Systems, Structure and Control 2012 Part 2

Tham khảo tài liệu 'systems, structure and control 2012 part 2', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Control Designs for Linear Systems Using State-Derivative Feedback 13 When 32 and 33 are feasible they can be easily solved using available softwares such as LMISol de Oliveira et al 1997 that is a free software or MATLAB Gahinet et al 1995 Sturm 1999 . These algorithms have polynomial time convergence. Remark 4. From the analysis presented in the proof of Theorem 2 after equation 36 note that when 32 and 33 are feasible the matrix A a defined in 25 has a full rank. Therefore A a with a full rank is a necessary condition for the application of Theorem 2. Moreover from 25 observe that for ai 1 and ak 0 i k i k 1 2 . ra then A a Ai. So if A a has a full rank then Aị i 1 2 . ra has a full rank too. Usually only the stability of a control system is insufficient to obtain a suitable performance. In the design of control systems the specification of the decay rate can also be very useful. Decay Rate Conditions Consider for instance the controlled system 31 . According to Boyd et al. 1994 the decay rate is defined as the largest real constant Y Y 0 such that x t 0 holds for all trajectories x t t 0 . One can use the Lyapunov conditions 29 to impose a lower bound on the decay rate replacing 29 by P 0 and An a p P PAN a fi -2yP . 38 where Y is a real constant Boyd et al. 1994 . Sufficient conditions for stability with decay rate for Problem 1 are presented in the next theorem Assunctio et al. 2007c . Theorem 3. The closed-loop system 31 given in Problem 1 has a decay rate greater or equal to Y if there exist a symmetric matrix Q e A nxn and a matrix Y e Rmxn such that Q 0 Q AịQ BjYAj AịY B Q YBj Q BjY -Q 2y 39 40 0 where i 1 . ra and j 1 . rb. Furthermore when 39 and 40 hold then a robust state-derivative feedback matrix is given by K YQ-1 . 41 Proof Following the same ideas of the proof of Theorem 2 multiply both sides of 40 by OCịPj for i 1 . ra and j 1 . rb and consider 26 to conclude that QA a A a Q B p YA a A a Y B p Q B p Y Q Y B P -Q 2y Now using the .

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