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210 CHAPTER 11. H 2 MODEL-ORDER REDUCTION FROM ORDER N TO N-3<br />
The two-parameter polynomial eigenvalue problem (11.3) involving the two matrices<br />
Ãã N−1<br />
(ρ 1 ,ρ 2 ) T and Ãã N−2<br />
(ρ 1 ,ρ 2 ) T can be jointly written as:<br />
[ ]<br />
ÃãN−1 (ρ 1 ,ρ 2 ) T<br />
ÃãN−2 (ρ 1 ,ρ 2 ) T<br />
v = A(ρ 1 ,ρ 2 ) v = 0 (11.39)<br />
where the matrix A(ρ 1 ,ρ 2 ) is rectangular and polynomial in both ρ 1 and ρ 2 . This<br />
matrix has 2 N columns and at most 2 N+1 (unique) rows. The <strong>to</strong>tal degree of the terms<br />
in ρ 1 and ρ 2 in the matrix A(ρ 1 ,ρ 2 ) is denoted by Ñ and, as stated in Corollary 10.2,<br />
is equal <strong>to</strong> N −1 (where N is the order of the original model H(s) <strong>to</strong> reduce). Because<br />
we are searching for blocks L T η 1<br />
<strong>to</strong> split off, we continue working with the transposed<br />
version of the matrix A(ρ 1 ,ρ 2 ). Thus, the matrix A(ρ 1 ,ρ 2 ) T can be written as:<br />
A(ρ 1 ,ρ 2 ) T =<br />
(<br />
)<br />
M 0 (ρ 2 )+ρ 1 M 1 (ρ 2 )+...+ ρÑ1 MÑ(ρ 2 )<br />
(11.40)<br />
where the matrices M i (ρ 2 ) are expanded as:<br />
M i (ρ 2 )=M i,0 + ρ 2 M i,1 + ρ 2 2 M i,2 + ...+ ρÑ−i 2 M i, Ñ−i , (11.41)<br />
for i =0,...,Ñ.<br />
The matrix A(ρ 1 ,ρ 2 ) T is rectangular of dimension m×n. Note that the dimensions<br />
of the matrix A(ρ 1 ,ρ 2 ) T are much smaller than the dimensions of the (linearized)<br />
matrices P (ρ 2 ) T and Q(ρ 2 ) T in the previous subsection. Therefore m and n denote<br />
different, smaller values here. Also in this non-linear case one should be able <strong>to</strong> split<br />
off singular parts. Assuming that there is dependency between the columns of the<br />
matrix A(ρ 1 ,ρ 2 ) T means that:<br />
A(ρ 1 ,ρ 2 ) T z(ρ 1 ,ρ 2 ) = 0 (11.42)<br />
has a non-zero solution z(ρ 1 ,ρ 2 ) depending on ρ 1 and ρ 2 . Take such a solution z in<br />
which the degree of ρ 1 is η 1 , the degree of ρ 2 is η 2 and where η 1 is as small as possible.<br />
Then z(ρ 1 ,ρ 2 ) can be written as:<br />
z(ρ 1 ,ρ 2 )=z 0 (ρ 2 ) − ρ 1 z 1 (ρ 2 )+ρ 2 1 z 2 (ρ 2 )+...+(−1) η1 ρ η1<br />
1 z η 1<br />
(ρ 2 ) (11.43)<br />
where z 0 (ρ 2 ),z 1 (ρ 2 ),...,z η1 (ρ 2 ) all polynomial in ρ 2 .<br />
Substituting such a solution z(ρ 1 ,ρ 2 ) in<strong>to</strong> (11.42) where the matrix A(ρ 1 ,ρ 2 ) T is<br />
written as in (11.40), yields the following when the relationships for the coefficients of<br />
the powers of ρ 1 are worked out and the resulting expressions are written in matrix-