Question

如果我有下图中红线所有点的坐标，那么如何在matlab中编写代码来使用这些坐标绘制如下图所示的行？

我的意思是因为一条线来自上轴，例如可以将坐标放在矩阵中，然后绘制它吗？对于所有线都来自左轴的情况，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

Answer 1

也许我误解了，但你可以在这样的情节中绘制多条线：

>> 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

找到一种情节

1 个答案: