重庆大学电气《自动控制理论》课后答案.

第2章习题解2.1(a)求图示RC电路的传递函数G(s)。21Uo2CRUiCR1+-+-(a)sC111RsCR221UrUo-I1I2I)1)(1()1)(1()(2211212211sCRsCRsCRsCRsCRsG2.2(a)是一反相比例运算电路，其12)(RRsG2.3试用复数阻抗法画出图E2.3所示电路的动态结构图，并求传递函数。、)()()(sUsUsGrc)()()(sUsUsGro-+UiU0R1R2R3(a)L1R2RruC-+cu+-uo图E2.3题2-3RLC电路uoucL1R2RruC-+cu+-uououcIR1ICILsC1111RR2Ls1UrUo--IR1ICUCIL-1212))(1()()()(RRLsCsRRsUsUsGroL1R2RruC-+cu+-uououcL1R2RruC--uc++I11RCsLsRCsLsR1)(1)(22UrUc-I2121212)()()()(RRsLCRRLCsRRLssUsUsGrcR0+-R0C+--+11R2R2C3R4Rrucu图E2.5题2-6图2-6试画出图E2.5所示系统的动态结构图，并求传递函数。)()(sUsUrcR0+-R0C+--+11R2R2C3R4Rrucu)1(101CsRRRsCR221UrUc-34RR411232041)1()()()(RRCsRsCRRRRRsUsUsGrc2.7系统动态结构图如图E2.6所示。试求传递函数)()(11sRsC)()(12sRsC)()(21sRsC)()(22sRsC、、、。1RG4G13G2GC12R2C图E2.6题2.7图1RG4G13G2GC12R2C图E2.6题2.7图43211111)()(GGGGGsRsC4321321121)()(GGGGGGGsRsC4321431211)()(GGGGGGGsRsC43213221)()(GGGGGsRsC-+R(s)n(s)GK1N(s)(s)C34KTs+1KK2s图E2.7题2.8图2.8系统结构图如图E2.7所示。求传递函数C(s)/R(s)和C(s)/N(s)。若要消除干扰对输出的影响（即C(s)/N(s)=0），问Gn(s)=?-+R(s)n(s)GK1N(s)(s)C34KTs+1KK2s图E2.7题2.8图解：1）令N(s)=0,则321321)1()()(KKKTssKKKsRsC2）令R(s)=0,则32143321)1()()(KKKTssSKKGKKKsNsCn-+R(s)n(s)GK1N(s)(s)C34KTs+1KK2sK413TsKsKK21GnN(s)C(s)-+-R(s)=0框图简化如下K413TsKsKK21GnN(s)C(s)-+-13TsKsKK21GnN(s)C(s)-+-K441K13TsKsKK21GnN(s)C(s)-+-K441K143TsKKsKKK421GnN(s)C(s)+--1143TsKKsKKK421GnN(s)C(s)+--1143TsKKsKKK421GnN(s)C(s)+--1sKKK421143TsKKsKKK421GnN(s)C(s)+--1sKKK421143TsKKN(s)C(s)+--1sKKK421sKKK421Gn143TsKKN(s)C(s)+--1sKKK421sKKK421Gn32143321)1()()(KKKTsssKKGKKKsNsCn要消除干扰对输出的影响，令C(s)/N(s)=0214)(KKsKsGn则2.9简化图E2.8中各系统结构图，并求出传递函数C(s)/R(s)。G(s)R1C(s)G23GG4++++(s)R3G12GG(s)C++（a）（b）图E2.8习题2-9图2G1G++(s)RC(s)+（c）G(s)R1C(s)G23GG4++++（a）G1+G2G3-G4-RC))((1)()()(432121GGGGGGsRsCsGG1G2-RCG2G3-++G1RCG2G2G3-++32211)()()(GGGGsRsCsG(s)R3G12GG(s)C++（b）2G1G++(s)RC(s)+（c）G1-RC221GG212211)1()()()(GGGGGsRsCsG2.10系统结构如图E2.9所示，试求出系统的传递函数。(s)R3G1G2GHG4++(s)C+++图E2.9题2.10系统结构图(s)R3G1G2GHG4++(s)C+++G1G2G3G4HHRC--G1G2G3G4HHRC--G1G3RCG2G2H-G4G4H-G1G3RCG2G2H-G4G4H-G1G3RCG2G2H-G4G4H-)(1)()()(424321HGHGGGGGsRsCsG2.11系统结构如图E2.10所示，试求出系统的传递函数。R1G321H2GHGCG4++++(a)图E2.10题2-11图R1G321H2GHGCG4++++(a)G2H1H1G4G1G3H2--++RCG2H1H1G4G1G3H2--++RCG2H1/G1G3H1/G3G4G1G3H2/G1--++RCG2H1/G1G3H1/G3G4G1G3H2/G1--++RCG1G2G3G4++RC3112311GGHHGHGG1G2G3G4++RC3112311GGHHGHGG4+RC23212121321HGGHGHGG-1GGG2321212132141)()()(HGGHGHGGGGGGsRsCsG2.12已知系统结构图如图E2.11所示，试写出系统在输入R(s)及扰动N(s)同时作用下输出C(s)的表达式。