Lecture Notes: Week 4b
Topic: Balanced model reduction
ECE/MAE 7360
Optimal and Robust Control
( Fall 2003 Offering)
Instructor: Dr YangQuan Chen, CSOIS, ECE Dept.,
Tel. : (435)797-0148.
E-mail: [email protected] or, [email protected]
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Balanced Model Reduction - Lecture Notes | MAE 7360, Study notes of Mechanical Engineering
Material Type: Notes; Class: Optimal and Robust Control; Subject: Mechanical & Aerospace Engr; University: Utah State University; Term: Fall 2003;
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Lecture Notes: Week 4b
Topic: Balanced model reduction
ECE/MAE 7360
Optimal and Robust Control
(Fall 2003 Offering)
Instructor : Dr YangQuan Chen, CSOIS, ECE Dept.,
Tel. : (435)797-0148.
E-mail: [email protected] or, [email protected]
Chapter 7: Balanced Model Reduction
- Balanced Realization
- Balanced Model Reduction
- Frequency Weighted Balanced Model Reduction
- Relative Reduction
Now partition the realization (A, B, C, D) compatibly with P as
A 11 A 12 B 1
A 21 A 22 B 2
C 1 C 2 D
^.
Then (^) ^ A^11 B^1 C 1 D
is also a realization of G. Moreover, (A 11 , B 1 ) is controllable if A 11 is stable. Proof Using 0 = AP + P A∗^ + BB∗ to get B 2 = 0 and A 21 = 0. Hence, part of the realization is not controllable:
A 11 A 12 B 1
A 21 A 22 B 2
C 1 C 2 D
^ =
A 11 A 12 B 1
0 A 22 0
C 1 C 2 D
^ =
^ A^11 B^1 C 1 D
(^).
- Let
^ A^ B C D
(^) be a state space realization of a (not necessarily stable)
transfer matrix G(s). Suppose that there exists a symmetric matrix
Q = Q∗^ =
Q^1 0 0
with Q 1 nonsingular such that QA + A∗Q + C∗C = 0. Now partition the realization (A, B, C, D) compatibly with Q as
A 11 A 12 B 1
A 21 A 22 B 2
C 1 C 2 D
^.
Then (^) ^ A^11 B^1 C 1 D
is also a realization of G. Moreover, (C 1 , A 11 ) is observable if A 11 is stable.
- Let P and Q be the controllability and observability Gramians,
AP + P A∗^ + BB∗^ = 0 A∗Q + QA + C∗C = 0. Suppose P = Q = Σ = diag(σ 1 , σ 2 ,... , σn) Then the state space realization is called internally balanced realiza- tion and σ 1 ≥ σ 2 ≥... ≥ σn ≥ 0, are called the Hankel singular values of the system. Two other closely related realizations are called input normal real- ization with P = I and Q = Σ^2 , and output normal realization with P = Σ^2 and Q = I. Both realizations can be obtained easily from the balanced realization by a suitable scaling on the states.
- Let P and Q be two positive semidefinite matrices. Then there exists a nonsingular matrix T such that
T P T ∗^ =
(T −^1 )∗QT −^1 =
respectively, with Σ 1 , Σ 2 , Σ 3 diagonal and positive definite.
Suppose G(s) =
^ A B C 0
(^) ∈ RH (^) ∞
is a balanced realization; that is, there exists
Σ = diag(σ 1 Is 1 , σ 2 Is 2 ,... , σN IsN ) ≥ 0
with σ 1 > σ 2 >... > σN ≥ 0, such that
AΣ + ΣA∗^ + BB∗^ = 0 A∗Σ + ΣA + C∗C = 0
Then
σ 1 ≤ ‖G‖∞ ≤
∫ (^) ∞ 0 ‖g(t)‖^ dt^ ≤^2
∑N i=
σi
where g(t) = CeAt^ B.
Proof.
x ˙ = Ax + Bw z = Cx.
(A, B) is controllable and (C, A) is observable.
d dt (x∗Σ−^1 x) = ˙x∗Σ−^1 x + x∗Σ−^1 x˙ = x∗(A∗Σ−^1 + Σ−^1 A)x + 2〈w, B∗Σ−^1 x〉
d dt
(x∗Σ−^1 x) = ‖w‖^2 − ‖w − B∗Σ−^1 x‖^2
Integration from t = −∞ to t = 0 with x(−∞) = 0 and x(0) = x 0 gives
x∗ 0 Σ−^1 x 0 = ‖w‖^22 − ‖w − B∗Σ−^1 x‖^22 ≤ ‖w‖^22
w∈L^ inf 2 [−∞,0)
{ ‖w‖^22
∣∣ ∣∣ x(0) = x 0
} = x∗ 0 Σ−^1 x 0.
