
Post by zedsnotdead on Jul 9, 2012 22:28:13 GMT 5
Hi all! This will probably be the most stupid/awkward question you will read in your life, but it's killing me! Ok, here it goes: so we know that every note on a scale have a frequency. For guitars with standard tuning and 22 frets, it means a range from 82Hz (low E) to around 2000Hz (string freted@ 22 fret, high E). Now... a pickup with tone control. Say it's a 500K pot with a 47nF capacitor. Don't mind the cable and amp load for this exercise. Let's say that the pickup has somewhat flat response, I mean, a regular pickup would be flat for frequencies below 3000Hz, isn't it?... So , I pluck the A note on high E, it will vibrate @ 440Hz... so the overall impedance of the tone circuit would be 500K+7,6K=507K with the pot @100%. But if I roll back the pot to 0%, it would be only 7,6K resistance to ground. And I will hear a bassy tone, it will "bleed" the high frequencies... Ok, but WHICH "high frequency"? I mean, the string is vibrating @ 440Hz, so the signal that the pickup is emitting is fundamentally 440Hz also. What frequency are we talking about? Another strange question is this: So suppose we have a volume pot, also 500K, and the tone pot is at 0%, so the signal will have a significant easier path to ground through the tone pot. I mean, just roll off the volume pot to say 70%, the signal will face 150K resistance at this pot. But only 7,6K in the tone pot. Shouldn't the signal be affected? I mean, shouldn't the signal volume drop significantly? You see, just silly questions. Probably i am just dumb... sorry



Post by newey on Jul 9, 2012 23:11:25 GMT 5
ZnD "Impedance" and "resistance" are two different things, at least in the setting of an AC guitar circuit. The tone control uses a capacitor which, in combination with the pickups and other components, has a property called "capacitive reactance" which must be considered. Uh, no. . It is wrong to think of the action of the tone pot as erecting "a wall" above which no frequency will pass. It is more of a changeable slope. Read this, it may clear up some of these issues, or it may leave you more confused, but it will show you that the subject is note quite as simple as you have made it out to be: ChrisK explains the passive hicut tone control



Post by zedsnotdead on Jul 10, 2012 0:50:56 GMT 5
"Impedance" and "resistance" are two different things, at least in the setting of an AC guitar circuit. The tone control uses a capacitor which, in combination with the pickups and other components, has a property called "capacitive reactance" which must be considered. Well, for what I know, reactance is the opposition that a capacitor and/or indutance do to a changing voltage (AC) and thus capacitive and/or inductive reactance. It is given in Ohms, and it varies with the frequency of the signal (voltage) applied to the circuit. Impedance is the absolut sum of resistance and the reactance. Also given in Ohms. So, for this given frequency of 440Hz, and disregarding the impedance of the pickup (resistance and reactance) just to isolate the tone circuit that we are studying, we have a parallel resistor (the tone pot) and capacitor (47nF) that will act as a 7,6K (capacitive reactance)+500K (tone pot) = 504,7K, in absolute values... If the resistor is turned to 0%, it only remains barely the reactance 7.6K... so the signal will have a much easier path to ground through the capacitor. All this in the 440Hz frequency range, of course... Am I wrong? Yes, i reckon that it's a slope directly related to frequency. More frequency, less reactance. That's why I said it "bleeds" the hifreqs to ground, and the remaining lowfreqs go to signal output, hence the "bassy" sound. Thanks newye! I will read ChrisK topic. I hope I can clarify things a little more.



Post by ashcatlt on Jul 10, 2012 1:30:48 GMT 5
The answer to your first question is mostly pick attack, which is a rapid acceleration of the string and includes a bunch of high frequency content. Also, the string almost always carries little ripples on top of its fundamental wave. Harmonics.
The answer to the question of T vs V control is that since the resistances are parallel to one another, the voltage across them is equal. Assuming that the source can provide enough current, you can change either or both without changing the voltage across either.
But you cannot isolate the passive tone control circuit. It all works together. I'm sure ChrisK's thing covers it, but I think it bears repeating.
It's really a matter of voltage division. The reactance of the pickup at the frequency in question is the "top half" of the divider. The resistance of the variable resistor plus the reactance of the cap are the "bottom half".



