By Martin W. Bowman

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66) which is unstable. 71%. 1. 5 2 Fig. 1. 2 Stable Model Reduction The method described in the preceding subsection cannot guarantee the preservation of stability of the reduced-order models. This is a typical problem associated with frequency domain methods. On the other hand, time domain model reduction is easier to deal with stability preservation. One sees that the step response of a stable G(s) will remain finite while that of an unstable ˆ G(s) will tend to infinity. The squared error between them will diverge, and ˆ unstable G(s) will be excluded from the solutions to the problem of minimizing such an error, in other words, the solution must be stable.

P (s) is said to be stable if det(P (s)) has all roots in C− , the open left-half of the complex plane. , there are integers ν1 , ν2 , · · · , νm such that for some Sr (s) = diag{sν1 , sν2 , · · · , sνm }, the limit lims→∞ Sr−1 (s)U (s)P (s) = Phr exists and is nonsingular. Without loss of generality, Phr can be assumed to be an identity matrix. 20) where ν = max{ν1 , ν2 , · · · , νm }. Also, let Dr (s) = diag (sν−ν1 , sν−ν2 , · · · , sν−νm ). 2 Polynomial Matrices 23 m where ∆ν = (ν − νi ). Introduce the constant matrix A as follows: i=1 0 0 −P0 Im −P1 A= ..

Suppose full rank of [DT (s) N T (s)]T . 29) (ii) [DT (s) N T (s)]T has full rank for every s ∈ C. Proof. 4, it is always possible to write a GCRD ˆ ˆ R(s) of N (s) and D(s) as R(s) = X(s)N (s) + Yˆ (s)D(s) for polynomial X ˆ and Y . Moreover, if N (s) and D(s) are coprime, R(s) must be unimodular so that ˆ I = R−1 (s)[X(s)N (s) + Yˆ (s)D(s)] = X(s)N (s) + Y (s)D(s), ˆ and Y (s) := R−1 (s)Yˆ (s) are both polynomial. 29). e. ˆ (s)R(s), N (s) = N ˆ D(s) = D(s)R(s). 29) then becomes ˆ (s) + Y (s)D(s) ˆ I = X(s)N R(s), yielding ˆ (s) + Y (s)D(s), ˆ R−1 (s) = X(s)N a polynomial matrix.