Page 156 - Modern Control of DC-Based Power Systems
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120                                 Modern Control of DC-Based Power Systems



                                     _ V              _ V     V
                         C eq L fi           P eq
                   d i 52       2        1       _ V 2  2
                          S i E i  R L C eq  C eq v 2  T f  C eq T f R L
                            P eq      V               2
                                                 _
                       2         2        2 2ξω 0 V 2 ω VÞ 1 w i
                          C eq T f V  C eq L eq       0
                                                            !
                           C eq L fi      1         1      2
                      52            2          2       2 ω    V
                                                           0
                            S i E i   C eq T f R L  C eq L eq
                                       !
                       1      1               P eq       P eq
                                         _
              1   2        2    2 2ξω 0 V 1        _ V 2      Þ 1 w i  (5.32)
                    R L C eq  T f            C eq v 2  C eq T f V
             Since the system described in [3] does not include a resistive load, the
          calculations have to be repeated including this load which yields the line-
          arized system (5.33). The detailed derivation of Eq. (5.33) is depicted in
          the Appendix.
                                                    !
                               €
                              V1      1   1  1  1 K 2 V _
                                   R L C eq  T f

                                                      !
                                   1         1
                            1           1        1 K 1 V
                                C eq T f R L  C eq L eq

                             w 1 E 1  w 2 E 2  w 3 E 3
                          5        1        1        1 K 1 V 0
                            C eq L f 1  C eq L f 2  C eq L f 3
                                           1         1
                                    2
                             K 1 5 ω 2          2
                                    0
                                       C eq T f R L  C eq L eq
                                            1      1
                               K 2 5 2ξω 0 2  2                       (5.33)
                                           T f   R L C eq
             Inserting the values of K 1 and K 2 , it can be seen that the systems are
          equal. Summarizing, the LSF can be split up into a linearizing part f l and
          a compensating part f c of (5.34).
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