电力拖动自动控制系统-第四版-课后答案

1、习题解答（供参考）习题二r minnnn s D (1s)1000 0.02 (10 0.98)2.04rpm2.04rpmn0 max1500r min n0min150r minnN15r minnmax nminnmaxn0 maxnN150015 1485nminn0minnN 150 15 135Dnmaxnmin1485 13511snNn015 15010%Ce(U NI N Ra ) nN(220378 0.023) 14300.1478V rpmnI N R Ce378(0.0230.022)0.1478 115rpmDnN Sn(1s)14300.2115(10.2)3.1

2、DnN Sn(1s)14300.3115(10.3)5.33nNSNnNnNI NR Ce 3050.18 0.2 274.5r / minSNnNn0 274.5 (1000 274.5) 21.5%nnN S D (1 s)1000 0.05 20 0.95 2.63r / minU u*8.8V K P2 K S15 U d U d U u*UdKpK U *(1KpKs)2158.8(12 15 0.7) 12VsuU d8.8215264VU u*U d (1K p K s)K p Ks12(12150.35) (2 15) 4.6Vs5%DnN s / nN 1s1015002%

3、/nN98%nN1500 2% / 98%10 3.06r / minKnop/ncl1100/ 3.06 1 31.7nop1Kncl1158128nclnop / 1K128 / 1304.13rpmncl1 /ncl 21.937nU NI NRa/ CeCe22012.51.5 /1500 201.25 /1500 0.134V min/ rnU NI NR/ CenopI NR / Ce12.5 3.3 / 0.134 307.836r / minnNnN s /D 1s1500 10% （/ 20*90% ） 8.33r / minncl8.33r / minnK p K sU n

4、I d R / Ce 1KKU n /1KI d R / Ce 1KKnop /ncl1307.836 / 8.33135.955150035.955 15 /135.95512.53.3 / 0.134 135.9550.0096V min/ rK * Ce35.955* 0.134K p14.34K s *35* 0.0096U nU nnU n*150.01n1500K pKC e / K sK p35.9550.134 /350.0113.76I dbl2I NI dcr1.2I NI dbl2I N25 A I dcr1.2I N15AI dcrU com / Rs15U com /

5、 RsIdblU *Ucom/ Rs25 15Ucom/ RR 1.5nssU com151.522.5V(R /3)1.01.50.8 /31.1,Rs(R /3)不符合要求，取 Rs1.1 ,需加电流反馈放大器Rs1.1U comI dcrRs151.116.5VI dI dcrn KpKU*/ Ce1 KKpKsKiRIdUcom/ C1 KRId/ Ce1 KsnseK p K s U n*K i U com / Ce 1KRK p K sK i RsI d / Ce 1KI dblK p K s U n*K iU com/RK p K sK i RsU n*K iU com/ K i

6、 Rs251516.5Ki/1.1 KiK i15 /22.513.51.36GD 2 =1.6Nm 2L50mH GD 21.6Nm2 R3.3Ce0.134V / rpmTlL / R0.05 / 3.30.015 sTGD 2R/375C Cm1.63.3 /3750.1340.13430 / 3.14me5.28 / 64.330.082sTs0.00333sKTm TlTs2/ TTls0.0820.015 0.003332（）Ts0.00333/0.0151* 0.003330.0015 0.003332/ 0.0000498330.52PN2.8kW U N220VI N15.6

7、 A nN1500 Ra RrecRL K s35D30 sD30 s10%D30 s10% U n*10V I dI N nn NK pC e22015.61.5 / 15000.1311V min/ rnopINR / Ce 15.63.3 / 0.1311392.68r / minnmin1500 / 3050snop /n0 min392.68 /392.68 5088.7%0.1nn /5/ 0.9n505.56r / minn K p K sU n* / C e 1 K R I d / Ce 1KKK p K s / Ce1500K p K sU n*/ Ce 1KR 15.6 /

8、 Ce 1 KKnop / ncl1297.48 / 5.561 52.5f 0 1MHz n1500r / min n150r / minQ60601.465r / minZTc1024 40.01n60M 1ZTcn1500r / min M 1nZTc15004 1024 0.0160601024n150r / min M 1nZTc150 41024 0.01102.460601500r / minmax %1100%10.098%M 1100%1024150r / minmax %110.98%100%100%M 1102.4Zn210244 15002n 1500r / min Q

