Turning a complex function into 3d graph

15 En. 2023
0 Respuestas
Actualizado a las 15 En. 2023
3 Visualizaciones (30 días)
Ran in:

0 votos

Ran in:
lambda= 0.1;
k= 2.*pi./lambda;
theta= 0:.01:2.*pi;
b= 4.*lambda;
r= 6.*lambda;
z= r.*cos(theta);
rho= r.*sin(theta);
f= (exp(-j.*k.*(z-(j.*b)))./(z-(j.*b)).*exp(j.*k.*sqrt(rho)./2.*(z-(j.*b)))).'
f =
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this is the function and I use a few method for 3D graph but I can't get enough result for my project. I think someting is missing. I don't like my results. I used meshgrid and surf command but results not good. Can anyone help me this one? Thank you.

9 comentarios
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Walter Roberson
Walter Roberson el 15 de En. de 2023
What are your independent axes? You have theta, but for a 3d plot you need a second independent axes.
Beyza Nur
Beyza Nur el 15 de En. de 2023
This is all the information I have. Can we use z and rho as axes? Cause they're depend on theta or maybe we can add phi. Is it possible?
Torsten
Torsten el 15 de En. de 2023
Editada: Torsten el 15 de En. de 2023
A 2d function (f is complex-valued) over a 2d region (x=r*cos(theta), y=r*sin(theta)) is 4d, not 3d.
Walter Roberson
Walter Roberson el 15 de En. de 2023
Your code defines values on a circle whose size is controlled by lambda. Should lambda be the other independent variable?
Beyza Nur
Beyza Nur el 15 de En. de 2023
@Torsten I don't quite understand what you mean in this case.Do you think this function can't be drawn in 3d?
@Walter Roberson lambda is a constant so we can't say like that, in my opinion. But I will keep trying changing the values.
thank you for your attention.
Torsten
Torsten el 15 de En. de 2023
Do you think this function can't be drawn in 3d?
Yes, that's what I think. Or you want to graph abs(f) instead of f. Or you want to graph real(f) and imag(f) separately in two plots.
Beyza Nur
Beyza Nur el 15 de En. de 2023
What should I write for abs(f) instead of f, I want to see the result.
Torsten
Torsten el 15 de En. de 2023
Editada: Torsten el 15 de En. de 2023
Ran in:
Ran in:
lambda= 0.1;
k= 2*pi/lambda;
theta= 0:.01:2*pi;
b= 4*lambda;
r= 6*lambda;
z= r*cos(theta);
rho= (r*sin(theta)).';
f= exp(-j.*k.*(z-(j.*b)))./(z-(j.*b)).*exp(j.*k.*sqrt(rho)./2.*(z-(j.*b)));
surf(z,rho,abs(f),"edgecolor","none")
Beyza Nur
Beyza Nur el 15 de En. de 2023
okey, thank you for your concern. If I get different information I will update this question.

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Beyza Nur
Beyza Nur
el 15 de En. de 2023

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Beyza Nur
Beyza Nur
el 15 de En. de 2023

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