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Post by ssstonelover on Jul 3, 2024 2:12:16 GMT -5
I'm trying to get a handle on the differences between an anti-mud mod to a pickup, and a bass cut set up on a pot (G&L or other type). Certainly putting a fixed value cap/resistor on a pickup is a dedicated mod, vs a variable cut as used on a bass cut pot which can apply to all pickups. That much is clear, but when I see the values used on the anti-mud (~.01uF, ~ 100K) and for the bass cut pot (~.0022uF, ~1M) they seem worlds apart. --Firstly what 'rules' underpin this, so I can make sense of it all? --Secondly I see these mods generally on the 'hot side', but in reality could they not be on the ground side? I believe I hear this is possible. I have made some possible alternatives for the anti-mud mod, but in theory or practice could this not also apply to the bass cut pot too right? (not illustrated) If so what are the trade-offs? To get things started, here are some sketches to put things in a graphic context  |
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Post by sumgai on Jul 3, 2024 15:45:43 GMT -5
Bethany, I'm surprised that no one has yet answered this for you. In essence, you are totally correct, the cap (and resistor) could go on the ground side of the pickup. What we're looking at is a simple circuit, just like the lesson in school about a battery, a switch, and a lamp. And where we learned that the switch could go on either side of the battery, it didn't matter so long as the flow of current was interrupted, thus allowing us to control the lamp being lit, or not. And as it happens, the 'standardized' Bass cut pot of a PTB could also go on the 'ground' side of the pickup, your intuition is correct about that one too. But let me ask you a question... beyond mere curiosity, why the interest? Do you have an actual need to perform alternate surgery, perhaps in order to better fit components in the cavity? Further discussion: As it happens, the labeling of a pickup's wires is totally arbitrary - we can just as easily assign a wire color to either hot or ground, the electrical result will remain exactly the same. Only further on in the circuit, namely the Vol and Tone controls, must we pay attention so as not to make a mess of things. But that's probably intuitive to most mildly interested modders, I'm sure. What's of interest here is, are we assuming, or assigning, too much value in a supposed "best practice" of doing the same way it's always done? Could be, could be. But I'm not so much interested in "doing it the right way" as I am in getting the needed results, with the minimum necessary hassle. In other words, one should always be ready to Think Outside Of The Box. It's a mantra that I've always hammered away at here in The Nutzhouse, and more than a few members have benefitted from it, I'm also sure of that. (But indeed, a few didn't need my prodding, it came naturally for them. I'm looking at YogiB in particular.  ) OK, enough of this. Carry on!  sumgai |
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Post by ssstonelover on Jul 3, 2024 18:54:55 GMT -5
Bethany, And as it happens, the 'standardized' Bass cut pot of a PTB could also go on the 'ground' side of the pickup, your intuition is correct about that one too. But let me ask you a question... beyond mere curiosity, why the interest? Do you have an actual need to perform alternate surgery, perhaps in order to better fit components in the cavity? As it happens, the labeling of a pickup's wires is totally arbitrary - we can just as easily assign a wire color to either hot or ground, the electrical result will remain exactly the same. Only further on in the circuit, namely the Vol and Tone controls, must we pay attention so as not to make a mess of things. But that's probably intuitive to most mildly interested modders, I'm sure. Actually, to quote Shakespeare (Hamlet), "there is method to my madness"..... In this thread you mentioned (reply 13) about using the S-1 switch as a terrific space-saver link-1 and JohnH chimed in, loving that idea, and added (reply 17) that adding a PTB (the bass cut part) would be killer if attached just on the ground side of the series (Brian May). He wrote, "... I'd consider putting it just into the series chain at the end of S3,4,5, so that it only operates in series mode That way it can be left preset to make a perfect series tone, but not affect the standard Strat settings. Then a quick move to a boosted series sound is just one switch press." A very insightful comment indeed. Well, I can only say I've never seen it done that way, but hey, it is genius and thinking outside the box. Reply 21, second drawing, is my attempt to design such a circuit (using John's words as my reference). You see any issue with it? I ask as I about to solder up the pickguard now, and would love to incorporate that feature. Also from that humble beginning, begat this newest partially related anti-mud thread, so yes, it's a continuation of a thought, but adding in other elements, making it a broader musing. Thanks for showing I'm not crazy, electricity does not care in these cases. Overall, I can see the advantage of the cap/resistor in series to a pickup as a permanent fix to mud, I suppose a treble bleed as yet another way (though just addressing another form of mud picked up by the volume pot), and the PTB is yet another way (like when facing series loading) all these tools are quite different yet all anti-mud. I suppose the application determines the method chosen (turning down a volume, hitting amp high gain, or series loading). |
