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Post by antigua on Dec 11, 2016 19:10:28 GMT -5
www.lacemusic.com/Hot_Golds_6K.phpA few months ago I bought a used Lace Sensor Hot Gold just to rip open and look at, and I bought another three to put in a guitar. In this review, I looked at the three that went in the guitar, and then further below I've reposted the dissection for the sake of comprehensiveness. Technically speaking, passive guitar pickups don't compress. There is no electrical mechanism by which that can happen, but having said that, Lace Sensor Hot Golds sound like compressed single coils. Something seems to happen that sounds audibly similar to compression. Billy Corgan used Lace Sensors in the 90's, and it seems especially easy to get his long, singing Big Muff sustain with Lace Sensors. On the clean side, I miss the apparent "dynamic range" of true single coils. Picking and strumming with gusto doesn't seem to pay off with Lace Sensor Hot Golds they way it does with true single coils. One reason might be that Lace Sensors produce a small fraction of the flux density. I measured only 260G at the center of the pickup. AlNiCo 2 or 3 would tend to measure between 600G and 700G, while AlNiCo 5 is typically around 1050G. I'm not a believer in the idea that string pull is a big issue, but technically there is a lot less flux density at the strings. Even if that were to explain a longer sustain, it wouldn't really explain a muted pick attack. Hopefully as more becomes known about the relationship between flux density and tonal outcomes, these things won't be so vague, and we'll have better command over them, too. Lace Sensor Hot Gold 2015, #1 DC Resistance: 5.50K Inductance: 2.242H Calculated C: 212pF (222-10)
Resonant Peak: dV: 8.3dB f: 7.13kHz (black) Loaded (200k & 470pF): dV: 5.0dB f: 3.85kHz (red)
Lace Sensor Hot Gold 2015, #2 DC Resistance: 5.53K Inductance: 2.227H Calculated C: 194pF (204-10)
Resonant Peak: dV: 9.3dB f: 7.46kHz (green) Loaded (200k & 470pF): dV: 5.0dB f: 3.85kHz (gray)
Lace Sensor Hot Gold 2015, #3 DC Resistance: 5.50K Inductance: 2.275H Calculated C: 200pF (210-10)
Resonant Peak: dV: 9.6dB f: 7.29kHz (pink) Loaded (200k & 470pF): dV: 5.0dB f: 3.85kHz (black)
Flux density: 260G at center, (450G if the cover were removed)
Bode plot: Lace specs the DC resistance for this set at 6.0K and 2.2 henries. My inductance measurement agrees, but I read 500 ohms less resistance across the board. They put the peak resonance at 3,600Hz, I show 7k unloaded, and so I believe Lace is measuring the peak resonance with the pickups "loaded" in a Strat. My loaded frequency was 3.85k for all three pickups, which places them right where the name says they should be, between vintage and hotter Strat sets. The peak resonance is nearly identical to the Lollar Blackface set. It makes for an interesting comparison, since the Blackface set features strong AlNiCo 5 poles, compared to the Hot Gold's weak comb-shaped rail. The unloaded peak shows only +10dB at the resonance, compared to a typical Strat pickup's ~15dB at the resonance, due to mild eddy current losses. The loaded peak shows +5dB, which is typical of most Strat pickups, including the Lollar Blackfaces. The capacitance working out to about 200pF is a high for a single coil, either owing to the metal construction, the very fine coil wire, or both. The 5.5k ohms resistance tells us that there are not as many turns on the little coil, since this same resistance is typically seen with Strat pickups, and they have much larger coils. The same number of winds with a finer wire will produce a much higher DC resistance. The 2.2H inductance is also typical of "vintage" Stratocaster pickups, and so the higher inductance probably owes to the mostly solid core and permeable return paths (side walls), where as a typical Strat pickup gets achieves inductance almost mostly through the coil windings alone. Lace has always portrayed the Lace Sensors as being sort of a futuristic pickup, and in their marketing materials it appears that they looked for synonyms to describe known concepts of pickup design as though it were something novel, for example "radiant field barriers", usually simply referred to as "shielding". The Lace Sensor is a standard single coil pickup in all the important respects. What makes it unique are the use of rubberized magnet (like a refrigerator magnet) that fills voids that normal machined magnets won't, a very tiny coil that is set very close to the strings, and a walls on each side that form a magnetic return path in addition to acting as shielding. The Fender Jaguar pickups feature the same sort of permeable side walls. The comb shaped rail that serves as the core of the tiny coil is not significantly different from the type of rail seen in "Hot Rails", there just happens to be one of them, and it's smooshed flat at the top, so that the plastic cover will conceal it. Pics: The Dissected Lace SensorI wish I had it hooked up, then I could of said "the vivisected Lace Sensor". So they apparently place the pickup into the plastic housing, then drown it in epoxy. This is how it looks wit the cover pried off. If you look real close, there are six circles in the epoxy, that comes from six faint circles on the inside of the pickup cover, that correspond to the six holes on a standard Strat pickups cover. I sanded off the epoxy as much as possible. What you're seeing there is not entirely plastic or epoxy between the metal. The black substance along the outer edges, outside of the comb's teeth, is rubber magnet, but the inner portion, directly under and in between the comb teeth, is the top of the plastic bobbin that holds the coil. The magnetic polarity of metal comb is "north", while the polarity of the rubber magnet strips (as well as the metal casing) the on the edge is "south", which makes for a very short magnetic return path, since the flux hops directly from the "north" over to the "south". That's one of the major design points that makes a Lace Sensor unique. It would be like taking a normal Strat pickup and taping magnets to the side of the pickup, upside down relative to the pole pieces, so that the magnetic lines of flux are