1、一阶电路习题课一阶电路习题课一、一阶电路全响应的计算一、一阶电路全响应的计算三要素:初始值、稳态值、时间常数三要素:初始值、稳态值、时间常数二、全响应的的两种分解方式:二、全响应的的两种分解方式:a a、零输入响应和零状态响应、零输入响应和零状态响应b b、稳态(强制)分量和暂态(自由)分量、稳态(强制)分量和暂态(自由)分量三、一阶电路冲激响应的计算三、一阶电路冲激响应的计算一、计算下图电路中电压、电流的初始值一、计算下图电路中电压、电流的初始值 1.S(t=0)100.1H2050.2H3VuL1uL22.S(t=0)460.60.1F1Ai1i2i3求求:i1(0+),i2(0+),i3
2、(0+)。3.S(t=0)20164016V400.5FicuL1H求求:uL(0+)和和ic(0+)。(t)A4.221Fuc求求:uc(0+)。一、计算下图电路中电压、电流的初始值一、计算下图电路中电压、电流的初始值 1.S(t=0)100.1H2050.2H3VuL1uL2iL1iL2解解:AiiLL1.0303)0()0(11 AiiLL0)0()0(22 0+电路电路 102053VuL1uL20.1AVuL21.020)0(2 VuL01.0)1020(3)0(1 uC求:求:i1(0+),i2(0+),i3(0+)。解:解:Vuucc616)0()0(2.S(t=0)460.60
3、.1F1Ai1i2i3Ai26/46.06)0(3 Aii2.1646)0()0(31 Aii8.0644)0()0(32 0+电路电路 460.6i1i2i36ViL3.S(t=0)20164016V400.5FicuL1Huc求求:uL(0+)和和ic(0+)。解解:Vuucc8)80/20(16)80/20(16)0()0(AiiLL0)0()0(0+电路电路20164016V40icuL8VVuL44040408)0(Aiiiic0 80820816816 )0(321 i2i3i1ic(t)A4.221Fuc求求:uc(0+)。解解:tddutdduCiccc )(22ttdduut
4、dduccc )(212tutdducc 两边积分两边积分,从从0-0+1)0()0(2 ccuuVuucc2121)0()0(方法二方法二0-到到0+,先求电容中电流先求电容中电流,此时此时uc(0-)=0。Attic)(21222)(VCtdiCuuccc21211)0()0(00 方法一方法一ic二、二、t=0时开关时开关S闭合,求闭合,求uc(t)并定性画出其波形并定性画出其波形。S(t=0)9V6V2k1k1k200Puc(t)用三要素解用三要素解Vuucc6)0()0(Vuc491315)(求求:2k1k1kR kR3513211/2)0()()0()()(teuuututcccc
5、)0(104 )4(64)(66103103 tVeVetuttc0t)(tuc 6V-4VsRC610310311021035 ic另求另求ic(t)并定性画出其波形并定性画出其波形S(t=0)9V6V2k1k1k200Puc(t)用三要素解用三要素解Vuucc6)0()0()0()()0()()(teiiititcccc)0(6 060)(66103103 tmAeetittc0t)(tic-6m A9V6V2k1k1k6Vic0 0+电路电路mAic66.36.32.132321932165.05.26)0(0)(ci三、三、t=0时开关时开关S闭合,求闭合,求u3(t)。解解:VuuC
6、c5)0()0(ViumAi10553)(110103 )(:加压求流加压求流需求需求求求iR kiuRiiui20205)31(0.13()10 15(0)tu teVt 500 F10k5k+uC+10V5V+i3i+u3S(t=0)VuuC5)0()0(3 10k5ki3i+u3+uisRC1010500102063 四、已知:四、已知:求求uc(t)。VuFCkRkRAttiVtucss50)0(,10,2,1)(314sin1.0,)(10021 usisR1R2Cuc求零输入响应:求零输入响应:uc1(t)求零状态响应:求零状态响应:uc2(t)SCRRVuVuuccc221111
7、032)/(0)(,50)0()0()0(50)(1501 tVetutc电压源单独作用电压源单独作用sccVuVu2221032,7.66)(,0)0()0()1(7.66)(1502 tVetutcusR1R2Cuc电流源单独作用电流源单独作用scVu221032,0)0(isR1R2CucVCjRRCjRRIUSC5.6427.28 )1()/()1)(/(21212 Vtuc)5.64314sin(7.28)(2 Vuc9.25)5.64sin(7.28)(02 Vettutc15029.25)5.64314sin(7.28)(完全响应:完全响应:)()()()()()(22121tu
8、tututututucccccc Veteettt1501501509.25)5.64314sin(7.28)1(7.6650 Vett1502.9)5.64314sin(7.2866.7 )0()()0()()(0 teuuututcccc已知:已知:uc1(0-)=0)30200sin(500100sin10021 tutusst=0时时k由由2合向合向1,求求 i。解解:(1)(1)求求uc2(0-)5001020100162CX457.701005005005002 jjUmCVuc50)0(2 (2)(2)求求i(0+)Vtuc)45100sin(7.702 i250V+-c2 50
