On Sat, 17 Jan 2004 17:27:19 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 22:33:13 +0000, Don Pearce
wrote:

On Sat, 17 Jan 2004 14:02:26 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 17:59:46 +0000, Don Pearce
wrote:

John, your total confusion over dB, dB(V) and dB(W) here surprises me
more than somewhat.

---
Hopefully my explanation of the notation I use clears it up for you. ;-)

I think the moral here is to stick to the notation that everybody else
understands.

---
I agree, and I'm sure you'll find that the accepted notation when
referring to an accepted reference level is to include the notation
referring to that level, without parentheses, as part of the argument.

For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I'm not confused, and I doubt anyone else is. I do *not* read dBV as
being the same as dB(V), or dBW as the same as dB(W), and I doubt that
any qualified engineer does. Of course, since there's absolutely no
need to use dB(whatever) at all, perhaps Svante should simply have
used dB in the first place!
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Sat, 17 Jan 2004 17:27:19 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 22:33:13 +0000, Don Pearce
wrote:

On Sat, 17 Jan 2004 14:02:26 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 17:59:46 +0000, Don Pearce
wrote:

John, your total confusion over dB, dB(V) and dB(W) here surprises me
more than somewhat.

---
Hopefully my explanation of the notation I use clears it up for you. ;-)

I think the moral here is to stick to the notation that everybody else
understands.

---
I agree, and I'm sure you'll find that the accepted notation when
referring to an accepted reference level is to include the notation
referring to that level, without parentheses, as part of the argument.

For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I'm not confused, and I doubt anyone else is. I do *not* read dBV as
being the same as dB(V), or dBW as the same as dB(W), and I doubt that
any qualified engineer does. Of course, since there's absolutely no
need to use dB(whatever) at all, perhaps Svante should simply have
used dB in the first place!
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Sat, 17 Jan 2004 17:27:19 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 22:33:13 +0000, Don Pearce
wrote:

On Sat, 17 Jan 2004 14:02:26 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 17:59:46 +0000, Don Pearce
wrote:

John, your total confusion over dB, dB(V) and dB(W) here surprises me
more than somewhat.

---
Hopefully my explanation of the notation I use clears it up for you. ;-)

I think the moral here is to stick to the notation that everybody else
understands.

---
I agree, and I'm sure you'll find that the accepted notation when
referring to an accepted reference level is to include the notation
referring to that level, without parentheses, as part of the argument.

For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I'm not confused, and I doubt anyone else is. I do *not* read dBV as
being the same as dB(V), or dBW as the same as dB(W), and I doubt that
any qualified engineer does. Of course, since there's absolutely no
need to use dB(whatever) at all, perhaps Svante should simply have
used dB in the first place!
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On Sat, 17 Jan 2004 17:27:19 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 22:33:13 +0000, Don Pearce
wrote:

On Sat, 17 Jan 2004 14:02:26 -0600, John Fields
wrote:

On Sat, 17 Jan 2004 17:59:46 +0000, Don Pearce
wrote:

John, your total confusion over dB, dB(V) and dB(W) here surprises me
more than somewhat.

---
Hopefully my explanation of the notation I use clears it up for you. ;-)

I think the moral here is to stick to the notation that everybody else
understands.

---
I agree, and I'm sure you'll find that the accepted notation when
referring to an accepted reference level is to include the notation
referring to that level, without parentheses, as part of the argument.

For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I'm not confused, and I doubt anyone else is. I do *not* read dBV as
being the same as dB(V), or dBW as the same as dB(W), and I doubt that
any qualified engineer does. Of course, since there's absolutely no
need to use dB(whatever) at all, perhaps Svante should simply have
used dB in the first place!
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.
--

Stewart Pinkerton | Music is Art - Audio is Engineering

On 18 Jan 2004 02:49:39 -0800, (Svante)
wrote:

chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

That is irrelevant, since a deciBel is simply one tenth of a Bel, the
original unit which was considered to be excessively large for
everyday use.

