Page 267 - Mathematical Models and Algorithms for Power System Optimization
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Optimization Method for Load Frequency Feed Forward Control 259

               its corresponding difference equation is:

                                                 ð
                                                          ðÞXk ðÞ
                                               Xk +1Þ ¼ ϕ τ 1
               where τ6¼τ 1 , that is, with different sampling intervals.
               The following two cases are to be considered, respectively:

               (1) τ 1 >τ, τ 1 is the integral multiple of τ, that is, τ 1 ¼nτ
                    From the definition of transition matrix, we know:
                                                           n
                                                    ðÞ
                                                  ϕ τ 1 ¼ ϕ τðÞ
                    ϕ(τ 1 ) can be solved using the above equation.

               (2) τ 1 <τ,  τ¼nτ 1
                                                                    Aτ
                    Take the first two terms of polynomial expansion of e
                                                             2 2
                                             e Aτ  ¼ I + Aτ +  1 A τ + ⋯                     (7.96)
                                                          2!
               to get the approximate equation of matrix ϕ(τ)
                                                   ϕτðÞ   I + Aτ                             (7.97)

               Hence,
                                                      1
                                                  A   ½ ϕτðÞ IŠ                              (7.98)
                                                      τ
               Because τ 1 ¼τ/n, correspond to Eq. (7.98), the approximate expression of A and the transition
               matrix ϕ τ 1 of τ 1 is:
                       ðÞ
                                                     1
                                                         ðÞ
                                                 A     ½ ϕ τ 1  IŠ                           (7.99)
                                                     τ 1
               Combining Eqs. (7.98) and (7.99), we know:
                                                      τ 1
                                                ðÞ
                                              ϕ τ 1 ¼   ½ ϕ τðÞ IŠ + I                     (7.100)
                                                      τ
               Apparently, ϕ(τ 1 ) obtained from Eq. (7.100) is very rough and needs to be further refined
               generally by iteration. Let the k-th iterative values ϕ(τ 1 ) and ϕ(τ)be ϕ k (τ 1 ) and ϕ k (τ). Let ϕ(τ 1 )
               solved from Eq. (7.100) be ϕ 1 (τ 1 ). Judging from the properties of transition matrix, we have
                                                   n
                                                    ðÞ
                                                  ϕ τ 1 ¼ ϕ τ ðÞ                           (7.101)
                                                   k        k
               In the k-th iteration, if the error lies between ϕ k (τ) and ϕ(τ), where the former is solved by
               Eq. (7.101) and the latter is the given one, is not satisfactory.
               ϕ k+1 (τ 1 ) can be rectified using the following:
                                         ϕ                                                 (7.102)
                                              τ 1 ¼ ϕ τ 1 + α ϕ τðÞ ϕ τðފ
                                          k +1  ðÞ   k  ðÞ  ½       k
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