I have been studying about boundary discontinuity in DFT and why it is not used as transform in image compression. What i understand is that DFT is N periodic and that causes a discontinuity at the boundary. This means that more number of terms are needed to represent the function and the rate of convergence is slow.

Can somebody explain as to what does it mean when they say that "more number of terms are needed to represent the function?" . And how does it help to use DCT which is 2N periodic?

Reference article is at : http://en.wikipedia.org/wiki/Discrete_cosine_transform



2 Answers 2


The lower frequency basis vectors of a finite size DFT have nearly equal values at the 2 edge points, thus are poor at encoding "smooth" data where the first and last element can be quite different, such as arbitrary image block edges, into a small number of basis vectors. (e.g. It takes a massive pile of vectors that are all nearly the same at the boundary points to represent any big difference between these two points, as well as properly transform encoding the rest of the data.).

Other transforms (DCT, etc.) include at least some smooth basis vector that can easily represent a huge difference between the first and last point in the window, thus possibly requiring a smaller number of basis vectors for a close approximation to an invertible transform of reasonably smooth data.


Because the left (top) edge of an image is unlikely to be a reflection of its right (bottom) edge, there are discontinuities all along the edges of an image when it is viewed as N-periodic. These discontinuities are represented in the frequency domain by high-frequency coefficients. From an image compression point of view, using the 2-D DFT as a transform would require to store these high frequency coefficients (or non-zero terms in the decomposition of the image as a sum of elementary components), which do not describe anything meaningful in the image itself, but are a nasty side-effect of an inappropriate choice of transform.


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