Post by 4real on Jul 10, 2012 4:47:10 GMT 5
en.wikipedia.org/wiki/Harmonic_series_(music)There are others who know more and can explain the filtering and maths of such things...but a grasp of how the string vibrates is an important concept to understand, especially in relation to an instruments intrinsic "tone" before the pickups, electronics and amps do their thing with it. Behind the 'note' which is 'fundamentally' the naming note..ie, the fundamental say, A=440. But behind that are a whole series of harmonics. Strike a harmonic on the 12th fret above a note and you will get an 'overtone' or 'harmonic' an octave higher. What you are doing is suppressing the fundamental to reveal the harmonic that is behind that. So it is with other frets and harmonics points of the nodes along the string. These harmonics in a balance are always present in various levels in the vibration of any string. The position of a pickup along the string sensing things can have a significant effect with the neck pickup having a smoother less complex harmonic structure sensed while a bridge pickup will contain a wider range in the higher order harmonics and so a brighter sound. Pickups themselves have characteristic filtering of things in themselves due to various factors like resonance. The passive tone can dial out the high frequencies and leave more of the lower frequencies, largely the fundamental. Again, pull the tone down and try the harmonic things and you will find these to be lacking. Pick attack can also provide a 'burst' of harmonics and frequencies as mentioned. Aggressive picking can even pull a string sharp before released and of course one can bring out 'pick harmonics' from where it is picked. There is more to 'tone' though of course, the attack. sustain and decay qualities too are important. Anyway, an apprecitaion of what the string is doing can help to understand the guitar and how it can be manipulated. Such harmonics and things like sustain or attack can be affected by technique, but a lot of it is intrinsic to the instrument. The 'mojo' can come in with claims to be able to bring things out that are perhaps not present in the vibration of the instrument. The construction and materials of the guitar do make a difference and can be a reasonably powerful 'filter' before it ever hits the electronics.



Post by Yew on Jul 10, 2012 13:34:05 GMT 5
With regards to the tone pot, the impedance of a capacitor for AC is defined as __1__ wc (j) ; where c is the capacitance, and w is the frequency  dont worry about the (j) for now, we will get to that laterSo if we have a frequency of 1hz, and a capacitor of 10 farads (very large) we get a resistance of ; ___1___ 1 x 10 (j) Ohms = 0.1(j) ohms
Now lets say theortically there is a harmonic at 10hz, using the same equation we get; ___1___ 10 x 10 (j) Ohms = 0.01(j) ohmsNow if there is a third harmonic at 100hz, we would get an impedance of 0.001(j) ohms. Note how every time we multiply the frequency by 10, the impedance reduces by 10. We could also change the capacitance of the capacitor to get the same effect. Now since the capacitor on a guitar circuit goes to ground, this means that Higher frequencies will find it easier to pass through the capacitor and reach the ground, yet lower frequancies will have to go through the amplifier. However, if we put a capacitor onto a circuit, we would have a simple on or off. If we want to vary the level of frequancies cut, we use a potentiometer, called a variable resistor. this is where the (j) notation comes in. Lets say we have a 1hz frequency, into a 1 farad capacitor, connected in series with a 0.1 ohm resistor, we would have Impedance = 0.1 ohm + 0.1(j) ohmNow unfortunately this does not add up to 0.2ohms, the (j) means it is something called an imaginary number. This requires some explaining; Let us say there is a force of magnitude f now if we want a force of f, ie going the other way, we would multiply f by 1 If we look at this then, we have rotated force f by 180 degrees, so it is facing the other direction, however what happens with 90 degrees? If we have the number 9, the square root of it is 3. and 3 x 3 = 9, this can also be written as sqrt(9) x sqrt(9) = 9 This means we can write (f x (1)) as [f x (sqrt(1)) x (sqrt(1))] However unfortunately we cannot calculate a real number for the square root of 1. however we can use an imaginary number on an imaginary axis, like this. Now in practice with an imaginary number you just put (j) in front of it, and manipulate it like any normal algebraic term. Now back to our circuit, if we have an impedance of 0.1 + 1(j), we know that the (j) term is at 90 degrees to the real term. So if we think of a map, the coordinate is one block to the right, and 0.1 block up (the red line). Now electricity will take the shortest possible path to get to this place, a straight line between zero and our point (the blue line). to solve this we use phthagoras theroem; 1^{2}+ 0.1^{2} = impedance^{2}this gives; Impedance = sqrt(1+0.01) = 0.141 ohms ; This means that the impedance is now dependant on three things,
 1) the value of the capacitor } the cap value and the frequency work together to control the j term
 2) the frequency
 3) the value of a resistor } this works on its own to control the real term
In reality, this means the the range of frequencies cut depends on the capacitor, and the amount cut depends on the resistor. If you put a smaller capacitor you can cut more of the midrange, but a cap will cut only the highs. Then you can control the amount of this frequency range by changing the resistor value. I hope this made some sense, Yew



Post by sumgai on Jul 10, 2012 14:30:47 GMT 5
This means we can write (f x (1)) as [f x (sqrt(1)) x (sqrt(1))]
A genuine NoPrize to the first poster who can spell out what's wrong with this equation. Though I gotta say, nice presentation, Yew! ;D (EDITed for spelling error.)