(s)RG12(s)G(s)H1(s)G3G4(s)CN++++++图E2.11题2-12系统结构图(s)RG12(s)G(s)H1(s)G3G4(s)CN++++++解：1）令N(s)=0,求出CR(s)G1G2H1G3-+-RCG1G2H1G3-+-RCG1-RCG3+1221HGG12213231HGGHGGG)(1)(211321311221132131sRGGHGGGGGHGGGHGGGGGsCR2）令R(s)=0,求出CN(s)(s)RG12(s)G(s)H1(s)G3G4(s)CN++++++G4N(s)C(s)-++G11221HGG+G3+G4N(s)C(s)-++G11221HGG+G3+G4N(s)C(s)-++G112213231HGGHGGG+G4N(s)C(s)-++G112213231HGGHGGG+G4N(s)C(s)++G112213231HGGHGGG+-1G4N(s)C(s)++G112213231HGGHGGG+-1G4N(s)C(s)++12211321311HGGGHGGGGG+-112211321311HGGGHGGGGGN(s)C(s)[phone]HGGGHGGGGG[phone])(HGGGHGGGGGG-G4N(s)C(s)++12211321311HGGGHGGGGG+-112211321311HGGGHGGGGGN(s)C(s)[phone]HGGGHGGGGG[phone])(HGGGHGGGGGG-2113213112211321314121)(1)()(GGHGGGGGHGGGHGGGGGGHGsNsC则)(1)(1)(211321311221132131412sNGGHGGGGGHGGGHGGGGGGHGsCNR、N同时作用时)()()(sCsCsCNR第3章习题解3-1系统在作用下。测得响应为，又知C(0–)=0，试求系统的传递函数。)(1)(1)(ttttrtettc109.0)9.0()(解:22111)(sssssR)10()1(10109.09.01)(22sssssssC1010)()()(ssRsCsG3-2已知惯性环节的传递函数为110)(ssG希望采用负反馈的方法将调节时间ts减小为原来的0.1倍，并保证总放大系数不变，试选择图E3-1中的K1和K2的值。G(s)K1K2–图E3-1题3-2的结构图G(s)K1K2–)(1)(21sGKsGK2110110KsK110110110221KsKK据题意:1.010111010110221KKK求得:K1=10,K2=0.9110)(ssG3-3假定温度计可以用传递函数来描述。如果用它来测容器中恒定的水温，需要1分钟才能指示出实际水温的98%的数值。如果给容器加热，使水温按10℃/分的速度线性变化。温度计的稳态指示误差有多大？11)(TssG解:1))%2(60误差带按sts15604TtTst/sC04T0.022)btsCttr)/(61min/10)(2)(sbsR则温度计传递函数可由如下框图构成11)(TssGTs1R(s)C(s)-E(s))(1)(111)(sRTsTssRTssE则1lim)(lim200ssrbTsbTsTsssEsess5.2)(15)/(61ssC3-4单位反馈系统的开环传递函数为试分别求出K=10和K=20时，系统的阻尼比和自然振荡角频变n，及单位阶跃响应的超调量%和峰值时间tp。并讨论K的大小对过渡过程性能指标的影响。)11.0()(ssKsG解：KssKKssKs1010101.0)(221）当K=10时，210.510ndnωstdp362.016.3%%%21e2）当K=20时，21020010n2nK354.042210102nn则%5.30%%21estdP237.03）当Kn%)(rPtt初始响应速度加快动态平稳性变差5n不变nst)4(33-6欲加负反馈来提高阻尼比，并保持总放大系数K和自然谐振角频率n不变，试确定H(s)(见图E3-2)。2222nnnSSKR(s)–C(s)H(s)图E3-2题3-6图解：)()()(sNsMsH设加负反馈后总的传递函数为)()()2()()(2222sMKsNsssNKsnnnn)()()2()()(2222sMKsNsssNKsnnnn据要求：N(s)=1，要使，则应增大分母一次项系数，所以M(s)应为一阶比例微分环节设M(s)=Ts+，则)()2()(22222nnnnnKsTKsKs要保持前后n不变，则=0TKnnn22'2又nKT)(2'nKsH)(2(s):'综合得3-7二阶系统的单位阶跃响应曲线如图E3-3所示。如果该系统属于单位反馈控制形式。试确定其开环传递函数。h(t)1.251.0000.2t(S)图E3-3题3-7题解：25.0%21e404.0则2.012nPt又16.17n)86.13(16.17)2()(22sssssGnn)()()(%CCtCP3-8试在s平面上绘出典型二阶系统满足下列条件的闭环极点可能位于的区域。1、0.7071n2秒–12、00.52n4秒–23、0.50.707n2秒–1解：典型二阶系统的闭环传递函数为2222)(nnnsss当0