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1、例9：正弦电源激励已知iL(0_)=1A，us=10sin(t+45)V，求开关闭合后 iL（t）,K(t=0),iL,+us,解：(1)iL(0+)=1A(2)求稳态解,1H,2,iL=4.47sin(t+18.43)A iL(0+)=4.47sin18.43=1.414A(3)=0.5s(4)由三要素法：iL(t)=4.47sin(t+18.43)+(11.414)e2t=4.47sin(t+18.43)0.414 e2t(A),The circuit looks like 2th-order but in fact two 1st-order.,Example 11-10 on page
2、 263,例：t0时电路处于稳态，求t0+后uL(t)和uC(t)。,3,1F,K(t=0),1,+12v,+uc,+10v,1 H,2,+uL,解：iL(0+)=i L(0_)=12/(1+3)=3A,uC(0+)=uC(0_)=3 3=9V iL()=(12-10)/1=2A,uC()=10V L=1/1=1s,C=2 1=2s所以：t0+后 iL(t)=2+(3-2)et A=2+et A uL(t)=1 d iL(t)/dt=et V uC(t)=10+(9 10)e0.5t=10 e0.5t V,iL,练习题：题1：图为一种测速装置原理图。A、B为金属丝，相距1m。当子弹匀速地击断A
3、再击断B时，测得uc=8V，求子弹的速度。,10,V,10 V,100F,A,B,解：设子弹击断A为t=0，子弹击断A后求三要素：uC(0+)=uC(0_)=0，uC()=10V，=1010-4=1ms击断B之前，uC(t)=10（1 e1000t）Vt=T时B被击断，uC(T)=10（1 e1000T）=8解得：T=1.6ms最后得子弹速度：,+,题2：t0时电路处于稳态，t=0时K闭合，求开关闭合后i(t)。,2,K(t=0),1,+4.5V,1.2 H,3,解：iL(0+)=i L(0_)=4.5/3+4.5/(1+2)=3A iL()=4.5/3=1.5A=1.2/(3|2)=1s所以
4、：iL(t)=1.5+(3-1.5)et A=1.5+1.5 et A,i,i L,解:(1)iL(0+)=i L(0_)=4.5/3+4.5/(1+2)=3A可用叠加定理、等效变换、支路电流法求i(0+)。2(i(0+)-4.5)+3(i(0+)-4.5+3)=4.5,i(0+)=3.6A(2)i()=4.5/1=4.5A(3)=1.2/(3|2)=1s所以：i(t)=4.5+(3.64.5)et A=4.50.9 et A,t0+,Method 2:,综合题1：如图所示电路，N为含源线性电阻网络，当t0时电路处于稳态，t=0时开关K闭合，已知uC(0-)=8V，R=1，C=0.25F，求
5、a、b间电压u(t),t0。,uC(0+)=8V，uC()=4V，0.5su(t)=4+2e-2t(V),综合题2：如图所示电路，NR为线性无源电阻网络（互易网络），换路前电路已处于稳态，t=0时开关K由1的位置打到2的位置，求换路后的电压uC(t)和电流i1(t)。已知C=0.05F，,，,。,uC(0+)=20V，uC()=25，0.5suC(t)=25-5e-2t(V)，i(t)=0.5e-2t-2(A)，,11.5 Step Response of First-Order Circuit,一、阶跃函数1、the unit step function 单位阶跃函数,(t),0 t 0-1
6、 t 0+,t,1,0,2、电压、电流阶跃幅值为K，记为K(t),t,1,0,t0,(t-t0),0 t t0-1 t t0+,2v,C,K(t=0),+2(t),C,f(t)=2(t)2(t3),f(t)=10(t)15(t2)+5(t3),to model the switch in practice表示工程中的开关,3、电路中引入阶跃函数的意义,可以起始任一连续函数，标定时间范围,t,0,f(t),t,0 6,f(t)(t6),t,0,u(t),0,3et(t),t 0的 u(t),t,求出阶跃响应就可以求冲激响应,3et,二、Step response 阶跃响应,1、单位阶跃响应：电路
7、对于单位阶跃输入的零状态响应，记为s(t)。,+(t),C,+uc(t),R,uc(t)的单位阶跃响应：,2、阶跃响应：A(t)-As(t),例1：求零状态响应u2,t/s,5,0 2,us/V,+us,10K,100pF,+u2,解：us=5(t)5(t210-6)V求u2的单位阶跃响应s(t)：u2(0+)=1V,u2()=0,=1 s,激励为us时，电路的响应为：,解：unit step response is:,激励为us=61(t)时，电路的全响应为：,例2：With the same initial condition,u=e100t V given that us=0;u=63e
