第五章一阶电路和二阶电路§5-1动态电路的方程及其初始条件一阶电路:用一阶微分方程描述的电路。一.换路:指电路中开关的突然接通或断开,元件参数的变化,激励形式的改变等。换路时刻0t(通常取0t=0),换路前一瞬间:0_t,换路后一瞬间:0t。二.换路定则c0c0()()ututL0L0()()ititC0C0()()itit,L0L0()()utut,R0R0()()itit,R0R0()()utut三.初始值的计算:1.求C0L0(),()utit:①给定C0L0(),()utit;②0tt时,原电路为直流稳态:C—断路L—短路③0tt时,电路未进入稳态:0C0C()()|ttutut,0L0L()()|ttitit2.画0t时的等效电路:C00()()utut,L0L0()()itit换路前后电压(流)不变的为电压(流)源C—电压源L—电流源C0()0ut,L0()0itC—短路L—断路3.利用直流电阻电路的计算方法求初始值。例1:已知:0t时,原电路已稳定,0t时,打开开关S。求:0t时,各物理量的初始值。•••R10ΩS(t=0)CuC+_iciR215ΩR2R110Ω_uR1+iLL_uL+10V解:1.求CL(0),(0)ui:0t时,CL(0)7.5V,(0)0.25Aui2.画0t时的等效电路:3.0t时:R1(0)0.25102.5VuR27.5(0)0.5A15iLR1C(0)(0)10(0)0uuu2CLR(0)(0)(0)0.25iiiA例2:已知:0t时,原电路已稳定,0t时,打开开关S。求:0t时,1(0),(0)ii。解:1.求C(0)u:0t时:C1111C(0)14(0)10(0)4(0)(0)(0)4(0)(0)2A(0)28Vuiiiiiiiu2.作0t时的等效电路:0t时:11(0)(0)414(0)7(0)28iiii184(0)A,(0)A33ii4Ω14Ω7Ω4ACS(t=0)uC(t)+_10i1+_i(t)i14A4Ωi1(0+)10i1(0+)+_i(0+)7Ω28V+_15Ω10Ω10ΩiL(0_)10VuC(0_)+_iR2(0+)uL(0+)_+10Ω10Ω15Ω0.25A7.5V+_iC(0+)10VuR1(0+)_+14Ω4A10i1()(0+)0i1(0+)10i1()+_i(0_)4Ωi1(0-)7Ω•+-uC(0-)-10i1(0-)+§5-2一阶电路的零输入响应RCSKVL:()()(0)utututCCCRCVAR:,duduiCuRiRCdtdtCCSC(0)(0)?duRCuutdtu零输入响应:指输入为零,初始状态不为零所引起的电路响应。零状态响应:指初始状态为零,而输入不为零所产生的电路响应。完全响应:指输入与初始状态均不为零时所产生的电路响应。一.RC放电过程已知:0t时,电容已充电至0U,0t时,S由a合向b。求0t后的CRC(),(),()ututit。1.定性分析:0t时,C0(0)uU,RS0(0)uUU,S0C(0)UUiR0t时,CC0(0)(0)uuUR0(0)uU0(0)CUiRUSRC_uR(t)+uC+_iC(t)S(t=0)S(t=0)baC_uR(t)+uC+_iC(t)•USC,tu,RC,ui;CRC,0,0,0tuui2.定量分析:0t时,CCC00(0)(0)duRCutdtuUC()etRCutK令0t,C0(0)1uKUC0()e(0)tRCutUtRC0()()e(0)tRCututUt0RC()()e(0)tRCUutittRR()(0)e(0)tRCftft3.时间常数:RCR:由动态元件看进去的戴维南等效电阻伏特库仑安培秒==秒安培伏特安培C0:()ut的物理意义衰减到36.8%C0()ut所需时间C0()e(0)tRCutUtC_uR(t)+uC+_iC(t)uC(t)uCU0-U0-U0/R•••uRiC(t)••Us-U0(Us-U0)/Rt0•t0t0+τtuC(t)U0uC(t0)uC(t0+τ)=36.8%uc(t0)00C00()etRCutu00C000C0()eee()0.368ttRCRCRCutuUutτ的几何意义:由0C0[,()]tut点作C()ut的切线所得的次切距。4t时,电路进入新的稳态,4C0C0C0(4)()e1.82%()0ututut211422()4eV(0)2s()4eV(0)4sttuttutt可见,时间常数反映了物理量的变化快慢,越小,物理量变化越快。二.RL放磁过程已知0t时,L0(0)iI,求0t时的LL(),()itut.利用对偶关系:LCLCiuLCuiGRRC串联:CCC00(0)(0)duRCutdtuURL并联:LLL00(0)(0)diGLitdtiIL0()e(0)tGLitItLGLRLRuR(t)_+uL(t)+_iL(t)•tu1u20u1,u24V0L()e(0)tGLIuttG()(0)e(0)tftft综上所述,一阶电路的零输入响应变化模式相同,即()(0)e(0)tftft故求一阶电路的零输入响应时,确定出(0)f和τ以后,就可以唯一地确定响应表达式。