I am trying to plot the following equation in MATLAB:
ratio = sqrt(1+1/(kr)^2)
With k and r on the x and y axes, and ratio on the z axis. I used meshgrid to create a matrix with values for x and y varying from 1 to 10:
[x,y] = meshgrid([1:1:10],[1:1:10]);
The problem now is to create values for z. I've tried to just type the whole equation in, but that gives this result:
>> Z = sqrt(1+1/(x .* y)^2)???
Error using ==> mldivide
Matrix dimensions must agree.
So what I did is go to through the whole process manually, which produces the right graph in the end:
z = z^2;
z = 1 ./ z;
z = 1 + z;
z = sqrt(z);
mesh(x,y,z)
Is there a more elegant way to do this? Or a way to type in the equation and let MATLAB handle the rest?
Try this:
Z = sqrt(1+1./(x .* y).^2);
surf(Z);
The problem that you had is related to using / instead of ./ and ^2 instead of .^2
Related
I am stuck with an apparently simple problem. I have to revolve of 360° a 2D curve around an axis, to obtain a 3D plot. Say, I want to do it with this sine function:
z = sin(r);
theta = 0:pi/20:2*pi;
xx = bsxfun(#times,r',cos(theta));
yy = bsxfun(#times,r',sin(theta));
zz = repmat(z',1,length(theta));
surf(xx,yy,zz)
axis equal
I now want to visualize the numerical values of the Z plane, stored in a matrix. I would normally do it this way:
ch=get(gca,'children')
X=get(ch,'Xdata')
Y=get(ch,'Ydata')
Z=get(ch,'Zdata')
If I visualize Z with
imagesc(Z)
I don't obtain the actual values of Z of the plot, but the "un-revolved" projection. I suspect that this is related to the way I generate the curve, and from the fact I don't have a function of the type
zz = f(xx,yy)
Is there any way I can obtain the grid values of xx and yy, as well as the values of zz at each gridpoint?
Thank you for your help.
Instead of bsxfun you can use meshgrid:
% The two parameters needed for the parametric equation
h = linspace(0,2) ;
th = 0:pi/20:2*pi ;
[R,T] = meshgrid(h,th) ;
% The parametric equation
% f(x) Rotation along Z
% ↓ ↓
X = sin(R) .* cos(T) ;
Y = sin(R) .* sin(T) ;
% Z = h
Z = R ;
surf(X,Y,Z,'EdgeColor',"none")
xlabel('X')
ylabel('Y')
zlabel('Z')
Which produce:
And if you want to extract the contour on the X plane (X = 0) you can use contour:
contour(Y,Z,X,[0,0])
Which produce:
I want to plot a two-dimensional function of the polar coordinates r and theta in three-dimensional cartesian coordinates. I have that (sorry about bad maths formatting, LaTeX not compatible, it seems)
f(r,theta) = r/2 * (cos(theta - pi/4) + sqrt(1 + 1/2 * cos(2*theta)))
Converting r and theta to cartesian coordinates
x = r * cos(theta), y = r * sin(theta)
Further, the domain is -1<r<1 and 0<theta<2 * pi, which I define by
r = -1:2/50:1;
and
theta = 0:2*pi/50:2*pi;
giving me two vectors of the same dimensions.
I can define the x and y values used for plotting as row vectors by
x = r. * cos(theta);
and
y = r. * sin(theta);
So now I need to define the z values, which will depend on the values of x and y. I thought I should make a 101x101 where each matrix element contains a data point of the final surface. But how should I do this? I thought about using a double for loop:
for i=1:numel(r)
for j=1:numel(theta)
z(i,j) = r(i)/2 .* cos(theta(j) - pi/4) + r(i).*sqrt(1 + 1/2 * cos(2.*theta(j)));
end
end
Then simply surf(z)
While this definitely gives me a surface, it gives me the incorrect surface! I don't know what is happening here. The incorrect surface is given in Figure 1, while the correct one is given in Figure 2. Can anyone help me out? For reference, the correct surface was plotted with GeoGebra, using
A = Function[<expression 1>, <Expresison 2>, <Expression 3>, <var 1>, <start>, <stop>, <var 2>, <start>, <stop>]
Figure 1. Incorrect surface.
Figure 2. Correct surface.
As others have said, you can use meshgrid to make this work.
Here's your example using gridded r and theta and an anonymous function to replace the double loop:
r = -1:2/50:1;
theta = 0:2*pi/50:2*pi;
% define anonymous function f(r,theta)
f = #(r,theta) r/2 .* (cos(theta - pi/4) + sqrt(1 + 1/2 .* cos(2.*theta)));
% generate grids for r and theta
[r, theta] = meshgrid(r,theta);
% calculate z from gridded r and theta
z = f(r,theta);
% convert r,theta to x,y and plot with surf
x = r.*cos(theta);
y = r.*sin(theta);
surf(x,y,z);
You need to use meshgrid to get matrix coordinates if you want to use surf. Taking your x and y (lower case), call
[X,Y] = meshgrid(x,y);
Then X and Y (upper case) will have the same values as you gave it, but laid out in the two-dimensional array as expected by surf. Loop over the indices here and compute your Z, which should have all(size(Z) == size(X)).
https://www.mathworks.com/help/matlab/ref/meshgrid.html
I have two variables x and y. I decide to plot the square root of their difference using contour as follows:
x=0:0.1:100;
y=0:0.1:100;
G=sqrt(x-y);
test2 = G;
test2(~(G<0)) = nan;
[C,h]=contourf(x,y,G,'ShowText','off');
set(gca,'FontSize',20)
However I get this error : Error using contourf (line 69)
Z must be size 2x2 or greater.