Given x(0) = x 0 and w = 0 for t ≥ 0, the norm of z(t) = CeAt^ x 0 can be found from ∫ (^) ∞ 0 ‖z(t)‖
(^2) dt = ∫^ ∞ 0 x
∗ 0 e
A∗t (^) C∗CeAt (^) x 0 dt = x∗ 0 Σx^0
To show σ 1 ≤ ‖G‖ (^) ∞, note that
‖G‖∞ = sup w∈L 2 (−∞,∞)
‖g ∗ w‖ 2 ‖w‖ 2 = sup w∈L 2 (−∞,∞)
√∫ −∞∞ ‖z(t)‖^2 dt √∫ −∞∞ ‖w(t)‖^2 dt
≥ sup w∈L 2 (−∞,0]
√∫ 0 ∞ ‖z(t)‖^2 dt √∫ −∞^0 ‖w(t)‖^2 dt^ = sup^ x^06 =
√√ √√ √ x
∗ 0 Σx 0 x∗ 0 Σ−^1 x 0
= σ 1
We shall now show the other inequalities. Since
G(s) :=
∫ (^) ∞ 0 g(t)e
−st (^) dt, Re(s) > 0 ,
by the definition of H∞ norm, we have
‖G‖∞ = sup Re(s)> 0
∥∥ ∥∥ ∫ (^) ∞ 0 g(t)e
−st (^) dt∥∥∥∥
≤ sup Re(s)> 0
∫ (^) ∞ 0
∥∥ ∥∥g(t)e−st
∥∥ ∥∥ dt
≤
∫ (^) ∞ 0 ‖g(t)‖^ dt. To prove the last inequality, let ei be the ith unit vector and define
E 1 =
[ e 1 · · · es 1
] ,... ,
EN =
[ es 1 +···+sN − 1 +1 · · · es 1 +···+sN
] .
Then ∑N i=
Ei E i∗ = I and ∫ (^) ∞ 0 ‖g(t)‖^ dt^ =^
∫ (^) ∞ 0
∥∥ ∥∥ ∥∥CeAt/^2
∑N i=
Ei E i∗ eAt/^2 B
∥∥ ∥∥ ∥∥ dt
≤ ∑N i=
∫ (^) ∞ 0
∥∥ ∥∥CeAt/^2 Ei E i∗ eAt/^2 B
∥∥ ∥∥ dt
∑N i=
∫ (^) ∞ 0
∥∥ ∥∥CeAt/^2 Ei
∥∥ ∥∥
∥∥ ∥∥E i∗ eAt/^2 B
∥∥ ∥∥ dt
∑N i=
√∫ (^) ∞ 0
∥∥ ∥CeAt/^2 Ei
∥∥ ∥^2 dt
√∫ (^) ∞ 0
∥∥ ∥E i∗ eAt/^2 B
∥∥ ∥^2 dt
≤ 2
∑N i=
σi
Balanced Model Reduction
G = Gr + ∆a , =⇒ inf deg(Gr)≤r ‖G − Gr‖∞.
- Suppose
G(s) =
A 11 A 12
A 21 A 22
B 1
B 2
C 1 C 2 D
is a balanced realization with Gramian Σ = diag(Σ 1 , Σ 2 )
AΣ + ΣA∗^ + BB∗^ = 0 A∗Σ + ΣA + C∗C = 0.
where
Σ 1 = diag(σ 1 Is 1 , σ 2 Is 2 ,... , σr Isr ) Σ 2 = diag(σr+1Isr+1, σr+2Isr+2,... , σN IsN )
and σ 1 > σ 2 > · · · > σr > σr+1 > σr+2 > · · · > σN where σi has multiplicity si , i = 1, 2 ,... , N and s 1 +s 2 +· · ·+sN = n. Then the truncated system
Gr(s) =
^ A^11 B^1 C 1 D
is balanced and asymptotically stable. Furthermore
‖G(s) − Gr(s)‖∞ ≤ 2(σr+1 + σr+2 + · · · + σN ).
- ‖G(s) − G(∞)‖∞ ≤ 2(σ 1 +... + σN ).
- ‖G(s) − Gn− 1 (s)‖∞ = 2σN.
