0
$\begingroup$

The frequency resolution of DFT is

$\Delta f = \frac{1}{T_{0}} = \frac{1}{NT} = \frac{f_{s}}{N}$

where $f_{s}$ is sampling frequency, $T_{0}$ is sampling time, $N$ is number samples, and $T$ is the sampling interval.

Where does this definition come from? Espescially the $1/T_{0}$, why this equality?

$\endgroup$
  • $\begingroup$ There are a lot of answered questions about freq. resolution already. Seee dsp.stackexchange.com/questions/17263/…, dsp.stackexchange.com/questions/8525/…, and the search function... $\endgroup$ – Deve Aug 19 '14 at 11:12
  • $\begingroup$ it's a huge avalanche of the questions about FFT/DFT resolution one the site lately... Maybe it should be included in "read first before you begin" topic? =) $\endgroup$ – Serj Aug 20 '14 at 2:43
  • $\begingroup$ The problem with many of the posts pointed to, is that they often confuse resolution with the DFT/FFT bin spacing - and they are not the same thing. Resolution is the ability to detect/distinguish (or resolve) to closely spaced equal strength signals. $\endgroup$ – David Aug 26 '14 at 14:31
0
$\begingroup$

The longer the total sample window (T0), the smaller the frequency difference between two sinusoids with exactly M periods and exactly M+1 periods (for integer M <= N/2) in that window, which is required for orthogonality between adjacent DFT bins.

Also, the bin spacing is not the resolution. The frequency resolution can be greater or less than the DFT bin spacing, depending on the S/N ratio and your resolution needs (separation or estimation).

$\endgroup$
-1
$\begingroup$

The resolution comes from the Rayleigh resolution - it is the distance from the peak of the main lobe of the response to the first null or zero in the response. In the continuous Fourier Transform this is a sinc() function.

Lets examine the DFT of a rectangular window (note - an arbitrary frequency can be created by multiplying by a complex exponential and using DFT transform properties i.e. frequency shifting).

For the time being lets ignore the sampling rate and just use indices in the time and frequency domain. So if we have a rectangular pulse of length $K$ samples and take a DFT of length $N$ then if $X(m)$ denotes the DFT of the rectangular pulse $x(n)$. Then (see here)

$$X(m) =e^{j\phi(m)} \frac{\sin(\pi mK/N)}{\sin(\pi m/N)}$$

The peak of the response occurs when $m=0$. The first null in the response occurs when $$\pi m K/N=\pi $$ or $$ m=N/K.$$ Now to convert this index into a frequency - we know that the DFT frequency bin spacing is given by $f_s/N$ where $f_s=1/T$ is the sampling frequency and $T$ is time domain sample spacing. This gives: $$ f_{res} = m\frac{f_s}{N} =\frac{N}{K} \frac{f_s}{N}=\frac{f_s}{K}=\frac{1}{KT}=\frac{1}{T_0},$$ where $T_0$ is the time duration of the signal (as per your original post).

Note - Using $\Delta f$ to denote resolution is problematic because it is also used to denote the DFT bin spacing - which is not the same thing as resolution.

Note - The phase term $\phi(m)$ is kind of complicated (see the link I provided), but it is irrelevant to this analysis because we are only interested in the magnitude of the response i.e. where is the peak and where is the first zero/null.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.