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Post by newey on Jul 3, 2024 20:10:08 GMT -5
You see any issue with it? I ask as I about to solder up the pickguard now, and would love to incorporate that feature. Haven't got a chance at the moment to go through the diagram. But it occurs to me that, with the amount of work that you do, you need a "test mule" that you could wire up in various configurations to see what works best, before investing time/energy/cost on a final product that maybe could have been tweaked a bit. ChrisK had one years ago, photos of it are kicking around here somewhere. |
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Post by ssstonelover on Jul 4, 2024 0:14:56 GMT -5
You see any issue with it? I ask as I about to solder up the pickguard now, and would love to incorporate that feature. Haven't got a chance at the moment to go through the diagram. But it occurs to me that, with the amount of work that you do, you need a "test mule" that you could wire up in various configurations to see what works best, before investing time/energy/cost on a final product that maybe could have been tweaked a bit. ChrisK had one years ago, photos of it are kicking around here somewhere. I guess you could call the guitar I'm working up now effectively a test mule. It's an old well-worn Samick (19 ply body) I picked up cheap with a (now) very generously-sized control and pickup cavities I've rerouted. I used a piece of clear window plexiglass to visualize and confirm routing and placements. Parts are mainly recycled with new low cost toggle switches). In terms of ultimate test mule,I could rout all the way to the back so I don't even have to pull off the pickguard, so could have rear access with a cavity cover on that side too, hm, something that could be pretty cool, though the body is a bit on the thin size (~39mm thick rather than 45mm, so not all 5-way switch types would fit, at least not with the cover closed or recessed flush, but that is a small issue. ChrisK likely had something more comprehensive, but this could be a good start. This will give an idea of what I have cooking, and a bit on my process.  |
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Post by sumgai on Jul 4, 2024 1:50:19 GMT -5
That's pretty cool, to be sure. Here's what JFrankParnell did, about 14 years ago:  You can see more images in this thread, starting at Post #25, here: guitarnuts2.proboards.com/post/45486/thread(If you click on each image, which are links, Photobucket will show you the original without the watermark.) Just for your information, of course. sumgai |
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Post by reTrEaD on Jul 4, 2024 9:44:36 GMT -5
but when I see the values used on the anti-mud (~.01uF, ~ 100K) and for the bass cut pot (~.0022uF, ~1M) they seem worlds apart. --Firstly what 'rules' underpin this, so I can make sense of it all? Anti-mud and bass-cut are the same concept. They are both a high-pass shelving equalizer. Let's analyze how they function and apply the component values you mentioned. If we only have a series capacitor, the bass would roll off at a rate of 6dB per octave below the corner frequency and continue to do so all the way down to DC. But the resistor in parallel with the capacitor limits the voltage division, so we have a shelf at lower frequencies. To begin, we'll determine the voltage division limit. In the case of the anti-mud, the 'top' resistor in the voltage divider is 100k. The bottom resistor is the parallel resistance of the volume pot and the amplifier input impedance. Unless we place the anti-mud between the treble-cut tone pot and the volume pot, we'll also need to add the resistance of the tone pot in parallel, unless it's a no-load pot. Not sure what your intentions are in that regard, so I'll do the calculations with just the volume pot and amplifier input resistance. 250k in parallel with 1M = 200k. So our voltage division limit is 200k / 300k. or .6667 this will translate to roughly -3.5dB. Not a big distance down to the lower shelf. To determine the corner frequency, we enter the 0.01uF and 200k into the formula 1 / (2πRC) or just plug it into an online calculator. Roughly 80Hz. With a 80HZ corner frequency and a maximum cut of 3.5dB, this very subtle. 3dB is slightly noticeable. The low E on a six string guitar in standard tuning is 83 Hz. In the case of the bass-cut we'll calculate the voltage division limit by using the maximum value of the bass pot (1Meg) and the 200k load. 200k / 1.2 Meg = .1667 or roughly -15.6 dB. Corner frequency is 362Hz. This is pronounced. Apparent volume is halved at -10db and quartered at -20dB. So a mathless estimate would be a reduction in volume to one-third, two octaves below the corner frequency. 362Hz is roughly the high e string fretted at the second fret. |
