similarity transformation of the state vector:
Q(γ) + U1X11V1T + V1X11U1T + V1X12U2T + U2X21V1T + V1X˙11V1T ≺ 0.
Note that X22 is absent in this inequality, and can be chosen arbitrarily at
every instant. Apply the elimination lemma on X12 = X21T, which results in
two conditions for its existence.
V1⊥TQ(γ)V1⊥ ≺ 0, U2⊥T Q(γ) + U1X11V1T + V1X11U1T + V1X˙11V1T
U2⊥ ≺ 0,
where V1⊥ and U2⊥ span the nullspace of V1T and U2T, respectively. The first constraint can be reformulated as a condition on γ, while the second has the same structure as the original problem. Thus recursion (X := X11, Q :=
U2⊥TQU2⊥, V := U2⊥TV1 and U := U2⊥TU1) can be applied until eventually V
gets full column rank. These steps correspond to the removal of Lj and Nj
blocks in the KCF approach.
Note that the condition V1⊥TQ(γ)V1⊥≺ 0, or equivalently, V⊥TQ(γ)V⊥ 1 ≺
0, only needs to be checked once. This condition can also be written as γ > ¯σ D11D21⊥ where D21 D21⊥T D
⊥
21= 0 I .
Step B) Since V has full column rank, assume that V = V1 0 , where V1 is nonsingular. Q11+ U1XV1T + V1XU1T + V1XV˙ 1T V1XQ12+ U2T Q21+ U2XV1T Q22 ≺ 0.
We already know that Q22 ≺ 0 since V⊥TQV⊥ ≺ 0, so we can apply the
Schur complement to obtain
Q11+ U1XV1T + V1XU1T + V1XV˙ 1T − V1XU2T + Q12 Q22−1 U2XV1T + Q21 ≺ 0.
This step corresponds to the removal of LTj blocks in the KCF approach. Step C) Only Jj(µ) blocks remain. Since V1 is non-singular, we obtain
the Riccati differential inequality X −I H I X + ˙X ≺ 0,
where the Hamiltonian is given by H = UT 1 V −T 1 0 −V1−1Q11V1−T −V −1 1 U1 + UT 2 V1−1Q12 Q−122 Q21V1−T U2 . (8)
The maximizing solution will be obtained by solving the corresponding Ric-cati differential equation.
3.3
The Hamiltonian DAE
We will now compare these steps with the following Hamiltonian differential algebraic equation (DAE), Hξ = L ˙ξ, or
A B1 0 C1 D11 −γI C2 D21 0 0 −AT −CT 1 −C2T γI −BT 1 −D11T −DT21 | {z } H ξ1 ξ2 ξ3 ξ4 ξ5 = I 0 0 I 0 0 | {z } L ˙ ξ1 ˙ ξ2 ˙ ξ3 ˙ ξ4 ˙ ξ5 (9) By making five reduction steps, we will eventually arrive at the Hamiltonian defined in (8). Step 1) Remove ξ4 = γ−1(C1ξ1+ D11ξ2): A B1 0 C2 D21 0 −Q11 −Q12 −AT −C2T −Q21 −Q22 −B1T −DT21 ξ1 ξ2 ξ3 ξ5 = I 0 I 0 ˙ ξ1 ˙ ξ2 ˙ ξ3 ˙ ξ5 . Step 2) Replace ξ1 ξ2 = NXζ: A B1 0 0 −Q11 −Q12 −AT −C2T −Q21 −Q22 −B1T −DT21 NXζ ξ3 ξ5 = I 0 I 0 d dt(NXζ) ˙ ξ3 ˙ ξ5 .
Pre-multiply with the non-singular I NT X UT X
, where UX spans the range
of C2 D21 T : UT 0 0 −Q −U 0 ∗ ∗ (∗) NXζ ξ3 ξ5 = d dt(V Tζ) V ˙ξ3 0
where we can choose ξ5 to satisfy the DAE, since (∗) has full row rank. Consequently, UT 0 −Q(γ) −U ζ ξ3 = " d dt(V Tζ) V ˙ξ3 # , which corresponds to (7).
