Here is the given system I want to plot and obtain the vector field and the angles they make with the x axis. I want to find the index of a closed curve.
I know how to do this theoretically by choosing convenient points and see how the vector looks like at that point. Also I can always use
to compute the angles. However I am having trouble trying to code it. Please don't mark me down if the question is unclear. I am asking it the way I understand it. I am new to matlab. Can someone point me in the right direction please?
This is a pretty hard challenge for someone new to matlab, I would recommend taking on some smaller challenges first to get you used to matlab's conventions.
That said, Matlab is all about numerical solutions so, unless you want to go down the symbolic maths route (and in that case I would probably opt for Mathematica instead), your first task is to decide on the limits and granularity of your simulated space, then define them so you can apply your system of equations to it.
There are lots of ways of doing this - some more efficient - but for ease of understanding I propose this:
Define the axes individually first
xpts = -10:0.1:10;
ypts = -10:0.1:10;
tpts = 0:0.01:10;
The a:b:c syntax gives you the lower limit (a), the upper limit (c) and the spacing (b), so you'll get 201 points for the x. You could use the linspace notation if that suits you better, look it up by typing doc linspace into the matlab console.
Now you can create a grid of your coordinate points. You actually end up with three 3d matrices, one holding the x-coords of your space and the others holding the y and t. They look redundant, but it's worth it because you can use matrix operations on them.
[XX, YY, TT] = meshgrid(xpts, ypts, tpts);
From here on you can perform whatever operations you like on those matrices. So to compute x^2.y you could do
x2y = XX.^2 .* YY;
remembering that you'll get a 3d matrix out of it and all the slices in the third dimension (corresponding to t) will be the same.
Some notes
Matlab has a good builtin help system. You can type 'help functionname' to get a quick reminder in the console or 'doc functionname' to open the help browser for details and examples. They really are very good, they'll help enormously.
I used XX and YY because that's just my preference, but I avoid single-letter variable names as a general rule. You don't have to.
Matrix multiplication is the default so if you try to do XX*YY you won't get the answer you expect! To do element-wise multiplication use the .* operator instead. This will do a11 = b11*c11, a12 = b12*c12, ...
To raise each element of the matrix to a given power use .^rather than ^ for similar reasons. Likewise division.
You have to make sure your matrices are the correct size for your operations. To do elementwise operations on matrices they have to be the same size. To do matrix operations they have to follow the matrix rules on sizing, as will the output. You will find the size() function handy for debugging.
Plotting vector fields can be done with quiver. To plot the components separately you have more options: surf, contour and others. Look up the help docs and they will link to similar types. The plot family are mainly about lines so they aren't much help for fields without creative use of the markers, colours and alpha.
To plot the curve, or any other contour, you don't have to test the values of a matrix - it won't work well anyway because of the granularity - you can use the contour plot with specific contour values.
Solving systems of dynamic equations is completely possible, but you will be doing a numeric simulation and your results will again be subject to the granularity of your grid. If you have closed form solutions, like your phi expression, they may be easier to work with conceptually but harder to get working in matlab.
This kind of problem is tractable in matlab but it involves some non-basic uses which are pretty hard to follow until you've got your head round Matlab's syntax. I would advise to start with a 2d grid instead
[XX, YY] = meshgrid(xpts, ypts);
and compute some functions of that like x^2.y or x^2 - y^2. Get used to plotting them using quiver or plotting the coordinates separately in intensity maps or surfaces.
Related
I have a 2D 2401*266 matrix K which corresponds to x values (t: stored in a 1*266 array) and y values(z: stored in a 1*2401 array).
I want to extrapolate the matrix K to predict some future values (corresponding to t(1,267:279). So far I have extended t so that it is now a 1*279 matrix using a for loop:
for tq = 267:279
t(1,tq) = t(1,tq-1)+0.0333333333;
end
However I am stumped on how to extrapolate K without fitting a polynomial to each individual row?
I feel like there must be a more efficient way than this??
