Since the power spectral density is just the squared of the fourier transform, why is it useful ? Can't I just replace every solution that requires the psd with the fourier transform ?


Power Spectral Density (PSD) is a theoretical concept, required to deal with random processes (signals). It's by definition the Fourier transform of the auto-correlation function (another theoretical concept) of the random processes.

Practical computation (estimation) of the PSD can be done in a number of ways and some of them utilize the way you expressed as the (scaled) square of the magnitude of the Fourier transform of an observed instance of the random process. Such an estimator of the PSD is called as the Periodogram (or its various variants).

Therefore, they are not substitudes for eachother. In practice you can only estimate the true PSD.

  • $\begingroup$ Hello thanks for the reply. But why not just take the Fourier transform of the random process ? What does squaring it tell me ? $\endgroup$ – Kong Oct 4 '18 at 22:51
  • $\begingroup$ if the process is random, you can't take the fourier transform because there's noise in the system. you can "estimate" it and that's essentially what the psd is doing. in fact, as Fat32 said, you're estimating the square of the magnitude of it. $\endgroup$ – mark leeds Oct 5 '18 at 4:49
  • $\begingroup$ @kong PSD is the FT of the autocorrelation function of the random process, not of the process itself. $\endgroup$ – AlexTP Oct 5 '18 at 8:01
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    $\begingroup$ @kong You cannot take the Fourier transform of a random process. That's why PSD is defined over its auto-correlation function. A random process is an ensemble of sample functions and its Fourier trasform is not unique. However its ACF is a deterministic unique function whose FT can be defined. Note that in practice you would be taking the square of its FT. Indeed for both random processes and deterministic signals the power spectrum of the signal is related to the square of FT. $\endgroup$ – Fat32 Oct 5 '18 at 8:30

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