I wanted to plot the above function on Matlab so I used the following code
ezplot('-log(x)-log(y)+x+y-2',[-10 10 -10 10]);
However I'm just getting a blank screen. But clearly there is at least the point (1,1) that satisfies the equation.
I don't think there is a problem with the plotter settings, as I'm getting graphs for functions like
ezplot('-log(y)+x+y-2',[-10 10 -10 10]);
I don't have enough rep to embed pictures :)
If we use solve on your function, we can see that there are two points where your function is equal to zero. These points are at (1, 1) and (0.3203 + 1.3354i, pi)
syms x y
result = solve(-log(x)-log(y)+x+y-2, x, y);
result.x
% -wrightOmega(log(1/pi) - 2 + pi*(1 - 1i))
% 1
result.y
% pi
% 1
If we look closely at your function, we can see that the values are actually complex
[x,y] = meshgrid(-10:0.01:10, -10:0.01:10);
values = -log(x)-log(y)+x+y-2;
whos values
% Name Size Bytes Class Attributes
% values 2001x2001 64064016 double complex
It seems as though in older versions of MATLAB, ezplot handled complex functions by only considering the real component of the data. As such, this would yield the following plot
However, newer versions consider the magnitude of the data and the zeros will only occur when both the real and imaginary components are zero. Of the two points where this is true, only one of these points is real and is able to be plotted; however, the relatively coarse sampling of ezplot isn't able to display that single point.
You could use contourc to determine the location of this point
imagesc(abs(values), 'XData', [-10 10], 'YData', [-10 10]);
axis equal
hold on
cmat = contourc(abs(values), [0 0]);
xvalues = xx(1, cmat(1,2:end));
yvalues = yy(cmat(2,2:end), 1);
plot(xvalues, yvalues, 'r*')
This is because x = y = 1 is the only solution to the given equation.
Note that the minimum value of x - log(x) is 1 and that happens when x = 1. Obviously, the same is true for y - log(y). So, -log(x)-log(y)+x+y is always greater than 2 except at x = y = 1, where it is exactly equal to 2.
As your equation has only one solution, there is no line on the plot.
To visualize this, let's plot the equation
ezplot('-log(x)-log(y)+x+y-C',[-10 10 -10 10]);
for various values of C.
% choose a set of values between 5 and 2
C = logspace(log10(5), log10(2), 20);
% plot the equation with various values of C
figure
for ic=1:length(C)
ezplot(sprintf('-log(x)-log(y)+x+y-%f', C(ic)),[0 10 0 10]);
hold on
end
title('-log(x)-log(y)+x+y-C = 0, for 5 < C < 2');
Note that the largest curve is obtained for C = 5. As the value of C is decreased, the curve also becomes smaller, until at C = 2 it completely vanishes.
Related
So essentially I have this code here that I can use to generate a 2D Random Walk discretely along N number of steps with M number of walkers. I can plot them all on the same graph here.
clc;
clearvars;
N = 500; % Length of the x-axis, also known as the length of the random walks.
M = 3; % The amount of random walks.
x_t(1) = 0;
y_t(1) = 0;
for m=1:M
for n = 1:N % Looping all values of N into x_t(n).
A = sign(randn); % Generates either +1/-1 depending on the SIGN of RAND.
x_t(n+1) = x_t(n) + A;
A = sign(randn); % Generates either +1/-1 depending on the SIGN of RAND.
y_t(n+1) = y_t(n) + A;
end
plot(x_t, y_t);
hold on
end
grid on;
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'Outerposition', [0, 0.05, 1, 0.95]);
axis square;
Now, I want to be able to Create graphs that show the distribution in space of the positions of a large number
(e.g. n = 1000) random walkers, at three different time points (e.g. t = 100, 200 and 300 or any three time points really).
I'm not sure how to go about this, I need to turn this into a function and iterate it through itself three different times and store the coordinates? I have a rough idea but iffy on actually implementing. I'd assume the safest and least messy way would be to use subplot() to create all three plots together in the same figure.
Appreciate any assistance!
