chapter7一阶电路v10

1、 Linear Circuit Analysis Timing is a critical aspect of a manufacturing process.In practice, the linear increase in voltage is approximated by the “linear” part of an exponential response of an RC circuit.Exponential approximationLinear sweep or sawtooth waveformtAll these applications utilize a sig

2、nal having sawtooth shape and called a linear voltage sweep.Signal generationFirst-Order RL and RC Circuits1. What is a first-order circuit?A first-order circuit is characterized by a first-order differential equation. It consists of resistors and the equivalent of one energy storage element.Typical

3、 examples of first-order circuits:(a) First-Order RL circuit(b) First-Order RC circuitUSLR+_CRUS+_+_SU1R2RL( )aSIRC( )b+_SURLC( )c+_SUR2iiC( )d+_SUR2L1L( )eL+_SURCInterconnections of sources, resistors, capacitors, and inductors lead to new and fascinating circuit behavior.CRUS+_i+_CuFirst-Order RC

4、circuitA loop equation leads toSCURiusinceCduiCdtCSCduURCudtor , equivalently,11CCSduuUdtRCRCThis equation is called Constant-coefficient first-order linear differential equationApply duality principle,11LLSdiiIdtGLGL2. Some mathematical preliminariesThe first-order RL and RC circuits have different

5、ial equation models of the form11CCSduuUdtRCRC11LLSdiiIdtGLGLRC circuit first-order differential equationRL circuit first-order differential equation00( )( )( ),( )dx tx tf tx txdtor , equivalently,00( )( )( ),( )dx tx tf tx txdtvalid for tt0, where x(t0)=x0 is the initial condition. The term f(t) d

6、enotes a forcing function. Usually, f(t) is a linear function of the input excitations to the circuit.The parameter denotes a natural frequency of the circuit.00( )( )( ),( )dx tx tf tx txdt00( )( )( ),( )dx tx tf tx txdtThe main purpose of this chapter is to find a solution to the differential equa

7、tion.orThe solution to the equation for tt0 has the form00()()0( )( )( )tt tttx tex tefd(1) Satisfies the differential equation(2) Satisfies the correct initial condition, x(t0)=x0 Integrating factor method00( )( )( ),( )dx tx tf tx txdtFirst step, multiply both sides of the equation by a so-called

8、integrating factor et.( )( )( )tttdx teex tef tdtBy the product rule for differentiation,( )tdex tdt( )( )ttdx teex tdt( )tef tIntegrate both sides of the equation from t0 to t as follows 0( )ttdexdd0( )ttex00( )( )ttex tex t0( )ttefd000( )( )( )ttttex tex tefd00()()0( )( )( )tt tttx tex tefd3. Sour

9、ce-free or zero-input responseLR( )aRC( )bA parallel connection of a resistor with an inductor or capacitor without a source.In these circuits, one assumes the presence of an initial inductor current or initial capacitor voltage.(a) KCL implies(b) KVL impliesRLii RiLi+_LuCi+_Cu+_RuLLRudiLiRR dtLLdiR

10、idtL RCuuCRCduui RRCdt 1CCduudtRC Both differential equation models have the same general form,( )1( )( )dx tx tx tdt 1LiGL ( )1( )( )dx tx tx tdt The solution to the equation for tt0 has the form0()0( )( )t tx tex t00( )t tex tWhere is a special constant called the time constant of the circuit. The

11、 response for tt0 of the undriven RL and RC circuit are, respectively, given by0()0( )( )Rt tLLLiteit01()0( )( )t tRCCCuteutRC circuitRL circuitRCLGLRThe time constant of the circuit is the time it takes for the source-free circuit response to drop to e1=0.368 of its initial value.For more general c

12、ircuits, those containing multiple resistors and dependent sources, it is necessary to use the Thevenin equivalent resistance seen by the inductor or capacitor in place of the R.LinearResistivecircuit NoindependentsourcesLiLLinearResistivecircuit NoindependentsourcesC+_CuThevenin equivalentLiLthR+_C

13、uCthRThevenin equivalent0()0( )( )tht tLLLRiteit01()0( )( )tht tCCCRuteutExample 1. For the circuit of the figure, find iL(t) and uL(t) for t 0 given that iL(0)=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5 resistor over the time interval 0.4, ). 8H+_Lu20Li50.4ts

