I have this function:
and I want plot it, I think the result is a periodic function...
I tried this but got only one point :(
x1=-50:0.1:50;
x2=-50:0.1:50;
plot(cos(sqrt(power(x1,2)+power(x2,2)))/(power(x1,2)+power(x2,2)));
where is my problem and what is the correct way?
appreciate any help.
You need to plot it as a 3-D surface. For example, use surf:
[X1, X2] = meshgrid(-5:0.25:5, -5:0.25:5);
F = cos(sqrt(X1 .^ 2 + X2 .^ 2)) ./ (X1 .^ 2 + X2 .^ 2 + 1);
surf(X1, X2, F)
Note two things:
You forgot the "+1" in the denominator.
I've reduced the range of x1 and x2 coordinates, for better visualization.
If the black edges look annoying and seem to clutter the plot, you can remove the edge lines by disabling the EdgeColor property (as user Shai pointed out):
surf(X1, X2, F, 'EdgeColor', 'None')
The final result should look something like this:
That's a 3d plot, since there are two inputs x1 and x2. So you've got to use plot3 (or surf as #EitanT points out, or any 3d plotting function).
You're now only plotting the pairs (-50;-50), (-49.9;-49.9),...,(50;50), because you start from two vectors, you probably want to cover all the combinations. Therefore, use meshgrid (for higher dimensions, there is also ndgrid):
x1=-50:0.1:50;
x2=-50:0.1:50;
[X1, X2] = meshgrid(x1,x2);
You now use matrix operations, read through this link and you'll see that you need elementwise operations: a.*b instead of a*b, etc. power(a,b) is already the element-wise operation (the same as a.^b), matrix equivalent is mpower(a,b) or a^b.
f = cos(sqrt(power(X1,2)+power(X2,2)))./(power(X1,2)+power(X2,2)+1);
plot3(X1,X2,f);
Related
I need to plot a parabola and ellipse. However the ellipse is giving me trouble. Can anyone help? The equations are: y = -5*x^2 + 2 and (x^2/16) + (y^2/2) = 4
I've tried this code but obviously I feel like like it isn't right.
x = linspace(-5, 5);
y1 = (x.^2/16) + (y.^2/2) - 1;
y2 = -5*x.^2 +2;
figure(1)
plot(x, y1)
hold on
plot(x, y2)
hold off
Firstly, you did not define a range variable x. Secondly, the ellipse won't pass the vertical line test and can't be plotted like a regular function f(x). Thirdly, your equation y1 = (x.^2/16) + (y.^2/2) - 1; is non-sensical because you have y on each side.
You could correct your method by defining a range variable x1 and x2 that each have appropriate ranges for the functions your plotting. What I mean by this is that you probably don't want the same range for each function, because the ellipse is undefined over most of the range that the parabola is defined. To plot the ellipse using f(x) you could observe that there are + and - values that are identical, using this fact you could plot your ellipse by two functions one to represent the top half and one to represent the bottom half, each of these would pass the vertical line test.
OR
You could utilize ezplot and have a nice time with it because it makes your life easier. Here is a solution.
ezplot('x^2/16+y^2/2-4'); axis equal; hold on
ezplot('-5*x^2+2-y')
There are multiple ways to plot an ellipse, e.g. you could also use a parametric representation of the equation.
In your approach though, when plotting functions using plot(x,y) command, you need to express your dependent variable (y) through independent variable (x). You defined the range for x, which is what you substitute into your equations in order to find y's. While for the parabola, the dependency of y from x is obvious, you forgot to derive such a relationship for the ellipse. In this case it will be +-sqrt((1 - x^2/16)*2). So in your approach, you'll have to take into account both negative and positive y's for the same value of x. Also there's discrepancy in your written equation for the ellipse (=4) and the one in Matlab code (=1).
x = linspace(-5, 5);
y1 = sqrt((1 - x.^2/16)*2);
y2 = -5*x.^2 +2;
figure(1)
plot(x, real(y1), 'r', x, -real(y1), 'r')
hold on
plot(x, y2)
hold off
Since the ellipse has real y's not on the whole x domain, if you want to plot only real parts, specify real(y1) or abs(y1) (even though Matlab does it for you, too). You can also dismiss complex numbers for certain x when computing y1, but you'll need a for-loop for that.
