link to my thesis

212 CHAPTER 11. H 2 MODEL-ORDER REDUCTION FROM ORDER N TO N-3
there is a polynomial solution z(ρ 1 ,ρ 2 )ofA(ρ 1 ,ρ 2 ) T z = 0 where the degree of ρ 1 is
η 1 = 1 and the degree of ρ 2 is η 2 =2.
Using (11.40), the expression A(ρ 1 ,ρ 2 ) T z(ρ 1 ,ρ 2 ) = 0 can be written as:
(
)
M 0 (ρ 2 )+ρ 1 M 1 (ρ 2 )+ρ 2 1 M 2 (ρ 2 )+ρ 3 1 M 3 (ρ 2 ) z(ρ 1 ,ρ 2 ) = 0 (11.47)
where:
M 0 (ρ 2 ) = M 0,0 + ρ 2 M 0,1 + ρ 2 2 M 0,2 + ρ 3 2 M 0,3
M 1 (ρ 2 ) = M 1,0 + ρ 2 M 1,1 + ρ 2 2 M 1,2
M 2 (ρ 2 ) = M 2,0 + ρ 2 M 2,1
M 3 (ρ 2 ) = M 3,0 ,
(11.48)
and where every matrix M i,j is of dimension m × n. Recall again that the dimensions
m and n are smaller than in the previous subsection because no linearization step is
involved here.
Every solution z(ρ 1 ,ρ 2 ) should take the form z 0 (ρ 2 ) − ρ 1 z 1 (ρ 2 ) because η 1 =1. If
such a solution is substituted into (11.47), and the relationships for the coefficients
of the powers of ρ 1 are worked out, we find:
⎛
⎜
⎝
M 0 (ρ 2 ) 0
M 1 (ρ 2 ) M 0 (ρ 2 )
M 2 (ρ 2 ) M 1 (ρ 2 )
M 3 (ρ 2 ) M 2 (ρ 2 )
0 M 3 (ρ 2 )
⎞
⎟
⎠
(
z0 (ρ 2 )
−z 1 (ρ 2 )
)
(
z0 (ρ
= M 1,2 (ρ 2 )
2 )
−z 1 (ρ 2 )
)
= 0 (11.49)
where the dimension of the block matrix M 1,2 (ρ 2 )is(η 1 + Ñ +1)m × (η 1 +1)n =
5m × 2n.
The degree of ρ 2 is η 2 = 2 and therefore z i (ρ 2 ) can be written as z i (ρ 2 ) =
z i,0 + ρ 2 z i,1 + ρ 2 2 z i,2 for i =0, 1. This can be substituted into (11.49). When
the relationships for the coefficients of the powers of ρ 2 are worked out again, the
following system of equations in matrix-vector form occurs:
⎛
M 1,2 ⎜
⎝
z 0,0
z 0,1
z 0,2
z 1,0
z 1,1
z 1,2
⎞
= 0 (11.50)
⎟
⎠
where the dimension of the matrix M 1,2 is (η 2 + Ñ + 1)(η 1 + Ñ +1) m × (η 1 + 1)(η 2 +