14、SUSC+_u(t)+_RK ( )/u tV/ t s0US过渡期为零过渡期为零USR+_u(t)K +_（1）问题的提出问题的提出( )/u tV/ t s0US换路换路换路换路一阶电路的初始条件一阶电路的初始条件充电过程充电过程放电过程放电过程（2）电路的初始条件）电路的初始条件 t0 和和 t 0 的概念的概念认为换路在认为换路在 t 0 时刻进行时刻进行换路前一瞬间换路前一瞬间换路后一瞬间换路后一瞬间t 0t 0初始条件初始条件为为 t 0 时，时， u、i 及其各阶导数的值。及其各阶导数的值。0t 0t 0t0t0t0t(0 ), (0 )ui(0 ), (0 )ui 换路定律换路

15、定律(0 )(0 )CCuu(0 )(0 )LLii换路瞬间，换路瞬间，若电容电流保持为有限值若电容电流保持为有限值，则电容电压换路前后保持不变。则电容电压换路前后保持不变。换路瞬间，换路瞬间，若电感电压保持为有限值若电感电压保持为有限值，则电感电流换路前后保持不变。则电感电流换路前后保持不变。反映了能量不能跃变反映了能量不能跃变例例1电路如图所示，试求电路如图所示，试求 。 (0 )Ci解：解：+_10V10kC40kK+_CuCi+_10V10k40k+_Cu+_10V10k+_Ci8V 由换路定律由换路定律 画出画出 电路，求出电路，求出 0(0 )Cu电路电路0电路电路0 画出画出 电

16、路，求出电路，求出 0(0 )Ci40(0 )1081040CuV(0 )(0 )8CCuuV3108(0 )0.210 10CimA(0 )0CiA(0 )(0 )CCii可见，可见，电容电容开路开路电容用电容用电压源电压源替代替代Ci+_10V14KL+_LuLi+_10V14+_LuLi+_10V14+_Lu2A例例2电路如图所示，试求电路如图所示，试求 。 (0 )Lu解：解： 由换路定律由换路定律 画出画出 电路，求出电路，求出 0(0 )Li电路电路0电路电路0 画出画出 电路，求出电路，求出 0(0 )Lu10(0 )214LiA(0 )(0 )2LLiiA(0 )4 28LuV

17、 (0 )0LuV(0 )(0 )LLuu可见，可见，电感电感短路短路电感用电感用电流源电流源替代替代求解初始条件的步骤求解初始条件的步骤 画画 0 等效电路，即换路前电路（等效电路，即换路前电路（稳定状态稳定状态），求），求 uC(0) 和和 iL(0)。 由换路定律得由换路定律得 uC(0+) 和和 iL(0+)。 画画 0+ 等效电路，即换路后的电路。等效电路，即换路后的电路。 由由 0+ 电路求所需各变量的电路求所需各变量的 0+ 值。值。电容用电压源来替代，大小为电容用电压源来替代，大小为 uC(0)电感用电流源来替代，大小为电感用电流源来替代，大小为 iL(0)。电压源和电流源的方

18、向均与原来的电压、电流方向一致。电压源和电流源的方向均与原来的电压、电流方向一致。其中其中电容相当于开路电容相当于开路电感相当于短路电感相当于短路其中其中SiLRC0t K+_LuLiCi+_CuSiRLi+_Cu+_LuRSi+_SRi例例3电路如图所示，试求电路如图所示，试求 ， 。 (0 )(0 )CLiu解：解： 由换路定律由换路定律 画出画出 电路，求出电路，求出 ， 0(0 )(0 )CLui电路电路0电路电路0 画出画出 电路，求出电路，求出 ， 0(0 )(0 )CLiu(0 )LSii A(0 )(0 )CCSuuRi V(0 )0SCSRiiiAR(0 )CSuRi VCi

19、(0 )(0 )LLSiii A(0 )LSuRi V +_48V2L23CK+_LuLi+_CuCii例例4电路如图所示，试求开关电路如图所示，试求开关K闭合瞬间各支路的电流和电感上的电压。闭合瞬间各支路的电流和电感上的电压。 解：解：电路电路0电路电路048(0 )(0 )1222LLiiA2(0 )(0 )482422CCuuV(0 )482 1224LuV 4824(0 )83CiA(0 )(0 )(0 )20LCiiiA+_48V223Li+_Cu+_Lu+_48V2312A+_24VLiiCi100LC100+_200V100K+_CuLi100100+_200V100+_100V

