I want to plot a function y[n] = x[n+2]. My problem is that it does not plot in right range or even does not draw the zero sample points.
n = 1:6;
x = 1:1:8;
f = figure;
subplot(1,2,1)
stem(n, x(n));
axis([-3,8, 0, 7]);
xlabel('n');
ylabel('x[n]');
title('Subplot 1')
subplot(1,2,2)
stem(n, x(n + 2));
xlabel('n');
ylabel('y[n]');
title('Subplot 2')
How to change the variables n or x to get the right plot?
In the end, it ought to look like this:
You are confusing the concept of indices with your dependent variable. You should construct a function x which transforms an input n using the relationship that you know
function y = x(n)
% Set all outputs to 0
y = zeros(size(n));
% Replace the values that fall between 0 and 6 with their same value
y(n >= 0 & n <= 6) = n(n >= 0 & n <= 6);
end
Then you should pass this function a range of n values to evaluate.
nvalues = -3:8;
yvalues = x(nvalues);
stem(nvalues, yvalues)
You can also apply a transformation to the n values
nvalues = -3:8;
yvalues = x(nvalues + 2);
stem(nvalues, yvalues)
Related
I have the following function :
ySol2=(2*(x^3 + 1)^(1/2))/cos(x) +2/cos(x)
my question is how can I draw it in the interval 1000;5000 with the constrains cos(x) != 0 and x> -1?
Something like this perhaps?
% your function. Note element-wise operations (.^, ./)
ySol2 = #(x) (2*(x.^3 + 1).^(1/2))./cos(x) +2./cos(x);
% your x interval, with step 0.1. Note also that the argument for 'cos' is in [rad].
x = 1000:0.1:5000;
% calculate function values
y = ySol2(x);
% keep only non-constrained values
ind = (cos(x)~=0) | (x>-1);
x = x(ind);
y = y(ind);
% plot
figure;
plot(x,y);
As shown by the solid and dashed line, I'd like to create a function where I set a threshold for y (Intensity) from that threshold it gives me corresponding x value (dashed line). Very simple but my while statement is off. Any help would be much appreciated!
%% Curve fit plotting %%
x1 = timeStamps(1:60); % taking timestamps from 1 - 120 given smoothed y1 values
y1 = smooth(tic_lin(1:60),'sgolay',1);
% Find coefficients for polynomial (order = 4 and 6, respectively)
fitResults1 = polyfit(x1',y1, 7);
% evaluate the fitted y-values
yplot1 = polyval(fitResults1,x1');
% interpolates to find yi, the values of the underlying function Y at the points in the vector or array xi. x must be a vector.
Time_points = interp1(yplot1, x1', yplot1);
figure( 'Name', 'Curvefit1_poly' );
h = plot(x1', y1);%smoothed-points
hold on;
plot(x1', yplot1);%polyfit points
hold on;
plot(Time_points, yplot1, '*r');%interpolated points of x given y
%given y-threshold, output x(corresponding time_point).
broken = false;
while broken == false
if yplot1 >= 2024671226502.99
index = find(yplot1);
xDesired = x1(index);
broken = true;
else
disp("next iteration through");
end
end
No while loop is needed here... You can do this with logical indexing for the threshold condition and find to get the first index:
% Start with some x and y data
% x = ...
% y = ...
% Get the first index where 'y' is greater than some threshold
thresh = 10;
idx = find( y >= thresh, 1 ); % Find 1st index where y >= thresh
% Get the x value at this index
xDesired = x( idx );
Note that xDesired will be empty if there was no y value over the threshold.
Alternatively, you already have a polynomial fit, so you could use fzero to get the x value on that polynomial for a given y (in this case your threshold).
% x = ...
% y = ...
thresh = 10;
p = polyfit( x, y, 3 ); % create polynomial fit
% Use fzero to get the root of y = a*x^n + b*x^(n-1) + ... + z when y = thresh
xDesired = fzero( #(x) polyval(p,x) - thresh, x(1) )
Note, this method may give unexpected results if the threshold is not within the range of y.
I'm trying to plot a piecewise function as an interpolation for the function f(x) = 1/(1+25x^2). This is how I plotted two functions previously when I wasn't dealing with piecewise.
z = linspace(-1,1,200);
yexact = 1./(1+25.*z.^2);
plot(z,yexact)
N=2;
x = size(N+1);
for i = 1:(N+1)
x(i) = -1+(1+cos(((2*i+1)*pi)/(2*(N+1))));
end
a = polyfit(x,1./(1+25.*x.^2),N);
yinter = polyval(a,z);
plot(z,yexact,z,yinter);
title('N = 2');
legend('exact','interpolation');
This was done for N = 2, 5, 10, 15, 20, 30. Now I need to change this to work for piecewise with the same N values. The x(i)'s are the intervals and the P(i)'s are the slopes of the piecewise function. So for N = 2, I need to plot P(1) from x(1) to x(2) and P(2) from x(2) to x(3).
N=2;
x = size(N+1);
P = size(N);
for i = 1:(N+1)
x(i) = -1 + 2*i/N;
end
for i = 1:N
P(i) = (1/(1+25*(x(i)).^2)) + ((i-1-x(i))/(x(i+1)-x(i)))*((1/(1+25*(x(i+1)).^2))-(1/(1+25*(x(i)).^2)));
end
All you have to do is to define your N values as a vector and, then, iterate over it. Into each iteration, the result returned by the computation over the current N value is plotted over the existing axis (for more information, visit this link of the official Matlab documentation).