§5-3一阶电路的零状态响应一.RC充电过程已知C(0)0u,求0t时的CRC,,uui。1.定性分析:0t时,(0)0CuRS(0)uUSC(0)UiRC,tu,RC,ui;CSRC,,0,0tuUui2.定量分析:CCSC(0)(0)0duRCuUtdtuCCpCh()()()utututCp()ut为非齐次微分方程任一特解,Ch()ut为对应齐次微分方程的通解,cpu—强制响应,与输入具有相同形式,cpS()utAAU,cpS()utU/ch()etRCutK—固有响应,与电路结构有关。CS()etRCutUK0t令=iCuR0tuC,uR,iCUSUS/RuCUS+_uR(t)_+uC(t)+_iC(t)CCUS+_uR(0+)_+iC(0+)CSS(0)0uUKKUCSSS()e(1e)(0)ttRCRCutUUUtRSCS()()etRCutUutU(0)tSRC()etRCUuitRR(0)tCCpChSC()()()e()(1e)(0)ttRCutututUKut其中:SU为稳态响应(C()u),etRCK为暂态响应(必将衰减为0)RC为时间常数0C0S()(1e)tutU0C0S()(1e)tutU001SSS(1e)(1e)(1e)ttUUUC0SC0()63.2%()utUut即充电过程中时间常数的物理意义为由初始值上升了稳态值与初始值差值的63.2%处所需的时间。4t时,电路进入新的稳态。3.充电效率()100%()()CRC63.2%(US-uC(t0))tuC(t)USuC(t0)uC(t0+τ)t0t0+τ0τUS-uC(t0)22CCS1()()22CWCuU222SRCS00()(e)2tRCUCWRidtRdtUR50%例:已知:0t时,原电路已稳定,0t时合上S,求0t时的C0(),()utut。解:已知(0)0Cu1.C()u:t时,C2()V3u2.求eq23R2s31.5C2()(1e)V(0)3tutt1.50C12()1()eV(0)33tututtt0uC,u00()utU0()uC(t)1V2/3V1/3VuR(∞)_+2Ω1V1Ω••1Ω2ΩReq1VuC(t)_+1F2ΩS(t=0)+0()ut-二.RL充磁过程已知:L(0)0i。求:0t时的L()it利用对偶关系RL充磁过程LLSL(0)(0)0diGLiItdtiLSL()(1e)()(1e)(0)ttGLitIit例:已知:0t时,原电路已稳定,0t时合上S,求0t时的Lo(),()itit解:已知L(0)0i1.求L()it时L()3Ai2.求102s5LR1Ω4Ω1.2Ω5Ω10H18VS(t=0)iL(t)i0(t)iL(∞)5A1Ω4Ω1.2Ω5Ω18V5Ω1Ω4Ω1.2ΩReq=5ΩUS+_S(t=0)RLiLISLRiL(t)IS=US/R(G)2L()3(1)A(0)titetLL2o410()20.5eA(0)6tdiidtitt§5-4一阶电路的完全响应已知C0(0)uU,0t时合上S,求0t时的C()utCCSC0(0)(0)duRCuUtdtuUCCpChS()()()eettRCRCutututAKUK令0t,CS00S(0)1uUKUKUUCS0S()()e(0)tRCutUUUt稳态响应暂态响应完全响应=稳态响应+暂态响应C0S()e(1e)ttRCRCutUU零输入响应零状态响应完全响应=零输入响应+零状态响应CCCC()()(0)()e(0)tRCutuuut一阶电路的三要素法:前提:①一阶电路②直流激励2.5A1A2A3A••••iL,i0iLt0S(t=0)RCuC(t)+_US+_U0USUSU0t0UCt0USU0USU0uCph()()()etftftftAK令t:()0()fAAf()()etftfK令0t:(0)()1ffK(0)()Kff()()(0)()e(0)tftffft一阶电路三要素公式(0)f-初始值CL(0),(0)ui——由0t的等效电路中求,CLRR(0),(0),(0),(0)iuiu必须由0t的等效电路求。0t时:C-电压源零状态下:C-短路L-电流源L-断路()f-稳态值t时,C-断路,L-短路-时间常数,,LRCR,R-由动态元件两端看进去的戴维南等效电阻。例:已知0t时已稳定,求0t时,Lo,ii解:1.求o(0),(0)Lii0t时,39A2823io9(0)A8iio2Ω2Ω1Ω3V-3/4A2Ω2Ω3Vio(0-)=-9/8AiiL(0-)iL(t)io(t)Sba3V3V1H2Ω1Ω2ΩL923(0)A834i0t时,o3133(0)()A4248i2.求Lo(),()ii。Lo39()A,()A48ii3.求。1s2LR2L33()eA42tit(0)t2o93()eA84tit(0)t三要素法:例1:已知:0t时原电路已稳定,0t时合上开关S。求0t时,C(),()utiti(t)10kΩ20kΩ10kΩ1mAS