20、1A+_LuCiKi例例5电路如图所示，试求开关电路如图所示，试求开关K闭合瞬间流过它的电流值。闭合瞬间流过它的电流值。 解：解：电路电路0电路电路0200(0 )(0 )1100100LLiiA(0 )(0 )100100CCLuuiV100100+_200V100+_CuLi200100(0 )12100100KiA 单位阶跃函数单位阶跃函数00( )10tu tt( )u tt01延时（延时（delayed）单位阶跃函数）单位阶跃函数0000()1ttu tttt( )u tt010t分段常量信号（分段常量信号（piecewise-constant signal）( )f tt00t(

21、)f tt010t02t03t（矩形）脉冲（矩形）脉冲（pulse）脉冲串（脉冲串（pulse train）1单位阶跃函数及分段常量信号单位阶跃函数及分段常量信号0( )( )()f tu tu tt000( )( )()(2 )(3 )f tu tu ttu ttu tt运用阶跃函数和延时阶跃函数，分段常量信号可以表示为一系列阶跃信号之和。运用阶跃函数和延时阶跃函数，分段常量信号可以表示为一系列阶跃信号之和。( )f tt00t1( )f tt00t02t03t10( )( )()f tu tu tt( )f tt00t02tAA000( )( )()(2 )(3 )f tu tu ttu

22、ttu tt00( )( )2()(2 )f tAu tAu ttAu ttSolutionExample 1. For the circuit of the figure, find iL(t) and uL(t) for t 0 given that iL(0)=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5 resistor over the time interval 0.4, ). 8H+_Lu20Li50.4tsSStep 1. With switch S open

23、, compute the response for 0t 0.4s. From the continuity property of the inductor current, (0 )(0 )10LLiiA( )(0 )thRtLLLitei2.510teAStep 2. With switch S closed, compute the response for t 0.4s. 20/54thR 0()0( )( )thRt tLLLiteit0.5(0.4)(0.4)tLie0.5(0.4)3.679teAStep 3. Write the complete response as a

24、 single expression.2.50.5(0.4)( )10 ( )(0.4)3.679(0.4)ttLiteu tu teu tAStep 4. Plot the complete response.( ),LitA, t s0.4s2sThe 0.4s time constant has a much faster rate of decay than the lengthy 2s time constant.Step 5. Compute uL(t).8H+_Lu20Li50.4tsS( )(0 )thRtLLLuteufor 0t 0.4s,in particular,(0

25、)20(0 )200LLuiA hence,2.5( )200tLuteV for t 0.4s,(0.4)( )(0.4)thRtLLLuteuin particular,(0.4)4(0.4 )14.716LLuiV 0.5(0.4)14.716teV Step 6. Compute energy dissipated in the 5 resistor over the time interval 0.4, ). 20.5(0.4) 2(0.4)5( ) 14.716(54)3.35ttLutepteW(0.4)50.443.34(0.4,)3.3tWedtJ Example 1. Fo

26、r the circuit of the figure, find iL(t) and uL(t) for t 0 given that iL(0)=10A and the switch S closes at t=0.4s. Then compute the energy dissipated in the 5 resistor over the time interval 0.4, ). 8H+_Lu20Li50.4tsS(0 )(0 )10LLiiA2.52.5( )(0 )10( )(0 )200ththRttLLLRttLLLiteieAuteueV 0t 0则有则有024UV又由已

27、知条件又由已知条件4 520RCs 20240tCueVt 201604tCuieAt利用并联分流，得利用并联分流，得 20212403tiieAt 20311203tiieAt+_24V44236H6(0)K t Li12+_Lu0t 6HR+_LuLi示例示例t = 0 时时 ，开关，开关K由由12，求求电感电压和电流。电感电压和电流。解：解：RL电路零输入响应问题电路零输入响应问题00RtLLiI eAt0(0 )(0 )LLIii246223/6436A(24)/636R 616LsR20tLie At120tLLdiuLe Vtdt 一阶电路的零输入响应是由储能元件的初值引起的响应一

