Accelerometer Noise PSD to Velocity Random Walk

schaffmc

I have the intrinsic noise PSD spec for Honeywell's QA3000 accels, and I need to find the velocity random walk due to noise to show that I'm within my allowable. The PSD is given in units of g^2/Hz vs Hz. I'm given a velocity random walk due to noise spec of 15 (micro-m/s)/rt(s) that I must be below.

If I was unaware of the specification, the way I would go about finding the velocity random walk is to integrate under the PSD curve and take the square root to find the acceleration noise level in g's, and then integrate that value with respect to my time period (.25s). That would give me random walk in micro-m/s, which of course is not the same units as the spec I've been given. I imagine that the reason velocity random walk is in these particular units is to manage its time-dependence, but I'm not sure how or why.

Can anybody provide insight into 1. why the units are what they are, and 2. how I would go about getting from my noise PSD to velocity random walk?

Thanks!
Melissa

schaffmc

My question has changed due to more research. Here it is:

I have a specification that says this: "The velocity random walk due to the white noise portion of the acceleration self-generated output noise PSD shall be less than 15 (micro-m/s)/sqrt(s)." I have the PSD for my accelerometer, in units of g^2/Hz vs frequency between 1E-6 and 1e4Hz. There are three white noise portions. [I can send you the spec if you want.] We're filtering out the noise above 300Hz (which eliminates one of the white noise portions, and is also in the midst of one)...the third white noise portion is between 1e-3 and 1Hz, as well.

My question is, how do I go about finding whether or not this accelerometer meets the criteria? After some research over the last couple weeks, it seems that the method I should use is find the PSD at our filtering frequency since it is in the midst of a white noise portion, multiply by 9.81^2 to get into m/s^2 units, and take the square-root. This gives me a velocity RW of 12.58 (micro-m)/sqrt(s), which is right below my spec. Is this the correct method?

D H

Another way to express that 15 microns/second/sqrt(second) requirement from post #1 is 15 microns/second²*sqrt(second). In terms of g (9.80665 m/s²), that is about 1.5 micro-g*sqrt(sec). You need an extremely sensitive and noise-free accelerometer here! This is verging on nano-g accelerometer territory.

Per Honeywell's specs (see http://www.inertialsensor.com/qa3000.shtml), the intrinsic noise from a QA 3000 is 7 micro-g*sqrt(sec) at low frequencies, 70 micro-g*sqrt(sec) at midlevel frequencies, and 1500 micro-g*sqrt(sec) at high frequencies. Not even close to the < 2 micro-g*sqrt(sec) your application needs, even if you can reliably filter out the high frequency components.

schaffmc

That was the initial way I went about looking at this once the units made sense to me. But, I happen to know this spec was written such that the QA3000 passes, so that drove me to keep researching.

I came across an IEEE standard (IEEE 952-1997, annex C.1.1, "angle random walk" -- it was written for gyros). It associates rate noise PSD (S(f)) with random walk (N) by the equation: S(f) = N². We're filtering out everything above 300Hz (and the PSD happens to have a zero-slope portion, ie white noise), so I found N like this: N = sqrt(S(f)) = sqrt(S(300Hz)) = 12.58 (micro-m/s)/sqrt(s).

It still doesn't entirely make sense why exactly it would work that way, but that's what I get from the IEEE standard... any ideas?