link to my thesis
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220 CHAPTER 11. H 2 MODEL-ORDER REDUCTION FROM ORDER N TO N-3<br />
31m × 6n:<br />
⎛<br />
M 1,2 =<br />
⎜<br />
⎝<br />
⎞<br />
B 0 0 0 0 0 0<br />
B 1 B 0 0 0 0 0<br />
B 2 B 1 B 0 0 0 0<br />
B 3 B 2 B 1 0 0 0<br />
0 B 3 B 2 0 0 0<br />
0 0 B 3 0 0 0<br />
C 1,0 0 0 B 0 0 0<br />
C 2,0 0 0 0 0 0<br />
C 1,1 C 1,0 0 B 1 B 0 0<br />
C 3,0 0 0 0 0 0<br />
C 2,1 C 2,0 0 0 0 0<br />
C 1,2 C 1,1 C 1,0 B 2 B 1 B 0<br />
0 C 3,0 0 0 0 0<br />
0 C 2,1 C 2,0 0 0 0<br />
0 C 1,2 C 1,1 B 3 B 2 B 1<br />
0 0 C 3,0 0 0 0<br />
0 0 C 2,1 0 0 0<br />
0 0 C 1,2 0 B 3 B 2<br />
0 0 0 0 0 B 3<br />
0 0 0 C 1,0 0 0<br />
0 0 0 C 2,0 0 0<br />
0 0 0 C 1,1 C 1,0 0<br />
0 0 0 C 3,0 0 0<br />
0 0 0 C 2,1 C 2,0 0<br />
0 0 0 C 1,2 C 1,1 C 1,0<br />
0 0 0 0 C 3,0 0<br />
0 0 0 0 C 2,1 C 2,0<br />
0 0 0 0 C 1,2 C 1,1<br />
0 0 0 0 0 C 3,0<br />
⎟<br />
0 0 0 0 0 C 2,1 ⎠<br />
0 0 0 0 0 C 1,2<br />
. (11.73)<br />
If the rank of this matrix M 1,2 is smaller than 6n, one can compute solutions<br />
z(ρ 1 ,ρ 2 )ofA(ρ 1 ,ρ 2 ) T z(ρ 1 ,ρ 2 ) = 0, by computing non-zero vec<strong>to</strong>rs in the kernel of<br />
this matrix. From the vec<strong>to</strong>rs z 0,0 ,z 0,1 ,z 0,2 ,z 1,0 ,z 1,1 ,z 1,2 , which are parts of a vec<strong>to</strong>r<br />
in the kernel of M 1,2 , one can construct the solution z as z(ρ 1 ,ρ 2 )=(z 0,0 + ρ 2 z 0,1 +<br />
ρ 2 2 z 0,2 ) − ρ 1 (z 1,0 + ρ 2 z 1,1 + ρ 2 2 z 1,2 ).<br />
The vec<strong>to</strong>rs z(ρ 1 ,ρ 2 ) are used <strong>to</strong> split off singular parts of the singular matrix pencil<br />
A(ρ 1 ,ρ 2 ) T = B(ρ 2 )+ρ 1 C(ρ 1 ,ρ 2 ): first construct transformation ( matrices V (ρ 2 )<br />
and W (ρ 2 ), second, perform the transformation W (ρ 2 ) −1 B(ρ 2 )+ρ 1 C(ρ 1 ,ρ 2<br />
)V (ρ 2 )<br />
and finally split off one or more singular parts of dimensions (η 1 +1)× η 1 with the<br />
same structure as in (11.14).