Hi, Cheng! The frequency resolution is defined as Fs/N in FFT. Where Fs is sample frequency, N is number of data points used in the FFT. For example, if the sample frequency is 1000 Hz and the number of data points used by you in FFT is 1000. Then the frequency resolution is equal to 1000 Hz/1000 = 1 Hz. 

Hello,

it's quite the other way round:

Your sampled signal defines the basic parameters for the FFT: the lowest frequency is the one defined by the period of the whole sample set, the highest frequency as well as the frequency is given by the sampling period (resp. the reciprocals of the periods).

If you are measuring continuously, the period of the sample set is to be replaced by the number of samples passed to the FFT times the sampling period.

Hi, Cheng! The frequency resolution is defined as Fs/N in FFT. Where Fs is sample frequency, N is number of data points used in the FFT. For example, if the sample frequency is 1000 Hz and the number of data points used by you in FFT is 1000. Then the frequency resolution is equal to 1000 Hz/1000 = 1 Hz. 

Further more, the problem become complex if you apply a window, such as Hanning window, to your data before doing FFT. Applying window will decrease frequency resolution of FFT result. Better window leads to worse frequency resolution. 

... where "Better window" = lower side lobes, so that weak components, that are far from strong components can still be observed.   

Thanks for the answers from everyone of you. Actually, yes, I use the Hanning window to tape my EEG data. But, how do I know how much frequency resolution I lost during the windowing process? 

Generate the frequency response of your Hanning window. Multiplication in the time domain is convolution in the frequency domain. So the true spectrum will be convolved with the spectrum of your window! So if the main lobe of your window fcn is 'smeared' over say 2.5 frequency bins, then a sinusoidal component (which should be an impulse in the frequency domain, in the absence of any window) will be 'blurred' over 2.5 bins.

The frequency resolution of FFT (or DFT)  is equal to the inverse of continuous  sampling time (s) :   Freq.res = 1/T

if you are sampling data for 1 second, the resolution is 1Hz. If you are sampling for 10 seconds, the resolution is  0.1Hz .

Tre frequency resolution is independent from the sampling frequency.

Like Rujun Chen said before, the frequency resolution is defined as Fs/N in FFT. Where Fs is sample frequency, N is number of data points used in the FFT. The make it clear, N is the number of data points used in each buffer for Fast Fourier Transform ( eg. 32, 64, 128, 256...131072 points).

my question is: the script I am using does not ask for the sampling frequency. it assumes that is 1 s or min or 1 h. In our case our sampling is 10 min. How I get the period related to a specific frequency in the spectra

i.e., if the f = 0.001038 its inverse is 963.39... what is the real period

96.339 (by dividing by 10) or 9633.9 (by multiplying by 10)

Any comment is welcome. thanks