What's the output of a record cartridge playing an out-of-speed record
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I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
I'm not considering the filters that the cartridge might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal stylus and cartridge.
brushless-dc-motor
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add a comment |
$begingroup$
I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
I'm not considering the filters that the cartridge might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal stylus and cartridge.
brushless-dc-motor
$endgroup$
add a comment |
$begingroup$
I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
I'm not considering the filters that the cartridge might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal stylus and cartridge.
brushless-dc-motor
$endgroup$
I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
I'm not considering the filters that the cartridge might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal stylus and cartridge.
brushless-dc-motor
brushless-dc-motor
edited 1 hour ago
user207421
7961617
7961617
asked 19 hours ago
Gabriel SantosGabriel Santos
313
313
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add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.
The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.
$endgroup$
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.
In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.
$endgroup$
1
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."
$endgroup$
– uhoh
13 hours ago
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Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
add a comment |
$begingroup$
Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.
A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.
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Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
3
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It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
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Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
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@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
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@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
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– Henning Makholm
3 hours ago
|
show 2 more comments
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To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any companding, but it answers your question).
Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.
If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.
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2
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I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
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– Toor
19 hours ago
2
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@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
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– Graham
10 hours ago
add a comment |
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I have practical experience with this--record players with variable speed drives used to exist. These were specialty systems intended for blind people--they allowed the listener to speed up the records. They were made variable because not everyone wanted the same speed.
Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players.
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Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
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– DJohnM
11 hours ago
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@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
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– Graham
10 hours ago
add a comment |
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The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.
Often 5 millivolts output at 5 centimeters per second needle movement. This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil.
The high resistance causes a high random thermal electron noise floor.
The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms.
But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. The output voltage is often 0.2 millivolts or even less,
at that stated needle velocity, at 1KHz.
[edit] Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.
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add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
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active
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votes
$begingroup$
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.
The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.
$endgroup$
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.
The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.
$endgroup$
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.
The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.
$endgroup$
Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?
The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.
The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.
edited 18 hours ago
K H
2,360215
2,360215
answered 19 hours ago
TransistorTransistor
88.3k785189
88.3k785189
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
What happens to the pitch is obvious, but it is not at all clear how the phase and amplitude responds. The wiggles of the groove feed into a mechanical system with both elastic and inertial parts, and somewhere in that system either the motion or the position is transduced to (I suppose, usually) voltage. Immediately I suppose the output signal ends up being some combination of the first and second derivatives of the groove position, but the coefficients will determine how the phase/amplitude changes differ at different frequencies.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.
In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.
$endgroup$
1
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."
$endgroup$
– uhoh
13 hours ago
$begingroup$
Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
add a comment |
$begingroup$
Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.
In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.
$endgroup$
1
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."
$endgroup$
– uhoh
13 hours ago
$begingroup$
Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
add a comment |
$begingroup$
Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.
In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.
$endgroup$
Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.
In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.
answered 17 hours ago
user217611
1
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."
$endgroup$
– uhoh
13 hours ago
$begingroup$
Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
add a comment |
1
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."
$endgroup$
– uhoh
13 hours ago
$begingroup$
Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
1
1
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."$endgroup$
– uhoh
13 hours ago
$begingroup$
+1
Bingo! This is the best answer for someone "...very interested in vinyl records and analog music."$endgroup$
– uhoh
13 hours ago
$begingroup$
Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
$begingroup$
Far and away the best answer. (As @uhoh says!)
$endgroup$
– Fattie
2 hours ago
add a comment |
$begingroup$
Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.
A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.
$endgroup$
$begingroup$
Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
3
$begingroup$
It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
$begingroup$
Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
$begingroup$
@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
$endgroup$
– Henning Makholm
3 hours ago
|
show 2 more comments
$begingroup$
Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.
A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.
$endgroup$
$begingroup$
Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
3
$begingroup$
It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
$begingroup$
Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
$begingroup$
@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
$endgroup$
– Henning Makholm
3 hours ago
|
show 2 more comments
$begingroup$
Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.
A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.
$endgroup$
Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.
A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.
answered 19 hours ago
Dave Tweed♦Dave Tweed
123k9152266
123k9152266
$begingroup$
Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
3
$begingroup$
It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
$begingroup$
Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
$begingroup$
@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
$endgroup$
– Henning Makholm
3 hours ago
|
show 2 more comments
$begingroup$
Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
3
$begingroup$
It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
$begingroup$
Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
$begingroup$
@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
$begingroup$
Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
$endgroup$
– Gabriel Santos
19 hours ago
3
3
$begingroup$
It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
$begingroup$
It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
$endgroup$
– Dave Tweed♦
19 hours ago
$begingroup$
Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
Is it clear that the amplitude will remain unchanged? Depending on technology the pickup might measure either the position of the needle (relative to the pickup arm) or its lateral speed or perhaps even its acceleration, or some combination of these. The amplitude of those signals will react differently to a slowdown.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
$begingroup$
@HenningMakholm: If it does any of those things without proper compensation, then that pickup will have serious frequency response issues, even when played at the correct speed. So yes, we can assume that changing the groove speed will not affect that.
$endgroup$
– Dave Tweed♦
3 hours ago
$begingroup$
@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
@DaveTweed: In order to produce a signal at all it will have to measure at least one of position, speed and acceleration. Your response leaves me none the wiser about which of those the output is supposed to correspond to. (I can figure out for myself that if it does something different from the standard, compensation will be needed. This insight does not tell me what the standard is).
$endgroup$
– Henning Makholm
3 hours ago
|
show 2 more comments
$begingroup$
To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any companding, but it answers your question).
Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.
If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.
