Matlab, DFT, FFT, frequency range; -

I really have a hard time understanding this in the catalog works in fft-function. According to:

The distance between the sample points in the FS / N spectrum, where FS is the sample frequency and the length of the n signal. The code they present is the removal of the frequencies (although I do not know), but according to the implementation of FFT, they exist:

Distance between points Spectrum should be: 1 / fs because the index includes J * T instead of J, where T = 1 / FS is sampling time and then you can calculate the distance between points, which is not to be FS / N needed.

I would be truly grateful if someone could explain to me what the distance between points in the frequency domain is and why it is :)

[edit]
It This is a problem about the relationship between Fourier-Transformations and Discrete-Fourier-Transformation and scaling / frequency of frequency-axis, not a Matlab-specific problem. A very good explanation can be found in page 3 in this PDF-document:

texas.math.ttu.edu/~gilliam/ttu/dft.pdf

[/ edit ]

The distance between the frequency points on the spectrum is fs / n Cases, and this is the correct value if we take the code inside, then we have:

  & gt; & Gt; FS = 1000; % Sampling frequency & gt; & Gt; T = 1 / FS; % Sample Time & gt; & Gt; L = 1000; Signal length & gt; & Gt; T = (0: L-1) * t; % Time vector & gt; & Gt; NFFT = 2 ^ next pose 2 (L); The next power of y is the power of 2 y & gt; & Gt; F = FS / 2 * Lynxpace (0,1, NFFT / 2 + 1); & Gt; & Gt; F (2) -f (1) ans = 0.97656 & gt; & Gt; FS / NFFT ans = 0.97656  

You can check all other f (n + 1) -f (n) , they are all the same.