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1、MATLAB 电路仿真实验报告学院班级姓名学号实验一 直流电路一、实验目的：1、加深对直流电路的节点电压法和网孔电流法的理解2、学习MATLAB的矩阵运算方法。二、实验示例1、节点分析示例一电路如图所示，求节点电压VI、V2和V3。MATLAB 求解：Y = 0.15 -0.1 -0.05; -0.1 0.145 -0.025;-0.05 -0.025 0.075 ;I = 5;0;2 ;fprintf(坯0卩吉V1,V2(V3:n)v = inv(Y)*I仿真结果：节点V1,V2和V3:404.2857350.0000412.85712、回路分析示例二使用解析分析得到通过电阻RB的电流。另外
2、，求10V电压源提供的功率。R11R2150VR330?1R430MATLAB 求解：Z = 40 -10 -30;-10 30 -5;-30 -5 65;V = 10 0 0;I = inv(Z)*V;IRB = I(3)-I(2);fprintf(the current through R is %8.3f Amps n ,IRB)PS = I(1)*10;fprintf(the power bupplied by 10V source is %8.4f watts n,PS) 仿真结果：the current through R is 0.037 Ampsthe power buppli
3、ed by 10V source is 4.7531 watts三、实验内容：1、电阻电路的计算如图，已知：R1=2, R2=6,R3=12,R4=8,R5=12,R6=4,R7=2 如 Us=10V,求 i3,u4,u7; 如 U4=4V,求 Us,i3,i7.R72求解（1）Z = 20 -12 0; -12 32 -12; 0 -12 18;V = 10 0 0;I = inv(Z)*V;i3 = I(1)-I(2);u4 = 8*I(2);u7 = 2*I(3);fprintf(i3=%f n,i3)fprintf(u4=%f n,u4)fprintf(u7=%f n,u7) 仿真结果
4、：i3=0.357143u4=2.857143u7=0.476190求解（2）0 8 0;-12 32 -12;0 -12 18;V = 4 0 0;I = inv(Z)*V;Us = 20*I(1)-12*I(2);i3 = I(1)-I(2);i7 = I(3);fprintf(Us=%f n,Us)fprintf(i3=%f n,i3)fprintf(i7=%f n,i7) 仿真结果：Us=14.000000i3=0.500000i7=0.3333332、求解电路里的电压，例如 V1,V2,V5.Y = 1 -1 2 -2 0;0 5 -13 8 0;2 0 4 -11 0;176 -5
5、 5 -196 0;0 0 0 0 1;I = 0 -200 -120 0 24;V = inv(Y)*I; fprintf(V1=%fVnV2=%fVnV3=%fVnV4=%fVnV5=%fVn,V(1),V(2),V(3),V(4),V(5)仿真结果：V1=117.479167VV2=299.770833VV3=193.937500VV4=102.791667VV5=24.000000V3、如图，已知 R1=R2=R3=4,R4=2控制常数 k1=05,k2=4,is=2求 i1 和 i2.Z = 1 0 0 0;-4 16 -8 -4;0 0 1 0.5;0 -8 4 6;V = 2 0
6、 0 0;I = inv(Z)*V;i1 = I(2)-I(3);i2 = I(4);fprintf(i1=%f Vni2=%f Vn,i1,i2)仿真结果：i1=1.000000 Vi2=1.000000 V四、实验总结仿真前需进行准确的计算，列出节点或回路表达式方可列出矩阵惊醒计算。实验二 直流电路（2）一、实验目的：1、加深对戴维南定律，等效变换等的了解。2、进一步了解MATLAB在直流电路的应用。二、实验示例1、戴维南定理如图所示电路，已知R1=4Q , R2=2Q , R3=4Q , R4=8Q ;is1=2A,is2=05A。(1)负载RL为何只是能获得最大功率？(2)研究RL在0
7、100范围内变化时，其吸收 功率的情况。MATLAB仿真：clear,format compactR1=4;R2=2;R3=4;R4=8;is1=2;is2=0.5;a11=1/R1+1/R4;a12=-1/R1;a13=-1/R4;a21=-1/R4;a22=1/R1+1/R2+1/R3;a23=-1/R4;a31=-1/R4;a32=-1/R3;a33=1/R3+1/R4;A=a11,a12,a13;a21,a22,a23;a31,a32,a33;B=1,1,0;0,0,0;0,-1,1;X1=AB*is1;is2;0;uoc=X1(3)X2=AB*0;0;1;Req=X2(3)RL=Re
