1.通过例子,掌握求解二阶电路的方法、步骤。2.通过例子得出二阶电路的一般规律。*本节讨论的具体结果只适用于本例,不能套用到其它电路。但得出的规律具有一般性。第七章二阶电路二阶电路的零输入响应,零状态响应和全响应重点掌握学习方法§7.1二阶电路的零输入响应CLuRiu0dddd22CCCutuRCtuLCptcAeu设012RCPLCP特征方程为LCLRRP2/422,1LCLRLR1)2(22uC(0+)=U0i(0+)=0已知求uC(t),i(t),uL(t).tuCiCddRLC+-iucuL+-(t=0)解2C2ddddtuLCtiLuL根的性质不同,响应的变化规律也不同二个不等负实根2CLR二个相等负实根2CLR二个共轭复根2CLRLCLRLRp1)2(222,1tptpCeeu2121AAtpCetu)AA(21)sin(tKeutC不等的负实根一,2.21ppCLRtptpCeeu2121AA0210AA)0(UUuC0AA0)0()0(dd2211PPCituC0121201221AAUPPPUPPP)(2112120tptpCePePPPUuRLC+-iucuL+-(t=0))(2112120tptpCePePPPUuU0tuc设|P2||P1|1202PPUP|P1|小1201PPUP|P2|大)()()()(21211202121120tptptptpCeePPLUeppeppPPCUdtduCit=0+i=0,t=i=0t=tm时i最大0ttmi增加,uL0ttmi减小,uL0t=2tm时uL极小)()(2121120tptpLePePPPUdtdiLu2tmuLtmiuL(0)=U0t2tmuL衰减加快012RCPLCPt0i0tU0uc)(2112120tptpCePePPPUu由uL=0可计算tm02121tptpepeptpptptpeeepp)(2211212)(1221lnpppptm由duL/dt可确定uL为极小值的时间t0212221tptpepeptpptptpeeepp)(211212mtppppt2)ln(12221)()(2121120tptpLePePPPUu能量转换关系0ttmuc减小,i增加。ttmuc减小,i减小.RLC+-RLC+-tU0uctmi0非振荡放电过阻尼2.CLR二特征根为一对共轭复根LCLRLRP1)2(22jP2LR令uC的解答形式:)(212121tjtjttptpCeAeAeeAeAu22022)2(1LRLC衰减因子固有振荡角频率(阻尼振荡角频率)0无阻尼振荡角频率虚数实数tjtetjtetjtjsincossincos]sin)AA(cos)AA[(2121tjtetp1,p2共轭A1,A2也共轭)AA(AA212121tjtjttptpCeeeeeu)sin(tKet)AA(21tjtjtCeeeu]sinBcosA[ttet01221AUPPP01212AUPPP式中0BA0)0(ddtuC由初始条件00A)0(UUuC]sinBcosA[tteutC00BAUU)sin(cos0ttUeutC或)sin(00teUt]osBinA[]sinBcosA[ddtctsettetuttC)sincos(0000ttUeutC0)sin(cos0tteUutC0teULteULCLteULCLCLteUCCtuCittttCsin1sin)(1sin)(1sin)(dd002002202)sin()cossin()cossin(1dd0000000teUtteUteteUtiLuttttLteLUtuCitCsindd0)sin(dd00teUtiLutLuL零点:t=,+,2+...n+i零点:t=0,,2...n,i极值点为uL零点。uLuC-2-uctU0teU00teU0002i+)sin(00teUutCuC零点:t=-,2-...n-,uC极值点为i零点。t--ti+uct-2-2U0teU00teU000uCRLC+-RLC+-能量转换关系0tRLC+-uC减小,i增大uC减小,i减小|uC|增大,i减小衰减振荡欠阻尼特例R=0LCutUu)2sin(00tLC+-02,1jp210,LC0tLUi000sinteLUtuCitCsindd0)sin(dd00teUtiLutL)sin(00teUutC0等幅振荡无阻尼LRPPP221)(21tAAeutC010)0(UAUuC解出0201UAUA)1(dddd)1(000teUtiLuetLUtuCietUutLtCtC相等的实根三,2.21ppCLR0)(0)0(dd21AAtuC由初始条件非振荡放电临界阻尼小结:非振荡放电过阻尼,2CLRtptpceAeAu2121振荡放电欠阻尼,2CLR)sin(tAeutc非振荡放电临界阻尼,2CLR)(21tAAeutc可推广应用于一般二阶电路定积分常数)0()0(dtduuCC由电路所示如图t=0时打开开关。