10 A Single Step Modal Parameter Estimation Algorithm: Computing. . . 99 obtained from the Nyquist criterion. Employing orthogonal polynomials [8] for frequency tranformation results is another approach to improve conditioning, but is computationally more expensive. Once the least squares solution for the rational fraction matrix-coefficients (both [αi] and [ ˆβi]) is computed, the residue and the numerator matrix-coefficients may be separated using a simple, fully determined de-convolution solution as shown in [12]. Hence for each model order iteration, the complete set of [α], [β], [ ˆβ] and[R] coefficients is available i.e. Equation 10.4 is fully defined. [H(s)]=[α(s)]−1[ β(s)]+[R(s)]=[α(s)]−1[ ˆβ(s)] (10.4) Where, [α(s)]= m i=0 [αi]s i No×No [β(s)] = m−1 i=0 [βi]s i No×Ni [R(s)]= nu i=−nl [Ri]s i No×Ni [ ˆβ(s)]= m−1 i=0 [I][βi]s i No×Ni + m i=0 [αi]s i No×No nu i=−nl [Ri]s i No×Ni Normally, the numerator matrix-coefficients remain unused and are discarded [8] or they are expressed in the form of denominator matrix-coefficients [7] to improve the speed of the least-squares solution. The denominator coefficients are used to construct a companion matrix, and its eigen-values and eigen-vectors (state-vectors of the order of the polynomial) are calculated. These poles and vectors are then used to build the stabilization chart. However, instead of utilizing anything but the denominator matrix-coefficients, all the available information is used to compute the residues. The algorithm from Vu [13, 14] allows the inverse of a square matrix-coefficient polynomial to be described as a ratio of its adjoint (matrix-coefficient) polynomial and the characteristic (monic) equation (Equation 10.5). [α]−1 No×No = (No−1)m i=0 [ α+i ]No×No si Nom i=0 dis i = [ α+] d (10.5) The algorithm requires recursive computation of a new set of matrices as defined in Equation 10.6 and coefficients as defined in Equation 10.7 for the nth model order. [Bd,c]= c v=1 d w=1 [Bd−w,c−v][αw][α0] v−1 (10.6) Where, c =1, 2, . . . ,No; d =0, 1, . . . ,No ∗n; and, [B0,c]=[α0] c bj,i = 1 j j c=1 i d=0 (−1)c−1 bj−c,i−d ∗trace([Bd,c]) (10.7) Where,i =0, 1, 2, . . . ,No ∗n
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