directed to come back, directly down towards the outer edge of the pickup, rather than shoot off into space before coming back to the pickup. The end result is that it produces a reasonably strong concentration of magnetic flux, despite the relative weakness of magnetized combs and the rubber magnet and magnetized side walls, compared to AlNiCo pole pieces. The two round holes correspond with where the old-style Lace Sensor's metal mounting bar attached to the chassis. The red and white wires connect directly to the coil, which takes up the top half of the pickup, and the green connects to the chassis. Everything below the coil, down to the metal frame is rubber magnet. The pickup still works despite the extreme rug burn. I pressed in on the rubber magnet underside with a little screw driver just to show how malleable the material is. The comb and the chassis are magnetic and electrically conductive, so they help direct the magnetic flux lines to be very close to the pickup, but also therefore contribute to a higher inductance and eddy current activity. The coil reads 6.0k ohms DC resistance, so it's surely wound with finer wire, such as 43 AWG, since 42AWG fills up an entire full sized Strat bobbin by itself. The coil must have far fewer turns than a typical Strat pickup in order to achieve only 6k ohms with a finer wire. Fewer turns would normally mean less inductance, and therefore be very bright sounding, but it's stands to reason that the increased magnetism brought about by all the ferrous metal surrounding the coil works to jack up the inductance and allow it to even out. The overall structure of the pickup is not far removed from a P-90, or a ceramic single coil, where it has a permanent magnet underneath, and ferrous metal in the middle of the coil that poke out of the top. The difference is that this bulks up on EMF shielding, while still providing a Strat-like electrical properties, despite having a lot less space to work with due to that shielding. Flux DensityA normal AlNiCo 5 Strat pickup reads over 1000 Gauss over the pole piece. See the comb measure 400-500 "north" And around 500 "south" at the edge, making for a tight magnetic return path. One of the selling points of the Lace Sensor is reduces string pull, but the sad truth is that cheap ceramic pickups you receive in import guitars have even less Gauss at the string. Stratitus is really only a problem that comes with classic AlNiCo pole piece pickups, while almost every other pickup design inflicts much less magnetic pull upon the strings. OTOH, Lace Sensor Golds sound a lot better than these things.
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Post by Deleted on Dec 12, 2016 0:26:01 GMT -5
One of the selling points of the Lace Sensor is reduces string pull, but the sad truth is that cheap ceramic pickups you receive in import guitars have even less Gauss at the string. Stratitus is really only a problem that comes with classic AlNiCo pole piece pickups, while almost every other pickup design inflicts much less magnetic pull upon the strings. OTOH, Lace Sensor Golds sound a lot better than these things. Somebody had to say it, It annoys me when techs/writers bring up stratitis as the reason for poor sustain in every single article they write. Nice review BTW, keep up the good work.
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Post by antigua on Dec 12, 2016 1:48:06 GMT -5
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Post by ms on Dec 12, 2016 7:19:22 GMT -5
"The magnetic polarity of metal comb is "north", while the polarity of the rubber magnet strips (as well as the metal casing) the on the edge is "south", which makes for a very short magnetic return path, since the flux hops directly from the "north" over to the "south"."
I think you cannot say very much about this without knowing the permeability of the rubber magnet material, at least in regards to the return path for the flux from the vibrating string.
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Post by unheardofinstruments on May 28, 2020 7:11:15 GMT -5
I love lace sensors but I wonder if the magnetic pull is that much lower given the sweet spot for tone is so much closer to the string, nevertheless I have respect for Don Lace and his innovations both for loving the tone and silence of the sensors way back when they came out and the clever use of the weaker south facing magnets to direct the flux, that and the great thread analysing the alumitones (they had me baffled when I got a set for a tapping stick) which made me realise how brilliant Don is now I see how the thinking behind those worked using a big solid single low impedance kind of half turn humbucking coil. Here's to Don!
Flux path is a very interesting variable. The Flynn flux switching discovery may have inspired him, who knows. Never a free lunch but always a new way to rearrange the box we have been thinking in.
Low impedence does create a kind of clarity like an allembic active bass pickup tried to do, I imagine the length of the flux paths being more equal due to the magnetic return path in the sensor and the voltage path being shorter from the coil being shorter and not going through so many different physically distinct paths in the coil space keeps everything in phase better. In effect the lace is amplified by having a sweet tone placed closer to the string so gets away with less magnetic pull and a different coil which both have advantages but would be unworkable in the standard format with alnico poles.
The alumitones transformer has two coils physically more proximal to the outer flux paths so I suspect it is wired to cancel as well as change the output voltage but in the patent drawings they are drawn the same size. Perhaps this helps them function more like a single coil shaped vertically stacked humbucker with one weaker/lower coil to help cancel hum but one to dominate and keep a single coil sound.
Aluminium seems to add a shrillness wherever you use it, especially in bridges/necks so perhaps the aluminium helped the muddiness of a weaker more distributed rubberised magnet field vs the more focused poles per string and this is how they found a way to make it sound good.
