电磁场数值分析《电磁场数值分析》（作业）--- 2016学年 ---学院：学号：姓名：联系方式：任课教师：2016年6月6日作业1一个二维正方形（边长a=10mm)的静电场区域，电位边界条件如图所示（单位：V)，求区域内的电位分布。

要求用超松弛迭代法求解差分方程组进行计算。

代码：hx=11;hy=11;v1=zeros(hy,hx);v1(hy,:)=ones(1,hx)*100;v1(1,:)=ones(1,hx)*50;for i=1:hy;v1(i,1)=0;v1(i,hx)=100;endw=2/(1+sin(pi/(hx-1)));maxt=1;t=0;v2=v1;n=0;while(maxt>1e-6)n=n+1;maxt=0;for i=2:hy-1;for j=2:hx-1;v2(i,j)=(1-w)*v1(i,j)+w*(v1(i+1,j)+v1(i,j+1)+v2(i-1,j )+v2(i,j-1))/4;t=abs(v2(i,j)-v1(i,j));if (t>maxt)maxt=t;endendendv1=v2;endsubplot(1,2,1)mesh(v2)axis([0,11,0,11,0,100])subplot(1,2,2)contour(v2,20)结果：coef1=dt/(mu0*dx);coef2=dt/(eps0*dx);coef3=(c*dt-dx)/(c*dt+dx);ezold=ez;for now=1:timestep;hx=hx-coef1*(ez(:,2:ymesh+1)-ez(:,1:ymesh));hy=hy+coef1*(ez(2:xmesh+1,:)-ez(1:xmesh,:));ez(2:xmesh,2:ymesh)=ez(2:xmesh,2:ymesh)-...coef2*(hx(2:xmesh,2:ymesh)-hx(2:xmesh,1:ymesh-1))-...coef2*(hy(2:xmesh,2:ymesh)-hy(1:xmesh-1,2:ymesh));ez(1,:)=ezold(2,:)+coef3*(ez(2,:)-ezold(1,:));ez(xmesh+1,:)=ezold(xmesh,:)+coef3*(ez(xmesh,:)-ezold (xmesh+1,:));ez(:,1)=ezold(:,2)+coef3*(ez(:,2)-ezold(:,1));ez(:,ymesh+1)=ezold(:,ymesh)+coef3*(ez(:,ymesh)-ezold (:,ymesh+1));ez(xmesh/2+1,ymesh/2+1)=sin(now*dt*2*pi*c/25.0); mesh(ez)pause(0.01)ezold=ez;end结果：作业3基于Pocklington方程用MoM分析半波对称振子天线：观察天线线径和分段数目分别取不同值对天线阻抗和辐射特性的影响（半径分别取0.001λ，0.0001λ，0.00001λ，分段数取11,21,31）代码：%%初始化参数c=3e-8;r=1;f=c/r;w=2*pi*f;e0=8.85e-12;u0=4*pi*1e-7;a=0.0001*r;L=0.5*r;k=2*pi/r;N=31;dl=L/(N+1);l=L/2-dl/2;lz=-l:dl:1;lzs=lz(1:N);lzm=lz(1:N)+dl/2;lze=lz(2:N+1);%%阻抗矩阵元素求解fi=log(dl/a)/(2*pi*dl)-k/(4*pi)*1j;fi_1=exp(-k*dl*1j)/(4*pi*dl);fi_2=exp(-k*2*dl*1j)/(8*pi*dl);z=ones(N,N);for m=1:Nfor n=1:Nif m==nfi1=fi;fi2=fi_1;fi3=fi_2;z(m,n)=((k^2*dl^2-2)*fi1+fi2+fi3);elseif abs(m-n)==1fi1=fi_1;fi2=fi;fi3=fi_2;z(m,n)=((k^2*dl^2-2)*fi+fi2+fi3);elsefi1=exp(-k*abs(m-n)*dl*1j)/(4*pi*abs(m-n)*dl);fi2=exp(-k*abs(m+1-n)*dl*1j)/(4*pi*abs(m+1-n)*dl);fi3=exp(-k*abs(n+1-m)*dl*1j)/(4*pi*abs(n+1-m)*dl); z(m,n)=((k^2*dl^2-2)*fi+fi2+fi3);endendend%%电压矩阵求解V=zeros(N,1);V((N+1)/2)=-1*(1j*w*e0);I=z\V;Z_in=1/I((N+1)/2);disp(['输入阻抗=',num2str(Z_in)])I_amp=abs(I);Max=max(I_amp);Iunit2=[0;I_amp/Max(1);0];figure(1)h=0:dl/r:L/r;Ithe=sin(pi*h*r/L);plot(h,Iunit2,'b',h,Ithe,'r','linewidth',2)legend('pocklinton','解析值')grid onxlabel('电长度')ylabel('归一化电流')%%方向图theta=0:0.01:2*pi;abs_f=zeros(1,length(theta));for n=1:1:Nabs_f=abs_f+I(n)*exp(k*(n*dl-L/2)*cos(theta)*1j);endabs_f=abs(sin(theta)*dl.*abs_f);Max_f=abs(sum(I)*dl);Far_patten2=abs_f/Max_f(1);theta_2=0:0.1:2*pi;Far_theory=abs((cos(k*(L/2)*cos(theta_2))-cos(k*L/2)) ./sin(theta_2));figure(2)polar(theta,Far_patten2,'-b')hold onpolar(theta_2,Far_theory,'or')hold offlegend('pocklinton','解析值')title('半波阵子天线E面方向图')figure(3)polar(theta,ones(1,length(theta)),'-b')title('半波阵子天线H面方向图')%%半波阵子增益I_in=I((N+1)/2);A=(w*u0)^2/(4*pi*sqrt(u0/e0)*real(Z_in)*(abs(I_in))^2 );G_theta=A*abs_f.^2;Max_gain=max(G_theta);Max_gain_dB=10*log10(Max_gain);disp(['半波阵子增益=',sprintf('%.4fdB',Max_gain_dB)])结果：作业4基于电场积分方程用MoM分析对称振子天线：计算振子总长度分别为0.25λ ，0.5λ，λ，1.5λ时，振子的输入阻抗和E面方向图。

