范世贵电路基础2版答案(第13章)

word文档精品文档分享第十三章一阶电路时域分析13-1图题13-1所示电路，t<0时K一直在0点。今从t=0时刻开场。每隔T秒，依次将K向左扳动，扳道4点是长期停住。试画出u(t)的波形，并用阶跃函数将u(t)表示出来。答案解：u(t)的波形如图13-1〔a〕所示。13-2粗略画出以下时间函数的波形。〔〕tU(t)；(2)tU(t+1);(3)(t-1)U(t-1);(4)-tU(t);(5)tU(t-1)word文档精品文档分享(6)U(t-1)U(t-2);(7)U(t)+U(t-2);(8)U(-t+3);(9)tU(3t+1(t)U(t));(10)(t)U(t1)(11);(12)5t(1)eUt;(13)U(t-1)-U(t-4)。答案解：各波形相应如图题13-2所示。word文档精品文档分享13-3求以下导数：dd(1〕；〔2〕；[u(t)U(t1)][u(t)U(t1)]dtdtddt5t(3)；(4)；[eU(t)][eU(t4)]dtdt(5)2d2[()]tUtdt答案1〕(t)(t1)；〔〕(t1)；word文档精品文档分享(3)；(4)；t5t5t(t)eU(t)e(t4)5eU(t4)(5)。(t)13-4写出下表格单一元件电路的单位阶跃响应i(t)u(t)的表达式。画出波形。电路响应的表达式形波1i(t)U(t)Lu(t)U(t)i(t)C(t)u(t)U(t)i(t)U(t)u(t)L(t)i(t)U(t)1u(t)U(t)Cword文档精品文档分享13-5写出下表中个单一元件电路的单位冲激响应i(t)、u(t)的表达式，画出波形。响电路应的表达式波形1i(t)U(t)Lu(t)(t)'i(t)C(t)u(t)(t)i(t)(t)'u(t)L(t)i(t)(t)word文档精品文档分享1u(t)U(t)C13-6写出下表中各一阶电路单位阶跃响应i(t)、u(t)的表达式，画出波形。电路响应的表达式形波11ti(t)(1e)U(t)R1tu(t)eU(t)L1ti(t)(1e(t)1tu(t)ReU(t)L11ti(t)eU(t)R1tu(t)(1e)U(t)C1ti(t)eU(t)1tu(t)R(1e)U(t)Cword文档精品文档分享i(t)13-7图题13-7电路，鼓励的波形如图所si(0)0示，。求响应。答案i(t)U(t)U(t2)Au1(t)解：U(t)产生的响应按图s题13-7(a)用三要素法求，即51Su1(0)5V，u1()0,5tu1(t)5eU(t)V故。i(t)故产生的响应u(t)为st(t1)(t2)5eU(t)5eU(t1)10eU(t2)Vu(t)=。word文档精品文档分享13-8一阶线性定常电路，在一样的初始条件下，当鼓励为f(t)时，[t<0时，f(t)=0]，其全响应为ty1(t)2ecos2tt0；当鼓励为2f(t)ty2(t)e2cos2t；t0求鼓励为4f(t)时的全响应y(t)。答案y(t)y1(t)f(t)的零状态响应为2f(t)xy1(t)产生的零状态响应为2。故有ty(t)y(t)2ecos2tx1ty(t)2y(t)e2cos2tx1ty(t)3eU(t)联解得，xtytet1()cos2鼓励为4f(t)时，全响应为tty(t)y(t)4y(t)3e4(ecos2t)x1t(e4cos2t)U(t)。13-9图题13-9电路，t<0时,K闭合电路已达稳定状态今于t=0时断开开关u(t)u(t)tK,求t>0时的全响应c及c经过零值得时刻o。word文档精品文档分享答案u(0)3V解:t<0时,K闭合,电路已达稳定状态,故有。t>0时K翻开,故有cu(0)u(0)u()RC2Sccc，。故得0.5tu(t)(13e)U(t)Vc又u(t)013eco0.5tot2.2S解得o。u(t)c的波形如图题所示。13-10图题13-10电路,t<0时K闭合电路已达稳定状态。