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#42
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Distorsion percentage, power or voltage?
John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. |
#43
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Distorsion percentage, power or voltage?
John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. |
#44
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Distorsion percentage, power or voltage?
John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. |
#45
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Distorsion percentage, power or voltage?
John Fields wrote in message . ..
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. |
#46
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Distorsion percentage, power or voltage?
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. "A dB is a dB". Or is it? Hmm, se below! 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. 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? I mean, the fundaments of dB assumes that we measure a power ratio. 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 simply assumes that the load resistenace is constant. Hmmm... Interesting. |
#47
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Distorsion percentage, power or voltage?
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. "A dB is a dB". Or is it? Hmm, se below! 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. 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? I mean, the fundaments of dB assumes that we measure a power ratio. 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 simply assumes that the load resistenace is constant. Hmmm... Interesting. |
#48
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Distorsion percentage, power or voltage?
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. "A dB is a dB". Or is it? Hmm, se below! 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. 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? I mean, the fundaments of dB assumes that we measure a power ratio. 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 simply assumes that the load resistenace is constant. Hmmm... Interesting. |
#49
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Distorsion percentage, power or voltage?
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. "A dB is a dB". Or is it? Hmm, se below! 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. 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? I mean, the fundaments of dB assumes that we measure a power ratio. 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 simply assumes that the load resistenace is constant. Hmmm... Interesting. |
#50
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Distorsion percentage, power or voltage?
(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... |
#51
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Distorsion percentage, power or voltage?
(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... |
#52
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Distorsion percentage, power or voltage?
(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... |
#53
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Distorsion percentage, power or voltage?
(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... |
#54
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Distorsion percentage, power or voltage?
(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. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? Exactly! That is (part of) why I like dBs! 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). Yes I am aware of this, but one cannot point it out too often. There is no more or less logic to doing it one way or the other, they are exactly equivalent. Yes, and the risk for misunderstanding speaks for measuring distortion in dBs rather than in %, IMO. Isn't it surprising that no marketers have found the opportunity to say that eg a loudspeaker has a distorsion of loudspeaker is 0.01% (power ratio) rather than 1% (voltage ratio). Both could in a sense be correct, and correspond to -40 dB. Oh, I better shut up. |
#55
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Distorsion percentage, power or voltage?
(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. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? Exactly! That is (part of) why I like dBs! 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). Yes I am aware of this, but one cannot point it out too often. There is no more or less logic to doing it one way or the other, they are exactly equivalent. Yes, and the risk for misunderstanding speaks for measuring distortion in dBs rather than in %, IMO. Isn't it surprising that no marketers have found the opportunity to say that eg a loudspeaker has a distorsion of loudspeaker is 0.01% (power ratio) rather than 1% (voltage ratio). Both could in a sense be correct, and correspond to -40 dB. Oh, I better shut up. |
#56
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Distorsion percentage, power or voltage?
(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. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? Exactly! That is (part of) why I like dBs! 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). Yes I am aware of this, but one cannot point it out too often. There is no more or less logic to doing it one way or the other, they are exactly equivalent. Yes, and the risk for misunderstanding speaks for measuring distortion in dBs rather than in %, IMO. Isn't it surprising that no marketers have found the opportunity to say that eg a loudspeaker has a distorsion of loudspeaker is 0.01% (power ratio) rather than 1% (voltage ratio). Both could in a sense be correct, and correspond to -40 dB. Oh, I better shut up. |
#57
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Distorsion percentage, power or voltage?
(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. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? Exactly! That is (part of) why I like dBs! 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). Yes I am aware of this, but one cannot point it out too often. There is no more or less logic to doing it one way or the other, they are exactly equivalent. Yes, and the risk for misunderstanding speaks for measuring distortion in dBs rather than in %, IMO. Isn't it surprising that no marketers have found the opportunity to say that eg a loudspeaker has a distorsion of loudspeaker is 0.01% (power ratio) rather than 1% (voltage ratio). Both could in a sense be correct, and correspond to -40 dB. Oh, I better shut up. |
#58
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Distorsion percentage, power or voltage?
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#59
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Distorsion percentage, power or voltage?
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#60
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Distorsion percentage, power or voltage?