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Post by ssstonelover on Jul 4, 2024 15:10:20 GMT -5
You see any issue with it? I ask as I about to solder up the pickguard now, and would love to incorporate that feature. Haven't got a chance at the moment to go through the diagram. The only part of the diagram I need a second set of eyes on concerns tone pot wiring (treble and bass pots), the other stuff (pickup selection) should line up well with previous iterations of the drawing. Thanks. |
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Post by ssstonelover on Jul 4, 2024 15:28:43 GMT -5
reTrEaD, thanks for the overview and math lesson, that explains the big differences. Good point about series vs parallel and to look at the total resistance chain including pots in parallel, meaning positioning is indeed important I have no immediate plan for the anti-mud mod but I do see it as useful for potentially curing muddy humbuckers (so new ones don't need to be substituted, like the nice new Seymour Duncan I used to replace an OEM Ibanez pup in a Gio, a change I did recently per client request). The PTB is something I have used and it's as you say, mathematically very powerful -- meaning a little goes a long way. I even tried it on a bass guitar once....way drastic there, so I'm thinking a less aggressive approach might work better there (than a C1M paired with something like a .0022uF) might make sense. What would theory tell us to get a gentler rollover especially on the volume side in such a case?
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Post by JohnH on Jul 5, 2024 0:07:36 GMT -5
Its been nice to pull the covers off GuitarFreak again on recent threads, and take her out for a spin on a sunny afternoon: Anti-mud and Bass-cutSo here's some plots on Bass-cut vs Anti-mud, based on the posts above. I agree that they are essentially the same thing, being an added series impedance of cap and resistor in parallel, forming a high-pass filter. They are both before the volume pot, and the volume pot is a key part of making them work. I'm testing them both as bass-cut circuits, and I don't think the slightly different positions in the circuits shown are significant, ie, between humbucker coils or fully after the pickup. To make that truer, Ive taken out the treble-cut pot. This is for an SD59 with 500k volume pot, all plots at full volume. Blue-dashed is no cut, as a basic tonal reference The Bass-cut is plotted with four variations, being a 1M bass pot at either 100% cut, or 20% cut (ie 200k active), and with either 2.2nF or 1nF caps The Anti-mud has a 10nF cap as recommended in the diagram, and 100k in parallel (like a 1M bass pot at 10% cut)  You can see how the bass cut, with its pot and small capacitors, can give a significant tonal change (or no change if set at no cut, as the reference) The Anti-mud (red trace) is a much more subtle change, and looks like it just trims a bit of lows, reducing low-end mud, as it should be expected to do. But I expect it's a case where you cant really choose optimum values without trying them, but this thinking and plotting can give insight into suitable values to try and what they might be expected to do. |
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Post by ssstonelover on Jul 5, 2024 20:08:04 GMT -5
The anti-mud sure looks subtle in the example JohnH, as it reduces volume just 2dB at 100Hz in this Guitarfreak simulation. Maybe that is more significient going through some effects and gain. I'm thinking the bass cut circuit, at the cost of more complexity gives more sculping, and thus is also a good "on the fly" flexible solution. I like how you tried 1nf and 2.2nf for the bass cut. At 1000Hz and 20% resistance cut they are quite different in how they cut that frequency, though 2Db is may be subtle but maybe it more than subtle overall. Hard of me to judge, what's the reality there? Somehting I'm missing? |
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Post by sumgai on Jul 5, 2024 21:06:17 GMT -5
Bethany,
Ordinarily, when we speak of dB in sonic terms, a 3dB difference is either half as loud, or twice as loud, depending on direction from the reference level.
I suspect that 2dB is enough difference to be noticable, without being overbearingly different. Give it a try, as John says, and see what you find out.
