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218 CHAPTER 11. H 2 MODEL-ORDER REDUCTION FROM ORDER N TO N-3<br />

terms in ρ 1 and ρ 2 are worked out again, the following system of equations emerges:<br />

⎛ ⎞<br />

z 0,0<br />

.<br />

z 0,η2<br />

M η1,η 2 .<br />

= 0 (11.67)<br />

z η1,0<br />

⎜<br />

⎝<br />

⎟<br />

. ⎠<br />

z η1,η 2<br />

where the matrix M η1,η 2<br />

is a block matrix containing the coefficient matrices B 0 ,...,<br />

BÑ, and C 0,0 ,...,CÑ−1,0 from the Equations (11.60) and (11.61). The dimension of<br />

this matrix is 1 2 (η 1 + 2)(η 2 + Ñ + 1)(η 2 + Ñ +2)m × (η 1 + 1)(η 2 +1)n. Note that<br />

this matrix is independent of ρ 1 and ρ 2 . Every non-zero vec<strong>to</strong>r in the kernel of the<br />

matrix M η1,η 2<br />

provides a vec<strong>to</strong>r z(ρ 1 ,ρ 2 ) which is a solution of (11.59).<br />

Remark 11.6. The system of equations (11.67) in general is <strong>to</strong>o restrictive in comparison<br />

with the approach of the previous section. This is due <strong>to</strong> the fact that the<br />

powers of ρ 1 and ρ 2 play a simultaneous role in the equations in (11.67). This is the<br />

reason why this approach does not work in general. However, this approach is useful<br />

because when the system of equations (11.65) admits a solution, we have a solution<br />

for the system of equations (11.44) of the previous subsection <strong>to</strong>o. In this way we can<br />

still split off a singular block of the matrix pencil. Note that the system of equations<br />

(11.44) may have solutions which are not of this more ‘restrictive’ form.<br />

Example 11.1 (continued). Using this approach, A(ρ 1 ,ρ 2 ) T z(ρ 1 ,ρ 2 ) = 0 is written<br />

as: (<br />

)<br />

B(ρ 2 )+ρ 1 C(ρ 1 ,ρ 2 ) z(ρ 1 ,ρ 2 ) = 0 (11.68)<br />

where<br />

and<br />

B(ρ 2 )=B 0 + ρ 2 B 1 + ρ 2 2 B 2 + ρ 3 2 B 3 (11.69)<br />

C(ρ 1 ,ρ 2 )= ( C 0,0 + ρ 2 C 0,1 + ρ 2 2 C 0,2<br />

)<br />

+ ρ1 (C 1,0 + ρ 2 C 1,1 )+ρ 2 1 (C 2,0 ) . (11.70)<br />

We know that every solution z(ρ 1 ,ρ 2 ) should take the form z 0 (ρ 2 ) − ρ 1 z 1 (ρ 2 )<br />

because η 1 is chosen as 1. If such a solution is substituted in<strong>to</strong> (11.68), and the<br />

relationships of the coefficients of the powers of ρ 1 are equated <strong>to</strong> zero, it can be<br />

written as:<br />

⎛<br />

⎞<br />

B(ρ 2 ) 0 ( )<br />

( )<br />

⎝ C(ρ 1 ,ρ 2 ) B(ρ 2 ) ⎠ z0 (ρ 2 )<br />

z0 (ρ<br />

= M 1,2 (ρ 1 ,ρ 2 )<br />

2 )<br />

= 0 (11.71)<br />

−z 1 (ρ 2 )<br />

−z 1 (ρ 2 )<br />

0 C(ρ 1 ,ρ 2 )<br />

where the dimension of the block matrix M 1,2 (ρ 1 ,ρ 2 )is(η 1 +2) m × (η 1 +1) n =<br />

3 m × 2 n.

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