第6章正弦电源激励下的零状态响应2

四.正弦电源激励下的零状态响应（以RL电路为例）iL(0-)=0iK(t=0)L+–uLRuS+-)sin(umstUui(0-)=0utuS求：i(t)接入相位角)sin(umtUdtdiLRi'iii强制分量(稳态分量)自由分量(暂态分量)teiARSUjL+-I22)(LRUImmRLarctg)sin('umtIi)sin(umstUuiL(0-)=0iK(t=0)L+–uLRuS+-用相量法计算稳态解itumAetIiii')sin()sin(umIAtumumeItIi)sin()sin(解答为讨论几种情况：1）合闸时u=，电路直接进入稳态，不产生过渡过程。2）u=±/2即u-=±/2)sin(umtIimIAtmeIi定积分常数AAIium)sin(0)0(由则A=0,无暂态分量0itmumeItIi)sin(u=-/2时波形为mIi2max最大电流出现在t=T/2时刻。iImitmmeItIi)2/sin(-ImiT/2ti0§6-6一阶电路的全响应全响应：非零初始状态的电路受到激励时电路中产生的响应一.一阶电路的全响应及其两种分解方式iK(t=0)US+–uRC+–uCRSCCUutuRCdd稳态解uC'=US解答为uC(t)=uC'+uCuC(0-)=U0非齐次方程=RCtSCeUuA1.全解=强制分量(稳态解)+自由分量(暂态解)暂态解tCeuAuC(0+)=A+US=U0A=U0-US由起始值定A强制分量(稳态解)自由分量(暂态解)0)(0teUUUutSSCuC-USU0暂态解uC'US稳态解U0uc全解tuc0iK(t=0)US+–uRC+–uCRuC(0-)=U0iK(t=0)US+–uRC+–uCR=uC(0-)=0+uC(0-)=U0C+–uCiK(t=0)+–uRR2.全响应=零状态响应+零输入响应零状态响应零输入响应)0()1(0teUeUuttSCtuc0US零状态响应全响应零输入响应U0二.三要素法分析一阶电路teffftf])()0([)()(0时间常数起始值稳态解三要素)0()(ff一阶电路的数学模型是一阶微分方程,解的一般形式为teftfA)()(令t=0+A)()0(0ff0)()0(ffA1A2例113F+-uCV2)0()0(CCuuV667.01122)(Cus2332CR等033.1667.0)667.02(667.05.05.0teeuttC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)32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)(5tetit0.2sA2)(s2.00)0(1iiAii5)(5.026.1)2.0(226.122)2.0(2.05eiA74.35)()2.0(2tetiit(s)0.25(A)1.262例3+-u(t)155HiL已知:u(t)如图示,iL(0)=0求:iL(t),并画波形.法一0t1iL(0+)=0t0iL(t)=0iL()=1AiL(t)=1-e-t/6A=5/(1//5)=6su(t)12120t(s)V+-155HiL1V0t11t2iL(1+)=iL(1-)=1-e-1/6=0.154AiL()=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=0.437AiL()=2AiL(t)=0.437e-(t-2)/6AiL(t)=0t01-e-t/6A0t12-1.846e-(t-1)/6A1t20.437e-(t-2)/6At2iL(t)=1-e-t/6A0t1155HiLt2+-155HiL2V1t2=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§6-7一阶电路的冲激响应零状态h(t))(t单位冲激响应：单位冲激激励在电路中产生的零状态响应一.由单位阶跃响应求单位冲激响应单位阶跃响应单位冲激响应h(t)s(t)单位冲激(t)单位阶跃(t)dttdt)()()()(tsdtdth零状态h(t))(t零状态s(t))(t证明：1f(t)t)(1)(1)(tttf)(1ts)(1ts)]()([1lim)(0tststh)(tsdtd1注意：s(t)定义在（-，）整个时间轴)(t先求单位阶跃响应令is(t)=)()1()(teRtuRCtCiCRisC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)(teiRCtc)()1(teRdtduRCtC)(t再求单位冲激响应令is(t)=)()1(teRRCt)(1teCRCt)(1teCRCt)()0()()(tfttf0)]([ddtetiRCtc)(1)(teRCteRCtRCt)(1)(teRCtRCtuCRt0iC1t0uCt0C1iCt(1)RC1冲激响应阶跃响应=1=01)(0000dttdtiqcCuc1)0()(tRudtduCcc000000)(dttdtRudtdtduCccuc不可能是冲激函数,否则KCL不成立二.