HTH
sumgai
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Post by ssstonelover on Jul 6, 2024 0:41:21 GMT -5
Hi sumgai, I was looking at dB definitions before the last posting, and what I found was this, " Although an increase of 3 dB represents a doubling of the sound pressure, an increase of about 10 dB is required before the sound subjectively appears to be twice as loud. The smallest change we can hear is about 3 dB." The way I'm understanding this is that 3 dB is the smallest perceptible change, though I'm sure you are 100% correct that the power needed is double. By this logic 2 dB may not be perceptible at all. But enough about definitions, what does the mean in the real world, as source difference compared to amplified differences may be quite different. Here is my question, does amplification (say a guitar amp) drastically accentuate a source difference of 3 dB / -3 dB? Does an amp magnify/diminish a 3/-3 dB source difference in a linear or in a 'log type' way as far as the human ear is concerned? I ask as I am having some difficulty finding anything on the web about this -- about how source dB difference scale at amp output -- and that would have a huge bearing on what is heard. I'm guessing it has more impact that one might suspect as first glance, or the anti-mud mod would not work in the real world. Not having worked on amps, I have to confess ignorance on this. |
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Post by JohnH on Jul 6, 2024 2:52:54 GMT -5
My belief is that 1dB is the smallest change we might hear, and 3dB is a distinct but quite small step in volume.
One of my ongoing projects is amplifier attenuators, and my design happens to step down in increments of 3.5dB, so very close to 3dB. You can definitely hear that difference clearly, and decide if you want one more, or one less of those steps. But unlikely to want to subdivide again.
Also, if listening to cleanly amplified signals, a step of a certain dB value probably sounds much the same at different overall volumes, so long as it's within the (wide) range of reasonable hearling. But in an amplified-overdriven guitar tone, a couple of dB could represent the difference between fairly clean and distinctly dirty, and so quite audible as a tonal difference as well as a volume change. With the anti-mud, this change might sometimes de-flub the bass.
And those plots are just a set of math results with one set of values. They can be tweaked to get more or less at any particular frequency, by changing values.
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Post by reTrEaD on Jul 6, 2024 10:29:55 GMT -5
My belief is that 1dB is the smallest change we might hear, and 3dB is a distinct but quite small step in volume. I mostly agree with that. If you're listening carefully, you might be able to hear a 1dB difference. If you're listening carefully, you will be able to hear a 2dB difference but if you're listening casually, it might fly under the radar. A 3dB difference won't escape notice, even if you're listening casually. JMO. A 10dB difference will seem twice as loud and a -10dB difference will seem half as loud. |
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Post by sumgai on Jul 6, 2024 22:44:13 GMT -5
So I misspoke a bit last time out, sorry 'bout that.  Here is my question, does amplification (say a guitar amp) drastically accentuate a source difference of 3 dB / -3 dB? Does an amp magnify/diminish a 3/-3 dB source difference in a linear or in a 'log type' way as far as the human ear is concerned? Ah, the beauty of ratios. For our purposes, a Bel, and perforce a deciBel, is the measure of a ratio, not a finite unit of some kind. As you might expect, the ratio is not disturbed by the amp, vis-a-vis the signal level. What you'll notice is that if you input a 1 volt signal, and that gets you 5 volts of output, then putting in 2 volts will get you 10 volts of output (assuming perfect linearity). The ratio remained the same, i.e. 1:5 in my example. By definition, an amplifier is supposed to be linear - what goes in is what comes out, with no distortion of any kind. In reality, that just doesn't happen. In fact, we musicians love certain kinds of distortion, hence the "near-mythical" linear amp is a non-starter for us. And yes, an amp can be designed, on purpose, to be non-linear in a logarithmic fashion. We often call them either compressors or expanders. In such cases, the ratio I just described is tossed out like so much bathwater, but we like it and use it, particularly where compression is desired. It's all done with smoke and mirrors, of course. HTH sumgai |