9、60 1 106171r / min60 f0 Zn1025 4 1500nZn21024415021.55r / min150r / min Q60 1 10660 f 0 Zn1024 4 150n60 f 0M 260 f0ZM 2Znn1500r / min60106M 249.7710241500n6010 697.7150r / min M 241024150n1500r / minmax %1100%1100%11.4%1M 29.771n150r / minmax %1100%1100%1%1M 297.71习题三U nm* nN I N R K s C e U n* I dL

10、 n U n U i* U i U c U i* U cU nm*/ nN15V / 1500rpm0.01V / rpm当 U n*5V ,转速nU n*5V500rpm0.01V / rpmU im*15V0.375V / AI dm40 AU i*I d0.375*103.75VU iU CU d 0EI dL RCenNI dL R0.127 *500 10 * 2K sK sK s4.175V20即n500rpm ,U n5V ,U i*U i3.75V ,U c4.175vU i*I dm15VU cU d 0CenI d RI dm R40*24VK sKsKs20PN U N

11、I N nN R U nm*U im*U cm I dm K sU d 0 , U i* ,U i , U cUim*8VU nm*8VI dm0.2V / AnN0.008V / rpm40A1000rpmU d 0EIdl RCe nNI dl R40A*1.560VU n*8V ,U n 0 U i*8V ,U i8V, UCU d 0 K S60 401.5VU im*U i* U c U cU im*8V0.1V / AU i*I dm80AUEIRC nIRU d 0CenI d RUC UCU cKsd 0dle NdlK s（ 1）（ 2）t s t r（ 3）t rKT0.6

12、9,0.6, % 9.5%(1)K0.69 / T 0.69 / 0.1 6.9(1/ s)(2)tr3.3T0.33 Sts36T 6 0.1 0.6sntr0.25s KT1,0.5 trK1/ T10Wobj( s)K110s 10.01 s1KT0.5,0.707, 查表 3-1，得 % 4.3%W (s)1,W (s)110,这样，T， K=10/10/K 10/50 0.2Sss (0.01s 1)0.01W (s)11s0.2sWobj ( s)K 110s(Ts1)s(0.02s1)WPI (s)K PI( s1) ,校正后系统的开环传递函数W (s)K PI ( s1)K 1

13、sss(Ts 1)K KPI K1/,hT , 选 h8， 查表 3-4，%=27.2%hT 8*0.020.16sKh181175.78,KPIK/ K1175.78*0.16 / 102.812*8 2*0.02 22h2T2PN60 U N220 I N308 nN1000 Ce R K s Tl Tm T0 i T0nin1.1I NkU im* / I dm8V / (1.1* I N )8V / 339 A0.0236V / A10 /10000.01V min/ ra)Ts0.00333sb)Toi0.0025sc)T iT0iTs0.00250.003330.00583sKi

14、(i S 1)i 5% WACR (S)i SiTl0.012s, 选 K I T i0.5, K I0.5 / T i85.76s 1KiK Ii R 85.760.012 0.18K s350.2240.0173ciK I85.76s 1R040KRiKi R00.22440K8.96 KCii / Ri0.012/ (9103 ) 1.33FC0i4T0i / R0 40.0025/ 40 1030.25 Fa) 1/ K IK I T i0.51/ KI2T i20.00583 0.01166sTon 0.015sc)T n1/ K ITon0.011660.0150.02666sWA

15、SR (s)Kn (n s1)n snhT n ,取 h5,nhT n 0.1333 sKNh12 526168.82s 22h2T 2n0.026662Kn(h 1)Ce Tm60.02360.1960.122hRT n250.016.940.18 0.02666cnK N/1K Nn168.820.133322.5s 1nhT n0.07998 sKN(h1)/ 2h2T 2n4/(2 90.026662 )312.656s 2Kn(h1)CeTm / 2hRT n40.0236 0.1960.12/ (23 0.01 0.18 0.02666) 7.6cnK N/1K Nn312.656