--

Stewart Pinkerton | Music is Art - Audio is Engineering

Svante wrote:
chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

It's defined in such a way so that voltage ratios in dB is consistent
with power ratios in dB.

Read any textbook. dB is always defined, not derived.

Svante wrote:
chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

It's defined in such a way so that voltage ratios in dB is consistent
with power ratios in dB.

Read any textbook. dB is always defined, not derived.

Svante wrote:
chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

It's defined in such a way so that voltage ratios in dB is consistent
with power ratios in dB.

Read any textbook. dB is always defined, not derived.

Svante wrote:
chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

It's defined in such a way so that voltage ratios in dB is consistent
with power ratios in dB.

Read any textbook. dB is always defined, not derived.

On Sun, 18 Jan 2004 15:46:42 +0000 (UTC), (Stewart
Pinkerton) wrote:

On Sat, 17 Jan 2004 09:25:45 -0600, John Fields
wrote:

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

Since this appears to be descending into pedantry, I should note that
the gram is certainly *not* a unit of weight, but of mass.

---
Ah, yessss!

But, descending even further, notice that my description of the events
leading up to the definition of the gram as being equal to the weight of
1 cubic centimeter of water was written in the past tense, indicating
that that decision was made 'way back when'. Sometime between 1791 and
1799, I believe, by the French Academy of Sciences.

In fact, not until 1901 was the gram adopted as the unit of mass, but
since we're mostly living in the present I guess I should have continued
with my pedantry and indicated that the weight of a body is the pull or
force due to gravity acting on that body. In cgs units, therefore, the
weight of a 1g mass here on Earth would be 1g * ~ 980cm/s² ~ 980 dynes.

Also, I believe that in order to make life easier, the weight of an
object with a mass of one gram is sometimes referred to as one
gram-weight. :-)

--
John Fields

On Sun, 18 Jan 2004 15:46:42 +0000 (UTC), (Stewart
Pinkerton) wrote:

On Sat, 17 Jan 2004 09:25:45 -0600, John Fields
wrote:

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

Since this appears to be descending into pedantry, I should note that
the gram is certainly *not* a unit of weight, but of mass.

---
Ah, yessss!

But, descending even further, notice that my description of the events
leading up to the definition of the gram as being equal to the weight of
1 cubic centimeter of water was written in the past tense, indicating
that that decision was made 'way back when'. Sometime between 1791 and
1799, I believe, by the French Academy of Sciences.

In fact, not until 1901 was the gram adopted as the unit of mass, but
since we're mostly living in the present I guess I should have continued
with my pedantry and indicated that the weight of a body is the pull or
force due to gravity acting on that body. In cgs units, therefore, the
weight of a 1g mass here on Earth would be 1g * ~ 980cm/s² ~ 980 dynes.

Also, I believe that in order to make life easier, the weight of an
object with a mass of one gram is sometimes referred to as one
gram-weight. :-)

--
John Fields

On Sun, 18 Jan 2004 15:46:42 +0000 (UTC), (Stewart
Pinkerton) wrote:

On Sat, 17 Jan 2004 09:25:45 -0600, John Fields
wrote:

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

Since this appears to be descending into pedantry, I should note that
the gram is certainly *not* a unit of weight, but of mass.

---
Ah, yessss!

But, descending even further, notice that my description of the events
leading up to the definition of the gram as being equal to the weight of
1 cubic centimeter of water was written in the past tense, indicating
that that decision was made 'way back when'. Sometime between 1791 and
1799, I believe, by the French Academy of Sciences.

In fact, not until 1901 was the gram adopted as the unit of mass, but
since we're mostly living in the present I guess I should have continued
with my pedantry and indicated that the weight of a body is the pull or
force due to gravity acting on that body. In cgs units, therefore, the
weight of a 1g mass here on Earth would be 1g * ~ 980cm/s² ~ 980 dynes.

