I have done a power spectrum density estimate using Bartlett's method. However, I was wondering if it is also possible to get a sense of the variation of the averaged spectra.

I have plotted the mean and std of the DFT amplitudes (no normalisation), which at least does not look completely off. Enter image description here

However, when I plot the power spectrum density estimate with mean and std in dB-scale, the signal looks of when the power density estimates is around 0dB. This makes mathematically sense, since everything around $log(1)$ will be zero. However, I lack an intuition about the interpretation of the plot.
I was wondering if one would actually plot it like this, since it is somehow misleading. Here is the plot:

enter image description here

Now calculating first the difference and than the dB scale, leads to the expected graph.

enter image description here

Thanks in advance!


1 Answer 1


The power is always a positive quantity, and assuming the noise on the signal is white Gaussian noise, the distribution of the power will be Ricean (and Rayleigh distributed when there is only noise).

To include the standard deviation in the dB plot, when the mean is significantly larger than the standard deviation (higher SNR conditions) the Ricean distribution approaches a non-zero mean Gaussian distribution in which case we can simply add and subtract it from the mean prior to computing dB, and then plot the 3 lines separately (I plot mean+sigma and then mean-sigma with dashed line)

$$\mu_{dB} = 10log10(\mu)$$

$$(\mu+\sigma)_{dB} = 10log10(\mu+\sigma)$$

$$(\mu-\sigma)_{dB} = 10log10(\mu-\sigma)$$

Where $\mu$ is the mean power level and $\sigma$ is one standard deviation (as a power quantity, hence 10Log).

For lower SNR conditions, refer to this post on the math exchange site that details computing the confidence interval for a Rayleigh (Chi Squared) distribution.

Related is this excellent and comprehensive paper detailing the computation of power spectral density using the DFT:


  • $\begingroup$ Oh man, that was actually obvious. Thanks! Is there a general way how to cope with the case $\sigma > \mu \rightarrow \mu - \sigma < 0$? Since the logarithm would not be defined in this case? $\endgroup$
    – Helmut
    Feb 21, 2023 at 14:42
  • $\begingroup$ Let me update my answer later - the power is always a positive quantity and the distribution will be Chi squared (Rayleigh) so based on that there is a better way to do what you want along with an offset error for the case of white noise that I believe I detailed in another post … let me find that $\endgroup$ Feb 21, 2023 at 15:01
  • $\begingroup$ Thank you so much for the effort! $\endgroup$
    – Helmut
    Feb 21, 2023 at 15:17
  • $\begingroup$ @Helmut all set, see my update $\endgroup$ Feb 21, 2023 at 15:25

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