同轴电缆的特性阻抗计算

同轴电缆的特性阻抗计算同轴电缆 特性阻抗 拉普拉斯方程 矩形网格同轴电缆的横截面可以看做是两个同心圆。外圆半径为2，内圆半径为1。外圆上的电势为1，内圆上的电势为0。我们依据这些条件，通过编写matlab程序来计算出同轴缆线的特性阻抗。首先介绍一下计算中所用到的物理学公式。特性阻抗的公式为如下所示，C为电容，C0为光速。 由这两个公式，我们可将求解阻抗的问题转化为求解电量的问题。此时我们可以使用高斯公式。 为了处理截面上的问题，我们将面积分化为线积分。 本次计算过程中编程采用的方法是逐次超松弛迭代法。先将同轴电缆的截面按矩形网格进行划分。由于同轴电缆截面具有对称性，为了缩短程序运行时间，我们可以先计算四分之一截面内的电位分布。电位的迭代公式如下。由于这个程序采用矩形网格来处理圆的问题，所以处理精度和处理速度都没有采用极坐标处理理想。如果希望得到跟极坐标情况下同样误差的结果，则需要耗费更多的计算时间。图一为基本算法。图二、图三、图四分别是将代误差率为百万分之一时的特性阻抗、电势分布图和电场分布图。在文章的最后附有程序的代码。图一图二图三图四程序代码：clcclear all;ticr1=2; r2=1; n=.01;c=299792458;%err=8.854e-12; wuchalv=.0001; x=-r1:n:r1;y=r1:-n:-r1;l=length(x);dones=ones(l+1)/2);dlens=n*dones; dianwei_1=NaN(l+1)/2);X,Y=meshgrid(x,y); for i=1:(l+1)/2 for j=1:(l+1)/2 if X(i,j)2+Y(i,j)2=1 dianwei_1(i,j)=0; else end endenddianwei_2=isnan(dianwei_1);len3=dlens;for i=1:(l+1)/2 for j=1:(l+1)/2-1 if dianwei_2(i,j)=1&dianwei_2(i,j+1)=0 len3(i,j+1)=abs(abs(sqrt(r12-Y(i,j+1)2)-abs(X(i,j+1); else end endendlen3(l+1)/2,1)=0;len2=len3;len1=dlens;for i=1:(l+1)/2 for j=1:(l+1)/2-1 if dianwei_2(i,j)=0&dianwei_2(i,j+1)=1 len1(i,j)=abs(abs(sqrt(r22-Y(i,j)2)-abs(X(i,j); else end endend len4=len1;c1=len1./n;c2=len2./n;c3=len3./n;c4=len4./n;dianwei_3=dianwei_1 dianwei_1(:,(l+1)/2);dianwei_1(l+1)/2,:) NaN;dianwei_4=dianwei_3;dianwei_5=dianwei_3;maxerl=1;en=1; while maxerl=0 for i=1:(l+1)/2 for j=1:(l+1)/2 if c1(i,j)=1&c2(i,j)=0&c3(i,j)=0&c4(i,j)=1 dianwei_3(i,j)=1; elseif c1(i,j)=1&c2(i,j)0&c3(i,j)=0&c4(i,j)=1 dianwei_3(i,j)=1; elseif c1(i,j)=1&c3(i,j)0&c2(i,j)=0&c4(i,j)=1 dianwei_3(i,j)=1; elseif c1(i,j)=0&c2(i,j)=1&c3(i,j)=1&c4(i,j)=0 dianwei_3(i,j)=0; elseif c1(i,j)=0&c2(i,j)=1&c3(i,j)=1&c4(i,j)0 dianwei_3(i,j)=0; elseif c1(i,j)0&c2(i,j)=1&c3(i,j)=1&c4(i,j)=0 dianwei_3(i,j)=0; end endendfor i=2:(l+1)/2 for j=2:(l+1)/2 %c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c3(i,j)*dianwei_3(i,j+1)+c1(i,j)*dianwei_3(i,j-1)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)*dianwei_3(i-1,j)+c2(i,j)*dianwei_3(i+1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); if c1(i,j)=1&c2(i,j)=1&c3(i,j)0&c4(i,j)=1 dianwei_4(i,j)=c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c3(i,j)*dianwei_3(i,j+1)+c1(i,j)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)*dianwei_3(i-1,j)+c2(i,j)*dianwei_3(i+1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); elseif c1(i,j)=1&c2(i,j)0&c3(i,j)=1&c4(i,j)=1 dianwei_4(i,j)=c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c3(i,j)*dianwei_3(i,j+1)+c1(i,j)*dianwei_3(i,j-1)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)+c2(i,j)*dianwei_3(i+1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); elseif c1(i,j)0&c2(i,j)=1&c3(i,j)=1&c4(i,j)=1 