Liquid Pickups got me pretty inspired, ferrofluid in interchangeable chambers with different materials and therefore tones, aluminium once again being on the shrill side, if only he could get some capital to get it into mass production...
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Post by aquin43 on May 28, 2020 8:35:32 GMT -5
So, a narrow pickup with a pair of tight flux return paths that normally works close to the string. Maybe its response is much more non-linear than the standard strat pickup because of a more rapid variation of the string magnetisation with vertical displacement.
Arthur
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Post by ms on May 28, 2020 11:03:24 GMT -5
It appears that the magnetic geometry would produce a stronger than normal component of the field along the string, pointing N to S in both directions away from the center line of the pickup. This would be effective at exciting magnetization along the string oppositely directed on the two sides of the center line, and we now know that it is this magnetization that makes a pickup work. That is, the component of the permanent field parallel to the axis of the coil should not be needed. Yet this is the component that should generate stratitis. So could it be that it magnetizes the string strongly in the required way without creating much stratits?
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Post by antigua on May 29, 2020 0:15:34 GMT -5
The magnetic pull of the Lace Sensors is so weak that it won't cause stratitus even if it were nearly touching the strings. I think most of the design aspects of the Lace Sensor amount to trivialities and that it's best described as a very tiny pickup in a relatively large housing. It has something in common with a Bill Lawrence Micro-coil, in that it localizes a small coil very close to the strings, and that improves the coupling coefficient between the whole of the coil and the guitar string.
Based on aquin43's model, I'd assume the "window" is more narrow, since the return path appears to be tighter for two reasons, due the coil being narrower and the magnetic path being guided in a tighter loop. I'll test this out the next time I have the experimental rig set up. ms suggested a test where the magnet is held outside of an magnetized pickup, but unfortunately there is absolutely no way to demagnetize a Lace Sensor.
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Post by aquin43 on May 29, 2020 3:40:25 GMT -5
... Technically speaking, passive guitar pickups don't compress. There is no electrical mechanism by which that can happen, but having said that, Lace Sensor Hot Golds sound like compressed single coils. Something seems to happen that sounds audibly similar to compression. Billy Corgan used Lace Sensors in the 90's, and it seems especially easy to get his long, singing Big Muff sustain with Lace Sensors. On the clean side, I miss the apparent "dynamic range" of true single coils. Picking and strumming with gusto doesn't seem to pay off with Lace Sensor Hot Golds they way it does with true single coils There is a built in non linearity in virtually all guitar pickups in that the magnetising field diminishes with distance from the pickup, while the primary sensitivity to string movement is along precisely this axis. The rate of variation will be least with a rail type pickup with long magnets, most particularly the Charlie Christian with its very long magnets which will make the pole piece resemble an isolated pole with B ~ 1/R^2, reduced to B~1/R because of the rail format. I would guess that the Strat pickup comes next with its fairly long magnets and, even without a rail structure, considerable contributions at each pole from adjacent magnets.
The Lace is a rail, of course, but the closeness of the flux path is likely to make the external flux variation tend to the dipole B~1/R^3, tempered to B~1/R^2 by the rail structure. Any such effect would be exacerbated by working with the pickup close to the string.
This is speculation, of course, but it would be interesting to know if the sensitivity does, in fact, fall off more quickly with distance than with an ordinary Strat pickup,
Arthur
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Post by ms on May 29, 2020 6:57:22 GMT -5
... Technically speaking, passive guitar pickups don't compress. There is no electrical mechanism by which that can happen, but having said that, Lace Sensor Hot Golds sound like compressed single coils. Something seems to happen that sounds audibly similar to compression. Billy Corgan used Lace Sensors in the 90's, and it seems especially easy to get his long, singing Big Muff sustain with Lace Sensors. On the clean side, I miss the apparent "dynamic range" of true single coils. Picking and strumming with gusto doesn't seem to pay off with Lace Sensor Hot Golds they way it does with true single coils There is a built in non linearity in virtually all guitar pickups in that the magnetising field diminishes with distance from the pickup, while the primary sensitivity to string movement is along precisely this axis. The rate of variation will be least with a rail type pickup with long magnets, most particularly the Charlie Christian with its very long magnets which will make the pole piece resemble an isolated pole with B ~ 1/R^2, reduced to B~1/R because of the rail format. I would guess that the Strat pickup comes next with its fairly long magnets and, even without a rail structure, considerable contributions at each pole from adjacent magnets.
The Lace is a rail, of course, but the closeness of the flux path is likely to make the external flux variation tend to the dipole B~1/R^3, tempered to B~1/R^2 by the rail structure. Any such effect would be exacerbated by working with the pickup close to the string.
This is speculation, of course, but it would be interesting to know if the sensitivity does, in fact, fall off more quickly with distance than with an ordinary Strat pickup,
Arthur
Compression from electronics tends to limit peaks, although not necessarily symmetrically. But the pickup is sensitive to the velocity of the string. The sensitivity is increased when the string is nearer the pickup and decreases when it is further away, but in both extremes, the velocity is small, and so it is the region near the zero crossing of the waveform that is most altered by the change in sensitivity. That is, the non-linear effect is significant where the waveform is small, not large. Peaks are not affected, and so I think the effect is very different from limiting in electronics, and I suspect that the effect is small in any case. Also, there is another factor that tends to reduce the non-linearity, by reducing the change in the effective field at different heights from the pole. The component along the string is what drives the string polarization, and the relative importance of this component increases further from the pole since the angle of the field changes. I do not know how important this is, but I think it would not be so hard to plot the sine of the angle of the field from the pole axis with distance from the pole.
