SODA PLAYER FOR MAC STUCK AT EXTRACTING ADDITIONAL DEPENDENCIES SERIES
This corresponds directly with the highest frequency (last) term of a series representation.Ī two dimensional Fourier (or Cosine) transformation of the image results in an array of values the same size as the image, representing the same information not as a function of space, but a function of 1/space.
The most abrupt transistion that can occur in the image is going from 0 (say black) to 255 (say white) over the distance of 1 pixel. To understand what spatial frequency is, consider an 8 bit grey level image and a pair of adjacent pixels. A function like that can be represented in terms of its spatial frequency. In other words, the image is a function of a 2D space, that being a rectangular region. An (static) image is a function that gives a numerical value for for every pair of row, column indeces. In the 1D sense, the square wave is merely a function that gives some numerical value for for an argument that we said was time. Second the notion of time makes no sense. Such things can be represented with finite series'.įor images, the mathematics are the same, but two things are different. Audio signals, for example occupy the range 20Hz to 20KHz, where Hz is 1/time. In the real world, there are no square waves. Sticking with the square wave as a function of time, and each series term a function of frequency there are two equivalent representations: a function of time and a function of frequency (1/time). These are from the increasing frequency of each term. Successive terms have more "up and down" transitions. Take another look at the plot of the series terms. To represent such a thing requires an infinite series. If you think of a square wave as a function of time, it goes from -1 to 1 in zero time. The reason for this is that the square wave is discontinuous. But, to get from an approximation to the exact signal, you have to sum an infinite number of terms. The more terms you add up, the better the approximation. Looking at it, note that if the functions for the terms are added together, they begin to approximate the shape of the square wave. The image here shows a square wave along with several of the first terms of an infinite series. The concepts underlying these kinds of transformations are more easily seen by first looking at a one dimensional case. If you find matches of low frequency values, you can increase into comparing the higher frequencies. This reduces the number of values you need to match against. One possible advantage of using the DCT to find similar regions is that you can do a first pass match on low frequency values (top-left corner). This is where the compression comes in as all those runs of zeros get packed down to almost nothing. To complete the compression it is usual to use a lossless compression over the DCT'd source. However, this does result in loss of information. As lassevk mentioned, it is usual to just zero out these higher frequencies as they typically constitute very minor parts of the source. The outcome of the DCT is usually a collection of amplitudes at the more usual lower frequencies (the top left quadrant) and less entries at the higher frequencies. The top left entry stores the "amplitude" the "base" frequency and frequency increases both along the horizontal and vertical axes. The result of a DCT is a transformation of the original source into the frequency domain. Now, I need to know how can I take advantage of this transformation? I'd like to detect other 8x8 blocks in the same image ( or another image ) that represent a good match.Īlso, What does this transformation give me? Why is the information stored in the top right of the converted image important? Into this an image with all the important information at the top right. Now I can transform an 8x8 section of a black and white image, such as: I based my algorithm on the ideas presented at the following URL:
This technique is used in MPEG video encoding. I've implemented an image/video transformation technique called discrete cosine transform.
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