8、q;P=uo”2*RL/(Req+RL)入2RL=0:10,p=(RL*uoc./(Req+RL).*uoc./(Req+RL),figure(1),plot(RL,p),gridfor k=1:21ia(k)=(k-1)*0.1;X=AB*is1;is2;ia(k);u(k)=X(3);endfigure(2),plot(ia,u,x),grid c=polyfit(ia,u,1);仿真结果：uoc =2.3333Req =3.6667P =0.3712RL =012345678910p =Columns 1 through 70 0.25000.3391 0.36750.37050.362
9、40.3496Columns 8 through 110.3350 0.3200 0.3054 0.2915三、实验内容1、在图2-3,当RL从0改变到50KG,绘制负载功率损耗。检验当RL=10KQ的最大功率损耗。R=10;U=10;RL=10;P=U2*(RL*1000)/(R+RL)*1000)入2RL=0:50;p=(RL*1000*U./(R+RL)*1000).*U./(R+RL)*1000) figure(1),plot(RL,p),grid程序运行结果：P =0.0025p =Columns 1 through 700.00080.00140.00180.00200.00220
10、.0023Columns 8 through 140.00240.00250.00250.00250.00250.00250.0025Columns 15 through 210.00240.00240.00240.00230.00230.00230.0022Columns 22 through 280.00220.00210.00210.00210.00200.00200.0020Columns 29 through 350.00190.00190.00190.00180.00180.00180.0018Columns 36 through 420.00170.00170.00170.001
11、60.00160.00160.0016Columns 43 through 490.00160.00150.00150.00150.00150.00140.0014Columns 50 through 510.0014 0.0014x 1042、在如图所示电路中，当R1取0,2,4,6,10,18,24,42,90和186 Q时，求RL的电压UL,电流IL和RL消耗的功率。75V52UsR420A=3/4 -1/2 0;1/2 -33/24 5/6;0 1 -1;I=15 0 0;U=inv(A)*I;us=U(3);R=6;Z=0 2 4 6 10 18 24 42 90 186;RL=Z(
12、1,:), i=us./(R+RL) u=us.*RL./(R+RL) p=(RL.*us./(R+RL).*us./(R+RL) figure(1),plot(RL,i),grid figure(2),plot(RL,u),grid figure(3),plot(RL,p),grid 仿真结果：RL =0246 101824 4290 186i =Columns 1 through 78.00006.00004.80004.00003.00002.00001.6000Columns 8 through 101.0000 0.50001 10.2500u =Columns 1 through
13、7012.000019.200024.000030.000036.000038.4000Columns 8 through 1042.0000 45.0000小 46.5000p =Columns 1 through 7072.000092.160096.000090.000072.000061.4400Columns 8 through 1042.0000 22.5000 11.6250四、实验总结1、对ATLAB变量生成的应用和戴维南等效变换有了 了解。2对ATLAB中图像的生成有了 了解。实验三 正弦稳态一、实验目的：1 学习正弦稳态电路的分析方法。2.学习MATLAB复数的运算方法。二、实验示例1、如图所示电路，已知 R=5Q, wL=3Q, 1/wC=2O, uc=10Z30 V,求 Ir, Ic, I和UL,Us。并画出其向量图。CMatlab 程序：Z1=3j;Z2=5;Z3=-2j;Uc=10*exp(30j*pi/180);Z23=Z2*Z3/(Z2+Z3);Z=Z1+Z23;Ic=Uc/Z3,Ir=Uc/Z2,I=Ic+Ir,U1=I*Z1,Us=I*Z; dis
