电气课件电路理论-chapter15the laplace trans

CHAPTER15THELAPLACETRANSFORMThelaplacetransformissignificant:1.Itcanbeappliedtoawidervarietyofinputs.2.Itprovidesaneasywaytosolvecircuitproblemsinvolvinginitialconditions.3.Itcanprovideusthecompleteresponsecomprisingboththenaturalandforcedresponses.CHAPTER15THELAPLACETRANSFORM15.1DefinitionoftheLaplaceTransform15.2PropertyoftheLaplaceTransform15.3TheInverseLaplaceTransform15.4ApplicationtoCircuits15.1DefinitionoftheLaplaceTransformGivenafunctionf(t),itsLaplacetransformation,denotedbyF(s),isgivenbyWheresisacomplexvariablegivenbys=σ+jω1.One_sidedLaplaceTransform2.Convergencecriterion:3.ALaplacetransformpair:f(t)F(S)4.TheLaplaceTransformofsometypicalfunctions:(2)Theunitstepfunction(单位阶跃函数):(1)Theexponentialfunction(指数函数):(3)Theimpulsefunction(冲激函数):=115.2PropertiesofTheLaplaceTransform:1.Linearity:2.Circuitelementsinsdomain:R：u=Ri1.Kirchhoff’slawsinsdomain:+u-iR+U(S)-I(S)R2.scaling:Example:3.Differentiation(1).TimeDifferentiationL:i+u-L+

-sLU(s)I(s)sL+-U(s)I(s)ML1L2i1i2+u1-+u2-L1i1(0-)Mi2(0-)Mi1(0-)L2i2(0-)+U2(S)-+U1(S)-I1(S)I2(S)SL1SL2+

-+-SMM:(2).FrequencyDifferentiation:4.Integration:(1)TimeIntegration:+u-iC:IC(S)1/sCuc(0-)/SUc(s)

1/sCCuc(0-)Ic(s)Uc(s)(1).Frequencyshift4.Shift4.Shift(2).Timeshift(延迟定理)f(t)(t)ttf(t-t0)(t-t0)t0f(t)(t-t0)tt0例1：1Ttf(t)TTf(t)?Tt例2：5.Timeperiodicity...tf(t)1T/2TIff

(t)(f1(t)(0,T))isaperiodicfunction6.InitialandFinalValues:Initial_valuetherom：f(t)在t=0处无冲激则Final_valuetheorem：例2：1(t)RC+u-校验：15.3.TheInverseLaplaceTransformGivenF(s),howtoobtainthecorrespondingf(t)：(1)Theinverselaplacetransform(2)MatchingentriesinTable15.2:(3)PartialfractionexpansiontobreakF(s)downintosimpletermswhoseinversetransformcanbegotfromTable15.2UsepartialfractionexpansiontoposeF(s)：1.Simplepoles:（洛比达法则）（分解定理）上例:一对共轭复根为k1,k2也是一对共轭复根2.Complexpoles:method2：completingthesquare(配方）3.Repeatedpoleas:频域延迟2.)求F（S）分母多项式等于零的根，将F（S）分解成部分分式之和3.)求各部分分式的系数4.)对每个部分分式和多项式逐项求拉氏反变换。小结：1.)

将F(S)化成最简真分式由F(S)求f(t)的步骤例2：1(t)RC+u-校验：15.4ApplicationtoCircuitsStepsinapplyingthelaplacetransform:Transformthecircuitfromthetimedomaintothesdomain.Solvethecircuitusingnodalanalysis,meshanalysis,sourcetransformation,superposition,oranycircuitanalysistechniquewithwhichwearefamiliar.Taketheinversetransformofthesolutionandthusobtainthesolutioninthetimedomain.2.Circuitelementsinsdomain:R：u=Ri1.Kirchhoff’slawsinsdomain:+u-iR+U(S)-I(S)RL:i+u-L+

-sLU(s)I(s)sL+-U(s)I(s)+u-iC:IC(S)1/sCuc(0-)/SUc(s)

