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- Lyapunov Equations
- Gramians
- Controllable Decomposition
- Observable Decomposition
- Balanced Realization
- Computing Balanced Realization
- Balanced Truncation and Matlab
- Bounds of Functions Norms
- Balanced Model Reduction
- Frequency Weighted Balanced Model
Reduction
- Relative Reduction
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- Consider the following Lyapunov equation
-
A*X+XA+Q = 0
- Assume that A is stable, then the following statements hold:
- X =
- X > 0 if Q>0 and X ³0 if Q³0.
- if Q³0, then (Q,A) is observable iff X > 0.
- Proof:
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- Suppose X is the solution of the Lyapunov equation, then
- Reli(A)£0 if X > 0 and Q³0.
- A is stable if X > 0 and Q
> 0.
- A is stable if X ³0, Q³0 and (Q, A) is
detectable.
- Proof: Let l be an
eigenvalue and v be a
corresponding eigenvector of A, I.e., Av=lv. Then
- 0=v*(A*X+XA)v+v*Qv=(l*+l)v*Xv+v*Qv=2 Re l v*Xv+v*Qv
- ³ 2 Re l v*Xv since Q³0. Thus Re l £0.
- If Re l =0, then
v*Qv=0. Thus Qv=0 and Av=lv. By PBH test, (Q, A) is not detectable.
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- Let A be stable. Then a pair(C,A) is observable iff the observability
Gramian Q > 0
- A*Q+QA+C*C=0.
- Similarly, (A,B) is controllable iff the controllability Gramian P >
0
- AP+PA*+BB*=0
- When A is not necessarily stable, the above Lyapunov equations may still
have solutions. But they are not gramians.
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- Let be a state space
realization of a (not necessarily stable) transfer matrix G(s). Suppose
that there exists a symmetric matrix
- (P is not necessarily the
Gramian) with P1 nonsingular such that
- AP+PA*+BB*=0
- Now partition the realization (A,B,C,D)
compatibly with P as
- Then
is also a realization of G.
- Moreover, (A11,B1) is controllable if A11
is stable.
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- Proof Using the partitioned equation 0=AP+PA*+BB* to get
- A11P1+P1A11*+B1B1*=0, P1A21*+B1B2*=0,
and B2B2*=0
- then we conclude that B2 = 0 and A21 = 0. Hence,
part of the realization is not controllable:
- P1>0 if A11 is stable. So (A11, B1)
is controllable.
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- Let be a state space
realization of a (not necessarily stable) transfer matrix G(s). Suppose
that there exists a symmetric matrix
- (Q is not necessarily the
Gramian) with Q1 nonsingular such that
- A*Q+QA+C*C=0
- Now partition the realization (A,B,C,D)
compatibly with Q as
- Then
is also a realization of G.
- Moreover, (C1 , A11) is observable if A11
is stable.
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- Let P and Q be the controllability and observability Gramians,
- AP+PA*+BB*=0, A*Q+QA+C*C=0
- Suppose P=Q=S=diag(s1, s2,
, sn). Then the state space realization
is called internally balanced realization and s1 ³ s2 ³
³sn ³0, are called the Hankel singular values of the
system.
- Two other closely related realizations are called
- input normal realization with P = I and Q = S2
- and
- output normal realization with P = S2 and Q = I.
- Both realizations can be obtained easily from the balanced realization
by a suitable scaling on the states.
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- In the special case where
is a minimal realization, a balanced realization can be obtained
through the following simplified procedure:
- 1. Compute P>0 and Q>0.
- 2. Find a matrix R such that P =R*R.
- 3. Diagonalize RQR* to get RQR*=U S2U*.
- 4. Let T 1=R*U S-1/2. Then T PT*=(T*)1QT 1=S and is balanced.
- In general, let P and Q be two positive semidefinite matrices. Then
there exists a nonsingular matrix T such that
- respectively with S1,
S2,
S3
diagonal and positive definite.
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- Suppose sr »
sr+1
for some r then the balanced realization implies that those states
corresponding to the singular values of sr+1,
,sn are less controllable and observable
than those states corresponding to s1,
,sr. Therefore, truncating those less
controllable and observable states will not lose much information about
the system.
