I'm having issues unterstanding the function fill in MATLAB , I have a PSD of a file the I want to change it background like :
[xPSD,f] = pwelch(x,hanning(4096),2048,4096*2 ,fs);
plot(f,10*log10(xPSD));
x= f(100:150);
y= 10*log10(xPSD(100:150))
fill(x,y,'y')
the result is in the right direction but not what I need :
I would like get the color tell x axis like :
is their a way to do this
A working solution is:
[xPSD,f] = pwelch(x,hanning(4096),2048,4096*2 ,fs);
plot(f,10*log10(xPSD));
hold on
x= f(100:150);
y= 10*log10(xPSD(100:150));
yMax = ylim;
yMax = yMax(2);
x = x'; % Use this line only in the case that the size(x, 1) > 1
X = [x fliplr(x)];
Y = [y' ones(1, length(y)) .* yMax];
fill(X, Y, 'y')
What you were missing is that fill method looks for an area to fill. In the above code the area is defined by pairs of points. That, for the first (i.e. lower part) of the area we have the x vector and the y points. The second area (i.e. the upper part) is defined by the reversed vector x (image your pencil to first starting drawing towards rights for the lower part and then for the upper going left) and the points of the upper limit of your axes.
Edit:
Minimal example with the handel data from MATLAB:
load handel;
x = y; % Just to be consistent with the OP
fs = Fs; % Just to be consistent with the OP
[xPSD,f] = pwelch(x,hanning(4096),2048,4096*2 ,fs);
plot(f,10*log10(xPSD));
hold on
x= f(100:150);
y= 10*log10(xPSD(100:150));
yMax = ylim;
yMax = yMax(2);
x = x'; % Use this line only in the case that the size(x, 1) > 1
X = [x fliplr(x)];
Y = [y' ones(1, length(y)) .* yMax];
fill(X, Y, 'y')
xlim([0 200]) % To focus on the result
The result is:
Yes, there is always a way ;)
In your case, you simply need to add two points in x and y that go to the top boudary of the plot:
x = f(100:150);
y = 10*log10(xPSD(100:150))
% Add two points
Y = ylim;
x = [x(1) ; x(:) ; x(end)];
y = [Y(2) ; y(:) ; Y(2)];
% Display filled area
fill(x,y,'y')
Best,
Related
I am trying to generate a semi ellipsoidal dome shape by using x, y, z values. In my code below, I define the x,y,z values but I am unable to assign those values to sphere.
How do I resolve this?
clc
x = [65 55.2125 50.8267 46.7398 42.9232 39.3476 35.9815 32.7882 29.7175 26.6833 23.4690 18.7605];
y = x;
z = [0.0,0.9,2.7,5.2,8.2,11.8,15.8,20.3,25.2,30.7,37.1,47.5]; % max height of dome is 47.5
[x,y,z] = sphere(20);
x = x(12:end,:);
y = y(12:end,:);
z = z(12:end,:);
r = 65; % radius of the dome
surf(r.*x,r.*y,r.*z);
axis equal;
It will be simpler or at least more elegant to use the parametric equations of an ellipsoide.
