1

 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

2

 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:

3


4

 Suppose X is the solution of the Lyapunov equation, then
 Rel_{i}(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.

5

 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.

6

 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 P_{1} 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, (A_{11},B_{1}) is controllable if A_{11}
is stable.

7

 Proof Using the partitioned equation 0=AP+PA*+BB* to get
 A_{11}P_{1}+P_{1}A_{11}*+B_{1}B_{1}*=0, P_{1}A_{21}*+B_{1}B_{2}*=0,
and B_{2}B_{2}*=0
 then we conclude that B_{2} = 0 and A_{21} = 0. Hence,
part of the realization is not controllable:
 P_{1}>0 if A_{11} is stable. So (A_{11}, B_{1})
is controllable.

8

 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 Q_{1} 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, (C_{1 }, A_{11}) is observable if A_{11}
is stable.

9

 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(s_{1}, s_{2},
, s_{n}). Then the state space realization
is called internally balanced realization and s_{1 }³_{ }s_{2 }³_{ }
³s_{n} 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 = S^{2}
 and
 output normal realization with P = S^{2} and Q = I.
 Both realizations can be obtained easily from the balanced realization
by a suitable scaling on the states.

10

 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 S^{2}U*.
 4. Let T ^{1}=R*U S^{1/2}. Then T PT*=(T*)^{1}QT ^{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 S_{1},
S_{2},
S_{3}
diagonal and positive definite.

11

 Suppose s_{r }»_{
}s_{r+1}
for some r then the balanced realization implies that those states
corresponding to the singular values of s_{r+1},
,s_{n} are less controllable and observable
than those states corresponding to s_{1},
,s_{r}. 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};
 >> [G_{b},sig] = sysbal(G);
 >> G_{r} = strunc(G_{b},2); %truncate to the
secondorder.

12

 Suppose
is a balanced realization: that is, there exists
 S=diag(s_{1 }I_{s}_{1},
s_{2 }I_{s}_{2},
,
s_{N }I_{s}_{N}) with
s_{1
}>_{ }s_{2 }>_{ }
>s_{N} ³0,
 such that
 AS+SA*+BB*=0, A*S+SA+C*C=0
 Then
 where g(t)=Ce^{At}B.

13

 Proof: Consider
 where (A,B) is controllable and (C,A) is observable.
 Integration from t= ₯ to t = 0
with x(₯ )=0
and x(0) = x_{0} gives
 Given x(0) = x_{0} and w = 0 for t³0, the norm of z(t)=Ce^{At }x_{0}
can be found from

14

 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.

15

 To prove the last inequality, let e_{i} be the ith unit vector
and define
 Then and
 where we have used CauchySchwarz inequality and the following
relations

16

 Additive Model Reduction
 Suppose
 is a balanced realization with Gramian S=diag (S_{1}, S_{2})
 AS+SA*+BB*=0, A*S+SA+C*C=0
 where
 S_{1 }=diag
(s_{1 }I_{s}_{1},
s_{2 }I_{s}_{2},
,
s_{r }I_{s}_{r})
, S_{2}=diag
(s_{r+1 }I_{s}_{r+1}
, s_{2 }I_{s}_{r+2},
,
s_{N }I_{s}_{N})
 and s_{1 }>_{ }s_{2 }>_{ }
> s_{r}>_{
}s_{r+1
}>_{ }s_{r+2 }>_{ }
> s_{N} ³0,

17

 where s_{i}
has multiplicity s_{i}, i
= 1,2,
,N and s_{1}+s_{2}+
+s_{N }= n.
 Then the truncated system
 is balanced and asymptotically stable. Furthermore
 G(s)G_{r}(s)_{₯} £ 2(s_{r+1}+s_{r+2}+
+s_{N}).
 G(s)G(₯)_{₯} £ 2(s_{1}+s_{2}+
+s_{N}).
 G(s)G_{N1}(s)_{₯} =2s_{N}.

18

 Proof. We shall first show the one step model reduction. Hence we shall
assume S_{2}=s_{N }I_{s}_{N
}. Define the approximation error
 Apply a similarity transformation T to the preceding statespace
realization with

19

 to get
 Consider a dilation of E_{11}(s):

20

 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.

21

 The remainder of the proof is achieved by using the order reduction by
onestep results and by noting that
 obtained by the kth order partitioning is internally balanced with
balanced Gramian given by S_{1 }=diag (s_{1 }I_{s}_{1},
s_{2 }I_{s}_{2},
,
s_{r }I_{s}_{r})
 Let E_{k}(s)=G_{k+1}(s)G_{k}(s) for k=1,2,
,N1
and let G_{N}(s)=G(s). Then
 Since G_{k}(s) is a reducedorder model obtained from the
internally balanced realization of G_{k+1}(s) and the bound for
onestep order reduction holds.
 Noting that
 by the definition of E_{k}(s), we have
 This is the desired upper bound.

22

 Bound can be tight. For example,
 with a_{i}>0 and b_{i}>0. Then

23

 Bound can also be loose for systems with Hankel singular values close to
each other. For example,
 with Hankel singular values given by
 s_{1 }=1,_{
}s_{2}=0.9977,
s_{3}=0.99570,
s_{4}=0.9952

24

 General Case: given weights W_{o} and W_{i}, find G_{r}
such that
 Let
 and solving the following Lyapunov equations

25

 The input/output weighted Gramians P and Q are defined by
 P and Q satisfy the following lower order equations
 If W_{o}=I, then P can be obtained from
 PA*+AP+BB*=0
 If W_{i }=I, then Q can be obtained from
 QA+A*Q+C*C=0

26

 Now let T be a nonsingular matrix such that
 (i.e., balanced) and partition the system accordingly as
 Then a reduced order model G_{r} is obtained as
 Works well but without guarantee in general.

27

 Consider a special frequencyweighted 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 frequencyweighted model reduction to get

28

 (a) Then the input/output weighted Gramians P and Q are given by
 PA*+AP+BB*=0
 Q(ABD^{1}C)+(ABD^{1}C)*Q+C*(D^{1})*D^{1}C=0
 (b) Suppose P and Q are balanced:
 P=Q=S=diag (S_{1}, S_{2})=diag (s_{1 }I_{s}_{1} ,
, s_{r }I_{s}_{r}
,s_{r+1 }I_{s}_{r+1}
, s_{N }I_{s}_{N})
 and let G be partitioned
compatibly with S_{1},
S_{2}
as
 Then is stable
and minimum phase. Furthermore
