Page 242 - Mathematical Models and Algorithms for Power System Optimization
P. 242

234 Chapter 7

            The corresponding autocorrelation function is:

                                     8
                                                  1               ð k ¼ 0Þ
                                     >
                                     >
                                        θ k + θ 1 θ k +1 + ⋯ + θ q k θ q
                                     <
                                 ρ ¼         1+ θ + ⋯ + θ 2     ð 0 < k   qÞ             (7.23)
                                  k
                                                 2
                                     >           1       q
                                     >
                                                  0               ð k > qÞ
                                     :
                                                          0
            Eq. (7.23) shows, when the steps between Z t and Z t are greater than q, they are no longer
            correlated to each other. Seen from ρ 0 , ρ 1 , …, ρ q , …, the origin part of such sequence is zero as
            from q, and this autocorrelation function is called truncation.
            (3) Partial autocorrelation function. The approximation of Z k is represented by a linear
                 combination of Z t 1 , …, Z t k and the coefficient ϕ kz is chosen so that:
                                                                ! 2
                                                        k
                                                       X
                                        minS ¼ EZ t       ϕ Z t j                        (7.24)
                                                           kj
                                                       j¼1
            To obtain ϕ kj , take partial derivative to be zero:
                                           ∂S
                                               ¼ 0 j ¼ 1, 2, …, kÞ                       (7.25)
                                                   ð
                                           ∂ϕ kj
            Expand S to get:
                                          k          k          k
                                        X    2      X         X
                                 S ¼ γ +    ϕ γ  2     ϕ γ +2     ϕ ϕ γ                  (7.26)
                                     0       kj 0       kj i       kj ki j i
                                         j¼1        j¼1        j>i
            Substitute it into Eq. (7.25) and then divide by γ 0 to get:
                                     2                32     3   2  3
                                       1   ρ 1  ⋯ ρ k 1   ϕ k1    ρ 1
                                       ρ   ⋮      ρ       ⋮        ⋮
                                     6  1             76     7   6  7
                                       ⋮   ⋮        ⋮     ⋮        ⋮
                                     6             k 2 76    7  ¼  6  7                  (7.27)
                                     4                54     5   4  5
                                       ρ ρ          ⋮     ϕ       ρ
                                        k  k 2             kk      k
            ϕ 11 , ϕ 22 ,…, ϕ kk is called partial autocorrelation function of Z k . The following shows how to
            solve ϕ kj , proceeding directly from the definition of S, substitute Eq. (7.13) into S to get:
                                  "                                         # 2
                                                                    k
                                                                   X
                             S ¼ E ϕ Z t 1 + ϕ Z t 2 + ⋯ + ϕ Z t p + a t
                                    1       2           p             ϕ Z t j
                                                                       kj
                                                                   j¼1
                                  "                                # 2
                                        p                 k
                                                         X

                                       X
                              ¼ Ea t +     ϕ  ϕ kj  Z t j    ϕ Z t j
                                            j
                                                              kj
                                       j¼1              j¼p +1                           (7.28)
                                      "                            # 2
                                        p                 k
                                                         X
                                  2

                                       X
                              ¼ σ + E      ϕ  ϕ Z t j        ϕ Z t j
                                                 kj
                                             j
                                  a
                                                              kj
                                       j¼1               j¼p +1
                                σ 2
                                  a
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