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1、MATLAB电路仿真实验报告实验一 直流电路（矩阵的基本运算）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V1、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(V1,V2V3: n)v = inv(Y)*I仿真结果：节点V1,V2和V3: v = 404.2857 350.0000 412.85712、回路分析示例二使用解析分析得到通过电阻RB
2、的电流。另外，求10V电压源提供的功率。MATLAB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 Amps the power bupplied by 10V sou
3、rce 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.(1) 如Us=10V,求i3,u4,u7;(2) 如U4=4V,求Us,i3,i7.（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：i3=0.357143 u4=
4、2.857143 u7=0.476190（2）Z = 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=14.000000 i3=0.500000 i7=0.333333 2、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 -19
5、6 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=0.5,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 0 0;
6、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1、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。2、熟练矩阵运算公式，即：V=inv（Y）*I实验二 直流电路（2）（基本二维图形的绘制）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=4，R2=2，R3=4，R4=8;is1=2A,is2=0.5A。（1）负载RL
7、为何只是能获得最大功率？（2）研究RL在010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
8、;0;1;Req=X2(3)RL=Req;P=uoc2*RL/(Req+RL)2RL=0:10,p=(RL*uoc./(Req+RL).*uoc./(Req+RL),figure(1),plot(RL,p),gridfor k=1:21 ia(k)=(k-1)*0.1; X=AB*is1;is2;ia(k); u(k)=X(3);endfigure(2),plot(ia,u,x),gridc=polyfit(ia,u,1);仿真结果：uoc = 2.3333Req = 3.6667P = 0.3712RL = 0 1 2 3 4 5 6 7 8 9 10p = Columns 1 throug
9、h 7 0 0.2500 0.3391 0.3675 0.3705 0.3624 0.3496 Columns 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50K，绘制负载功率损耗。检验当RL=10K的最大功率损耗。R=10;U=10;RL=10;P=U2*(RL*1000)/(R+RL)*1000)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 th
10、rough 7 0 0.0008 0.0014 0.0018 0.0020 0.0022 0.0023 Columns 8 through 14 0.0024 0.0025 0.0025 0.0025 0.0025 0.0025 0.0025 Columns 15 through 21 0.0024 0.0024 0.0024 0.0023 0.0023 0.0023 0.0022 Columns 22 through 28 0.0022 0.0021 0.0021 0.0021 0.0020 0.0020 0.0020 Columns 29 through 35 0.0019 0.0019
11、0.0019 0.0018 0.0018 0.0018 0.0018 Columns 36 through 42 0.0017 0.0017 0.0017 0.0016 0.0016 0.0016 0.0016 Columns 43 through 49 0.0016 0.0015 0.0015 0.0015 0.0015 0.0014 0.0014 Columns 50 through 51 0.0014 0.00142、在如图所示电路中，当R1取0,2,4,6,10,18,24,42,90和186时，求RL的电压UL，电流IL和RL消耗的功率。A=3/4 -1/2 0; 1/2 -33/2
12、4 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(1,:),i=us./(R+RL)u=us.*RL./(R+RL)p=(RL.*us./(R+RL).*us./(R+RL)figure(1),plot(RL,i),gridfigure(2),plot(RL,u),gridfigure(3),plot(RL,p),grid仿真结果：RL = 0 2 4 6 10 18 24 42 90 186i = Columns 1 through 7 8.0000 6.0000 4.8000 4.
13、0000 3.0000 2.0000 1.6000 Columns 8 through 10 1.0000 0.5000 0.2500u = Columns 1 through 7 0 12.0000 19.2000 24.0000 30.0000 36.0000 38.4000 Columns 8 through 10 42.0000 45.0000 46.5000p = Columns 1 through 7 0 72.0000 92.1600 96.0000 90.0000 72.0000 61.4400 Columns 8 through 10 42.0000 22.5000 11.6250四、实验总结1、经过这次实
