电路原理课件讲义英文版-CHAPTER-8

Ifalloftheinitialconditionshavezerovalue,thenthecircuitissaidtobeinthezero-state,andthesolutiontononzeroinputsforthecircuitisknownasthezero-stateresponse.(零状态响应)1．Zero-StateResponseOfAnRCCircuitK+uCACiCabt=0t=0－：uC(0－)=0+uCUsCiCabR0CCSduRCuUtdtTheZero-StateResponseuC(0+)=uC(0)=0=US+AA=US0tCCpChSuuuUAetThegeneralsolution：(1)0tCSuUett2uC/Us0.6321.00.8650)1)((tCCeuuTheforcedresponserepresentswhatthecircuitisforcedtodobytheinputexcitation.Itisalsoknownasthesteady-stateresponse.(强制响应或稳态响应)Wheredoestheenergygo?1(1)0ttSCCSSSUuiRUUeRUetRtiC0US/R0.368US/R222001(0,)()2tSRCSUwRidtRedtCUR2211()()22CCSwCUCUTimeconstant=RC(Turnoffallindependentsources)t0+2345…∞uC/US00.6320.8650.950.9820.993…1Whetherthetimeconstantissmallorlarge,thecircuitreachessteadystateinfivetimeconstants.iSKR1R2C+uCFig.(a)Example:R1=1Ω,R2=2Ω,C=3F,iS=1A.Assumethattheswitchhasbeenclosedforalongtime.Att=0,theswitchisopened,anditisclosed6slater.FinduCfort0.Solution：0t6s：iSR1R2C+uC(1)Fig.(b)1=R2C=6s6(1)2(1)tCueV)1)((tCCeuut6s：2=RC=2sR1R2C+uC(2)RFig.(c)122//3RRR6(1)16(6)2(1)2(1)1.26tCtueeV(2)(1)(6)(6)1.26CCuuV6622(2)(2)(6)1.266ttCCuueeVtt(s)uC(V)61.26uC(2)uC(1)2．Zero-StateResponseOfAnRLCircuitK+uLAiLabLt=0iL(0)=0+uLUsLiLabR0LLSdiLRiUtdttLLpiiAeTheparticularsolution：SLpUiRTimeconstant：LR(0)0SSLUUiAARR(1),0ttSLLLSUdiieuLUetRdtt2iL0.632Us/R0tuL0US0.368+uL2USiRL2L1K+uL1Example:L1=1mH,L2=2mH,R=3kΩ,US=3V.Att=0,theswitchisclosed.Findthezero-stateresponseuL1anduL2fort0.L=L1+L2=3mH6110LsRSolution：)1)((tLLeiimHRUiSL1)(61010tiemAt6610101122,20ttLLdidiuLeVuLeVtdtdtThesolutionwithnonzeroinitialconditionsandnonzeroinputsforacircuitisknownasthecompleteresponse.(全响应)Thecompleteresponseofthecircuitisthesumofthenaturalresponseandtheforcedresponse.1．CompleteResponseOfAnRCCircuitK+uCACiCt=0uC(0－)=U0+uCUsCiCR,0CCSCSduRiuURCuUtdtCompleteResponseOfRCAndRLCircuituCp=uC(∞)=US，uCp(0+)=US，uC(0+)=uC(0－)=U0，τ=RC(0)(0)0tCCpCCpuuuuetThecompleteresponse：0()0tCSsuUUUet00()0ttCSCSSduUUdiCCUUUeetdtdtRCompleteresponse=zero-inputresponse+zero-stateresponse.uCtU0US0零输入响应零状态响应全响应00()(1)0tCSSttSuUUUeUeUetZero-inputresponse：0tUeZero-stateresponse：(1)tSUeThegeneralform：2．三要素法Theinitialvaluef(o+)Thefinalvaluef(∞)Thetimeconstantτf(t)=f(∞)+[f(o+)-f(∞)]e-t/τt0f(t)=f’(t)+[f(o+)-f’(0+)]e-t/τt0IfitisDCsource,UsK1R1C+uCK2R2Fig.