View Full Version : Re: Comments on resampling.

Leonard G. Caillouet
08-14-1998, 08:08 AM
Dr. Giakas,

I must disagree that resampling is necessary. Simply low pass filter the
data, then differntiate. As you noted, he will do so anyway. You are
trying to correct a lack of precision in sampling level by increasing the
quantization error by lowering the sampling rate. You simply will not
improve the limit in precision of the measured levels by resampling.
Filtering will result in an approximated signal anyway. Why not get that
approximation from the version of the signal with the least quantization
noise possible?


At 01:51 PM 8/14/98 +0100, Giannis Giakas wrote:
>Dear all
>The data collected by Greg Kawchuk are really low frequency data.
Therefore the sampling with 1000 Hz with his A/D collected a series of
repeated values (because of the A/D resolution). As a result I suggested to
him to resample his existing data. So the main point was to find the
appropriate sampling frequency.
>My suggestion was/is to check the power spectrum of the signal and then by
finding the frequency corresponding to an appropriate x% of cumulative
spectrum (e.g.99.9%), to multiply this with a factor of 4 (so to cover the
Niquist problem) and this could serve as the new frequency. Of course if
this procedure will give a value, for example, of 107 Hz, I would "play" it
safe and select the new frequency as 125 Hz or 200 Hz for obvious reasons
(so 8*125 =1000, or 5*200 = 1000).
>Comments on the responses
>1. Why to filter the signal before resampling ? I do not favour this
procedure because the new signal will need filtering anyway, so why
filtering twice the 0th derivative ?
>2. I support the process of averaging than collecting the 1st, 6th, 11th
.. points. This is because I believe it will provide a more representative
value of the actual data point although it will have a small smoothing effect.
>3. The solution of Ran Ravhon to contruct a derivative filter is not
really clear to me, because I can not see the difference with a usual
filter in the 0th derivative (given the fact that you will have the same
cut-off and same type of filtering) and then differentiation.
>4. The comment of Oyvind Stavdahl that the first derivative of the signal
will have increased power at higher frequencies and therefore the frequency
including the 99% of the signal will be shifted to the right is correct and
obvious. He suggests
>> 2. Take the derivative of the digitized signal, and determine the frequency
>> where x% of this signal is contained.
>Why to do this though ? Similarly, if I want to calculate the 2nd
derivative of that signal then by using this approach the 99% of the signal
would be contained in the highest frequency (because due to noise
amplification, the highest frequency will probably have the maximum power).
Does that mean that in order to calculate the 2nd derivative for this low
frequency signal I need to sample at 1000 Hz ?
>5. Tony Hodgson is correct on his comment that the percentage of the
energy in the signal will depend on how noisy your signal is, so this will
affect the selection of sampling frequency. My reply is that as I suggested
above you do not need to be exact on the sampling frequency, because you
don't know the real Niquist frequency anyway. If the "proper" (say
PROPER= 4 * "NIQUIST") sampling is 107 Hz, then I would be also happy to
use 200 Hz instead of 125Hz. In practice the noise of the signal will not
make such great differences in the calculation of the new sampling
frequency (compared to "my" bandwidth).
>Apologies for not replying before (although the original post mentioned my
name !) but I am on holidays and I do not check my Email regularly.
>Thank you for your time
>Giannis Giakas PhD
>Sport Health and Exercise
>School of Health
>Staffordshire University
>Stoke-on-Trent ST4 2DF
>Tel : +44 1782 294292
>Fax : +44 1782 294321
>Email: g.giakas@staffs.ac.uk
>webmaster sportscience list
>BASES lists moderator
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