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I'm trying to deduce the white noise and rate random walk for an IMU. Some sources recommend deducing those values by fitting a line to the corresponding part of the log-log plot, and then reading the values from those lines at respectively \$t\$ and \$3t\$ where \$t\$ corresponds to about \$1s\$. For example, this is what they recommend here.

My question is about how this values are usually determined, and why it makes sense to read them as described. For example, take this figure that is taken from the aforementioned source:

enter image description here

I think that I vaguely understand how Allan Variance relates to the Power Spectral Density. If I'm not mistaken, the part of the plot that has approximately a slope of \$0.5\$ is where the data behaves like a process in which random walk noise is dominant, and the blue line is fitted (more or less) to that part. But then, why would we read its value at \$t=3s\$, where the signal behaves as a process that is tainted by white noise?

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  • \$\begingroup\$ By the way, ASH, what is the current art of information-theory in images, particularly for self-driving cars? \$\endgroup\$ Dec 27 '17 at 3:13
  • \$\begingroup\$ @analogsystemsrf Hmmm... Sorry I'm not sure I understand the question... Could you please elaborate? \$\endgroup\$
    – Ash
    Dec 27 '17 at 8:10
  • \$\begingroup\$ @analogsystemsrf I imagine this is a joke comment, but... Well it flies a bit above my head. \$\endgroup\$
    – Ash
    Dec 27 '17 at 8:16
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In IEEE Std 952-1997 (Annex C) we find the answer to this. They have a fancy integral relation between Allan Variance (\$\sigma^{2}\$) and Power Spectral Density. If \$K\$ is the random walk coefficient of which you speak, they show that

$$\sigma^{2}(\tau) = \frac{K^{2} \tau}{3}$$

So

$$\sigma(\tau) = \frac{K \tau^{0.5}}{3^{0.5}}$$

The 0.5 power on \$\tau\$ is why you want the part of slope with 0.5 gradient, the fact that when \$\tau\$ is 3 we are left with

$$\sigma(3) = K$$

is why we care about the intercept with \$\tau==3\$.

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