eaintheaccidentinvestigation,managementandreporting,eachpostshouldbedevelopedunderthissystemspecialistscheck,cleartheexaminatons,time,cyclesandotherrelevantregulations.Strengtheningsitesupervisionandexamination,todetectandinvestigateillegalcommand,illegaloperationsandviolationsofoperatingrules.Secondsafetyreferstotheproductionsite,technologymanagement,equipment,facilities,andsooncanleadtoaccidentsrisksexist.1,accordingtotheextentofthesecurityrisks,solvingisdividedintoa,b,andclevelsofdifficulty;A-level:difficult,miningdifficulties,shallbereportedtothecompany'sproblems.B-class:difficulttoresolvedif图2

解：①建模。如图2，建立网孔电流方程： (R1+jwL1)I1 + jwMI2 = US jwMI1 + (R2+jwL2+RL)I2= 0 ②利用MATLAB编程： clear，format compact % 输入元件参数值

R1=20；R2=0.08；RL=42；L1=3.6 ；L2=0.06； M =0. 4650；w=10 *pi；Us=115； % 按Z*X=Y 列写电路的矩阵方程，其中 X=[I1; I2]

z11=R1+j *w*L1；z12=j*w*M； z21=j*w*M； z22=R2+j*w*L2+RL； %以上设置系数矩阵X的元素

Z=[z11 , z12 ; z21 , z22]；Y=[Us ; 0]; X=Z \\Y ; %求解结果 I1=X(1)；I2=X(2)； I1，I2 ，%输出结果

r1=abs (I1)， angle1=[angle (I1 ) * 180]/pi, r2=abs (I2)， angle2=[angle (I2 ) * 180]/pi 运行程序得：I1 =0.0470 - 0.1000 , I2 = -0.3501 - 0.0063I； r1 = 0.1105，angle1 = -64.8385； r2 = 0.3502，angle2 = -178.9683

上式中r1, r2分别代表电流I1，I2的模，angle1, angle2分别代表电流I1，I2的辐角。 1.2 动态电路分析

描述动态电路是用微分方程，一阶动态电路是一阶微分方程来描述的，可以直接求解微分方程，但也可以应用三要素法求解。当应用三要素法求解时，电路

iculties,shallconsistofminingorganizationstosolveproblems.C-class:fromsegmentsandbusinessrisksthatmustbeaddressedintheDepartment.2,open-pitmineunsafetypesinclude:electrical,transport,blasting,fire,andotherslope.3,accordingtotheseverityofthehazardfilledinbyunittroubleshooting,registrationform(seeatteaintheaccidentinvestigation,managementandreporting,eachpostshouldbedevelopedunderthissystemspecialistscheck,cleartheexaminations,time,cyclesandotherrelevantregulations.Strengtheningsitesupervisionandexamination,todetectandinvestigateillegalcommand,illegaloperationsandviolationsofoperatingrules.Secondsafetyreferstotheproductionsite,technologymanagement,equipment,facilities,andsooncanleadtoaccidentsrisksexist.1,accordingtotheextentofthesecurityrisks,solvingisdividedintoa,b,andclevelsofdifficulty;A-level:difficult,miningdifficulties,shallbereportedtothecompany'sproblems.B-class:difficulttoresolvedif的全响应=零输入响应+零状态响应，或者电路的全响应=稳态响应+暂态响应，用公式表示为f ( t)= [f (∞)－f (0+)]exp（-t /τ）+ f (∞)，该式作为其数学模型。二阶动态电路用二阶微分方程来描述，要应用拉氏变换来建立S域的数学模型，限于篇幅仅以一阶电路为例运用MATLAB进行分行计算。

例 ：如图3所示电路，已知R=２Ω，C=0.5F，电容初始电压Uc (0+)=４V，激励为正弦电压US (t)=8cos4t，当t=0时，开关Ｓ闭合，求电容电压的全部响应，并绘出波形图。

图3

解：①建模。当t≥0时，表征电容电压的微分方程为： dUc /dt+Uc /RC=US /RC 若用三要素法求解，其解为：

Uc(t)=Uc(∞) +[Uc(0+)－Uc(∞)] exp(- t/τ)式中，Uc(0)为电容的初始电压，Uc(∞)为电容的稳态值，τ为时间常数。但由于激励为正弦电压， 上述公式适当修改为：