求:电容电压uC,并画波形图。解(1)uc(0-)=25ViL(0-)=5A特征方程为50P2+2500P+106=013925jP0dd25]dd[dd.50)3(CCCutuCtuCt)139sin(25tKeutC例15ΩμF20Ω10Ω10Ω0.5H10050V+-uc+-iL5Ω20Ω10Ω10Ω50V+-iL+uC-0-电路20Ω10Ω10Ω+-25V5AiC0+电路(2)uc(0+)=25ViC(0+)=-5A20Ω10Ω10Ω+-uCLCt0电路)139sin(25tKeutC4105sin25cos13925sinKKK176358,K0V)176139sin(35825tteutC(4)525)0(dtduCuCC由uCt035825左图为有源RC振荡电路,讨论k取不同值时u2的零输入响应。节点A列写KCL有:dtduCRui111211111)(1)(uudtdtduCRuCdtduCRuRKVL有:0)3(2211212CRudtduRCkdtud整理得:例2u2u1ku1i2i3i1RCRCA++1111111KuuudtuRCdtduRCu特征方程013222CRPRCkP0)3(2211212CRudtduRCkdtud22)1()23(23RCRCkRCkP特征根(1)特征根为实数即23)1()23(22kRCRCk时为非振荡过程和即51kk)sin(1tAeut1k30衰减振荡3k50增幅振荡k=3=0等幅振荡)sin(01tKu(2)特征根为共轭复数即23)1()23(22kRCRCk时为振荡过程即51k22)1()23(23RCRCkRCkP令RCk23求所示电路中电流i(t)的零状态响应。i1=i-0.5u1=i-0.52(2-i)=2i-2由KVLidtdidtiii262)2(211整理得:1212dd8dd22ititi二阶非齐次常微分方程解:第一步列写微分方程2-ii1+u1-0.5u12W1/6F1Hk2W2W2Ai§7.2二阶电路的零状态响应和全响应一.零状态响应第二步求通解iP1=-2,P2=-6iii解答形式为:1212822idtdidtid第三步求特解i”稳态模型i=0.5u1u1=2(2-0.5u1)u1=2Vi()=1AtteAeAi6221'+u1-0.5u12W1/6F1Hk2W2W2AiP2+8P+12=0+u1-0.5u12W2Wi2AtteAeAi62211得零状态响应第四步求初值)0(1dd0)0()0(0LuLtiiiVuuuL825.0)0(11+u1-0.5u12W1/6F1Hk2W2W2Ai0+电路模型:0.5u+u1-12W2W2A+uL-V4221u第五步定常数tteAeAi62211212162810AAAA5.15.021AA0A5.15.01)(62teetitt+u1-0.5u12W1/6F1Hk2W2W2Ai8)0(1)0(dd0)0()0(LuLtiii二.全响应已知:iL(0)=2AuC(0)=0R=50W,L=0.5H,C=100F求:iL(t),iR(t)。解(1)列微分方程50dddd22LLLRitiLtiRLCtuCiRtiLLLdddd-50C4422102102dd200ddLLLititiRLCiRiLiC50Vt=0+-uL+-uCtiLuuLLCdd(2)求通解(自由分量)0200002002PP特征方程特征根P=-100j100)100sin(1)(100tKetitL全解(3)求特解(强制分量,稳态解)AiL14422102102dd200ddLLLititi)100sin()(100tKetitL通解(4)求全解(4)由初值定积分常数0cos100sin10002sin12)0(0KKdtdiKiLLo452K得0)45100sin(21)(100tAtetitLiL(0+)=2A,uC(0+)=0(已知))100sin(1)(100tKetitL全解0)0(1)0(1dd0CLLuLuLti)100cos(100)100sin(100dd100100tKetKetittL(5)求iR(t))100sin(1)(100tKetitR解答形式为:由初始值定积分常数150)0(50)0(CRuiRuicR50)0(1CiRCRLCiRiLiC50Vt=0+-uL+-uCRtutiCR00dddd0+电路RiR50V2AiC200101005016R=50WC=100FAiC1)0((5)求iR1)0(Ri200dd0tiR200sin100cos1001sin1KKK20K0100sin21)(100tAtetitR)100sin(1100tKeitR1.一阶电路是单调的响应,可用时间常数表示过渡过程