# Resolution of Discrete Fourier Transform is 1/T - Mathematical proof?

In many articles I see that the frequency resolution of the Discrete Fourier Transform (DFT) equals Fs/N where Fs is the sampling rate and N is the total number of samples. Fs/N is equivalent to 1/T where T is the total time the signal was sampled.

So the longer the time of sampling took place the better the resolution of the spectrum (allowing to separate different tones)

I'm looking for the mathematical proof of the 1/T resolution as I do not see it directly from the definition of the DFT.

• Is it possible to provide a reference that states this equivalence? – A_A May 3 '19 at 7:57
• "T" is a bad label for the entire length (in units of real time, like seconds) of the input buffer. we usually say that $T$ is the sampling period which is the reciprocal of the sample rate, $f_\mathrm{s}$. you should say $NT$ is the total time the signal was sampled and that $\frac{1}{N} f_\mathrm{s}$ is the frequency resolution. – robert bristow-johnson May 3 '19 at 8:28
• When you divide Fs [samples/sec] by N [ samples] you get units of [1/sec]. – D.Cohen May 3 '19 at 8:29
• $N$ is dimensionless. like radian, a sample is a dimensionless unit. (so also is the neper which is about 8.685889638 decibels.) $\frac{1}{N} f_\mathrm{s}$ is Hz, just like $f_\mathrm{s}$ is Hz. – robert bristow-johnson May 3 '19 at 8:32
• BTW, the DFT does not know about the sample rate, $f_\mathrm{s}$. but if you use the DFT of length $N$ to analyze sampled data sampled at rate $f_\mathrm{s}$ then $\frac{1}{N} f_\mathrm{s}$ is the resolution of your frequency analysis that is using the DFT. – robert bristow-johnson May 3 '19 at 8:37

$$\frac{Fs}{N}$$ is not equivalent to $$\frac{1}{T}$$, since $$Fs = \frac{1}{T}$$, $$\frac{1}{T N} \neq \frac{1}{T}$$ for $$N>1$$ (as it is implied here of course).

The point about the resolution of the Discrete Fourier Transform follows from its definition where an $$N$$ point transform, "divides the circle" in $$N$$ discrete points. As you can see $$k n$$ (the discrete frequency and sample "instance" variables) are multiplied by the constant $$\frac{i 2 \pi}{N}$$.

Here, it is still possible to "sample" a non-integer harmonic (say $$k = 1.267$$) but that would not improve the situation if you only have a limited number of true $$n$$ (samples).

Hope this helps.

The resolution of a DFT is not 1/T (or Fs/N). That is just the spacing of the DFT basis vectors or the DFT output bins for a DFT of length N. Or perhaps by making some crazy assumptions, such as the signal is infinitely long and perfectly periodic in T beyond the life of the universe (and therefore test equipment).

However to separate the magnitude peaks two arbitrary frequency (not-just-integer-periodic-in-aperture) windowed sinusoids (or other very narrow-band spectra), you need a gap in between peaks, thus the peak separation resolution is closer to 2/T (or 2Fs/N) (for a 3 dB gap), or a bit more, and greater in noise, or if the magnitudes of the adjacent peaks are different, or if there are more than 2 peaks, or if you use a window function different from the default rectangular window of a DFT.

To graphically resolve the frequency peak of an isolated windowed sinusoid in low noise and interference, the graphical resolution is much finer than the DFT bin spacing, perhaps to a small fraction of Fs/N via Sinc interpolation if the peak is surrounded by a very low noise floor.