Brett,
All of the spectral density specifications are of a 'specific' type,=
and the folks who 'set the stage' of convention (because they were first) =
were the electrical engineers. It is easy to understand their case, becaus=
e the power in watts dissipated by a resistive load R (unit of ohms) is giv=
en by the square of the voltage appearing across the resistor divided by R=
.. The conventional specification avoids the use of the R value by making t=
he PSD 'specific', corresponding to the amount of power dissipated by a one=
ohm load. If they didn't actually use a one ohm load, they divide their =
result by the ohmic size of the load used (in keeping with the calculation =
of specific quantities generally). Anything other than one ohm requires th=
at the PSD calculation be divided by the value of R in ohms. In turn, to g=
et the actual PSD from the specific one requires one to multiply the specif=
ic PSD by the conductance of the load, which is the reciprocal of resistanc=
e. The bottom line is the following-what the electrical engineers label W/=
Hz should really be W/Hz/mho, where mho is the unit of conductance. Nobody=
actually includes 'mho' in the specification, but it is not really a huge =
source of misunderstanding; because it is understood by the EE's that the (=
unspecified) load of their statement corresponds to a one ohm load. What =
is a big deal is that the convention leads mechanical oscillator users, who=
follow the EE lead, into a place of great confusion. Insofar as your comm=
ent about the squared terms-the only self consistent way to satisfy Parseva=
l's theorem is to work with the square of the modulus of the individual com=
ponents that result from taking the Fourier transform of the voltage appear=
ing across the one ohm load. Since the power in the time domain involves V=
squared, the power in the frequency domain must involve the modulus square=
d; i.e., the sum of the squares of the real and imaginary components of the=
FFT computed using that load voltage.
So what about the mechanical case? It cannot avoid the (inertial) mas=
s of the oscillator. The power that figures into the calculation of the s=
pecific PSD of mechanical type is the work per unit time done against the d=
amping (friction) force of a 1 kg oscillator. As with the EE convention, t=
he power depends on the 'amount' of the quantity against which the work is =
done. Since the damping friction depends on the size of the mass, the actu=
al PSD is proportional to the mass in kg. As with the EE's using one ohm, =
the ME's have to work with one kg-except that we want something that applie=
s to more seismometers than just those with an exact one kg inertial mass. =
Since the power associated with work against the damping friction is given=
by the product of force (in Newtons =3D kg m/s^2) and velocity (in m/s), w=
e see that the actual power must have units of kg m^2/s^3. This is made of=
specific form by dividing by kg to yield the result m^2/s^3, which differs=
from m^2/s^4. Which of these can represent power involving a 1 kg oscilla=
tor? The answer is-only m^2/s^3 which is equivalent to W/kg =3D J/s/kg =3D=
N m/s/kg. When one computes the frequency domain quantity corresponding t=
o this, it yields W/kg/Hz, or in most every actual FFT algorithm case, W/kg=
/bin width that can be converted to m^2/s^3/Hz. It is amazing to me how an=
ybody can argue against this result, if they ever passed a university physi=
cs course. Also, engineering professors likewise stress the critical impor=
tance of dimensional analysis. Simply put, if you call something power, an=
d the units are not in terms of watts (for the system international convent=
ion)-then there is something wrong with your nomenclature. The problem wit=
h the g^2/Hz =3D m^2/s^4/Hz is that it is not W/kg/Hz, but rather W/kg/s/Hz=
.. I have no idea what the power per second specified by these units corres=
ponds to; since it corresponds to the product of 'jerk' (derivative of acce=
leration) and velocity.
Acceleration, and acceleration only, is the only quantity to which a =
seismometer (or accelerometer) responds. To calculate the PSD, one first c=
alculates the Fourier transform of the output from the instrument, specifie=
d (by means of the calibration constant) in terms of acceleration in m/s^2.=
As with the electrical case, the modulus squared is next calculated, corr=
esponding to m^2/s^4/Hz. Finally, every one of these components must then =
be divided by its associated frequency to obtain the necessary m^2/s^3/Hz-w=
hen specifying the PSD in linear frequency units. When one specifies the P=
SD in log frequency (or log scale of frequency) the transformation to the =
log form automatically introduces a reciprocal frequency multiplier so that=
one does not have to otherwise do that division. The result that follows =
is m^2/s^3/octave. The reason is what I mentioned in a previous mailing-be=
cause the differential of ln f is df/f. Unfortunately nearly everybody bel=
ieves (erroneously) that the log transformation does nothing other than pro=
vide a convenient 'compression' of the graph-so that several decades of fre=
quency variation can be accommodated by a single graph.
I am compelled to also add the following:
When one is interested only in the eigenmodes (frequencies of spectral line=
s) that characterize the dynamics of a system, the preceding comments are t=
ypically of little importance. The actual 'heights' above noise of the lin=
es (using a log ordinate scale) are not usually considered in a significant=
quantitative sense-only their frequency values in a relative sense. Shoul=
d they be considered, however, with the units corresponding to m^2/s^4/Hz-w=
rong total power (and in turn energy) estimates will be generated from the =
data.
It is high time that seismologists and mechanical engineers rise above=
what amounts to a customary trivial inspection of spectral data. Absolute=
quantitative frequency domain information, which is meaningful only when p=
roper units are incorporated-would greatly facilitate the quest for better =
understanding of the world around us.
