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  #81   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote:

John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800,
(Dick Pierce)
wrote:

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


And, if you describe the ratio of the original to its distortion
products in dB rather than in percentage, what do you care?
A THD of 1% means the amplitudes differ by a factor of 100:1,
their powers differ by 10000:1, but it's 40 dB in both cases.
That's the entire point: A RATIO between two signals is EXACTLY
the same whether you describe the RATIO of the voltage, the RATIO
of the power or the LOGS of those ratios (assuming the impedance
is the same in all cases, which is a prefectly reasonable assumption).


---
Except that decibels describing the ratio of one power to another

P1
is dB = 10 log ---- , while for voltages or currents it's _20_ times
P2
the log of the ratio.


But WHY? It's simply because power is proportional to the square of
the voltage, thus since:

dB = 10 log P1/Pref

and since

P prop V^2

and thus:

dB = 10 log V1^2/Vref^2

or

dB = 10 log (V1/Vref)^2

and

dB = 2 * 10 log (V1/Vref)

thus

dB = 20 log V1/Vref

QED.

So, it stands that 40 dB is 40 dB, whether we started with the
ratio of two voltages, or the ratio of the equivalent powers
of those two voltages.


---
Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB.


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

--
John Fields
  #86   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................


What I meant about the kilogram is that it feel somewhat akward that
there should be the prefix "kilo" included in the base unit. Ah,
well...


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..."
  #87   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................


What I meant about the kilogram is that it feel somewhat akward that
there should be the prefix "kilo" included in the base unit. Ah,
well...


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..."
  #88   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................


What I meant about the kilogram is that it feel somewhat akward that
there should be the prefix "kilo" included in the base unit. Ah,
well...


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..."
  #89   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

(Stewart Pinkerton) wrote in message ...
On 17 Jan 2004 02:06:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


Because science and engineering are made much simpler when there is an
international agreement on units of mass, time and length. We did
indeed once use grams in the cgs system, when oi were nobut a lad, but
the ISO standard is based on the mks system, so that everyone (except
some of our more red-necked colonial cousins! :-)) is singing from the
same hymn sheet. Note that recent embarrassments in celestial
navigation could have been avoided if the US hadn't been so far behind
the rest of the world in this respect.......................


What I meant about the kilogram is that it feel somewhat akward that
there should be the prefix "kilo" included in the base unit. Ah,
well...


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..."
  #90   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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
(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*.



  #91   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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
(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*.

  #92   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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
(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*.

  #93   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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
(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*.

  #94   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

Svante wrote:

chung wrote in message ervers.com...
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.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage.


That's exactly why you want to express in voltage ratios, since the load
impedance would need to be constant if you want to use power ratios.

"A dB is a dB". Or is
it? Hmm, se below!


A dB is a dB. That was what we drilled into the heads of new graduates
at HP, way back when.


2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


No. A dB is not fundamentally intended to measure power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


You can have a choice of either using dBV or dBm as units for the
y-axis. In the latter case, a standard impedance, such as resistive 50
ohms at all frequencies, is assumed.

You can also have dB as a unit, in which case it is simply a ratio of
voltages or powers assuming constant impedance.

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.

It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.


In case of doubt, a dB is a dB!

  #95   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

Svante wrote:

chung wrote in message ervers.com...
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.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage.


That's exactly why you want to express in voltage ratios, since the load
impedance would need to be constant if you want to use power ratios.

"A dB is a dB". Or is
it? Hmm, se below!


A dB is a dB. That was what we drilled into the heads of new graduates
at HP, way back when.


2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


No. A dB is not fundamentally intended to measure power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


You can have a choice of either using dBV or dBm as units for the
y-axis. In the latter case, a standard impedance, such as resistive 50
ohms at all frequencies, is assumed.

You can also have dB as a unit, in which case it is simply a ratio of
voltages or powers assuming constant impedance.

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.

It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.


In case of doubt, a dB is a dB!



  #96   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

Svante wrote:

chung wrote in message ervers.com...
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.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage.


That's exactly why you want to express in voltage ratios, since the load
impedance would need to be constant if you want to use power ratios.

"A dB is a dB". Or is
it? Hmm, se below!


A dB is a dB. That was what we drilled into the heads of new graduates
at HP, way back when.


