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当前位置:首页 > 电子/通信 > 综合/其它 > 湖北工业大学电工电子实验教学示范中心
LinearCircuitAnalysisInanoscilloscopeatimingsignalcalledahorizontalsweepactsasatimebase,whichallowsonetoviewmeasuredinputsignalsasafunctionoftime.Inpractice,thelinearincreaseinvoltageisapproximatedbythe“linear”partofanexponentialresponseofanRCcircuit.First-OrderRLandRCCircuits1.Whatisafirst-ordercircuit?Afirst-ordercircuitischaracterizedbyafirst-orderdifferentialequation.Itconsistsofresistorsandtheequivalentofoneenergystorageelement.Typicalexamplesoffirst-ordercircuits:(a)First-OrderRLcircuit(b)First-OrderRCcircuitUSLR+_CRUS+_Interconnectionsofsources,resistors,capacitors,andinductorsleadtonewandfascinatingcircuitbehavior.CRUS+_i+_CuFirst-OrderRCcircuitAloopequationleadstoSCURiusinceCduiCdtCSCduURCudtor,equivalently,11CCSduuUdtRCRCThisequationiscalledConstant-coefficientfirst-orderlineardifferentialequationApplydualityprinciple,11LCSdiiIdtGLGL2.SomemathematicalpreliminariesThefirst-orderRLandRCcircuitshavedifferentialequationmodelsoftheform11CCSduuUdtRCRC11LCSdiiIdtGLGLRCcircuitfirst-orderdifferentialequationRLcircuitfirst-orderdifferentialequation00()()(),()dxtxtftxtxdtor,equivalently,00()()(),()dxtxtftxtxdtvalidfort≥t0,wherex(t0)=x0istheinitialcondition.Thetermf(t)denotesaforcingfunction.Usually,f(t)isalinearfunctionoftheinputexcitationstothecircuit.Theparameterλdenotesanaturalfrequencyofthecircuit.00()()(),()dxtxtftxtxdt00()()(),()dxtxtftxtxdtThemainpurposeofthischapteristofindasolutiontothedifferentialequation.orThesolutiontotheequationfort≥t0hastheform00()()0()()()tttttxtextefd(1)Satisfiesthedifferentialequation(2)Satisfiesthecorrectinitialcondition,x(t0)=x0Integratingfactormethod00()()(),()dxtxtftxtxdtFirststep,multiplybothsidesoftheequationbyaso-calledintegratingfactore-λt.()()()tttdxteexteftdtBytheproductrulefordifferentiation,[()]tdextdt()()ttdxteextdt()teftIntegratebothsidesoftheequationfromt0totasfollows0[()]ttdexdd0()ttex00()()ttextext0()ttefd000()()()ttttextextefd00()()0()()()tttttxtextefd3.Source-freeorzero-inputresponseLR()aRC()bAparallelconnectionofaresistorwithaninductororcapacitorwithoutasource.Inthesecircuits,oneassumesthepresenceofaninitialinductorcurrentorinitialcapacitorvoltage.