If that is resolved, I want to reach my goal and plot the actual function which relies on x, y and G itself as follows:
Function = 2 sqrt(x) / G * acoth((sqrt(x) + y/2 )/G )
I give an example
x=0:0.1:100;
y=0:0.1:100;
[X, Y]=meshgrid(x,y);
G=sqrt(X-Y);
test2 = G;
test2(~(G<0)) = nan;
[C,h]=contourf(X,Y,abs(G),'ShowText','off');
set(gca,'FontSize',20)
Input for contour should be 2D array, but your arrays are 1D.
Here, G is complex number. When you plot G, you should plot absolute G.
The result will be like below.
Regarding your function,
H=((2*sqrt(X))./G).*acoth((sqrt(X) + Y/2 )./G );
I was trying to plot the above surface in octave/matlab and ran into the this problem.
My code is as follows:
x = linspace(-sqrt(3),sqrt(3),1000);
y = linspace(-sqrt(2),sqrt(2),1000);
z = sqrt(6-2*x.^2-3*y.^2);
surf(x,y,z)
I got the error:
error: mesh: X, Y, Z arguments must be real.
I understand this was because some (x,y)s would result in negative 6-2*x.^2-3*y.*2, but I don't know how to tackle this because I can't trim either part of x or y. Any one can help? Thanks
It depends what you want to do with the non-real values of z.
One thing you could do is to set all these values to zero or NaN (as per #hbaderts' comment):
z = sqrt(6-2*x.^2-3*y.^2);
z( imag(z)~=0 ) = NaN;
One more thing though: your code might have a problem because z is a length-1000 vector, and you want it to be a 1000x1000 matrix. You should use meshgrid() on x and y to get two-dimensional matrices everywhere:
x = linspace(-sqrt(3),sqrt(3),1000);
y = linspace(-sqrt(2),sqrt(2),1000);
[xx,yy] = meshgrid(x,y);
z = sqrt(6-2*xx.^2-3*yy.^2);
z( imag(z)~=0 ) = NaN;
surf(xx,yy,z,'edgecolor','none')
(thanks #LuisMendo for the 'edgecolor','none' suggestion for better visualization.)
Running the above piece of code on octave gives this plot:
I am trying to graphically find the intersections between two surfaces and the x-y plane. (Intersection of surface z1 with the x-y plane and intersection z2 with the x-y plane)
I have created arrays representing the surfaces z1 = 3+x+y and z2 = 4-2x-4y and the z3 for the x-y plane using meshgrid. Looking everywhere, the only command that seems I can use to find the intersections between arrays is the intersect(A,B) command where A and B are arrays. When I enter intersect(z1,z3) however, I get the error "A and B must be vectors, or 'rows' must be specified." When I try intersect (z1,z2,'rows'), I am returned a 0-by-21 empty matrix. What am I doing wrong here?
My code:
x = -10:10;
y = -10:10;
[X,Y] = meshgrid(x,y);
z1 = 3+X+Y;
z2 = 4-2.*X-4.*Y;
z3 = 0.*X+0.*Y; %x-y plane
surf(X,Y,z1)
hold on
surf(X,Y,z2)
surf(X,Y,z3)
int1 = intersect(z1,z3,'rows');
int2 = intersect(z2,z3,'rows');
It sounds like you want the points where z1 = z2. To numerically find these, you have a couple options.
1) Numerical rootfinding: fsolve is capable of solving systems of equations. You can formulate the surfaces as functions of one vector, [x;y] and solve for the vector that makes the two surfaces equal. An example using the initial guess x=1, y=1 follows:
z1 = #(x) 3 + x(1) + x(2);
z2 = #(x) 4 - 2*x(1) - 4*x(2);
f = #(x) z1(x) - z2(x);
x0 = [1;1]
intersect = fsolve(#(x) f(x), x0);
2) Minimizing the error: If you are stuck with discrete data (arrays instead of functions) you can simply find the points where z1 - z2 is closest to zero. An easy starting point is to take the arrays Z1 and Z2 and find all points where the difference nears zero:
tol = 1e-3;
near_zero = abs(Z1 - Z2) < tol;
near_zero is going to be a logical array that is true whenever the difference between Z1 and Z2 is small relative to tol. You can use this to index into corresponding meshgrid arrays for X and Y to find the coordinates of intersection.
a simple way (no major function calls) to solve this is as follows:
x = -10:.1:10;
y = -10:.1:10;
[X,Y] = meshgrid(x,y);
z1 = 3+X+Y;
z2 = 4-2.*X-4.*Y;
z3 = z1 - z2;
[~,mn] = min(abs(z3));
the intersection is defined as (x, y(mn)).
This, of course is a numerical approximation (since you wanted a numerical method), subject to boundary condition which I haven't explored (you'll need to disregard values far from zero when performing the minimum function)
Note: if you're looking for an equation, consider performing a least squares approximation on the resulting data.