求t>0u2(t)时的响应，并画出波形。word文档精品文档分享答案解：t<0时，K闭合，电路已达稳定状态，故有Usi(0)R1。t>0时,K翻开,故有Usi(0)i(0)R1R2u(0)Ri(0)U22R1sR2u()U2RR12s故11RRtt22u(t)U(1e)U(t)UeU(t)V2ssRRR121。13-11图题13-11电路，t<0时K翻开，电路已达稳定状态。今于t=0时闭合ut()()iti(t)K，求t>0时的响应、、，画出波形。cLword文档精品文档分享答案解：t<0时，K翻开，电路已达稳定，故有i(0)0u(0)Lc，。t>0时，K闭合，此时有i(0)i(0)0uc(0)uc(0)LL，；2i()2AL1u()0，；c1212S210.50.5S1t故得i(t)(1e(t)A1L1tu(t)2e)U(t2cdu(t)c2ti(t)C2eACdt1t2ti(t)i(t)i(t)[2(12e)2e)U(t)A1LCword文档精品文档分享u2(0)013-12图题13-12电路，t<0时K翻开，电路已达稳定状态，且设。u2(t)今于t=0时闭合K，求t>0时的响应，画出波形。答案解:t<0时K翻开,电路已达稳定状态,故有u1(0)5Vu2(0)0；。t>0时K闭合,此时有u1(0)u2(0)，C1u1(0)C2u2(0)C1u1(0)C2u2(0)。uuV1(0)2(0)3联解并代入数据得。5R(CC)S12u1()u2()5V3又，。0.6tututeUtV1()2()(52)()故得。其波形图如图题13-12(a)所示。word文档精品文档分享i(0)0i(t)13-13图题13-13所示电路。求,画出波形图。答案解：求i(t)的等效电路电压源电路如图题13-13(a)所示。其中u(t)56U(t)ocR014。7ti(t)4(1e)U(t)A于是用三要素法可求得。word文档精品文档分享it1()i(0)2A13-14图题13-14所示电路，。求u(t)和,画出波形图。答案i(0)i(0)2Ai()0解：;由图题有：U3I(0.5UI)1ssssULsR080.5SIR0故得s，。故得2ti(t)2eU(t)Adi(t)2tu(t)L16eU(t)Vdt2ti(t)0.5u(t)i(t)10eU(t)A1word文档精品文档分享u(0)013-15图题13-15所示电路，，求u(t)。答案u(t)12U(t)V解：求u(t)的等效电压源电路如图题所示。其中，oc010。于是用三要素法可求得10tu(t)12(1e)U(t)V。13-16图题13-16所示电路，N内部只含有点源与电阻，零状态响应110.25tu(t)(e)U(t)'ou(t)?28伏。今把2F得电容换成2H电感,问响应o提示:接电感时的初始值与接电容式的稳态值相等。word文档精品文档分享答案解:因有'1u(0)u()Voo2'115u()u(0)Voo2881RC0.25又因有故得R10.25C2L2'R2故又得1S故得1t''''51t'u(t)u()[u()u(0)]e(e)U(t)Voooo88word文档精品文档分享13-17图题13-17所示电路,求响应i(t)。答案u(t)14(t)V解：求i(t)的等效电压源电路如图题所示,其中，oc014故得1tL117ti(t)i(0)e14eU(t)AR7SL0。其中u(0)2i1(t)i(t)u(t)13-18图题13-18所示电路,c求全响应、C、C。word文档精品文档分享答案u(t)R解：按图题13-18(a)和(b)分别求等效电压源的电压和,即oc0u(t)4U(t)ocUsR2.4I，。其等效电压源电路如图题s所示。于是用三要素法可求得1tu(t)u()[u()u(0)]ecccc1t(42e(t)V其中66RCS02.41102.410又1du(t)tci(t)C0.8333eU(t)ACdti1(t)12iC(t)0.25u1(t)10iC(t)0.25i(t)1故1t1(t)(80.667e)U(t)A得。