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I want to know how i can generate different colours and use those in my figure and then place on the figure that what colours represents what. My figure will look like the attached image, its only a 3 case example but my final image consits of large numb...
Though this does work in generating equal separation, I'm looking for how I can get N number of coordinates (say 80 coordinates) all over a square (say 20 by 20), and if you were to join the coordinates with their neighbors, you would get an equilateral triangle.編...
numSides = 6; % <=== CHANGE THIS NUMBER theta = linspace(0, 2*pi, numSides + 1); % Rotate the shape by subtracting an offset. theta = theta - pi/3; radius = 5; x = radius * cos(theta) + xCenter; y = radius * sin(theta) + yCenter; plot(x, y); axis square; xlim(...
% ex_2ndOrder_filter_test%% Define representative inputsN = 256;% Number of pointst = linspace(0,1,N);% Time vector from 0 to 1 secondf1 = N/2;% Target frequency of chirp set to Nyquistx_chirp = sin(pi*f1*t.^2);% Linear chirp from 0 to Fs/2 Hz in 1 secondx_step = ...
sin(2*pi*(0:511)/512)(default) |vector of real values Frequency—Frequency of generated signal (Hz) 100(default) |real scalar Amplitude—Amplitude of generated signal 1(default) |real scalar PhaseOffset—Normalized phase offset of generated signal ...
Open in MATLAB Online Hello, I have created a synthetic time series (x) over time (tsyn), but I need to remake this so that the spacing between points intsynis both uneven and random between 1 and 5. tsyn = 1:1600; x = 2*sin(2*pi*tsyn/100)+1*sin(2*pi*tsyn/41)+0.5*si...
radius = 5.0;% radius in meters% Set the number of steps to divide the circular pathnumSteps = 100.0; numPoints = radius * numSteps;% Create [x,y] points along a circle with specified radiusxpoints = radius * sin(linspace(-pi,pi,numPoints)); ypoints = radius * cos(linspace(-pi,pi...
y = yc + r*sin(t); plot(x,y,'yo') In case you feel adventurous and want to place a large number of such random points within the circle just to test their uniformity, do this: % The random points n = 8192; r = R*sqrt(rand(n,1...
Looking at your sample picture it is easy to see that the infringing portion of circle 2 into circle 1 are points that are within the radius of circle 1. (vice versa for circle 2) With this we can calculate the distance of all points of circle 2 ...