1/sCCuc(0-)Ic(s)Uc(s)ML1L2i1i2+u1-+u2-L1i1(0-)Mi2(0-)Mi1(0-)L2i2(0-)+U2(S)-+U1(S)-I1(S)I2(S)SL1SL2+

-+-SMM:+u1-+u2-Ri1u1受控源：(s)U+1(s)-m

RI1(S)+U2-U1(S)+u-iRLC+U(S)-I(S)RSL1/SCImpedanceinthesdomainOhm’slawinsdomain3.Acircuitmodelinsdomain:运算电路RRLSL1/SCI1(S)E/SI2(S)RRLLCi1i2Ee(t)时域电路1.Transformthesourcesintosdomain.2.Transformthecircuitelementsintosdomain.3.Considertheinitialconditionofthestorageelements.Thecircuitintimedomain例5Ω1F20Ω10Ω10Ω0.5H50V+-uc+

-iLt=0,opentheswitchuc(0-)=25ViL(0-)=5At>0,circuitmodelinsdomain200.5s-++-1/s25/s2.5V5IL(S)UC(S)4.Laplacetransformapplicationincircuits4.steps：1.Determineuc(0-),iL(0-)。2.Setupacircuitmodelinsdomain.3.Solvethecircuit.4.Taketheinversetransformofthesolution.Example1：200V30Ω0.1H10Ω-uc+1000μFiLWhent=0,kcloses,CalculateiL，uL.200V30Ω0.1H10Ω-uc+1000μFiL(2)Setupthecircuitmodelinsdomain200/SV300.1s0.5V101000/S100/SVIL(S)I2(S)200/SV300.1s0.5V101000/S100/SVIL(S)I2(S)(4)Inversetransformation:求UL(S)UL(S)?200/SV300.1s0.5V101000/S100/SVIL(S)I2(S)RC+ucis例2：求冲激响应R1/SC+Uc(S)Is(s)tuc(V)0tic例.

+-UskR1L1L2R2i1i20.3H0.1H10V2Ω3Ωt=0时打开开关k,求电流i1,i2。10/SV20.3S1.5V30.1SI1(S)ti1523.750UL1(S)10/SV20.3S1.5V30.1SI1(S)uL1-6.56t-0.375(t)0.375(t)uL2t-2.19ti1523.750磁链守恒：i2(0–)=i

(0–)=2AThecircuitisshownintheFig.,whent=0,theswitchisopen,calculatei.**–+1H20V2H1H2H10102i1i2i–+20V10102i1i2ii1(0–)=0i2(0–)=i

(0–)=2A**–+1H20V2H1H2H10102i1i2ii1(0–)=0**–+S2SS1010i++––2220S20SI(S)=+45S+20I(S)=5S(S+4)4S+20I(S)=S1S+40.2–i(t)=1–0.2e–4t小结：1、运算法直接求得全响应3、运算法分析动态电路的步骤2、用0-初始条件，跳变情况自动包含在响应中1).由.换路前电路计算uc(0-),iL(0-)。2).画运算电路图3).应用电路分析方法求象函数。4).反变换求原函数。15.5TRANSFERFUNCTIONSThetransferfunctionH(s)istheratiooftheoutputresponseY(s)totheinputexcitationX(s),assumingallinitialconditionsarezero.H(S)H(j)H

(j)=UR.U.+-U.RjC1jL+-UR.+-RSC1SL+-U(S)UR(S)H

(S)=UR(S)U(S)RR+SL+

SC1=jCRR+jL+

1=与对应正弦稳态响应的关系H(S)H(j)1.Sinusoidalresponse:

iL(0-)=0iK(t=0)L+–uLRuS+-i(0-)=0utuScalculate：i

(t)

ADISCUSSION:PHASORDOMAINANDSDOMAIN强制分量(稳态分量)自由分量(暂态分量)

iL(0-)=0iK(t=0)L+–uLRuS+-Poticularsolution:定积分常数A由

iL(0-)=0I(s)K(t=0)SL+–uL(s)RuS(S)+-Phasormethod:It’saconvenientwaytogetthesinusoidalsteady_stateresponse.Laplacetransform:apowerfultooltoanalysedynamicsofthehigher_ordercircuits.Itcanprovideusthecomplete