% semi axis parameters
a = 65; % x-axis
b = 65; % y-axis
c = 47.5; % z-axis
%% Parametrisation
%
% To reach each point of the ellipsoide we need two angle:
% phi ∈ [0,2𝜋]
% theta ∈ [0, 𝜋]
%
% But since we only need half of an ellipsoide we can set
% theta ∈ [0,𝜋/2]
[theta,phi] = ndgrid(linspace(0,pi/2,50),linspace(0,2*pi,50));
x = a*sin(theta).*cos(phi);
y = b*sin(theta).*sin(phi);
z = c*cos(theta);
%plot
surf(x,y,z)
axis equal
Result:
You can remove the bottom half of the sphere by assigning NaNs to the approptiate elements of x and y:
[x,y,z] = sphere(20);
I = (z<0);
x(I) = NaN;
y(I) = NaN;
surf(x,y,z)
I'm using random points to determine the area below a curve (Monte-Carlo):
X: 1xn vector of x values for the function
Y: 1xn vector of y = f(x)
RP: mxn matrix of m random y for each x
I would like to split RP into RPA and RPB depending on it being above or below the curve. The idea is then to plot RPA and RPB against X, in different colors. This code doesn't work because RPA and RPB number of columns is not the same than X:
clf
f = #(x) sin(x/10) + cos(x/60); % Function
xMin = 1; xMax = 100; % x interval
X = [xMin:xMax];
Y = f(X);
plot(X,Y), hold on % Plot function
yMin = min(Y); yMax = max(Y); % Axes limits
set(gca, 'xlim', [xMin, xMax], 'ylim', [yMin, yMax])
m = 20; % Random points per x value
RP = rand(m, columns(X)) .* (yMax-yMin) .+ yMin;
% Split points (doesn't work)
RPA = RP(RP>Y);
RPB = RP(RP<=Y);
br = size(RPB) / size(RP) % Ratio of points below
a = (xMax - xMin) * (yMax - yMin) * br % Area below
% Plot points
plot(X, RPA, 'r.') % Above
plot(X, RPB, 'g.') % Below
Is there a possibility to build RPA and RPB so that they are the same size than RP, with the excluded points y being NaN or something similar, which can be counted and plotted?
You gave a good answer yourself. You can build RPA and RPB with strategic NaNs:
% Split points (works!)
RPA = RP;
RPA(RP<=Y) = NaN;
RPB = RP;
RPB(RPB > Y) = NaN;
And than calculating the ration as the not-NaN:
br = sum(~isnan(RPB)) / sum(~isnan(RP)) % Ratio of points below
I get this nice image:
I'm looking for a simple way for creating a random unit vector constrained by a conical region. The origin is always the [0,0,0].
My solution up to now:
function v = GetRandomVectorInsideCone(coneDir,coneAngleDegree)
coneDir = normc(coneDir);
ang = coneAngleDegree + 1;
while ang > coneAngleDegree
v = [randn(1); randn(1); randn(1)];
v = v + coneDir;
v = normc(v);
ang = atan2(norm(cross(v,coneDir)), dot(v,coneDir))*180/pi;
end
My code loops until the random generated unit vector is inside the defined cone. Is there a better way to do that?
Resultant image from test code bellow
Resultant frequency distribution using Ahmed Fasih code (in comments).
I wonder how to get a rectangular or normal distribution.
c = [1;1;1]; angs = arrayfun(#(i) subspace(c, GetRandomVectorInsideCone(c, 30)), 1:1e5) * 180/pi; figure(); hist(angs, 50);
Testing code:
clearvars; clc; close all;
coneDir = [randn(1); randn(1); randn(1)];
coneDir = [0 0 1]';
coneDir = normc(coneDir);
coneAngle = 45;
N = 1000;
vAngles = zeros(N,1);
vs = zeros(3,N);
for i=1:N
vs(:,i) = GetRandomVectorInsideCone(coneDir,coneAngle);
vAngles(i) = subspace(vs(:,i),coneDir)*180/pi;
end
maxAngle = max(vAngles);
minAngle = min(vAngles);
meanAngle = mean(vAngles);
AngleStd = std(vAngles);
fprintf('v angle\n');
fprintf('Direction: [%.3f %.3f %.3f]^T. Angle: %.2fº\n',coneDir,coneAngle);
fprintf('Min: %.2fº. Max: %.2fº\n',minAngle,maxAngle);
fprintf('Mean: %.2fº\n',meanAngle);
fprintf('Standard Dev: %.2fº\n',AngleStd);
%% Plot
figure;
grid on;
rotate3d on;
axis equal;
axis vis3d;
axis tight;
hold on;
xlabel('X'); ylabel('Y'); zlabel('Z');
% Plot all vectors
p1 = [0 0 0]';
for i=1:N
p2 = vs(:,i);
plot3ex(p1,p2);
end
% Trying to plot the limiting cone, but no success here :(
% k = [0 1];
% [X,Y,Z] = cylinder([0 1 0]');
% testsubject = surf(X,Y,Z);
% set(testsubject,'FaceAlpha',0.5)
% N = 50;
% r = linspace(0, 1, N);
% [X,Y,Z] = cylinder(r, N);
%
% h = surf(X, Y, Z);
%
% rotate(h, [1 1 0], 90);
plot3ex.m:
function p = plot3ex(varargin)
% Plots a line from each p1 to each p2.
