In the theory of tomography imaging a sinogram is recorderded, which is series of projections at different angles of the sample. Taking FFT of this projections gives a slice in polar coordinates of the sample in the frequency space.
The command [X,Y] = pol2cart(THETA,RHO) will not do it automatically. So, how is the polar to cartesian grid interpolation implemented numerically in 2D in Matlab?
You need to do a phase transformation:
theta = 0:0.1:2*pi;
rho = linspace(0,1,numel(theta));
[x,y] = pol2cart(-theta+pi/2,rho);
figure;
subplot(1,2,1);
polar(theta,rho);
subplot(1,2,2);
plot(y,x);
axis([-1 1 -1 1]);
grid on;
The function [X,Y] = pol2cart(THETA,RHO) only performs the coordinate value conversion, i.e., X = RHO * cos(THETA) and Y = RHO * sin(THETA). However, what you need is the conversion of the data array, thus pol2cart() can do nothing to your problem.
You can refer to the function interp2(). On the other hand, since this problem is the interpolation with COMPLEX data, I am NOT sure whether interp2() could do the job directly. I also need the theory for complex interpolation.
Related
I'm trying to fit a shape perserving interpolation curve to my angular data (r/phi).
However, as I have repeated x-values when I transform the datapoints to (x/y), I can not simply use pchip.
I know for spline interpolation, there is cscvn and fnplt, is there anything similar for pchip?
Furthermore, there is an example of spline fitting to angular data in the matlab documentation of "spline", but I don't quite get it how I could adapt it to pchip and different data points.
I also found the interparc-function by John d'Errico, but I would like to keep my datapoints instead of having equally spaced ones.
To make it clearer, here a figure of my datapoints with linear (blue) and spline interpolation (black). The curve I'd like to get would be something in between this two, without the steep edges in the linear case but with less overshoot than in the spline case....
Thanks for your help!
use 1D parametric interpolation:
n = 20;
r = 1 + rand(n-1,1)*0.01;%noisy r's
theta = sort(2*pi*rand(n-1,1));
% closing the circle
r(end+1) = r(1);
theta(end+1) = theta(1);
% convert to cartesian
[x,y] = pol2cart(theta,r);
% interpolate with parameter t
t = (1:n)';
v = [x,y];
tt = linspace(1,n,100);
X = interp1(t,v,tt,'pchip');
% plot
plot(x,y,'o');
hold on
plot(X(:,1),X(:,2));
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)
I'm trying to create a surface plot from (x,y,z) data on an irregular grid. The datasets are located along diagonal lines of positive gradient in the (x,y) plane. The method is illustrated below
xi = linspace (min(x), max(x), 1000);
yi = linspace (min(y), max(y), 1000);
zi = linspace (min(z), max(z), 400);
[XI YI]=meshgrid(xi,yi);
F = TriScatteredInterp (x,y,z);
Vi = F(XI,YI);
surf(Xi,Yi,Vi);
shading interp;
view(2)
An example result is shown below:
The data should be smooth in the x direction (so at y=860 there should be a single, continuous blue dip). I think the problem is that the interpolation is being carried out well along each dataset but poorly between them as the spacing between the datasets is far larger than between points within a dataset.
What would be the best way to deal with this?
I'm currently frustrated by the following problem:
I've got trajectory data (i.e.: Longitude and Latitude data) which I interpolate to find a linear fitting (using polyfit and polyval in matlab).
What I'd like to do is to rotate the axes in a way that the x-axis (the Longitude one) ends up lying on the best-fit line, and therefore my data should now lie on this (rotated) axis.
What I've tried is to evaluate the rotation matrix from the slope of the line-of-fit (m in the formula for a first grade polynomial y=mx+q) as
[cos(m) -sin(m);sin(m) cos(m)]
and then multiply my original data by this matrix...to no avail!
I keep obtaining a plot where my data lay in the middle and not on the x-axis where I expect them to be.
What am I missing?
Thank you for any help!
Best Regards,
Wintermute
A couple of things:
If you have a linear function y=mx+b, the angle of that line is atan(m), not m. These are approximately the same for small m', but very different for largem`.
The linear component of a 2+ order polyfit is different than the linear component of a 1st order polyfit. You'll need to fit the data twice, once at your working level, and once with a first order fit.
Given a slope m, there are better ways of computing the rotation matrix than using trig functions (e.g. cos(atan(m))). I always try to avoid trig functions when performing geometry and replace them with linear algebra operations. This is usually faster, and leads to fewer problems with singularities. See code below.
This method is going to lead to problems for some trajectories. For example, consider a north/south trajectory. But that is a longer discussion.
Using the method described, plus the notes above, here is some sample code which implements this:
%Setup some sample data
long = linspace(1.12020, 1.2023, 1000);
lat = sin ( (long-min(long)) / (max(long)-min(long))*2*pi )*0.0001 + linspace(.2, .31, 1000);
%Perform polynomial fit
p = polyfit(long, lat, 4);
%Perform linear fit to identify rotation
pLinear = polyfit(long, lat, 1);
m = pLinear(1); %Assign a common variable for slope
angle = atan(m);
%Setup and apply rotation
% Compute rotation metrix using trig functions
rotationMatrix = [cos(angle) sin(angle); -sin(angle) cos(angle)];
% Compute same rotation metrix without trig
a = sqrt(m^2/(1+m^2)); %a, b are the solution to the system:
b = sqrt(1/(1+m^2)); % {a^2+b^2 = 1}, {m=a/b}
% %That is, the point (b,a) is on the unit
% circle, on a line with slope m
rotationMatrix = [b a; -a b]; %This matrix rotates the point (b,a) to (1,0)
% Generally you rotate data after removing the mean value
longLatRotated = rotationMatrix * [long(:)-mean(long) lat(:)-mean(lat)]';
%Plot to confirm
figure(2937623)
clf
subplot(211)
hold on
plot(long, lat, '.')
plot(long, polyval(p, long), 'k-')
axis tight
title('Initial data')
xlabel('Longitude')
ylabel('Latitude')
subplot(212)
hold on;
plot(longLatRotated(1,:), longLatRotated(2,:),'.b-');
axis tight
title('Rotated data')
xlabel('Rotated x axis')
ylabel('Rotated y axis')
The angle you are looking for in the rotation matrix is the angle of the line makes to the horizontal. This can be found as the arc-tangent of the slope since:
tan(\theta) = Opposite/Adjacent = Rise/Run = slope
so t = atan(m) and noting that you want to rotate the line back to horizontal, define the rotation matrix as:
R = [cos(-t) sin(-t)
sin(-t) cos(-t)]
Now you can rotate your points with R