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206 CHAPTER 11. H 2 MODEL-ORDER REDUCTION FROM ORDER N TO N-3
and Q(ρ 2 ) T = C T are plugged in too, the following system of equations arises:
⎛ ⎞
z 0,0
z 0,1
z 0,2
M 1,2 = 0 (11.29)
z
⎜ 1,0
⎟
⎝ z 1,1 ⎠
z 1,2
where the dimension of M 1,2 is (η 1 +2) (η 2 +2) m × (η 1 +1) (η 2 +1) n =12m × 6 n
and where M 1,2 is structured as:
⎛
B T ⎞
0 0 0 0 0
D T B T 0 0 0 0
0 D T B T 0 0 0
0 0 D T 0 0 0
C T 0 0 B T 0 0
M 1,2 =
0 C T 0 D T B T 0
0 0 C T 0 D T B T
. (11.30)
0 0 0 0 0 D T
0 0 0 C T 0 0
⎜
0 0 0 0 C T 0
⎟
⎝ 0 0 0 0 0 C T ⎠
0 0 0 0 0 0
If the rank of this matrix M 1,2 is smaller than 6n, one can compute solutions
z(ρ 1 ,ρ 2 )of(B T + ρ 1 C T + ρ 2 D T ) z(ρ 1 ,ρ 2 ) = 0, by computing non-zero vectors in the
kernel of this matrix: a non-zero vector in this kernel is structured as [z0,0, T z0,1, T z0,2,
T
z1,0, T z1,1, T z1,2] T T and from these (sub)vectors, one can construct the solution z(ρ 1 ,ρ 2 )
as z(ρ 1 ,ρ 2 )=(z 0,0 + ρ 2 z 0,1 + ρ 2 2 z 0,2 ) − ρ 1 (z 1,0 + ρ 2 z 1,1 + ρ 2 2 z 1,2 ).
Once a solution z(ρ 1 ,ρ 2 )=z 0 (ρ 2 ) − ρ 1 z 1 (ρ 2 )+...+(−1) η1 ρ η1
1 z η 1
(ρ 2 ) of (11.22)
has become available, it can be used to split off one or more singular parts. As stated
in Proposition 11.1, the only singular parts that occur in such a pencil are L T η (11.14)
of dimension (η 1 +1)×η 1 . Because we are working with the transposes of the matrices
P (ρ 2 ) and Q(ρ 2 ), we find the transposed version of the singular blocks L T η . Singular
blocks (11.13) of dimension ε 1 × (ε 1 + 1) can not occur here because they would allow
for an infinite number of solutions to the original problem.
If there exists a solution z(ρ 1 ,ρ 2 ), where the degree of ρ 1 is η 1 , the degree of ρ 2
is η 2 and η 1 is as small as possible, then the matrix P (ρ 2 )+ρ 1 Q(ρ 2 ) is equivalent to
a matrix of the form: ( L
T
η1
0
0 ˜P (ρ2 )+ρ 1 ˜Q(ρ2 )
The singular block L T η 1
)
. (11.31)
has dimension (η 1 +1)× η 1 and it corresponds to unwanted
indeterminate eigenvalues. Therefore it can legally be split off from the pencil