二、三要素法分析一阶电路(inspectionmethod)(书§7.6)这种方法是求解一阶电路的简便方法。它可用于求解电路任一变量(状态和非状态)的零输入响应和直流作用下的零状态响应和全响应。这一方法的背景:1、一阶电路中的响应是按指数规律变化的,且有它的初始值和稳态值,其变化过程唯一地由时间常数决定;2、可直接应用置换定理,不必等状态变量求得后再运用。te)](y)0(y[)(y)t(y时间常数初始值稳态解三要素)0(y)(y一阶电路的数学模型是一阶微分方程,解的一般形式为te)(y)t(yA令t=0+A)(y)0(y)(y)0(yA三要素法是分解方法和叠加方法的结合,便于对电路的估算和运算。即:其响应是按指数变化的,且有它的初始值和稳态值(平衡值)。设已知电容电压初始值uC(0+)或电感电流初始值iL(0+)。用三要素法解题步骤:1、用电压为uC(0)的直流电压源置换电容或用电流为iL(0)的直流电流源置换电感,得到t=0时刻的直流等效电阻电路,由此可求出电路中任一电压或电流的初始值ujk(0)或ij(0)。2、用开路代替电容或用短路代替电感,所得为一t=时的直流等效电阻电路,由此可求出电路中任一电压或电流的初始值:ujk()或ij()。3、用戴维南或诺顿等效电路计算时间常数=R0C或L/R0。4、依三要素,得任一电压或电流的解答。1A2例113F+-uCV2)0()0(CCuuV667.01122)(Cus2332CR等0tVe33.1667.0e)667.02(667.0ut5.0t5.0C已知:t=0时合开关求换路后的uC(t)。解:tuc2(V)0.6670例2i10V1Hk1(t=0)k2(t=0.2s)32已知:电感无初始储能t=0时合k1,t=0.2s时合k2求两次换路后的电感电流i(t)。解:0t0.2sA22)(5tetit0.2sA2)(is2.01*)32(10)0(i1AisAi5)()(5.0)(26.1)2.0(2)(26.122)2.0(2.05AeiA74.35)()2.0(2tetiit(s)0.25(A)1.262例3+-u(t)155HiL已知:u(t)如图示,iL(0)=0求:iL(t),并画波形.法一0t1iL(0+)=0t0iL(t)=0iL()=1AiL(t)=1-e-t/6(A)=5/(1//5)=6su(t)12120t(s)V+-155HiL1V0t11t2iL(1+)=iL(1-)=1-e-1/6=0.154AiL()=0iL(t)=2+[0.154-2]e-(t-1)/6=2-1.846e-(t-1)/6At2iL(2+)=iL(2-)=2-1.846e-(2-1)/6=0.437AiL()=2AiL(t)=0.437e-(t-2)/6AiL(t)=0t01-e-t/6A0t12-1.846e-(t-1)/6A1t20.437e-(t-2)/6At2iL(t)=1-e-t/6A0t1155HiLt2+-155HiL2V1t2=6s=6su(t)=(t)+(t-1)-2(t-2)u(t)12120t(s)(t)(1-e-t/6)(t)(t-1)(1-e-(t-1)/6)(t-1)-2(t-2)-2(1-e-(t-2)/6)(t-2)iL(t)=(1-e-t/6)(t)+(1-e-(t-1)/6)(t-1)-2(1-e-(t-2)/6)(t-2)00.1540.43712t(s)iL(t)A法二+-u(t)155HiL例7-10在右边所给的电路中,已知:is=2A、t0;is=0、t0,r=2。求i(t),t0。解:t0is=0A)0(0)0(CCuu]4)0(22[21)0(ii解得:A8.0)0(i的等效电路0t的等效电路tA2)(i所以t为无穷大时刻,应用电源等效变换下面求时间常数:用外加电压法求等效电阻11111i10i2i4i4u10i/uR110s1.001.010CR0所以:0tA)e2.12(A)]e1)(8.02(8.0[)t(it10t10i(t)的波形:iC+–uCRuC(0-)=0)(t)()1()(tetuRCtC)(1)(teRtiRCttuc1)(teiRCt和0teiRCt的区别t0R1i§7-7阶跃响应及分段常量信号响应因为动态电路的叠加性,只研究本节单位阶跃函数和下一节单位冲激函数作用下的零状态响应。)5.0(10)(10ttuS例求图示电路中电流iC(t)10k10k+-ic100FuC(0-)=0)(10t10k10k+-ic100FuC(0-)=0)5.0(10t10k10kus+-ic100FuC(0-)=00.510t(s)us(V)0依电路的叠加定理:+-ic100FuC(0-)=05k)(5ts5.010510100RC3610k10k+-ic100FuC(0-)=0)5.0(10tmA)5.0()5.0(2teitCmA)t(e)]t(5[e51it2t2CmA)5.0()()5.0(22teteittC10k10k+-ic100FuC(0-)=0)(10t等效分段表示为:s)0.5(mA0.632-s)5.0(0mA)(0.5)-2(-2tetetittCt(s)iC(mA)01-0.6320.5波形0.368)5.0