How do i plot a discontinuous function?
12 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
abril
el 4 de Mayo de 2020
Comentada: Star Strider
el 5 de Oct. de 2020
Hi i need to plot in 3d the following function in matlab:
f(x,y)= (x^2)/(x^2 - y^2) when |x| ≠ |y|
0 when |x| = |y|
how should i do it? thx
0 comentarios
Respuesta aceptada
Star Strider
el 4 de Mayo de 2020
Try this:
f = @(x,y) ((x.^2)./(x.^2 - y.^2)) .* (abs(x) ~= abs(y))
[X,Y] = ndgrid(linspace(-1,1,150));
figure
surf(X, Y, f(X,Y))
grid on
shading('interp')
It will automatically be 0 when the logical condition is not met, so no specific test need be added for the equality condition.
.
2 comentarios
Nathan Shapiro
el 5 de Oct. de 2020
I'm having a similar problem and tried your approach, but for some reason it is giving me an error message ("Matrix is singular to working precision"). Do you know what is causing the problem?
clear
clc
r=100;
x1 = linspace(-10,10,r);
y2 = linspace(-10,10,r);
[x,y] = meshgrid(x1,y2);
z=(sin(x)+sin(y))/(x.*y) .* (x~=0 & y~=0); %% the only discontinuity is when x or y equals 0
figure
surf(x,y,z)
xlabel('x');
ylabel('y');
zlabel('z');
grid on
shading interp
colorbar
Star Strider
el 5 de Oct. de 2020
‘Do you know what is causing the problem?’
Yes!
Use element-wise (dot-operator) division:
z=(sin(x)+sin(y))./(x.*y) .* (x~=0 & y~=0); %% the only discontinuity is when x or y equals 0
↑ ← HERE
and the problem no longer exists.
Más respuestas (0)
Ver también
Categorías
Más información sobre Surface and Mesh Plots en Help Center y File Exchange.
Productos
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!