Also, I believe that in order to make life easier, the weight of an
object with a mass of one gram is sometimes referred to as one
gram-weight. :-)

--
John Fields

On Sun, 18 Jan 2004 15:46:42 +0000 (UTC), (Stewart
Pinkerton) wrote:

On Sat, 17 Jan 2004 09:25:45 -0600, John Fields
wrote:

Now, since water is/was ubiquitous on the surface of the earth and,
presumably, weighed the same everywhere, it was decided that a certain
volume of water (the 'cubic centimeter', a cube one centimeter on an
edge) would become the standard of weight and was called the 'gramme'.

The prefix 'kilo', indicating that a multiplication of the quantity
following it by 1000 is required, means "1000 grams" when appended with
'gram'. Hence, kilo+gram = kilogram = 1000 * 1 gram = 1000 grams.

Since this appears to be descending into pedantry, I should note that
the gram is certainly *not* a unit of weight, but of mass.

---
Ah, yessss!

But, descending even further, notice that my description of the events
leading up to the definition of the gram as being equal to the weight of
1 cubic centimeter of water was written in the past tense, indicating
that that decision was made 'way back when'. Sometime between 1791 and
1799, I believe, by the French Academy of Sciences.

In fact, not until 1901 was the gram adopted as the unit of mass, but
since we're mostly living in the present I guess I should have continued
with my pedantry and indicated that the weight of a body is the pull or
force due to gravity acting on that body. In cgs units, therefore, the
weight of a 1g mass here on Earth would be 1g * ~ 980cm/s² ~ 980 dynes.

Also, I believe that in order to make life easier, the weight of an
object with a mass of one gram is sometimes referred to as one
gram-weight. :-)

--
John Fields

Hi,

In message , Stewart Pinkerton
writes

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

I'm similar, but not a Scot :-)

We bought new weighing scales that only work in kilos, and I'm tempted
to throw them at the wall. I can only think in stones, despite knowing
the conversion factor very well. I think I'll be that way until I die.

It makes me laugh when the large print on milk bottles says "1.136
litres", with "2 pints" written underneath in a minuscule typeface. It's
so obviously designed for people who think in imperial measures, but the
law says it has to be sold in metric.

Slightly more on topic, I understand there is a move afoot to replace
the decibel with the 'neper' (an SI unit). I hope it doesn't catch on;
I'm only just beginning to get a proper understanding of the old system.

--
Regards,
Glenn Booth

Hi,

In message , Stewart Pinkerton
writes

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

I'm similar, but not a Scot :-)

We bought new weighing scales that only work in kilos, and I'm tempted
to throw them at the wall. I can only think in stones, despite knowing
the conversion factor very well. I think I'll be that way until I die.

It makes me laugh when the large print on milk bottles says "1.136
litres", with "2 pints" written underneath in a minuscule typeface. It's
so obviously designed for people who think in imperial measures, but the
law says it has to be sold in metric.

Slightly more on topic, I understand there is a move afoot to replace
the decibel with the 'neper' (an SI unit). I hope it doesn't catch on;
I'm only just beginning to get a proper understanding of the old system.

--
Regards,
Glenn Booth

Hi,

In message , Stewart Pinkerton
writes

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

I'm similar, but not a Scot :-)

We bought new weighing scales that only work in kilos, and I'm tempted
to throw them at the wall. I can only think in stones, despite knowing
the conversion factor very well. I think I'll be that way until I die.

It makes me laugh when the large print on milk bottles says "1.136
litres", with "2 pints" written underneath in a minuscule typeface. It's
so obviously designed for people who think in imperial measures, but the
law says it has to be sold in metric.

Slightly more on topic, I understand there is a move afoot to replace
the decibel with the 'neper' (an SI unit). I hope it doesn't catch on;
I'm only just beginning to get a proper understanding of the old system.