2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


No. A dB is not fundamentally intended to measure power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


You can have a choice of either using dBV or dBm as units for the
y-axis. In the latter case, a standard impedance, such as resistive 50
ohms at all frequencies, is assumed.

You can also have dB as a unit, in which case it is simply a ratio of
voltages or powers assuming constant impedance.

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.

It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.


In case of doubt, a dB is a dB!

  #97   Report Post  
chung
 
Posts: n/a
Default Distorsion percentage, power or voltage?

Svante wrote:

chung wrote in message ervers.com...
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.


Several reasons:

1. 40 dB down is 40 dB down, whether you're talking about voltage or
power, assuming constant load impedance. If the 2nd harmonic is 40 dB
down, it means the voltage ratio is 1%, and the ratio of delivered power
is 0.01%. A dB in voltage is a dB in power!


Yes, I am aware of this, and perhaps that is why I like expressing
distorsion in dB rather than as a percentage.


That's exactly why you want to express in voltage ratios, since the load
impedance would need to be constant if you want to use power ratios.

"A dB is a dB". Or is
it? Hmm, se below!


A dB is a dB. That was what we drilled into the heads of new graduates
at HP, way back when.


2. Audio amplifiers are voltage devices. The actual power delivered to
the load depends on the load impedance. For example, let's say an
amplifer has 1% 2nd harmonic distortion in voltage. How much power is
delivered to the load at that 2nd harmonic frequency? The answer depends
on the load impedance at that frequency. It is not unusual for a
speaker's impedance to change substantially over an octave. So in this
case, the power ratio may not be 0.01%.


That is the best explanation so far, and the only one that doesn't
rely on a convention, but rather a physical fact (a frequency
dependent load resistance). However, this would actually speak against
using dB as a measure of distorsion, since dB is fundamentally
intended to measure a POWER ratio.


No. A dB is not fundamentally intended to measure power ratio.


3. The measuring equipment measures ratios of voltages. It does not
measure power delivered to the load.


So... A spectral display based on voltage measurement should not
really be allowed to display "dB" on the y axis, unless we know that
we have a constant, resistive load?


You can have a choice of either using dBV or dBm as units for the
y-axis. In the latter case, a standard impedance, such as resistive 50
ohms at all frequencies, is assumed.

You can also have dB as a unit, in which case it is simply a ratio of
voltages or powers assuming constant impedance.

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.

It simply assumes that the load
resistenace is constant.

Hmmm... Interesting.


In case of doubt, a dB is a dB!

  #98   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote in message . ..
On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

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.


Also, note the new binary prefixes:

From
http://physics.nist.gov/cuu/Units/binary.html

Prefixes for binary multiples
--------------------------------------------------------------------------------

Factor Name Symbol Origin Derivation
2^10 kibi Ki kilobinary: (2^10)^1 kilo: (10^3)^1
2^20 mebi Mi megabinary: (2^10)^2 mega: (10^3)^2
2^30 gibi Gi gigabinary: (2^10)^3 giga: (10^3)^3
2^40 tebi Ti terabinary: (2^10)^4 tera: (10^3)^4
2^50 pebi Pi petabinary: (2^10)^5 peta: (10^3)^5
2^60 exbi Ei exabinary: (2^10)^6 exa: (10^3)^6


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

Examples and comparisons with SI prefixes
one kibibit 1 Kibit = 210 bit = 1024 bit
one kilobit 1 kbit = 103 bit = 1000 bit
one mebibyte 1 MiB = 220 B = 1 048 576 B
one megabyte 1 MB = 106 B = 1 000 000 B
one gibibyte 1 GiB = 230 B = 1 073 741 824 B
one gigabyte 1 GB = 109 B = 1 000 000 000 B

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

Now I think this thread is drifting off the original subject... :-)
  #99   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote in message . ..
On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

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.


