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Do the result of DFT (Discrete Fourier Transform) of a 1D sequence order as low frequency to high frequency? and why is that?

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2 Answers 2

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By convention, the output of a DFT routine is a vector that starts at frequency = 0 and goes up from there.

That's just by convention, however. If you write an FFT algorithm for $N = 2^m$ points, one "natural" ordering for its output is at frequencies in the sequence $0, \frac{F_s}{2}, \frac{F_s}{4}, \frac{3F_s}{4}, \cdots, \frac{(N - 1)F_s} N$ (i.e., as if it were addressed by an $m$-bit word in bit-reversed order).

Note that it's a pretty strong convention. All of the common math packages will take such any such penultimate result and reorder it into the conventional ordering. I don't think I've run across one that doesn't, and I don't think that one that did go against this established convention would gain much traction in the marketplace of ideas.

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  • $\begingroup$ is it same convention when doing Wavelet Transform or other transform? $\endgroup$
    – holder hé
    Commented Dec 31, 2022 at 10:53
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Yes, the result of the DFT of a 1D sequence is commonly ordered from low frequency to high frequency.  The signal's slowest parts are represented by the low-frequency components, while the high-frequency components represent its fastest-changing parts. Hence, arranging this way makes the resulting frequency spectrum easier to interpret.

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