$endgroup$
2
$begingroup$
I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
$endgroup$
– Toor
19 hours ago
2
$begingroup$
@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any companding, but it answers your question).
Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.
If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.
$endgroup$
2
$begingroup$
I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
$endgroup$
– Toor
19 hours ago
2
$begingroup$
@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any companding, but it answers your question).
Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.
If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.
$endgroup$
To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any companding, but it answers your question).
Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.
If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.
edited 17 hours ago
answered 19 hours ago
TimWescottTimWescott
6,6441416
6,6441416
2
$begingroup$
I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
$endgroup$
– Toor
19 hours ago
2
$begingroup$
@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
$endgroup$
– Graham
10 hours ago
add a comment |
2
$begingroup$
I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
$endgroup$
– Toor
19 hours ago
2
$begingroup$
@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
$endgroup$
– Graham
10 hours ago
2
2
$begingroup$
I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
$endgroup$
– Toor
19 hours ago
$begingroup$
I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
$endgroup$
– Toor
19 hours ago
2
2
$begingroup$
@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
$endgroup$
– Graham
10 hours ago
$begingroup$
@Toor And where it doesn't is exactly why the vinyl crowd are wrong on all levels. The reason older recordings are bass-light is to stop the player needle being kicked so hard it leaves the record. Because motors never spin round perfectly linearly and the record is never perfectly centred, you get the distinctive "wow" of record players. And even after that, the electronics can't give you the same signal-to-noise as CDs. None of this is opinions - it's measurable. Basically it was the best they could design at the time, and it was very clever, but its time is past.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
I have practical experience with this--record players with variable speed drives used to exist. These were specialty systems intended for blind people--they allowed the listener to speed up the records. They were made variable because not everyone wanted the same speed.
Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players.
$endgroup$
$begingroup$
Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
$endgroup$
– DJohnM
11 hours ago
$begingroup$
@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
I have practical experience with this--record players with variable speed drives used to exist. These were specialty systems intended for blind people--they allowed the listener to speed up the records. They were made variable because not everyone wanted the same speed.
Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players.
$endgroup$
$begingroup$
Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
$endgroup$
– DJohnM
11 hours ago
$begingroup$
@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
I have practical experience with this--record players with variable speed drives used to exist. These were specialty systems intended for blind people--they allowed the listener to speed up the records. They were made variable because not everyone wanted the same speed.
Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players.
$endgroup$
I have practical experience with this--record players with variable speed drives used to exist. These were specialty systems intended for blind people--they allowed the listener to speed up the records. They were made variable because not everyone wanted the same speed.
Obviously, for music this would be insane but these units were intended for playing voice--magazines read aloud onto special 8 1/3 rpm 9" flexible plastic records. They were not durable at all (but neither are magazines) but did their job at a much lower cost than other technologies of the day. Other than the variable speed drive, the low speed settings (their highest was 33 1/3), and the ability to survive being mailed as is they were ordinary players.
answered 13 hours ago
Loren PechtelLoren Pechtel
22328
22328
$begingroup$
Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
$endgroup$
– DJohnM
11 hours ago
$begingroup$
@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
$endgroup$
– DJohnM
11 hours ago
$begingroup$
@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
$endgroup$
– Graham
10 hours ago
$begingroup$
Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
$endgroup$
– DJohnM
11 hours ago
$begingroup$
Other use for variable speed drives: 1) to play along, tune the whole orchestra to match your personal instrument
$endgroup$
– DJohnM
11 hours ago
$begingroup$
@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
$endgroup$
– Graham
10 hours ago
$begingroup$
@DJohn Also popular for dance teachers. They could dial down the speed while the class got to grips with the moves.
$endgroup$
– Graham
10 hours ago
add a comment |
$begingroup$
The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.
Often 5 millivolts output at 5 centimeters per second needle movement. This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil.
The high resistance causes a high random thermal electron noise floor.
The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms.
But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. The output voltage is often 0.2 millivolts or even less,
at that stated needle velocity, at 1KHz.
[edit] Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.
$endgroup$
add a comment |
$begingroup$
The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.
Often 5 millivolts output at 5 centimeters per second needle movement. This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil.
The high resistance causes a high random thermal electron noise floor.
The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms.
But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. The output voltage is often 0.2 millivolts or even less,
at that stated needle velocity, at 1KHz.
[edit] Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.
$endgroup$
add a comment |
$begingroup$
The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.
Often 5 millivolts output at 5 centimeters per second needle movement. This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil.
The high resistance causes a high random thermal electron noise floor.
The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms.
But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. The output voltage is often 0.2 millivolts or even less,
at that stated needle velocity, at 1KHz.
[edit] Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.
$endgroup$
The output voltage of record cartridges is given at 1KHz at some standard velocity of the needle.
Often 5 millivolts output at 5 centimeters per second needle movement. This would be for MOVNG-MAGNET, with thousands of ohms because of the very tiny wires in the FIXED coil.
The high resistance causes a high random thermal electron noise floor.
The lowest noise cartridges are MOVING_COIL, often with resistance under 10 ohms.
But serious amplification, at low noise, and low VDD trash injection, and thorough shielding, is needed. The output voltage is often 0.2 millivolts or even less,
at that stated needle velocity, at 1KHz.
[edit] Some RIAA preamplifiers use common-source JFET amplifiers, and those type of gain stages have ZERO power supply rejection. Thus large RC filters are needed for VDD inputs, and regulators are SHUNT designs, with approximately 1nanoVolt/rootHertz noise density.
edited 12 hours ago
answered 13 hours ago
analogsystemsrfanalogsystemsrf
16k2822
16k2822
add a comment |
add a comment |
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