You can use cumsum to linearize the process. Basically you only want to cumsum a random matrix composed of [-1 and 1].
clc;
close all;
M = 50; % The amount of random walks.
steps = [10,200,1000]; % here we analyse the step 10,200 and 1000
cc = hsv(length(steps)); % manage the color of the plot
%generation of each random walk
x = sign(randn(max(steps),M));
y = sign(randn(max(steps),M));
xs = cumsum(x);
xval = xs(steps,:);
ys = cumsum(y);
yval = ys(steps,:);
hold on
for n=1:length(steps)
plot(xval(n,:),yval(n,:),'o','markersize',1,'color',cc(n,:),'MarkerFaceColor',cc(n,:));
end
legend('10','200','1000')
axis square
grid on;
Results:
EDIT:
Thanks to #LuisMendo that answered my question here, you can use a binomial distribution to get the same result:
steps = [10,200,10000];
cc = hsv(length(steps)); % manage the color of the plot
M = 50;
DV = [-1 1];
p = .5; % probability of DV(2)
% Using the #LuisMendo binomial solution:
for ii = 1:length(steps)
SDUDx(ii,:) = (DV(2)-DV(1))*binornd(steps(ii), p, M, 1)+DV(1)*steps(ii);
SDUDy(ii,:) = (DV(2)-DV(1))*binornd(steps(ii), p, M, 1)+DV(1)*steps(ii);
end
hold on
for n=1:length(steps)
plot(SDUDx(n,:),SDUDy(n,:),'o','markersize',1,'color',cc(n,:),'MarkerFaceColor',cc(n,:));
end
legend('10','200','1000')
axis square
grid on;
What is the advantage ? Even if you have a big number of steps, let's say 1000000, matlab can handle it. Because in the first solution you have a bruteforce solution, and in the second case a statistical solution.
If you want to show the distribution of a large number, say 1000, of these points, I would say the most suitable way of plotting is as a 'point cloud' using scatter. Then you create an array of N points for both the x and the y coordinate, and let it compute the coordinate in a loop for i = 1:Nt, where Nt will be 100, 200, or 300 as you describe. Something along the lines of the following:
N = 500;
x_t = zeros(N,1);
y_t = zeros(N,1);
Nt = 100;
for tidx = 1:Nt
x_t = x_t + sign(randn(N,1));
y_t = y_t + sign(randn(N,1));
end
scatter(x_t,y_t,'k*');
This will give you N x and y coordinates generated in the same way as in the sample you provided.
One thing to keep in mind is that sign(0)=0, so I suppose there is a chance (admittedly a small one) of not altering the coordinate. I am not sure if you intended this behaviour to be possible (a walker standing still)?
I will demonstrate the 1-dimensional case for clarity; you only need to implement this for each dimension you add.
Model N steps for M walkers using an NxM matrix.
>> N = 5;
>> M = 4;
>> steps = sign(randn(N,M));
steps =
1 1 1 1
-1 1 -1 1
1 -1 -1 -1
1 1 -1 1
1 -1 -1 -1
For plotting, it is useful to make a second NxM matrix s containing the updated positions after each step, where s(N,M) gives the position of walker M after N steps.
Use cumsum to vectorize instead of looping.
>> s = cumsum(steps)
s =
1 1 1 1
0 2 0 2
1 1 -1 1
2 2 -2 2
3 1 -3 1
To prevent plot redraw after each new line, use hold on.
>> figure; hold on
>> plot(1:N, s(1:N, 1:M), 'marker', '.', 'markersize', 20, 'linewidth', 3)
>> xlabel('Number of steps'); ylabel('Position')
The output plot looks like this: picture
This method scales very well to 2- and 3-dimensional random walks.
I would like to create a density plot of this function:
In Maple, one could use the densityplot function to achieve this (code at the end), which gives:
However, I am not sure what to use for plotting a similar figure in MATLAB.
Here is my current MATLAB code:
x = [0:10:100];
y = [-50:10:50];
s = [10, 0];
i = [50,25];
for ii = 1 : length(x)
sir(ii) = -10 * 9.8 * log10((power((x(ii) - s(1)),2) + power((y(ii) - s(2)),2)) / (power((x(ii) - i(1)),2) + power((y(ii) - i(2)),2)));
end
Could someone suggest an equivalent in MATLAB?
For the density plot in Maple, I used
densityplot(sir(x,y), x=0..100, y=-50..50, axes=boxed, style=patchnogrid, scaletorange=-5..50, colorscheme = [black, "green", "white"])
You can use surf (a 3D surface plot) to achieve this, but you will need a finer grid than steps of 10 for it to look good!
