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I have read the following quote from a different post:

If the noise is white (spread evenly across frequency), then the power will be distributed equally to each bin, and as we increase the number of bins, the amount in each bin will reduce accordingly (this is the "FFT noise floor").

FFT is the discrete sampling of the DTFT at frequencies $\omega = 2π\frac{k}{N}$, so how can the PSD change (therefore the DFT) if the DTFT is invariant to the DFT length?

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  • $\begingroup$ The PSD is defined as the Fourier transform of the autocorrelation sequence. The autocorrelation of white noise is a delta, which means the spectrum is flat. Additionally, if the noise is white, the statistical second moments of the signal are zero everywhere except for when there is no time-lag. This means the correlation matrix is an identity matrix scaled by the variance of the noise. This implies the flat spectrum is scaled by the variance. $\endgroup$
    – Baddioes
    Dec 17, 2023 at 22:09
  • $\begingroup$ There is not necessarily a direct relationship between the power spectrum (power in the signal) and the PSD (Fourier transform of the signal’s autocorrelation sequence). $\endgroup$
    – Baddioes
    Dec 17, 2023 at 22:10

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The PSD doesn't change – the D stands for Density, i.e. "how much power is there per bandwidth), and with an increased DFT length, there's less bandwidth per bin.

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  • $\begingroup$ So this means the power itself changes? and the PSD doesnt change because some sort normalization? do we normalize by the total sum of the power? $\endgroup$ Dec 15, 2023 at 20:17
  • $\begingroup$ No, as said in my answer. No, as said in my answer. No, as said in my answer! $\endgroup$ Dec 15, 2023 at 20:23
  • $\begingroup$ The PSD simply does not change because you're doing something different with the signal (just as the size of the world doesn't change just because you're looking through a looking glass). The DFT's output changes – but that's simply not the same as the PSD! $\endgroup$ Dec 15, 2023 at 20:25
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Think of it as total energy not changing, but since DFT length dictates frequency resolution, the power of each band will be less dense.

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    $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Dec 17, 2023 at 12:06

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