潘双来电路基础第二版课后习题答案与分析第八章作业

图题8-1(a)-+Cu1i-+S01F.042A)0(t412i8-1(1)电路如图(a)所示，试列出求)(Ctu的微分方程。tuCidd2C1tuCidd2C1042)4(2111iiuiC12010ddCutuC解：A5)0()0(LLii17V1231154)0()0(CCuu-+LuCiS35A)0(t41-+CuLi2HF10-+12V8-3(c)换路后：解：)0(312)0()0(4)0(CCLLiuiuV90121754tiLuddLLS/A5.4921)0(1ddL0LuLti-+LuCi35A4-+CuLi2HF10-+12V8-3(c)055)0(5)0(LCii8-3(c)-+LuCi35A4-+CuLi2HF10-+12V换路后：解：S/A5.4921)0(1ddL0LuLtituCiddCC0)0(1ddC0CiCtu055)0(5)0(LCii8-6(b)解：-+ui-+SF23k9V)0(t6k3k6ks10910210263363V3ee)0()(31091ttutu)0(t)0()0(uu3V363963698-6(b)解：-+ui-+SF23k9V)0(t6k3k6kV3ee)0()(31091ttutu)0(tA)e1091(106dd)(3109136ttuCtimAe3231091t)0(t8-8(b)解：)0(t-+uLi0.1H2L2i1-+ui2i21iiiiu5)2(25iuRs02.051.0RLAe2e)0()(05LLttitiVe6)(3)(05Lttitu8-11(b)解：)0(t-+oCU1i-+42A412i-+u1i-+4412iiRV12624111ociiiU111102)44(iiiu101iiuRs10iCR)Ve1(12)e(1)(1.0ocCttUtu8-17解：oCUI-+100V20V50-+-+205V7550202050100205020OCIU1520//205iRA51575)(iOCLRUis152/iRL8-17A51575)(iOCLRUis152/iRLLu-+S100V2H20)0(tV50Li-+-+205A4520100)0()0(LLiiAe5))]e()0([)()(5.7LLLLttiiitiVe15e5.72d)(d)(5.75.7LLttttiLtu8-20解：-+Cu-+SA2F5.0)0(t1201V10-+CuA2F5.001A1V10)0()0(CCuuV10)21(10)(Cus55.010RCVe2010))]e()0([)()(2.0CCCCttuuutu换路后：8-28（a）解：(a))(tfo1t1223(a)3)(2)()1()()(tttttfo1t1)(tf(b)(b))1()1()()(tttttf8-38（1）0)0(Cu)()(sttu1.图示电路中，，试求：（1）时，电流i2的阶跃响应。VRRRuC212)(CsRRRC)2//1(VteututCC)()1)(()(VtReuRtutitCc)()1)(()()(2228-47电路在开关S闭合前已达稳态，已知uc(0_)=-100V,求电流0),(ttiLdtduCuiuiccccL10102210dtudLCdtduLdtdiLuCCLLcCCCcudtudLCdtduLdtduCu)10()10(3020022200404.010*4224cCCudtdudtud200404.010*4224cCCudtdudtud20021ppA)(520021tLetKKi55)0(1KiL1500)0(200)0('21LuKKiLL0tA,15005200tLtei8-41（a）试判断图示两电路的过渡过程是欠阻尼还是过阻尼的。LCLRLRp12222,1821012LCLR过渡过程是临界阻尼。su1u15ucuLi64H1F18-56写出标准形式的状态方程LccsidtduCuu4))(5(6csLLcuuidtdiLuSLCLC3041631141dddduiutitususicuLi2R1RLC8-57写出标准形式的状态方程LccsidtduCRuu1)(2sLLciiRdtdiLussiuLRCRiuLRLCCRtitu21LC21LC001111dddd原8-57写出状态方程和输出方程。解：选uC、iL1和iL2作为状态变量。C1R1Lsu-+2L2RCu-+L2iL1iIII1-+-+ou-+)(ddL2L1CiituCL2L1C11ddiCiCtuL11SCL11ddiRuutiLS1L111C1L111dduLiLRuLti解：标准形式状态方程S1L2L1C222111L2L1C0100101110dddddduLiiuLRLLRLCCtititu8-57写出状态方程和输出方程。解：以uo为输出量的输出方程C1R1Lsu-+2L2RCu-+L2iL1iIII1-+-+ou-+L22oiRu8-57写出状态方程和输出方程。