As my question states, I want to calculate the Fourier transform F(q) of a radial function f(r) (defined on [0,infinity[ and which decays like an exponential exp(-Ar +b) at large r) as accurately as possible in Fortran. The function values come from a data file (which I can easily interpolate through cubic interpolation for example and extrapolate since the behaviour at large r is known).
I'm using the "physics" definition of the Fourier transform in 3D, which gives (because f is radial) :
I first tried to calculate this integral for some chosen values of q by using Gauss-Legendre quadrature, by generating some 60 or 100 abscissas and weights via the NAG routine D01BCF (D01BCF link). In the case of Gauss Legendre quadrature, the problem is to choose the interval [0,B] on which to integrate. While the function f loses 4 to 5 orders of magnitude from r=10 to r=20 (example), the choice of B as a strong influence on the result of the calculation... When I compared the result I get to a "nearly exact" calculation (made with matlab but with a veeeery long computation time), I saw that in fact this was only valid for small values of q (of the order of 5, when I have to deal with values as large as 150). A Gauss-Laguerre quadrature does not give any better result, probably because of the oscillatory part of the integrand.
I then tried to compute this Fourier transform for some given values of q with the routine D01ASF (D01ASF link). It is a "one-dimensional quadrature, adaptive, semi-infinite interval, weight function cos(ωx) or sin(ωx) ", which is exactly what I need. The results are quite convincing for q up to 80 or 100 if I input absolute error tolerances of 10E-5. Problems are : I would need to go at larger q, and the Fourier transform F(q) oscillates with a magnitude of ~ 10E-6 at such q's. Lowering the tolerance to 10E-5 already takes some time and even makes the whole thing to output some error message from the subroutine so I don't know if 10E-6 would be feasible.
I'm thus currently wondering if trying to calculate this Fourier transform with FFT wouldn't be a good idea ? The problems I face are that I don't know how to calculate radial wave functions with FFT (and also that I don't even know how to use FFT properly either since the definition of the transform is not even the same (exponent sign and argument) and that I never used it before).
Would you have ideas ? :)
EDIT 2 : I tried by FFT (using the routine C06FAF from NAG library). It works quite well up to some large values of q. The problem I face is that there is always some constant normalising factor to account for. I don't get why. This normalising factor evolves with the number N of points used in the mesh. It has the for of a power law : Normalising Factor F = N^(-0.5) x exp(9.9) approximately (see figure where the black line is the "exact" Fourier Transform and the green, magenta, blue, red and yellow lines are the FFT calculated for different values of N)
EDIT3 : I found the factor to be A*N^(-0.5) where A is the length of the integration mesh
Related
After applying a Fourier transform to a signal, the energy of a single sine wave is often spread out across multiple bins (aka smearing). Have a look at the right side of the image below for an illustration:
I want to extract a list of peak frequencies. Just finding the highest bin is easy. But after that smearing becomes a problem.
I would like to have a heuristic which tells me if the magnitude of a specific bin is possibly the result of smearing or if there has to be another peak frequency in order to explain the signal. (It is better if I miss some than have false positives)
My naive approach would be to just calculate a few thousand examples and take the maximum of these to get an envelope curve so that any smearing is likely below that envelope.
But is there a better way to do this?
The FFT result of any rectangularly windowed pure unmodulated sinusoid is a Sinc function. This Sinc (sin(pix)/(pix)) function is only zero for all bins except one (at the peak magnitude) when the frequency of the input sinusoid has exactly an integer number of periods in the FFT width.
For all other frequencies that are not at the exact center of an FFT result bin, if you know that exact frequency and magnitude (which won't be in any single FFT bin), you can calculate all the other bins by sampling the Sinc function.
And, of course, if the input sinusoid isn't perfectly pure, but modulated in any way (amplitude, frequency or phase), this modulation will produce various sidebands in the FFT result as well.
I am trying to extract the curvature of a pulse along its profile (see the picture below). The pulse is calculated on a grid of length and height: 150 x 100 cells by using Finite Differences, implemented in C++.
I extracted all the points with the same value (contour/ level set) and marked them as the red continuous line in the picture below. The other colors are negligible.
Then I tried to find the curvature from this already noisy (due to grid discretization) contour line by the following means:
(moving average already applied)
1) Curvature via Tangents
The curvature of the line at point P is defined by:
So the curvature is the limes of angle delta over the arclength between P and N. Since my points have a certain distance between them, I could not approximate the limes enough, so that the curvature was not calculated correctly. I tested it with a circle, which naturally has a constant curvature. But I could not reproduce this (only 1 significant digit was correct).