dianwei_4(i,j)=c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c1(i,j)*dianwei_3(i,j-1)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)*dianwei_3(i-1,j)+c2(i,j)*dianwei_3(i+1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); elseif c1(i,j)=1&c2(i,j)=1&c3(i,j)=1&c4(i,j)0 dianwei_4(i,j)=c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c3(i,j)*dianwei_3(i,j+1)+c1(i,j)*dianwei_3(i,j-1)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)*dianwei_3(i-1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); elseif c1(i,j)=1&c2(i,j)0&c3(i,j)0&c4(i,j)=1 dianwei_4(i,j)=c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c3(i,j)*dianwei_3(i,j+1)+c1(i,j)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)+c2(i,j)*dianwei_3(i+1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); elseif c1(i,j)0&c4(i,j)0&c2(i,j)=1&c3(i,j)=1 dianwei_4(i,j)=c1(i,j)*c2(i,j)*c3(i,j)*c4(i,j)*(c1(i,j)*dianwei_3(i,j-1)/(c1(i,j)*c3(i,j)*(c1(i,j)+c3(i,j)+(c4(i,j)*dianwei_3(i-1,j)/(c2(i,j)*c4(i,j)*(c2(i,j)+c4(i,j)/(c1(i,j)*c3(i,j)+(c2(i,j)*c4(i,j); elseif c1(i,j)=c2(i,j)=c3(i,j)=c4(i,j) dianwei_4(i,j)=0.25*(dianwei_3(i-1,j)+dianwei_3(i+1,j)+dianwei_3(i,j+1)+dianwei_3(i,j-1); end endenddianwei_4(l+1)/2+1,:)=dianwei_3(l+1)/2-1,:);dianwei_4(:,(l+1)/2+1)=dianwei_3(:,(l+1)/2-1); dianwei_5=dianwei_4; dianwei_4=dianwei_3; dianwei_3=dianwei_5;er=abs(dianwei_3-dianwei_4);maxer=max(max(er);q,w=find(er=maxer);e=length(q); erl=zeros(1,e);for o=1:e erl(1,o)=er(q(o),w(o)-(wuchalv)*dianwei_3(q(o),w(o);endmaxerl=max(max(erl); for i=2:(l-1)/2 p(i-1)=(dianwei_3(i-1,i-1)-dianwei_3(i,i)/(n*sqrt(2)*2*pi*(2-(i-1)*n)*sqrt(2);endk1=1;for k=1:(l-1)/2-1 if isnan(p(k)=1 Q(k1)=p(k); k1=k1+1; end end Q1=mean(Q); for i=2:(l-1)/4 p1(i)=(dianwei_3(l+1)/2,i-1)-dianwei_3(l+1)/2,i)/(n)*2*pi*(2-(i-1)*n);endP1=mean(p1); R1=Q1 P1;dianrong=mean(R1)*err;Z(en)=1/(c*dianrong);en=en+1;end plot(Z);hold onM=1/c/(2*pi*err/log(r1/r2);plot(M*ones(1,length(Z),-r);xlabel(迭代次数);ylabel(特性阻抗);text(1000,M,理论值)hold off dianwei_6_1=fliplr(dianwei_3);dianwei_6_2=dianwei_3;dianwei_6_3=flipud(dianwei_3);dianwei_6_4=fliplr(dianwei_6_3); figure(2)dianwei_6=dianwei_6_2(1:(l+1)/2,1:(l+1)/2) dianwei_6_1(1:(l+1)/2,3:(l+1)/2+1);dianwei_6_3(3:(l+1)/2+1,1:(l+1)/2) dianwei_6_4(3:(l+1)/2+1,3:(l+1)/2+1);contourf(X,Y,dianwei_6); figure(3)cc ch=contour(X,Y,dianwei_6,15);clabel(cc);hold onFX,FY=gradient(dianwei_6,1,-1);quiver(X(1:20:401,1:20:401),Y(1:20:401,1:20:401),-FX(1:20:401,1:20:401),-FY(1:20:401,1:20:401);hold off toc个人总结a) 本次作业的主要目的是练习一下用计算机处理FDM。本题主要是处理与同心圆有关的数学问题，如果在计算中采用极坐标的话，能够有效的提高运算精度，并且大幅减小迭代次数。下图是其他同学使用拉普拉斯方程的极坐标形式得出的结果。由图可见，迭代次数在2000次时就可以得出很不错的数值解。所以，在处理物理问题时选择合适的坐标系来进行数学运算往往能够事半功倍。b) Matlab更擅长处理矩阵运算，而在处理循环运算时效率较低耗时较长。所以在Matlab编程中要多用矩阵运算来代替循环运算。c) 在编程过程中要养成良好的编程习惯。变量名称一定要便于记忆，这样才不会在后面的工作中混淆变量。d) 通过这次实践，对有限差分法有了更深的认识，为今后处理更复杂的物理问题打下了基础。9