There are countless of extrapolation methods in the literature, "fitting a polynomial to each row" would be just one of them, not necessarily invalid, not sure why you mention that you do no wan't to do it. For 2D data perhaps fitting a surface would lead to better results though.
However, if you want an easy, simple way (that might or might not work with your problem), you can always use the function interp2, for interpolation. If you chose spline or makima as interpolation functions, it will also extrapolate for any query point outside the domain of K.
Due to the nature of my problem, I want to evaluate the numerical implementations of the Radon transform in Matlab (i.e. different interpolation methods give different numerical values).
while trying to code my own Radon, and compare it to Matlab's output, I found out that my radon projection sizes are different than Matlab's.
So a bit of intuition of how I compute the amount if radon samples needed. Let's do the 2D case.
The idea is that the maximum size would be when the diagonal (in a rectangular shape at least) part is proyected in the radon transform, so diago=sqrt(size(I,1),size(I,2)). As we dont wan nothing out, n_r=ceil(diago). n_r should be the amount of discrete samples of the radon transform should be to ensure no data is left out.
I noticed that Matlab's radon output is always even, which makes sense as you would want a "ray" through the rotation center always. And I noticed that there are 2 zeros in the endpoints of the array in all cases.
So in that case, n_r=ceil(diago)+mod(ceil(diago)+1,2)+2;
However, it seems that I get small discrepancies with Matlab.
A MWE:
% Try: 255,256
pixels=256;
I=phantom('Modified Shepp-Logan',pixels);
rd=radon(I,pi/4);
size(rd,1)
s=size(I);
diagsize=sqrt(sum(s.^2));
n_r=ceil(diagsize)+mod(ceil(diagsize)+1,2)+2
rd=
367
n_r =
365
As Matlab's Radon transform is a function I can not look into, I wonder why could it be this discrepancy.
I took another look at the problem and I believe this is actually the right answer. From the "hidden documentation" of radon.m (type in edit radon.m and scroll to the bottom)
Grandfathered syntax
R = RADON(I,THETA,N) returns a Radon transform with the
projection computed at N points. R has N rows. If you do not
specify N, the number of points the projection is computed at
is:
2*ceil(norm(size(I)-floor((size(I)-1)/2)-1))+3
This number is sufficient to compute the projection at unit
intervals, even along the diagonal.
I did not try to rederive this formula, but I think this is what you're looking for.
This is a fairly specialized question, so I'll offer up an idea without being completely sure it is the answer to your specific question (normally I would pass and let someone else answer, but I'm not sure how many readers of stackoverflow have studied radon). I think what you might be overlooking is the floor function in the documentation for the radon function call. From the doc:
The radial coordinates returned in xp are the values along the x'-axis, which is
oriented at theta degrees counterclockwise from the x-axis. The origin of both
axes is the center pixel of the image, which is defined as
floor((size(I)+1)/2)
For example, in a 20-by-30 image, the center pixel is (10,15).
This gives different behavior for odd- or even-sized problems that you pass in. Hence, in your example ("Try: 255, 256"), you would need a different case for odd versus even, and this might involve (in effect) padding with a row and column of zeros.
Say for example I have data which forms the parabolic curve y=x^2, and I want to read off the x value for a given y value. How do I go about doing this in MATLAB?
If it were a straight line, I could just use the equation of the line of best fit to calculate easily, however I can't do this with a curved line. If I can't find a solution, I'll solve for roots
Thanks in advance.
If all data are arrays (not analytical expressions), I usually do that finding minimal absolute error
x=some_array;
[~,ind]=min(abs(x.^2-y0))
Here y0 is a given y value
If your data are represented by a function, you can use fsolve:
function y = myfun(x)
y=x^2-y0
[x,fval] = fsolve(#myfun,x0,options)
For symbolic computations, one can use solve
syms x
solve(x^2 - y0)
Assuming your two curves are just two vectors of data, I would suggest you use Fast and Robust Curve Intersections from the File Exchange. See also these two similar questions: how to find intersection points when lines are created from an array and Finding where plots may cross with octave / matlab.