word文档精品文档分享13-19图题13-19所示电路,t<0时K在a点,电路已达稳定.今于t=0时将K扳道u0(t)b点，求t>0时的全响应。答案解t<0时,K在a点,电路已达稳定.1U0Vs21ZR6j6jC6245U1sU0.545VCZjCword文档精品文档分享故u(t)0.52cos(2t45)Vc故u(0)0.52cos(45)0.5Vct>0时,K扳到b点.用三要素法可求得:uuuV0(0)c(0)c(0)0.5u0()uc()3V，1RCS12故得12tu0(t)3(30.5)(32.5e)U(t)V13-20图题13-20所示电路,求u(t)。答案解:求u(t)的等效电路如图题13-10(a)所示。136u(0)10V0.10.2SCu()036故，，。1t5tu(t)u(0)e10eU(t)V故得。word文档精品文档分享13-21图题13-21u(t)。答案解：求u(t)、i(t)的等效电压源电路如图题13-21(a)所示。其中u(t)(t)Voc01，。故有L11Si(0)1ALi()0R0，，。故得1tti(t)i(0)eU(t)A故di(t)tu(t)(t)eU(t)Vdtword文档精品文档分享13-22图题13-22所示电路，求u(t)。答案解：t>0时的等效电路如图题13-22(a)所示。1u(0)0.2C1C1，u2(0)u2(0)0CCu(0)u(0)u(0)C1C2u()0，CC12R6SCC121t6u(t)1eU(t)V故。word文档精品文档分享13-23图题13-23所示电路，t<0K接在＂１＂，电路已达稳定。今于t=0时将K接道＂２＂。今欲使t>0时电路中的电流只存在正弦稳态响应，u(t)10cos2tV问应选多达数值？。答案解：t<0时K故有1i(0)AR11i(t)t>0时K在２。先求零输入响应：x1i(0)i(0)AxR11L0.5Si()0x，R212ti(t)eU(t)AxR1故得1。i(t)f再求零状态响应：10UV2ZRjLj211245word文档精品文档分享UI545AZi(t)52cos(2t45)A故得正弦稳态响应为s。1t2ti(t)BeBeU(t)A其瞬态响应为t。2ti(t)i(t)i(t)52cos(2t45)Befst故零状态响应为。i(0)052cos(45)B故f，B5故得故2ti(t)52cos(2t45)5eU(t)Af故全响应为12t2ti(t)i(t)i(t)e52cos(2t45)5exfR111050i(t)52cos(2t45)A11可见，欲使，那么必须有，01故得。13-24图题13-24所示电路，为使全响应电路i(t)中的瞬态响应分量为零。求i(0)的。答案1t2ti(t)i(0)ei(0)eU(t)A解：零输入响应为xxx，word文档精品文档分享L50.5SR10其中。i(t)零状态响应为f：10UVs2Z10jL11.1826.57UsI0.63326.57AZ故得正弦稳态响应为i(t)0.6332cos(t26.57)As自由响应为2ti(t)BeU(t)Af故得零状态响应为2ti(t)i(t)i(t)Be0.6332cos(t26.57)Afts故i(0)0B0.6332cos(26.57)fB0.797故得故得2ti(t)0.797e0.6332cos(t26.57)Af故得全响应为2ti(t)i(t)i(t)i(0)e0.7970.6332cos(t26.57)Axfi(t)0.6332cos(t26.57)Ai(0)0.797故欲使，那么必须有A。u(t)u(t)13-25图所示电路，e(t)的波形如图(b)所示。求零状态响应和。CRword文档精品文档分享答案解：单位冲激响应th(t)eU(t)故u(t)h(t)e(t)