()5.0()5.0()()5.0(2222teteteteittttC)5.0()5.0()]5.0()([)5.0(2)5.0(212teteettettt)5.0(632.0)]5.0()([)5.0(22tettett又可做下面的变换:mA)5.0()()5.0(22teteittC零状态h(t))(t单位阶跃响应:单位阶跃激励在电路中产生的零状态响应一、单位阶跃响应(unitstepresponse)和单位冲激响应(unitimpulseresponse)单位冲激响应h(t)单位阶跃响应s(t)单位冲激(t)单位阶跃(t)dttdt)()()()(tsdtdth§7-8一阶电路的冲激响应单位冲激响应:单位冲激激励在电路中产生的零状态响应零状态s(t))t(线性时不变电路一个重要性质对于线性时不变电路,若激励x产生响应y,则:激励dx/dt产生的响应为dy/dt;激励xdt产生的响应为ydt+K。零状态h(t))(t零状态s(t))(t证明:1f(t)t)(1)(1)(tttf)(1ts)(1ts)]()([1lim)(0tststh)(tsdtd1注意:s(t)定义在(-,)整个时间轴考虑f(t)作用于零状态电路的响应?当时:f(t)(t)响应)(t解:先求单位阶跃响应令is(t)=)()1()(teRtuRCtCiCRisC例1+-uCuC(0+)=0uC()=R=RC0)0(cu已知:求:is(t)为单位冲激时电路响应uC(t)和iC(t)iC(0+)=1iC()=0)t(eiRCtc)()1(teRdtduRCtC)(t再求单位冲激响应令is(t)=)()1(teRRCt)(1teCRCt)(1teCRCt0与电路的零输入响应相同,表明电路的冲激响应是电路本身固有性质的反映。)]([ddtetiRCtc)(1)(teRCteRCtRCt)(1)(teRCtRCtuCRt0iC1t0uCt0C1iCt(1)RC1冲激响应阶跃响应电容上的电压已经发生了突变。与前面的换路定则矛盾吗?此时,电路电流有界的前提已经不存在了。二.分二个时间段来考虑冲激响应0-0+0+t零输入响应iCRisC+-uCuC(0-)=0电容充电uCt0C1iCt(1)RC1dt]Ru)t([dtiq00c00cCuc1)0(1)]0()0([ccuuC)0(Cu电容中的冲激电流使电容电压发生跳变电荷转移为定值=1=0)(tRudtduCcc000000)(dttdtRudtdtduCccuc不可能是冲激函数,否则KCL不成立1.t在0-0+间iCRisC+-uCuC(0-)=01dt)t(002.t0+零输入响应(RC放电)Cuc1)0(icRC+uc-01teCuRCtc01teRCRuiRCtccuCt0C1iCt(1)RC1)(1)()(1teRCtiteCuRCtcRCtc一.在冲激激励下,电容电压或电感电流初值的跃变tiCuutCccd1)0()0(0CAuC)0(iCC)(tA例1+-uCtucCAuC(0-)uC(0+)0tic)(tA0§7-9电容电压或电感电流初值的跃变二.换路后电路有纯电容(或纯电容和电压源)构成的回路,且构成回路的电容和电压源0-初值的电压代数和不为零,电容电压初值发生跃变。tuLiitLLLd1)0()0(0LAiL)0(tuL)(tAuL+-iL)(tA例2+-tiLLAiL(0-)iL(0+)0合闸后由KVLuc(0+)=E)(tEuc)(ddtCEtuCiCcEiCC0)0(cuk(t=0)例1uctE0ict(CE)0已知:E=1V,R=1,C1=0.25F,C2=0.5F,C2的初始值为零,t=0时合开关k。求:uC1,uC2。解:V1)0(1EuC0)0(2Cu电容电压初值一定会发生跃变。)0()0()0(21CCCuuu合k前合k后tuCtuCiCCdddd2211ttuCtuCtiCCd)dddd(d00221100)]0()0([)]0()0([0222111CCCCuuCuuC例2ERC1C2+-uc1+-uc2k(t=0)iiC1iC2节点电荷守恒V315.025.0125.0)0(211CCECuC可解得)0()()0()0(212211CCCuCCuCuCq(0+)=q(0-)uC()=1V0321)131(1)(3434teetuttC=R(C1+C2)s43ERC1C2+-uc1+-uc2k(t=0)iiC1iC2上面的等式是t=0+时间域的,若对整个时间域,等式应怎样写?ut0)(92)(61)](98)()321()([41dd3434111tettetttuCittC0)0(1)0(021CCuut)t()e321()t()t(ut341C)()321()(342tetutC1/3uC21uC1tuCiCdd222it-1/62/9iC14/91/6iC2)(94)(6134tett)](98)()321[(2134tett三.换路后电路有纯电感(或纯电感和电流源)构成的割集,且构成割集的电感和电流源0-初值的电流代数和不为零,电感电流初值发生跃变。k例+-10V20.3H0.1Hi1i23+u1-+u2-A5)0(1i0)0(2i)0()0(21ii而电感电流发生跃变已