TAILIEUCHUNG - Time Delay Systems Part 4

Tham khảo tài liệu 'time delay systems part 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Stability of Linear Continuous Singular and Discrete Descriptor Time Delayed Systems 49 Now we proceed to develop delay independent criteria for finite time stability of system under consideration not to be necessarily asymptotic stable . so we reduce previous demand that basic system matrix A0 should be discrete stable matrix. Theorem Suppose the matrix I - AT1 A 1 0 . System given by 69 is finite time stable with respect to such that k0 KN aa P ll til2 Ị. a p if there exist a positive real number p p 1 II x k -1 12 p2 x k 2 Vk eKN Vx k eSp 104 and if the following condition is satisfied Nestorovic et al. 2011 Lx p Vk N 105 where 2max 2max aT0 i - AT1A1 A 0 p2l . Proof. Now we consider again system given by 69 . Define V x k xT k x k xT k -1 x k -1 as a tentative Lyapunov-like function for the system given by 69 . Then the AV x k along the trajectory is obtained as AV x k V x k 1 - V x k xT k 1 x k 1 - xT k -1 x k -1 xT k At0 a 0 x k 2 xT k AT0 A 1 x k -1 xT k -1 AT1 A 1 x k -1 - xT k -1 x k -1 From 108 one can get xT k 1 x k 1 xT k A a 0 x k 2xT k At0 a 1 x k -1 xT k -1 AT1 A 1 x k -1 Using the very well known inequality with choice r I - AT1A 1 0 I being the identity matrix it can be obtained xT k 1 x k 1 xT k AT0a 0 x k xT k A 1 I - AT1A 1 a x k xT k -1 x k -1 and using assumption 104 it is clear that 111 reduces to 106 107 108 109 110 111 50 Time-Delay Systems xT k 1 x k 1 xT k AT ự I - A 1 A p2I J A 0 x k Amax A 0 A1 p x k x k 112 where Amax A0 A1 p Amax I A I - A1 A -1 A 0 p21 113 with obvious property that gives the natural sense to this problem Amax A 0 A 1 p 0 when I - A 1 A 0. Following the procedure from the previous section it can be written lnxT k 1 x k 1 - lnxT k x k lnAmax . 114 k0 k-1 By applying the sum on both sides of 112 for vk e KN one can obtain j k0 k k-1 lnxT k0 k x k0 k ln n Amax lnAmax lnx k0 x k0 vk e KN 115 j k0 Taking into account the fact that x0 2 a and condition of Theorem 105 one can get lnxT k0 k x k0 k 111

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