--
Regards,
Glenn Booth

Hi,

In message , Stewart Pinkerton
writes

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

I'm similar, but not a Scot :-)

We bought new weighing scales that only work in kilos, and I'm tempted
to throw them at the wall. I can only think in stones, despite knowing
the conversion factor very well. I think I'll be that way until I die.

It makes me laugh when the large print on milk bottles says "1.136
litres", with "2 pints" written underneath in a minuscule typeface. It's
so obviously designed for people who think in imperial measures, but the
law says it has to be sold in metric.

Slightly more on topic, I understand there is a move afoot to replace
the decibel with the 'neper' (an SI unit). I hope it doesn't catch on;
I'm only just beginning to get a proper understanding of the old system.

--
Regards,
Glenn Booth

John Fields wrote in message . ..
On 18 Jan 2004 02:49:39 -0800, (Svante)
wrote:

chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

---
The original unit was the BEL, but it proved to be too large for
convenient use so it was divided by 10 to yield the deci-BEL, now noted
as the 'dB', so what dB = 20 log (V/Vref) is saying is that 20 times
the log of the voltage (or current) ratio is equal to one deci-BEL.

You miss my point; I know what "deci" means; a tenth. Now if dB is
DEFINED as logarithmic voltage ratio, without any association to
power, WHY ON EARTH would it be defined with TWENTY before the
logarithm, why not TEN?

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

John Fields wrote in message . ..
On 18 Jan 2004 02:49:39 -0800, (Svante)
wrote:

chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

---
The original unit was the BEL, but it proved to be too large for
convenient use so it was divided by 10 to yield the deci-BEL, now noted
as the 'dB', so what dB = 20 log (V/Vref) is saying is that 20 times
the log of the voltage (or current) ratio is equal to one deci-BEL.

You miss my point; I know what "deci" means; a tenth. Now if dB is
DEFINED as logarithmic voltage ratio, without any association to
power, WHY ON EARTH would it be defined with TWENTY before the
logarithm, why not TEN?

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

John Fields wrote in message . ..
On 18 Jan 2004 02:49:39 -0800, (Svante)
wrote:

chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

---
The original unit was the BEL, but it proved to be too large for
convenient use so it was divided by 10 to yield the deci-BEL, now noted
as the 'dB', so what dB = 20 log (V/Vref) is saying is that 20 times
the log of the voltage (or current) ratio is equal to one deci-BEL.

You miss my point; I know what "deci" means; a tenth. Now if dB is
DEFINED as logarithmic voltage ratio, without any association to
power, WHY ON EARTH would it be defined with TWENTY before the
logarithm, why not TEN?

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

John Fields wrote in message . ..
On 18 Jan 2004 02:49:39 -0800, (Svante)
wrote:

chung wrote in message rvers.com...
Svante wrote:

chung wrote in message ervers.com...
Svante wrote:
I mean, the fundaments of dB
assumes that we measure a power ratio.

No such assumption.

The equation for "voltage dBs"
(20*log(u/uref)) is a derivation based on that p~u^2 neglecting the
effects of varying load resistance.

It is a definition, not a derivation.

So, if it is a definition, free from association with the power ratio,
why does it say "20" times the logarith of the ratio. deci would mean
ten (or a tenth).

---
The original unit was the BEL, but it proved to be too large for
convenient use so it was divided by 10 to yield the deci-BEL, now noted
as the 'dB', so what dB = 20 log (V/Vref) is saying is that 20 times
the log of the voltage (or current) ratio is equal to one deci-BEL.

You miss my point; I know what "deci" means; a tenth. Now if dB is
DEFINED as logarithmic voltage ratio, without any association to
power, WHY ON EARTH would it be defined with TWENTY before the
logarithm, why not TEN?

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 08:46:44 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...

But yes, I still *think* in feet, not metres................

Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

Three centimetres shorter than myself... :-) I have the exact same
feeling, except it is the other way around. There ARE things that I
feel more comfortable using non-SI units, like the diameter of a
loudspeaker (inches) or the speed of my car (km/h, not m/s) or the
power of my car engine (horsepowers, not kW).