16、0.0799825s1a)1/ 3( K I/ T i )1/21/ 3(85.76/ 0.00583) 1/240.43s 1cnb)1/ 3( KI/ Ton )1/21/3(85.76 / 0.015) 1/225.2 s 1cnn272.2%1.1(3080.18 / 0.1961000)(0.02666 / 0.12)9.97%10%R040K RnK nR0 7.640304KCnn / Rn0.07998/ 3101030.258FCon4Ton / R0 40.015/ 40 1031.5Fn %272.2%(1.10.4)(3080.18 / 0.196100)(0.0266

17、6 / 0.12)63.5%根据电机运动方程： GD 2 dnTe TL ,dnCm ( Idm I dL )R(I dm IdL )(I dmIdL ) R375 dtdtGD 2GD2RCeTm375Ce 375CmCeCeTmn*0.196*0.12*10000.385st(1.1*3080)*0.18(I dm I dL )RTl Tm T0iT0nincicnU im*100.00877V / AI dm1.5*760U nm*100.0267V min/ rnN375a)TS0.00176sb)Toi0.002 sc)T i0.00367 sT l/T i=10i=T l =KiK

18、 I TI R / K s136.240.0310.14 / 750.008770.899RiKi R00.899 40 35.96CiTi / Ri0.031/ 36 1030.86FC0i4T0 i / R0 4 0.002/ 401030.2 f确定时间常数：电流环等效时间常数 1/K I ：因为 K ITi= 则 1/K I=2Ti=2*=b)Ton=c)Tn=1/K I+Ton=+=速度调节器结构确定：按照无静差的要求，应选用PI 调节器，W ASR(s)=Kn(ns+1)/ns速度调节器参数确定：n=hTn，选 h=5，则 n=hTn=,KN =(h+1)/（2h2 T2n） =6

19、/2*25*= s -2Kn =(h+1)CeTm/（2hRTn）= 6*2*5*=可见满足近似等效条件。速度调节器的实现：选R0=40K，则 Rn=Kn*R 0=*40=420K由此 Cn=n/Rn=420*103=F取 FC0n=4T0n/R0=4*40*10 3=2F电流环的截止频率是： ci=K I= s-1速度环的截止频率是： cn= s-2从电流环和速度环的截止频率可以看出，电流环比速度环要快，在保证每个环都稳定的情况下，再求系统的快速性，充分体现了多环控制系统的设计特点。I dL I d n Tmnmax tvK N n s1Wn s2 T n s1s习题五U d2ABO CU

20、d2U dSx12uxU dSx02uxU d (2Sx 1)2O us(uAuB ejuC ej 2 )SASBSCu AuBuCusU dU dU du0222U dU dU du1222U dU dU du2222U dU dU du3222U dU dU du4222U dU dU du5222U dU dU du6222U dU dU du7222u3u2u4u0 , u7u1u5u6u AO uBO uCO u AO u BOu CO u suAOuAOuBOuBOe juCOu ej 2CO2U d32U d e j 3322U d ej332U d ej342U d ej335

21、2U d ej 33usu AO uBOuCO uAO uBO e juCO e j 2B( e j)usu COuCOu BOu BOA(e j 0 )u AOC (e j 2 )u s s u AO uBO uCO u s susdsRs i sdtdsusdtsusdtse j ( 1t)sj ( 1t)us1 se2u s1us 21s1us1s 41u s4u s4us 2us3u s3ud开关周期 T0,输出频率 w1,3Nw1t1 u1t2 u2t1 U dt2 U d e 3, t1, t2jT0T0T0T0T0T0t1 , t2 ,Tt1 t2u 2t 2 u 2u sT 0

22、3u 1t1 u 1T 0u s ud习题六2111C3/ 222333022i AI mco s(t ) i BI mcos( t2I m cos(t2is) iC) i s3311i A11is21222i A2 i B2 iCiBis3333330iC022i BiC2223 iA002303 iB3 iC22i AiB iC03 iA003t )iscos(2 22 I m2is333233202i B2i C2co s( t3)cos( t3)3t)cos(2I m22222j ( t)j ( t)333j ( t)j ( t)3 eee3e32223 cos(t )2I m2j222233 e3e 3j te