2011-02-13 94 views
0

如果我在下图中有红色线条的所有点的坐标, 那么我怎样才能在matlab中编写代码来使用这些坐标绘制如下图的线条?找到一种方法来绘制

我的意思是因为一条线来自上轴,是否有可能将坐标放入矩阵中,然后绘制它?对于所有行都来自左轴的情况,我可以使用矩阵绘制它们。

enter image description here

看看这个

clc; clear; 
format long e 
s=0; 
lmin=0.8;  lmax=2.5; 
bmin=1.0;   bmax=1.5; 
lam=linspace(lmin,lmax,100); 
for n=1:length(lam) 
    increm*emphasized text*ent=0.0001; tolerence=1e-14; xstart=bmax-increment; 
    x=xstart; 
    dx=increment; 
    m=0; 
    while x > bmin 
     while dx/x >= tolerence 
      if sign(fTE(lam(n),x,s))*sign(fTE(lam(n),x-dx,s))<0 
       dx=dx/2; 
      else 
       x=x-dx; 
      end 
     end 

     if abs(fTE(lam(n),x,s)) < 1e-2 
      m=m+1; 
      if n>1 && m==1 && (x-max(r(1,:))) > 1e-4 
       b=zeros(1,n-1); 
       r=[b;r]; 
      end 
      r(m,n)=x; 
     end 
     dx=increment; 
     x=0.999*x; 
    end 
end 
switch s 
    case 0  
     figure 
    hold on,plot(lam,r(1,:),'b') 
    text(1.6,1.98,'TE(0)','FontSize',10), 
    hold on,plot(lam,r(2,:),'c') 
    text(1.8,1.89,'TE(2)','FontSize',10), 
    xlim([lmin,lmax]);ylim([bmin,bmax]), 
    xlabel('\lambda(\mum)'),ylabel('\beta-bar') 
    case 1 
     figure 
    hold on,plot(lam,r(1,:),'m') 
    %text(1.4,1.9,'TE(1)','FontSize',10), 
    hold on,plot(lam,r(2,:),'r') 
    %text(1.37,1.54,'TE(3)','FontSize',10), 
    xlim([lmin,lmax]);ylim([bmin,bmax]), 
    xlabel('\lambda(\mum)'),ylabel('\beta-bar') 
end 

和FTE是:

function y=fTE(lambda,betab,s) 

n1=2; n2=1.5; n3=1; 
z0=120*pi; 
d1=1; d2=1; d3=1; a=1; 
k0=2*pi/lambda; 

ub= sqrt(n1^2-betab^2); 
vb= sqrt(n2^2-betab^2); 
w= sqrt(betab^2-n3^2); 

Ub=k0*ub*d1; 
Vb=k0*vb*d2; 
W=k0*w*d3; 

z1=z0/ub; z1b=z1/z0; 

a0b=tan(Vb)/(w*ub)+tan(Ub)/(vb*w)-tanh(W)/(vb*ub)-tan(Vb)*tanh(W)*tan(Ub)/w^2; 
b0b=tan(Vb)*tan(Ub)/(vb^2*w)-tan(Vb)*tanh(W)/(vb^2*ub)+tanh(W)*tan(Ub)/(vb*w^2)- tan(Vb)*tan(Ub)/(w*ub^2)-tan(Vb)*tanh(W)/(w^2*ub)+tanh(W)*tan(Ub)/(vb*ub^2); 
c0b=tan(Ub)/(vb*w*ub^2)+tanh(W)/(vb*w^2*ub)+tan(Vb)*tanh(W)*tan(Ub)/(vb^2*ub^2)+tan(Vb)/(vb^2*w*ub); 

U0= k0*ub*a; m=s; 

y=(a0b*z1b^2+c0b)+(a0b*z1b^2-c0b)... 
    *cos(2*U0+m*pi)-b0b*z1b*sin(2*U0+m*pi); 
end 

回答

2

也许我误会,但你可以绘制多条线路到这样一个情节:

>> x = 1:0.2:3; 
>> y = sin(x); 
>> y2 = cos(x); 
>> [x; y; y2] 

ans = 

    Columns 1 through 9 

    1.0000 1.2000 1.4000 1.6000 1.8000 2.0000 2.2000 2.4000 2.6000 
    0.8415 0.9320 0.9854 0.9996 0.9738 0.9093 0.8085 0.6755 0.5155 
    0.5403 0.3624 0.1700 -0.0292 -0.2272 -0.4161 -0.5885 -0.7374 -0.8569 

    Columns 10 through 11 

    2.8000 3.0000 
    0.3350 0.1411 
    -0.9422 -0.9900 

>> plot(x,y) 
>> hold on 
>> plot(x,y2) 

The result from my example

+0

我没有任何功能,只是点。你看,每个x我有两个或更多的y。 – Alireza 2011-02-13 19:38:04

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