Prev: Help on MATLAB debugging
Next: Speech Recognition
From: Dinne on 15 Feb 2010 03:27 Dear all, >Your sample rate seems skimpy. With these sensors the noise output is defined as x degree/sec/sqr(Hz). This means that a higher samplerate will introduce more noise into the signal. It is the question whether the increase in differentiator accuracy outweighs the decrease in signal accuracy. Probably this will only be determined after some field tests. I eventually implemented a differentiator as described in: http://www.holoborodko.com/pavel/?page_id=245 I chose this approach because the filter design accommodates for changes in the length of the filter (N). The 6Hz cutoff is achieved by choosing the filter with parameters n=4, N=9 regards, Dinne
From: Jerry Avins on 15 Feb 2010 20:34
Dinne wrote: > Dear all, > >> Your sample rate seems skimpy. > > With these sensors the noise output is defined as x degree/sec/sqr(Hz). > This means that a higher samplerate will introduce more noise into the > signal. The Hz above refers to sensor bandwidth, not to sample rate. Rethink. > It is the question whether the increase in differentiator accuracy > outweighs the decrease in signal accuracy. Probably this will only be > determined after some field tests. > > I eventually implemented a differentiator as described in: > > http://www.holoborodko.com/pavel/?page_id=245 > > I chose this approach because the filter design accommodates for changes in > the length of the filter (N). The 6Hz cutoff is achieved by choosing the > filter with parameters n=4, N=9 Jerry -- Engineering is the art of making what you want from things you can get. ����������������������������������������������������������������������� |