Also you will need meshgrid to get all combinations of the x and y coordinates.
Please see the comments for further details.
% Set up grid points
x = 0:0.1:100;
y = -50:0.1:50;
[x,y] = meshgrid(x,y);
% Set up parameters i, s and g
i = [50 25]; s = [10 0]; g = 9.8;
% Work out density
% - no need for loop if we use element-wise operations ./ and .^
% - power(z,2) replaced by z.^2 (same function, more concise)
% - You forgot the sqare roots in your question's code, included using .^(1/2)
% - line continuation with "...", could remove and have on one line
sir = -10*g*log10( ((x-s(1)).^2 + (y-s(2)).^2).^(1/2) ./ ...
((x-i(1)).^2 + (y-i(2)).^2).^(1/2) );
% Plot, and set to a view from above
surf(x,y,sir,'edgecolor','none','facecolor','interp');
view(2);
% Change the colour scheme
colormap('bone')
Result:
Matching your example
You used the Maple command scaletorange=-5..50. This limits the scale between -5 and 50 (docs), so since sir is our scale variable, we should limit it the same. In MATLAB:
% Restrict sir to the range [-5,50]
sir = min(max(sir,-5),50);
% Of course we now have to replot
surf(x,y,sir,'edgecolor','none','facecolor','interp');
view(2);
Now, if you wanted the black/green colours, you can use a custom colormap, this would also smooth out the banding caused by the 'bone' colormap only having 64 colours.
% Define the three colours to interpolate between, and n interpolation points
black = [0 0 0]; green = [0 1 0]; white = [1 1 1];
n = 1000;
% Do colour interpolation, equivalent to Maple's 'colorscheme = [black, "green", "white"]'
% We need an nx3 matrix of colours (columns R,G,B), which we get using interp1
colormap(interp1(1:3, [black; green; white], linspace(1,3,n)));
With g=3.5 (not sure what you used), we get an almost identical plot
I have a function which is basically recreating the freqz command in matlab. I have figured out how to plot the entire transform of my frequency response, but I only need half of it, and I need to normalize it from pi to 1 (where 0:pi represents my x axis, and I want that to go to 0:1). The code is here:
function freqrespplot(b, a)
hb = fft(b,512);
ha = fft(a,512);
magh = (abs(hb) ./ abs(ha));
angh = angle(hb ./ ha);
x = linspace(0,2*pi, 512);
subplot(2,1,1);
plot(x,magh,'g');
subplot(2,1,2);
plot(x,angh,'r');
end
In the example of b = [1 2 2], a = [0 1 .8], plots like the following:
freqrespplot([1 2 2], [0 1 .8]);
Magnitude
and my corresponding phase plot is
I want the x-axis (omega) to go from 0 to 1 in both cases, and correspond to 0 to pi by taking only half of the graph, like this if possible:
You only need some small changes, marked with comments in the code below:
function freqrespplot(b, a)
hb = fft(b,512);
ha = fft(a,512);
magh = (abs(hb) ./ abs(ha));
angh = angle(hb ./ ha);
x = linspace(0,2, 513); %// 2, not 2*pi. And 513. Last value will be discarded
x = x(1:end-1); %// discard last value to avoid having 0 and 2*pi (they are the same)
subplot(2,1,1);
plot(x(1:end/2),magh(1:end/2),'g'); %// plot only half
subplot(2,1,2);
plot(x(1:end/2),angh(1:end/2),'r'); %// plot only half
end
With your example b and a this produces
You may want to include one additional sample to reach the right edge of the graph. I comment only differences with the above code:
function freqrespplot(b, a)
hb = fft(b,512);
ha = fft(a,512);
magh = (abs(hb) ./ abs(ha));
angh = angle(hb ./ ha);
x = linspace(0,2, 513);
x = x(1:end-1);
subplot(2,1,1);
plot(x(1:end/2+1),magh(1:end/2+1),'g'); %// plot only lower half plus one value
subplot(2,1,2);
plot(x(1:end/2+1),angh(1:end/2+1),'r'); %// plot only lower half plus one value
end
Compare the resulting graph (rightmost part):
I'm trying to represent a the intersection of two fuzzy sets as a 3d mesh in MatLab.