(a)KCLimplies(b)KVLimpliesRLiiRiLi+_LuCi+_Cu+_RuLLRudiLiRRdtLLdiRidtLRCuuCRCduuiRRCdt1CCduudtRCBothdifferentialequationmodelshavethesamegeneralform,()1()()dxtxtxtdt()1()()dxtxtxtdtThesolutiontotheequationfort≥t0hastheform0()0()()ttxtext00()ttextWhereτisaspecialconstantcalledthetimeconstantofthecircuit.Theresponsefort≥t0oftheundrivenRLandRCcircuitare,respectively,givenby0()0()()RttLLLiteit01()0()()ttRCCCuteutRCcircuitRLcircuitRCLGLRThetimeconstantofthecircuitisthetimeittakesforthesource-freecircuitresponsetodroptoe-1=0.368ofitsinitialvalue.Formoregeneralcircuits,thosecontainingmultipleresistorsanddependentsources,itisnecessarytousetheTheveninequivalentresistanceseenbytheinductororcapacitorinplaceoftheR.LinearResistivecircuitNoindependentsourcesLiLLinearResistivecircuitNoindependentsourcesC+_CuTheveninequivalentLiLthR+_CuCthRTheveninequivalent0()0()()thttLLLRiteit01()0()()thttCCCRuteutExample1.Forthecircuitofthefigure,findiL(t)anduL(t)fort≥0giventhatiL(0-)=10AandtheswitchSclosesatt=0.4s.Thencomputetheenergydissipatedinthe5Ωresistoroverthetimeinterval[0.4,∞).8H+_Lu20Li50.4tsSUSC+_u(t)+_RK()/utV/ts0US过渡期为零USR+_u(t)K+_(1)问题的提出()/utV/ts01t2t3tUS第一个稳定状态第二个稳定状态第三个稳定状态过渡状态过渡状态瞬态(transientstate)稳态(steadystate)换路换路代数方程描述微分方程描述0t一阶电路的初始条件(2)电路的初始条件①t=0-和t=0+的概念认为换路在t=0时刻进行换路前一瞬间换路后一瞬间t=0+t=0-00(0)lim()ttfft00(0)lim()ttfft初始条件为t=0+时u、i及其各阶导数的值②电容的初始条件C+_Cui1()()tCutidC0011()()tididCC01(0)()tCuidC001(0)(0)()CCuuidCt=0+时当()i为有限值(0)(0)CCuuCqCu(0)(0)qq电荷守恒换路瞬间,若电容电流保持为有限值,则电容电压(电荷)换路前后保持不变。③电感的初始条件L+_uLi1()()tLitudL0011()()tududLL01(0)()tLiudL001(0)(0)()LLiiudLt=0+时当()u为有限值(0)(0)LLiiLLi(0)(0)磁链守恒换路瞬间,若电感电压保持为有限值,则电感电流(磁链)换路前后保持不变。④换路定律(0)(0)CCuu(0)(0)qq(0)(0)LLii(0)(0)换路瞬间,若电容电流保持为有限值,则电容电压(电荷)换路前后保持不变。换路瞬间,若电感电压保持为有限值,则电感电流(磁链)换路前后保持不变。反映了能量不能跃变例1电路如图所示,试求。(0)Ci解:+_10V10kC40kK+_CuCi+_10V10k40k+_Cu+_10V10k+_Ci8V②由换路定律①画出电路,求出0(0)Cu电路0电路0③画出电路,求出0(0)Ci40(0)1081040CuV(0)(0)8CCuuV3108(0)0.21010CimA(0)0CiA(0)(0)CCii可见,电容开路电容用电压源替代Ci+_10V14KL+_LuLi+_10V14+_LuLi+_10V14+_Lu2A例2电路如图所示,试求。(0)Lu解:②由换路定律①画出电路,求出0(0)Li电路0电路0③画出电路,求出0(0)Lu10(0)214LiA(0)(0)2LLiiA(0)428LuV(0)0LuV(0)(0)LLuu可见,电感短路电感用电流源替代求解初始条件的步骤①画0-等效电路,即换路前电路(稳定状态),求uC(0-)和iL(0-)。②由换路定律得uC(0+)和iL(0+)。③画0+等效电路,即换路后的电路。④由0+电路求所需各变量的0+值。电容用电压源来替代,大小为uC(0+)电感用电流源来替代,大小为iL(0+)。电压源和电流源的方向均与原来的电压、电流方向一致。其中电容相当于开路电感相当于短路其中SiLRC0tK+_LuLiCi+_CuSiRLi+_Cu+_LuRSi+_SRi例3电路如图所示,试求,。(0)(0)CLiu解:②由换路定律①画出电路,求出,0(0)(0)CLui电路0电路0③画出电路,求出,0(0)(0)CLiu(0)LSiiA(0)(0)CCSuuRiV(0)0SCSRiiiAR(0)CSuRiVCi(0)(0)LLSiiiA(0)LSuRiV+_48V2L23
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