Page 265 - Mathematical Models and Algorithms for Power System Optimization
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Optimization Method for Load Frequency Feed Forward Control 257

               Definition 1 Let
                                                    ϕ τðÞ ¼ e Aτ
                                                                                             (7.80)


               By contrast with the logarithmic definition of real field, the logarithm matrix is defined as:
                                                   lnϕ τðÞ ¼ Aτ                              (7.81)


               Definition 2 Define the series expansion of logarithmic matrix as:

                                                            h                    i 3
                                                        1  1                    1
                           lnϕ τðÞ ¼ 2 ½ ϕ τðÞ IŠ ϕ τðÞ + IŠ  +  ð ϕ τðÞ IÞ ϕ τðÞ + IÞ  + ⋯
                                               ½
                                                                      ð
                                                           3
                                                                                             (7.82)

                                       1   h                   1  i 2k 1
                                   +        ð ϕ τðÞ IÞ ϕ τðÞ + IÞ   + ⋯    ð k ¼ 1, 2, ⋯Þ
                                                     ð
                                     2k  1
               where the real part of eigenvalue of matrix ϕ(τ) is greater than 0.
               Theorem 3 If the real parts of all eigenvalues of matrix ϕ(τ) are positive, according to
               Eqs. (7.81) and (7.82), then matrix A can be expressed as:

                                 2                   1   1  h                1  i 3
                                                                    ð
                                            ½
                             A ¼    ½ ϕ τðÞ IŠ ϕ τðÞ + IŠ  +  ð ϕ τðÞ IÞ ϕ τðÞ + IÞ  + ⋯
                                 τ                       3
                                                                                             (7.83)

                                     1  h                   1 i 2k 1
                                 +       ð ϕ τðÞ IÞ ϕ τðÞ + IÞ    + ⋯   ð k ¼ 1, 2, ⋯Þ
                                                  ð
                                  2k  1
               Proof Using the proof by contradiction, that is, assuming Eq. (7.83) is satisfied, then the real
               parts of all eigenvalues of ϕ(τ) are positive.
               Let
                                                                   1
                                                          ½
                                                 ½
                                             B ¼ ϕ τðÞ IŠ ϕ τðÞ + IŠ                         (7.84)
               Judging from Eq. (7.81), the necessary and sufficient condition for matrix A to exist is:
                                                         k
                                                    lim B ¼ 0                                (7.85)
                                                    k!∞
                                                                                     0
               It means that the modulo of eigenvalue of matrix B must be less than 1. Let λ represent the
               eigenvalues of matrix B, then:
                                                       0
                                                      λ jj < 1                               (7.86)
               Based on Eq. (7.84), the characteristic equation of matrix B is
                                                                 1

                                                ½
                                                        ½
                                           λ I   ϕ τ ðÞ IŠ ϕ τ ðÞ + IŠ     ¼ 0               (7.87)
                                            0
               or

                                         0            0               1
                                                            ½
                                       ð j  λ +1ÞI   1 λð  Þϕ τðÞj ϕ τðÞ + IŠ     ¼ 0        (7.88)
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