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PStamler PStamler is offline
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On Nov 21, 10:41*pm, John O'Flaherty wrote:

There is a sense in which calling a voltage gain of 10 a gain of 20 dB
does refer to power. In a circuit in which nothing is changed but that
gain (including output loading and input signal level), if that gain
is reduced to 0 dB, the output power level will be reduced by a factor
of 100.


Huh-uh. We had this discussion years ago.

Given a non-inverting opamp with high input impedance and negligible
output impedance, a 1k resistor from the + input to ground, and a 100k
resistor from the output to ground (in other words a 100k load).
Assume the feedback resistance network is high enough that it draws
negligible current.

1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB. The usage, however,
is nearly universal, so speaking as a descriptivist who believes that
dictionaries should reflect how people actually use the language,
perhaps the official definition needs revision to take into account
the dual usage of the term. I've suggested "dBG" as an indicator that
voltage gain is being discussed rather than power gain, and perhaps
that's the way to go.

Peace,
Paul

[other things snipped[


I believe these examples show the power nature of dB measurements.

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On Nov 22, 7:14*am, PStamler wrote:
On Nov 21, 10:41*pm, John O'Flaherty wrote:

There is a sense in which calling a voltage gain of 10 a gain of 20 dB
does refer to power. In a circuit in which nothing is changed but that
gain (including output loading and input signal level), if that gain
is reduced to 0 dB, the output power level will be reduced by a factor
of 100.


Huh-uh. We had this discussion years ago.

Given a non-inverting opamp with high input impedance and negligible
output impedance, a 1k resistor from the + input to ground, and a 100k
resistor from the output to ground (in other words a 100k load).
Assume the feedback resistance network is high enough that it draws
negligible current.

1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB.

But, if I were to double the input voltage I would see a 6dB increase
in power output would I not? In context, the use of dB for voltage
gain is entirely correct and acceptable.

The usage, however,
is nearly universal, so speaking as a descriptivist who believes that
dictionaries should reflect how people actually use the language,
perhaps the official definition needs revision to take into account
the dual usage of the term. I've suggested "dBG" as an indicator that
voltage gain is being discussed rather than power gain, and perhaps
that's the way to go.

I think the world has already decided, so it's a bit late for a new
definition an in any case everyone would think that dBG was dB
referred to 1G. What's that? Gravity?
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davew wrote:

On Nov 22, 7:14*am, PStamler wrote:

[...]
1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB.

But, if I were to double the input voltage I would see a 6dB increase
in power output would I not? In context, the use of dB for voltage
gain is entirely correct and acceptable.


It can be used that way as long as the two voltages being compared are
across the same impedance and both relate to the power in the same way.
This was the case in your example.

You can compare voltages at two different places as long as the
impedances at those places are the same. So an amplifier working with
600-ohm input and output terminations could have its power gain
expressed in dB by just measuring the voltage gain.

What you must not do (which is all-too frequently done) is to use the
same calculation to compare voltages in different impedances. For that,
you must use the voltages to calculate the power at each point and then
compare the powers to get a result in dB. If you want to use a
logarithmic scale to compare voltages, call it something different (dBu,
dBV) because it is not "dB".

The convenient shortcut dB formula relating voltages is often taught
with insufficient emphasis being placed on the conditions for which it
is valid. After a while, people begin to believe that the voltage
formula is the true representation of dB until a situation arises where
it generates nonsense results - then they get confused and sometimes
adamant. At least one well-known textbook has got it wrong.



--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
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On Mon, 22 Nov 2010 12:41:10 +0000,
lid (Adrian Tuddenham) wrote:

davew wrote:

On Nov 22, 7:14*am, PStamler wrote:

[...]
1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB.

But, if I were to double the input voltage I would see a 6dB increase
in power output would I not? In context, the use of dB for voltage
gain is entirely correct and acceptable.


It can be used that way as long as the two voltages being compared are
across the same impedance and both relate to the power in the same way.
This was the case in your example.

You can compare voltages at two different places as long as the
impedances at those places are the same. So an amplifier working with
600-ohm input and output terminations could have its power gain
expressed in dB by just measuring the voltage gain.

What you must not do (which is all-too frequently done) is to use the
same calculation to compare voltages in different impedances. For that,
you must use the voltages to calculate the power at each point and then
compare the powers to get a result in dB. If you want to use a
logarithmic scale to compare voltages, call it something different (dBu,
dBV) because it is not "dB".

The convenient shortcut dB formula relating voltages is often taught
with insufficient emphasis being placed on the conditions for which it
is valid. After a while, people begin to believe that the voltage
formula is the true representation of dB until a situation arises where
it generates nonsense results - then they get confused and sometimes
adamant. At least one well-known textbook has got it wrong.


The full equation is

20 log(v2/v1 *sqrt(z1/z2))

That gets the power gain right for any impedance. (V1 is across
impedance z1, and v2 is across z2).

d
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On Nov 22, 12:41*pm, (Adrian
Tuddenham) wrote:
davew wrote:
On Nov 22, 7:14*am, PStamler wrote:

[...]
1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB.

But, if I were to double the input voltage I would see a 6dB increase
in power output would I not? *In context, the use of dB for voltage
gain is entirely correct and acceptable.


It can be used that way as long as the two voltages being compared are
across the same impedance and both relate to the power in the same way.
This was the case in your example.

You can compare voltages at two different places as long as the
impedances at those places are the same. *So an amplifier working with
600-ohm input and output terminations could have its power gain
expressed in dB by just measuring the voltage gain.

What you must not do (which is all-too frequently done) is to use the
same calculation to compare voltages in different impedances. *For that,
you must use the voltages to calculate the power at each point and then
compare the powers to get a result in dB. *If you want to use a
logarithmic scale to compare voltages, call it something different (dBu,
dBV) because it is not "dB".

The convenient shortcut dB formula relating voltages is often taught
with insufficient emphasis being placed on the conditions for which it
is valid. *After a while, people begin to believe that the voltage
formula is the true representation of dB until a situation arises where
it generates nonsense results - then they get confused and sometimes
adamant. *At least one well-known textbook has got it wrong.

--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)www.poppyrecords.co.uk


This is all well and good, but in a lot of small signal territory and
in DSP, we don't care about power or impedances. We may worry about
output impedance and power drain of a small signal stage, for example,
but we would usually say that a given stage has a voltage gain, not a
power gain i.e. where voltage gain, not power gain, is the object.
So, a gain of 10 in the signal chain would result in an end output
power increase (assuming we were feeding some form of power amplifier)
of 20dB.

Consider a FET source follower as an example. Tremendous gain i.e.
power gain in dB but voltage gain of a little under unity or 0dB. Now
you ask a hundred analogue electronics engineers what the gain is in
dB and (yes it's a trick question) 99/100 will say a bit less than
0dB.


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On 11/22/2010 7:41 AM, Adrian Tuddenham wrote:

The convenient shortcut dB formula relating voltages is often taught
with insufficient emphasis being placed on the conditions for which it
is valid.


This is why it's only been used over the last 30 or maybe 40
years, while the "power dB" goes back 80 years or so. It's
only been since the adoption of the nearly universal
voltage-not-power scheme of interfacing. This is primarily a
result of the use of solid state electronics which
characteristically have nearly zero output (source)
impedance but without the ability to supply any significant
amount of current. It's just how things work now.

After a while, people begin to believe that the voltage
formula is the true representation of dB until a situation arises where
it generates nonsense results


Not a lot of confusion, because, in general, audio people
TODAY work with dB of gain or voltage levels, and absolute
(not relative) watts when dealing with power levels. Someone
will understand a 100W or 200W amplifier, but won't think of
one having 3 dB more (power) gain than the other.

--
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operated without a passing knowledge of computing, although
it seems that it can be operated without a passing knowledge
of audio." - John Watkinson

http://mikeriversaudio.wordpress.com - useful and
interesting audio stuff
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"Mike Rivers" wrote in message


Someone will understand a 100W or 200W amplifier, but
won't think of one having 3 dB more (power) gain than the
other.


That would be their loss. If you compare amp power in dBs, and remember that
it takes about 10 dB to create the perception of "twice as loud", then a lot
of things in the real world that may seem mysterious become understandable.


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davew wrote:

On Nov 22, 12:41*pm, (Adrian
Tuddenham) wrote:
davew wrote:
On Nov 22, 7:14*am, PStamler wrote:

[...]
1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB.
But, if I were to double the input voltage I would see a 6dB increase
in power output would I not? *In context, the use of dB for voltage
gain is entirely correct and acceptable.


It can be used that way as long as the two voltages being compared are
across the same impedance and both relate to the power in the same way.
This was the case in your example.

You can compare voltages at two different places as long as the
impedances at those places are the same. *So an amplifier working with
600-ohm input and output terminations could have its power gain
expressed in dB by just measuring the voltage gain.

What you must not do (which is all-too frequently done) is to use the
same calculation to compare voltages in different impedances. *For that,
you must use the voltages to calculate the power at each point and then
compare the powers to get a result in dB. *If you want to use a
logarithmic scale to compare voltages, call it something different (dBu,
dBV) because it is not "dB".

The convenient shortcut dB formula relating voltages is often taught
with insufficient emphasis being placed on the conditions for which it
is valid. *After a while, people begin to believe that the voltage
formula is the true representation of dB until a situation arises where
it generates nonsense results - then they get confused and sometimes
adamant. *At least one well-known textbook has got it wrong.




This is all well and good, but in a lot of small signal territory and
in DSP, we don't care about power or impedances.


There is nothing wrong with that approach, but you must not then express
the ratios as 'dB'.

You may include the letters 'dB' to suggest a logarithmic scale as long
as they also have some sort of qualifier prefix or suffix to warn the
student or the pedant (or anyone in between) that you are not comparing
powers and these measurements do not follow the correct definition of a
decibel.

Much confusion has been caused over the years by the omission of that
qualifier. Try explaining to a student who is used to thinking of dB in
terms of voltage alone where the gain occurs in the following example:

http://www.poppyrecords.co.uk/other/images/Gain.gif


--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk
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In article , "Arny Krueger" wrote:
"Mike Rivers" wrote in message


Someone will understand a 100W or 200W amplifier, but
won't think of one having 3 dB more (power) gain than the
other.


That would be their loss. If you compare amp power in dBs, and remember that
it takes about 10 dB to create the perception of "twice as loud", then a lot
of things in the real world that may seem mysterious become understandable.


You got to consider that.
But, for understanding basics, knowing 6 dB is twice the voltage is what
people should know and use, and talk about. The only thing power
dB is good for is speaker dissapation and cost of amplifiers, and
indirectly, mains power. Forget volume levels.

greg
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GS wrote:

In article ,
"Arny Krueger" wrote:
"Mike Rivers" wrote in message


Someone will understand a 100W or 200W amplifier, but
won't think of one having 3 dB more (power) gain than the
other.


That would be their loss. If you compare amp power in dBs, and remember that
it takes about 10 dB to create the perception of "twice as loud", then a lot
of things in the real world that may seem mysterious become understandable.


You got to consider that.
But, for understanding basics, knowing 6 dB is twice the voltage is what
people should know....


They should first know that dB is a power measurement, this is the
fundamental fact on which the rest is based. When they have grasped the
basics, then they can be shown that the voltage is a handy way of
comparing two power levels in the right circumstances (and they can
remember some handy voltage ratios if appropriate).


--
~ Adrian Tuddenham ~
(Remove the ".invalid"s and add ".co.uk" to reply)
www.poppyrecords.co.uk


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On Sun, 21 Nov 2010 23:14:27 -0800 (PST), PStamler
wrote:

On Nov 21, 10:41*pm, John O'Flaherty wrote:

There is a sense in which calling a voltage gain of 10 a gain of 20 dB
does refer to power. In a circuit in which nothing is changed but that
gain (including output loading and input signal level), if that gain
is reduced to 0 dB, the output power level will be reduced by a factor
of 100.


Huh-uh. We had this discussion years ago.

Given a non-inverting opamp with high input impedance and negligible
output impedance, a 1k resistor from the + input to ground, and a 100k
resistor from the output to ground (in other words a 100k load).
Assume the feedback resistance network is high enough that it draws
negligible current.

1V into 1k, at the input, means 1mA, so the power is 1mW. 10V into
100k at the output means 0.1mA, so the power is 1mW again. There's 0dB
power gain, but there's voltage gain of 10x, which is coded +20dB in
the voltage-gain realm of decibel calculation. This deviates from the
"real" standard of what decibels are, by divorcing the voltage and
power gains, but using the same unit for them, dB. The usage, however,
is nearly universal, so speaking as a descriptivist who believes that
dictionaries should reflect how people actually use the language,
perhaps the official definition needs revision to take into account
the dual usage of the term. I've suggested "dBG" as an indicator that
voltage gain is being discussed rather than power gain, and perhaps
that's the way to go.


My point, though, was that it's not a comparison of input to output
power, but of the difference in output power that holds between an
amplifier with 20 dB gain vs. one with 19 dB or 0 dB, all else being
equal. If it ever gets to, say, a loudspeaker, there will be that
difference. I incline to descriptivism too, but it just seems to me
that this actual usage of "dB" does refer to power, in the sense I
meant.
--
John
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On Mon, 22 Nov 2010 06:53:48 GMT, (Eric
Jacobsen) wrote:

On Sun, 21 Nov 2010 15:39:11 -0600, John O'Flaherty
wrote:


{trimmed}

Get a group of comm engineers in a room and see if anybody agrees on
the definition of SNR. Hint: don't get people started. There is no
single definition. deciBels are a similar animal. e.g., what is
power? What kind of power? RMS? Peak? Which is appropriate for
dB?


