PSN-L Email List Message
Subject: Re: FFT bin magnitude question
From: Brett Nordgren brett3nt@.............
Date: Fri, 13 Jul 2012 09:25:17 -0400
Randall,
No challenge, just a suggestion as to the source=20
of the confusion regarding units.
Power Spectral Density plots can be found which=20
are plotted both "per Hz" and "per 1/6 decade" or=20
"per octave", etc. There is a subtle distinction=20
regarding the units. Octaves and decades, being=20
ratios, are dimensionless. Hz, on the other hand, has units of 1/seconds.
Considering an acceleration PSD, given as=20
m^2/sec^4 per Hz, and applying the units for Hz=20
we get just m^2/sec^3. On the other hand, the=20
units m^2/sec^3 per octave are also correct. I=20
believe these are consistent, both with what I=20
see in most published plots, and with what you were suggesting.
When working with Hz, I tend to prefer the form=20
m^2/sec^4 per Hz as it is more obviously an=20
acceleration-power density with respect to=20
frequency, but that's mostly a matter of style.
Regards,
Brett
At 01:35 PM 7/11/2012, you wrote:
> I will probably meet with debate over this=20
> posting, since I have gone =91round and round=92=20
> for more than a decade with the professional=20
> seismologists about their need to start working=20
> with a better set of power spectral density=20
> units. To satisfy both Parseval=92s theorem and=20
> to also work with something that makes formal=20
> physics sense in terms of power=ADthere can be=20
> only one set of PSD units that are reasonable=20
> in terms of the whole world=92s recognition of the proper unit of power=
(watts).
>That set of units can only be W/kg/Hz which=20
>equals m^2/s^3/Hz and not the long-standing=20
>seismology and mechanical engineering tradition of m^2/s^4/Hz (or g^2/Hz).
>Some of them more correctly label their graphs=20
>with the words acceleration spectral=20
>density. When one computes the FFT of the=20
>output from a seismometer (accelerometer in=20
>general) in the units corresponding to the only=20
>thing to which the instrument responds; i.e.,=20
>acceleration; then the ASD is simply=20
>m^2/s^4/Hz. In other words, the ASD is very=20
>close to the single sided Fourier transform=20
>spectrum discussed above, if properly normalized=20
>to satisfy Parseval=92s theorem. One must be=20
>very, very careful, however, to recognize=20
>another subtlety that most everybody=20
>ignores. The units as generated are really=20
>m^2/s^4/bin width. The bin width is never=20
>actually one Hz, and so a multiplicative factor=20
>must be considered if one is to actually use m^2/s^4/Hz, properly computed.
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