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....................... But yes, I still *think* in feet, not metres................ -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Distorsion percentage, power or voltage?
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....................... But yes, I still *think* in feet, not metres................ -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Distorsion percentage, power or voltage?
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Distorsion percentage, power or voltage?
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Distorsion percentage, power or voltage?
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....................... But yes, I still *think* in feet, not metres................ -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Distorsion percentage, power or voltage?
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....................... But yes, I still *think* in feet, not metres................ -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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Distorsion percentage, power or voltage?
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. -- John Fields |
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Distorsion percentage, power or voltage?
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. -- John Fields |
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Distorsion percentage, power or voltage?
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. -- John Fields |
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Distorsion percentage, power or voltage?
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. -- John Fields |
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Distorsion percentage, power or voltage?
(Stewart Pinkerton) wrote in message ...
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. Nope. Think a bit. "deci" is a tenth. Why would mister Bell have defined a Bel as TWO times the logarithm of the ratio between two voltages/pressures/currents? The original definition of Bel is simply (ONE time) the tenth logarithm of a power ratio. Then, if we want to end up with the same number measuring voltages, we will have to take the logarithm of the SQUARE of the voltage ratio, which is the same as TWO times the logarithm of the voltage ratio. Or: Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref) or: deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref) ....given that pref and uref corresponds to the same power in the resistance in question. The "2 (* log...)" is a simple by-product of the square in P=U^2/R So I guess it is safe to assume that the fundament of the deciBel rests on a 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? Not at all, since a voltage ratio of 2:1 is approximately 6dB, regardless of current flow. This is why voltage, not power, is used as a standard measure of speaker sensitivity, since it is independent of the load impedance. Not if the decibel indicates the power ratio. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. Indeed, the Bel is a useful logarithm of a power ratio. 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 simply assumes that the load resistenace is constant. It assumes no such thing. If you get back to the fundaments, it does. But I know what you mean, I am just trying to point out that on the way to the voltage ratio, somewhere, the resistance has been ignored. |
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Distorsion percentage, power or voltage?
(Stewart Pinkerton) wrote in message ...
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. Nope. Think a bit. "deci" is a tenth. Why would mister Bell have defined a Bel as TWO times the logarithm of the ratio between two voltages/pressures/currents? The original definition of Bel is simply (ONE time) the tenth logarithm of a power ratio. Then, if we want to end up with the same number measuring voltages, we will have to take the logarithm of the SQUARE of the voltage ratio, which is the same as TWO times the logarithm of the voltage ratio. Or: Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref) or: deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref) ....given that pref and uref corresponds to the same power in the resistance in question. The "2 (* log...)" is a simple by-product of the square in P=U^2/R So I guess it is safe to assume that the fundament of the deciBel rests on a 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? Not at all, since a voltage ratio of 2:1 is approximately 6dB, regardless of current flow. This is why voltage, not power, is used as a standard measure of speaker sensitivity, since it is independent of the load impedance. Not if the decibel indicates the power ratio. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. Indeed, the Bel is a useful logarithm of a power ratio. 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 simply assumes that the load resistenace is constant. It assumes no such thing. If you get back to the fundaments, it does. But I know what you mean, I am just trying to point out that on the way to the voltage ratio, somewhere, the resistance has been ignored. |
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Distorsion percentage, power or voltage?
(Stewart Pinkerton) wrote in message ...
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. Nope. Think a bit. "deci" is a tenth. Why would mister Bell have defined a Bel as TWO times the logarithm of the ratio between two voltages/pressures/currents? The original definition of Bel is simply (ONE time) the tenth logarithm of a power ratio. Then, if we want to end up with the same number measuring voltages, we will have to take the logarithm of the SQUARE of the voltage ratio, which is the same as TWO times the logarithm of the voltage ratio. Or: Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref) or: deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref) ....given that pref and uref corresponds to the same power in the resistance in question. The "2 (* log...)" is a simple by-product of the square in P=U^2/R So I guess it is safe to assume that the fundament of the deciBel rests on a 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? Not at all, since a voltage ratio of 2:1 is approximately 6dB, regardless of current flow. This is why voltage, not power, is used as a standard measure of speaker sensitivity, since it is independent of the load impedance. Not if the decibel indicates the power ratio. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. Indeed, the Bel is a useful logarithm of a power ratio. 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 simply assumes that the load resistenace is constant. It assumes no such thing. If you get back to the fundaments, it does. But I know what you mean, I am just trying to point out that on the way to the voltage ratio, somewhere, the resistance has been ignored. |
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Distorsion percentage, power or voltage?