Step 3) Assume that V = V1 0 , and we obtain
U1T 0 0 UT 2 0 0 −Q(γ) −U1 −U2 ζ ξ3 ξ4 = d dt VT 1 0 ζ V1 0 ˙ ξ3 ˙ ξ4 = d dt(V T 1 ζ) 0 V1ξ˙3 , which yields UT 2 ζ = 0 and, consequently, ζ = U2⊥ζ1. UT 1U ⊥ 2 0 0 −U⊥T 2 QU2⊥ −U2⊥TU1 0
−U2kTQU2⊥ −U2kTU1 −U kT 2 U2 ζ1 ξ3 ξ4 = d dt(V T 1 U ⊥ 2 ζ1) U2⊥TV1ξ˙3 U2kTV1ξ˙3 ,
where ξ4 can be chosen so that the last row is always satisfied (U kT
2 U2 has
full row rank). Then " U1TU2⊥ 0 −U⊥T 2 QU ⊥ 2 −U ⊥T 2 U1 # ζ1 ξ3 = " d dt(V T 1 U ⊥ 2 ζ1) U2⊥TV1ξ˙3 # .
This step corresponds to replacing Lj → Lj−1 and Nj → Nj−1. Apply this
step recursively (ζ := ζ1, Q := U2⊥TQU2⊥, V := U2⊥TV1 and U := U2⊥TU1)
until V gets full column rank (removal of Lj and Nj).
Step 4) Assume than V = V1 0 , where V1 is non-singular: UT 1 U2T 0 −Q11 −Q12 −U1 −Q21 −Q22 −U2 ζ1 ζ2 ξ3 = d dt VT 1 0 ζ1 ζ2 V1 0 ˙ ξ3 = d dt V T 1 ζ1 V1ξ˙3 0 ,
which yields ζ2 = −Q−122(Q21ξ1+ U2ξ3), if we assume that Q22is non-singular.
Then U1T 0 −Q11 −U1 + U2T −Q12 Q−122 −Q21 −U2 ζ1 ξ3 = d dt(V T 1 ζ1) V1ξ˙3
This step corresponds to removing LTj blocks.
Step 5) Use the fact that V1 is non-singular, we obtain H
ξ1 ξ3 = ˙ ξ1 ˙ ξ3
where H is defined in (8) and ξ1 = V1Tζ1.
If we assume that ξ1 = V1Tζ1 is non-singular we can use X = ξ3ξ1−1 as a
solution to the LMI, since V1XV˙ 1T = V1ξ˙3− V1ξ3(V1Tζ1)−1V1Tζ˙1 (V1Tζ1)−1VT = −Qζ1− U ξ3 − V1ξ3(V1Tζ1)−1UTζ1 (V1Tζ1)−1VT = − Q + V1XUT + U XV1T ζ1(V1Tζ1)−1V1T and consequently, Q + V1XUT + U XV1T + V1XV˙ 1T = 0.
Note that the steps performed in this approach correspond to the steps in the Guptri algorithm [1, 2] used to partition H −λL into its KCF structure. Table 1 compares the steps in the different algorithms.
Table 1: Comparison of algorithmic steps. In Step 5) Re µ < 0 implies that ξ1 = 0 in the LTI case, which corresponds to the truncation of the Jj(µ)
block in the KCF approach.
KCF LTV DAE
(4a) → (7) Step 1)
NX (7) Step 2)
Lj and Nj blocks Step A) Step 3)
LT
j blocks Step B) Step 4)
Jj(µ) blocks Step C) Step 5)
3.4
The X and Y conditions
The condition
X I
I Y
0
where Xξ1 = ξ3 and Y υ1 = υ3, need only to be verified in the subspace of ξ1
and υ1: ξT 1Xξ1 ξ1Tυ1 υT 1ξ1 υ1TY υ1 = ξT 1ξ3 ξ1Tυ1 υT 1ξ1 υ1Tυ3 = ξ1 0 0 υ1 T ξ3 υ1 ξ1 υ3 0. (10)
Also,
ξ1T(0)ξ3(0) = ξ1T(0)P0ξ1(0) and υ1T(T )υ3(T ) = υ1T(T )P −1 T υ1(T )
need to be verified.
