2
$\begingroup$

I don't completely understand the term " frequency" with respect to digital image processing. for 1D electrical signals , its pretty easy to understand , visually too. higher the frequency - more packed( or closer) the signal looks in a given time interval than its lower frequency counterpart.

But when it comes to images , with parameters like intensity, 8-bit gray scale, pixel value,etc. - what does frequency mean ?

Improve this question
$\endgroup$
1
  • $\begingroup$ The 2D frequency domain of the image assumes that we can get a good approximation of the image by combining sine (cosine) waves. For analysis purposes in the frequency domain the image it self is replicated. You could get information about this approach to analyzing images by searching the WWW. $\endgroup$
    – Moti
    Apr 5, 2015 at 6:06

3 Answers 3

Reset to default
1
$\begingroup$

Digital images are composed of Spatial Frequencies which describe "change" with respect to position in space. For more information please see: http://en.wikipedia.org/wiki/Spatial_frequency

In fact, the two-dimensional Fourier Transform can decompose a given image to its spatial frequencies which means that it can decompose an image to a set of "sinusoidal plates". Think of these plates like egg-cartons (http://www.asia.ru/images/img/alibaba/img/product/11/24/61/11246140.jpg) of different density (egg positions per unit of space). That is, sinusoids in 2D. For more information, please see: http://cns-alumni.bu.edu/~slehar/fourier/fourier.html and http://homepages.inf.ed.ac.uk/rbf/HIPR2/fourier.htm

As a rule of thumb, keep in mind that small features (i.e. object details) occupy high spatial frequencies while large features (i.e. object form) occupy low spatial frequencies.

A comprehensive example of this would be high-pass filtering of an image which would preserve the edges of represented objects (an edge being a very sudden change in light reflectance across a direction) but completely lose all information about the rough form of the image. The opposite of this would be low-pass filtering which would completely wipe out the details (sharp transitions of reflectance across a direction) but preserve the very slow transitions of reflectance within an image. For more information on shaping the spatial content of an image please see: http://en.wikipedia.org/wiki/Kernel_%28image_processing%29

Hope this helps.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Wow, Thanks for the links. really good! So , here's my understanding- 1) a sinusoidal wave in case of images is basically the white and black strips. So 1 period of my sine wave is 1 black stripe and 1 white stripe merging into each other whose intensities are analogous to amplitude values of sine wave. 2) more no. of strips in a given area , higher is my frequency. $\endgroup$
    – That-Kickass-GirL
    Apr 9, 2015 at 12:15
  • $\begingroup$ and 3) if in my fourier domain , all the dots are in the horizontal axis -or datum- then my stripes are vertically alligned. However if they are located at (x,y) point in the Fourier domain , and make some angle with the dc origin , then in my original image the stripes rotate by the same angle. 4)fourier coefficients , control how bright or dull the strips look ( white looks little gray-ish)in spatial domain. 5) These sine waves or set of stripes with different frequencies, angles or phases add up (superimpose) to give my resultant image ! So have I stated my understanding correctly ? $\endgroup$
    – That-Kickass-GirL
    Apr 9, 2015 at 12:17
0
$\begingroup$

In images, intensity means amplitude and you can see it as in 1D case.
More packed (or change of intensity) in 2D area is higher frequency.
Note that, this way is just visual approximation by human's eyes.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I don't understand what you mean by the second line . "More packed" with respect to image meaning ? $\endgroup$
    – That-Kickass-GirL
    Apr 5, 2015 at 8:08
  • $\begingroup$ I used your word. I thought this will help you understand easily. To be clear, "More change of intensity in given Pixel block(time interval in 1D)" Does this help? $\endgroup$
    – pakornosky
    Apr 5, 2015 at 8:26
  • $\begingroup$ yeah i got it now . just to be clear , what if i have a pixel block within which there is a straight sharp line that separates a darker object from another lighter one, then how would that affect my frequency domain image? $\endgroup$
    – That-Kickass-GirL
    Apr 5, 2015 at 8:37
  • $\begingroup$ This is basically considered as the high frequency part. $\endgroup$
    – pakornosky
    Apr 5, 2015 at 8:44
0
$\begingroup$

Consider a photo of a long picket fence (white pickets against a dark background). The closer together the pickets are in the image, the higher the frequency in "horizontal-pickets-per-1024-pixels", or whatever. That would be a large component of the frequency in the Y direction for a 2D photo.

Improve this answer
$\endgroup$

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