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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.

Unfortunately, I have not found a method for this problem, so any reference is most welcome!

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, I am not sure how I would normalise the std values for the power spectrum. Also, is there any useful interpretation of these values?

Thanks in advance!

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  • $\begingroup$ What do you mean by "variation"? Maybe what you have in mind is the spectrogram ? $\endgroup$
    – Jdip
    Feb 21, 2023 at 1:52
  • $\begingroup$ Yes, this goes in the right direction. However, I have only blocks and not the the "continuous" time as in the spectrogram method. And although it is possible to see variations graphically, I would like if I can use the standard variation in a more quantitative way? $\endgroup$
    – Helmut
    Feb 21, 2023 at 11:11
  • $\begingroup$ Based on Dan’s answer and your comments, it’s safe to say the spectrogram ISN’T what you want! Dan’s is the correct answer :) $\endgroup$
    – Jdip
    Feb 21, 2023 at 14:53

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The standard deviation is an indication of how accurate the given mean is to the "true" underlying power spectral density. What the Bartlett Method does (as well as the related Welch Method) is reduce that uncertainty (lower standard deviation) at the expense of frequency resolution. To see this first hand, compare the result above to the FFT of the entire sequence (instead of breaking into smaller blocks and averaging the power spectrums derived from the block FFTs as done here). The result of this will have a much wider standard deviation, or uncertainty as to what the actual power spectral density at any given frequency.

The standard deviation from all the blocks for the power spectral density at any given frequency gives an indication of the confidence for the estimated mean value (see https://en.wikipedia.org/wiki/Confidence_interval). This is analogous to estimating the average from the samples of a stationary non-zero random process by computing the mean over $N$ samples. If all samples are independent, then the standard deviation of the resulting mean will be $1/\sqrt{N}$ (reducing the noise on the estimate of the mean). Similarly to the extent the same bin in each FFT over all chunks are independent, (and to the extent the time domain waveform is stationary over the duration of all chunks), by increasing the number of chunks (smaller FFTs), we increase $N$ and get the $1/\sqrt{N}$ improvement in accuracy. However smaller FFT's will have a wider frequency resolution, which is the trade the Bartlett method makes (increase the resolution bandwidth, but improve accuracy of the resulting power spectral density over the bandwidth given by the resolution bandwidth).

This article at the link copied below provides more complete analytic details on deriving the confidence of the power spectral density estimate using Bartlett's (and Welch's) method and confirms the $1/\sqrt{N}$ relationship I have provided above: https://www.researchgate.net/publication/323678279_Precision_analysis_of_the_noise_power_spectrum_estimate_in_radiography_imaging#pf4

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  • $\begingroup$ Hmm, my original signal is really long. I broke it down into smaller pieces for a) computational reasons and b) for reduction in uncertainty of the power estimate as you said. However, I am still interested in the variation between the blocks to get an overall feeling about the quality of my estimate derived by the Bartletts method. Is there a way to use this? $\endgroup$
    – Helmut
    Feb 21, 2023 at 11:09
  • $\begingroup$ Yes the standard deviation would be just that metric. I added some further details that may help you. $\endgroup$ Feb 21, 2023 at 11:17
  • $\begingroup$ Thank you! The last paper helped me alot! I edited my question, since I am starting now to understand what I actually wanted to ask. Is it still reasonable to scale psd-variance in dB-log scale? $\endgroup$
    – Helmut
    Feb 21, 2023 at 12:31
  • $\begingroup$ Did :) I would appreciate, if you could have a look at the new question Unfortunately I have too few reputation points to upvote :/ $\endgroup$
    – Helmut
    Feb 21, 2023 at 13:23

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