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Page 6 of 14                    Ma et al. Complex Eng Syst 2023;3:10  I http://dx.doi.org/10.20517/ces.2023.14


               Case 2 In the converse case    2 ≤ 1, one has


                                             ¤                 1                                       (14)
                                                2 ≤ −2   12 |   1 ||   1 | erf(|   1 |) + 2   11    1
                                                                                   2     −   0
                        2                                                           0    1 =    2   . Considering the
               As 1 +    0    ≥ 1 and |   1 | ≤ 1, it can be obtained that min |   1 |     1  ≥ min |   1 |
                        1
               Lemma 3, then (14) is converted into the following form
                                            ¤                 1
                                              2 ≤ −2   12 |   1 ||   1 | tanh(|   1 |) + 2   11    1
                                                            1         1
                                              ≤ 2   12 |   1 ||   1 |  + 2   12 |   1 |    2 + 2   11    1
                                                              1
                                              ≤ −2   12 |   1 ||   1 |  + 2   12    2 + 2   11    1
                                                           1 +1
                                              ≤ −2   12 |   1 |  + 2   11    1 + 2   12    2
                                                                                                       (15)
                                                      −   0
                                              ≤ −2   12    2   |   1 | + 2   11    1 +    12    2
                                                     1
                                                          
                                                     2
                                              ≤ −   1    + ˜
                                                    2
                                                       1             1
                                                                     2
                                                                          
                                                       2
                                              ≤ −   13    1    − (1 −    13 )   1    + ˜
                                                       2            2
                                                                                   1
                            −   0
               with    1 = 2   12    2   , and ˜ = 2   11    1 +2   12    2. When 0 <    13 < 1, and ˜−(1−   13 )   1    ≥ 0, (15) can be simplified
                                                                        
                                     
                                                                                   2
                                                                                  2
                             1
                  ¤
                             2
               as    2 ≤ −   13    1    . Then, the solution of    2 will reach a small set Δ 1, which is defined as Δ 1 =    1 |   1 (   1 ) ≤
                             2
                   ˜     2 	                      2
               (       ) within a settling time       2 ≤  .
                   1 (1−   13 )                     1    13

               In view of the above two cases, the auxiliary variable    1 will converge into a small set Δ 1 =     1 |   1 (   1 ) ≤
                   ˜     2
               (       ) within settling time       =       1 +       2.
                   1 (1−   13 )
               Then, the disturbance observation error
                                            ˜       ˆ
                                              1 =    1 −    1
                                                                                                       (16)

                                                                        1
                                             =    1 −       11 erf(   1 ) +    12 |   1 | erf(   1 )
               The disturbance    1 is bounded according to Assumption 1. Thus, the disturbance observer (11) can estimate
                                                                               n
                                                  ˜

                  1 accurately, and the observation error    1 can remain in a small set Δ 2 =    1 |   1 | ≤    1   +       11 erf(Δ 1 ) +
                              o
                                                            0 Δ 2
                  12 |Δ 1 | erf(Δ 1 ) after a fixed time, where ¯   1 =  1  .
                     ¯    1
                                                        1+   0 Δ 2
                                                            1
               3.1.2. Fixed-time sliding mode controller
               For the subsystem (7), define       =    −      . A fixed-time integral sliding mode surface is introduced as
               follows [22]
                                           ¹    

                                                                                    2
                                                         1       1           2  + d      c ) d         (17)
                                      1 =       +     11 (d      c  + d      c ) +    12 (d      c
                                            0
               with 0 <       < 1,       > 1, and (   = 1, 2). For any    ∈ R,    ∈ R , the notation is defined as d  c = |  | sign(  ).
                                                                                               
                                                                                                     
                                                                   +
               Based on the sliding mode surface as (17), the fixed-time controller is designed as follows:

                                                                                                   ˆ
                                    1      1           2       2         3        3  +    3 erf(   1 ) − ¤      −    1 (18)
                     1 = −      11 (d      c  + d      c ) +    12 (d      c  + d      c ) +    1 d   1 c  +    2 d   1 c
                                                                         1   . In addition,       ,       , (   = 1, 2, 3) are all
               where       ,    1   , (   = 1, 2) are positive constants,    3 satisfies    3 ≥    
               positive odd integers with 0 <       < 1,       > 1.
               Theorem 2 For the second-order system (7), if the fixed-time controller is constructed in the form of (18), then
               the real sliding mode variable will converge into a small set within a fixed time.
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