이 질문을 팔로우합니다.
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Turning a complex function into 3d graph
조회 수: 2 (최근 30일)
이전 댓글 표시
Beyza Nur
2023년 1월 15일
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.
댓글 수: 9
Walter Roberson
2023년 1월 15일
What are your independent axes? You have theta, but for a 3d plot you need a second independent axes.
Beyza Nur
2023년 1월 15일
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?
Walter Roberson
2023년 1월 15일
Your code defines values on a circle whose size is controlled by lambda. Should lambda be the other independent variable?
Beyza Nur
2023년 1월 15일
@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
2023년 1월 15일
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
2023년 1월 15일
okey, thank you for your concern. If I get different information I will update this question.
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