Also, note the new binary prefixes:

From
http://physics.nist.gov/cuu/Units/binary.html

Prefixes for binary multiples
--------------------------------------------------------------------------------

Factor Name Symbol Origin Derivation
2^10 kibi Ki kilobinary: (2^10)^1 kilo: (10^3)^1
2^20 mebi Mi megabinary: (2^10)^2 mega: (10^3)^2
2^30 gibi Gi gigabinary: (2^10)^3 giga: (10^3)^3
2^40 tebi Ti terabinary: (2^10)^4 tera: (10^3)^4
2^50 pebi Pi petabinary: (2^10)^5 peta: (10^3)^5
2^60 exbi Ei exabinary: (2^10)^6 exa: (10^3)^6


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

Examples and comparisons with SI prefixes
one kibibit 1 Kibit = 210 bit = 1024 bit
one kilobit 1 kbit = 103 bit = 1000 bit
one mebibyte 1 MiB = 220 B = 1 048 576 B
one megabyte 1 MB = 106 B = 1 000 000 B
one gibibyte 1 GiB = 230 B = 1 073 741 824 B
one gigabyte 1 GB = 109 B = 1 000 000 000 B

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

Now I think this thread is drifting off the original subject... :-)
  #100   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote in message . ..
On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

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.


Also, note the new binary prefixes:

From
http://physics.nist.gov/cuu/Units/binary.html

Prefixes for binary multiples
--------------------------------------------------------------------------------

Factor Name Symbol Origin Derivation
2^10 kibi Ki kilobinary: (2^10)^1 kilo: (10^3)^1
2^20 mebi Mi megabinary: (2^10)^2 mega: (10^3)^2
2^30 gibi Gi gigabinary: (2^10)^3 giga: (10^3)^3
2^40 tebi Ti terabinary: (2^10)^4 tera: (10^3)^4
2^50 pebi Pi petabinary: (2^10)^5 peta: (10^3)^5
2^60 exbi Ei exabinary: (2^10)^6 exa: (10^3)^6


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

Examples and comparisons with SI prefixes
one kibibit 1 Kibit = 210 bit = 1024 bit
one kilobit 1 kbit = 103 bit = 1000 bit
one mebibyte 1 MiB = 220 B = 1 048 576 B
one megabyte 1 MB = 106 B = 1 000 000 B
one gibibyte 1 GiB = 230 B = 1 073 741 824 B
one gigabyte 1 GB = 109 B = 1 000 000 000 B

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

Now I think this thread is drifting off the original subject... :-)


  #101   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote in message . ..
On 17 Jan 2004 02:06:26 -0800, (Svante)
wrote:

(Stewart Pinkerton) wrote in message ...
On 16 Jan 2004 14:40:26 -0800,
(Svante)
wrote:

(Stewart Pinkerton) 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.

Think what you like, voltage is the standard.

Yes I will think what I like, and I know voltage is the standard, that
should be clear from my post. But I asked about the REASON for the
standard.

Why does it matter? Why is the kilogram the standard for mass?


To annoy the americans? :-) You might not find it interesting, and
that is OK. I did find it interesting and posted the question to see
if anyone else had thought about it too. I bet you could find people
interested in why a kilogram is a kilogram too. Why isn't it a gram?
Hmm...


---
To obtain a standard of length a quadrant of the earth (one-fourth of a
circumference) was surveyed from Dunkirk to Barcelona along the meridian
that passes through Paris. The distance from the pole to the equator was
divided into ten million parts to constitute the meter (spelled metre in
some countries).

Further subdivisions in the length of the meter, by orders of magnitude,
into more convenient-to-use units for some applications led to the
naming of the decimeter (one-tenth of a meter), the centimeter (one
one-hundredth of a meter), the millimeter (one one-thousandth of a
meter) and so on. Note that by defining the unit of length the
definition of the unit of volume followed automatically.

Note also the curious coincidence(?) of units in the metric system being
divisible everywhere by ten and the fact that we have ten digits on our
hands.

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.


