As I understand, the DFT of a signal $x$ is a representation of this signal in the basis

$$ \{ e^{j2\pi kn/N} \}_{k = 0, 1, \dots, N-1}$$

Is it possible to form a base of such discrete complex exponentials but with different frequencies ? ($\{ e^{j2\pi f_i n} \}$)

Why did we choose these $\frac{k}{N}$ frequencies ?

In fact, my question is more about why we evaluate the DFT at these specific frequencies and not others.

  • 3
    $\begingroup$ The DFT basis consists of all possible complex exponentials which have period N. $\endgroup$
    – Juancho
    Apr 10, 2022 at 16:42
  • $\begingroup$ IIRC the frequencies you choose will not be orthogonal. $\endgroup$ Apr 11, 2022 at 11:08

1 Answer 1


$k/N$ with $N$ bases is the only basis which is all of:

  1. Orthogonal & invertible. Means we don't lose any information. Invertibility can be seen to follow from the DFT being a square (N x N) matrix of rank N (orthogonal = all columns independent), so we can solve $Ax = y$ with $x = A^{-1}y$.
  2. $N$-periodic. Every basis will repeat after $N$ uniform samples. This makes the DFT a uniform & $N$-periodic sampling of one period of DTFT which yields many useful properties, and enables circular convolution.
  3. Minimal. Maps out the greatest continuous-time bandwidth per unit sample, meaning it's max information-dense. This follows the sampling theorem, which demands a minimum of $N$ samples to capture a bandwidth up to (but not including) $N/2$ (in physical terms, sampling at rate of $f_s$ to capture $f_s/2$). Other bases will take $N$ or more samples (if succeeding at all, see example).

$2k/N$ with $N$ bases is orthogonal and $N$-periodic, but not minimal, nor invertible. In fact it's not invertible for any number of bases, ironically due to periodicity. $Gk/N$, where $G$ is non-integer, is none of those things with $N$ bases. I'll also informally add,

  1. Exact. Follows from 1. Non-orthogonality implies a base has non-zero correlation with other bases, meaning e.g. frequency of 1.5 means "1.5, and a bit of 2.5, and a bit of 0.5, and ...". Though, the part in quotes is a bit tricky depending on interpretation, since it happens no matter what for a realistic input (e.g. spectral leakage).
  • 1
    $\begingroup$ It would be great if you could include the proof, or cite a proof, that there is no other basis that satisfies all the four properties. $\endgroup$
    – AlexTP
    Apr 11, 2022 at 7:34
  • $\begingroup$ @AlexTP Fair, wouldn't know from top of my head though. Think it could be shown by considering every possible $N$-base cisoid basis that's orthogonal and $N$-periodic, then checking which also meets sampling theorem criteria, so a sort of system of equations. $\endgroup$ Apr 11, 2022 at 17:24

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