控制工程基础习题解

第二章第三章第四章第五章习题解第六章第七章第二章习题第二章习题解2-4：对于题图2-4所示的曲线求其拉氏变化sessUttu3102.036)()102.0(16)(则：解：0.206t/msu/V2-5：求输出的终值和初值2332:123322332100030222033000000ssssYssssXsssYssXxysXxssXsYyssYiiiiii或代入上式，得：又：解：322332limlimlims0s00tssssYty初值：232332limlimlim0s00s0tssssYty终值：第二章习题解2-6：化简方块图，并确定其传递函数。+-G1G2G3H1H3H2XiX0+-+-（a）第一步：消去回路+-G1G2G31+G3H3XiX0+-H1H2①③②③②①第二章习题解第二步：消去回路+-G1G2G31+G3H3+G2G3H2XiX0H1第三步：消去回路G1G2G31+G3H3+G2G3H2+G1G2G3H1XiX01321232333211)(HGGGHGGHGGGGsG③②③第二章习题解+-G1G2G3H1G4H2XiX0+-+-（b）第一步：回路的引出点前移+++①③②④②+-G1G2G3G2H1G4H2XiX0+-+-+++①③②④第二章习题解第二步：消去并联回路，回路的引出点后移②+-G1G2G3+G4G2H1G2G3+G4H2XiX0+-+-①③②④第三步：消去回路+-G1G2H1G2G3+G4XiX0+-③②④G2G3+G4(G2G3+G4)H2第二章习题解第四步：消去回路+-XiX0+-②④G1（G2G3+G4)1+(G2G3+G4)H2+G1G2H1第五步：消去回路XiX0④G1（G2G3+G4)1+(G2G3+G4)H2+G1G2H1+G1（G2G3+G4)4321121232443211)(GGGGHGGHGGGGGGGsG第二章习题解+G1G2G3H1G4H2XiX0+-+-（c）第一步：回路的引出点后移①③②④-+G1G2G3H1G4H2XiX0+-+-①③②④-③1/G3++第二章习题解第二步：先后消去回路G4XiX0+-①②④③G1G2G31+（1-G1）G2H1+G2G3H2第三步：消去并联回路④423212132111)(GHGGHGGGGGsG第二章习题解+G1G2H1H3H2XiX0++-第一步：利用加法交换律和结合律对回路进行整理①③②-③+（d）-+G1G2H1H3H2XiX0+-①③②-①+第二章习题解+H3XiX0③-第二步：先后消去回路①②G11+G1H1G21+G2H2XiX0③第二步：消去回路G1G21+G1H1+G2H2+G1G2H3+G1G2H1H2)21213212211211)(HHGGHGGHGHGGGsG第二章习题解2-7：(1)求X0(s)和Xi2(s)之间的闭环传递函数；(2)求X0(s)和Xi1(s)之间的闭环传递函数；+-G1G2G3H1H3H2Xi1X0+-+-（1）解：第一步，回路后移①③②Xi2+++-G1G2G3H1H3H2Xi1X0+-+-①③②②1/G3第二章习题解第二步，只有一个前向通道，且具有公共的传递函数G3，则系统传递函数为：322313213211)(HGHGHGGGGGGsG（2）解：第一步，方框图整理：+-G1G2G3-H1H3H2Xi2X0+++-①③②第二章习题解第二步，回路的相加点前移：+-G2G3-G1H1H3H2Xi2X0+++-①③②①G2第二步，消去回路：+G3Xi2X0+③②①11+G2H3-（G1G2H1+H2）1321233232311)(HGGGHGHGGGGsG第二章习题解2-8：对于题图2-8所示系统，分别求出+G1G2G3H1H2Xi1X01+-+-Xi2++)()()()()()()()(120201220101sXsXsXsXsXsXsXsXiiii，，，X02G4G5G6第二章习题解1)：求出+G1G2G3H1H2Xi1X01+-++)()(101sXsXiG4G5-解：第一步，方框图整理+G1G2G3Xi1X01++-第二步，消去回路，对回路整理得：①③②①③②2154142121443211)1()(HHGGGGGGGGGGGGGsGG4G5H1H21+G4②第三步，二个回路具有公共的传递函数G1，由梅逊特殊公式求得第二章习题解2)：求出)()(202sXsXi解：第一步，方框图整理+G4G5G6Xi2X02++-第二步，消去回路，对回路整理得：①③②21541421214216541)1()(HHGGGGGGGGGGGGGGsGG1H1H21+G1G2②-+Xi2+X02G4G5G6H2H1-++G1