同轴电缆的特性阻抗计算.docx

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，C为光速。 00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。 1Q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。 Q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。 DdsQ,Q Z,C, 1 ,Q,Eds,cc31,，cc42,，nn，1n(),ds,11(,1)(,ijij，,1)(1,)iji,，(1,)j()dl,CC,U,|,|0.000001,|s,0,ijijij,，,，,(,)ijs,n,slcccc1324cccc13(13)cccc24(24)，建立一个所有元素均是nan的矩阵U在U中将1/4个圆环离散化(圆环所包括的点取0)U(i,j)为边界点没有计算c1 c2 c3 c4中U(i,j)不为边界有c1=c2=c3=c4=1不等于1的值U(i,j)=0时上下左右是否有nan将所有点的c1 c2 c3 c4分别存入四个与U同维的矩阵C1 C2 C3 C4中将边界上的电势值和C1 C2 C3 C4带入迭代公式开始反复迭代矩阵U图一若干次迭代后便得出在四分之一个圆环内的电势分布图二 2 图三 图四 3 程序代码: clc clear all; tic r1=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 end end dianwei_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 end end len3(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 4 end end end 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 end end for 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); 5 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) 6 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 end end dianwei_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); end maxerl=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); end k1=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); end P1=mean(p1); R1=Q1 P1; 7 dianrong=mean(R1)*err; Z(en)=1/(c*dianrong); en=en+1; end plot(Z); hold on M=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 on FX,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 8 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