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I read one equation from a technical document, i.e., assuming the time-domain white Gaussian noise samples with variance $\sigma_{TD}^{2}$ in dB are sampled using ADC with sampling rate $f_{s}^{ADC}$, then if they are downsampled to the sampling frequency $f_{s}$ (e.g., with a downsampling rate of 2 and appropriate filtering, $f_{s} = 0.5 \cdot f_{s}^{ADC}$), then the frequency-domain power in dB becomes $$\sigma_{FD}^{2} = \sigma_{TD}^{2} + 10 \cdot \log_{10}\left(\frac{f_{s}}{f_{s}^{ADC}}\right)$$

This implies that with downsampling of 2, the frequency-domain power will be 3 dB less than the time-domain power.

I have read some documents in order to figure out why. But I am not convinced myself by any explanation yet.

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  • $\begingroup$ "... from a technical document ..." please edit your question to properly cite this technical document, or excerpt the section that develops that argument, or both. If you just sample a signal without any anti-aliasing filter, then the noise power remains the same, so there is clearly context missing. $\endgroup$
    – TimWescott
    Aug 22, 2023 at 3:04
  • $\begingroup$ Hi, Tim, thanks a lot for the comment, I have added some additional comments in the questions. Regarding your comment that "the noise power stays the same without anti-aliasing filter", can you see my comments in DamienBradley's answer. $\endgroup$
    – Vic
    Aug 22, 2023 at 14:24

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White gaussian noise has constant power spectral density to frequency, in downsampling by 2 you are removing half of the available frequency band.

Integration of power spectral density thereafter gives an average power that has decreased by a factor of half. $10\log(0.5) = -3\texttt{dB}$

This is assuming of course that you use an anti-aliasing filter before downsampling the signal - this is the real source of the loss in average power.

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  • $\begingroup$ The basic idea and the math is correct but you don't need the Gaussian assumption. In practice it's a little more complicated. ADC quantization noise is usually not white. At high SNR, the quantization noise tends to be correlated, so you tend not to see the 3 dB improvement. At lower SNRs, the white noise begins to dominate and you'll begin to approach 3dB improvement. $\endgroup$
    – David
    Aug 21, 2023 at 19:32
  • $\begingroup$ @DamienBradley ,thanks. another question is : is the fitlering that TimWescott mentioned in his comments required for this 3 dB noise reduction? Tim mentions that without filtering, the noise power stays the same. my understanding is: the PSD is the same before and after downsampling, and the frequency band has decreased with the downsampling, so even without filtering, the power should still be decreased with the downsampling; but to avoid aliasing and get the SNR improvement, the anti-aliasing filtering is required. $\endgroup$
    – Vic
    Aug 22, 2023 at 14:21
  • $\begingroup$ @David, when I searched for the answer for my questions, I have also learnt that the quantization noise decreased by 3 dB by doubling the ADC sampling frequency. But is the SNR improvement from downsampling and filtering operation due to less thermal noise power or less quantization noise power or both? $\endgroup$
    – Vic
    Aug 22, 2023 at 14:28
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    $\begingroup$ See this post deriving a constant PSD for downsampling (without filtering) a wide-sense stationary signal. $\endgroup$
    – Ash
    Aug 22, 2023 at 14:34
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    $\begingroup$ If the quantization noise is white, then the quantization noise power will be spread across the entire frequency sampling bandwidth i.e. $F_s$. By filtering out half the frequency band, the noise power will be reduced by a factor of 2 - since we assumed it was spread equally across $F_s$. If there is no power loss in the signal due to the filtering, then you will see a 3dB gain in the SNR. $\endgroup$
    – David
    Aug 22, 2023 at 19:40

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