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Post by Yogi B on Jul 7, 2024 0:14:38 GMT -5
To determine the corner frequency, we enter the 0.01uF and 200k into the formula 1 / (2πRC) or just plug it into an online calculator. Roughly 80Hz. That calculation is mostly* irrelevant to a shelving filter: imagine if the 100k resistor is even smaller, then the shelf may not be as low as −3dB point of the cut-off frequency. *: in this specific case it's slightly relevant — the magnitude of the transfer function is sqrt(2)/2, −3dB, at the frequency given by that formula when the ratio of R 1 (the upper resistor, 100k in this example) to R 2 (the lower resistor, 200k) is either: ∞:1, i.e. the usual use of that formula, a non-shelving filter, when R 1 isn't included; or, coincidentally, 1:2. That's it (well, technically that's the only real solution — if we were to allow complex resistances, it's x + (1 − x)j : 2 for any real number x). And though that happens to be true, the −3dB point has little to do with the shape of the shelving filter. What is more useful is the centre frequency — the frequency at which half of the maximum change occurs. That's half in logarithmic units (i.e. decibels), but a square root in linear units: 0.5 log(x) = log(sqrt(x)). The largest reduction is R 2 / (R 1 + R 2), as stated. This occurs at 0Hz when the impedance of C 1 is (ideally) infinite. But, in general (i.e. at other frequencies), the voltage division is given by:
\frac{V_\text{out}}{V_\text{in}}
= \frac{R_2}{R_2 + (R_1 \parallel \frac{1}{2 \pi j f C_1})}
= \frac{R_2}{R_2 + \frac{R_1}{1 + 2 \pi j f R_1 C_1}}
from which the centre frequency can be found via a few different ways. Most obviously from setting the absolute value of the above equal to 'half' the max change — i.e. sqrt(2/3) — and solving to find f, but it can get quite verbose separating out the real and imaginary parts in order to find the absolute value. Alternatively, for the calculation of the centre frequency (and only the centre frequency), the previous can be greatly simplified by replacing the imaginary impedance of the capacitor with its (real valued) reactance. (I'll expand on, and hopefully clarify, why this method works a little later.) However, the most direct method is probably via the geometric mean of the pole and zero frequencies of the above transfer function. Assuming R 1, R 2 & C 1 are all positive, real, and finite. \begin{array}{c}
\begin{aligned}
0 &= \frac{R_2 + \frac{R_1}{1 + 2 \pi j f R_1 C_1}}{R_2}
\\[2ex]
&= R_2 + \frac{R_1}{1 + 2 \pi j f_p R_1 C_1}
\\[2ex]
-R_2 (1 + 2 \pi j f_p R_1 C_1) &= R_1
\\[2ex]
-2 \pi j f_p R_1 R_2 C_1 &= R_1 + R_2
\\[2ex]
f_p &= \frac{R_1 + R_2}{-2 \pi j R_1 R_2 C_1}
\\[3ex]
&= \frac{(R_1 + R_2)j}{2 \pi R_1 R_2 C_1}
\end{aligned}
\\[3ex] \\ \hdashline \\
\begin{aligned}
0 &= \frac{R_2}{R_2 + \frac{1}{\frac{1}{R_1} + 2 \pi j f_z C_1}}
\\[3ex]
0 &= \frac{1}{R_1} + 2 \pi j f_z C_1
\\[3ex]
-2 \pi j f_z C_1 &= \frac{1}{R_1}
\\[3ex]
f_z &= \frac{1}{-2 \pi j R_1 C_1}
\\[3ex]
&= \frac{j}{2 \pi R_1 C_1}
\end{aligned}
\\[3ex] \\ \hdashline \\
\begin{aligned}
f_c &= \sqrt{f_p \cdot f_z}
\\[2ex]
&= \sqrt{
\frac{(R_1 + R_2)j}{2 \pi R_1 R_2 C_1}
\cdot \frac{j}{2 \pi R_1 C_1}
}
\\[3ex]
&= \frac{j\sqrt{R_1 + R_2}}{2 \pi R_1 \sqrt{R_2} C_1}
\end{aligned}
\end{array}
Since the pole-zero pair are at imaginary frequencies (which we should expect since the response doesn't shoot up to infinity or down to zero when plotted over real frequencies), their geometric mean is also imaginary. However it is the same magnitude as the real centre frequency. For the given values this is: sqrt(375000)/pi ≈ 195Hz.
