Variance of PSD estimates

Now, we've said that we aren't measuring PSD, we are only estimating it. Without proving it, I think you should see that if you take a longer sample of data, you might get a better estimate of the PSD (assuming stationary vibrations). As a vibrations engineer, sometimes you will be handed a set of data and told, ``see if you can make something out of this''. In this case you have to hope there is enough data to get a good estimate. Other times, you get to decide how much data you want to take. Now, obviously if you don't take enough you won't get a good estimate. But you can't just guess a really large number either, because your boss will point out that it costs money to take all this data and he'll ask if you really need that much or if you could get away with a third as much. If you answer ``Gee I don't know'' guess how your boss is going to react. You need to be able to calculate a priori how much data you really need and be able to defend it.

So how do we quantify this? Obviously, if you take an PSD estimate today and a PSD estimate tomorrow you'll get different answers. If you take a whole bunch of estimates, you'll get a whole bunch of answers. What we are going to quantify is the standard deviation of the estimate. But of course, the standard deviation by itself isn't the number we want. That is, is a 1 ${\frac{{g^{2}}}{{Hz}}}$ standard deviation or a small one? If we have a mean PSD of 100, then 1 is pretty small. But if we have a mean of 2, then 1 is pretty big. So we quantify standard deviation divided by mean, or $\sigma$/$\mu$. Now here's the formula we want [4]

$${\frac{{\sigma}}{{\mu}}} \approx {\frac{{1}}{{\sqrt{BT}}}}$$ (1)

where B is the bandwidth (Hz) of the measurement (it may be helpful to think FFT bin width, although B may not technically be exact the FFT bin width), and T is the time (s) of the measurement. So, for example, if we want a variance of 1% of the mean, and we need a 1 Hz bandwidth (e.g. if we have two signals that are about 2 Hz apart), then we need 10,000s or about 2.78 hours worth of data. Depending on how expensive your data is, your boss might be thrilled that it's only 3 hours, or he might decide that 5% variance will be good enough.

" . $row['Name'] . " Posted on " . $row['DateTime']; echo "
"; echo $row['Comment']; echo "

"; } echo ""; ?> Leave a comment on this page:

Name: (optional)
To prove you are not a robot, what is 2+3?

This work by Daniel Kiracofe (daniel dot kiracofe at gmail dot com) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License./' $I