Integration over two variable dependent function
Show older comments
Dear all,
I have function which generates 3d plot. I change the function to obtain data dependent on the two variables then function is giving zero except last term after integration.
The code is as given below. This is working as i want and giving the correct plot.
clear
Nx=50;
Ny=50;
Mx=20e-9;
My=20e-9;
x=linspace(-Mx/2,Mx/2,Nx);
y=linspace(-My/2,My/2,Ny);
[X,Y]=meshgrid(x,y);
Vb1=0.228;
Rx1=5e-9;
Ry1=5e-9;
alfa=0e-9; w=1E+12; T=2*pi/w;
idx=(X/Rx1).^2 + (Y/Ry1).^2 < 1;
V0=(idx)*0 + (1-idx)*Vb1;
surf(x*1e9,y*1e9,V0)
However when i change the code as given below, it is not working, making all rows 0 except last one.
How can i correct it?
clear
Nx=50;
Ny=50;
Mx=20e-9;
My=20e-9;
x=linspace(-Mx/2,Mx/2,Nx);
y=linspace(-My/2,My/2,Ny);
Vb1=0.228;
Rx1=5e-9;
Ry1=5e-9;
alfa=0e-9; w=1E+12; T=2*pi/w;
T = 1.0;
for i=numel(x)
for j = 1:numel(y)
fx =@(t) x(i) + alfa.*sin(w.*t);
idx=@(t) (fx(t)./Rx1).^2 + (y(j)/Ry1).^2 < 1;
Vb = @(t) idx(t).*0 + (1-idx(t)).*Vb1;
V0(i,j) = (1/T).*integral(Vb,0,T,'ArrayValued',true);
end
end
surf(x*1e9,y*1e9,V0)
3 Comments
Simon Chan
on 9 Mar 2022
for i=1:numel(x)
not for i=numel(x)
Torsten
on 9 Mar 2022
I must admit that I don't understand the function you are trying to integrate.
This gives you a code to integrate your original problem. Is it that what you want in a t-dependent form somehow ?
Vb1 = 0.228;
Rx1 = 5e0;
Ry1 = 5e0;
idx = @(x,y)(x/Rx1).^2 + (y/Ry1).^2 < 1;
V0 = @(x,y) idx(x,y)*0 + (1-idx(x,y))*Vb1;
V = integral2(V0,-10,10,-10,10)
Özgür Alaydin
on 11 Mar 2022
Answers (0)
Categories
Find more on Condensed Matter & Materials Physics in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