% Inputs:
% p1 3xN
% p2 3xN
% args plot3 configuration string
% NOTE: p1 and p2 number of points can range from 1 to N
% but if the number of points are different, one must be 1!
% PVB 2016
p1 = varargin{1};
p2 = varargin{2};
extraArgs = varargin(3:end);
N1 = size(p1,2);
N2 = size(p2,2);
N = N1;
if N1 == 1 && N2 > 1
N = N2;
elseif N1 > 1 && N2 == 1
N = N1
elseif N1 ~= N2
error('if size(p1,2) ~= size(p1,2): size(p1,2) and/or size(p1,2) must be 1 !');
end
for i=1:N
i1 = i;
i2 = i;
if i > N1
i1 = N1;
end
if i > N2
i2 = N2;
end
x = [p1(1,i1) p2(1,i2)];
y = [p1(2,i1) p2(2,i2)];
z = [p1(3,i1) p2(3,i2)];
p = plot3(x,y,z,extraArgs{:});
end
Here’s the solution. It’s based on the wonderful answer at https://math.stackexchange.com/a/205589/81266. I found this answer by googling “random points on spherical cap”, after I learned on Mathworld that a spherical cap is this cut of a 3-sphere with a plane.
Here’s the function:
function r = randSphericalCap(coneAngleDegree, coneDir, N, RNG)
if ~exist('coneDir', 'var') || isempty(coneDir)
coneDir = [0;0;1];
end
if ~exist('N', 'var') || isempty(N)
N = 1;
end
if ~exist('RNG', 'var') || isempty(RNG)
RNG = RandStream.getGlobalStream();
end
coneAngle = coneAngleDegree * pi/180;
% Generate points on the spherical cap around the north pole [1].
% [1] See https://math.stackexchange.com/a/205589/81266
z = RNG.rand(1, N) * (1 - cos(coneAngle)) + cos(coneAngle);
phi = RNG.rand(1, N) * 2 * pi;
x = sqrt(1-z.^2).*cos(phi);
y = sqrt(1-z.^2).*sin(phi);
% If the spherical cap is centered around the north pole, we're done.
if all(coneDir(:) == [0;0;1])
r = [x; y; z];
return;
end
% Find the rotation axis `u` and rotation angle `rot` [1]
u = normc(cross([0;0;1], normc(coneDir)));
rot = acos(dot(normc(coneDir), [0;0;1]));
% Convert rotation axis and angle to 3x3 rotation matrix [2]
% [2] See https://en.wikipedia.org/wiki/Rotation_matrix#Rotation_matrix_from_axis_and_angle
crossMatrix = #(x,y,z) [0 -z y; z 0 -x; -y x 0];
R = cos(rot) * eye(3) + sin(rot) * crossMatrix(u(1), u(2), u(3)) + (1-cos(rot))*(u * u');
% Rotate [x; y; z] from north pole to `coneDir`.
r = R * [x; y; z];
end
function y = normc(x)
y = bsxfun(#rdivide, x, sqrt(sum(x.^2)));
end
This code just implements joriki’s answer on math.stackexchange, filling in all the details that joriki omitted.