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 08:46:44 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...

But yes, I still *think* in feet, not metres................

Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

Three centimetres shorter than myself... :-) I have the exact same
feeling, except it is the other way around. There ARE things that I
feel more comfortable using non-SI units, like the diameter of a
loudspeaker (inches) or the speed of my car (km/h, not m/s) or the
power of my car engine (horsepowers, not kW).

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 08:46:44 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...

But yes, I still *think* in feet, not metres................

Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

Three centimetres shorter than myself... :-) I have the exact same
feeling, except it is the other way around. There ARE things that I
feel more comfortable using non-SI units, like the diameter of a
loudspeaker (inches) or the speed of my car (km/h, not m/s) or the
power of my car engine (horsepowers, not kW).

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 08:46:44 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...

But yes, I still *think* in feet, not metres................

Should I say that I'm a Swede, if you wonder. I was raised with the
metric system, and appreciate it a lot and cannot understand how feet
could possibly be more intuitive than metres. Heck, even our mile is
metric (10 000 metres).
I can't help quoting my mechanics teacher who was a member of the
swedish standards committe: "England is going metric, inch by inch..."

I am a Scot, but I'm 56 years old, and I was brought up with feet and
inches, and with pounds, shillings and pence. Yes, the metric system
is simpler, but this doesn't affect how we *think*. I am 6 feet 3
inches tall, I know how tall is someone who is 6 feet tall, but I have
no idea how tall is someone who is 1.83 metres tall.................

Three centimetres shorter than myself... :-) I have the exact same
feeling, except it is the other way around. There ARE things that I
feel more comfortable using non-SI units, like the diameter of a
loudspeaker (inches) or the speed of my car (km/h, not m/s) or the
power of my car engine (horsepowers, not kW).

(Stewart Pinkerton) wrote in message ...
On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.

I was joking. But if decibels were to be applied to DISTANCES would the equation be:
10*log(x/xref)
or:
20*log(x/xref)

My bet would go for the 20, since distance is closely related to velocity.
In that case I have 86 dB(m) to work... :-)

(Stewart Pinkerton) wrote in message ...
On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.

I was joking. But if decibels were to be applied to DISTANCES would the equation be:
10*log(x/xref)
or:
20*log(x/xref)

My bet would go for the 20, since distance is closely related to velocity.
In that case I have 86 dB(m) to work... :-)

(Stewart Pinkerton) wrote in message ...
On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.

I was joking. But if decibels were to be applied to DISTANCES would the equation be:
10*log(x/xref)
or:
20*log(x/xref)

My bet would go for the 20, since distance is closely related to velocity.
In that case I have 86 dB(m) to work... :-)

(Stewart Pinkerton) wrote in message ...
On 18 Jan 2004 02:52:37 -0800, (Svante)
wrote:

John Fields wrote in message . ..
For example, 0dBm (not 0dB(m)) identifies the reference level as being
one milliwatt.

I guess that 0 dB(m) would be one meter then... ;-)

No, the correct term is dBm, not dB(m), which is meaningless.

I was joking. But if decibels were to be applied to DISTANCES would the equation be:
10*log(x/xref)
or:
20*log(x/xref)

My bet would go for the 20, since distance is closely related to velocity.
In that case I have 86 dB(m) to work... :-)

chung wrote in message rvers.com...
Svante wrote:

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800, (Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.

Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.

Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

In the standard distortion analyzers, like the HP 339, Soundtech 1700,
etc., they either use a rms circuit or an average detector to measure

^^^^^^^ Really???

(a) the level of the signal under test, and (b) the level of that signal
after the fundamental has been nulled out. No one uses a power sensor to
measure distortion or signal power.

You can also use a spectrum analyzer to measure distortion. In that
case, a narrow-band measurement is made at each harmonic frequency, but
the detector used is still either a rms detector or an average detector.