Power is rate of transfer of energy, and its time distribution, its
form, and its location of measurement require further specification,
but I don't see why dB shouldn't be applicable to all cases.


It is applicable, but it's not as clearly defined as some think or are
at least expressing here. Power measurement, as you just said,
requires integration over time. How much time? It is often
(usually) not specified, so there's already ambiguity in the
"definition" or "standard". "Instantaneous power" is a hand-wavy way
around that, but you can't measure that practically, so time
integration is required. How much is up to the implementer.


But the measurement procedure is a separate question from what power
is and what dB means.

{trimmed again}

Are there any actual examples of the use of dBFS that don't relate
to a full-scale voltage or current? Of course, the FS has to be
defined- voltage current, pressure. But I bet that anyone who was
using a full scale defined in terms of power would use a formula with
a factor of 10, not 20.


Actually, dBFS implies a digital number scale system, so the
traditional notions of voltage or current or power don't really even
apply any more. The analysis is performed on a numeric sequence,
which could represent anything. A single sample can be taken from
the numeric sequence, say X, and dBFS could be computed as either

ans = 10*log(X/FS) if one were interested in interpreting X as an
instantaneous power measurement (and ADCs often have internal
integration over some fraction of the sample period so that can be
argued). This follows the definition of RMS for a numeric sequence
when n = 1, as long as X is positive.

or

ans = 10*log(X/FS) if one were interested in interpreting X as a
voltage.


I'm assuming you meant 20 in that sentence. If so, I don't think we
disagree much. As you point out, when it's numeric, the use depends on
the interpretation. But if X _is_ being interpreted as a voltage, and
it is squared, then "ans" will be indistinguishable from a power
ratio.

I'd suggest, though, that one use whatever is consistent with the rest
of the analysis being performed.

There's nothing magical about the factor of 10 or 20. As always, one
just has to keep track of what one is doing and be consistent to get a
useful result.


Hard to argue with that!

{trimmed yet again}

Yes, dB per se is unitless but dBm and dBW aren't. +20 dB has no
units, but +20 dBm means 100 milliwatts. If you append RMS to dB,
that's a procedural specification, and you can have +10 dBVRMS, where
a unit is specified as well as the measurement procedure.
I agree that everything should be specified; nevertheless, if dB is
used for something that is not power, or not directly relatable to
power, I think it's being misused.


dBm and dBW are, actually, strictly speaking, still unitless or
dimensionless. The units cancel in the ratio of the reference and the
measurement, which HAVE to have the same units to get a meaningful
result. dBm and dBW (and others, but definitely not all) have the
odd property that they completely define a dimension, despite being
dimensionless. They still carry or reflect (or whatever) the
indicated dimensional unit with the quantity conveyed. Sort of.
IMHO, that's actually a hint that one has to pay attention to what one
is doing to get usable results.


Well, good point. You can't plug dBm into an ordinary equation and
have the units cancel correctly, so they just carry the information
about the unit.

--
John
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On 11/22/2010 11:36 AM, Adrian Tuddenham wrote:

They should first know that dB is a power measurement, this is the
fundamental fact on which the rest is based. When they have grasped the
basics, then they can be shown that the voltage is a handy way of
comparing two power levels in the right circumstances (and they can
remember some handy voltage ratios if appropriate).


But most people working in audio don't need to compare power
levels, they need to compare voltage levels. I'd teach them
the "20" formula first. But then you may be more into theory
and less into practice than I am.

--
"Today's production equipment is IT based and cannot be
operated without a passing knowledge of computing, although
it seems that it can be operated without a passing knowledge
of audio." - John Watkinson

http://mikeriversaudio.wordpress.com - useful and
interesting audio stuff
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Mike Rivers wrote:

On 11/22/2010 11:36 AM, Adrian Tuddenham wrote:

They should first know that dB is a power measurement, this is the
fundamental fact on which the rest is based. When they have grasped the
basics, then they can be shown that the voltage is a handy way of
comparing two power levels in the right circumstances (and they can
remember some handy voltage ratios if appropriate).


But most people working in audio don't need to compare power
levels, they need to compare voltage levels. I'd teach them
the "20" formula first. But then you may be more into theory
and less into practice than I am.