28、阶电路的零输入响应是由储能元件的初值引起的响应, 都是由初始值都是由初始值衰减为零的指数衰减函数。衰减为零的指数衰减函数。 衰减快慢取决于时间常数衰减快慢取决于时间常数 。 同一电路中所有响应具有相同的时间常数。同一电路中所有响应具有相同的时间常数。 一阶电路的零输入响应和初始值成正比，称为零输入线性。一阶电路的零输入响应和初始值成正比，称为零输入线性。小结小结( )(0 )0tx txetRC 电路电路(0 )(0 )CCuuRL 电路电路(0 )(0 )LLiiRC 电路电路RCRL 电路电路LGLRR 为与动态元件相连的一端口电路的等效电阻。为与动态元件相连的一端口电路的等效电阻。4.

29、DC or step response of first-order circuitsThis section takes up the calculation of voltage and current responses when constant voltage or constant current sources are present.LinearResistivecircuitWithindependentsourcesLiLLinearResistivecircuit WithindependentsourcesC+_CuThevenin equivalentThevenin

30、 equivalentC+_thRocU+_CuCithRL+_ocULi+_LuC+_thRocU+_CuCithRL+_ocULi+_LuDeriving the differential equation models characterizing each circuits voltage and current responses.By KVL and Ohm law,Locth LuUR iLdiLdt1thLLocRdiiUdtLL By KCL and Ohm law,ocCCthUuiRCduCdt11CCocththduuUdtR CR C Exercise. Consta

31、nt differential equation models for the parallel RL and RC circuits of the figure. Note that these circuits are Norton equivalents of those in the figure. Again choose iL(t) as the response for the RL circuit and uC(t) as the response for the RC circuit.LiLthR+_LuscIthRscI+_CuCCiAnswers:ththLLscRRdi

32、iIdtLL 11CCscthduuIdtR CC 1thLLocRdiiUdtLL 11CCocththduuUdtR CR C ththLLscRRdiiIdtLL 11CCscthduuIdtR CC Observe that four differential equations have the same structure:( )1( )dx tx tFdt where,ththLG LR,thR CocscthUI RFLLscocthIUFCR CAnd the general formula for solving such a differential equation:0

33、0()()0( )()( )tt tttx tex tefdfor RC casefor RL case00()()0( )()( )tt tttx tex tefdwhere 1, (0 )(0 )xxas long as x(t) is a capacitorvoltage or inductor current, and f()=F is a constant (nonimpulsive) forcing function.000( )()qt ttttx tex tedqF000()(1)t tt tex tFe00 ()t tFx tFeWhich is valid for tt0.

34、 After some interpretation, this formula will serve as a basis for computing the response to RL and RC circuits driven by constant sources.00( ) ()t tx tFx tFeif0,then00( )lim ( )lim ()t tttxx tFx tFe FocscthUIRocUfor RC casefor RL caseHence, the solution of the differential equation given constant

35、or dc excitation becomes 00( )( ) ()( )t tx txx txe 0()0( )( ) ()( )thRt tLLLLLitiitie 00( )( )()( )tht tR CCCCCutuutue for RC casefor RL caseExample 4. For the circuit of the figure, suppose a 10V unit step excitation is applied at t=1 when it is found that the inductor current is iL(1)=1A. The 10V

36、 excitation is represented mathematically as uin(t)=10u(t1)V for t1. Find iL(t) and uL(t) for t1.5R +_Li+_Lu( )inut2LHSolutionStep 1. Determine the circuits differential equation model.1thLLocRdiiUdtLL 2.55 (1)Liu t where the time constant0.4sStep 2. Determine the form of the response.1( ) ( ) (1 )(

37、 ) (1)tLLLLitiiieu tA 5R +_Li+_Lu( )inut2LHStep 3. Compute iL() and set forth the final expression for iL(t).replace the inductor by a short circuit,( )2inLuiAR scIIt follows that,2.5(1)( )2(12) (1)tLiteu tAand since(1 )(1 )1LLiiAStep 4. Plot iL(t).2.5(1)2 (1)teu tAStep 5. Compute uL(t).5R +_Li+_Lu(