In order to make things simpler, you can check the function fimplicit, ezplot is not recommended according to Matlab's documentation. Or if you want to plot the ellipse in a parametric way, fplot will work, too.
Another (more classic) approach for parametric plotting is given here already, then you don't need any other functions than what you already use. I think it is the simplest and most elegant way to plot an ellipse.
You will not be able to generate points for the ellipse using a function f(x) from a Cartesian linspace range. Instead, you can still use linspace but for the angle in a polar notation, from 0 to 2*pi. You should also be able to easily adjust radius and offset on both axis on the cos and sin expressions.
x = linspace(-5, 5);
y2 = -5*x.^2 +2;
figure(1)
clf;
plot(x, y2)
hold on
a = linspace(0,2*pi);
x2 = 4*cos(a);
y2 = sqrt(2)*sin(a);
plot(x2, y2)
xlim([-5,5]);
ylim([-5,5]);
hold off
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
I am writing a small code in Matlab to achieve following:
X, Y, and Z are all vectors and correspond to different physical quantities.
x1 and x2 are two different vectors made from allowed values which X can have.
Now I have Y as function of x1
Y=f(x1)
and Z as function of x2
Z=g(x2)
I do not know the functions 'f' and 'g' explicitly but I have a table which gives me Y and Z corresponding to x1 and x2 respectively.
So I can plot Y=f(x1) and Z = g(x2) in MATLAB using plotyy function using a common x-scale but 2 different y axis.
I want to plot Y as function of Z. What will be the most efficient way to do this? I guess I might need to use interpolation somewhere in this.
Thanks in advance !
As mentioned by #mikkola, interpolation would be the way to go. You could interpolate Y onto the domain of Z, Z onto the domain of Y, or Y and Z onto a completely different domain. Do be careful though, as you want to make sure you interpolate, not extrapolate. The default for interp1 is linear interpolation, but you can get it to implement other types of interpolation by passing in an optional parameter.
The following is an example with two different domains x1 and x2, corresponding to functions Y=sin(x) and Z=cos(x). These are interpolated onto a different domain x3 so that Y and Z can be plotted against each other, yielding a circle as we would expect.
% Two different domains
x1 = 0:0.02:2*pi;
x2 = 0:0.03:2*pi;
% Yielding two different functions
Y = sin(x1);
Z = cos(x2);
% Interpolate them both onto a common domain (x3)
x3 = 0:0.01:2*pi;
Yi = interp1(x1,Y,x3);
Zi = interp1(x2,Z,x3);
% Plot - circle, as expected
figure;
plot(Yi,Zi)
I have solved the heat equation in octave via finite difference and produced the following 3-D plot whose point colors correspond to the temperatures in each element of my three dimensional hamburger.
My computational resources limit the resolution at which I may solve my burger. Thus the only way to get the plot I want is to make my scatter3 points huge blobs of color and it looks kind of bad.
[x,y,z] = meshgrid(1:nx,1:ny,1:nz) % Defines a grid to plot on
scatter3(x(:), y(:), z(:), 40, burgermatrix(:), 's', 'filled')% Point color=value
What I want is a nice gorgeous smooth rectangular prism like this:
So I figure I need to somehow interpolate between the 3D points that I have. Can anyone help me figure out how to do this?
I might be missing something obvious, but here's the example from octave's help slice:
[x, y, z] = meshgrid (linspace (-8, 8, 32));
v = sin (sqrt (x.^2 + y.^2 + z.^2)) ./ (sqrt (x.^2 + y.^2 + z.^2));
slice (x, y, z, v, [], 0, []);
[xi, yi] = meshgrid (linspace (-7, 7));
zi = xi + yi;
slice (x, y, z, v, xi, yi, zi);
shading interp; %addition by me
Isn't this exactly what you need? You have your grid (x,y,z), your solutions (T), so you just need to plot it slicing along [0 0 1] etc. Something like
[xi yi]=meshgrid(unique(x),unique(y));
slice (x, y, z, T, xi, yi, max(z(:))*ones(size(xi)));
and the same for cuts along the two other axes. (Obviously the unique calls should be substituted with the vectors you already have from which you constructed the 3d mesh in the first place.)