分二个时间段来考虑冲激响应iCRisC+-uC0-0+0+t零输入响应uC(0-)=0电容充电1.t在0-0+间1)]0()0([ccuuC)0(Cu电容中的冲激电流使电容电压发生跳变（转移电荷）2.t0+零输入响应（RC放电）Cuc1)0(icRC+uc-01teCuRCtc01teRCRuiRCtccuCt0C1iCt(1)RC1)(1)()(1teRCtiteCuRCtcRCtc)(tdtdiLRiLLiL不可能是冲激dttdtdtdiLdtRiLL000000)(100LdiLLiL1)0(定性分析：)(tuL100duLLLiiLL1)0()0(LduLiiLLL11)0()0(001.t在0-0+间L+-iLR)(t例20)0(Li+-uL2.t0+RL放电RL01teLitL0teLRRiutLL)(1teLitL)()(teLRtutLRuLiLLiL1)0(+-tiL0L1tuL)(tLR§6-8卷积积分一.卷积积分定义设f1(t),f2(t)t0均为零dtfftftft)()()(*)(20121性质1)(*)()(*)(1221tftftftfdtfftftft)()()(*)(20121)d)(()(021tftftdftf021)()(证明令=t-：0t:t0)(*)(12tftf性质2)(*)()(*)()]()([*)(3121321tftftftftftftf二.卷积积分的应用线性网络零状态e(t))(th(t))(*)()(thtetrr(t)即dthetrt)()()(0物理解释将激励e(t)看成一系列宽度为，高度为e(k)矩形脉冲叠加的。性质4dtftftttf)()()(*)()(*)(筛分性性质3)](*)([*)()(*)](*)([321321tftftftftftf=f(t))()(*)()(*)(000ttftftttttf单位脉冲函数的延时t0)(tee(0)2k(k+1))]2()()[()]()()[0()(ttettete)])1(()([1)(0ktktkek)()(0ktpkek)])1(()()[(0ktktkek第1个矩形脉冲)()0()()0(tpetpeh若单位脉冲函数p(t)的响应为hp(t)第k个矩形脉冲)()()()(ktpkektpkeht0)(tee(0)2k(k+1)t时刻观察到的响应应为0~t时间内所有激励产生的响应的和ttk:脉冲作用时刻t：观察响应时刻t02k(k+1)r(t))()()(0ktpketek激励0)()()(kpkthketr响应脉冲响应kk,d,当tdthetr0)()()(0)()(lim)(kpkkthketr响应k当脉冲)()(lim)(0ktpketekk激励)(t冲激)(th冲激响应积分积分变量（激励作用时刻）t参变量(观察响应时刻)解：先求该电路的冲激响应h(t)uC()=0V)()(2tethts5.0101050063RC00d1)0(tiCusC例1RCiS+–uC已知：R=500k,C=1F,uC(0)=0求：uC(t)A)(2teits00d)(1ttCV1106CA)(tis再计算时的响应uC(t)Ateits)(2d)()()(*)()(0thithtitutSSCV)()22(2teettttdee0)(26102tttteedee02626)1(102102例2)(*)()()()]1()([2)(2121tftftetftttft求dtfftftft)()()(*)(20121解被积函数积分变量参变量f2(t-)10tf2(-)10f1()201f1()f2(t-)021t1f1(t)*f2(t)0t1t’t’t’f2()10f1(-)201-1f1(t)*f2(t)0t1t’t’-1t’卷移乘积f1(t-)01-1t2f2()f1(t-)01-1t21tetttftf*)]1()([2)(*)(21由图解过程确定积分上下限2011e-(-)t01e-(t-)t0)(0tftttteetft22d2)(100)(ttteeetft22d2)(1)1(10)(tt0)(0tfttteetft