武大電氣matlab電路仿真實驗報告(多份)

1、武漢大學電氣工程學院MATLAB電路仿真實驗報告姓名：XXX學號：2014302540XXX班級：X班目錄 TOC o 1-3 h z u HYPERLINK l _Toc438668305 實驗一 直流電路 實驗一 直流電路一、實驗目的：1、加深對直流電路的節(jié)點電壓法和網(wǎng)孔電流法的理解。2、學習MATLAB的矩陣運算方法。二、實驗示例1、節(jié)點分析示例一電路如圖所示，求節(jié)點電壓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

2、 = inv(Y)*I仿真結(jié)果：節(jié)點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過電阻RB的電流。另外，求10V電壓源提供的功率。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

3、watts n,PS)仿真結(jié)果：the current through R is 0.037 Amps the power bupplied by 10V source is 4.7531 watts 三、實驗內(nèi)容：1、電阻電路的計算如圖，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%通過R3的電流i3的計算i3=I(1)-I(2);fprintf(th

4、e i3 is %8.4f An,i3)%R4的電壓u4的計算u4=8*I(2);fprintf(the u4 is %8.4f Vn,u4)%R7的電壓u7的計算u7=2*I(3);fprintf(the u7 is %8.4f Vn,u7)仿真結(jié)果：thei3 is 0.3571 Atheu4 is 2.8571 Vtheu7 is 0.4762 V（2）Z = 20 0 -1; -12 -12 0; 0 18 0;V = 6 -16 6;H = inv(Z)*V%電壓源的電壓Us的計算Us=H(3);fprintf(the Us is %8.4f Vn,Us)%通過R3的電流i3的計算i

5、3=H(1)-0.5;fprintf(the i3 is %8.4f An,i3)%R7的電壓i7的計算i7=H(2);fprintf(the i7 is %8.4f Vn,i7)仿真結(jié)果：the us is 14.0000 Vthe i3 is 0.5000 Vthe i7 is 0.3333 A2、求解電路里的電壓，例如V1,V2,V5.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

6、=%fVnV4=%fVnV5=%fVn,V(1),V(2),V(3),V(4),V(5)仿真結(jié)果：V1=117.479167VV2=299.770833VV3=193.937500VV4=102.791667VV5=24.000000V3、如圖，已知R1=R2=R3=4,R4=2,控制常數(shù)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;I = inv(Z)*V;i1 = I(2)-I(3);i2 = I(4);fprintf(i1=%f Vni2=%f Vn,i1,i2)仿真結(jié)果

7、：i1=1.000000 Vi2=1.000000 V實驗二 直流電路（2）一、實驗目的：1、加深對戴維南定律，等效變換等的了解。2、進一步了解MATLAB在直流電路的應(yīng)用。二、實驗示例1、戴維南定理如圖所示電路，已知R1=4，R2=2，R3=4，R4=8;is1=2A,is2=0.5A。（1）負載RL為何只是能獲得最大功率？（2）研究RL在010范圍內(nèi)變化時，其吸收功率的情況。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;a2

8、2=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=1:21ia(k)=(k-1)*0.1; X=AB*is1;is2;

9、ia(k);u(k)=X(3);endfigure(2),plot(ia,u,x),gridc=polyfit(ia,u,1);仿真結(jié)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 0.29三、實驗內(nèi)容1、在圖2-3，當RL從0改變到50K，繪制負載功率損耗。檢驗當RL=10K的最大功率損耗。US=10

10、;RS=10000;RL=0:100:50000;p=(RL*US./(RS+RL).*US./(RS+RL);figure(1),plot(RL,p),gridMaximun power occurs at 10000.00 OhmsMaximun power dissipation is 0.0025 Watts2、在如圖所示電路中，當R1取0,2,4,6,10,18,24,42,90和186時，求RL的電壓UL，電流IL和RL消耗的功率。R1=5;R2=20;R3=2;R4=24;R5=1.2;is=15;k=0;a11=1/R1+1/R2+1/R3; a12=-1/R3;a21=-1/

11、R3; a22=1/R3+1/R4+k/R5;A=a11,a12;a21,a22;B=1,0;0,1;X1=AB*is;0;uoc=X1(2);k=1;a22=1/R3+1/R4+k/R5;A=a11,a12;a21,a22;X2=AB*is;0;Req=R5*uoc/X2(2);Z=0 2 4 6 10 18 24 42 90 186;RL=Z(1,:),i=uoc./(Req+RL)u=uoc.*RL./(Req+RL)p=(RL.*uoc./(Req+RL).*uoc./(Req+RL)figure(1),plot(RL,i),gridfigure(2),plot(RL,u),gridf

12、igure(3),plot(RL,p),grid仿真結(jié)果：RL = 0 2 4 6 10 18 24 42 90 186i = 1 至 7 列 8.0000 6.0000 4.8000 4.0000 3.0000 2.0000 1.6000 8 至 10 列 1.0000 0.5000 0.2500u = 1 至 7 列 0 12.0000 19.2000 24.0000 30.0000 36.0000 38.4000 8 至 10 列 42.0000 45.0000 46.5000p = 1 至 7 列 0 72.0000 92.1600 96.0000 90.0000 72.0000 61

