Plotting of a surface

4 ビュー (過去 30 日間)
Andrei
Andrei 2023 年 2 月 23 日
コメント済み: William Rose 2023 年 2 月 24 日
Currently I am working on creating ring potential with cosine-modulated depth. Since I know, that modulation is radial invariant I split my hypothetic function on radial and angular components (like variable separation). Radial part defines mu ring with central radius r0, width w and magnitude of potential is some A:
R(r)= -A*exp(-(r-r0).^2/w.^2)
Next I need cosine modulation. I tried just multiplying on cos(atan2(y,x)) but it causes wrong plot - magnitude of a function at line x = 0 goes to some value >> A. I think that happends because multiplication of small and big power exponents (just guessing). Is there any way to define such potential? P.S. I work in cartesian coordinates, so r = sqrt(x.^2+y.^2) in R(r) function.
  3 件のコメント
Andrei
Andrei 2023 年 2 月 23 日
Thanks Walter on another reply. cos(atan2(y, x)) function is specific on x=0 axis. It is not smooth there
Andrei
Andrei 2023 年 2 月 23 日

サインインしてコメントする。

採用された回答

William Rose
William Rose 2023 年 2 月 24 日
My understanding is that you want to create a potential function which is the product of a radial component and an angular component.
Here is the radial part:
[X,Y]=meshgrid(-4:.04:4);
r=(X.^2+Y.^2).^0.5;
A=1; r0=2; w=2;
R=-A*exp(-(r-r0).^2/w^2);
mesh(X,Y,R);
colorbar; view(0,90); axis equal
xlabel('X'); ylabel('Y'); zlabel('R'); title('R(r)')
Now compute and plot the angular part, Phi(theta). Phi() is undefined at the origin, because theta is undefined, because atan2(0,0) is undefined.
theta=atan2(X,Y);
Phi=cos(4*theta); %4 to get 4 cycles
mesh(X,Y,Phi);
colorbar; view(0,90); axis equal
xlabel('X'); ylabel('Y'); zlabel('Phi'); title('\Phi(\theta)')
If you compute Z(x,y)=R(r)*Phi(theta), (where r and theta are functions of x,y, as shown above), you get an ugly surface, that is especially ugly at the origin.
Another idea: multiply Phi(theta) by a function R2(r), which is is zero at the origin and 1 far from the origin. This will suppress the ugly behavior of Phi() at the origin. w2 is a radial scaling factor that determines how rapidly R2 turns on.
w2=0.7; R2=(1-exp(-r/w2)).^2;
mesh(X,Y,R2); xlabel('X'); ylabel('Y'); zlabel('R2'); title('R2(r)')
Next, compute the modulation function A(r,theta)=(1+ a2*R2( r)*Phi(theta)). a2 (0<a2<1) is a scalar that sets the size of the angular modulation.
a2=0.3;
A=1+a2.*R2.*Phi;
mesh(X,Y,A); colorbar; view(0,90); axis equal
xlabel('X'); ylabel('Y'); zlabel('A'); title('A(r,\theta)')
The plot of A(r,thata) above looks similar to Phi(theta), but it is not discontinuous at the origin, and the scale is different, as the colorbars indicate.
Now compute Z(r,theta)=R(r)*A(r,theta)
Z=R.*A;
mesh(X,Y,Z); colorbar; view(0,90); axis equal
xlabel('X'); ylabel('Y'); zlabel('Z'); title('Z(r,\theta)')
This may not be quite what you are looking for, but it gives you some ideas.
Good luck.
  3 件のコメント
Andrei
Andrei 2023 年 2 月 24 日
Thanks a lot. You did too much, I am really grateful. I wanted just a hint, but your solution is absolutely right! Thamks!
William Rose
William Rose 2023 年 2 月 24 日
You're welcome, @Andrei.

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeOrange についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by