9、0-+-50VAi6.0500300)0(us212k+-us1+-c1c2f 20 500i五五.f 20)()3(i求求us212k+-us1+-c1c2f 20 500if 20 25010202001621CCXX75707.050050030500 jImAti)75200sin(707.0)()30200sin(5002 tusus2k+-c1c2f 20 500if 20us212k+-us1+-c1c2f 20 500if 20(4)(4)由初值定常数由初值定常数tAeti200)75200sin(707.0 0829.075sin707.06.0 AAAti)75200sin
10、(707.0)(Ai6.0500300)0(Aetit2000829.0)75200sin(707.0 求求i 电路电路求求i 电路电路t=0时,闭合开关求时,闭合开关求 i。iii +-6H10V6V6 2 iLi”AiL3)0(+-5F4Ais+-10V2 i Vuc16)0(Aeeiiitt25.01.075.6594 ii”+-5F4Ais+-10V2+-6H6V6 2 iL六六.i0t+-1Fici2 2 2 4(t)i(t-3)七七.+-0)0(cu求求ic ,并画出并画出ic的变化曲线的变化曲线。(转折点值表清楚转折点值表清楚)解解:0 t 3264.1)1(2)3()3(33
11、ttcceuuU0ViiU120 1Fi2 2 2 i+-UIAAAAAIIIIU25212 5.2等等R0 t 3Aeitc3 32 Veutc)1(23 利用戴维南等效求利用戴维南等效求ic+-ic1V2.5 1F264.1)3(cu+-Aic0.1060.1062.52.51.2641.2641 1)(3(30)(ci15.2 /)3(106.0 tceiAeitc3 32 0 t 3t(s)ic(A)0.245-0.1060.6673iL+-uL2(t)A1 1 2 1H0)0(Li求求 iL和和uL的冲激响应的冲激响应。解:解:-+(t)1 1H+-uLiL1.t在在 0-_ 0+间
12、间)(tidtdiLL 000000)(dttdtidtdtdiLL iL不可能是冲激函数不可能是冲激函数1)0()0(LLiiL1)0(Li八八.2.t 0+零输入响应零输入响应-+(t)11H+-uLiL1)0(Li)(teitL )(teutL )()(tetutL )(teitL 讨论讨论:AiL1)0(-+(t)1 1H+-uLiL1.t 在在 0-_ 0+间间)(tuL dtuLiiLLL 001)0()0(AiL2)0(2.t 0+)0(2 teitL)(2teutL 注意注意:三要素法应用于直流或正弦电源激励三要素法应用于直流或正弦电源激励电路,其余激励源一般需解非齐次方程。电
13、路,其余激励源一般需解非齐次方程。代入原式代入原式特解为特解为全解全解由由UC(0+)=UC(0-)=0得得有有()StUtUe(0)0CU(指数激励指数激励),),求求K(t=0时时)闭合后的闭合后的()CUt。九九.,通解特解通解特解t-ttCRCCCCdURCUUeUkeUAedt/,1得得tttURCAeAeUeA=-RC1-tCUUeRC/1ttRCCCCUUUUeke-RC/-1-Uk=RC()=1-ttRCCUUteeRCUsRCUcK十:十:如图电路,如图电路,R=1=1,C=1=1F,IS=1=1A,=0=0.5 5,电路已达稳态。,电路已达稳态。求当求当 突变为突变为1 1
14、.5 5后的电容电压。后的电容电压。ISiiRRRucC C解:用三要素法求解解:用三要素法求解1 1)电容电压初始值)电容电压初始值(0)CU(0.5)2SIIII0.43SIIA(0)0.4CURIV2 2)电容电压稳态值)电容电压稳态值(1.5)2SIIII233SIIA23CPURIV3 3)时间常数)时间常数iiRRR(b)iiRR(a)图图(a)(a)电路的入端电阻电路的入端电阻(2)0.5RR图图(b)(b)电路的入端电阻电路的入端电阻013RRRRR013CR电容电压为电容电压为322()0.433CtUte0309LLRL 0.3,LH P30()37.5tu te190CF
15、()U t十一:十一:为电阻网络,开关闭合后,为电阻网络,开关闭合后,零状态响应零状态响应现把现把L换为换为,则零状态响应,则零状态响应为多少?为多少?解:解:L时间常数时间常数001,30LLRR1 1为为 端入端电阻。端入端电阻。得:得:0110CR C()(0)ctU tUUUe稳稳10(0)4.5,(0)3()4.57.5tUUUU te 稳换为电容后换为电容后。1 1讨论:讨论:L时的稳态值等于时的稳态值等于C时初始值(时初始值(短路)短路)1 1L时的初始值等于时的初始值等于C时的稳态值(时的稳态值(开路)开路)设设:C时时1 11LUsPu(t)十二、已知:十二、已知:VURRH