So, in all cases, you are still measuring voltages, not powers, for the
simple reason that you are supplying a signal in the form of *voltage*.

So let's say that you have only two distorsion products, 2% of second
harmonic and 1% of third harmonic. How much total harmonic distortion
do you have then?

My response would be sqrt(0.02^2+0.01^2)=0.022 or 2.2%. The equation
can be seen as "calulate the powers, add them up, and convert back to
voltage".
An RMS measurement of the residual voltage would yield this value, but
if an analyser would measure AVERAGE (absolute) voltage of the
residual, the result would come out wrong, right?

And, in some sense, by measuring RMS voltage the power is involved.
Not directly though, of course.

chung wrote in message rvers.com...
Svante wrote:

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800, (Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.

Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.

Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

In the standard distortion analyzers, like the HP 339, Soundtech 1700,
etc., they either use a rms circuit or an average detector to measure

^^^^^^^ Really???

(a) the level of the signal under test, and (b) the level of that signal
after the fundamental has been nulled out. No one uses a power sensor to
measure distortion or signal power.

You can also use a spectrum analyzer to measure distortion. In that
case, a narrow-band measurement is made at each harmonic frequency, but
the detector used is still either a rms detector or an average detector.

So, in all cases, you are still measuring voltages, not powers, for the
simple reason that you are supplying a signal in the form of *voltage*.

So let's say that you have only two distorsion products, 2% of second
harmonic and 1% of third harmonic. How much total harmonic distortion
do you have then?

My response would be sqrt(0.02^2+0.01^2)=0.022 or 2.2%. The equation
can be seen as "calulate the powers, add them up, and convert back to
voltage".
An RMS measurement of the residual voltage would yield this value, but
if an analyser would measure AVERAGE (absolute) voltage of the
residual, the result would come out wrong, right?

And, in some sense, by measuring RMS voltage the power is involved.
Not directly though, of course.

chung wrote in message rvers.com...
Svante wrote:

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800, (Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.

Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.

Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

In the standard distortion analyzers, like the HP 339, Soundtech 1700,
etc., they either use a rms circuit or an average detector to measure

^^^^^^^ Really???

(a) the level of the signal under test, and (b) the level of that signal
after the fundamental has been nulled out. No one uses a power sensor to
measure distortion or signal power.

You can also use a spectrum analyzer to measure distortion. In that
case, a narrow-band measurement is made at each harmonic frequency, but
the detector used is still either a rms detector or an average detector.

So, in all cases, you are still measuring voltages, not powers, for the
simple reason that you are supplying a signal in the form of *voltage*.

So let's say that you have only two distorsion products, 2% of second
harmonic and 1% of third harmonic. How much total harmonic distortion
do you have then?

My response would be sqrt(0.02^2+0.01^2)=0.022 or 2.2%. The equation
can be seen as "calulate the powers, add them up, and convert back to
voltage".
An RMS measurement of the residual voltage would yield this value, but
if an analyser would measure AVERAGE (absolute) voltage of the
residual, the result would come out wrong, right?

And, in some sense, by measuring RMS voltage the power is involved.
Not directly though, of course.

chung wrote in message rvers.com...
Svante wrote:

(Dick Pierce) wrote in message . com...
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800, (Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.

---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.

Precisely. ALL of the THD meters in existance are essentially
voltmeters with filters. Going from the ancient and venerable
General Radio 1936 or Hewlett-Packard 330, through the HP 334,
through the Sound Technology 1700, Amber, Marconi and so on,
they are ALL voltmeters that have a narrow-band notch filter
in them.

Hmm, just curious, did they have circuitry to measure RMS voltage,
rather than average voltage? I guess so, because that would be crucial
to end up with a correct result. The point here is that the only
reasonable way to add up harmonics is to add their powers. That is why
the *RMS* voltage of the (sum of all) harmonics would have to be
measured.

