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#82
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Distorsion percentage, power or voltage?
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#83
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Distorsion percentage, power or voltage?
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#84
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Distorsion percentage, power or voltage?
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#86
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 |
#110
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Distorsion percentage, power or voltage?
Hi,
In message , Stewart Pinkerton writes On 17 Jan 2004 02:02:57 -0800, (Svante) wrote: However, this would actually speak against using dB as a measure of distorsion, since dB is fundamentally intended to measure a POWER ratio. The dB was originally a measure of sound pressure level, and the logarithmic scale is used simply becuause our ears respond to sound in a logarithmic fashion. I won't attempt to refute this statement, but I was under the impression that the decibel (or rather, the Bel) was originally created in order to simplify the calculation of relative electrical power levels in telecommunication circuits. I think I read such in the Yamaha Sound Reinforcement Handbook. Since I have no idea how accurate that source is, I would be interested in finding out the real history. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. I can't bring to mind any meaningful use of the decibel that is not derived from power measurements, including dBSPL. -- Regards, Glenn Booth |
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Distorsion percentage, power or voltage?
Hi,
In message , Stewart Pinkerton writes On 17 Jan 2004 02:02:57 -0800, (Svante) wrote: However, this would actually speak against using dB as a measure of distorsion, since dB is fundamentally intended to measure a POWER ratio. The dB was originally a measure of sound pressure level, and the logarithmic scale is used simply becuause our ears respond to sound in a logarithmic fashion. I won't attempt to refute this statement, but I was under the impression that the decibel (or rather, the Bel) was originally created in order to simplify the calculation of relative electrical power levels in telecommunication circuits. I think I read such in the Yamaha Sound Reinforcement Handbook. Since I have no idea how accurate that source is, I would be interested in finding out the real history. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. I can't bring to mind any meaningful use of the decibel that is not derived from power measurements, including dBSPL. -- Regards, Glenn Booth |
#112
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Distorsion percentage, power or voltage?
Hi,
In message , Stewart Pinkerton writes On 17 Jan 2004 02:02:57 -0800, (Svante) wrote: However, this would actually speak against using dB as a measure of distorsion, since dB is fundamentally intended to measure a POWER ratio. The dB was originally a measure of sound pressure level, and the logarithmic scale is used simply becuause our ears respond to sound in a logarithmic fashion. I won't attempt to refute this statement, but I was under the impression that the decibel (or rather, the Bel) was originally created in order to simplify the calculation of relative electrical power levels in telecommunication circuits. I think I read such in the Yamaha Sound Reinforcement Handbook. Since I have no idea how accurate that source is, I would be interested in finding out the real history. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. I can't bring to mind any meaningful use of the decibel that is not derived from power measurements, including dBSPL. -- Regards, Glenn Booth |
#113
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Distorsion percentage, power or voltage?
Hi,
In message , Stewart Pinkerton writes On 17 Jan 2004 02:02:57 -0800, (Svante) wrote: However, this would actually speak against using dB as a measure of distorsion, since dB is fundamentally intended to measure a POWER ratio. The dB was originally a measure of sound pressure level, and the logarithmic scale is used simply becuause our ears respond to sound in a logarithmic fashion. I won't attempt to refute this statement, but I was under the impression that the decibel (or rather, the Bel) was originally created in order to simplify the calculation of relative electrical power levels in telecommunication circuits. I think I read such in the Yamaha Sound Reinforcement Handbook. Since I have no idea how accurate that source is, I would be interested in finding out the real history. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. I can't bring to mind any meaningful use of the decibel that is not derived from power measurements, including dBSPL. -- Regards, Glenn Booth |
#114
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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
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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
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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
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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
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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
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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
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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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