# Signal power depending on FFT bin size

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?

• 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. Dec 17, 2023 at 22:09
• 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). Dec 17, 2023 at 22:10