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

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.
  #10   Report Post  
Dick Pierce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.

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).

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.


  #11   Report Post  
Dick Pierce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.

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).

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.
  #12   Report Post  
Dick Pierce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.

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).

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.
  #13   Report Post  
Dick Pierce
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.

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).

There is no more or less logic to doing it one way or the other,
they are exactly equivalent.
  #14   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.
  #15   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


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

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.
  #17   Report Post  
Svante
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.
  #22   Report Post  
John Fields
 
Posts: n/a
Default Distorsion percentage, power or voltage?

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:

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.

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.

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

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:

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.

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.

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

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:

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.

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.

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

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:

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.

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.

--
John Fields


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

On 16 Jan 2004 14:33:31 -0800, (Svante)
wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


---
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.

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

On 16 Jan 2004 14:33:31 -0800, (Svante)
wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


---
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.

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

On 16 Jan 2004 14:33:31 -0800, (Svante)
wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


---
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.

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

On 16 Jan 2004 14:33:31 -0800, (Svante)
wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


---
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.

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

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!

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%.

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



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

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!

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%.

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

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

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!

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%.

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

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

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!

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%.

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

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

Svante wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.


Only if the load impedance is constant over that frequency range.

Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


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

Svante wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.


Only if the load impedance is constant over that frequency range.

Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.




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

Svante wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.


Only if the load impedance is constant over that frequency range.

Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


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

Svante wrote:

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.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.


Only if the load impedance is constant over that frequency range.

Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


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