So I have a Matlab function that creates a series of square impulses, then I apply a noise over them and a filter, the problem is I need to change the form of the impulses into triangular form:
x = zeros(1,1000)
x(100:200) = 1
x(400:500) = 1
x(700:800) = 1
plot(x)
Try to define a function that creates a triangle wave and then use it (or define it by hand everytime). Something like this should work fine:
x = zeros(1,1000);
tri = #(x) [(0:(floor(x/2)-1))/floor(x/2),1,((floor(x/2)-1):-1:0)/floor(x/2)];
x(100:200) = tri(101);
x(400:500) = tri(101);
x(700:800) = tri(101);
plot(x)
If this is not what you was asking for, eg if you meant that you want a sawtooth wave, then you should check out the sawtooth function, try:
x = 0:0.1:15;
y=sawtooth(x,0.5);
plot(x,y);
However, I would encourage you to change the name of the question, which is really about what to plot and not which plot function you are supposed to use.
Related
I have made this matlab script for a potential flow around a cylinder and I would like to add a point source. See the definition of point source in picture. How can you define theta in matlab? And plot this its streamlines Psi?
clear
% make axes
xymax = 2;
x = linspace(-xymax,xymax,100);
y = linspace(-xymax,xymax,100);
% note that x and y don't include 0
[xmesh,ymesh] = meshgrid(x,y);
x_c=0;
y_c=0;
q=1;
U=1
r = sqrt((xmesh-x_c).^2+(ymesh-y_c).^2);
sin_th= ((ymesh-y_c)./r)
%(ymesh-y_c)./r = sin(teta)
%(xmesh-x_c)./r = cos(teta)
psi1 = -q./r.*((ymesh-y_c)./r);
psi2 = r.*sin_th;
psi=psi1+psi2;
figure
contour(xmesh,ymesh,psi,[-xymax:.25:xymax],'-b');
To plot the streamlines in potential flow you are correct that you have to plot contour lines of constant stream function.
In the case of a point source, if you are plotting in cartesian coordinates in MATLAB you have to convert theta to cartesian coordinates using arctangent as follows: theta = arctan(y/x). In MATLAB use the function atan2 which will limit the number of discontinuities to between -PI and PI: https://www.mathworks.com/help/matlab/ref/atan2.html
Your code should read: psi2 = ( m/(2*PI) ) * atan2(y,x)
For more information on plotting elements of potential flow in 2D cartesian coordinates, see more information here:
https://potentialflow.com/flow-elements
https://potentialflow.com/equations
I am using the following MATLAB code to perform a Fourier transformation of a normal density function:
N=100;
j=0:(N-1);
a=-5;
b=5;
dx = (b-a)/N;
x = a+j*dx;
dt = 2*pi/(N*dx);
f1 = -N/2*dt;
f2 = N/2*dt;
t= f1+ j*dt;
GX = normpdf(x,0,1);
fft_GX = real(fft(GX))';
However, I do not get the expected bell shaped curve when I try to plot fft_GX.
The Fourier transformation of a normal density has the form of e^(-t^2/2). Can someone please help as to what I am doing incorrect?
Trying using abs instead of real.
Another helpful function to recenter the frequency domain is fftshift. Otherwise you will see the plot from 0 to 2*pi I believe, instead of the more recognizable view from -pi to pi.
fft_GX = abs(fftshift((fft(GX))');
plot(fft_GX);
You may need to do some further normalization based on the number of samples you have, but it looks more like the expected bell curve than what you were seeing originally.
I want to plot smooth contour plot from X Y Z matrix.
sf = fit([X Y] Z, 'poly23');
plot(sf);
I have not enought smooth curve..
What I need?
You can use the functions such as griddata and csaps. Together they will lead you to the result as smooth as you wish. The first function adds additional points to your data matrix set. The second one makes the result smoother. The example of the code is below. In the example smoothing is done first in X direction and then in Y direction. Try to play around with the resolution and smoothing_parameter (the current set of these parameters should be OK though).
x = min_x:step_x:max_x;
y = min_y:step_y:max_y;
resolution = 10;
xg = min_x:(step_x/resolution):max_x;
yg = min_y:(step_y/resolution):max_y;
[X,Y] = meshgrid(x,y);
[XG,YG] = meshgrid(xg,yg);
smoothing_parameter = 0.02;
fitted = griddata(X,Y,Z,XG,YG,'cubic');
fitted_smoothed_x = csaps(xg,fitted,smoothing_parameter,xg);
fitted_smoothed_xy = csaps(yg,fitted_smoothed_x',smoothing_parameter,yg);
surf(XG,YG,fitted_smoothed_xy');
EDIT: If you want to get just a contour plot, you can do, for example, as presented below. As I don't have the real data, I will use build-in function peaks to generate some.
[X,Y,Z] = peaks(30);
figure
surfc(X,Y,Z)
view([0 90])
zlim([-10 -8])
Here you just look at your contour plot from above being below the surface.
MATLAB's surf command allows you to pass it optional X and Y data that specify non-cartesian x-y components. (they essentially change the basis vectors). I desire to pass similar arguments to a function that will draw a line.
How do I plot a line using a non-cartesian coordinate system?
My apologies if my terminology is a little off. This still might technically be a cartesian space but it wouldn't be square in the sense that one unit in the x-direction is orthogonal to one unit in the y-direction. If you can correct my terminology, I would really appreciate it!