At least one of the solution vectors, ξ or υ, needs to be stored as a function of time; the other one can be verified during integration. If ξ has k columns we need 2nk real values at each time instant. We can reduce this by a non-singular transformation removing k2 variables. Also, we can use the fact that ξT
3ξ1 is symmetric and remove another 12k(k − 1) variables. This results in 1
2(4n − 3k + 1)k variables.
4
Examples
4.1
The NN3 example
This is an academic example from the COMPleib’s benchmarks as examples
[7]. It has four states and is defined by
G := A B C D = 0.5 1 1.5 1 0 0 −1 3 2.1 2 0 0 1 −1 −0.6 1 1 0 −2 2 −1 1 0 1 1 0 0 0 0 0 0 0 0 1 0 0
Even if this is defined as an LTI system, we can regard it as an LTV problem over a finite time horizon, say t ∈ [0, 10], where the terminal constraints are defined by P0 and PT.
The KCF structure of FX is N2, J1(0.7), J1(4) and of FY is N2, J1(−2.6), J1(1.5).
The optimal gain of the LTI problem is γ = 7.9019, which we will also use for the LTV case.
The eigenvalues of H − λL for FX are ±0.7 and ±4. For FY the
eigenvalues are ξ = 0.0933 0.0013 0.0746 0.0021 −0.0373 0.0016 0 0 −0.0671 0.0081 0.1600 0.0591 −0.1347 −0.0542 −0.5306 0.0640 −0.0960 0.1923 0.0118 0.0002 0.8033 −0.9759 , υ = 0.0376 0 −0.0216 0 0.0054 0 0 0 0.0108 0 0.0850 0 0.0543 0.0352 −0.3213 0.1409 0.2578 −0.2466 0.0007 0 −0.9045 0.9582 ,
where ξ and υ satisfy (10).
4.2
The REA2 example
This is an reactor model with four states, also from the COMPleib’s
bench-marks.
The FX problem has the KCF structure of U − λV is 2LT1, 2N1 and
for H − λL is given by 8N1, 2N3, J2(0), 2J1(±6.76924). If we select γ just
marginally above its LTI optimal value, γ∗ = 1.13409, we obtain two J1blocks
instead of J2(0), with eigenvalues in ±. The eigenvectors corresponding to
± will approach each other as → 0. If we select γ < γ∗, we obtain two
imaginary eigenvalues. For the LTI case this FX condition alone limits the
performance γ. In the Riccati approach this corresponds to the transition from real eigenvalues, ± to imaginary ones, ±j.
If we consider this as an LTV problem with fixed ABCD-matrices over a limited time interval [0, T ], we can reduce γ arbitrarily close to 0, if we at the same time choose the terminal constraints appropriately. For instance, for γ = 0.5, we obtain T = 0.9027, for γ = 1, T = 4.589, and for γ = 1.1, T = 11.285.
4.3
A launcher model
Consider the following LTV system defined in the interval t ∈ [0, 15]:
G := −0.013 t −0.020 t −25 − 0.2 t2 −25 −0.003 t −0.022 t −0.055 t2 −18 0 1 0 0 0 0 1 0 .
where the states are lateral velocity, vz, pitch angular rate, ˙θ, and pitch
angle, θ. This is a simplified model of a rocket with thrust vector control, which is valid during the first 15 seconds of its flight. At liftoff, t = 0, the dynamics is composed of a double and a single integrator in parallel, for which there exists no stabilizing controller. As the velocity increases the aerodynamics becomes gradually more and more important as the dynamic pressure increases (defined by the t and t2 terms).
We augment the system in a loop shaping fashion: G = I 0 0 0 0 I I 0 0 + I 0 I W2G d W1 0 I I ,
where W1 = 2 + 14t and W2 = 1. In order to account for computational
delays and servo dynamics, a first-order Padé delay of 50 ms, d(s) = 40−s40+s, implemented as
"
−40 4√5 4√5 −1
#
, is also included. This means that G has four states.
A quasi-stationary LTI design yields a γ between 3.06 and 3.16 for t > 0; we specify a constant γ = 3.19 to hold over the entire interval [0, 15]. We also take PT as the inverse of the quasi-stationary solution of FY at t = 15,
and P0 = 2I.
