Giannis Giakas

08-13-1998, 10:51 PM

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

Sincerely

Giannis

--

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

http://www.staffs.ac.uk/health/sport/giannis/index.htm

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http://www.sportsci.org

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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

Sincerely

Giannis

--

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

http://www.staffs.ac.uk/health/sport/giannis/index.htm

webmaster sportscience list

http://www.sportsci.org

BASES lists moderator

-------------------------------------------------------------------

To unsubscribe send UNSUBSCRIBE BIOMCH-L to LISTSERV@nic.surfnet.nl

For information and archives: http://www.bme.ccf.org/isb/biomch-l

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