(a)Example:US=6V,R1=R2=1Ω,C=0.5F.Assumethattheswitch1andtheswitch2hasbeenopenedforalongtime.Att=0,theswitch1isclosed.FinduCfort0.Andtheswitch2isclosed2slater.FinduCfort2s.uC(∞)=US=6V，uC(0+)=uC(0－)=0，τ1=C(R1+R2)=0.5×2=1sUsR1C+uCR2Fig.(b)Solution：uC=uC(∞)+[uC(0+)-uC(∞)]e-t/τ1=6(1-e-t)V2t0t=2s：uC(2－)=66e－2=5.19VuCp=uC(∞)=6V，uC(2+)=uC(2－)=5.19V，τ2=R1C=0.5sUsK1R1C+uCK2R2Fig.(a)UsR1C+uCFig.(c)uC=uC(∞)+[uC(2+)-uC(∞)]e-(t-2)/τ2=6-0.81e-2(t-2))Vt2解：例：R1=1Ω，R2=2Ω，C=3F，US=4V，IS=5A，K打开前电路处稳态，t=0时K打开，计算uC。Us+uCR1CR2IsKUs+uCR1R2Ist=0112(0)611SSCUIRuVRRuC(∞)=US+R1IS=9V，τ=R1C=3sUs+uCR1Ist=∞3930tCueVtuC=uC(∞)+[uC(0+)-uC(∞)]e-t/τt01．Theunitstepfunction(单位阶跃函数)001()10ttt7-4SingularityFunctions(奇异函数)Singularityfunctionsarefunctionsthateitherarediscontinuousorhavediscontinuousderivatives.0100)(ttt10tAnequivalentcircuit：1VKRL1(t)RLTherelationshipbetweentheunitrampfunctionandtheunitstepfunction：00()0trttt0t110t11(t)0()1(),()1()tdrttrtddtTheunitstepfunctiondelayedbyt0(时移)0t10t1t0001()10ttt000001()10ttttttInversion(倒置)0t10t10001()100tttttThemultiplicationoftwofunctions0tE010tEE1(t)0tE010tt0t0EE1(tt0)0tf(t)010tf(t)f(t)1(t)0tf(t)010tf(t)t0t0f(t)1(tt0)ThesumoftwofunctionsEt0EEt0E1(t)E1(tt0)2．Theunitimpulsefunction(单位冲激函数)00()0ttt()1tdt0t(t)0011()0,10tfttt0t1/Therelationshipbetweentheunitimpulsefunctionandtheunitstepfunction：001()()1(),()10ttdttdttttdtSifting(sampling)property：(筛分)00()()(0),()()()fttdtfftttdtftf(t)δ(t)=f(0)δ(t)；f(t)δ(tt0)=f(t0)δ(tt0)3．ImpulseResponseUnitimpulseresponseh(t)：thezero-stateresponsewiththeunitimpulsefunctionexcitation.0t1/p(t)01()lim(),()[1()1()]tptpttt01()lim[1()1()]ttt01()lim[()()]htstst()()dsthtdt1(t)s(t)；(t)h(t)Example:FindtheunitimpulseresponsesuCandiC.Solution：Method1:(t)+uCRCiCiR0)0(Cu)(tiiKCLiRC)(tRudtduCiCCdttdtRudtdtduCiCC000000)(1)]0()0([CCuuCCCuuCC11)0()0(t≥0+zero-inputresponse)(11teCutCAteRCtitiC)(11)((t)1(t)+uCRCiC()(1)1()tRCCustRet0()()[(1)1()]111()(1)()1()tRCCtttRCRCRCtdstduhtRetdtdteteRtetCCSolution：Method2:RCtRuC),(1)(0()11()[1()]()1()1()tttCRCRCRCCttRCdutdihtCCeteetdtdtCRCtetRCtuC01/CiCt(t)1/RCExample:FindtheunitimpulseresponsesiLanduL.R＋－uS(t)＋－uLLiL)(tRidtdiLuLL0)0(Li000000)(dttdtidtdtdiLuLL1)]0()0([LLiiLLLiiLL11)0()0(RLteLittL)(110