The eight discrete eigenvalues of H − λL of both FX and FY consist of
a real pair ±40 (delay), a complex quadruple roughly around ±6 ± 6j, and a slow real pair ±, where goes from 0 to 0.132. All eigenvalues but this slow pair are fast compared to the time variability of the system; they can be regarded as almost time invariant. The slow mode in ±, on the other hand, needs to be treated as time varying, since it has a long memory extending throughout the entire time interval.
In order to solve this problem numerically, we start by generating H(t) and L(t) based on G for t ∈ [0, 15]. After reordering the states according to L, we get H11 H12 H21 H22 | {z } H x1 x2 = I 0 0 0 | {z } L ˙x1 ˙x2 = ˙x1 0 , which results in ¯ Hx1 = H11− H12H22−1H21 x1 = ˙x1.
The solution can be obtained by solving ˙ X(t) = −X(t) I ¯H(t) I X(t)
numerically. The solution, Y (t), to FY is generated similarly. We can now
verify that miniλi(XY ) ≥ 1 holds for all t ∈ [0, 15], see Figure 2. Thus,
there exists an LTV controller that satisfies the H∞ design constraints (γ,
P0, PT).
Note that the quasi-stationary LTI design is singular for t = 0.
Figure 2: Minimum eigenvalue, miniλi(XY ) ≥ 0
4.4
Numerical issues
Even if the linear differential equation ˙ξ = Hξ looks like a relatively simple problem, it may still cause some significant numerical issues. Fast modes in H (with eigenvalues that have large real part) will cause rounding errors to increase exponentially and eventually drown the slower modes.
The standard Riccati equation, ˙X = −X I H
I X
, and ˙ξ = Hξ are connected by X = ξ2ξ1−1, where we can regard T = ξ
−1
1 as a
transfor-mation matrix. Let ζ = ξξ1−1 = ξT and, consequently, ˙ζ = Hζ + ξ ˙T = Hζ + ζT−1T :˙ d dt I X = H I X − I X I 0 H I X = 0 0 −X 1 H I X .
or ˙ X = −X I AT R −Q −A | {z } H I X = −XAT − AX − Q − XRX which is stable if A + RX is Hurwitz.
For the general case we have
L ˙ζ = LζT−1T + Hζ.˙
Assume that T−1T = −KHζ for some matrix K. Then,˙ L ˙ζ = (I − LζK) Hζ,
which has a Riccati type structure with a quadratic term. Let ζ + δ be a disturbed solution:
L ˙δ = (I − LζK)Hδ − LδKHζ,
The stationary solution (I − LζK)Hζ = 0 requires Hζ = LζKHζ and that L ˙δ = Hδ − LζKHδ − LδKHζ becomes stable.
5
Conclusions
The conditions for the existence of a controller for the LTV H∞ synthesis
problem can be formulated as two time-varying LMIs in X and Y , connected by a third LMI and a rank constraint. For the existence problem, only the extreme solutions, X and Y , are needed. Alternatively, these can solved using two differential algebraic equations (DAE).
There are still open problems with this approach. One important issue is how to solve the DAE in a numerically reliable way. Another is how to recover the controller when X and Y are available.
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Avdelning, Institution Division, Department
Division of Automatic Control Department of Electrical Engineering
Datum Date 2005-05-20 Språk Language 2 Svenska/Swedish 2 Engelska/English 2 Rapporttyp Report category 2 Licentiatavhandling 2 Examensarbete 2 C-uppsats 2 D-uppsats 2 Övrig rapport 2
URL för elektronisk version http://www.control.isy.liu.se
ISBN — ISRN
—
Serietitel och serienummer Title of series, numbering
ISSN 1400-3902
LiTH-ISY-R-3085
Titel Title
Generalized Riccati Equations for H∞Synthesis
Författare Author
Anders Helmersson
Sammanfattning Abstract
Conditions for the existence of controllers for linear time-varying (LTV) systems is given. The closed loop performance is specified in terms of energy gain, which also includes terminal constraints. The conditions can be formulated either as linear matrix inequalities (LMIs) or as solutions to Riccati differental equations with algebraic constraints.