Uc(t)=Ucp(t) + [Uc(0+)- Ucp(0+)] exp(-t /τ)式中，Ucp(t)为电容的稳态值, 它应是和电源同频率的正弦电压，只是其幅值和相位不同，记为Ucp(t)=Ucpm * cos(ωt +θ), U cp(0+)为 0+时刻的Ucp(t)值，即Ucp(0+)=Ucpm * cos(θ)。 ②编程。 clear all

R=2; C= 0.5;T =R * C; Uc0=4；%输入元件参数 Usm=8; w=4; Zc=1/(j * w * C)； t=0:0.1:10；

Us=Usm * cos(w * t)；%输入激励信号 Ucp=Us * Zc/(R+Zc)；%计算稳态分量 Ucp0=Ucp(1)；%计算稳态分量的初始值

iculties,shallconsistofminingorganizationstosolveproblems.C-class:fromsegmentsandbusinessrisksthatmustbeaddressedintheDepartment.2,open-pitmineunsafetypesinclude:electrical,transport,blasting,fire,andotherslope.3,accordingtotheseverityofthehazardfilledinbyunittroubleshooting,registrationform(seeatteaintheaccidentinvestigation,managementandreporting,eachpostshouldbedevelopedunderthissystemspecialistscheck,cleartheexaminatons,time,cyclesandotherrelevantregulations.Strengtheningsitesupervisionandexamination,todetectandinvestigateillegalcommand,illegaloperationsandviolationsofoperatingrules.Secondsafetyreferstotheproductionsite,technologymanagement,equipment,facilities,andsooncanleadtoaccidentsrisksexist.1,accordingtotheextentofthesecurityrisks,solvingisdividedintoa,b,andclevelsofdifficulty;A-level:difficult,miningdifficulties,shallbereportedtothecompany'sproblems.B-class:difficulttoresolvedifUct=[Uc0-Ucp0 ]* exp(-t/T)；%计算暂态分量 Uc=Uct+Ucp；%计算电路的全响应

plot (t ,Uc,'-g',t,Uct,'+r',t,Ucp,'* b' )，grid% 绘制 稳态分量，暂态分量，全响应的波形图(如图4) legend('Uc','Uct','Ucp')%波形注释

运行程序，得到仿真结果：

Uc：电路的全响应 Uct：电路的暂态分量 Ucp：电路的稳态分量

2 小结

电路分析的基本方法是建立数学模型（一般是方程或者已知电路方程组），并求解方程组，得到各支路电压和电流。当电路规模较大时，求解很复杂，借助计算机可以大大简化计算量，以前有FORTRAN和BASIC语言编写的程序，但一般程序较大，较复杂，而利用MATLAB则要简单得多，而且还可以进行仿真，除编写专用程序外，可以建立通用的电路分析程序。以上三个题例，均是采用编程的方式，其实也可以用MATLAB的命令方式求解，这样更简单。 参考文献：

[1] 邱关源．电路（第四版）［M］．北京：高等教育出版社，1999．

iculties,shallconsistofminingorganizationstosolveproblems.C-class:fromsegmentsandbusinessrisksthatmustbeaddressedintheDepartment.2,open-pitmineunsafetypesinclude:electrical,transport,blasting,fire,andotherslope.3,accordingtotheseverityofthehazardfilledinbyunittroubleshooting,registrationform(seeatteaintheaccidentinvestigation,managementandreporting,eachpostshouldbedevelopedunderthissystemspecialistscheck,cleartheexaminations,time,cyclesandotherrelevantregulations.Strengtheningsitesupervisionandexamination,todetectandinvestigateillegalcommand,illegaloperationsandviolationsofoperatingrules.Secondsafetyreferstotheproductionsite,technologymanagement,equipment,facilities,andsooncanleadtoaccidentsrisksexist.1,accordingtotheextentofthesecurityrisks,solvingisdividedintoa,b,andclevelsofdifficulty;A-level:difficult,miningdifficulties,shallbereportedtothecompany'sproblems.B-class:difficulttoresolvedif[2] 周金萍等．MATLAB6实践与提高［M］．北京：中国电力出版社，2001． [3] 陈怀琛等．MATLAB及在电子信息课程中的应用[M]．北京：电子工业出版社，2002．

iculties,shallconsistofminingorganizationstosolveproblems.C-class:fromsegmentsandbusinessrisksthatmustbeaddressedintheDepartment.2,open-pitmineunsafetypesinclude:electrical,transport,blasting,fire,andotherslope.3,accordingtotheseverityofthehazardfilledinbyunittroubleshooting,registrationform(seeatt