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Post by aquin43 on May 29, 2020 8:07:38 GMT -5
There is a built in non linearity in virtually all guitar pickups in that the magnetising field diminishes with distance from the pickup, while the primary sensitivity to string movement is along precisely this axis. The rate of variation will be least with a rail type pickup with long magnets, most particularly the Charlie Christian with its very long magnets which will make the pole piece resemble an isolated pole with B ~ 1/R^2, reduced to B~1/R because of the rail format. I would guess that the Strat pickup comes next with its fairly long magnets and, even without a rail structure, considerable contributions at each pole from adjacent magnets.
The Lace is a rail, of course, but the closeness of the flux path is likely to make the external flux variation tend to the dipole B~1/R^3, tempered to B~1/R^2 by the rail structure. Any such effect would be exacerbated by working with the pickup close to the string.
This is speculation, of course, but it would be interesting to know if the sensitivity does, in fact, fall off more quickly with distance than with an ordinary Strat pickup,
Arthur
Compression from electronics tends to limit peaks, although not necessarily symmetrically. But the pickup is sensitive to the velocity of the string. The sensitivity is increased when the string is nearer the pickup and decreases when it is further away, but in both extremes, the velocity is small, and so it is the region near the zero crossing of the waveform that is most altered by the change in sensitivity. That is, the non-linear effect is significant where the waveform is small, not large. Peaks are not affected, and so I think the effect is very different from limiting in electronics, and I suspect that the effect is small in any case. Also, there is another factor that tends to reduce the non-linearity, by reducing the change in the effective field at different heights from the pole. The component along the string is what drives the string polarization, and the relative importance of this component increases further from the pole since the angle of the field changes. I do not know how important this is, but I think it would not be so hard to plot the sine of the angle of the field from the pole axis with distance from the pole. That is a good point, that the non linearity is acting on the differential of the output signal. I wonder how that affects the intermodulation between the harmonics and the fundamental. One could imagine the harmonics being alternately compressed and expanded as the string moves through a fundamental cycle.
Regarding your second point; in the 2D case of the ribbon over a rail, virtually all of the flux entering the ribbon seems to end up directed along the ribbon, regardless of its original orientation.
Arthur
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Post by antigua on May 29, 2020 8:26:49 GMT -5
Another cause of pseudo-compression IMO is from pickups having a rolled off treble, because the transient pick attack is harmonically rich for the first cycle or two, and if the pickup has a substantial roll off, that transient sound blunted compared the the decay period of string movement, but the Lace Sensor appears to have a response curve very similar to an AlNiCo Strat pickup. I think the minimal amount of steel and the large amount of rubber ferrite keep the eddy currents low and the Q factor relatively high. I don't even know if there is really a difference in the dynamic behavior of Lace Sensors, it's only been observed by ear and not by any scientific instrument, although I'm not the first person to make that observation. If someone says a high inductance pickup sounds "compressed" I'm tempted to chalk it up to the suppression of transient harmonics, but the Lace Sensors earn a similar adjective even with a higher resonant peak and Q factor. When the string is first plucked, it's a maximum of displacement and it moves towards a minimum. What I think this might come down to is if there is a design aspect of a pickup that can cause a given maximum of displacement to produce less voltage, relative to a given minimum. I know of at least one way that is possible, maybe this is what aquin43 was referring to with the magnetic field equations, is that if the string movement is very wide, wider than the discrete pole pieces, there will more more 2nd order harmonic effect from the symmetrical string movement, and that's a non linearity in that the string's spacial movement starts out wide, then becomes small, all the while the size of the pole piece remains the same, and of course with a rail style pickup, that doesn't happen at all, and a Lace Sensor is a rail. Maybe another factor to consider is the tight magnetic return path of the Lace Sensor, the string might be more weakly magnetized at it's maximum. I don't personally believe the rail geometry is a strong contender for cause and effect because I have other rail pickups, and when split, such that they have a higher resonant peak, I don't personally perceive that "compressed" effect to be too profound in that case. Just a moment ago, I played by Strat with these Choppers in it guitarnuts2.proboards.com/thread/8502/dimarzio-chopper-analysis-review , and when in series, they sound compressed due to a loaded resonant peak in the 2kHz range blunting the transient harmonics, but split, with the resonant peak now only in the mid 3kHz range, it sounds plenty bright, and not especially compressed. Also I have Burns Tri-Sonics guitarnuts2.proboards.com/thread/8438/burns-sonic-strat-analysis-review, which are ceramic bars under the hood, and they sound a lot like the split Chopper, they have a high resonant peak and don't sound especially compressed, either. That's just subjective hearing though, very far from being a definitive statement of fact. I'm a little interested in trying out the Lace Sensor Golds, whatever their lowest inductance model is. Maybe that won't sound as "compressed". As far as practical testing goes, producing a consistent string pluck is a lot harder than producing a sustained wave, but I've had luck with a method that involves pulling a release pin, I'll try to put together a test soon. This would be a great phenomena to capture, if it really happens.