38、 )inut2LH( )LLdiutLdt2.5(1)5(1)teu tV1 (1 )( )(1)tLLLiieu t Exercise. Verify that in example 4, uL(t) can be obtained without differentiation by ( )( )Locth LutUR itExercise. In example 4, suppose R is changed to 4. Find iL(t) at t=2s. Specifically, we need only compute , , and the time constant or

39、. 0()x t( )x thLRthR CExample 5. The source in the circuit of the figure furnishes a 12V excitation for t10.16Rk+_( )inut23Rk10tsC 0.5mF+_Cu103(10 )(10 )2412CCuueStep 4. Find uC(t) for t10.23.57VThevenin equivalent2thRk+_8ocUVC 0.5mF+_Cu1 ,thR Cs( )8CuV 10( )( )(10 )( )thtR CCCCCutuuue teV(10)815.57

40、Step 5. Set forth the complete response using step functions.tCuteu tu teu t10(10)3( )(2412) ( )(10) 815.57 (10)Step 6. Plot uC(t).3s1sExercise. Suppose the switch in example 5 opens again at t=20s. Find uC(t) at t=25s.+_100V50010 F+_CuKi示例示例t = 0 时时 , 开关开关K 闭合，已知闭合，已知 uC(0) = 0V，求求（1）电容电压和电流；）电容电压和

41、电流；（2）uC80V 时的充电时间时的充电时间 t 。解：解： （1） RC 电路零状态响应问题电路零状态响应问题53500 105 10RCs 200(1)100(1)0t tCSuUe eV t 2000.20ttCSduUiCeeA tdtR （2）设经过）设经过 t 秒，秒，uC80V200( )100(1)80tCute 8.045tms 10A2H1R20030080+_LuKLi10A2HR+_LuLi例例1t = 0 时时 , 开关开关K 打开，求打开，求 t 0 后后 iL，uL 的变化规律的变化规律 。 解：解： RL 电路零状态响应问题，电路零状态响应问题，先化简电路先

42、化简电路80200/300200R 0t 20.01200LsR ()10SSLUIiAR 10010(1)0tLieAt 10010010 20020000ttLueeVt2HR+_LuLi+_SU2A2H51010+_LuKLi+_u例例2t = 0 时时 , 开关开关K 打开，求打开，求 t 0 后后 iL，uL以及电流源两端的电压以及电流源两端的电压 u 。 解：解： RL 电路零状态响应问题，电路零状态响应问题，先化简电路先化简电路101020R 0t 2 1020SUV20.120LsR ()1SLUiAR 10(1)tLieA 101020ttLSuU eeV105102010t

43、SLLuIiueV全响应的两种分解方式全响应的两种分解方式 根据电路的两种工作状态根据电路的两种工作状态0()0tCSSuUUUet稳态分量稳态分量暂态分量暂态分量全响应全响应/CuV/ t s0USU0US稳态分量稳态分量暂态分量暂态分量U0物理概念清晰物理概念清晰 根据激励与响应的因果关系根据激励与响应的因果关系0(1)0ttCSuU eUet零输入响应零输入响应零状态响应零状态响应全响应全响应便于叠加计算便于叠加计算/CuV/ t s0USU0零输入响应零输入响应零状态响应零状态响应全响应全响应全响应全响应+_24V40.6H8(0)K t Li+_Lu例例1t = 0 时时 , 开关开

44、关K 打开，求打开，求 t 0 后的后的 iL 。 解：解： RL 电路全响应问题电路全响应问题24(0 )(0 )64LLiiA0 611220L.sR 零输入响应零输入响应206tLieA 零状态响应零状态响应202024(1)2(1)12ttLieeA 全响应全响应20202062(1)24tttLLLiiieeeA+_10V111F+_Cu1A1+_uK例例2t = 0 时时 , 开关开关 K 闭合，求闭合，求 t 0 后的后的 iC，uC 以及电流源两端的电压以及电流源两端的电压 u ，已知已知 uC(0+) = 1V。 解：解：RC电路零状态响应问题电路零状态响应问题稳态分量稳态分

