Talk:DFT matrix

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Roots of unity

The article says:

The matrix W of size nxn, may be described as a Vandermonde matrix:

$W=V(w)$ where w is a vector $w_{i}=\omega _{n}^{i}$ , the nth root of unity.

This strikes me as badly written... surely the vector w isn't the nth root of unity, but rather the vector consisting of each of the nth roots of unity beginning with 1 and progressing around the origin of the complex plane in an anti-clockwise direction? I'm not quite sure that this is right, so don't want to fix it. JulesH 20:09, 20 November 2006 (UTC)[reply]

Exactly, your interpretation is correct, if you look at the page history I've written this line with the comment "should be word'd better", I do apologise. Anyway, I ment to write 'w' is a vector, whose ith coord (hence

w_{i}

), is the nth root of unity. Maybe adding the word "where" after "vector" helps, anyway, please do word this better, I just have to add this line, as the page had the 2,4,8 instances of the matrix with no reference to the general form (which is simpler to look at, well sortof) Oyd11 02:36, 28 December 2006 (UTC)[reply]

Normalization

It seems like this article uses the 1/sqrt(N) convention in front, which makes it inconsistent with the article on DFTs! Could someone w/ expertise comment? EdenEH (talk) 22:56, 27 January 2010 (UTC)[reply]

There are multiple conventions in widespread use; no single scaling is "correct". In many practical situations it is convenient to put all of the 1/N scaling on one of the transforms (e.g. this makes the convolution theorem a bit nicer, as well as being computationally convenient, and FFT algorithms are marginally cleaner to derive without the extra constant factor running around). On the other hand, from a linear-algebra viewpoint in which the DFT appears as a matrix, it is conceptually convenient to scale it so as to be unitary, since it makes it easier to apply well-known properties of unitary matrices. — Steven G. Johnson (talk) 03:18, 28 January 2010 (UTC)[reply]

Image

the image that shows the graphical interpretation of each of the modes is slightly incorrect: for X[4], the Nyquist (highest) frequency, the imaginary part is zero everywhere and not a sine function.

Actually, the sin function should be drawn, specifically $\sin(2\pi {\frac {4n}{8}})$ has been shown such that all the sampled points happen to be 0. The complex exponential will still have 2 curves, it is not a constant function being sampled. If you note that is exactly what is shown in the image, all the sampling lines hit the sin function at 0. Maybe this graph will help you. Aditya — Preceding unsigned comment added by Aditya8795 (talk • contribs) 17:49, 20 November 2018 (UTC)[reply]