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#1
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Distorsion percentage, power or voltage?
Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the VOLTAGES that are compared (in the electrical case) not the POWERS. So if we have a second harmonic 40 dB down, the second harmonic distorsion is 1 %, not 0.01 %. (In this case the voltage of the harmonic is 1% of the fundamental, and its power is 0.01% of the fundamental) What is the reason for this convention? I'd think that power would be more logical. |
#2
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Distorsion percentage, power or voltage?
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#4
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Distorsion percentage, power or voltage?
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? -- Stewart Pinkerton | Music is Art - Audio is Engineering |
#5
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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... |
#6
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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 |
#7
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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 |
#8
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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 |
#9
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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 |
#10
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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 |
#11
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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 |
#12
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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 |
#13
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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 |
#14
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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... |
#15
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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... |
#16
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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... |
#17
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Distorsion percentage, power or voltage?
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? -- Stewart Pinkerton | Music is Art - Audio is Engineering |
#18
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Distorsion percentage, power or voltage?
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? -- Stewart Pinkerton | Music is Art - Audio is Engineering |
#19
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Distorsion percentage, power or voltage?
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? -- Stewart Pinkerton | Music is Art - Audio is Engineering |
#20
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Distorsion percentage, power or voltage?
(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. |
#21
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Distorsion percentage, power or voltage?
(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. |
#22
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Distorsion percentage, power or voltage?
(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. |
#23
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Distorsion percentage, power or voltage?
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#24
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Distorsion percentage, power or voltage?
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#26
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Distorsion percentage, power or voltage?
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#27
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Distorsion percentage, power or voltage?
John Fields wrote in message . ..
On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: Harmonic distorsion is expressed as the ratio between the distorsion components and the fundamental. What surprises me is that it is the VOLTAGES that are compared (in the electrical case) not the POWERS. So if we have a second harmonic 40 dB down, the second harmonic distorsion is 1 %, not 0.01 %. (In this case the voltage of the harmonic is 1% of the fundamental, and its power is 0.01% of the fundamental) What is the reason for this convention? I'd think that power would be more logical. --- It is _much_ easier to notch out the fundamental(s) and measure the voltage of the remaining component(s) than it is to measure their power. Then, knowing the voltage of the offending component(s) appearing across the load and the impedance of the load at the various frequencies at which distortion rears its ugly head, their contribution to the power being dissipated in/by the load can be easily calculated. Precisely. ALL of the THD meters in existance are essentially voltmeters with filters. Going from the ancient and venerable General Radio 1936 or Hewlett-Packard 330, through the HP 334, through the Sound Technology 1700, Amber, Marconi and so on, they are ALL voltmeters that have a narrow-band notch filter in them. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). There is no more or less logic to doing it one way or the other, they are exactly equivalent. |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: Harmonic distorsion is expressed as the ratio between the distorsion components and the fundamental. What surprises me is that it is the VOLTAGES that are compared (in the electrical case) not the POWERS. So if we have a second harmonic 40 dB down, the second harmonic distorsion is 1 %, not 0.01 %. (In this case the voltage of the harmonic is 1% of the fundamental, and its power is 0.01% of the fundamental) What is the reason for this convention? I'd think that power would be more logical. --- It is _much_ easier to notch out the fundamental(s) and measure the voltage of the remaining component(s) than it is to measure their power. Then, knowing the voltage of the offending component(s) appearing across the load and the impedance of the load at the various frequencies at which distortion rears its ugly head, their contribution to the power being dissipated in/by the load can be easily calculated. Precisely. ALL of the THD meters in existance are essentially voltmeters with filters. Going from the ancient and venerable General Radio 1936 or Hewlett-Packard 330, through the HP 334, through the Sound Technology 1700, Amber, Marconi and so on, they are ALL voltmeters that have a narrow-band notch filter in them. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. -- John Fields |
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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. |
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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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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?