NOTE: By the way, you should really consider changing the default (jet) colormap. I was yesterday enlightened by a colleague about the viridis colormap made by the SciPy people, see for instance this post and video link therein. Their reasoning is overwhelming, and their colormap is beautiful. This should define it: viridis, although I haven't tried it myself yet.
(If it wasn't for jet, I'd tell you that your temperature profile seems strongly 1d. Do you happen to have periodic boundary conditions along the vertical walls and homogeneous (i.e. constant) boundary conditions along the horizontal ones?)
I want to plot relations like y^2=x^2(x+3) in MATLAB without using ezplot or doing algebra to find each branch of the function.
Does anyone know how I can do this? I usually create a linspace and then create a function over the linspace. For example
x=linspace(-pi,pi,1001);
f=sin(x);
plot(x,f)
Can I do something similar for the relation I have provided?
What you could do is use solve and allow MATLAB's symbolic solver to symbolically solve for an expression of y in terms of x. Once you do this, you can use subs to substitute values of x into the expression found from solve and plot all of these together. Bear in mind that you will need to cast the result of subs with double because you want the numerical result of the substitution. Not doing this will still leave the answer in MATLAB's symbolic format, and it is incompatible for use when you want to plot the final points on your graph.
Also, what you'll need to do is that given equations like what you have posted above, you may have to loop over each solution, substitute your values of x into each, then add them to the plot.
Something like the following. Here, you also have control over the domain as you have desired:
syms x y;
eqn = solve('y^2 == x^2*(x+3)', 'y'); %// Solve for y, as an expression of x
xval = linspace(-1, 1, 1000);
%// Spawn a blank figure and remember stuff as we throw it in
figure;
hold on;
%// For as many solutions as we have...
for idx = 1 : numel(eqn)
%// Substitute our values of x into each solution
yval = double(subs(eqn(idx), xval));
%// Plot the points
plot(xval, yval);
end
%// Add a grid
grid;
Take special care of how I used solve. I specified y because I want to solve for y, which will give me an expression in terms of x. x is our independent variable, and so this is important. I then specify a grid of x points from -1 to 1 - exactly 1000 points actually. I spawn a blank figure, then for as many solutions to the equation that we have, we determine the output y values for each solution we have given the x values that I made earlier. I then plot these on a graph of these points. Note that I used hold on to add more points with each invocation to plot. If I didn't do this, the figure would refresh itself and only remember the most recent call to plot. You want to put all of the points on here generated from all of the solution. For some neatness, I threw a grid in.
This is what I get:
Ok I was about to write my answer and I just saw that #rayryeng proposed a similar idea (Good job Ray!) but here it goes. The idea is also to use solve to get an expression for y, then convert the symbolic function to an anonymous function and then plot it. The code is general for any number of solutions you get from solve:
clear
clc
close all
syms x y
FunXY = y^2 == x^2*(x+3);
%//Use solve to solve for y.
Y = solve(FunXY,y);
%// Create anonymous functions, stored in a cell array.
NumSol = numel(Y); %// Number of solutions.
G = cell(1,NumSol);
for k = 1:NumSol
G{k} = matlabFunction(Y(k))
end
%// Plot the functions...
figure
hold on
for PlotCounter = 1:NumSol
fplot(G{PlotCounter},[-pi,pi])
end
hold off
The result is the following:
n = 1000;
[x y] = meshgrid(linspace(-3,3,n),linspace(-3,3,n));
z = nan(n,n);
z = (y .^ 2 <= x .^2 .* (x + 3) + .1);
z = z & (y .^ 2 >= x .^2 .* (x + 3) - .1);
contour(x,y,z)
It's probably not what you want, but I it's pretty cool!