电路理论 华科电气Chapter 7 First Order Circuits 2

《电路理论 华科电气Chapter 7 First Order Circuits 2》由会员分享，可在线阅读，更多相关《电路理论 华科电气Chapter 7 First Order Circuits 2（38页珍藏版）》请在装配图网上搜索。

1、2021/6/16 1 7.5 The Zero-State Response If all of the initial conditions have zero value, then the circuit is said to be in the zero-state, and the solution to nonzero inputs for the circuit is known as the zero-state response. (零 状 态 响 应 )Zero-State Response Of An RC Circuit K +u CA C iCa b t=0t=0

2、： uC(0 )=0 +uCUs CiCabR 0C C SduRC u U tdt 2021/6/16 2 The particular solution：It is usually called the forced response or the steady-state response.uC ()=US ， uC p = uC ()=US R UsC +uC iThe homogeneous solution：0C CduRC udt The characteristic equation： RCS+1=0The characteristic value： 1S RC tRCStCh

3、u Ae Ae The homogeneous solution is usually called the natural or transient response. 2021/6/16 3 uC (0+)=uC (0)=0=US +A A= US 0,)( tevvtv tssc teyyty )()()( 0tC Cp Ch Su u u U Ae t The general form of the step response of RC circuits： 2021/6/16 4 US-US USR i uC0 tuC=US(1- e )-RCt （ t0）i= -RCtUSRe （

4、 t0）1 zero-state response of DC input:u C= US US e-RCt （ t0）特 解 齐 次 解稳 态 分 量 暂 态 分 量 +- uC+- CRiUS（ 2） Discussion 2021/6/16 5 7.6 Step Response Circuitn When the dc source is suddenly applied, the signal of the source can be modeled as a step function, the response is known as a step response.The st

5、ep response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. 2021/6/16 6 7.6.1 Step Response of RC CircuitsK +uCA CiCa b t=0 +uCUs CiCa b RUs1(t)t=0 ： uC(0 )=U0 0C C SduRC u U tdt Its solution includes two parts : the particular solut

6、ion and the homogeneous solution.Complete response: external source and initial condition of the storage element 2021/6/16 7 The particular solution：It is usually called the forced response or the steady-state response.uC ()=US ， uC p = uC ()=US R UsC +uC iThe homogeneous solution：0C CduRC udt The c

7、haracteristic equation： RCS+1=0The characteristic value： 1S RC tRCStChu Ae Ae The homogeneous solution is usually called the natural or transient response. 2021/6/16 8 uC (0+)=uC (0)=U0=US +A A= U0 US 0,)()( 0 teuuutu tssc teyyyty )()0()()( 0tC Cp Ch Su u u U Ae t The general form of the step respon

8、se of RC circuits： 2021/6/16 9 Complete Response例 u cUs C1 2 R +-U0 US U0 uc(0+)=U0RC =USduCdt + （ t0）uC USuC=(U0 US)e-RCt + There are two ways to excite the circuit: By initial conditions of the storage elements in the circuits. By independent sources. 2021/6/16 10 1、 Two ways to determine complete

9、 response:（ 1） to linear circuit: teyyyty )()0()()( teyyty )()()( Complete responseZero_state responseZero_input response teyty )0()( Complete response = zero-input response + zero-state response.Zero_input response is a special complete response when y( )=0;Zero_state response is a special complete

10、 response when y( 0 )=0; 2021/6/16 11 y(t) = yh(t) + yp(t)自 由分 量 强 制分 量 ty(t)= ke + yp(t)暂 态响 应 稳 态响 应(2) To linear circuit, complete response=natural response+forced response teyyyty )()0()()( 2021/6/16 12 teyyyty )()0()()(Thus, to find the step response requires three things: The initial value y(0

11、+); The final value y( );1. The time constant ; 2021/6/16 13 （ 2） 与 输 入 无 关 ， 归 结 为 求 由 电 容 元 件 或 电 感 元 件 观 察 的 入 端 电 阻 R LR=RC =关 于 uc(0-)， iL(0-) 和 uc(0+)， iL(0+) （ 3） y(0+) uc(0-)， iL(0-) uc(0+)， iL(0+)确 定 其 它 变 量 的 初 始 值（ 1） y() 归 结 为 求 解 电 阻 网 络 （ 电 容 元 件 相 当 于 开 路 ， 电 感 元 件 相 当 于 短 路 ）Three fa

