Page 474 - Mathematical Techniques of Fractional Order Systems
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460  Mathematical Techniques of Fractional Order Systems


               This function results in control functions given by Eq. (15.21) to give
            multiswitching synchronization with switch 2.


            15.4.3 Case 3
            For Switch 3 defined by Eq. (15.13), the time derivative of the errors is
            given by
                            q
                           d e 31
                                5 y 2 1 u 1 ðtÞ
                            dt q
                                5 e 32 1 x 1 2 x 3 1 u 1 ðtÞ
                            q
                           d e 32      3
                                5 y 1 2 y 2 αy 2 1 fcosωt 1 u 2 2 x 2
                                       1
                            dt q
                                5 e 31 2 αe 32 1 f 31 ðx; yÞ 1 u 2
            where f 13 are nonliear terms in e 11 and e 12 given as
                                         3
                                 f 13 52 y 2 αx 1 1 fcosωt
                                         1
            Theorem 3: If the control function u 1 ðtÞ and u 2 ðtÞ are chosen such that

                             u 1 ðtÞ 52 x 1 1 x 3 1 k 1 e 31 1 e 32
                                                                     ð15:25Þ
                             u 2 ðtÞ 52 f 13 ðx; yÞ 1 e 31 1 ðk 2 2 αÞe 32
            then the drive system 15.9 will achieve multiswitching synchronization with
            the response system 15.10
            Proof: The controller is defined as

                                  u 1 ðtÞ 5 V 31 ðtÞ 2 x 1 1 x 3
                                  u 2 ðtÞ 52 f 13 ðx; yÞ 1 V 32
            where V 31 ðtÞ and V 32 ðtÞ are virtual control functions to be determined.
               Following the same procedure as in Switch 1, we have

                                     V 31 ðtÞ    e 31
                                           5 C
                                     V 32 ðtÞ    e 32
               Matrix C is chosen as

                                        k 1    1
                                   C 5                               ð15:26Þ
                                         1  ðk 2 2 αÞ
               This yields
                                     V 31 5 k 1 e 31 1 e 32          ð15:27Þ

                                  V 32 5 e 31 1 ðk 2 2 αÞe 32        ð15:28Þ
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