13、.4400 8 至 10 列 42.0000 22.5000 11.6250實驗三 正弦穩(wěn)態(tài)一、 實驗目的：1 學習正弦穩(wěn)態(tài)電路的分析方法。2 學習MATLAB復數(shù)的運算方法。二、實驗示例1、如圖所示電路，已知R=5，wL=3，1/wC=2，uc=1030V，求Ir，Ic，I和UL,Us。并畫出其向量圖。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(UcIrIc I u1 Us)disp(),dis

14、p(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)仿真結(jié)果：Ic = -2.5000 + 4.3301iIr = 1.7321 + 1.0000iI = -0.7679 + 5.3301iU1 = -15.9904 - 2.3038iUcIrIc 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 98.198

15、6 -171.8014 159.80562、如圖所示電路，已知C1=0.5F,R2=R3=2，L4=1H;Us(t)=10+10cost,Is(t)=5+5cos2t,求b，d兩點時間的電壓U（t）。MATLAB仿真：clear,formatcompactw=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 )

16、disp(w,abs(U),angle(U)*180/pi)仿真結(jié)果：w Um phi 0.0000 10.0000 01.0000 3.1623 -18.43492.0000 7.0711 -8.13013、含受控源的電路：戴維南定理如圖所示電路，設(shè)Z1=-j250，Z2=250，Is=20A，球負載ZL獲得最大功率時的阻抗值及其吸收功率。clear,formatcompactZ1=-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

17、=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)仿真結(jié)果：Uoc = 5.0000e+002 -1.0000e+003iZeq = 5.0000e+002 -5.0000e+002iPLmax = 625三、實驗內(nèi)容1、如圖所示電路，設(shè)R1=2,R2=3,R3=4,jxl=j2,-jXC1=-j3,-jXC2=-j5,Us1=80V,Us2=60,Us3=0,Us4=150,求各電路的電流相量和電壓向量。

18、R1=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/R2+1/ZC1;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)

19、ua=I1/Y1ub=I3/(-Y3)I1R=ua/R1I1L=ua./ZLI2R=(Us2+ub-ua)/R2I2C=(ua-ub)./ZC1I3R=(Us3-ub)/R3I3C=(Us4-ub)./ZC2ha=compass(ua,ub,I1R,I1L,I2R,I2C,I3R,I3C);set(ha,linewidth,3)程序運行結(jié)果：I1 = 1.2250 - 2.4982iI2 = -0.7750 + 1.5018iI3 = -0.7750 - 1.4982iua = 3.7232 - 1.2732iub = 4.8135 + 2.1420iI1R = 1.8616 - 0.6366

20、iI1L = -0.6366 - 1.8616iI2R = 2.3634 + 1.1384iI2C = 1.1384 - 0.3634iI3R = 0.7966 - 0.5355iI3C = 0.4284 + 2.0373i2、含電感的電路：復功率如圖，已知R1=4,R2=R3=2,XL1=10,XL2=8,XM=4,Xc=8,Us=100V,Is=100A.求電壓源，電壓源發(fā)出的復功率。R1=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/(R

21、2+j*(XL2-XM);Y5=1/R3;a11=1/Y1+1/Y2+1/Y3;a12=-1/Y3;a21=-1/Y3;a22=1/Y3+1/Y4+1/Y5;A=a11,a12;a21,a22;B=US/(Y1*R1);-IS/Y5;I=inv(A)*B;ua=US/(R1*Y1)-I(1)/Y1;uc=IS/Y5+I(2)/Y5;Pus=US*(US-ua)/R1Pis=uc*IS程序運行結(jié)果：Pus = -4.0488 + 9.3830iPis = 1.7506e+02 + 3.2391e+01i3、正弦穩(wěn)態(tài)電路：求未知參數(shù)如圖3-6所示電路，已知Us=100V,I1=100Ma,電路吸收

22、功率P=6W，XL1=1250，Xc=750，電路呈感性，求R3及XL3。ZL1=1250;ZC=750;US=60+j*80;I1=0.1;U1=I1*j*ZL1;U2=US-U1;I2=U2/(-j*ZC);U3=U2;I3=I1-I2;Z3=U3/I3R3=real(Z3)XL3=imag(Z3)程序運行結(jié)果：Z3 = 7.5000e+02 + 3.7500e+02iR3 = 750XL3 = 3754、正弦穩(wěn)態(tài)電路，利用模值求解如圖所示電路，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

23、(-150j*pi/180);200*exp(-30j*pi/180);I=(U-200)./(-j*XC);X=200./(I-10);XL=imag(X)仿真結(jié)果：XL =5.359074.6410實驗四 交流分析和網(wǎng)絡(luò)函數(shù)一、實驗目的1、學習交流電路的分析方法。2、學習交流電路的MATLAB分析方法。二、實驗示例1、如圖，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;-0.002

24、5*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)仿真結(jié)果：voltage v3, magnitude: 1.850409 voltage v3, angle in degree:-72.453299三、實驗內(nèi)容1、電路顯示如圖所示，求電流i1(t)和電壓uc(t)Y

25、=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)voltageit,magnitude: 0.387710 volta

26、geit,angle in degree: 15.019255 fprintf(voltage Vc,magnitude: %f n voltage Vc,angle in degree: %f ,Vc_abs,Vc_ang)voltageVc,magnitude: 4.218263 voltageVc,angle in degree: -40.861691 2、如圖，顯示一個不平衡wye-wye系統(tǒng)，求相電壓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

27、(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)