Question

= θ = turret angle ω-turret angular rate x2 = angular acceleration produced by hydraulic drive x3 x-q = hydraulic servo valve displacement K, u-control input to servo valve Ku = servo motor gan Jturret inertia Ω.-motor natural frequency The quantities d, d, and d represent disturbances, including effects of nonlinearities not accounted for by the K-differential pressure feedback coefficient lincarized model K servo valve gairn After simplification by combining the feedback loops around the integrators, the equivalent block diagram, with the disturbances omitted 1/s2 And the transfer function from the system input u to the angle θ is ls times p(su(s). Thus K,K./J K, = 94.3 Lu-1.0J-7900 Kmー8.46 × 10" 100 980 45.9 and Ksp 6.330H(s)- ss140.2s210 449s 100980) we find numerically that H(s) 10 s(s3 140s210449s + 105) Problem 6.2 Hydraulically actuated gun turret It is desired to increase the bandwidth of the hydraulically actuated gun turret of Example 4E by use of state-variable feedback. The dominant poles, le, those closest to the orign, are to be moved to s--10VX1 ± 1). The other poles (at s64.5+69.6) are already in suitable locations, but they can be moved in the interest of simplifying the feedback law by eliminating feedback paths. (a) Determine the regulator gains for which the closed-loop poles are at s-102(1 ± 1) and at s-64.569.6 (b) For simplicity, only two nonzero regulator gains are pemitted: the gain from and one other gain, either from x2 = ω orfrom x) =p. Is it possible with a gain matrix of the form g. = [81,82, 0,0] g.-Lei, 0, g, 0] or to place the dominant poles ats-10v2(1 l) and still keep the fast" poles at their approximate locations? If both g, and g, can achieve this requirement, which is the better choice? Explain

Answer 1

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