Proof. We shall first show the one step model reduction. Hence we shall assume Σ 2 = σN IsN. Define the approximation error
E 11 :=
A 11 A 12
A 21 A 22
B 1
B 2
C 1 C 2 D
^ −
^ A^11 B^1 C 1 D
A 11 0 0 B 1
0 A 11 A 12 B 1
0 A 21 A 22 B 2
−C 1 C 1 C 2 0
Apply a similarity transformation T to the preceding state-space realiza- tion with
T =
I/ 2 I/ 2 0
I/ 2 −I/ 2 0
0 0 I
^ ,^ T^
I I 0
I −I 0
0 0 I
to get
E 11 =
A 11 0 A 12 / 2 B 1
0 A 11 −A 12 / 2 0
A 21 −A 21 A 22 B 2
0 − 2 C 1 C 2 0
Consider a dilation of E 11 (s):
E(s) =
E^11 (s)^ E^12 (s) E 21 (s) E 22 (s)
A 11 0 A 12 / 2 B 1 0
0 A 11 −A 12 / 2 0 σN Σ− 1 1 C 1 ∗ A 21 −A 21 A 22 B 2 −C 2 ∗ 0 − 2 C 1 C 2 0 2 σN I − 2 σN B 1 ∗ Σ− 1 1 0 −B 2 ∗ 2 σN I 0
A^ ˜^ B˜ C^ ˜ D˜
Let Ek(s) = Gk+1(s) − Gk(s) for k = 1, 2 ,... , N − 1 and let GN (s) = G(s). Then σ [Ek(jω)] ≤ 2 σk+
since Gk(s) is a reduced-order model obtained from the internally balanced realization of Gk+1(s) and the bound for one-step order reduction holds. Noting that G(s) − Gr(s) =
N∑− 1 k=r
Ek(s)
by the definition of Ek(s), we have
σ [G(jω) − Gr(jω)] ≤
N∑− 1 k=r
σ [Ek(jω)] ≤ 2
N∑− 1 k=r
σk+
This is the desired upper bound. 2
- bound can be tight. For example,
G(s) = ∑n j=
b (^) i s + ai
−a 1
b 1 −a 2
b 2
... ... −an
√ b^ n b 1
b 2 · · ·
b (^) n 0
with ai > 0 and b (^) i > 0. Then P = Q =
bibj ai+aj
(^) and
‖G(s)‖∞ = G(0) = ∑n i=
b (^) i ai
= 2trace(P ) = 2 ∑n i=
σi
- bound can also be loose for systems with Hankel singular values close to each other. For example,
G(s) =
with Hankel singular values given by
σ 1 = 1, σ 2 = 0. 9977 , σ 3 = 0. 9957 , σ 4 = 0. 9952.
r 0 1 2 3 ‖G − Gr‖∞ 2 1.996 1.991 1. Bounds: 2 ∑^4 i=r+1 σi 7.9772 5.9772 3.9818 1. 2 σr+1 2 1.9954 1.9914 1.
Now let T be a nonsingular matrix such that
T P T ∗^ = (T −^1 )∗QT −^1 =
Σ^1 Σ 2
(i.e., balanced) and partition the system accordingly as
^ T AT^ −^1 T B CT −^1
(^) =
A 11 A 12 B 1
A 21 A 22 B 2
C 1 C 2 0
^.
Then a reduced order model Gr is obtained as
Gr =
^ A^11 B^1 C 1 0
(^).
Works well but with guarantee.
Relative Reduction
Gr = G(I + ∆rel), =⇒ inf deg(Gr)≤r
∥∥ ∥∥G−^1 (G − Gr)
∥∥ ∥∥ ∞
and a related problem is
G = Gr(I + ∆mul)
Let G(s) =
^ A^ B C D
(^) ∈ RH (^) ∞ be minimum phase and D be nonsingular.
Then Wo = G−^1 (s) =
^ A^ −^ BD−^1 C^ −BD−^1 D−^1 C D−^1
.
(a) Then the input/output weighted Gramians P and Q are given by P A∗^ + AP + BB∗^ = 0 Q(A − BD−^1 C) + (A − BD−^1 C)∗Q + C∗(D−^1 )∗D−^1 C = 0. (b) Suppose P and Q are balanced: P = Q = diag(σ 1 Is 1 ,... , σr Isr , σr+1Isr+1,... , σN IsN ) = diag(Σ 1 , Σ 2 ) and let G be partitioned compatibly with Σ 1 and Σ 2 as
G(s) =
A 11 A 12 B 1
A 21 A 22 B 2
C 1 C 2 D
^.
Then Gr(s) =
^ A^11 B^1 C 1 D
is stable and minimum phase. Furthermore
‖∆rel‖∞ ≤ ∏N i=r+
( 1 + 2σi(
√ 1 + σ i^2 + σi)
) − 1
‖∆mul‖∞ ≤
∏N i=r+
( 1 + 2σi(
√ 1 + σ i^2 + σi)
) − 1.