To try and make better sense of what we're doing in the above, following is a plot of the shelving filters transfer function over the s-domain (complex-valued frequency domain). Note that, since s = σ + ωj (where ω is angular frequency = 2 πf), 'normal' real-valued frequency lies on the imaginary axis and imaginary-valued frequencies lie on the real axis. This is why the above calculated imaginary pole & zero frequencies end up plotted in the negative (real) region of the graph (j × j = −1). The vertical axis is the magnitude in decibels, whereas the phase is represented by the colouring of the surface (roughly: red = 0°, green = 90°, cyan = 180°, blue = 270°). As noted, the imaginary axis (which runs from the bottom left to the top right) is the frequency axis that we're used to seeing on frequency response plot, but usually only the positive half. As such the black line (which lies along this axis / is coplanar with the vertical slice where σ = 0) shows the response of the shelving high-pass filter, being close to 0dB for large frequencies and dropping by a handful of decibels for frequencies closer to zero. (As I say, we don't usually see this plotted into the negative, where the filter also lets high negative frequencies pass.) The other horizontal axis is proportional to σ (the real part of s). I don't know if there's an "official" term for the quantity represented by σ. I've seen damping and decay suggested, but I'd want to add angular to each of those so there's not an implied factor of 2 π difference between the real & imaginary parts. Though, since the other axis is commonly just the "(angular) frequency axis", this could be the (negative) imaginary (angular) frequency axis, where we're effectively using the names as though s were divided though by j — but that'd be highly confusing. Despite the log-log scaling distorting them into their apparent squarish shape, the blue, green & red lines are circles (when viewed from above) centred on s = 0 with radii respectively equal to the magnitudes of the pole, centre & zero frequencies. An important observation is that magnitude of the transfer function is equal to sqrt(2/3) at every point along the circumference of green circle AND every point where magnitude of the transfer function is equal to sqrt(2/3) is on that circumference. This is why it is possible to calculate the centre frequency by either: replacing the impedance of the capacitor with its reactance; or as the geometric mean of the pole & zero frequencies. These calculations are finding the intersections of the green circle with (respectively, the positive & negative halves of) the white line and, being a circle, a radius that is instead measured along the frequency axis is still equal. 
Ordinarily, when we speak of dB in sonic terms, a 3dB difference is either half as loud, or twice as loud, depending on direction from the reference level. A 10dB difference will seem twice as loud and a -10dB difference will seem half as loud. When applied to power, a multiplication by 10 times is represented by a 10dB increase — thus a doubling is: 10 log 10(2) ≈ 3dB. When applied to a 'root-power' quantity such as voltage or sound pressure level (what our ears actually sense) it is expected that the complementary variable (current in the case of voltage, and particle velocity in the case of SPL) also experiences a change proportional to the measured quantity. For example, increasing voltage ten times also increases current tenfold (since V = I Z, and assuming constant impedance) in this case a 10 times increase in voltage relates to 100 times (or 20dB) increase in power. So for doubling of a 'root-power' quantity we have 10 log 10(2 2) = 20 log 10(2) ≈ 6dB. |
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Post by ssstonelover on Jul 7, 2024 1:21:07 GMT -5
.... But in an amplified-overdriven guitar tone, a couple of dB could represent the difference between fairly clean and distinctly dirty, and so quite audible as a tonal difference as well as a volume change. With the anti-mud, this change might sometimes de-flub the bass. Yup makes sense |
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Post by ssstonelover on Jul 7, 2024 1:29:56 GMT -5
As you might expect, the ratio is not disturbed by the amp, vis-a-vis the signal level. What you'll notice is that if you input a 1 volt signal, and that gets you 5 volts of output, then putting in 2 volts will get you 10 volts of output (assuming perfect linearity). The ratio remained the same, i.e. 1:5 in my example. In reality, that just doesn't happen. In fact, we musicians love certain kinds of distortion, hence the "near-mythical" linear amp is a non-starter for us. And yes, an amp can be designed, on purpose, to be non-linear in a logarithmic fashion. We often call them either compressors or expanders. In such cases, the ratio I just described is tossed out like so much bathwater, but we like it and use it, particularly where compression is desired. It's all done with smoke and mirrors, of course. Yup, just twiddle the presets on an amp (especially a modeling amp) to alter the dB spreads at the different frequencies, get tube saturation, etc..... Fun and games, real world vs theory. |
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Post by JohnH on Jul 7, 2024 17:54:46 GMT -5
Each of us trying to describe how these circuits work mostly seems to show how its actually really quite difficult to tease apart the various things that are going on at each frequency.
Each interaction is mostly quite 'soft', as such passive circuits tend to create soft rounded transitions that spread and overlap and blur into each other. Parts of the typical response curves are affected more by certain components, but its all interactive and everything affects all of it. But to try....
At very low frequencies, the caps let little through, the inductance of the pickup has very low impedance. So cross all those out and we just have resistances left, pickup, bass-cut resistor, feeding into the volume pot and the output, as a simple voltage divider. So the lower region of frequency looks like a flat response. This is best seen on the bass-cut plots at 200k.
Then gradually as frequency rises, the bass-cut cap let's more through and we have a rising response.
This then continues up to a region where the bass-cap is letting almost everything through, flattening-off the curve, but often in the same frequency range, the pickup inductance is interacting with the cable, to we get the high treble peak, which can get trimmed if the bass cap is very small.
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