While studying spectrum analyzers I ran into the concept of RMS averaging as a mean to reduce dispersion of data without affecting power spectral density (PSD).

What is not really clear to me is how this is achieved: as per the calculation, I found that it is obtained by averaging the magnitude of the squared values of each FFT's frequency bin over a certain number of acquired traces. However, while the definition itself is not the problem and it is obvious that the dispersion of data will be reduced (it's an average), I don't see how the PSD can stay the same without being affected by such a computation.


Here follows all I found on the subject (professor's online notes):

`` • Change of point of view: now we want to measure the power spectral density of noise without reducing it

• Since we processed the input signal by truncating and sampling, also white noise trace is displayed with a «noisy» spectrum

• In order to quantify the noise power density, it is useful to introduce the RMS averaging, that is the average on the magnitude of the square value of each frequency bin of the FFT performed on Np traces

• This technique reduces the dispersion of data displayed without changing the PSD''

  • $\begingroup$ Can you describe what happens in formulas or give an appropriate reference where the technique is described? $\endgroup$ – Maximilian Matthé Dec 26 '16 at 16:12
  • $\begingroup$ Unfortunately, I don't have any formulas. It was something just mentioned during a lecture and that "explanation" of the calculation was found in the professor's online notes, as it is reported here. I know it's not much, but I thought it was a renown concept... $\endgroup$ – DavideM Dec 27 '16 at 10:10
  • $\begingroup$ I believe with this less information its hard to answer. I personally, dont know exactly, how the averaging is done, what you mean by "dispersion" and what you mean by PSD staying the same. Can you mathematically write the process you describe in your second paragraph of the question? $\endgroup$ – Maximilian Matthé Dec 27 '16 at 19:43
  • $\begingroup$ @DavideM It's not clear what you mean by the computation affecting the PSD. The PSD of the underlying signal is what it is. A computational routine will produce an estimate of this PSD. A straightforward way of estimating the PSD is to compute its magnitude squared DFT (a.k.a. periodogram). To reduce the estimator variance ("dispersion"??), you can split the signal into overlapping segments, compute the periodogram of each segment, and average the periodograms. I believe this is what a spectrum analyzer might be computing. (cf. Welch Method) $\endgroup$ – Atul Ingle Dec 27 '16 at 22:49
  • $\begingroup$ MaximilianMatthé, @AtulIngle; I edited the question in order to provide the "full" explanation I got on the topic. It seems to me quite close to what AtulIngle proposed in his comment. $\endgroup$ – DavideM Dec 29 '16 at 14:45

You can check out the manual for the HP3582A spectrum analyzer or the manual for the SR785 dynamic signal analyzer, which is basically a copy of the former instrument. My guess is that the definition used in these manuals is the closest to what your lecturer is referring to.

In this case, the power spectral density RMS average is found by taking successive averages of the product between the FFT and its complex conjugate (the absolute value), i.e.: $$ \hat{S} = \left< FFT^{*}(\text{windowed time-series segment}) \cdot FFT(\text{windowed time-series segment}) \right> $$

According to the manual:

[...] RMS averaging reduces fluctuations in the data but does not reduce the actual noise floor (squared values never cancel). With a sufficient number of averages, a very good approximation of the actual noise can be obtained.

I belive this should be equivalent to the Bartlett method (periodogram averaging), mentioned by Atul Ingle. The salient point regarding periodogram averaging is the fact that taking the discrete Fourier transforms (DFTs) of a long time-series as a PSD estimate will not lead to a low variance of the PSD estimate. However, segmenting the time-series and taking the average of the DFTs of the segments will reduce the variance of the resulting PSD estimate.

A very good reference with regards to PSD estimation can be found in S. M. Kay and S. L. Marple, "Spectrum analysis—A modern perspective," in Proceedings of the IEEE, vol. 69, no. 11, pp. 1380-1419, Nov. 1981.

This would be in contrast to syncronous averaging, or triggered averaging, used for periodic signals, where time-series segments are set to an integer multiple of the period of a signal of interest. The time-series segments are then averaged before the DFT is found. This operation removes random noise, and hence will not provide a PSD estimate of any random process. The signal-to-noise ratio for the periodic signal of interest is however drastically improved.


A straightforward way of estimating the PSD is to compute magnitude squared DFT (a.k.a. periodogram).

Suppose we have an $N$ point sequence $x(n)$ and we would like to estimate its power spectral density (PSD) $P_{xx}(f)$.

Method 1: Periodogram

We compute the magnitude squared DFT. No averaging. $$ \hat{P}_{xx}(f) = \frac{1}{N}\left | \sum_{i=0}^{N-1} x(n) e^{-j2\pi fn/N} \right |^2 $$

Method 2: Periodogram Averaging

We instead use $M$ different realizations of the $N$-point sequence $x(n)$ giving us $M$ sequences $x_i(n)$, $0\leq i\leq M-1$. We compute the periodogram estimate $\hat{P}_{x_ix_i}(f)$ of each $x_i$ and average them to get the final PSD estimate: $$ \hat{P}(f) = \frac{1}{M}\sum_{i=0}^{M-1} \hat{P}_{x_ix_i}(f). $$

It can be shown that the variance of this averaged periodogram estimate is lower:

$$ \mbox{Var}[\hat{P}(f)] = \frac{1}{M} \mbox{Var}[\hat{P}_{xx}(f)]. $$

Note that it is a little unfair to compare Methods 1 and 2 because Method 2 uses more data ($MN$ data samples, instead of just $N$). If all you have is the original $N$ point sequence $x(n)$, you end up making a tradeoff between PSD estimator's variance and frequency resolution.

The PSD of the underlying signal is what it is. A computational routine will produce an estimate of this PSD. Whether a PSD estimate is good or bad will depend on your definition of good or bad. One commonly used metric variance. In that sense, periodogram averaging estimate $\hat{P}$ is a better estimator of the PSD $P_{xx}$ than $\hat{P}_{xx}$.

cf. Bartlett Method, Welch Method

Reference: Digital Signal Processing, Proakis and Manolakis, Ch. 14 (4th ed.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.