Here’s a script that shows how to use it.
clearvars
coneDir = [1;1;1];
coneAngleDegree = 30;
N = 1e4;
sol = randSphericalCap(coneAngleDegree, coneDir, N);
figure;plot3(sol(1,:), sol(2,:), sol(3,:), 'b.', 0,0,0,'rx');
grid
xlabel('x'); ylabel('y'); zlabel('z')
legend('random points','origin','location','best')
title('Final random points on spherical cap')
Here is a 3D plot of 10'000 points from the 30° spherical cap centered around the [1; 1; 1] vector:
Here’s 120° spherical cap:
Now, if you look at the histogram of the angles between these random points at the coneDir = [1;1;1], you will see that the distribution is skewed. Here’s the distribution:
Code to generate this:
normc = #(x) bsxfun(#rdivide, x, sqrt(sum(x.^2)));
mysubspace = #(a,b) real(acos(sum(bsxfun(#times, normc(a), normc(b)))));
angs = arrayfun(#(i) mysubspace(coneDir, sol(:,i)), 1:N) * 180/pi;
nBins = 16;
[n, edges] = histcounts(angs, nBins);
centers = diff(edges(1:2))*[0:(length(n)-1)] + mean(edges(1:2));
figure('color','white');
bar(centers, n);
xlabel('Angle (degrees)')
ylabel('Frequency')
title(sprintf('Histogram of angles between coneDir and random points: %d deg', coneAngleDegree))
Well, this makes sense! If you generate points from the 120° spherical cap around coneDir, of course the 1° cap is going to have very few of those samples, whereas the strip between the 10° and 11° caps will have far more points. So what we want to do is normalize the number of points at a given angle by the surface area of the spherical cap at that angle.
Here’s a function that gives us the surface area of the spherical cap with radius R and angle in radians theta (equation 16 on Mathworld’s spherical cap article):
rThetaToH = #(R, theta) R * (1 - cos(theta));
rThetaToS = #(R, theta) 2 * pi * R * rThetaToH(R, theta);
Then, we can normalize the histogram count for each bin (n above) by the difference in surface area of the spherical caps at the bin’s edges:
figure('color','white');
bar(centers, n ./ diff(rThetaToS(1, edges * pi/180)))
The figure:
This tells us “the number of random vectors divided by the surface area of the spherical segment between histogram bin edges”. This is uniform!
(N.B. If you do this normalized histogram for the vectors generated by your original code, using rejection sampling, the same holds: the normalized histogram is uniform. It’s just that rejection sampling is expensive compared to this.)
(N.B. 2: note that the naive way of picking random points on a sphere—by first generating azimuth/elevation angles and then converting these spherical coordinates to Cartesian coordinates—is no good because it bunches points near the poles: Mathworld, example, example 2. One way to pick points on the entire sphere is sampling from the 3D normal distribution: that way you won’t get bunching near poles. So I believe that your original technique is perfectly appropriate, giving you nice, evenly-distributed points on the sphere without any bunching. This algorithm described above also does the “right thing” and should avoid bunching. Carefully evaluate any proposed algorithms to ensure that the bunching-near-poles problem is avoided.)
it is better to use spherical coordinates and convert it to cartesian coordinates:
coneDirtheta = rand(1) * 2 * pi;
coneDirphi = rand(1) * pi;
coneAngle = 45;
N = 1000;
%perfom transformation preventing concentration of points around the pole
rpolar = acos(cos(coneAngle/2*pi/180) + (1-cos(coneAngle/2*pi/180)) * rand(N, 1));
thetapolar = rand(N,1) * 2 * pi;
x0 = rpolar .* cos(thetapolar);
y0 = rpolar .* sin(thetapolar);
theta = coneDirtheta + x0;
phi = coneDirphi + y0;
r = rand(N, 1);
x = r .* cos(theta) .* sin(phi);
y = r .* sin(theta) .* sin(phi);
z = r .* cos(phi);
scatter3(x,y,z)
if all points should be of length 1 set r = ones(N,1);
Edit:
since intersection of cone with sphere forms a circle first we create random points inside a circle with raduis of (45 / 2) in polar coordinates and as #Ahmed Fasih commented to prevent concentration of points near the pole we should first transform this random points, then convert polar to cartesian 2D coordinates to form x0 and y0
we can use x0 and y0 as phi & theta angle of spherical coordinates and add coneDirtheta & coneDirphi as offsets to these coordinates.
then convert spherical to cartesian 3D coordinates
I like to visualize conics with Matlab or Octave. The (general) conic is given by the equation 0 = ax² + bxy + cy² +dxz +eyz+f*z² for a point p=(x,y,z). How can I plot this with Matlab or octave if I know the parameters a,b,c,d,e and f? Or respectively, how can I find the points that satisfy this equation?