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Post by antigua on May 29, 2020 8:57:20 GMT -5
A more condensed way of putting the pseudo-compression issue, it's been said that one way of defining a pickup as a "variable reluctance sensor" en.wikipedia.org/wiki/Variable_reluctance_sensor which is astonishingly on the nose; the potential for a large transient amplitude would rely on having a magnetic circuit where the reluctance delta is very high, a big maximum and a small minimum. Maybe something about the Lace Sensor's reluctance path between pickup and string limits the minimum and maximum of reluctance.
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Post by aquin43 on May 29, 2020 10:35:25 GMT -5
I have been looking at the delay line Spice model of the string, where the pluck is a burst of acceleration and we integrate to get the velocity for the pickup output. Adding a further stage of integration gives the displacement. In the early part of the waveform, the acceleration consists of pairs of spikes which, on integration give an alternating series of wider single spikes. Integrating these gives a series of double ramps. The sloping portions of the these double ramps coincide with the velocity spikes. In other words, the string displacement takes place during the velocity spikes and any variation in the pickup gain due to the displacement will modulate the shape of the spikes. This means that the pickup non-linearity will be manifest during the early few tenths of a second of the note, i.e. during the period that most defines its character.
Arthur
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Post by ms on May 29, 2020 11:48:24 GMT -5
I have been looking at the delay line Spice model of the string, where the pluck is a burst of acceleration and we integrate to get the velocity for the pickup output. Adding a further stage of integration gives the displacement. In the early part of the waveform, the acceleration consists of pairs of spikes which, on integration give an alternating series of wider single spikes. Integrating these gives a series of double ramps. The sloping portions of the these double ramps coincide with the velocity spikes. In other words, the string displacement takes place during the velocity spikes and any variation in the pickup gain due to the displacement will modulate the shape of the spikes. This means that the pickup non-linearity will be manifest during the early few tenths of a second of the note, i.e. during the period that most defines its character.
Arthur My understanding is this: The pick draws the string away from equilibrium. Assume the pick is at the middle of the string. Just before the pick releases the string, the displacement is maximum and the velocity is zero. The pick releases the string and the restoring force increases the velocity from zero in the direction of equilibrium. As the displacement passes through zero, the restoring forces also passes through zero and increases in the opposite direction. The string decelerates and the velocity goes to zero when the displacement is maximum in the opposite direction from the start. This is more complicated if you pick toward one end, but I think maximum displacement with zero velocity is an inevitable boundary condition.
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Post by antigua on May 29, 2020 12:11:25 GMT -5
Someone on another forum opined that the maximum string displacement constituted a limit / compression, because at a certain point, strumming harder fails to result in more voltage output, but that is independent of the guitar pickup.
Messing around with a loose Lace Sensor and the WT10A, it appears the Lace Sensor's magnetic field has a more upwards orientation, where as a typical Strat pickup is more outward. Maybe an FEM model could help visualize it.
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Post by aquin43 on May 29, 2020 12:43:33 GMT -5
I have been looking at the delay line Spice model of the string, where the pluck is a burst of acceleration and we integrate to get the velocity for the pickup output. Adding a further stage of integration gives the displacement. In the early part of the waveform, the acceleration consists of pairs of spikes which, on integration give an alternating series of wider single spikes. Integrating these gives a series of double ramps. The sloping portions of the these double ramps coincide with the velocity spikes. In other words, the string displacement takes place during the velocity spikes and any variation in the pickup gain due to the displacement will modulate the shape of the spikes. This means that the pickup non-linearity will be manifest during the early few tenths of a second of the note, i.e. during the period that most defines its character.
Arthur My understanding is this: The pick draws the string away from equilibrium. Assume the pick is at the middle of the string. Just before the pick releases the string, the displacement is maximum and the velocity is zero. The pick releases the string and the restoring force increases the velocity from zero in the direction of equilibrium. As the displacement passes through zero, the restoring forces also passes through zero and increases in the opposite direction. The string decelerates and the velocity goes to zero when the displacement is maximum in the opposite direction from the start. This is more complicated if you pick toward one end, but I think maximum displacement with zero velocity is an inevitable boundary condition. This is all happening on a transmission line. When the pick releases the string the restoring force due to the tension in the displaced string is no longer balanced by the pick force so there is an impulse of force at the pick position. The string impedance is essentially resistive so there is an immediate pair of acceleration spikes that travel down the string in both directions. These spikes reflect inverted from the bridge and nut because the impedance at those boundaries is much higher than the string impedance. This pattern of impulses travels up and down the string, diminishing and rounding off slowly as time passes because of slight imperfections in the reflections and frictional losses with the air and inside the string. The wave velocity on the string is also slightly dispersive, so the impulses tend to spread with time.
There is no way that the string can immediately respond as a whole to the pluck. There is a wave velocity sqrt(T/m), where T is tension and m the mass per unit length. This is equal to 2 * L * f0 (L = string length, f0 = frequency). No disturbance can travel down the string faster than that.
The Spice model reproduces much of this tolerably well and adding the extra integration step reveals how the displacement over the pickup changes during the velocity spike.