2) Second derivative of the line parametrized by arclength
I calculated the first derivative of the line with respect to arclength, smoothed with a moving average and then took the derivative again (2nd derivative). But here I also got only 1 significant digit correct.
Unfortunately taking a derivative multiplies the already inherent noise to larger levels.
3) Approximating the line locally with a circle
Since the reciprocal of the circle radius is the curvature I used the following approach:
This worked best so far (2 correct significant digits), but I need to refine even further. So my new idea is the following:
Instead of using the values at the discrete points to determine the curvature, I want to approximate the pulse profile with a 3 dimensional spline surface. Then I extract the level set of a certain value from it to gain a smooth line of points, which I can find a nice curvature from.
So far I could not find a C++ library which can generate such a Bezier spline surface. Could you maybe point me to any?
Also do you think this approach is worth giving a shot, or will I lose too much accuracy in my curvature?
Do you know of any other approach?
With very kind regards,
Jan
edit: It seems I can not post pictures as a new user, so I removed all of them from my question, even though I find them important to explain my issue. Is there any way I can still show them?
edit2: ok, done :)
There is ALGLIB that supports various flavours of interpolation:
Polynomial interpolation
Rational interpolation
Spline interpolation
Least squares fitting (linear/nonlinear)
Bilinear and bicubic spline interpolation
Fast RBF interpolation/fitting
I don't know whether it meets all of your requirements. I personally have not worked with this library yet, but I believe cubic spline interpolation could be what you are looking for (two times differentiable).
In order to prevent an overfitting to your noisy input points you should apply some sort of smoothing mechanism, e.g. you could try if things like Moving Window Average/Gaussian/FIR filters are applicable. Also have a look at (Cubic) Smoothing Splines.
I have a current implementation of Gaussian Blur using regular convolution. It is efficient enough for small kernels, but once the kernels size gets a little bigger, the performance takes a hit. So, I am thinking to implement the convolution using FFT. I've never had any experience with FFT related image processing so I have a few questions.
Is a 2D FFT based convolution also separable into two 1D convolutions ?
If true, does it go like this - 1D FFT on every row, and then 1D FFT on every column, then multiply with the 2D kernel and then inverse transform of every column and the inverse transform of every row? Or do I have to multiply with a 1D kernel after each 1D FFT Transform?
Now I understand that the kernel size should be the same size as the image (row in case of 1D). But how will it affect the edges? Do I have to pad the image edges with zeros? If so the kernel size should be equal to the image size before or after padding?
Also, this is a C++ project, and I plan on using kissFFT, since this is a commercial project. You are welcome to suggest any better alternatives. Thank you.
EDIT: Thanks for the responses, but I have a few more questions.
I see that the imaginary part of the input image will be all zeros. But will the output imaginary part will also be zeros? Do I have to multiply the Gaussian kernel to both real and imaginary parts?
I have instances of the same image to be blurred at different scales, i.e. the same image is scaled to different sizes and blurred at different kernel sizes. Do I have to perform a FFT every time I scale the image or can I use the same FFT?
Lastly, If I wanted to visualize the FFT, I understand that a log filter has to be applied to the FFT. But I am really lost on which part should be used to visualize FFT? The real part or the imaginary part.
Also for an image of size 512x512, what will be the size of real and imaginary parts. Will they be the same length?
Thank you again for your detailed replies.
The 2-D FFT is seperable and you are correct in how to perform it except that you must multiply by the 2-D FFT of the 2D kernel. If you are using kissfft, an easier way to perform the 2-D FFT is to just use kiss_fftnd in the tools directory of the kissfft package. This will do multi-dimensional FFTs.
The kernel size does not have to be any particular size. If the kernel is smaller than the image, you just need to zero-pad up to the image size before performing the 2-D FFT. You should also zero pad the image edges since the convoulution being performed by multiplication in the frequency domain is actually circular convolution and results wrap around at the edges.
So to summarize (given that the image size is M x N):
come up with a 2-D kernel of any size (U x V)
zero-pad the kernel up to (M+U-1) x (N+V-1)
take the 2-D fft of the kernel
zero-pad the image up to (M+U-1) x (N+V-1)
take the 2-D FFT of the image
multiply FFT of kernel by FFT of image
take inverse 2-D FFT of result
trim off garbage at edges
If you are performing the same filter multiple times on different images, you don't have to perform 1-3 every time.
Note: The kernel size will have to be rather large for this to be faster than direct computation of convolution. Also, did you implement your direct convolution taking advantage of the fact that a 2-D gaussian filter is separable (see this a few paragraphs into the "Mechanics" section)? That is, you can perform the 2-D convolution as 1-D convolutions on the rows and then the columns. I have found this to be faster than most FFT-based approaches unless the kernels are quite large.