Since you are asking for the conics, I understand that you are probably referring to the 2D contours of the general conic equation. I will also cover how to visualize this equation in other different ways.
For all the following examples, I have set the conic constants so that I obtain a hiperboloid.
My code is written in MATLAB syntax. If you are using Octave, it might differ slightly.
Visualizing 2D Conics with CONTOUR
I have isolated z in terms of x and y, from the general conic equation:
z = (1/2)*(-d*x-e*y±sqrt(-4*a*f*x.^2-4*b*f*x.*y-4*c*f*y.^2+d^2*x.^2+2*d*e*x.*y+e^2*y.^2))/f;
Since z is a piecewise function due to (± sqrt), I need to make sure that I plot both hemispheres. I designate z1 for +sqrt, and z2 for -sqrt.
Finally, I plot the contours for z1 and z2 that will yield the set of conics in 2D. This conics will be circles of different radius.
Code:
clear all;
clc;
% Conic constants.
a = 1;
b = 0;
c = 1;
d = 0;
e = 0;
f = -1;
% Value for x and y domain.
v = 10;
% Domain for x and y.
x = linspace(-v,v);
y = linspace(-v,v);
% Generate a 2D mesh with x and y.
[x,y] = meshgrid(x,y);
% Isolate z in terms of x and y.
z1 = (1/2)*(-d*x-e*y+sqrt(-4*a*f*x.^2-4*b*f*x.*y-4*c*f*y.^2+d^2*x.^2+2*d*e*x.*y+e^2*y.^2))/f;
z2 = (1/2)*(-d*x-e*y-sqrt(-4*a*f*x.^2-4*b*f*x.*y-4*c*f*y.^2+d^2*x.^2+2*d*e*x.*y+e^2*y.^2))/f;
% Find complex entries in z.
i = find(real(z1)~=z1);
j = find(real(z2)~=z2);
% Replace complex entries with NaN.
z1(i) = NaN;
z2(j) = NaN;
figure;
subplot(1,2,1);
% Draw lower hemisphere.
contour(x,y,z1,'ShowText','on');
% Adjust figure properties.
title('2D Conics: Lower hemishphere');
xlabel('x-axis');
ylabel('y-axis');
axis equal;
grid on;
box on;
axis([-10 10 -10 10]);
subplot(1,2,2);
% Draw upper hemisphere.
contour(x,y,z2,'ShowText','on');
hold off;
% Adjust figure properties.
title('2D Conics: Upper hemishphere');
xlabel('x-axis');
ylabel('y-axis');
axis equal;
grid on;
box on;
axis([-10 10 -10 10]);
Output:
Visualizing 3D Conics with CONTOUR3
Same as on the previous example, but now we plot the set of conics in 3D.
Code:
clear all;
clc;
% Conic constants.
a = 1;
b = 0;
c = 1;
d = 0;
e = 0;
f = -1;
% Value for x and y domain.
v = 10;
% Domain for x and y.
x = linspace(-v,v);
y = linspace(-v,v);
% Generate a 2D mesh with x and y.
[x,y] = meshgrid(x,y);
% Isolate z in terms of x and y.
z1 = (1/2)*(-d*x-e*y+sqrt(-4*a*f*x.^2-4*b*f*x.*y-4*c*f*y.^2+d^2*x.^2+2*d*e*x.*y+e^2*y.^2))/f;
z2 = (1/2)*(-d*x-e*y-sqrt(-4*a*f*x.^2-4*b*f*x.*y-4*c*f*y.^2+d^2*x.^2+2*d*e*x.*y+e^2*y.^2))/f;
% Find complex entries in z.
i = find(real(z1)~=z1);
j = find(real(z2)~=z2);
% Replace complex entries with NaN.
z1(i) = NaN;
z2(j) = NaN;
% Lower hemisphere. Draw 20 conics.
