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. (零状态响应零状态响应)Zero-State Response Of An RC Circuitt=0：uC(0)=02021/6/161The particular solution：It is usually called the forced response or the steady-state response.uC ()=US ，uC p = uC ()=US The homogeneous solution：The characteristic equation： RCS+1=0The characteristic value：The homogeneous solution is usually called the natural or transient response.2021/6/162uC (0+)=uC (0 )=0=US +A A= USThe general form of the step response of RC circuits：2021/6/163US-US USRiuC0tuC=US(1- e )-RCt（t 0）i=-RCtUSRe（t 0）1 zero-state response of DC input:uC= US US e-RCt（t 0）特解特解齐次解齐次解稳态分量稳态分量暂态分量暂态分量+-uC+-CRiUS（2）Discussion2021/6/1647.6 Step Response CircuitnWhen 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 step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.2021/6/1657.6.1 Step Response of RC CircuitsUs1(t)t=0：uC(0)=U0 Its solution includes two parts : the particular solution and the homogeneous solution.Complete response: external source and initial condition of the storage element 2021/6/166The particular solution：It is usually called the forced response or the steady-state response.uC ()=US ，uC p = uC ()=US The homogeneous solution：The characteristic equation： RCS+1=0The characteristic value：The homogeneous solution is usually called the natural or transient response.2021/6/167uC (0+)=uC (0 )=U0=US +A A= U0 USThe general form of the step response of RC circuits：2021/6/168Complete Response例例ucUsC12R+-U0US U0uc(0+)=U0RC=USduCdt+（t 0）uCUSuC=(U0 US)e-RCt+There are two ways to excite the circuit:1)By initial conditions of the storage elements in the circuits.2)By independent sources.2021/6/1691、Two ways to determine complete response:（1）to linear circuit:Complete responseZero_state responseZero_input responseComplete 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 response when y( 0 )=0;2021/6/1610y(t) = yh(t) + yp(t)自由自由分量分量强制强制分量分量 ty(t)= ke+ yp(t)暂态暂态响应响应稳态稳态响应响应(2) To linear circuit, complete response=natural response+forced response2021/6/1611Thus, to find the step response requires three things:1.The initial value y(0+);2.The final value y( );3.The time constant ;2021/6/1612（2） 与输入无关，归结为求由电容元件或电感元件观与输入无关，归结为求由电容元件或电感元件观 察的入端电阻察的入端电阻RLR=RC=关于关于uc(0-)，iL(0-) 和和 uc(0+)，iL(0+) （3） y(0+) uc(0-)，iL(0-) uc(0+)，iL(0+)确定其它确定其它变量的初始值变量的初始值（1） y( ) 归结为求解电阻网络（电容元件相当于开路，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）电感元件相当于短路）Three factors:2021/6/1613Switch s is closed at t=0, uc(0-)=2V, calculate the response uc.+u-iCi12i1+2V+1 1 1 0.8FuC SuC (V)t1.5OExample:Method 1:2021/6/1614Reduce the circuit at first.i12i1+2V+1 1 1 0.8FuC S+1.5V+0.25 1 0.8FuC SMethod 2: ty(t)= y(0+) - y( )e+ y( )2021/6/1615L=L1+L2=3mH2021/6/1616Drill 1. Calculate Uc.3A100 50 0.3F+-uCuC(0)=150VuC(0+)=150V=0.3 5000/150=10SuC=100 + 50e0.1t (t 0)uC( )=3 5000/150=100V ty(t)= y(0+) - y( )e+ y( )2021/6/161710mA1k +-1k 2k 10V+-uLiL1HDrill 2. Calculate Ul.iL(0)=5mAiL(0+)=5mAiL( )=5+5=10mAuL(0+)=5VuL( )=0 =103SiL=10 5e1000t mAuL=5e1000t V5mA+-2k 2k 10V+-uL+-10V(t=0+) ty(t)= y(0+) - y( )e+ y( )2021/6/1618+-15 10 30V+-uCiL3 FiK2mH5 Drill 3. Calculate Uc, il, ik.iuC(0)=20ViL(0)=1AuC(0+)=20ViL(0+)=1Ai (0+)=20/15=4/3AuC( )=1.2 15=18Vi( )=30/25=1.2AiL( )=0 C=3 106 150/25=18 106 S L=2 103/5=1/2500 SuC=18+2e V106t18iL=e2500t Ai k=1.2+ e e2500t A215106t18 ty(t)= y(0+) - y( )e+ y( )2021/6/16190t2, equivalent circuit2021/6/1620solution：uC( )=US=6V，uC (0+)=uC (0)=0，1=C(R1+R2)=0.52=1s2021/6/1621t=2s：uC (2)=66e2=5.19 VuC ( )=6V，uC (2+)=uC (2)=5.19 V，2=R1C=0.5s2021/6/1622+U-cN+Us-+U*-LN+Us-Fig.aFig.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*.2021/6/1623+-RC1C2+-u1(0) u2(0)u1u2+-i1USRSRL2L1i2Special Cases:2021/6/1624已知：已知：解：此题为初始值跃变情况解：此题为初始值跃变情况电流发生跃变电流发生跃变S(t =0)R1R2L1L2i1i2US回路磁链守恒回路磁链守恒2021/6/162512i20ti132021/6/1626 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 合合k前前合合k后后 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。 节点电荷守恒节点电荷守恒q(0+)= q(0-)ERC1C2+-uC1+-uC2k(t=0)iiC1iC2+2021/6/1627uC( ) = 1V = R (C1+C2) 2021/6/1628uC2ut01/31uC1-1/6 2/9iC1it 4/91/6iC2iiC1iC2i 无冲激无冲激2021/6/16297.7 First_order OP AMP CircuitsAn op amp circuit containing a storage element will exhibit first_order behavior.Examples: Differentiators and integrators C+_uo_+ +_uiRiCi- -u- -iRR+_uo_+ +_uiCiRi- -u- -iC2021/6/1630Analyze first_order op amp circuits nClassical way: write the circuit equation in the first_order differential form at first, get the solution.nUse the general form of the complete response of the first_order circuits. 2021/6/1631Drill 1: For the op amp circuit , find vo(t) for t0, given that v(0)=3V.-+vo(t) -5F20k80kSolution: vo(0+)=12V, Vo()=0V. Req=20K =ReqC=0.1s2021/6/16327.7 First_order OP AMP Circuits+Vo-+-t=010K20K20K50K+3V-1F+ v -+V1-Drill 2: Determine v(t) and vo(t).2021/6/1633Drill 3: For the op amp circuit , find vo(t) for t0, given that vi=21(t)V.-+-vo(t)+2F10k10k50k20k+vi-2021/6/16341. Sinusoidal response: iL(0-)=0iK(t=0)L+uLRuS+-i (0-)=0 utuScalculate：i (t) 2021/6/1635强制分量强制分量(稳态分量稳态分量)自由分量自由分量(暂态分量暂态分量) iL(0-)=0iK(t=0)L+uLRuS+-Poticular solution:2021/6/1636定积分常数定积分常数A由2021/6/1637 结束语结束语若有不当之处，请指正，谢谢！若有不当之处，请指正，谢谢！