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Ma et al. Complex Eng Syst 2023;3:10  I http://dx.doi.org/10.20517/ces.2023.14    Page 7 of 14


                                                                     and refer to Lemma 2 to Lemma 4, the time
               Proof of Theorem 2 Choose a Lyapunov function as    3 =  1 2
                                                                 2 1
               derivative of    3 is

                               ¤
                                                           1
                                 3 =    1 ¤      +    11 (d      c     1  + d      c ) +    12 (d      c     2  + d      c    2

                                      1
                                                                       1
                                                                                          2
                                 =    1  (   1 +    1 ) − ¤      +    11 (d      c     1  + d      c ) +    12 (d      c     2  + d      c )
                                        

                                                                     1                                 (19)
                                                3         3            ˜
                                 =    1 −    1 d   1 c  −    2 d   1 c  −    3 erf(   1 ) +     1
                                                                       
                                                                      1
                                            3 +1       3 +1                 ˜
                                 ≤ −   1 |   1 |  −    2 |   1 |  −    3    1 tanh(   1 ) +  |   1 ||   1 |
                                                                        
                                 ≤ −2    1     ¯    3  − 2    2     ¯    3  ¯
                                      ¯    3
                                                ¯    3
                                          2         2  +    1
                                        ¯
                                
                       
               where ¯ 3 =     3 +1 , ¯ 3 =     3 +1 ,    1 =    3    1 with    1 being a positive constant. By using Lemma 4, the second-
                           2         2
               order system (7) is fixed-time stable. The sliding mode surface    1 will converge into a small region Δ 3 =
               n                                o
                                      1       1
                   1 |  (   1 ) ≤ min     2  ¯    3 ,     2  ¯    3  around the origin in a fixed time       1  , which is determined by       1  ≤
                                           ¯   
                                  ¯   
                                  1 2 3     2 2 3
                    1            1    . Then, one can obtain that variables       and       converge to zero along the real
                  ¯       +    ¯   
                         
                                    
                  1 2 3    1 (1− ¯ 3 )     2 2 3    1 ( ¯ 3 −1)
               sliding mode in a fixed time [23] .
               3.2. Tracking control laws design for the third-order subsystem
               After the angular error       converges to zero according to Theorem 2, one can obtain that sin       equals zero,
               and cos       equals 1. The system (8) can be simplified as
                                                    ¤      =             −    +      
                                                   
                                                   
                                                   
                                                      ¤       = −                                      (20)
                                                   
                                                   
                                                   
                                                       ¤   =    2 +    2
                                                   
               3.2.1. Fixed-time disturbance observer
               Introduce the following auxiliary variable for the simplified third-order subsystem (20)
                                                                                                       (21)
                                                           2 =    −    2
               where    2 satisfies
                                                 1
                                            ¤    2 =     2 +    21 erf(   2 ) +    22 |   2 | erf(   2 )  (22)
                                                                         2
                                                  
                              3     2
               where    2 =  1  , and    3 and    3 are integers satisfying the constraints: 0 <    3 < 1, 1 +    3 <    3. The
                          1+   3    2 1
               parameters    21 and    22 are positive constants with    21 >    2   and    22 > 0.
               Theorem 3 For the simplified third-order subsystem (20), a fixed-time disturbance observer is developed in the
               form of

                                             ˆ                         2                               (23)
                                                2 =       21 erf(   2 ) +    22 |   2 | erf(   2 )
                                                                              ˆ
                                                                     ˜
               then it can estimate    2 in a fixed time, and the observation error    2 =    2 −    2 can converge into a small region
                                                .
               around the origin within a fixed time       2
               Proof of Theorem 3 Similar to the proof of Theorem 1.
               3.2.2. Fixed-time sliding mode controller
               For the third-order subsystem (20), introduce the following auxiliary variable:
                                    ¹
                                         
                                                                                           
                                                                                   3
                                =       +     1 erf(      ) −    2 erf(      ) +    3             erf(      ) +    1 |      | erf  d    (24)
                                      0                                                  3
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