45、量()10111CuV (1 1) 12RCs 全响应全响应0.511tCuKeV(0 )( )10CCKuu 0.511 10tCueV0.55tCCduiCeAdt0.51 1 1125tCCuiueV 故故小结小结状态变量状态变量uCiL零输入响应零输入响应零状态响应零状态响应 全响应全响应 0tRCCuU e(1)tRCCSuUe0()tRCCSSuUUUe00RtLLRtLUieRI e(1)(1)RtSLLRtLSUieRIe00()RtSSLLRtLSSUUUieRRIIIe5. Superposition and linearityProvided one properly a

46、ccounts for initial conditions, superposition still apply when capacitors and inductors are added to the circuit.CCduiCdt1212,CCCCduduiCiCdtdt12()CCCdiCuudtsupposeand12CCCuuufor a capacitor12CCduduCCdtdt12CCiiBy the same arguments, the current due to the input excitation1122CCCua ua uis .1122CCCia i

47、a iOne the other hand, suppose112211( ),( )ttCCCCuiduidCCand1122CCCia ia i1( )tCCuidC112211( )( )ttCCa ida idCC1122CCa ua uThus linearity and, hence, superposition hold.Arguments analogous to the preceding imply that a relaxed inductor satisfies a linear relationship, and thus superposition is valid

48、, whether the inductor is excited by currents or by voltages.11221( )( )tCCa ia idCFor a general linear circuit, one can view each initial condition as being set up by an input which shuts off the moment the initial condition is established.This mean that when using superposition on a circuit, one f

49、irst looks at the effect of each independent source on a circuit having no initial conditions. Then one sets all independent sources to zero and computes the response due to each initial condition with all other initial conditions set to zero. The sum of all the responses to each of the independent

50、sources plus the individual initial condition responses yields the complete circuit response, by the principle of superposition.Example 6. The linear circuit of the figure has two source excitations applied at t=0, as indicated by the presence of the step functions. The initial condition on the indu

51、ctor current is iL(0)=1A. Determine the response iL(t) for t0 using superposition.SolutionStep 1. Compute the part of the circuit response due only to the initial condition, with all independent sources set to zero.2( )(0 )( )thRttLLLiteieu t A +_10 ( )SUu t V110R Li2LH+_1Ru2203R 2H4Li2 ( )SIu t ASt

52、ep 2. Determine the response due only to US=10u(t)V.+_10 ( )SUu t V110R Li2LH+_1Ru2203R 2 ( )SIu t A1( )1 ,SLUiAR 12/4thRRR (0 )0 ,LiA( )( ) (0 )( )( )thRtLLLLLitiiieu t 2(1) ( )teu t AStep 3. Compute the response due only to the current source IS=2u(t)A.( )2 ,LiA 12/4thRRR (0 )0 ,LiA( )( ) (0 )( )(

53、 )thRtLLLLLitiiieu t 22(1) ( )teu t AStep 4. Apply the principle of superposition. ( )( )( )( )LLLLitititit222( )(1) ( )2(1) ( )ttteu teu teu t A due to initial condition due to source US due to source IS 2(34) ( )teu t AQuestion 1. What is the new response if the initial condition is changed to iL(

54、0)=5A?Question 2. What is the new response if the voltage source US is changed to 5u(t)V, with all other parameters held constant at their original values?Question 3. What is the new response if the initial condition is changed to 5A, the voltage source US is changed to 5u(t)V, and the current sourc

55、e IS is changed to 8u(t)A? The question still remains as to why the superposition principle holds an advantage over the Thevenin equivalent method. This allows one to explore easily a circuits behavior over a wide range of excitations and initial conditions. 6. Response classificationsZero-input res

56、ponse The zero-input response of a circuit is the response to the initial conditions when the input is set to zero.Zero-state response The zero-state response of a circuit is the response to a specified input signal or set of input signals given that the initial conditions are all set to zero.Comple

57、te response By linearity, the sum of the zero-input and zero-state responses is the complete response of the circuit.Natural response The natural response is that portion of the complete response that has the same exponents as the zero-input response.Forced response The forced response is that porti

58、on of the complete response that has the same “exponent” as the input excitation provided the input excitation has exponents different from that of the zero-input excitation.7. Further points of analysis and theoryNot only a capacitor voltage or an inductor current, it turns that any voltage or current in an RC or RL first-order linear circuit with constant input has the form00( )( ) (