8、100t V given that us=12(t)V时.Find the complete response of u if us=6(t)V.,线性RC网络,+u0,+us,12-7 Impulse Response 一阶电路的冲激响应,一、Impulse function冲激函数1、The unit impulse function单位冲激函数,t,0,(t),1,2、What Impulse voltage or current means?i=5(t)冲激电流强度为5库仑 u=8(t)冲激电压强度为8韦伯,例1：已知uc(0-)=0，求uc(0+),C,(t),+uc(t),解：uc
9、(0+)=1/c,3.The relationship between the unit impulse function and the unit step function：,4.Sifting(sampling)property：(筛分),f(t)(t)=f(0)(t)；f(t)(tt0)=f(t0)(tt0),t,0 2 3,f(t),1、unit impulse response 单位冲激响应 h(t)the zero-state response with the unit impulse function excitation.2、求h(t)的方法：方法一：(1)在0-，0+，u
10、C、iL跃变，求出 uC(0+)、iL(0+)(2)在0+，激励为零，求零输入响应。方法二：1(t)s(t)；(t)h(t),二、impulse response 冲激响应,L,iL,+(t),R,方法一：（1）t在0-，0+,（2）t在0+，,+uL(t),方法二：,求冲击响应iL,If unit step response is:,Then unit impulse response is:,例5：,解：（1）求iL的单位阶跃响应SiL(t)：iL(0+)=0,iL()=iC(0+)=1/6 A,1=RC=1/25 s,NR,+(t),(t),10Fic,4H,(3)求uL的单位冲激响应h
11、uL(t)：,(2)求uL的单位阶跃响应SuL(t)：,12-8 Second-Order Circuits二阶动态电路,+uc,C,R,+Us,K(t=0),i(t),uL+,以uC为变量：以uL为变量：,A second-order circuit is characterized by a second-order differential equation.It consists of resistors and the equivalent of two energy storage elements.,Find uC,i,iL,if us=0，zero-input response
12、if uC(0+)=0，iL(0+)=0，zero-state response,uC，i，iL=particular solution+homogeneous solution,Steady responseForced response,Transient responseNatural response,The characteristic equation：LCS2+RCS+1=0The natural frequencies（固有频率）:,二、RLC串联电路动态过程性质,t,过阻尼非振荡情况和临界阻尼非振荡情况：uC(0+)=2V，i(0+)=0,+uc,0.33F,4,K(t=0)
13、,i(t),+uL,1H,拐点,Zero-input response,+uc,K(t=0),i(t),+uL,欠阻尼振荡情况：uC(0+)=2V，i(0+)=0,0.4,0.96F,1H,C放电L充电,C放电L放电,C充电L放电,C放电L充电,C充电L放电,C充电L放电,Zero-input response,+uc,C,R,+Us,K(t=0),i(t),uL+,Zero-state response,t,uC,underdamped case,控制系统建模，研究稳定性，上升时间，超调量。,例1：某电路的微分方程为则电路的固有频率为：,A 1，3B-1，-3C-1，3D-1j3,通解形式为
14、：A Ke3tB KetC K1 et+K2e3t D Ket cos（3t+）,答案：B、C,要确定u：u=u+K1 et+K2e3t u(0+)=(),例2:LC振荡回路，L=1/16H，C=4F，uC(0+)=1V，iL(0+)=1A，求零输入响应uC(t)及iL(t),C,iL,+uc,解：,电路为欠阻尼振荡，设uC(t)=Asin(2t+),初值uC(0+)=1V，uC(0+)=iL(0+)/C=0.25V/s代入初值：,Asin=12Acos=0.25,所以,A=1.01=83,所以,This is an undamped response case when p1,2 are pure imaginary.,C,bi,+(t),L,i,例3：已知R=2，L=1H，C=0.01F，问b为何值时电路为临界阻尼情况？此时动态特性如何？,R,解：,时，为临界阻尼非振荡,bi,i,R,+u,i,
