电力拖动自动控制系统习题答案_第四版

运动控制系统习题答案第二章2-1答:在制动阶段,VT1始终不导通。VT2导通期间,并能耗制动;VT2不导通期间,VD1续流,并回馈制动。2-2100010100nNs10000.02nclnN2.04rpmD(1s100.98D2-3已知n0maxnmaxnN1500rpmn0minnminnN150rpmnNnmaxnN15rpm所以Dnmax15001511nmin15015nND1115s0.1nNnND148511152-4Ce(UNINRa/nN0.1478vmin/rnopIN(RaRs/Ce3780.045/0.1478115.1rpmnNs1430*0.23.1nN(1s115.1*0.8nNs1430*0.3D5.32nN(1s115.1*0.7D2-5nopINR/Ce3050.18/0.2274.5rpmsNnop/(nNnop27.45%nclnNnNs10000.052.63rpmD(1s200.952-6UdclKpKsUuKpKs112vUdopKpKsUu264Udop/Udcl22UuKpKs1KpKsUdcl4.6v2-7D107-1-
运动控制系统习题答案nclnNKnopnclnNs15000.057.9rpmD(1s100.95100111.667.912-8ncl180rpmK15所以:nopncl1(K18016rpm1280rpm如果:K30,则:ncl2D1nopK11280rpm/3141.29rpmnNsnNsncl1D80/41.292D2则:2ncl1(1sncl2(1sD1ncl22-91Ce(UNINRa/nN0.1342vmin/rnopIN(RaRrecRL/Ce12.53.3/0.1342307.4rpmnNs15000.18.33rpmD(1s200.9nop246.891135.9所以Kncl8.33ncl2)系统原理图和静态结构框图见书中;34)方法一:由UnUnn,所以Unm/nN15/15000.01vmin/r**KpKCe/Ks13.7734)方法二:由Kp[Ce(K1nNINR]/KsUnm14.2271*KCe/KpKs0.00972-10已知:IdcrUcom/Rs*Idbl(UcomUn/RsIdcrUcom/Rs1.2IN**Idbl(UcomUn/Rs2IN两式相加可得:Un/Rs0.8IN进一步:RsUn/0.8IN15/(0.812.51.5所以:UcomRs1.2IN22.5v要求Rs(RaRrecRL/31.1,所以:设置电流反馈系数Kfi1.5/1.11.364所以:Ucom22.5v1.1/1.516.5vRs1.1电流截止负反馈的单环速度闭环系统原理图和静态结构图略。另一种方法求Rs:由静特性可知:*n[KpKsUnmKpKs(IdblRsIdcrRsRIdbl]/Ce(K10,堵转时可知,*Rs[Unm(RIdbl/KpKs]/(IdblIdcr,与前方结果一致!*2-11TmGD2R/375CeCm0.0819sTlL/R0.0152sTs0.00333s所以:Kcr[Tm(TlTsTs]/TlTs30.2。因为稳态要求K35.9,而稳定性要求27-2-
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