Also, note the new binary prefixes:

From
http://physics.nist.gov/cuu/Units/binary.html

Prefixes for binary multiples
--------------------------------------------------------------------------------

Factor Name Symbol Origin Derivation
2^10 kibi Ki kilobinary: (2^10)^1 kilo: (10^3)^1
2^20 mebi Mi megabinary: (2^10)^2 mega: (10^3)^2
2^30 gibi Gi gigabinary: (2^10)^3 giga: (10^3)^3
2^40 tebi Ti terabinary: (2^10)^4 tera: (10^3)^4
2^50 pebi Pi petabinary: (2^10)^5 peta: (10^3)^5
2^60 exbi Ei exabinary: (2^10)^6 exa: (10^3)^6


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

Examples and comparisons with SI prefixes
one kibibit 1 Kibit = 210 bit = 1024 bit
one kilobit 1 kbit = 103 bit = 1000 bit
one mebibyte 1 MiB = 220 B = 1 048 576 B
one megabyte 1 MB = 106 B = 1 000 000 B
one gibibyte 1 GiB = 230 B = 1 073 741 824 B
one gigabyte 1 GB = 109 B = 1 000 000 000 B

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

Now I think this thread is drifting off the original subject... :-)
  #102   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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. ;-)

--
John Fields
  #103   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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. ;-)

--
John Fields
  #104   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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. ;-)

--
John Fields
  #105   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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. ;-)

--
John Fields


  #106   Report Post  
Don Pearce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

d

_____________________________

http://www.pearce.uk.com
  #107   Report Post  
Don Pearce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

d

_____________________________

http://www.pearce.uk.com
  #108   Report Post  
Don Pearce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

d

_____________________________

http://www.pearce.uk.com
  #109   Report Post  
Don Pearce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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

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

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

d

_____________________________

http://www.pearce.uk.com
  #114   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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:

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

--
John Fields
  #115   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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:

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

--
John Fields


  #116   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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:

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

--
John Fields
  #117   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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:

On Sat, 17 Jan 2004 11:17:03 -0600, John Fields
wrote:

Yes. Perhaps yet another way to skin the same poor cat...

Since

dB(V) = 20 log (V/Vref),

No it doesn't. dB(V) = 20 log(V). The equation you give above is
simply for dB. dB(V) is dB relative to 1V, not some arbitrary other
voltage.


---
Hmmm...

Perhaps the confusion arises from the notation.

I use dBV to refer to 1 volt as the reference, and dB(V) to indicate
that the units being compared are arbitrary volts.
---


Then, since 6dB represents about a 2:1 change in voltage and

P = EČ/R,

a 6dB change in voltage will double the E term, quadrupling the P term
for the same R.

Now, since

dB(W) = 10 log (P1/Pref)

Again here, this is an equation giving dB, NOT dB(W)


---
As previous, the notation I use in order to try to alleviate confusion
is dBW when the reference power is one watt and dB(W) when the units
being compared are arbitrary watts.
---

and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB.

So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage
effects a 6dB change in power, doubling the signal voltage into an
amplifier will quadruple the output power into the same load.

dB(V) does not equal dB(W) except when the load resistor is 1 ohm. We
can safely assume that this is rarely the case.


---
dBV = dBW when the load resistance is one ohm, but dB(V) = dB(W) for
any arbitrary choice of load.
---

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.

--
John Fields
  #118   Report Post  
Bazza
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote ---
---
When measuring _total_ harmonic distortion, the contribution of each of
the individual partials is immaterial in that what's being determined is
the contribution to distortion that _all_ of the harmonics due to the
fundamental's presence contribute.


Furthermore, even if the contributions of the individual partials were
to be measured, their voltages would each be measured using a tuned
voltmeter and then the process of determining their contribution
determined mathematically. As a matter of fact, in order to measure the
power directly, the normal load would have to be disconnected and a
bolometer with precisely the same impedance as the load substituted for
the load. Expensive and more than just a _bit_ awkward.


You raise a good point John.
I was going to post something to the same effect
which goes to the heart of the original question by Svante

In the 'measurement' of distortion, putting aside the quantitative accuracy
problem for the moment, it was customary for measurements to be done in two
ways.

- The nulling of the fundamental and then measuring the rest of the garbage.
This also included a noise component which might (not) be significant.
It is, (here I don a flame suit at this moment), the more common way.
Quite often a HP machine (334) is a typical device for this work

- The wave analyser approach whereby distortion products are discretely and
separately measured. This I remember as being the older method. A person then
calculated distortion by a formula which (correct me if wrong) was along the
lines of ...

1 / square root of the sum of the individual voltages (squared).
There was a variant of this formula but others can quibble if need be.

Of course. If you _missed_ some by-products as the result of 'mixing' i.e. by
assuming only harmonic products, then accuracy suffered, but hey, "the world
ain't poifect anyway". You _could_ , with equal validity, have measured the
products as currents (but voltmeters were more common).