12、ctors: 2021/6/16 14 Switch s is closed at t=0, uc(0-)=2V, calculate the response uc.+u- iC CC utuCu dd Ciuiu 1212 1 64dd4 CC utu 1044 ppV5.14/6 Cu tc Aeu tC Au e5.1 0)( V e5.05.1 tu tC i1 2i1+2V +1 1 10.8F uC S u C (V) t1.5O Example:Method 1: 2021/6/16 15 Reduce the circuit at first.s 18.0)25.01( RC

13、 V 5.1）（ Cu 0)( V e5.05.1 tu tC i1 2i1+2V +1 1 10.8F uC S + 1.5V +0.25 10.8F uC SMethod 2: ty(t)= y(0+) - y()e + y() 2021/6/16 16 +uL2US iR L2L1K + uL1 例 ： L1=1mH， L2=2mH， R=3k ， US=3V， t=0时 K 闭 合 ， 求零 状 态 响 应 uL1与 uL2。L=L1+L2=3mH 61 10L sR 6101 0ti e mA t 6 610 101 1 2 2, 2 0t tL Ldi diu L e V u L

14、e V tdt dt mAil 1)( 2021/6/16 17 Drill 1. Calculate Uc.3A 10050 0.3F +-uCuC(0)=150V uC(0+)=150V=0.3 5000/150=10Su C=100 + 50e0.1t (t 0)uC()=35000/150=100V ty(t)= y(0+) - y()e + y() 2021/6/16 18 10mA1k+- 1k2k10V +-uL iL1HDrill 2. Calculate Ul.iL(0)=5mAiL(0+)=5mAi L()=5+5=10mAuL(0+)=5V uL()=0=103SiL=1

15、0 5e1000t mAuL=5e1000t V 5mA+- 2k2k10V +-uL +-10V(t=0+) ty(t)= y(0+) - y()e + y() 2021/6/16 19 +- 151030V +-uC iL3F iK 2mH5Drill 3. Calculate Uc, il, ik.i uC(0)=20ViL(0)=1AuC(0+)=20ViL(0+)=1Ai (0+)=20/15=4/3AuC()=1.215=18Vi()=30/25=1.2AiL()=0C=3106 150/25=18 106 S L= 03/5=1/2500 S uC=18+2e V106t18iL

16、=e2500t Ai k=1.2+ e e2500t A215 106t18 ty(t)= y(0+) - y()e + y() 2021/6/16 20 例 ： US=6V， R1=R2=1 ， C =0.5F， 原 K1、 K2打 开 ， C 上 无 电 荷 ；t=0时 K1闭 合 ， 求 uC (t)， 又 当 t=2s时 ， K2又 闭 合 ， 求 uC(t)。 Us K1 R1 C+ u C K2 R2图 (a) Us R1 C+ u C R2图 (b)0t2, equivalent circuit 2021/6/16 21 solution：u C( )=US=6V，uC (0+)

17、=uC (0 )=0，1=C(R1+R2)=0.5 2=1s UsK1 R1 C+ uC K2 R2图 (a)Us R1 C+ uC R2图 (b) 02),1(6)()0()()( teeuuutu ttcccc 2021/6/16 22 t=2s： uC (2 )=66e 2=5.19 VuC ( )=6V，uC (2+)=uC (2 )=5.19 V，2=R1C=0.5s UsK1 R1 C+ uC K2 R2图 (a)Us R1 C+ uC 图 (c) 2,81.06)()2()()( )2(2 teeuuutu ttcccc 2021/6/16 23+U-cN+Us- +U*- LN

18、+Us-Fig.a Fig.bN is a linear resistive circuit, in Fig.a, c=1F, Uc(0 )=0, Us=U0, u=(0.75-e-2t)V;In Fig.b, L=1H, iL(0 )=0, calculate u*. 2021/6/16 24 +- RC1 C2+-u1(0) u2(0)u1 u2+- i1USRS RL2L1 i2Special Cases: 2021/6/16 25 已 知 ： VURRHLHL S 6,1,2,02.0,01.0 2121 并 定 性 画 出 波 形 。求 ： )(),( 21 titi解 ： 此 题