(Stewart Pinkerton) wrote in message ...
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. Nope. Think a bit. "deci" is a tenth. Why would mister Bell have defined a Bel as TWO times the logarithm of the ratio between two voltages/pressures/currents? The original definition of Bel is simply (ONE time) the tenth logarithm of a power ratio. Then, if we want to end up with the same number measuring voltages, we will have to take the logarithm of the SQUARE of the voltage ratio, which is the same as TWO times the logarithm of the voltage ratio. Or: Bel = log(p/pref) = log((u/uref)^2 = 2 * log(u/uref) or: deciBel= 10 * log(p/pref) = 10 * log((u/uref)^2 = 20 * log(u/uref) ....given that pref and uref corresponds to the same power in the resistance in question. The "2 (* log...)" is a simple by-product of the square in P=U^2/R So I guess it is safe to assume that the fundament of the deciBel rests on a 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? Not at all, since a voltage ratio of 2:1 is approximately 6dB, regardless of current flow. This is why voltage, not power, is used as a standard measure of speaker sensitivity, since it is independent of the load impedance. Not if the decibel indicates the power ratio. I mean, the fundaments of dB assumes that we measure a power ratio. No, it doesn't. It is simply a useful logarithmic ratio. Indeed, the Bel is a useful logarithm of a power ratio. 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 simply assumes that the load resistenace is constant. It assumes no such thing. If you get back to the fundaments, it does. But I know what you mean, I am just trying to point out that on the way to the voltage ratio, somewhere, the resistance has been ignored. |
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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....................... 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..." |
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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....................... 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..." |
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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....................... 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..." |
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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....................... 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..." |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. --- Yes. Perhaps yet another way to skin the same poor cat... Since dB(V) = 20 log (V/Vref), if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB. Then, since 6dB represents about a 2:1 change in voltage and P = EČ/R, a 6dB change in voltage will double the E term, quadrupling the P term for the same R. Now, since dB(W) = 10 log (P1/Pref) and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB. So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage effects a 6dB change in power, doubling the signal voltage into an amplifier will quadruple the output power into the same load. -- John Fields |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. --- Yes. Perhaps yet another way to skin the same poor cat... Since dB(V) = 20 log (V/Vref), if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB. Then, since 6dB represents about a 2:1 change in voltage and P = EČ/R, a 6dB change in voltage will double the E term, quadrupling the P term for the same R. Now, since dB(W) = 10 log (P1/Pref) and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB. So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage effects a 6dB change in power, doubling the signal voltage into an amplifier will quadruple the output power into the same load. -- John Fields |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 19:04:44 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 14:32:02 -0800, (Dick Pierce) wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. But WHY? It's simply because power is proportional to the square of the voltage, thus since: dB = 10 log P1/Pref and since P prop V^2 and thus: dB = 10 log V1^2/Vref^2 or dB = 10 log (V1/Vref)^2 and dB = 2 * 10 log (V1/Vref) thus dB = 20 log V1/Vref QED. So, it stands that 40 dB is 40 dB, whether we started with the ratio of two voltages, or the ratio of the equivalent powers of those two voltages. --- Yes. Perhaps yet another way to skin the same poor cat... Since dB(V) = 20 log (V/Vref), if V = 2Vref, then dB(V) = 20 log (2/1) ~ 6dB. Then, since 6dB represents about a 2:1 change in voltage and P = EČ/R, a 6dB change in voltage will double the E term, quadrupling the P term for the same R. Now, since dB(W) = 10 log (P1/Pref) and P1 = 4 Pref, we can say dB(W) = 10 log (4/1) ~ 6dB. So, even though (or because) dB(V) = dB(W), and a 6dB change in voltage effects a 6dB change in power, doubling the signal voltage into an amplifier will quadruple the output power into the same load. -- John Fields |
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