I'm new to Matlab can anyone explain why the circle is measuring as 10 on the graph double to the radius. I expected it to be 5 matching the radius
xCenter = 5;
yCenter = 5;
theta = 0 : 0.01 : 2*pi;
radius = 5;
x = radius * cos(theta) + xCenter;
y = radius * sin(theta) + yCenter;
plot(x, y);
axis square;
xlim([0 10]);
ylim([0 10]);
grid on;
Thanks
If you want the circle to reach 5 on the x and y axes, then it should be centered at the origin.
Related
How can I flip the array data values (based on Y values) so when I am plotting it will be like a mirror plot? (instead of looking like a “mountain” it will look like a “valley”)
Code:
clc
clear
close all
y = [4 5 6 9 10 20 22 25 22 20 15 10 0];
x = 0:12;
data = rot90(cat(1, x, y));
flipData = flip(data);
figure('Name','Data','NumberTitle','off');
plot(data(:,1),data(:,2),'r','LineWidth',2);
figure('Name','Flip Data','NumberTitle','off');
plot(flipData(:,1),flipData(:,2),'r','LineWidth',2);
You can plot max(y) - y:
y2 = max(y) - y;
plot(x, y2, 'r', 'LineWidth', 2);
You can reverse the direction of the axis, so that your graph is upside down, but Y-values are still correct.
y = [4 5 6 9 10 20 22 25 22 20 15 10 0];
x = 0:12;
data = rot90(cat(1, x, y));
figure('Name','Data','NumberTitle','off');
plot(data(:,1),data(:,2),'r','LineWidth',2);
figure('Name','max(y) - y','NumberTitle','off');
y2 = max(y) - y;
plot(x, y2, 'r', 'LineWidth', 2);
figure('Name','ax.YDir reverse','NumberTitle','off');
plot(data(:,1),data(:,2),'r','LineWidth',2);
ax = gca;
ax.YDir = 'reverse';
I have two vectors as follows:
x = 0:5:50;
sir_dB = [50 20 10 5 2 0 -5 -10 -20 -20 -20]
Where x denotes the distance on the x-axis and sir_dB the SNR. For this, I need to generate a color map for a grid of 50 x 60m something similar to this:
based on the value of sir_dB.
I tried the following:
sir_dB = [50 20 10 5 2 0 -5 -10 -20 -20 -20];
xrange = 0:50;
yrange = -30:30;
% create candidate set
[X, Y] = ndgrid(xrange, yrange); % grid of points with a spacing of 1.
candidate_set = [X(:), Y(:)];
test_pt = [0 30];
radius = 5;
% find which of these are within the radius of selected point:
idx = rangesearch(candidate_set, test_pt, radius );
neighborhood = candidate_set(idx{1}, :);
Once I have the neighbors at a radius of 5m, I need to color that part of the grid based on the sir_dB value for a corresponding x value.
I need to have the plot in such a way that for all values of sir_dB greater than 15, the grid should be colored green, yellow for y greater than 0 and red for y greater than -20.
Could someone provide me inputs of how to do this best?
Im not sure exactly what you want, but this should get you started with contourf. I increased the granularity of xrange and yrange to make the radius more smooth but you can change it back if you want.
x = 0:5:50;
sir_dB = [50 20 10 5 2 0 -5 -10 -20 -20 -20];
xrange = 0:0.1:50;
yrange = -30:0.1:30;
% create candidate set
[X, Y] = ndgrid(xrange, yrange); % grid of points with a spacing of 1.
candidate_set = [X(:), Y(:)];
test_pt = [0 30];
r = sqrt((test_pt(1)-X(:)).^2 + (test_pt(2)-Y(:)).^2);
idx = r>5;
snr = nan(size(X));
snr(idx) = interp1(x,sir_dB,X(idx),'linear');
% Some red, yellow, green colors
cmap = [0.8500 0.3250 0.0980;
0.9290 0.6940 0.1250;
0 0.7470 0.1245];
figure();
colormap(cmap);
contourf(X,Y,snr,[-20,0,15],'LineStyle','none');
Plotting the the contour plot alongside the original sir_dB we see that it lines up (assuming you want linear interpolation). If you don't want linear interpolation use 'prev' or 'next' for the interp1 method.
figure();
colormap(cmap);
subplot(2,1,1);
contourf(X,Y,snr,[-20,0,15],'LineStyle','none');
subplot(2,1,2);
plot([0,50],[-20,-20],'-r',[0,50],[0,0],'-y',[0,50],[15,15],'-g',x,sir_dB);
Here is another suggestion, to use imagesc for that. I nothed the changes in the code below with % ->:
x = 0:5:50;
sir_dB = [50 20 10 5 2 0 -5 -10 -20 -20 -20];
xrange = 0:50;
yrange = -30:30;
% create candidate set
[X, Y] = ndgrid(xrange, yrange); % grid of points with a spacing of 1.