19、为 初 始 值 跃 变 情 况AiAi 0)0(,3)0( 21 )0()0( 21 ii 电 流 发 生 跃 变S(t =0)R1 R2L1 L2i1 i2US )0()0()0()0( 22112211 iLiLiLiL 回 路 磁 链 守 恒ALL iLiLii 103.0 301.0)0()0()0()0( 21 221121 sRR LLAii 01.0,2126)()( 21 2121 2021/6/16 26 Aetiti t10021 2)()( 12i2 0 ti13 VettdidLu tL 100111 )(02.0 VettdidLu tL 100222 2)(02.0

20、 SLL UuuiRiR 212211 62)(02.0)(02.0 224 100100 100100 tt tt etet ee 2021/6/16 27 电 容 电 压 初 值 一 定 会 发 生 跃 变 。解 V1)0(1 EuC 0)0(2 Cu合 k前 )0()0()0( 21 CCC uuu合 k后 已 知 图 中 E=1V , R=1 , C1=0.25F , C2=0.5F 。求 ： uC1 、 uC2 、 iC1 和 iC2 并 画 出 波 形 。 )0()0()0()0( 22112211 CCCC uCuCuCuC )0()()0()0( 212211 CCC uCCu

21、CuC 节 点 电 荷 守 恒q(0+)= q(0-) E R C1 C2+-uC1 +-uC2k(t=0)i iC1 iC2+ 2021/6/16 28 V315.025.0 125.0)0( 21 1 CC ECuC可 解 得 uC() = 1V 0321)131(1)( 3434 teetu ttC = R (C1+C2) s43 0)0(1)0(0 21 CC uut )(92)(61)(98)()321()(41 )(98)()321()(41dd 3434 3434111 tettett tetettuCi tt ttC )()321()()( 341 tettu tC )()32

22、1()( 342 tetu tC 2021/6/16 29 uC2u t01/31 uC1-1/6 2/9 iC1i t 4/91/6 iC2tuCiC dd 222 )()321()( 342 tetu tC )()321()()( 341 tettu tC )(92)(61 341 teti tC )(94)(61 34 tet t )(98)()321(21 34 tet t i iC1 iC2i 无 冲 激 2021/6/16 30 7.7 First_order OP AMP CircuitsAn op amp circuit containing a storage element

23、 will exhibit first_order behavior.Examples: Differentiators and integrators C +_u o_+ +_ui R iCi-u-iR tuRCu d1 io R +_uo_+ +_ui C iRi-u-iC dtdvRCv io 2021/6/16 31 Analyze first_order op amp circuits n Classical way: write the circuit equation in the first_order differential form at first, get the s

24、olution.n Use the general form of the complete response of the first_order circuits. teyyyty )()0()()( 2021/6/16 32 Drill 1: For the op amp circuit , find vo(t) for t0, given that v(0)=3V.-+ +vo(t) -5F20k 80kSolution: vo(0+)=12V, Vo()=0V. Req=20K =ReqC=0.1s 10( ) ( ) (0) ( )( ) 12 V, 0 ttoy t y y y

25、ev t e t 2021/6/16 33 7.7 First_order OP AMP Circuits +V o-+-t=010K 20K 20K50K+3V- 1F+ v -+V1-Drill 2: Determine v(t) and vo(t). 2021/6/16 34 Drill 3: For the op amp circuit , find vo(t) for t0, given that vi=21(t)V.-+ -vo(t)+2F10k10k50k20k+vi- 2021/6/16 35 1. Sinusoidal response: iL(0-)=0 iK(t=0) L

26、 +uLRuS+- )sin( ums tUu i (0-)=0u tuScalculate： i (t) 2021/6/16 36 )sin( um tUdtdiLRi iii 强 制 分 量 (稳 态 分 量 )自 由 分 量 (暂 态 分 量 ) tei A 22 )( LR UI mm RL arctg )sin( um tIi )sin( ums tUu iL(0-)=0 iK(t=0) L +uLRuS+-Poticular solution: i 2021/6/16 37 tum AetIiii )sin( )sin( umIA tumum eItIi )sin()sin( 定 积 分 常 数 AAIi um )sin(0)0( 由 若 有 不 当 之 处 ， 请 指 正 ， 谢 谢 ！