Randall
Brett,
All of the s=
pectral density specifications are of a ‘specific’ type, and th=
e folks who ‘set the stage’ of convention (because they were fi=
rst) were the electrical engineers. It is easy to understand their ca=
se, because the power in watts dissipated by a resistive load R (unit of oh=
ms) is given by the square of the voltage appearing across the resist=
or divided by R. The conventional specification avoids the use of the=
R value by making the PSD ‘specific’, corresponding to the amo=
unt of power dissipated by a one ohm load. If they didn’t=
actually use a one ohm load, they divide their result by the ohmic size of=
the load used (in keeping with the calculation of specific quantities gene=
rally). Anything other than one ohm requires that the PSD calculation=
be divided by the value of R in ohms. In turn, to get the actual PSD=
from the specific one requires one to multiply the specific PSD by the con=
ductance of the load, which is the reciprocal of resistance. The bott=
om line is the following—what the electrical engineers label W/Hz sho=
uld really be W/Hz/mho, where mho is the unit of conductance. Nobody =
actually includes ‘mho’ in the specification, but it is not rea=
lly a huge source of misunderstanding; because it is understood by the EE&#=
8217;s that the (unspecified) load of their statement corresponds to a one =
ohm load. What is a big deal is that the convention leads mecha=
nical oscillator users, who follow the EE lead, into a place of great confu=
sion. Insofar as your comment about the squared terms—the only =
self consistent way to satisfy Parseval’s theorem is to work with the=
square of the modulus of the individual components that result from taking=
the Fourier transform of the voltage appearing across the one ohm load.&nb=
sp; Since the power in the time domain involves V squared, the power in the=
frequency domain must involve the modulus squared; i.e., the sum of the sq=
uares of the real and imaginary components of the FFT computed using that l=
oad voltage.
&n=
bsp; So what about the mechanical case? It cannot avoid the (ine=
rtial) mass of the oscillator. The power that figures into the =
calculation of the specific PSD of mechanical type is the work per unit tim=
e done against the damping (friction) force of a 1 kg oscillator. As =
with the EE convention, the power depends on the ‘amount’ of th=
e quantity against which the work is done. Since the damping friction=
depends on the size of the mass, the actual PSD is proportional to the mas=
s in kg. As with the EE’s using one ohm, the ME’s have to=
work with one kg—except that we want something that applies to more =
seismometers than just those with an exact one kg inertial mass. Sinc=
e the power associated with work against the damping friction is given by t=
he product of force (in Newtons =3D kg m/s^2) and velocity (in m/s), we see=
that the actual power must have units of kg m^2/s^3. This is made of=
specific form by dividing by kg to yield the result m^2/s^3, which differs=
from m^2/s^4. Which of these can represent power involving a 1 kg os=
cillator? The answer is—only m^2/s^3 which is equivalent to W/k=
g =3D J/s/kg =3D N m/s/kg. When one computes the frequency domain qua=
ntity corresponding to this, it yields W/kg/Hz, or in most every actual FFT=
algorithm case, W/kg/bin width that can be converted to m^2/s^3/Hz. =
It is amazing to me how anybody can argue against this result, if they ever=
passed a university physics course. Also, engineering professors lik=
ewise stress the critical importance of dimensional analysis. Simply =
put, if you call something power, and the units are not in terms of watts (=
for the system international convention)—then there is something wron=
g with your nomenclature. The problem with the g^2/Hz =3D m^2/s^4/Hz =
is that it is not W/kg/Hz, but rather W/kg/s/Hz. I have no idea what =
the power per second specified by these units corresponds to; since it corr=
esponds to the product of ‘jerk’ (derivative of acceleration) a=
nd velocity.
&n=
bsp; Acceleration, and acceleration only, is the only quantity t=
o which a seismometer (or accelerometer) responds. To calculate the P=
SD, one first calculates the Fourier transform of the output from the instr=
ument, specified (by means of the calibration constant) in terms of acceler=
ation in m/s^2. As with the electrical case, the modulus squared is n=
ext calculated, corresponding to m^2/s^4/Hz. Finally, every one of th=
ese components must then be divided by its associated frequency to obtain t=
he necessary m^2/s^3/Hz—when specifying the PSD in linear frequency u=
nits. When one specifies the PSD in log frequency (or log scale=
of frequency) the transformation to the log form automatically introduces =
a reciprocal frequency multiplier so that one does not have to otherwise do=
that division. The result that follows is m^2/s^3/octave. The =
reason is what I mentioned in a previous mailing—because the differen=
tial of ln f is df/f. Unfortunately nearly everybody believes (errone=
ously) that the log transformation does nothing other than provide a conven=
ient ‘compression’ of the graph—so that several decades o=
f frequency variation can be accommodated by a single graph.
I am compelled to also =
add the following:
When one is intereste=
d only in the eigenmodes (frequencies of spectral lines) that characterize =
the dynamics of a system, the preceding comments are typically of little im=
portance. The actual ‘heights’ above noise of the lines (=
using a log ordinate scale) are not usually considered in a significant qua=
ntitative sense—only their frequency values in a relative sense. =
; Should they be considered, however, with the units corresponding to m^2/s=
^4/Hz—wrong total power (and in turn energy) estimates will be genera=
ted from the data.
&n=
bsp; It is high time that seismologists and mechanical engineers=
rise above what amounts to a customary trivial inspection of spectral data=
.. Absolute quantitative frequency domain information, which is meanin=
gful only when proper units are incorporated—would greatly facilitate=
the quest for better understanding of the world around us.
=
=
Randall
=