Page 92 - Adaptive Identification and Control of Uncertain Systems with Nonsmooth Dynamics
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84   Adaptive Identification and Control of Uncertain Systems with Non-smooth Dynamics

                                               √
                                         T

                        where y =[z 1 ,z 2 ,r , Q, P],and U 1 and U 2 are defined as
                                                       2             2
                                           U 1 (y) = λ 1 	y	 ,U 2 (y) = λ 2 	y	     (5.36)
                        with λ 1 and λ 2 being positive constants.
                           Thetimederivativeof V L can be expressed as

                                              ˙
                         ˙ V L = z 1 ˙z 1 + z 2 ˙z 2 + r˙r + Q + ˙ P
                            = z 1 (z 2 − k 1z 1 ) + z 2 (r − k 2z 2 ) + r[ ˜ N + N − (k s + 1)r − β 1sgn(z 2 ) − z 2 ]
                                ˜  T  −1 ˙ ˜                                  2
                              +       − r[N B + N d − β 1sgn(z 2 )]− Hsgn(z 2 ) − β 2z − z 2N B
                                                                              2
                                  2     2                2     2
                            =−k 1z − k 2z + z 1z 2 − (k s + 1)r − β 2z + r ˜ N − z 2N B
                                  1
                                        2
                                                               2
                                 T
                                    −1 ˙
                                ˜
                              +       − Hsgn(z 2 )                                  (5.37)
                                      ˜
                                                               −1 ˙
                                                            ˜ T
                                                                  ˜
                        Substituting the adaptive law (5.21)into      , then one may verify that
                                                       T
                                             T
                                               −1 ˙
                                                                   T
                                                 =    (z)z 2 − σ                    (5.38)
                                                                     ˆ
                                                  ˜
                                                      ˜
                                           ˜
                                                                  ˜
                                                                ˜ T
                                               ˜ T
                        Then, we consider N B =    (Z), and have    (z)z 2 + z 2N B = 0. Con-
                        sequently, it follows
                                                                       2
                                                2
                                          2
                                                                2
                                ˙ V L =− k 1z − k 2z + z 1z 2 − (k s + 1)r + β 2z + r ˜ N
                                          1     2                      2            (5.39)
                                                  T
                                          T
                                     − σ    + σ    − Hsgn(z 2 )
                                         ˜
                                           ˜
                                                 ˜
                           By using Young’s inequality, one has:
                                                       1  2  1  2
                                                 z 1z 2 ≤ z + z 2                   (5.40)
                                                          1
                                                       2     2
                                                                     2       2
                                        T       T               	 	     σ
                                   −σ  ˜       ˜           ˜          +             (5.41)
                                      ˜
                                           + σ    ≤−σ 	 	−
                                                                 2         4
                           Substituting (5.40)and (5.41)into(5.39)results in
                                    1   2       1      2          2          	 	   2
                                                                         ˜
                         ˙ V L ≤−(k 1 − )z − (k 2 −  − β 2 )z − (k s + 1)r − σ 	 	−  + r ˜ N
                                                       2
                                        1
                                    2           2                              2
                                                      2


                                    2                       2
                                            ˜
                            ≤−λ 3 	z	 − σ 	 	−         +[k sr − η(	z	)	r		z	]       (5.42)
                                                  2
                                                 1
                                           1
                        where λ 3 = min{k 1 − ,k 2 − − β 2 ,1}.
                                           2     2
                           After completing the squares for the third term in (5.42), it follows
                                             2      2
                                            η (z)	z	                2
                                         2
                                                           ˜
                              ˙ V L ≤−λ 3 	z	 +      − σ 	 	−         ≤−U(y)        (5.43)
                                               4k s              2
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