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. |
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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. |
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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. |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: Harmonic distorsion is expressed as the ratio between the distorsion components and the fundamental. What surprises me is that it is the VOLTAGES that are compared (in the electrical case) not the POWERS. So if we have a second harmonic 40 dB down, the second harmonic distorsion is 1 %, not 0.01 %. (In this case the voltage of the harmonic is 1% of the fundamental, and its power is 0.01% of the fundamental) What is the reason for this convention? I'd think that power would be more logical. --- It is _much_ easier to notch out the fundamental(s) and measure the voltage of the remaining component(s) than it is to measure their power. Then, knowing the voltage of the offending component(s) appearing across the load and the impedance of the load at the various frequencies at which distortion rears its ugly head, their contribution to the power being dissipated in/by the load can be easily calculated. Precisely. ALL of the THD meters in existance are essentially voltmeters with filters. Going from the ancient and venerable General Radio 1936 or Hewlett-Packard 330, through the HP 334, through the Sound Technology 1700, Amber, Marconi and so on, they are ALL voltmeters that have a narrow-band notch filter in them. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. -- John Fields |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: Harmonic distorsion is expressed as the ratio between the distorsion components and the fundamental. What surprises me is that it is the VOLTAGES that are compared (in the electrical case) not the POWERS. So if we have a second harmonic 40 dB down, the second harmonic distorsion is 1 %, not 0.01 %. (In this case the voltage of the harmonic is 1% of the fundamental, and its power is 0.01% of the fundamental) What is the reason for this convention? I'd think that power would be more logical. --- It is _much_ easier to notch out the fundamental(s) and measure the voltage of the remaining component(s) than it is to measure their power. Then, knowing the voltage of the offending component(s) appearing across the load and the impedance of the load at the various frequencies at which distortion rears its ugly head, their contribution to the power being dissipated in/by the load can be easily calculated. Precisely. ALL of the THD meters in existance are essentially voltmeters with filters. Going from the ancient and venerable General Radio 1936 or Hewlett-Packard 330, through the HP 334, through the Sound Technology 1700, Amber, Marconi and so on, they are ALL voltmeters that have a narrow-band notch filter in them. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. -- John Fields |
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Distorsion percentage, power or voltage?
On 16 Jan 2004 14:32:02 -0800, (Dick Pierce)
wrote: John Fields wrote in message . .. On 16 Jan 2004 06:40:38 -0800, (Svante) wrote: Harmonic distorsion is expressed as the ratio between the distorsion components and the fundamental. What surprises me is that it is the VOLTAGES that are compared (in the electrical case) not the POWERS. So if we have a second harmonic 40 dB down, the second harmonic distorsion is 1 %, not 0.01 %. (In this case the voltage of the harmonic is 1% of the fundamental, and its power is 0.01% of the fundamental) What is the reason for this convention? I'd think that power would be more logical. --- It is _much_ easier to notch out the fundamental(s) and measure the voltage of the remaining component(s) than it is to measure their power. Then, knowing the voltage of the offending component(s) appearing across the load and the impedance of the load at the various frequencies at which distortion rears its ugly head, their contribution to the power being dissipated in/by the load can be easily calculated. Precisely. ALL of the THD meters in existance are essentially voltmeters with filters. Going from the ancient and venerable General Radio 1936 or Hewlett-Packard 330, through the HP 334, through the Sound Technology 1700, Amber, Marconi and so on, they are ALL voltmeters that have a narrow-band notch filter in them. And, if you describe the ratio of the original to its distortion products in dB rather than in percentage, what do you care? A THD of 1% means the amplitudes differ by a factor of 100:1, their powers differ by 10000:1, but it's 40 dB in both cases. That's the entire point: A RATIO between two signals is EXACTLY the same whether you describe the RATIO of the voltage, the RATIO of the power or the LOGS of those ratios (assuming the impedance is the same in all cases, which is a prefectly reasonable assumption). --- Except that decibels describing the ratio of one power to another P1 is dB = 10 log ---- , while for voltages or currents it's _20_ times P2 the log of the ratio. -- John Fields |
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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. |
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