16、LHLS6,1,2,02.0,01.02121 并并定定性性画画出出波波形形。求求:)(),(21titi解:此题为初始值跃变情况解:此题为初始值跃变情况AiAi0)0(,3)0(21 )0()0(21 ii电流发生跃变。电流发生跃变。S(t=0)R1R2L1L2i1i2USSUtdidLtdidLiRiR 221122110)0()0(22)0()0(112211 iiiiidLidL000两边积分:两边积分:002002100120021001tdUtdtdidLtdtdidLtdiRtdiRS0)0()0()0()0(222111 iiLiiL)0()0()0()0(22112211 i
17、LiLiLiL回路磁链守恒回路磁链守恒)0()0()0()0(22112211 iLiLiLiL)0()0(21 ii(KCL)(磁链守恒磁链守恒))0()0()0()(2211121 iLiLiLLALLiLiLii103.0301.0)0()0()0()0(21221121 sRRLLAii01.0,2126)()(212121 Atettit)()2()(3)(100112i20ti13V V)()(tet0.02tdidLu100t11L1V)()(t2et0.02 tdidLu100t22L2SLLUuuiRiR 21221162)(02.0)(02.0224100100100100
18、 tttteteteeAtetit)()2()(1002 电容电压初值一定会发生跃变。电容电压初值一定会发生跃变。解解 V1)0(1 EuC0)0(2 Cu合合k前前)0()0()0(21 CCCuuu合合k后后tuCtuCiCCdddd2211 ttuCtuCtiCCd)dddd(d00221100 )0()0()0()0(0222111 CCCCuuCuuC十四十四.已知图中已知图中E=1V,R=1 ,C1=0.25F,C2=0.5F。求:求:uC1、uC2、iC1 和和 iC2 并画出波形并画出波形。)0()0()0()0(22112211 CCCCuCuCuCuC)0()()0()0(
19、212211 CCCuCCuCuC节点电荷守恒节点电荷守恒q(0+)=q(0-)iiC1iC2ERC1C2+-uC1+-uC2k(t=0)iiC1iC2+V315.025.0125.0)0(211CCECuC可可解解得得uC()=1V 0321)131(1)(3434teetuttC =R(C1+C2)s43 0)0(1)0(021 CCuut)(92)(61)(98)()321()(41)(98)()321()(41dd34343434111tettetttetettuCittttC )()321()()(341tettutC )()321()(342tetutC uC2ut01/31uC1
20、-1/6 2/9iC1it 4/91/6iC2tuCiCdd222)()321()(342tetutC )()321()()(341tettutC )(92)(61341tetitC )(94)(6134tett )(98)()321(2134tett iiC1iC2i 无冲激无冲激十五、十五、求电感电流的零状态响应求电感电流的零状态响应iL(t)t0 (t)1 1/3 0.5F0.5HiL+iCiR解解dtdidtdiiLLR5.1315.0 15.05.011 dtdidtiiLCC代入、整理代入、整理 得得0222 LLLidtdidtid特征方程特征方程 p2+2p+1=0特征根特征根
21、 p=p1=p2=-1初值初值iL(0+)=0,uL(0+)=0.25VtLetAAti )()(21A1=0,A2=0.505.0)(tAtetitLiL(0+)=05.05.0/25.00 dtdiLLRCiii ttLeAetAAtdid 221)(十七、十七、已知零状态电路中已知零状态电路中,电流源电流源is=2sin0.5t (t)A,电压源电压源 us=5(t)V,求求:(1)电容电压电容电压uC(t);(2)电压源电压源us单独作用时电容电流单独作用时电容电流iC(t)。解解1)一阶电路用叠加定理进行分析一阶电路用叠加定理进行分析is单独作用时电容电压单独作用时电容电压2 2 1
22、FiS+uCiCuC1(0+)=0 =2s45222)2(20221 jjUC)455.0sin(22)(1 ttutC02)455.0sin(22)455.0sin(220)455.0sin(22)(2201 tVetetttutttC2 2 1FiSuS+uCiC2 1FuS+uCiCus单独作用单独作用)(5222tdtduuCC 5)0()0(222 CCuuuC2不是冲激不是冲激uC2(0+)=2.5VVtetutC)(5.2)(212 05.4)455.0sin(22)()()(221 tVettutututccC2)求求us单独作用时电容电流单独作用时电容电流iC(t)Atett
23、edtddtduCtittCC)(25.1)(5.2)(5.2)(5.05.0 十八、十八、已知线性网络零输入响应为已知线性网络零输入响应为8e-t。当激励当激励e(t)=(t)作用时作用时,网络产生的响应为网络产生的响应为4(1+e-t)t0。求当激励求当激励e(t)=0.5 e-3t (t)作用作用时网络产生的响应。时网络产生的响应。解解:单位阶跃响应:单位阶跃响应(零状态零状态)为为0)44(8)1(4 teeetttttee3 deeuthtytts)(0345.0*)()(零状态零状态)0(98)(33 teeeeetyttttt)(4)()44()(tetedtdthtt 单位冲激响应单位冲激响应(零状态零状态)为为激励激励e(t)=0.5 e-3t (t)作用时网络产生的作用时网络产生的零状态响应零状态响应全响应全响应