Arthur
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Post by perfboardpatcher on May 29, 2020 14:07:53 GMT -5
I have been looking at the delay line Spice model of the string, where the pluck is a burst of acceleration There is no way that the string can immediately respond as a whole to the pluck. when I tremolo pick my low E string and I avoid the release phase, i.e. I "tremolo scrape", I would be surprised to get anywhere passed 10 Herz. I believe there is only 1 kink in the string after the pluck even at higher frequencies for the pluck but anyway with 10 Herz the extra 2 kinks won't even fit on the string.
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Post by ms on May 29, 2020 14:40:19 GMT -5
My understanding is this: The pick draws the string away from equilibrium. Assume the pick is at the middle of the string. Just before the pick releases the string, the displacement is maximum and the velocity is zero. The pick releases the string and the restoring force increases the velocity from zero in the direction of equilibrium. As the displacement passes through zero, the restoring forces also passes through zero and increases in the opposite direction. The string decelerates and the velocity goes to zero when the displacement is maximum in the opposite direction from the start. This is more complicated if you pick toward one end, but I think maximum displacement with zero velocity is an inevitable boundary condition. This is all happening on a transmission line. When the pick releases the string the restoring force due to the tension in the displaced string is no longer balanced by the pick force so there is an impulse of force at the pick position. The string impedance is essentially resistive so there is an immediate pair of acceleration spikes that travel down the string in both directions. These spikes reflect inverted from the bridge and nut because the impedance at those boundaries is much higher than the string impedance. This pattern of impulses travels up and down the string, diminishing and rounding off slowly as time passes because of slight imperfections in the reflections and frictional losses with the air and inside the string. The wave velocity on the string is also slightly dispersive, so the impulses tend to spread with time.
There is no way that the string can immediately respond as a whole to the pluck. There is a wave velocity sqrt(T/m), where T is tension and m the mass per unit length. This is equal to 2 * L * f0 (L = string length, f0 = frequency). No disturbance can travel down the string faster than that.
The Spice model reproduces much of this tolerably well and adding the extra integration step reveals how the displacement over the pickup changes during the velocity spike.
Arthur
Yes, I think looking at the problem as a transmission line is a great way to do it. But looking at the standing wave solutions should work as well. I considered only the fundamental in my previous post. Really you should represent the shape of the string stretched by the pick just before release as a Fourier series of fundamental and overtones and then run the solution forward from there. But either way, the relationship between an extreme of displacement and the velocity is the same: it is where the string stops and then starts moving in the other direction. And so this is where the pickup, sensitive to the velocity, has zero output.
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Post by aquin43 on May 29, 2020 17:49:48 GMT -5
This is all happening on a transmission line. When the pick releases the string the restoring force due to the tension in the displaced string is no longer balanced by the pick force so there is an impulse of force at the pick position. The string impedance is essentially resistive so there is an immediate pair of acceleration spikes that travel down the string in both directions. These spikes reflect inverted from the bridge and nut because the impedance at those boundaries is much higher than the string impedance. This pattern of impulses travels up and down the string, diminishing and rounding off slowly as time passes because of slight imperfections in the reflections and frictional losses with the air and inside the string. The wave velocity on the string is also slightly dispersive, so the impulses tend to spread with time.
There is no way that the string can immediately respond as a whole to the pluck. There is a wave velocity sqrt(T/m), where T is tension and m the mass per unit length. This is equal to 2 * L * f0 (L = string length, f0 = frequency). No disturbance can travel down the string faster than that.
The Spice model reproduces much of this tolerably well and adding the extra integration step reveals how the displacement over the pickup changes during the velocity spike.
Arthur
Yes, I think looking at the problem as a transmission line is a great way to do it. But looking at the standing wave solutions should work as well. I considered only the fundamental in my previous post. Really you should represent the shape of the string stretched by the pick just before release as a Fourier series of fundamental and overtones and then run the solution forward from there. But either way, the relationship between an extreme of displacement and the velocity is the same: it is where the string stops and then starts moving in the other direction. And so this is where the pickup, sensitive to the velocity, has zero output. I think that where we seem to disagree is in the "run the solution forward from there" part. That is what is commonly done, but it is a quasi equilibrium solution and does not address the transient behaviour of the string. It is my understanding that the transient at the start of a note is the part that crucially defines the tone. Many instruments have very similar steady state harmonic spectra but they are differentiated by their starting transients.
I accept that your description is correct in the steady state.
Arthur
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Post by aquin43 on May 30, 2020 5:23:05 GMT -5
Actually, thinking a bit deeper about the Spice model, it models the transient behaviour OK and reproduces the sound of the string, but it doesn't consider the displacement of the string before the pluck release. This would have to be modelled as a charge distribution on the delay lines but there is no mechanism within Spice for doing that with the lossless delay lines.
There will obviously be distortion produced by the pickup non-linearity. Even the simple fundamental will be multiplied by a non-linear version of itself phase shifted by 90 degrees but this seems only to alter the shape from about half way up the sloping edges without compressing or expanding the peak.
Arthur
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Post by ms on May 30, 2020 14:26:35 GMT -5
Actually, thinking a bit deeper about the Spice model, it models the transient behaviour OK and reproduces the sound of the string, but it doesn't consider the displacement of the string before the pluck release. This would have to be modelled as a charge distribution on the delay lines but there is no mechanism within Spice for doing that with the lossless delay lines.
