[信息与通信]matlab仿真报告.doc

MATLAB电路仿真实验报告电气工程学院班级：2011级1班姓名：商思远学号：2011302540022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 412.8571 2、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，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 source is 4.7531 watts 三、实验内容： 1、（1）%该程序确定电流%R4的电压u4和R7的电压u7%R是阻抗矩阵%V是电压向量%初始化矩阵R和向量VR=20 -12 0; -12 32 -12; 0 -12 18;V=10 0 0%解答回路电流I=inv(R)*V;%通过R3的电流计算i3=I(1)-I(2);fprintf(the current through R3 is %9.4f Ampsn,i3)%通过R4的电压的计算u4=I(2)*8fprintf(the power through R4 is %9.4f Wattsn,u4)%通过R7的电压计算u7=I(3)*2fprintf(the power through R7 is %9.4f Wattsn,u7)V = 10 0 0the current through R3 is 0.3571 Ampsu4 = 2.8571the power through R4 is 2.8571 Wattsu7 = 0.4762the power through R7 is 0.4762 Watts（2）%此程序确定电源电压%通过R3的电流i3和通过R7的电流i7%R是关于电流与电压的系数矩阵%V是另一系数矩阵%初始化矩阵R和向量VR= 20 0 -1; 12 12 0; 0 18 0;V= 6 16 6;%解答回路电流i=inv(R)*V;%通过R3的电流计算i3=i(1)-0.5;fprintf(the current through R3 is %9.4f Ampsn,i3)%通过R7的电流计算i7=i(2);fprintf(the current through R7 is %9.4f Ampsn,i7)%电源电压计算Us=i(3);fprintf(the power of battery is %9.4f Wattsn,Us)the current through R3 is 0.5000 Ampsthe current through R7 is 0.3333 Ampsthe power of battery is 14.0000 Watts（2）Y = 1 -1 2 -2 0; 0 5 -13 8 0; 2 0 4 -11 0; 176 -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)V1=117.479167VV2=299.770833VV3=193.937500VV4=102.791667VV5=24.000000V（3）%该实验确定电路里的电压%支路电流i1和i2%V为节点电压矩阵%R为系数矩阵%初始化矩阵V和RV=1 -1; 1 -3;R=4 0;%解答节点电压U=inv(V)*R;%通过R2的电流计算i1=U(1)-U(2)/4;fprintf(the current through R2 is %9.4f Ampsn,i1);%通过R4的电流计算i2=U(2)/2;fprintf(the current through R4 is %9.4f Ampsn,i2);the current through R2 is 1.0000 Ampsthe current through R4 is 1.0000 Amps 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进行计算。 2、对于矩阵运算公式，即：V=inv（Y）*I要熟练掌握。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=4，R2=2，R3=4，R4=8;is1=2A,is2=0.5A。（1）负载RL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;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 through 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.%求负载功率损耗%RL为负载电阻，PL为负载功率损耗%U为电源电压，R为内电阻%设RL为一数组、求出的负载功率为另一数组，画出曲线找极大值uoc=10;R=10000;RL=R;PL=uoc2*RL/(R+RL)2 %求最大负载功率%设RL序列，求其功率RL=0:50000;PL=(RL*uoc./(R+RL).*uoc./(R+RL);figure(1),plot(RL,PL),gridPL =0.0025PL = Columns 1 through 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 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 -33/24 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(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 4.0000 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、经过这次实验基本了解了MATLAB变量生成的应用。