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Turning a complex function into 3d graph
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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 =
1.0e-06 *
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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 Comments
Walter Roberson
on 15 Jan 2023
What are your independent axes? You have theta, but for a 3d plot you need a second independent axes.
Walter Roberson
on 15 Jan 2023
Your code defines values on a circle whose size is controlled by lambda. Should lambda be the other independent variable?
Beyza Nur
on 15 Jan 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
on 15 Jan 2023
Edited: Torsten
on 15 Jan 2023
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")

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