This "voltage" approach is probably (well, I would contend that it is) the
single most important influence on the definition and approach to
measuring (and defining) distortion. All of this is predicated on the purity
of the original signal source (sine) which, in itself, is a separate subject
-)

.... Stands back .... prepares grapeshot to repel the oncoming attack ...
  #119   Report Post  
Bazza
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote ---
---
When measuring _total_ harmonic distortion, the contribution of each of
the individual partials is immaterial in that what's being determined is
the contribution to distortion that _all_ of the harmonics due to the
fundamental's presence contribute.


Furthermore, even if the contributions of the individual partials were
to be measured, their voltages would each be measured using a tuned
voltmeter and then the process of determining their contribution
determined mathematically. As a matter of fact, in order to measure the
power directly, the normal load would have to be disconnected and a
bolometer with precisely the same impedance as the load substituted for
the load. Expensive and more than just a _bit_ awkward.


You raise a good point John.
I was going to post something to the same effect
which goes to the heart of the original question by Svante

In the 'measurement' of distortion, putting aside the quantitative accuracy
problem for the moment, it was customary for measurements to be done in two
ways.

- The nulling of the fundamental and then measuring the rest of the garbage.
This also included a noise component which might (not) be significant.
It is, (here I don a flame suit at this moment), the more common way.
Quite often a HP machine (334) is a typical device for this work

- The wave analyser approach whereby distortion products are discretely and
separately measured. This I remember as being the older method. A person then
calculated distortion by a formula which (correct me if wrong) was along the
lines of ...

1 / square root of the sum of the individual voltages (squared).
There was a variant of this formula but others can quibble if need be.

Of course. If you _missed_ some by-products as the result of 'mixing' i.e. by
assuming only harmonic products, then accuracy suffered, but hey, "the world
ain't poifect anyway". You _could_ , with equal validity, have measured the
products as currents (but voltmeters were more common).

This "voltage" approach is probably (well, I would contend that it is) the
single most important influence on the definition and approach to
measuring (and defining) distortion. All of this is predicated on the purity
of the original signal source (sine) which, in itself, is a separate subject
-)

.... Stands back .... prepares grapeshot to repel the oncoming attack ...
  #120   Report Post  
Bazza
 
Posts: n/a
Default Distorsion percentage, power or voltage?

John Fields wrote ---
---
When measuring _total_ harmonic distortion, the contribution of each of
the individual partials is immaterial in that what's being determined is
the contribution to distortion that _all_ of the harmonics due to the
fundamental's presence contribute.


Furthermore, even if the contributions of the individual partials were
to be measured, their voltages would each be measured using a tuned
voltmeter and then the process of determining their contribution
determined mathematically. As a matter of fact, in order to measure the
power directly, the normal load would have to be disconnected and a
bolometer with precisely the same impedance as the load substituted for
the load. Expensive and more than just a _bit_ awkward.


You raise a good point John.
I was going to post something to the same effect
which goes to the heart of the original question by Svante

In the 'measurement' of distortion, putting aside the quantitative accuracy
problem for the moment, it was customary for measurements to be done in two
ways.

- The nulling of the fundamental and then measuring the rest of the garbage.
This also included a noise component which might (not) be significant.
It is, (here I don a flame suit at this moment), the more common way.
Quite often a HP machine (334) is a typical device for this work

- The wave analyser approach whereby distortion products are discretely and
separately measured. This I remember as being the older method. A person then
calculated distortion by a formula which (correct me if wrong) was along the
lines of ...

1 / square root of the sum of the individual voltages (squared).
There was a variant of this formula but others can quibble if need be.

Of course. If you _missed_ some by-products as the result of 'mixing' i.e. by
assuming only harmonic products, then accuracy suffered, but hey, "the world
ain't poifect anyway". You _could_ , with equal validity, have measured the
products as currents (but voltmeters were more common).

This "voltage" approach is probably (well, I would contend that it is) the
single most important influence on the definition and approach to
measuring (and defining) distortion. All of this is predicated on the purity
of the original signal source (sine) which, in itself, is a separate subject
-)

.... Stands back .... prepares grapeshot to repel the oncoming attack ...
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