In the standard distortion analyzers, like the HP 339, Soundtech 1700,
etc., they either use a rms circuit or an average detector to measure

^^^^^^^ Really???

(a) the level of the signal under test, and (b) the level of that signal
after the fundamental has been nulled out. No one uses a power sensor to
measure distortion or signal power.

You can also use a spectrum analyzer to measure distortion. In that
case, a narrow-band measurement is made at each harmonic frequency, but
the detector used is still either a rms detector or an average detector.

So, in all cases, you are still measuring voltages, not powers, for the
simple reason that you are supplying a signal in the form of *voltage*.

So let's say that you have only two distorsion products, 2% of second
harmonic and 1% of third harmonic. How much total harmonic distortion
do you have then?

My response would be sqrt(0.02^2+0.01^2)=0.022 or 2.2%. The equation
can be seen as "calulate the powers, add them up, and convert back to
voltage".
An RMS measurement of the residual voltage would yield this value, but
if an analyser would measure AVERAGE (absolute) voltage of the
residual, the result would come out wrong, right?

And, in some sense, by measuring RMS voltage the power is involved.
Not directly though, of course.

Hi,

In message , Svante
writes

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

That was my reasoning also. The factor of 2 is only necessary to account
for the squared term in the relationship between power and voltage (or
their equivalents). However, having checked a few links from Google, it
seems far from clear - there are many conflicting opinions. For example:

http://www.madengineer.com/blunders/decibels.htm
Claims the decibel was originally defined to relate pressures.

http://www.sizes.com/units/decibel.htm
Claims that the decibel originated to relate powers.

Using dB for power relationships seems mathematically clear and
intuitive - the maths needs to be massaged in order to compare voltages,
for example. The same goes for sound power, and sound pressure (pressure
being the mechanical analog of voltage).


So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

It does seem logical; unfortunately, I can't find any definitive
reference.

--
Regards,
Glenn Booth

Hi,

In message , Svante
writes

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

That was my reasoning also. The factor of 2 is only necessary to account
for the squared term in the relationship between power and voltage (or
their equivalents). However, having checked a few links from Google, it
seems far from clear - there are many conflicting opinions. For example:

http://www.madengineer.com/blunders/decibels.htm
Claims the decibel was originally defined to relate pressures.

http://www.sizes.com/units/decibel.htm
Claims that the decibel originated to relate powers.

Using dB for power relationships seems mathematically clear and
intuitive - the maths needs to be massaged in order to compare voltages,
for example. The same goes for sound power, and sound pressure (pressure
being the mechanical analog of voltage).


So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

It does seem logical; unfortunately, I can't find any definitive
reference.

--
Regards,
Glenn Booth

Hi,

In message , Svante
writes

My answer to this is that the original definition is for the power
ratio, and the logarithm of that power ratio was taken as a BEL. The
deci was introduced, just as for the decimeter, and we ended up with a
TEN before the log. To measure a power level difference by means of
voltages, given constant load resistance, we would have to take the
log of the SQUARE of the voltage ratio, since power is proportional to
voltage squared. Simple math makes us then realise that we can skip
the square if we put TWENTY before the log instead.

That was my reasoning also. The factor of 2 is only necessary to account
for the squared term in the relationship between power and voltage (or
their equivalents). However, having checked a few links from Google, it
seems far from clear - there are many conflicting opinions. For example:

http://www.madengineer.com/blunders/decibels.htm
Claims the decibel was originally defined to relate pressures.

http://www.sizes.com/units/decibel.htm
Claims that the decibel originated to relate powers.

Using dB for power relationships seems mathematically clear and
intuitive - the maths needs to be massaged in order to compare voltages,
for example. The same goes for sound power, and sound pressure (pressure
being the mechanical analog of voltage).


So in my mind there is no doubt that the original (deci-)bel
definition is for a power ratio, and that the equation for a voltage
ratio is derived from that.

It does seem logical; unfortunately, I can't find any definitive
reference.

--
Regards,
Glenn Booth