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Multiswitching Synchronization Chapter | 15  459


             15.4.2 Case 2
             For Switch 2 defined by Eq. (15.12), the time derivative of the errors is
             given by
                  q
                 d e 21
                     5 y 2 2 x 2 1 u 1 ðtÞ
                  dt q
                     5 e 22 1 x 3 2 x 2 1 u 1 ðtÞ
                  q
                 d e 22     3                                          3
                     5 y 1 2 y 2 αy 2 1 fcosωt 1 u 2 ðtÞ 2 β x 1 1 β x 2 1 β x 3 2 β x
                                                          2
                                                                3
                                                                      4 1
                            1
                                                    1
                  dt  q
                     5 e 21 2 αe 22 2 αx 3 1 x 1 1 f 21 ðx; yÞ 1 u 2
             where f 21 are nonliear terms in e 11 and e 12 given as
                                3
                         f 21 52 y 1 fcosωt 2 β x 1 1 β x 2 1 β x 3 2 β x 3
                                1           1     2     3     4 1
             Theorem 2: If the control function u 1 ðtÞ and u 2 ðtÞ are chosen such that
                          u 1 ðtÞ 52 x 3 1 x 2 1 k 1 e 21 1 e 22
                                                                      ð15:21Þ
                          u 2 ðtÞ 5 αx 3 2 x 1 2 f 21 ðx; yÞ 1 e 21 1 ðk 2 2 αe 22 Þ
             then the drive system 15.9 will achieve multiswitching synchronization with
             the response system 15.10

             Proof: The method of active control is employed to prove theorem 2.
                We redefine the controller in order to eliminate nonlinear terms in e 21
             and e 22 , so,
                              u 1 ðtÞ 52 x 3 1 x 2 1 V 21 ðtÞ
                              u 2 ðtÞ 5 αx 3 2 x 1 2 f 21 ðx; yÞ 1 V 22 ðtÞ
             where V 21 ðtÞ and V 22 ðtÞ are virtual control functions to be determined.
                Following the same procedure as in Switch 1, we have

                                     V 11 ðtÞ    e 11
                                            5 A
                                     V 12 ðtÞ    e 12
                Matrix B is chosen as

                                         k 1    1
                                    B 5                               ð15:22Þ
                                         1   ðk 2 2 αÞ
                This yields
                                      V 21 5 k 1 e 21 1 e 22          ð15:23Þ

                                   V 22 5 e 21 1 ðk 2 2 αÞe 22        ð15:24Þ
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