I know that the frequency resolution in a DFT is given by sampling rate upon the total number of samples. $$\Delta f = \frac{f_s}{N}$$ But, what I want to know is where does this relation comes from? Or What criteria(like Nyquist criteria limiting the maximum meaningful frequency) governs it?

  • $\begingroup$ I don't understand the question. What frequency has the first bin of an $N$-point DFT? what about the second? What about the $\frac N2$th one? $\endgroup$ Commented May 31, 2016 at 19:17
  • $\begingroup$ @MarcusMüller: see the answer in this link: dsp.stackexchange.com/questions/8525/… From where did that frequency spacing formula came from? $\endgroup$ Commented May 31, 2016 at 19:29
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    $\begingroup$ @SaravanaKumar It's part of the definition of the DFT. Think of the DFT as samples of the "true" spectrum (which can be calculated with the DTFT). The values are chosen so that the DFT can easily and quickly be calculated with an FFT. $\endgroup$
    – MBaz
    Commented May 31, 2016 at 21:20

3 Answers 3


The result of an $N$-point DFT is an $N$ element complex vector. For strictly real data, only the first $1+N/2$ bins matter, as the others are a redundant complex conjugate image. For properly bandlimited data sampled at $F_s$, the frequency range is $0$ to $F_s/2$. Divide $F_s/2$ by $N/2$, and you get $F_s/N$ as the frequency spacing of $1+N/2$ equally spaced DFT results, including both end points, DC and $F_s/2$.

So $F_s/N$ is the DFT result bin spacing of an $N$-point DFT.

However, the actual resolution can range from a fraction of the DFT bin spacing to $2$ or more DFT bins of separation, depending on the signal-to-noise ratio and what kind of resolution you want: frequency peak estimation, or peak separation (with a gap between peaks, typically of $3\textrm{ dB}$ or more).

In low noise (high signal-to-noise ratio), an isolated frequency peak can be resolved, by $\mathrm{Sinc}$ interpolation, to a much higher resolution than the DFT bin spacing, due to added information from the a-priori assumption (or a good guesstimate) of that signal-to-noise ratio and peak isolation.

  • $\begingroup$ might wanna be careful about using "$N$" for two different purposes. $\endgroup$ Commented Jun 2, 2016 at 6:16

The resolution of FFT is the reciprocal of the amount of time the signal is observed. Interpolating won't increase the resolution in any way. Higher spectral resolution algorithms do exist in the field of spectral estimation. (e.g. MUSIC, ARMA etc. )


You can think of the $N$ point DFT as $N$ points spaced in $0 - 2\pi$ range. Therefore the spacing between $2$ successive samples is $(2\pi)/N$. This is equal to $(2\pi\Delta F) / F_s$. Which is $\Delta F /F_s = 1/N$.


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