EDIT:
Below better demonstrates what I mean:
The commands:
datA=1:10;
datB=1:10;
X=cosd(8*datA)'*datB;
Y=datA'*log10(datB*3);
Z=ones(size(datA'))*cosd(datB);
XX=X./(1+Z);
YY=Y./(1+Z);
surf(XX,YY,eye(10)); view([0 0 1])
produces the following graph:
Here, the X and Y dimensions are not orthogonal nor equi-spaced. One unit in x could correspond to 5 cm in the x direction but the next one unit in x could correspond to 2 cm in the x direction + 1 cm in the y direction. I desire to replicate this functionality but drawing a line instead of a surf For instance, I'm looking for a function where:
straightLine=[(1:10)' (1:10)'];
my_line(XX,YY,straightLine(:,1),straightLine(:,2))
would produce a line that traced the red squares on the surf graph.
I'm still not certain of what your input data are about, and what you want to plot. However, from how you want to plot it, I can help.
When you call
surf(XX,YY,eye(10)); view([0 0 1]);
and want to get only the "red parts", i.e. the maxima of the function, you are essentially selecting a subset of the XX, YY matrices using the diagonal matrix as indicator. So you could select those points manually, and use plot to plot them as a line:
Xplot = diag(XX);
Yplot = diag(YY);
plot(Xplot,Yplot,'r.-');
The call to diag(XX) will take the diagonal elements of the matrix XX, which is exactly where you'll get the red patches when you use surf with the z data according to eye().
Result:
Also, if you're just trying to do what your example states, then there's no need to use matrices just to take out the diagonal eventually. Here's the same result, using elementwise operations on your input vectors:
datA = 1:10;
datB = 1:10;
X2 = cosd(8*datA).*datB;
Y2 = datA.*log10(datB*3);
Z2 = cosd(datB);
XX2 = X2./(1+Z2);
YY2 = Y2./(1+Z2);
plot(Xplot,Yplot,'rs-',XX2,YY2,'bo--','linewidth',2,'markersize',10);
legend('original','vector')
Result:
Matlab has many built-in function to assist you.
In 2D the easiest way to do this is polar that allows you to make a graph using theta and rho vectors:
theta = linspace(0,2*pi,100);
r = sin(2*theta);
figure(1)
polar(theta, r), grid on
So, you would get this.
There also is pol2cart function that would convert your data into x and y format:
[x,y] = pol2cart(theta,r);
figure(2)
plot(x, y), grid on
This would look slightly different
Then, if we extend this to 3D, you are only left with plot3. So, If you have data like:
theta = linspace(0,10*pi,500);
r = ones(size(theta));
z = linspace(-10,10,500);
you need to use pol2cart with 3 arguments to produce this:
[x,y,z] = pol2cart(theta,r,z);
figure(3)
plot3(x,y,z),grid on
Finally, if you have spherical data, you have sph2cart:
theta = linspace(0,2*pi,100);
phi = linspace(-pi/2,pi/2,100);
rho = sin(2*theta - phi);
[x,y,z] = sph2cart(theta, phi, rho);
figure(4)
plot3(x,y,z),grid on
view([-150 70])
That would look this way
It seems to be very basic question, but I wonder when I plot x values against y values, what interpolation technique is used behind the scene to show me the discrete data as continuous? Consider the following example:
x = 0:pi/100:2*pi;
y = sin(x);
plot(x,y)
My guess is it is a Lagrangian interpolation?
No, it's just a linear interpolation. Your example uses a quite long dataset, so you can't tell the difference. Try plotting a short dataset and you'll see it.
MATLAB's plot performs simple linear interpolation. For finer resolution you'd have to supply more sample points or interpolate between the given x values.
For example taking the sinus from the answer of FamousBlueRaincoat, one can just create an x vector with more equidistant values. Note, that the linear interpolated values coincide with the original plot lines, as the original does use linear interpolation as well. Note also, that the x_ip vector does not include (all) of the original points. This is why the do not coincide at point (~0.8, ~0.7).
Code
x = 0:pi/4:2*pi;
y = sin(x);
x_ip = linspace(x(1),x(end),5*numel(x));
y_lin = interp1(x,y,x_ip,'linear');
y_pch = interp1(x,y,x_ip,'pchip');
y_v5c = interp1(x,y,x_ip,'v5cubic');
y_spl = interp1(x,y,x_ip,'spline');
plot(x,y,x_ip,y_lin,x_ip,y_pch,x_ip,y_v5c,x_ip,y_spl,'LineWidth',1.2)
set(gca,'xlim',[pi/5 pi/2],'ylim',[0.5 1],'FontSize',16)
hLeg = legend(...
'No Interpolation','Linear Interpolation',...
'PChip Interpolation','v5cubic Interpolation',...
'Spline Interpolation');
set(hLeg,'Location','south','Fontsize',16);
By the way..this does also apply to mesh and others
[X,Y] = meshgrid(-8:2:8);
R = sqrt(X.^2 + Y.^2) + eps;
Z = sin(R)./R;
figure
mesh(Z)
No, Lagrangian interpolation with 200 equally spaced points would be an incredibly bad idea. (See: Runge's phenomenon).
The plot command simply connects the given (x,y) points by straight lines, in the order given. To see this for yourself, use fewer points:
x = 0:pi/4:2*pi;
y = sin(x);
plot(x,y)