% -> create a map for plotting
Signal_map = nan(size(Y));
candidate_set = [X(:), Y(:)];
test_pt = [10 20];
radius = 35;
% find which of these are within the radius of selected point:
idx = rangesearch(candidate_set,test_pt,radius);
neighborhood = candidate_set(idx{1}, :);
% -> calculate the distance form the test point:
D = pdist2(test_pt,neighborhood);
% -> convert the values to SNR color:
x_level = sum(x<D.',2);
x_level(x_level==0)=1;
ColorCode = sir_dB(x_level);
% -> apply the values to the map:
Signal_map(idx{1}) = ColorCode;
% -> plot the map:
imagesc(xrange,yrange,rot90(Signal_map,2))
axis xy
% -> apply custom color map for g-y-r:
cmap = [1 1 1 % white
1 0 0 % red
1 1 0 % yellow
0 1 0];% green
colormap(repelem(cmap,[1 20 15 35],1))
c = colorbar;
% -> scale the colorbar axis:
caxis([-21 50]);
c.Limits = [-20 50];
c.Label.String = 'SNR';
The result:
I am facing some problem in creating the meshed rectangle. I know the max and min coordinate values (xmin, xmax, ymin, ymax, zmin, zmax). So, I can easily find the vertex of the rectangle which are
[xmin ymin zmin;
xmax ymin zmin;
xmax ymax zmin;
xmin ymax zmin;
xmin ymin zmax;
xmax ymin zmax;
xmax ymax zmax;
xmin ymax zmax]
Now how can I create the rectangle with the surface mesh?
You can do this using a patch in which you define vertices and the patch faces. For a rectangular prism, you can create your vertices and faces in the following way:
[xx,yy,zz] = ndgrid([xmin, xmax], [ymin, ymax], [zmin zmax]);
vertices = [xx(:), yy(:), zz(:)];
% Each row corresponds to a face of the prism and the values are indices into vertices
faces = [1 2 6 5
2 4 8 6
4 3 7 8
3 1 5 7
1 2 4 3
5 6 8 7];
p = patch('Faces', faces, ...
'Vertices', vertices, ...
'FaceColor', [0, 0.4470, 0.7410], ...
'FaceAlpha', 0.2); % Set transparency so we can see it
This question already has answers here:
How do I double the size of a vector in MATLAB with interpolation?
(2 answers)
Closed 8 years ago.
I have a row vector like
x = [ 0.125 0.25 0.5 0.75 1];
And I would like to expand it to 100 points with interpolation between points. Ho can i do it so that at the end I have equally spaced points but a length of 100 points?
Thanks
Solution
xi = [0 25 50 75 100];
yi = [0.125 0.25 0.5 0.75 1];
x = 1:1:100;
y = interp1(xi, yi, x);
Solution is y.
Explanation:
I consider your vector [0.125 0.25 0.5 0.75 1] as the result of a function F such that F(xi) = yi where xi = [0 25 50 75 100] and yi = [0.125 0.25 0.5 0.75 1].
I create x a 100-size vector using the same interval as xi;
I compute the interpolation of x based on the relationship between xi and yi;
I have gone through the related solutions to "Visualizing a toroidal surface in Matlab" but my problem is a little different.I want to be able to visualize segmented toroidal surfaces arranged with vertices attached to each other.Just like a stuck of semi circles with axis along the y-axis in the x-y plane.
`th = linspace( pi/2, -pi/2, 100);
R = 1; %or whatever radius you want
x = -R*cos(th) + 4;
y = -R*sin(th) + 7;
plot(x,y); axis equal;
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) + 9;
plot(x,y); axis equal;
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) + 11;
plot(x,y); axis equal;
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) + 5;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) + 3;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) + 1;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) -1;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) -3;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) -5;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) -11;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) -9;
plot(x,y); axis equal
hold on
x = -R*cos(th) + 4;
y = -R*sin(th) -7;`
plot(x,y); axis equal`