Here are my sets of vectors:
x = [0.3 0.5 0.7]
y = [0.5 0.7 0.1]
Followed by these statements:
[u,v] = meshgrid(x,y)
w = min(u,v)
mesh(u,v,w)
The x and y ticks seem to be all over the place and do not correlate to the actual index number of each vector i.e. 1 to 3, and the graph should represent the shape of a small triangle/T-norm.
At the moment it looks like this:
Here is an example out of my book I'm following:
Ignore what looks like fractions, they are delimiters. Here is the resulting graph:
After looking up fuzzy sets and intersections, here's what I've come up with. First, let's reproduce the textbook example:
% possible values and associated degrees of truth for F
Fv = 1 : 5;
Ft = [0 0.5 1 0.5 0];
% possible values and associated degrees of truth for D
Dv = 2 : 4;
Dt = [0 1 0];
% determine degrees of truth for fuzzy intersection
It = bsxfun(#min, Ft', Dt);
% plot
h = mesh(Dv, Fv, It);
set(h, 'FaceColor', 'none')
set(h, 'EdgeColor', 'k')
xlim([0 4.5])
ylim([0 5])
xlabel D
ylabel F
view(37.5, 30)
The result is:
Not as pretty as in your book, but the same thing.
Applying the same code to your example yields:
Via the arguments u,v you are telling mesh to use the values in them, i.e. the values from x and y, for the positioning of the data points and corresponding ticks. If you just want positions and ticks at 1, 2, 3, leave these arguments out.
mesh(w)
I have two vectors, c and d, whose histogram I need to plot side by side in the same figure in matlab. when i do
hist(c);
hold on;
hist(d)
the scale changes and I cant see the histogram of c vector. Where am i going wrong? Any help will be appreciated.
If you want the two to be in the same figure, you could try adjusting the X and Y limits to suit your needs (try help xlim and help ylim). However plotting them in the same figure might not always suit your needs, as a particular plot has to of course maintain a certain limit for X and Y.
If displaying them side by side in different figures would suffice however, you could consider using subplot():
>> A=[1 1 1 2 2];
>> B=[1 2 2 2 2];
>> figure(1);
>> hold on;
>> subplot(1,2,1);
>> hist(A);
>> subplot(1,2,2);
>> hist(B);
Resultant figure:
Notice how the different axis limits are maintained.
You can use axis([xmin xmax ymin ymax]) to control the x and y axis and select a range that will display both histograms. Depending on what you want your plot to look like, you may also want to try using nelements = hist(___) to get the number of elements in each bin and then plot them using bar(x,nelements) to control the location of each bar.
hist assumes you want to divide the range into 10 equal sized bins by default. If you want to use the same bins for both histograms, first find the range of your values and make a set of bin centers (e.g. binCenters = linspace(min(x), max(x), 15)'), then callhist(x, binCenters)`.
I use MATLAB histograms quite frequently and have wrote this small matlab script to plot two histograms (first one red and second blue) in one figure. The script is quite simple but the important thing is that the histograms should be comparable (i.e. equally spaced frequency bins).
function myhist(varargin)
% myhist function to plot the histograms of x1 and x2 in a single figure.
% This function uses the same xvalue range and same bins to plot the
% histograms, which makes comparison possible.
if nargin<2
x1 = cell2mat(varargin(1));
x2 = x1;
res = 100;
elseif nargin==2
x1 = cell2mat(varargin(1));
if length(cell2mat(varargin(2)))==1
res = cell2mat(varargin(2));
x2 = x1;
else
x2 = cell2mat(varargin(2));
res = 100;
end
elseif nargin>2
x1 = cell2mat(varargin(1));
x2 = cell2mat(varargin(2));
res = cell2mat(varargin(3));
end
if numel(x1)~=length(x1) || numel(x2)~=length(x2)
error('Inputs must be vectors.')
return
end
xrangel = max(min(x1),min(x2));
xrangeh = min(max(x1),max(x2));
x1_tmp = x1(x1>=xrangel & x1<=xrangeh);
x2_tmp = x2(x2>=xrangel & x2<=xrangeh);
xbins = xrangel:(xrangeh - xrangel)/res:xrangeh;
hist(x1_tmp,xbins)
hold on
h = findobj(gca,'Type','patch');
set(h,'FaceColor','r','EdgeColor','w');
hist(x2_tmp,xbins)