Is there any instance where Fourier transforms of color images have been used in conjunction with other object recognition method?

Any instance of usage of Fourier transforms in color images? I think there is great potential in the usage of Fourier transforms of color images.

  • 1
    $\begingroup$ Have you looked for such examples and not found them? I would imagine using FFT in object recognition would be pretty common. $\endgroup$
    – Phonon
    Dec 4, 2012 at 2:13
  • $\begingroup$ I haven't found any and that's precisely why I am asking this. Most object recognition methods are spatially based. $\endgroup$
    – Naresh
    Dec 4, 2012 at 2:15

1 Answer 1


Okay I found some ideas about using Fourier Transforms in Object Recognition. What I am unable to understand is however, its downsides. Why was the method discarded.

For those who are interested in reading about how Fourier Transforms were applied in Object Recognition , look for Fourier Mellin Transforms.

The Fourier-Mellin transform is a useful mathematical tool for image recognition because its resulting spectrum is invariant in rotation, translation and scale. The Fourier Transform itself (FT) is translation invariant and its conversion to log-polar coordinates converts the scale and rotation differences to vertical and horizontal offsets that can be measured. A second FFT, called the Mellin transform (MT) gives a transform-space image that is invariant to translation, rotation and scale.

The steps: The Fourier Transform

The Discrete Fourier Transform (DFT) is given by the following expression: Discrete Fourier Transform

This is often computed using the Fast Fourier Transform algorithm to speed up the time needed for the calculation.

Cartesian to Log-Polar conversion The FFT is projected onto the log-polar plane by the coordinate transform shown below.

Rectangular co-ordinatesPolar co-ordinates

For the conversion from Cartesian coordinates to Log-Polar coordinates the following equation is true:

$$r = \sqrt{x^2+y^2}$$

The origin $(m_o, n_o)$ should be at the center of the image matrix to ensure the maximum number of pixels is included. If the image consists of a square $N \times N$ matrix then the coordinates of the origin are:

Polar Transforms 2

The maximum sampling radius for the conversion can now be calculated as: Polar Transforms 3

If an inscribed circle is chosen as the conversion boundary, some pixels that lie outside the circle will be ignored. If a circumscribed circle is chosen, all pixels will be taken in account, but some invalid pixels will also be included (pixels falling inside the circle but outside of the image matrix).

Since the pixels in Cartesian coordinates cannot be mapped one-to-one onto pixels in the Log-Polar coordinate space, an average of the surrounding pixels needs to be calculated. The standard methods to do this includes nearest neighbor, bilinear and bicubic resampling.

The relationship between the polar coordinates ($\rho$, $\theta$) used to sample the input image and the polar coordinates of the log-polar image (r , $\theta$) can be described by:

$$(\rho,\theta) = (e^r, \theta) $$

To map the input image pixels imagein(xi,yi) onto the output image pixels imageout(rm, $\theta$m), the co-ordinates xi, yi are computed using:

Transforms 5

where $$(\rho,\theta) = (e^r, \theta) $$

The next step is to get a transform-space image that is a rotation and scale invariant representation of the original image. This is done using the Mellin transform which can be expressed as:

Mellin Transform 1

Converting to polar coordinates, we have:

Mellin Transform 2

Now using $$(\rho,\theta) = (e^r, \theta) $$

we get:

Mellin Transform 3

The Fourier Mellin transform has been used for Image Recognition in R. Moller, H. Salguero, E. Salguero: “Image recognition using the Fourier-Mellin Transform”, LIPSE-SEPI-ESIME-IPN, Mexico

  • 2
    $\begingroup$ if you found something useful to you, we'd all like to hear it. Provide some links. If you went through the reference(s), you could explain the approach in a few basic lines... in short, there are some really amazing self-answers, no reason for you not to do that ;) $\endgroup$
    – penelope
    Dec 4, 2012 at 9:27
  • 2
    $\begingroup$ Yes. I'll do that. Though it will take some time. It seems to be an old research topic. $\endgroup$
    – Naresh
    Dec 4, 2012 at 10:52
  • $\begingroup$ I hope these negative votes can be reversed now? $\endgroup$
    – Naresh
    Dec 13, 2012 at 5:38
  • $\begingroup$ You don't have any from me (I got notified for some reason), but since the edit, hopefully people will see it and upvote what's good. $\endgroup$
    – penelope
    Dec 13, 2012 at 9:31
  • $\begingroup$ Hi Naresh - nice work! in the future, try to use math (latex) formatting - it's a lot clearer than images and moreover is searchable. $\endgroup$ Jun 21, 2013 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.