代码：lamda=1;a=0.0001;me=8.85e-12;mu=4*pi*(1e-7);arg=2*pi*(3e-8)/lamda;L=0.2*lamda;k=2*pi/lamda;N=21;dL=L/(N+1);l=L/2-dL/2;lz=-l:dL:1;lzs=lz(1:N);lzm=lz(1:N)+dL/2;lze=lz(2:N+1);for m=1:Nfor n=1:Nif n==mFmnmm=(1/(2*pi*dL))*log(dL/a)-1j*k/(4*pi); Fmnee=(1/(2*pi*dL))*log(dL/a)-1j*k/(4*pi); Fmnss=(1/(2*pi*dL))*log(dL/a)-1j*k/(4*pi); Fmnse=exp(-1j*k*dL)/(4*pi*dL);Fmnes=exp(-1j*k*dL)/(4*pi*dL);elseif abs(n-m)==1Fmnmm=exp(-1j*k*dL)/(4*pi*dL);Fmnee=exp(-1j*k*dL)/(4*pi*dL);Fmnss=exp(-1j*k*dL)/(4*pi*dL);if n>mFmnse=exp(-1j*k*2*dL)/(4*pi*2*dL);Fmnes=exp(1/(2*pi*dL))*log(dL/a)-1j*k/(4*pi);elseFmnes=exp(-1j*k*2*dL)/(4*pi*2*dL);Fmnse=exp(1/(2*pi*dL))*log(dL/a)-1j*k/(4*pi);endelsenum=abs(n-m);Fmnmm=exp(-1j*k*num*dL)/(4*pi*num*dL);Fmnee=exp(-1j*k*num*dL)/(4*pi*num*dL);Fmnss=exp(-1j*k*num*dL)/(4*pi*num*dL);if n>mFmnse=exp(-1j*k*(num+1)*dL)/(4*pi*(num+1)*dL);Fmnes=exp(-1j*k*(num-1)*dL)/(4*pi*(num-1)*dL);elseFmnes=exp(-1j*k*(num+1)*dL)/(4*pi*(num+1)*dL);Fmnse=exp(-1j*k*(num-1)*dL)/(4*pi*(num-1)*dL);endendz(m,n)=1j*arg*mu*dL*dL*Fmnmm+(1/(1j*arg*me))*(Fmnee-F mnes-Fmnse-Fmnss);endendV=zeros(N,1);fedp=(N+1)/2;V(fedp)=1;I=linsolve(z,V);Z=V(fedp)/I(fedp);theta=0:pi/100:2*pi;ftheta=0;for m=1:length(theta)F(m)=0;for n=1:NF(m)=F(m)+I(n)*exp(1j*k*(n-fedp)*dL*cos(theta(m)));endendF=abs(F);F1=F/max(F);polar(theta,F1,'b.-')结果：作业5对课程的建议、自己的收获等。