heperformanceoffeedbackcontrolsystems反饋控制系統(tǒng)的性能(編輯修改稿)
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【文章內(nèi)容簡介】 ce indices can be represented by the ones of the second order system. In the case, the poles of the second order system are called dominant poles of the system. %1 0 0.. 21 ?? ?? ???eOPnsT ???44 ??Notice: the above results is only for a transfer function without finite zeros. Simulation results for ζ= 2222))(/()(nn ssasasTn????????When the transfer function has a zero, ? For the given system, select the gain K and the parameter p so that the timedomain specifications will be satisfied. ? .≤5% ? Ts ≤4s ζ=, .=% Ts=4/ζωn≤4, ζωn≥1 Chose r12=1177。j, then .=% Ts=4s ζ=, ωn=1/ζ= 222)(22 ?????? ssKpssKsTIf a ζωn and τ 1/ζωn, the pole and zero will have little effect on the step response. .=55% according to Fig (a) Ts=4/3= )1)(2())(/()()()(222sssasasTsRsYnnn???????????Using a puter simulation for the actual thirdorder system, we find that the percent overshoot is equal to 38% and the settling time is seconds. Thus, the effect of the third pole of T(s) is to dampen the overshoot and increase the settling time (hence the real pole cannot be neglected). ? According to the percent overshoot . ? According to the number of cycles of the damped sinusoid during Ts %1 0 0.. 21 ?? ?? ???eOP1( ) 1 sin ( )n tny t e t?? ? ? ???? ? ?The frequency of the damped sinusoidal term for ζ 1 is 21nn? ? ? ? ?? ? ?The number of cycles in 1 second is ω/2π. The time constant for the exponential decay is τ = l/ζωn in seconds. The number of cycles of the damped sinusoid during one time constant is ( c y c l e s / s e c ) 222 nnn?????? ? ? ? ? ? ? ?? ? ? ?Assuming that the response decays in n visible time constants. For the secondorder system, the response remains within 2% of the steadystate value after four time constants (4 τ, ., n = 4) 2414 0 . 5 5c y c l e s v is ib l e 0 . 2 0 . 622 fo r?? ?? ? ? ? ??? ? ? ? ?? Example: examine the response shown in Figure for ζ = . Use y(t) = 0 as the first minimum point and count cycles visible (until the response settles with 2% of the final value). Then we estimate 0 . 5 5 0 . 5 5 0 . 3 91 . 4c y c le s? ? ? ?)()()(1 1)( sEsRsGsE a???Ea(s), actuating signal, which is a measure of the system error. E(s)=R(s)Y(s), the actual system error. When H(s)=1, N is call the type of systems. N=0, type0 system。 N=1, type1 system。 N=2, type2 system。 For a type0 system Define as position error constant. For N≥1 Step input
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