Please How I can get the figure like in the Picture below

1 次查看(过去 30 天)
hello community I look to get the figure like in the picture below using contourf and Thank you
clc
clear all
x=[]; y=[]; z=[];
for n=1:1001
x1=0.01*(n-1);
x2=0.01*(n-4);
B=[0,x1,4,x2;x1,3,x2,x2;x1,0,5,x2;0,5,x2,x1];
Pd=eig(B);
if max(real(Pd))<0
disp('fail');
disp(n);
end
B1=[B(1,1),B(1,2);B(2,1),B(2,2)];
B2=[B(3,3),B(3,4);B(4,3),B(4,4)];
B3=[B(1,3),B(1,4);B(2,3),B(2,4)];
Sum=det(B1)+det(B2)+2.*det(B3);
Et=sqrt(Sum-sqrt(Sum.^2-4.*det(B)))./sqrt(2);
E=2*max(0,real(2*Et));
x(n)=x1; y(n)=x2; z(n)=E;
n=n+1;
end
[X,Y] = meshgrid(x,y);
contourf(X,Y,Z,100, 'edgecolor','none');
plot(x,y)
  2 个评论
Rik
Rik 2022-8-12
You should calculate a z for each pair of x and y. I suspect the easiest way to do this is to use the meshgrid before your loop. That way you can also easily pre-allocate your arrays.
Abdelkader Hd
Abdelkader Hd 2022-8-12
@Rik thank you for your response, Please if you can write how I can do this, because I beginner use of matlab

请先登录,再进行评论。

采纳的回答

Rik
Rik 2022-8-12
You first need to define your variables:
n=(1:1001);
x=0.01*(n-1);
y=0.01*(n-4);
Now we have vectors, but you want the 2D grid they define:
[X,Y] = meshgrid(x,y);
Now we can create a Z array of the correct size to hold the output and loop through all elements of these arrays by using linear indexing.
Z=zeros(size(X));
for n=1:numel(X)
Z(n)=YourCode(X(n),Y(n));
end
contourf(X,Y,Z,100, 'edgecolor','none');
function E=YourCode(x1,x2)
% Don't forget to write comments to explain what this code does. You will
% have forgotten in 6 months, making it impossible to find any bugs.
B=[0,x1,4,x2;x1,3,x2,x2;x1,0,5,x2;0,5,x2,x1];
Pd=eig(B);
if max(real(Pd))<0
disp('fail');
disp(n);
end
B1=[B(1,1),B(1,2);B(2,1),B(2,2)];
B2=[B(3,3),B(3,4);B(4,3),B(4,4)];
B3=[B(1,3),B(1,4);B(2,3),B(2,4)];
Sum=det(B1)+det(B2)+2.*det(B3);
Et=sqrt(Sum-sqrt(Sum.^2-4.*det(B)))./sqrt(2);
E=2*max(0,real(2*Et));
end

更多回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Matrix Indexing 的更多信息

产品


版本

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by