Here is an example:
z = linspace(-1,1,200);
yexact = 1./(1+25.*z.^2);
plot(z,yexact);
hold on;
for n = [2 5 10]
x = size(n+1);
for i = 1:(n+1)
x(i) = -1+(1+cos(((2*i+1)*pi)/(2*(n+1))));
end
a = polyfit(x,1./(1+25.*x.^2),n);
yinter = polyval(a,z);
plot(z,yinter);
end
hold off;
And here is the resulting output:
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
Kind all,
I am working in MATLAB and I'm using Monte Carlo techniques to fit a model. Basically, if we assume that my model is a simple function such as
y=m*x^2+c
And that both my parameters m and c vary between 0.5 and 10, I may randomly draw from such a parameter space and obtain each time a new y. If I plot all my realizations of y I obtain something similar to the following figure:
Is there a way to represent the DENSITY of the realizations? I mean, is there a way (instead of plotting all the realizations) to obtain some kind of contour plot that lies between the minimum of my iterations and the maximum for which its color represents the amount of realizations that fall within a certain interval?
Thanks all!
This isn't very pretty, but you could vary the parameters and play with the scatter/plotting, to make it a bit more visually appealing.
Also I assumed a gaussian distribution instead of your random one (totally random coloring will give you a uniform density). Also this code could be optimized for speed.
n = 1000;
l = 100;
x = linspace(1, 10, l);
y = repmat(x.^2, n, 1);
c = repmat(normrnd(1, 1, n, 1), 1, l);
m = repmat(normrnd(1, 1, n, 1), 1, l);
y = y.*m + c;
p = plot(y', '.');
figure; hold on;
for i = 1:l
[N,edges,bin] = histcounts(y(:, i));
density = N./sum(N);
c = zeros(n, 3);
for j = 1:n
c(j, :) = [1-density(bin(j))/max(density), 1-density(bin(j))/max(density), 1-density(bin(j))/max(density)];
end
scatter(ones(n, 1)*x(i),y(:, i),[],c,'filled');
end
Gives
This creates a histogram of y values for every position in x, then calculates the probability density for each y-value and colors in the points. Here, the y-values for every position x are normalized individually, to color the points according to the overall density of the plot you need to renormalize.
You can calculate y for discrete points of x, while setting random values for c and m. Then using hist function you can find a "not-normalized density" of function values for a given x. You can then normalize it to get a real density of the values, so that the overall area under the distribution curve sums up to 1.
In order to visualize it, you construct a mesh grid [X, Y] along the values of x and y and put the density values as Z. Now you can either plot the surf or its contour plot.
Here is the code:
clear;
n = 1000000; %number of simulation steps
%parameter ranges
m_min = 0.5; m_max = 10;
c_min = 0.5; c_max = 10;
%x points
x_min = 1; x_max = 4; x_count = 100;
x = linspace(x_min, x_max, x_count);
x2 = x.^2;
y_min = 0; y_max = m_max*x_max*x_max + c_max; y_step = 1;
m = rand(n, 1)*(m_max - m_min) + m_min;
c = rand(n, 1)*(c_max - c_min) + c_min;
c = repmat(c, 1, x_count);
y = m*x2 + c;
x_step = (x_max- x_min)/(x_count-1);
[X, Y] = meshgrid(x_min:x_step:x_max, y_min-y_step:y_step:y_max+y_step);
Z = zeros(size(X));
bins = y_min:y_step:y_max;
for i=1:x_count
[n_hist,y_hist] = hist(y(:, i), bins);
%add zeros on both sides to close the profile
n_hist = [0 n_hist 0];
y_hist = [y_min-y_step y_hist y_max+y_step];
%normalization
S = trapz(y_hist,n_hist); %area under the bow
n_hist = n_hist/S; %scaling of the bow
Z(:, i) = n_hist';
end
surf(X, Y, Z, 'EdgeColor','none');
colormap jet;
xlim([x_min x_max]);
ylim([y_min y_max]);
xlabel('X');
ylabel('Y');
figure
contour(X,Y,Z, 20);
colormap jet;
colorbar;
grid on;
title('Density as function of X');
xlabel('X');
ylabel('Y');
Another interesting view is a plot of each section depending on the x value:
Here is the code for this plot:
clear;
n = 1000000; %number of simulation steps
%parameter ranges
m_min = 0.5; m_max = 10;
c_min = 0.5; c_max = 10;
%x points
x_min = 1; x_max = 4; x_count = 12;
x = linspace(x_min, x_max, x_count);
x2 = x.^2;
m = rand(n, 1)*(m_max - m_min) + m_min;
c = rand(n, 1)*(c_max - c_min) + c_min;
c = repmat(c, 1, x_count);
y = m*x2 + c;
%colors for the plot
colors = ...
[ 0 0 1; 0 1 0; 1 0 0; 0 1 1; 1 0 1; 0 0.75 0.75; 0 0.5 0; 0.75 0.75 0; ...
1 0.50 0.25; 0.75 0 0.75; 0.7 0.7 0.7; 0.8 0.7 0.6; 0.6 0.5 0.4; 1 1 0; 0 0 0 ];
%container for legend entries
legend_list = cell(1, x_count);
for i=1:x_count
bin_number = 30; %number of histogramm bins
[n_hist,y_hist] = hist(y(:, i), bin_number);
n_hist(1) = 0; n_hist(end) = 0; %set first and last values to zero
%normalization
S = trapz(y_hist,n_hist); %area under the bow
n_hist = n_hist/S; %scaling of the bow
plot(y_hist,n_hist, 'Color', colors(i, :), 'LineWidth', 2);
hold on;
legend_list{i} = sprintf('Plot of x = %2.2f', x(i));
end
xlabel('y');
ylabel('pdf(y)');
legend(legend_list);
title('Density depending on x');
grid on;
hold off;