# How to analyze the image with Fast Fourier Transform method?

I am learning about analyzing images with the method of FFT(Fast Fourier Transform). The image I am analyzing is attached below:

And the result of the FFT analysis of this picture is presented below:

On the FFT image, the low frequency are is in the center of the image and the high frequency areas are at the corners of the image. Can someone tell me the formation of the FFT image? For example, why there is a horizontal white line passing the center? Why is the FFT image like a "sun" emitting beams?

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Remember that the result of a Fourier transform is complex--it has both real and imaginary parts. I think you've plotted the magnitude of the FFT results, which hides the phase information. The phase information is at least as critical as the magnitude in terms of carrying image data; see imagemagick.org/Usage/fourier/#fft_partial for an example. In a simple sense, it's the phase of the FFT that tells you where features occur in the original image. – coneslayer Mar 28 '12 at 16:29
@coneslayer Thanks a lot. – Chuck Wang Mar 29 '12 at 2:06

You have a function of the spatial coordinates (x, y), the coordinates of the original image. Suppose, for clarity, that we are talking about a value from 0 to 255 for each (x, y) point in your original image. The transform is a function, again from 0 to 255, of the momentum coordinates (k1, k2) . The point (0, 0) - the sun - corresponds to the intensity of the constant part of the original function. Don't think, for a moment, to the fact that it represents an image, think of it like... a 2d bar chart or something like that. The constant is the average over the (periodically arranged) image. As you progress from the center you are sampling at higher frequencies (with sinusoidal and cosinusoidal function of increasing frequency). Given the spatial resolution of the details of your original image, you can see that the corners (high k1 frequency, high k2 frequency) are black (that is, the intensity of the transfor is low), and the central zone, lighter, correesponds to the "typical" spatial lenght of the details of your image. If you had took a picture of a more regular object (a grid?) you would have found a "typical" k corresponding to your "typycal" lenght (for example, this is the process that is used in physics to reconstructs the features of cristals).

The central line corresponds to the average values along the y direction for the various sampling frequencies along the x direction. It is roughly constant: this means that the average value of the image along the short side, independently of the frequency of sampling along the long side, is the same. This should be because the image exhibits a symmetry (the horizon) with a single feature (the girl) in a very concentrated region of space. It is relatively bright because the average value is influenced by the sky, which is mostly uniform and bright.

As an exercise, you could try to take a picture of a single/a few light object against a dark background and compare the results.

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 Thanks for reply. I will try it by myself. – Chuck Wang Mar 28 '12 at 1:17

I'm pretty sure the question will be closed, it's way too broad.

If you want to learn about Fourier Transform Image Processing, you should start with learning about basic Fourier Transforms (time domain to frequency domain mapping) and then you can go on to 2-dimensional Fourier Transforms.

Any number of pages will give you an overview, e.g.:

http://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm

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 I can only say that I hope it will not be closed :) – Francesco Mar 27 '12 at 21:17 @Ward Thanks for your reply and I am studing the stuff on that website. – Chuck Wang Mar 28 '12 at 1:20 @Ward Another question. Can you tell me the area where can I apply this technique ? – Chuck Wang Mar 28 '12 at 1:23 @ChuckWang I have no idea... I remember doing experiments w/ FT optics in a Physics class in University. I forget the setup, but with a laser as light source and the right lens arrangement, you can put a screen in a position and see the FT of the image. Then you can do some image processing on the image, e.g. filter out dust. – Ward Mar 28 '12 at 3:46 @Ward Thanks for your reply. – Chuck Wang Mar 28 '12 at 7:19