2、经过这次实验更加深刻了戴维南等效电路的原理。3、了解了MATLAB中图像的生成。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=5，wL=3，1/wC=2，uc=1030V，求Ir，Ic，I和UL,Us。并画出其向量图。Matlab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;disp(Uc Ir Ic I u1 Us)disp(),disp(abs(Uc,Ir,Ic,I,U1,Us)disp(),disp(angle(Uc,Ir,Ic,I,U1,Us)*180/pi)ha=compass(Uc,Ir,Ic,I,Us,Uc);set(ha,linewidth,3) 仿真结果：Ic = -2.5000 + 4.3301iIr = 1.7321 + 1.0000iI = -0.7679 + 5.3301iU1 = -15.9904 - 2.3038iUc Ir Ic I u1 Us幅值 10.0000 2.0000 5.0000 5.3852 16.1555 7.8102相角 30.0000 30.0000 120.0000 98.1986 -171.8014 159.80562、如图所示电路，已知C1=0.5F,R2=R3=2，L4=1H;Us(t)=10+10cost,Is(t)=5+5cos2t,求b，d两点时间的电压U（t）。MATLAB仿真：clear,format compactw=eps,1,2;Us=10,10,0;Is=5,0,5;Z1=1./(0.5*w*j);Z4=1*w*j;Z2=2,2,2;Z3=2,2,2;Uoc=(Z2./(Z1+Z2)-Z4./(Z3+Z4).*Us;Zeq=Z3.*Z4./(Z3+Z4)+Z1.*Z2./(Z1+Z2);U=Is.*Zeq+Uoc;disp( w Um phi )disp(w,abs(U),angle(U)*180/pi) 仿真结果：w Um phi 0.0000 10.0000 01.0000 3.1623 -18.43492.0000 7.0711 -8.13013、含受控源的电路：戴维南定理如图所示电路，设Z1=-j250，Z2=250，Is=20A，球负载ZL获得最大功率时的阻抗值及其吸收功率。clear,format compactZ1=-j*250;Z2=250;ki=0.5;Is=2;a11=1/Z1+1/Z2;a12=-1/Z2;a13=0;a21=-1/Z2;a22=1/Z2;a23=-ki;a31=1/Z1;a32=0;a33=-1;A=a11,a12,a13;a21,a22,a23;a31,a32,a33;B=1,0;0,1;0,0;X0=AB*Is;0;Uoc=X0(2),X1=AB*0;1;Zeq=X1(2),PLmax=(abs(Uoc)2/4/real(Zeq)仿真结果：Uoc = 5.0000e+002 -1.0000e+003iZeq = 5.0000e+002 -5.0000e+002iPLmax = 625三、实验内容1、如图所示电路，设R1=2,R2=3,R3=4,jxl=j2,-jXC1=-j3,-jXC2=-j5,Us1=80V,Us2=60,Us3=0,Us4=150,求各电路的电流相量和电压向量。clear,format compactR1=2;R2=3;R3=4;ZL=2*j;ZC1=-3*j;ZC2=-5*j;US1=8;US2=6;US3=8;US4=15;Y1=1/R1+1/ZL;Y2=1/ZC1+1/R2;Y3=1/R3+1/ZC2;a11=1/Y1;a12=1/Y2;a13=1/Y3;a21=0;a22=-1;a23=1;a31=-1;a32=1;a33=0;b1=0;b2=US2/R2-US3/R3-US4/ZC2;b3=-US1/ZL-US2/R2;A=a11,a12,a13;a21,a22,a23;a31,a32,a33;B=b1;b2;b3;I=inv(A)*B;I1=I(1),I2=I(2),I3=I(3),ua=I1/Y1,ub=I3./(-Y3),I1R=ua/R1,I1L=(US1-ua)./ZL,I2R=(US2-ua+ub)/R2,I2C=(ua-ub)./ZC1,I3R=(US3-ub)/R3,I3C=(US4-ub)./ZC2 程序运行结果：I1 = 1.2250 - 2.4982iI2 = -0.7750 + 1.5018iI3 = -0.7750 - 1.4982iua = 3.7232 - 1.2732iub = 4.8135 + 2.1420iI1R = 1.8616 - 0.6366iI1L = 0.6366 - 2.1384iI2R = 2.3634 + 1.1384iI2C = 1.1384 - 0.3634iI3R = 0.7966 - 0.5355iI3C = 0.4284 + 2.0373i2、含电感的电路：复功率如图，已知R1=4,R2=R3=2,XL1=10,XL2=8,XM=4,Xc=8,Us=100V,Is=100A.求电压源，电压源发出的复功率。