contour3(x,y,z1,20);
hold on;
% Upper hemisphere. Draw 20 conics.
contour3(x,y,z2,20);
hold off;
% Adjust figure properties.
title('3D Conics');
xlabel('x-axis');
ylabel('y-axis');
zlabel('z-axis');
axis equal;
grid on;
box on;
axis([-10 10 -10 10 -10 10]);
Output:
Visualizing Quadrics with ISOSURFACE
I have isolated f in terms of x, y and z, from the general conic equation, and renamed it to f_eq:
f_eq = -(a*x.^2+b*x.*y+c*y.^2+d*x.*z+e*y.*z)./z.^2;
Finally, I obtain the set of points that satisfy the equation f_eq = f, which is in fact an isosurface that yields a quadric; in this example a hiperboloid.
Code:
clear all;
clc;
% Conic constants.
a = 1;
b = 0;
c = 1;
d = 0;
e = 0;
f = -1;
% Value for x, y and z domain.
v = 10;
% Domain for x ,y and z.
x = linspace(-v,v);
y = linspace(-v,v);
z = linspace(-v,v);
% Generate a 3D mesh with x, y and z.
[x,y,z] = meshgrid(x,y,z);
% Evaluate function (3D volume of data).
f_eq = -(a*x.^2+b*x.*y+c*y.^2+d*x.*z+e*y.*z)./z.^2;
% Draw the surface that matches f_eq = f.
p = patch(isosurface(x,y,z,f_eq,f));
isonormals(x,y,z,f_eq,p)
p.FaceColor = 'red';
p.EdgeColor = 'none';
% Adjust figure properties.
title('Quadric');
xlabel('x-axis');
ylabel('y-axis');
zlabel('z-axis');
axis equal;
grid on;
box on;
axis([-10 10 -10 10 -10 10]);
camlight left;
lighting phong;
Output:
Im new to matlab and I want to plot a ContourPlot together with two inequalities but I dont know how. The function that I want to plot its ContourPlot is something like this:
Z = (X-2).^2 + (Y-1).^2;
and here are the two inequalities:
ineq1 = X.^2 - Y <= 2;
ineq2 = X + Y <= 2;
this is what I have dodne so far:
[X,Y] = meshgrid(-4:.2:4,-4:.2:4);
Z = (X-2).^2 + (Y-1).^2;
[C,h] = contour(X,Y,Z);
clabel(C,h)
ineq1 = X.^2 - Y <= 2;
ineq2 = X + Y <= 2;
range = (-4:.2:4);
hold on
plot(range,ineq1, range, ineq2)
hold off
but this does not feel right.
What I want to do is to visualize an optimization problem. First I want to plot the ContourPlot of the function and then the possible areas in that same plot, It would be great if I could show the intersection areas too.
If you want to draw the boundaries of the inequalities onto the contour plot, you can do this with line.
[X,Y] = meshgrid(-4:.2:4,-4:.2:4);
Z = (X-2).^2 + (Y-1).^2;
[C,h] = contour(X,Y,Z);
clabel(C,h)
x_vect = (-4:.05:4);
y_ineq1 = x_vect.^2 - 2;
y_ineq2 = 2 - x_vect;
line(x_vect, y_ineq1);
line(x_vect, y_ineq2);
Coloring the intersection area is a bit more tricky, and I'm not sure if it's exactly what you want to do, but you could look into using patch, something like
% Indexes of x_vect, y_ineq1 and y_ineq2 for region where both satisfied:
region_indexes = find(y_ineq1<y_ineq2);
region_indexes_rev = region_indexes(end:-1:1);
% To plot this area need to enclose it
% (forward along y_ineq1 then back along y_ineq2)
patch([x_vect(region_indexes),x_vect(region_indexes_rev)],...
[y_ineq1(region_indexes),y_ineq2(region_indexes_rev)],...
[1 0.8 1]) % <- Color as [r g b]
% May or may not need following line depending on MATLAB version to set the yaxes
ylim([-4 4])
% Get the original plot over the top
hold on
[C,h] = contour(X,Y,Z);
clabel(C,h)
hold off