There will obviously be distortion produced by the pickup non-linearity. Even the simple fundamental will be multiplied by a non-linear version of itself phase shifted by 90 degrees but this seems only to alter the shape from about half way up the sloping edges without compressing or expanding the peak.
Arthur
Just for fun, a measurement. The setup uses a short cable to the scope; the neck pickup of the guitar is heavily loaded to give response nearly flat passed 10 KHz. The E6 string is picked at the fifth fret, making the initial displacement over the pickup small. The blue line shows pickup output (velocity). The other is its "integral", the displacement, assuming zero initial displacement. Horizontal scale is time in seconds; vertical scale is arbitrary units, with data scaled so that both lines fit on the same arbitrary scale.
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Post by antigua on May 30, 2020 22:21:55 GMT -5
The integral is a result of a calculation from the known velocity? Is the reason why the high E and the low E have a similar output/loudness because both strings displace at a similar velocity?
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Post by ms on May 31, 2020 6:08:48 GMT -5
The integral is a result of a calculation from the known velocity? Is the reason why the high E and the low E have a similar output/loudness because both strings displace at a similar velocity? Yes, the displacement is calculated from the measured velocity. Point zero in the displacement is point one of the velocity minus point zero of the velocity. (Should multiply by the time increment between samples, but that is just assumed to be one since we scale this to fit on the plot anyway. This simple process also does not properly center the results, but the sampling rate is high enough so that this is not too important.) Point one of the displacement is point zero of the displacement plus the difference between point 2 and point 1 of the velocity, and to get the next value of displacement we add to the current value the next velocity difference times the time increment. This is an approximate numerical integration. If we had the displacement, we would estimate the velocity by subtracting neighboring values and dividing by the time increment. (Or we could use a more complicated process using more than two values of displacement to get a properly centered velocity.) Integration just goes the other way to get displacement from velocity. The integration starts at some particular time, and we should use the value of the displacement in the calculation of the zeroth point. We do not know this, but assume it is zero. This is the reason for picking towards the nut end of the string so that the initial displacement over the pickup is not very large. This seems to work out: when the displacement is at an extreme (positive or negative), the velocity is zero. I think the high E and low E are very different. With the high E only a quarter as many harmonics fit into the bandwidth of the system, and the string is plain rather than wrapped. I used the low E (6) because the velocity has so much detail because of this.
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Post by antigua on May 31, 2020 13:08:53 GMT -5
The integral is a result of a calculation from the known velocity? Is the reason why the high E and the low E have a similar output/loudness because both strings displace at a similar velocity? Yes, the displacement is calculated from the measured velocity. Point zero in the displacement is point one of the velocity minus point zero of the velocity. (Should multiply by the time increment between samples, but that is just assumed to be one since we scale this to fit on the plot anyway. This simple process also does not properly center the results, but the sampling rate is high enough so that this is not too important.) Point one of the displacement is point zero of the displacement plus the difference between point 2 and point 1 of the velocity, and to get the next value of displacement we add to the current value the next velocity difference times the time increment. This is an approximate numerical integration. If we had the displacement, we would estimate the velocity by subtracting neighboring values and dividing by the time increment. (Or we could use a more complicated process using more than two values of displacement to get a properly centered velocity.) Integration just goes the other way to get displacement from velocity. The integration starts at some particular time, and we should use the value of the displacement in the calculation of the zeroth point. We do not know this, but assume it is zero. This is the reason for picking towards the nut end of the string so that the initial displacement over the pickup is not very large. This seems to work out: when the displacement is at an extreme (positive or negative), the velocity is zero. I think the high E and low E are very different. With the high E only a quarter as many harmonics fit into the bandwidth of the system, and the string is plain rather than wrapped. I used the low E (6) because the velocity has so much detail because of this. Thanks for the explanation, it's very clear and complete. If the voltage is proportionate to rate of change of flux, and we suppose that the difference in thickness and permeability of the steel strings isn't a significant factor, then by process of elimination, would that imply that they produce a similar voltage due to the string displacing at a similar velocity, or is there another factor that might account for unequal loudness between the high and low E strings? Spectrograms of guitar strumming actually shows the wound string fundamentals being the loudest, with the amplitude dropping off at frequency, but I think guitar amps and/or speakers either purposefully of naturally to attenuate low end, which offsets the bass heaviness.
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Post by aquin43 on May 31, 2020 16:35:45 GMT -5
... If the voltage is proportionate to rate of change of flux, and we suppose that the difference in thickness and permeability of the steel strings isn't a significant factor, then by process of elimination, would that imply that they produce a similar voltage due to the string displacing at a similar velocity, or is there another factor that might account for unequal loudness between the high and low E strings? Spectrograms of guitar strumming actually shows the wound string fundamentals being the loudest, with the amplitude dropping off at frequency, but I think guitar amps and/or speakers either purposefully of naturally to attenuate low end, which offsets the bass heaviness. The difference in thickness and permeability of the steel strings is a significant factor.
Rearranging the basic string equation gives f^2 * m = T/(4 * l^2), where m is the mass per unit length of string, which is also an indicator of how much metal there is over the pickup. Assuming for simplicity an equal tension set of strings, f^2 * m would be constant. Considering the standing wave on the string, however, the velocity for a fixed amplitude is proportional to frequency. Between the top and bottom open strings with a 4:1 frequency ratio and with the same amplitude of vibration, the bottom string has 1/4 the velocity and 16 times the mass so the increased mass of metal can more than make up for the lower frequency.