I remember the struggles I had in my earlier days, trying to get a grasp
on how to use the dB system. There was so much misinformation and so
many things that didn't add up when you came to match theory with
practice. Nobody seemed to be able to explain what was really going on.

Now most of my recording chain is of my own manufacture and I design
specialist equipment for others. I am in a good position to look back
at my earlier mistakes and misunderstandings and see where I went wrong
(and where I was sent up the wrong path by others). I am also in a good
position to spot when other people are in the same quandry for the same
reasons as I was.

That's why I tend to be a bit pedantic about the use of dB; I know from
experience that a little pedantry at the outset will pay-off later, when
the student moves on and becomes a true professional.


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Chris Bore wrote:

Having written at length, I reluctantly consulted Wikipedia and to my
amazement found it contained useful information.

http://en.wikipedia.org/wiki/DBFS

It does refer to the ambiguity, and in fact refers to an AES standard
(0 dBFS is rms of a full scale sine wave) and to a practial standard
(Euphonix sound level meters) that has 0 dBFS as the rms of a full
scale square wave (equivalent to the 'instantaneous' definition that I
suggested).


I find the former of these two definitions more widely used than the
latter. Any signal whose RMS value is the same as that of a full-scale
non-clipping sine wave is 0 dBFS. So a clipping square wave is
+3 dBFS.

Steve


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(Steve Pope) writes:

Chris Bore wrote:

Having written at length, I reluctantly consulted Wikipedia and to my
amazement found it contained useful information.

http://en.wikipedia.org/wiki/DBFS

It does refer to the ambiguity, and in fact refers to an AES standard
(0 dBFS is rms of a full scale sine wave) and to a practial standard
(Euphonix sound level meters) that has 0 dBFS as the rms of a full
scale square wave (equivalent to the 'instantaneous' definition that I
suggested).


I find the former of these two definitions more widely used than the
latter. Any signal whose RMS value is the same as that of a full-scale
non-clipping sine wave is 0 dBFS. So a clipping square wave is
+3 dBFS.


Yeah, that bothers me. I like the reference level at FS square, so that
no signal EVER goes about 0 dBFS. But "like" and "define" are two
totally different things...

PS: Chris, thanks for pitching in!
--
Randy Yates % "And all you had to say
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Steve Pope Steve Pope is offline
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Randy Yates wrote:

(Steve Pope) writes:


I find the former of these two definitions more widely used than the
latter. Any signal whose RMS value is the same as that of a full-scale
non-clipping sine wave is 0 dBFS. So a clipping square wave is
+3 dBFS.


Yeah, that bothers me. I like the reference level at FS square, so that
no signal EVER goes about 0 dBFS. But "like" and "define" are two
totally different things...


One problem with the square wave is the measurment ends up bandwidth-
dependent. So it wouldn't be exactly 3.01 dB higher than the full-scale
sine wave; it would be a little less than 3.01 dB, depending. This would
make it harder to calibrate a level meter with repeatable results.

Steve
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Arny Krueger Arny Krueger is offline
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"Mike Rivers" wrote in message

On 11/22/2010 11:36 AM, Adrian Tuddenham wrote:

They should first know that dB is a power measurement,
this is the fundamental fact on which the rest is based.
When they have grasped the basics, then they can be
shown that the voltage is a handy way of comparing two
power levels in the right circumstances (and they can
remember some handy voltage ratios if appropriate).


But most people working in audio don't need to compare
power levels, they need to compare voltage levels. I'd
teach them the "20" formula first. But then you may be
more into theory and less into practice than I am.


Agreed, since in my world, voltmeters vastly outnumber true watt meters.

QED.


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davew davew is offline
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On Nov 23, 2:00*pm, (Steve Pope) wrote:
Chris Bore wrote:

Having written at length, I reluctantly consulted Wikipedia and to my
amazement found it contained useful information.


http://en.wikipedia.org/wiki/DBFS


It does refer to the ambiguity, and in fact refers to an AES standard
(0 dBFS is rms of a full scale sine wave) and to a practial standard
(Euphonix sound level meters) that has 0 dBFS as the rms of a full
scale square wave (equivalent to the 'instantaneous' definition that I
suggested).


I find the former of these two definitions more widely used than the
latter. * Any signal whose RMS value is the same as that of a full-scale
non-clipping sine wave is 0 dBFS. *So a clipping square wave is
+3 dBFS.

Steve

And a digital implementation of a VU meter reads +4dBFS for a full
scale square wave or 0dBFS depending on whether the ratio of peak to
mean absolute value is taken into account. We add 4dB to make a sine
wave read the same value as the peak sample value.
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