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What is the relation between DFT and PSD of a signal

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Mathematically the PSD of signal x(t) is the Fourier transform of Autocorrelation function of x(t)
2) But In MATLAB it is seen that the POWER SPECTRAL DENSITY (PSD) is directly obtained from the FFT of a signal x(t) as follows.
N = Number of data points or length of signal x(t), Nf = 2^nextpow2(N), Xk = fft (x, Nf), PSD = [ Xk * conj(Xk) ] / Nf, fs = Sampling frequency, f = fs * linspace (0, 1, Nf), Creates frequency vector, The graph of PSD Vs f is called PSD curve.
Does it mean that the DFT of x(t) directly gives PSD of x(t) ?
  1 Comment
Image Analyst
Image Analyst on 7 Sep 2012
No. Look: Xk * conj(Xk) - that's the square of the discrete Fourier Transform, not the Fourier Transform itself.

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Accepted Answer

Honglei Chen
Honglei Chen on 7 Sep 2012
Edited: Honglei Chen on 7 Sep 2012
It can be shown that PSD can alternatively be estimated by the square of magnitude of FFT. You should be able to find it in most spectral analysis text books, e.g., Kay's Modern Spectral Estimation, equation 4.2, pp 65

More Answers (1)

Azzi Abdelmalek
Azzi Abdelmalek on 7 Sep 2012
Edited: Azzi Abdelmalek on 8 Sep 2012
  1. The PSD shows how the power of your signal is distributed over your frequencies
  2. the FFT shows the amplitude and phase of each harmonic component of your signal



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