Arthur
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Post by antigua on May 31, 2020 21:46:06 GMT -5
... If the voltage is proportionate to rate of change of flux, and we suppose that the difference in thickness and permeability of the steel strings isn't a significant factor, then by process of elimination, would that imply that they produce a similar voltage due to the string displacing at a similar velocity, or is there another factor that might account for unequal loudness between the high and low E strings? Spectrograms of guitar strumming actually shows the wound string fundamentals being the loudest, with the amplitude dropping off at frequency, but I think guitar amps and/or speakers either purposefully of naturally to attenuate low end, which offsets the bass heaviness. The difference in thickness and permeability of the steel strings is a significant factor.
Rearranging the basic string equation gives f^2 * m = T/(4 * l^2), where m is the mass per unit length of string, which is also an indicator of how much metal there is over the pickup. Assuming for simplicity an equal tension set of strings, f^2 * m would be constant. Considering the standing wave on the string, however, the velocity for a fixed amplitude is proportional to frequency. Between the top and bottom open strings with a 4:1 frequency ratio and with the same amplitude of vibration, the bottom string has 1/4 the velocity and 16 times the mass so the increased mass of metal can more than make up for the lower frequency.
Arthur
That makes sense, I tuned a high E to be the same pitch as a low E and they were about equal loudness despite having the same frequency, but I think the loose high E was displacing a greater distance since it was under so much less tension, and that boosted the output in lieu of the higher frequency. Is there any reason whatsoever to suspect a Lace Sensor might in fact "compress", and by "compress" I mean any phenomena that could be perceived as such to the ear. I got the Digilent Analog 2 working, and I have the means to put this to the test, but I'd feel more comfortable with a hypothesis in front of the test. I can put a Lace Sensor and a regular Strat pickup, put them side by side under a string, record one into channel 1, the other into channel 2, and compare their magnitude difference between the attack and the decay. If the Lace Sensors did in fact compress, the rate of voltage drop would be lower for the Lace Sensor. But I can't think of any hypothetical reason for why that should happen.
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Post by wgen on Jun 1, 2020 2:23:26 GMT -5
I've been reading all the time, all of this is so interesting. This note just to ask you if, in the next test, there is a way to analyze the higher harmonics readings at the transients versus the lower harmonics/fundamental, how much difference there is in amplitude and length of the decay of the higher harmonics between the two different pickups. I remember a thread of 1 or 2 years ago, it seemed to underline more length of the decay of the higher harmonics when the magnetic strength was stronger, but this was so slight of a difference..
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Post by aquin43 on Jun 1, 2020 4:01:04 GMT -5
... Is there any reason whatsoever to suspect a Lace Sensor might in fact "compress", and by "compress" I mean any phenomena that could be perceived as such to the ear. I got the Digilent Analog 2 working, and I have the means to put this to the test, but I'd feel more comfortable with a hypothesis in front of the test. I can put a Lace Sensor and a regular Strat pickup, put them side by side under a string, record one into channel 1, the other into channel 2, and compare their magnitude difference between the attack and the decay. If the Lace Sensors did in fact compress, the rate of voltage drop would be lower for the Lace Sensor. But I can't think of any hypothetical reason for why that should happen. Not much, it seems. Any non linearity, as ms pointed out, is of the integral of the waveform not the waveform itself. This will modify the spectrum but not compress the peaks. In an extreme case small amplitude harmonics could be modulated to some extent by the fundamental. I would be surprised of any easily interpreted difference could be observed, especially given the essentially pulse like shape of the waveform in it earlier, stronger part.
Arthur
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Post by antigua on Jun 1, 2020 16:41:36 GMT -5
... Is there any reason whatsoever to suspect a Lace Sensor might in fact "compress", and by "compress" I mean any phenomena that could be perceived as such to the ear. I got the Digilent Analog 2 working, and I have the means to put this to the test, but I'd feel more comfortable with a hypothesis in front of the test. I can put a Lace Sensor and a regular Strat pickup, put them side by side under a string, record one into channel 1, the other into channel 2, and compare their magnitude difference between the attack and the decay. If the Lace Sensors did in fact compress, the rate of voltage drop would be lower for the Lace Sensor. But I can't think of any hypothetical reason for why that should happen. Not much, it seems. Any non linearity, as ms pointed out, is of the integral of the waveform not the waveform itself. This will modify the spectrum but not compress the peaks. In an extreme case small amplitude harmonics could be modulated to some extent by the fundamental. I would be surprised of any easily interpreted difference could be observed, especially given the essentially pulse like shape of the waveform in it earlier, stronger part.
Arthur
What if there were a weaker magnetic attraction between the string and the pickup? People talk about sustain suffering as a result of magnetic pull, but it might just be that the transient amplitude is lower with less magnetic pull (if any effect really exists at all). it might be possible that a stronger magnetic pull promotes a more vertical and elliptical movement pattern when the string isat maximum displacement, and gradually tends towards a more averaged, circular movement, where as a weaker magnetic pull causes less interference, resulting in the more circular pattern from the beginning. If that's the case, my idea of mounting the pickups side by side under the same string wound be invalid, since the effects of magnetic pull on the strings wouldn't be isolated to each pickup's output.
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