Adaptive Failure Compensation of Actuators in Controlling Servo System Driven by Twin Motors

Adaptive Dynamic Surface Control for Servo System Driven by Twin Motors With Unknown Actuator Failures

“…The structure of the servo system is shown in figure 1. The mathematical model and states selection are the same as [18]. Let…”

“…To improve the performance of control system, we should consider the influence of unknown actuator failures in the design of adaptive control scheme. Like [14] and [18], due to the loss rate of actuator effectiveness is unknown, the adaptive control method can be effectively used for actuator failures. In view The associate editor coordinating the review of this article and approving it for publication was Ding Zhai.…”

“…Compared with [18], in order to overcome the computational complexity problem in backstepping design, we introduce the dynamic surface technology on the basis of traditional backstepping motivated by previous research [19], [20]. In this paper, an adaptive dynamic surface controller for the servo system driven by twin motors with unknown actuator failures has been proposed.…”

“…The main contributions of this paper are provided as follows: (1) The control problem is investigated for servo system with unknown system parameters and unknown disturbances driven by twin motors instead of traditional single motor; (2) The effect caused by unknown actuator failures are considered during the controller design; (3) Based on [18], Dynamic Surface Control (DSC) is introduced to solve ''explosion of complexity'' problem in the traditional backstepping control scheme.…”

Improved Adaptive Hybrid Compensation for Compound Faults of Non-Gaussian Stochastic Systems

“…l ) − 11.5, Z 5 = −(ρ l + c l )2 + 9(ρ l + c l ) − 19.5,Z 6 = −(ρ l + c l ) 2 + 11(ρ l + c l )−29.5, Z 7 = 0.5(ρ l +c l −5) 2 , Z 8 = 0.5(ρ l +c l −6) 2 , Z 9 = 0.5(ρ l + c l − 7)2 , c l satisfies: c l = 0.05, and I O satisfies:I O = 1 ρ l ∈ [O + 1, O + 2] 0 otherwise.where O = 1, 2, 3, 4, 5. The ideal initial values are: x initial = [0.10.10.1] T .…”