- MATLAB:
- >>[Ab,Bb,Cb,sig,Tinv]=balreal(A,B,C); %sig is a vector of Hankel
singular values and Tinv=T-1;
- >> [Gb,sig] = sysbal(G);
- >> Gr = strunc(Gb,2); %truncate to the
second-order.
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- Suppose
is a balanced realization: that is, there exists
- S=diag(s1 Is1,
s2 Is2,
,
sN IsN) with
s1
> s2 >
>sN ³0,
- such that
- AS+SA*+BB*=0, A*S+SA+C*C=0
- Then
- where g(t)=CeAtB.
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- Proof: Consider
- where (A,B) is controllable and (C,A) is observable.
- Integration from t= -₯ to t = 0
with x(-₯ )=0
and x(0) = x0 gives
- Given x(0) = x0 and w = 0 for t³0, the norm of z(t)=CeAt x0
can be found from
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- To show note
that
- We shall now show the other inequalities. Since
- by the definition of H₯ norm, we have
- See next page for the proof of the last inequality.
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- To prove the last inequality, let ei be the ith unit vector
and define
- Then and
- where we have used Cauchy-Schwarz inequality and the following
relations
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- Additive Model Reduction
- Suppose
- is a balanced realization with Gramian S=diag (S1, S2)
- AS+SA*+BB*=0, A*S+SA+C*C=0
- where
- S1 =diag
(s1 Is1,
s2 Is2,
,
sr Isr)
, S2=diag
(sr+1 Isr+1
, s2 Isr+2,
,
sN IsN)
- and s1 > s2 >
> sr>
sr+1
> sr+2 >
> sN ³0,
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- where si
has multiplicity si, i
= 1,2,
,N and s1+s2+
+sN = n.
- Then the truncated system
- is balanced and asymptotically stable. Furthermore
- ||G(s)-Gr(s)||₯ £ 2(sr+1+sr+2+
+sN).
- ||G(s)-G(₯)||₯ £ 2(s1+s2+
+sN).
- ||G(s)-GN-1(s)||₯ =2sN.
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- Proof. We shall first show the one step model reduction. Hence we shall
assume S2=sN IsN
. Define the approximation error
- Apply a similarity transformation T to the preceding state-space
realization with
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- to get
- Consider a dilation of E11(s):
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- Then it is easy to verify that
- Using these equations, we have
- where a second quality is obtained by applying a similarity
transformation
- Hence
which is the desired result.
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- The remainder of the proof is achieved by using the order reduction by
one-step results and by noting that
- obtained by the kth order partitioning is internally balanced with
balanced Gramian given by S1 =diag (s1 Is1,
s2 Is2,
,
sr Isr)
- Let Ek(s)=Gk+1(s)-Gk(s) for k=1,2,
,N-1
and let GN(s)=G(s). Then
- 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
- by the definition of Ek(s), we have
- This is the desired upper bound.
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- Bound can be tight. For example,
- with ai>0 and bi>0. Then
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- Bound can also be loose for systems with Hankel singular values close to
each other. For example,
- with Hankel singular values given by
- s1 =1,
s2=0.9977,
s3=0.99570,
s4=0.9952
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- General Case: given weights Wo and Wi, find Gr
such that
- Let
- and solving the following Lyapunov equations
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- The input/output weighted Gramians P and Q are defined by
- P and Q satisfy the following lower order equations
- If Wo=I, then P can be obtained from
- PA*+AP+BB*=0
- If Wi =I, then Q can be obtained from
- QA+A*Q+C*C=0
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- Now let T be a nonsingular matrix such that
- (i.e., balanced) and partition the system accordingly as
- Then a reduced order model Gr is obtained as
- Works well but without guarantee in general.
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- Consider a special frequency-weighted model reduction problem:
- This is a relative error approximation problem.
- Consider also a related multiplicative error reduction problem
- Let
be minimum phase and D be nonsingular.
- Then
- Now apply the frequency-weighted model reduction to get
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- (a) Then the input/output weighted Gramians P and Q are given by
- PA*+AP+BB*=0
- Q(A-BD-1C)+(A-BD-1C)*Q+C*(D-1)*D-1C=0
- (b) Suppose P and Q are balanced:
- P=Q=S=diag (S1, S2)=diag (s1 Is1 ,
, sr Isr
,sr+1 Isr+1
, sN IsN)
- and let G be partitioned
compatibly with S1,
S2
as
- Then is stable
and minimum phase. Furthermore
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