clear,format compactR1=4;R2=2;R3=2;XL1=10;XL2=8;XM=4;XC=8;US=10;IS=10;Y1=1/R1+1/(-j*XC);Y2=1/(j*(XL1-XM);Y3=1/(j*XM);Y4=1/(j*(XL2-XM)+R2);Y5=1/R3;a11=1;a12=-1;a13=0;a14=0;a15=0;a21=0;a22=0;a23=0;a24=1;a25=-1;a31=0;a32=1;a33=-1;a34=-1;a35=0;a41=1/Y1;a42=1/Y2;a43=1/Y3;a44=0;a45=0;a51=0;a52=0;a53=-1/Y3;a54=1/Y4;a55=1/Y5;A=a11,a12,a13,a14,a15;a21,a22,a23,a24,a25;a31,a32,a33,a34,a35;a41,a42,a43,a44,a45;a51,a52,a53,a54,a55;B=-US/R1;-IS;0;0;0;I=inv(A)*B;I1=I(1);I2=I(2);I3=I(3);I4=I(4);I5=I(5);ua=-I1/Y1;ub=I3/Y3;uc=I5/Y5;Ii=US/R1+ua/R1;Pus=US*IiPis=uc*IS 程序运行结果： Pus = 54.0488 - 9.3830iPis = 1.7506e+002 +3.2391e+001i4、正弦稳态电路，利用模值求解如图所示电路，已知IR=10A，Xc=10，并且U1=U2=200V，求XL。clearU2=200;IR=10;R=U2/IR;XC=10;U=200*exp(-150j*pi/180);200*exp(-30j*pi/180);I=(U-200)./(-j*XC);X=200./(I-10);XL=imag(X)仿真结果：XL =5.359074.6410四、实验总结初步了解MATLAB向量图的绘制，有了基本的了解。实验四 交流分析和网络函数一、实验目的1、学习交流电路的分析方法。2、学习交流电路的MATLAB分析方法。二、实验示例1、如图，如果R1=20，R2=100，R3=50，并且L1=4H，L2=8H以及C1=250Uf，求v3（t），其中w=10rad/s。y=0.05-0.0225*j 0.025*j -0.0025*j; 0.025*j 0.01-0.0375*j 0.0125*j; -0.0025*j 0.0125*j 0.02-0.01*j;c1=0.4*exp(pi*15*j/180);i=c1 0 0;v=inv(y)*i;v3_abs=abs(v(3);v3_ang=angle(v(3)*180/pi;fprintf(voltage v3, magnitude:%f n voltage v3,angle in degree:%f,v3_abs,v3_ang)仿真结果：voltage v3, magnitude: 1.850409 voltage v3, angle in degree:-72.453299三、实验内容1、电路显示如图所示，求电流i1(t)和电压uc(t)Y=1 1 -1;6-5*j 0 4-2.5*j;6-5*j -10-8*j 0; c2=5;c3=2*exp(pi*75*j/180); v=0;c2;c3; i=inv(Y)*v;it_abs=abs(i(3); it_ang=angle(i(3)*180/pi;Vc_abs=abs(i(1)*-10*j); Vc_ang=angle(i(1)*-10*j)*180/pi;fprintf(voltage it,magnitude: %f n voltage it,angle in degree: %f ,it_abs,it_ang)voltage it,magnitude: 0.387710 voltage it,angle in degree: 15.019255 fprintf(voltage Vc,magnitude: %f n voltage Vc,angle in degree: %f ,Vc_abs,Vc_ang)voltage Vc,magnitude: 4.218263 voltage Vc,angle in degree: -40.861691 2、如图，显示一个不平衡wye-wye系统，求相电压VAN，VBN和VCN。Y=6+13*j 0 0;0 4+6*j 0;0 0 6-12.5*j;c1=110;c2=110*exp(pi*(-120)*j/180);c3=110*exp(pi*120*j/180);v=c1;c2;c3;i=inv(Y)*v;Van_abs=abs(i(1)*(5+12*j);Van_ang=angle(i(1)*(5+12*j)*180/pi;Vbn_abs=abs(i(2)*(3+4*j);Vbn_ang=angle(i(2)*(3+4*j)*180/pi;Vcn_abs=abs(i(3)*(5-12*j);Vcn_ang=angle(i(3)*(5-12*j)*180/pi;Y=6+13*j 0 0;0 4+2*j 0;0 0 6-12.5*j;c1=110;c2=110*exp(pi*(-120)*j/180);c3=110*exp(pi*120*j/180);i=inv(Y)*v;Van_abs=abs(i(1)*(5+12*j);Van_ang=angle(i(1)*(5+12*j)*180/pi;Vbn_abs=abs(i(2)*(3+4*j);Vbn_ang=angle(i(2)*(3+4*j)