Response to Edit
If the input is real, the output will still be complex except for rare circumstances. The FFT of your gaussian kernel will also be complex, so the multiply must be a complex multiplication. When you perform the inverse FFT, the output should be real since your input image and kernel are real. The output will be returned in a complex array, but the imaginary components should be zero or very small (floating point error) and can be discarded.
If you are using the same image, you can reuse the image FFT, but you will need to zero pad based on your biggest kernel size. You will have to compute the FFTs of all of the different kernels.
For visualization, the magnitude of the complex output should be used. The log scale just helps to visualize smaller components of the output when larger components would drown them out in a linear scale. The Decibel scale is often used and is given by either 20*log10(abs(x)) or 10*log10(x*x') which are equivalent. (x is the complex fft output and x' is the complex conjugate of x).
The input and output of the FFT will be the same size. Also the real and imaginary parts will be the same size since one real and one imaginary value form a single sample.
Remember that convolution in space is equivalent to multiplication in frequency domain. This means that once you perform FFT of both image and mask (kernel), you only have to do point-by-point multiplication, and then IFFT of the result. Having said that, here are a few words of caution.
You probably know that in digital signal processing, we often use circular convolution, not linear convolution. This happens because of curious periodicity. What this means in simple terms is that DFT (and FFT which is its computationally efficient variant) assumes that you signal is periodic, and when you filter your signal in such manner -- suppose your image is N x M pixels -- that it takes pixel at (1,m) to the the neighbor or pixel at (N, m) for some m<M. You signal virtually wraps around onto itself. This means that your Gaussian mask will be averaging pixels on the far right with pixels on the far left, and same goes for top and bottom. This might or might not be desired, but in general one has to deal with edging artifacts anyway. It is however much easier to forget about this issue when dealing with FFT multiplication because the problem stops being apparent. There are many ways to take care of this problem. The best way is to simply pad your image with zeros and remove the extra pixels later.
A very neat thing about using a Gaussian filter in frequency domain is that you never really have to take its FFT. It si a well-know fact that Fourier transform of a Gaussian is a Gaussian (some technical details here). All you would have to do then is pad you image with zeros (both top and bottom), generate a Gaussian in the frequency domain, multiply them together and take IFFT. Then you're done.
Hope this helps.
I'm studying the code behind the convolutionFFT2D example of the Nvidia CUDA sdk, but I don't get the point of this line:
cufftPlan2d(&fftPlan, fftH, fftW/2, CUFFT_C2C);
Apparently this initializes a complex plane for the FFT to be running in, but I don't see the point of dividing the plan width by 2.
Just to be precise: the fftH and fftW are rounded values for imageX+kernelX+1 and imageY+kernelY+1 dimensions (just for speed reasons). I know that in the frequency domain you usually have a positive component and a symmetric negative component of the same frequency.. but this sounds like cutting half of my image data away..
Can someone explain this to me a little better? I've never used a FFT (I just know the theory behind a fourier transformation)
When you perform a real to complex FFT half the frequency domain data is redundant due to symmetry. This is only the case in one axis of a 2D FFT though. You can think of a 2D FFT as two 1D FFT operations, the first operates on all the rows, and for a real valued image this will give you complex row values. In the second stage you apply a 1D FFT to every column, but since the row values are now complex this will be a complex to complex FFT with no redundancy in the output. Hence you only need width / 2 points in the horizontal axis, but you still need height pointe in the vertical axis.
In computer vision, we often want to remove noise from an image. We can do this by getting an image and replacing distorted pixels with an average of its neighbours. I have no trouble understanding this but what are all the variables in the following equation meant to be? I've just found it in some slides but it doesn't come with any explanation:
The (i,j) is probably a given pixel and its neighbour, but what is the function f, the Omega, and the w? Any guesses?!
Cheers guys.
This is way too vague. Notation changes between papers and different approaches. Generally speaking that formula is doing some averaging within a neighbouring set of the i,j point (defined by the points in \Omega_{ij}) w is some normalization constant and f(m,n) is some function which typically assigns a value to m,n proportional to its distance from i,j
As I said your question is a bit too vague to say anything else...
This looks similar to motion prediction in video encoding.
g(i,j) is likely the ith, jth pixel in a block / screen. whose value is the weighted sum of another heuristic function taking the neighbor positions (m,n)
Since I see Omega I suspect you are working in signal space. This might filter out high frequencies not found in our neighbors m,n