RMS averaging in spectrum analyzers 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.

------EDIT------

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''
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.)