①②G2第三步，二个回路具有公共的传递函数G4，由梅逊特殊公式求得第二章习题解3)：求出)()(201sXsXi解：第一步，方框图整理第二步，消去回路，得：①③215414212141543211)(HHGGGGGGGGGHGGGGGsGG41+G4-+Xi2+X01G4G5G3H2H1-++G1①②G2第三步，二个回路具有公共的传递函数G1，由梅逊特殊公式求得③+Xi2X01G5G3H2H1-++G1②G2第二章习题解4)：求出)()(102sXsXi解：第一步，方框图整理第二步，消去回路，得：①③21541421214265411)(HHGGGGGGGGGHGGGGsGG11+G1G2-+Xi1+X02G4G5G6H2H1-++G1①②G2第三步，二个回路具有公共的传递函数G4，由梅逊特殊公式求得③+Xi1X02G4G5G6H2H1-++②2-9：试求题图2-9所示机械系统的传递函数。第二章习题解sDDmssDsGsXmstsXDssXssXDtxmtxDtxtxDaii212102020100201)(由拉氏变换，得：解：212110201102011)(kkDskkDsksGsXksXsXDskDsktxktxtxkDskDskkbii由拉氏变换，得：解：第二章习题解2120002101)()()(kkDsmssFsYsGtymtykktyDtFeii解：21211102201102012222211111)(kksDsDsDksGsXsDksXsXsDktxktxtxksDkkDksDkkDkdii由拉氏变换，得：的等效刚度为：、的等效刚度为：、解：21102010201)(kkDsDsksGsXksXsXDsktxktxtxkDskkcii由拉氏变换，得：解：第二章习题解2121222001022222211111)(kksDsDmssDksGtxmtxktxtxksDkkDksDkkDkfi的等效刚度为：、的等效刚度为：、解：2122213202i0220201:)(kkDskskkmmDsksGtxtxktxmtftxkkktxtxtxktxkkDsktxgaaaaa又：解：设中间变量xa(t)x0(t)k1Dk2mfi(t)第二章习题解2-10：试求题图2-10所示无源电路网络的传递函数。111111)(212020212021CsRRCsRsUsUsGsIRsICssUsIRsICssIRsUtiRdttiCtutiRdttiCtiRtuaiii解：111111)(20020RCsLssUsUsGsICssUsIRsICssLsIsUdttiCtutRidttiCtidtdLtubiii解：第二章习题解sUsUsURRsIsIsLRsCRsIsLRsIsIsIsIsIsIsIRssILsIsCsUssILsIRsUsUtitititititiRtidtdLdttiCtutidtdLtiRtutuciii002171222271117652172625202111076521726252021110,1111)(又：解：21212121222121221212)(LLRRsLLRRsCLLRRLRRsLLRsG第二章习题解2-11：试求题图2-11所示有源电路网络的传递函数。CsRRRsUsUsGsICssUdttiCtusIRsURsUtiRtuRtuaicccci2120002010201111)(解：1111)(4212402020201401CsRRRCsRRsUsUsGsICsRsUdttiCtiRtusIRsURsUtiRtuRtubiii解：第二章习题解sUCsRsUtiRtutisIRsURsUsUsICssURsURsUtiRtutudttiCtuRtuRtutiticAcAcAAcAAiAcAAic202122021120211211)(的方向和解：关键是确定2122120CsRRRsUsUsGi第二章习题解11:)(1)(1)()()(1)(1)()()(2411252415122215452420454244550121454244550121sCRsCRsCRCRCRCRsCCRRRRRRsUsUsGsIsIsIsIsCsIRsUsIRsUsUsIsCsIRsUsIRsUtititidttiCtiRtutiRtutudttiCtiRtutiRtudiAAAiAAAi联立上述方程可求得解：第二章习题解2-12：试求题图2-12所示机械系统的传递函数。第二章习题解21212122321002200101121020010111)(JJDksJJJJksJDssJJksTssGssJsDsssksskssJsTtJtDttkttktJtTai