Its just a basic question. I am fitting lines to scatter points using polyfit.
I have some cases where my scatter points have same X values and polyfit cant fit a line to it. There has to be something that can handle this situation. After all, its just a line fit.
I can try swapping X and Y and then fir a line. Any easier method because I have lots of sets of scatter points and want a general method to check lines.
Main goal is to find good-fit lines and drop non-linear features.
First of all, this happens due to the method of fitting that you are using. When doing polyfit, you are using the least-squares method on Y distance from the line.
(source: une.edu.au)
Obviously, it will not work for vertical lines. By the way, even when you have something close to vertical lines, you might get numerically unstable results.
There are 2 solutions:
Swap x and y, as you said, if you know that the line is almost vertical. Afterwards, compute the inverse linear function.
Use least-squares on perpendicular distance from the line, instead of vertical (See image below) (more explanation in here)
(from MathWorld - A Wolfram Web Resource: wolfram.com)
Polyfit uses linear ordinary least-squares approximation and will not allow repeated abscissa as the resulting Vandermonde matrix will be rank deficient. I would suggest trying to find something of a more statistical nature.
If you wish to research Andreys method it usually goes by the names Total least squares or Orthogonal distance regression http://en.wikipedia.org/wiki/Total_least_squares
I would tentatively also put forward the possibility of detecting when you have simultaneous x values, then rotating your data about the origin, fitting the line and then transform the line back. I could not say how poorly this would perform and only you could decide if it was an option based on your accuracy requirements.
Starting from the plot of one curve, it is possible to obtain the parametric equation of that curve?
In particular, say x={1 2 3 4 5 6....} the x axis, and y = {a b c d e f....} the corresponding y axis. I have the plot(x,y).
Now, how i can obtain the equation that describe the plotted curve? it is possible to display the parametric equation starting from the spline interpolation?
Thank you
If you want to display a polynomial fit function alongside your graph, the following example should help:
x=-3:.1:3;
y=4*x.^3-5*x.^2-7.*x+2+10*rand(1,61);
p=polyfit(x,y,3); %# third order polynomial fit, p=[a,b,c,d] of ax^3+bx^2+cx+d
yfit=polyval(p,x); %# evaluate the curve fit over x
plot(x,y,'.')
hold on
plot(x,yfit,'-g')
equation=sprintf('y=%2.2gx^3+%2.2gx^2+%2.2gx+%2.2g',p); %# format string for equation
equation=strrep(equation,'+-','-'); %# replace any redundant signs
text(-1,-80,equation) %# place equation string on graph
legend('Data','Fit','Location','northwest')
Last year, I wrote up a set of three blogs for Loren, on the topic of modeling/interpolationg a curve. They may cover some of your questions, although I never did find the time to add another 3 blogs to finish the topic to my satisfaction. Perhaps one day I will get that done.
The problem is to recognize there are infinitely many curves that will interpolate a set of data points. A spline is a nice choice, because it can be made well behaved. However, that spline has no simple "equation" to write down. Instead, it has many polynomial segments, pieced together to be well behaved.
You're asking for the function/mapping between two data sets. Knowing the physics involved, the function can be derived by modeling the system. Write down the differential equations and solve it.
Left alone with just two data series, an input and an output with a 'black box' in between you may approximate the series with an arbitrary function. You may start with a polynomial function
y = a*x^2 + b*x + c
Given your input vector x and your output vector y, parameters a,b,c must be determined applying a fitness function.
There is an example of Polynomial Curve Fitting in the MathWorks documentation.
Curve Fitting Tool provides a flexible graphical user interfacewhere you can interactively fit curves and surfaces to data and viewplots. You can:
Create, plot, and compare multiple fits
Use linear or nonlinear regression, interpolation,local smoothing regression, or custom equations
View goodness-of-fit statistics, display confidenceintervals and residuals, remove outliers and assess fits with validationdata
Automatically generate code for fitting and plottingsurfaces, or export fits to workspace for further analysis