What is the relation between the 1/f spectral noise density and the Allan deviation noise floor? (I found a value of $\frac{5}{3}$)

Question

I have some noise density that consists of 1/f noise and white noise like so:

I then select an arbitrary sampling frequency $f_S$ and data length $n$ and compute a time domain noise data vector like this (for $f_S=$12.5 Hz and $n=$1e6):

Next I compute the Allan deviation for a bunch of different $\tau$ values (this is the overlap Allan deviation, OADEV). The points are the actual OADEV values. The line is an analytic result derived from the known amplitude noise density spectrum.

So far so good. The OADEV becomes constant at about the $\tau$, where I can see the noise corner frequency in the noise spectrum.

My question:

My question is about the height of the Allan noise floor. In this example I found it empirically to be at $\approx \frac{5}{3}$.

In general, when the 1/f spectral noise density is $\frac{A}{\sqrt f}$, then the Allan deviation floor seems to be at $\approx \frac{5}{3} A$.

Is there an analytical relation between the 1/f spectral noise density and the Allan deviation noise floor?

Answer 1

A very useful chart for ADEV and related measurements is Enrico Rubiola’s “Enrico’s Chart of Phase Noise and Two-Sample Variances”.

To be clear with understanding this chart, this chart is depicting the traditional use of ADEV as a measure of frequency stability, which is therefore sensitive to the frequency (and therefore then phase as well) fluctuations in the signal and is not sensitive to amplitude fluctuations. Typically units of fractional frequency are used (frequency error normalized to the center frequency of an oscillator). The OP however is using the ADEV computations more generally in substituting an amplitude variable for fractional frequency (I believe). This is ok to do, and ultimately would come down to the purpose an interest in using ADEV in the first place- it is a great generalized tool for evaluating non-stationary signals. That said, keep in mind when reviewing the charts we are instead considering for the OP the stability due to amplitude fluctuations in the signal (instead of frequency fluctuations).

We notice in the OP’s plot that the amplitude is going down as a magnitude quantity at $1\sqrt{f}$, which is consistent with a power spectral density going down at $1/f$ since the power quantity would be the magnitude quantity squared. It is the power spectral densities that are shown in Enrico’s chart and provide us the direct parameters to determine the ADEV floor.

With that, in Enrico’s chart at the bottom of Page 1 on the left hand side are two charts next to each other as Frequency Fluctuation PSD and Allan Variance (AVAR which is ADEV squares d). From this we see the 1/f noise, given as $S_y(f) = h_{-1}/f$ is related to the AVAR floor as:

$$\sigma_y^2(\tau) = 2\ln(2)h_{-1}$$

Given this, we can work the OP’s result backwards from the floor to get the 1/f noise in the same units to see if this makes sense:

From OP:

$$\sigma_y(\tau)= 5/3$$

$$\sigma_y^2(\tau) = 25/9$$

$$h_{-1}= \frac{\sigma_y^2(\tau)}{2\ln(2)} = \frac{25}{18\ln(2)} \approx 2$$

So therefore to have an ADEV floor of 5/3 the PSD due to fractional frequency fluctuations would be $S_y(f) = 2/f$.

The square root of this is $\sqrt{2}/\sqrt{f}$ while the OP’s sloped region in the first chart shown of amplitude noise density is clearly $1/\sqrt{f}$. Our results are off by a factor of $\sqrt{2}$.