01-29-2008 05:25 PM
04-05-2009 06:00 PM
Your approach is perfectly correct, if the mean acceleration were not zero, the object would be (in average) accelerating and moving from its rest position! In fact, piezoelectric accelerometers cannot measure zero frequency acceleration (steady acceleration) so the mean of the signal has to be zero. For the same reason, you can set the mean velocity to zero when integrating to displacement. Another approach to avoid drifting is to integrate in the frequency domain.
Hope this helps
11-28-2009 06:43 AM
I am trying to get displacement RMS value from a signal obtained from an accelerometer. This accelerometer is connected to a shaker that vibrates at 159,2 HZ with an acceleration RMS value of 3,16 m/s2. At that frequency it's supposed to obtain the same RMS value for velocity (3,16 mm/s) anf for displacement (3,16 um).
I have used the SVT RMS level.vi using the signal obtained from accelerometer as input. I obtain the correct aceleration RMS value. Then I double integrate that signal using SVT Intgration.vi and apply SVT RMS level.vi, but the value I obtain doesn't mach with the expected one(3,16 um).
Should I eliminate DC componet of the signal before integrating it? Or before using SVT RMS level?
12-21-2013 03:53 AM
can you explain with some example?
12-23-2013 12:26 PM
There was code posted in this thread. If you go back a few messages (it appears on the second page for me), there is a post date/time 01-28-2008 05:09 AM with prova_integraz1.vi attached. That's the code that the users were discussing in this thread.
Additionally, this thread is older (original post 2007, last post 2009). You may want to create a new thread that references this conversation to get more traffic flow. There is a chance that the users on this thread are still subscribed, but it's hard to say.
12-23-2013 12:29 PM - edited 12-23-2013 12:29 PM
Also, if it is the actual physics behind the conversion from acceleration to displacement, I did find this website:
http://www.cbmapps.com/docs/28
Personally, I'm not familiar with the physics behind this conversion.