Post by D2o on Jul 10, 2012 20:36:37 GMT 5
I'm guessing he's inadvertantly changed the 1 to a +1 with that (neg) * (neg) maneuver?
[edit] Actually, that would only be the case if the sqrt of 1 produced a minus ... there cannot be a sqrt of 1.



Post by zedsnotdead on Jul 10, 2012 22:31:08 GMT 5
This means we can write (f x (1)) as [f x (sqrt(1)) x (sqrt(1))]
A genuine NoPrize to the first poster who can spell out what's wrong with this equation. Though I gotta say, nice presentation, Yew! ;D (EDITed for spelling error.)1=jÂ² = j.j = sqrt(1).sqrt(1) = sqrt (1.1) = sqrt (1) = +1 or 1 sqrt rules only applies to real numbers >0. I just forgot about this, had to search the wiki (shame on me). Is this it?



Post by zedsnotdead on Jul 10, 2012 22:34:13 GMT 5
Also, I'd like to thank all of you for your effort explaining me these things. I really appreciated. Thanks a lot!!!



Post by sumgai on Jul 11, 2012 2:03:27 GMT 5
Looks like I better drum up two NoPrizes! ;D I wondered if anyone would catch the tripleerror, and out of two replies, two of the errors have been exposed. All in all, not a bad exercise in applied mathematics, eh?
Yew, as noted: a) You shouldn't use socalled imaginary numbers that lead to ambiguous results (1 and +1 both can be squared to the same answer);* b) You can't multiply two negative numbers and come up with another negative number.... at least not in this universe; and c) You can't derive the root of a negative number, because division obeys the same rule about multiplying two negative numbers. But even so, I have to say that it's obvious that school is doing you some good  you've come a long ways since you first arrived here in The NutzHouse! ;D sumgai * Shortcuts are fine, so long as they are spelled out either explicity or implicitly. When used in vector analysis, the letter j is usually understood to mean a linear quantity comprised of two components, the x and y coordinates on a graph. In general, these two coordinates are also taken to be positive (in the first quadrant) because they represent absolute values of some quantity that can be observed and measured. Calculated values are fine, and can certainly be used to modify j, but in general, j should not be expressed as a factored quantity.



Post by cynical1 on Jul 11, 2012 7:21:33 GMT 5
a) You shouldn't use socalled imaginary numbers that lead to ambiguous results (1 and +1 both can be squared to the same answer);* b) You can't multiply two negative numbers and come up with another negative number.... at least not in this universe; and c) You can't derive the root of a negative number, because division obeys the same rule about multiplying two negative numbers. It appears somebody has never seen how Government works from the inside... I was in a meeting yesterday where two of those three very points detailed above were proclaimed to be inconsequential to the decision making process and summarily disregarded. Take heart, Yew...you have a future in Government... Vote early, vote often. HTC1



Post by reTrEaD on Jul 11, 2012 11:11:03 GMT 5
Selected phrases that (imho) get right to the heart of the matter: It's really a matter of voltage division. The reactance of the pickup at the frequency in question is the "top half" of the divider. The resistance of the variable resistor plus the reactance of the cap are the "bottom half". Behind the 'note' which is 'fundamentally' the naming note..ie, the fundamental say, A=440.
But behind that are a whole series of harmonics. Let's go back to the OP and see what else we might elaborate on. You see, just silly questions. Probably i am just dumb... sorry Not silly and I think you know it wasn't. Just a matter of putting all the pieces in the right places. You did your homework and made a concerted effort. But you had a few misconceptions. Let's say that the pickup has somewhat flat response, I mean, a regular pickup would be flat for frequencies below 3000Hz, isn't it?... Not so much, really. The frequency response of a pickup often is not flat below 3kHz. We set up spice models that look that way, but in reality there is a ton more happening in a pickup than a source, a single internal resistance, inductance, and capacitance. Still, the simplistic model that we use functions reasonably well. We can get a decent approximation of how external loading will affect the output. So , I pluck the A note on high E, it will vibrate @ 440Hz... so the overall impedance of the tone circuit would be 500K+7,6K=507K with the pot @100%. But if I roll back the pot to 0%, it would be only 7,6K resistance to ground. And I will hear a bassy tone, it will "bleed" the high frequencies... Ok, but WHICH "high frequency"? I mean, the string is vibrating @ 440Hz, so the signal that the pickup is emitting is fundamentally 440Hz also. What frequency are we talking about? I think the discussion above regarding harmonics does a nice job of handling this part. The affect on the fundamental (440hz) will be noticeable, but not severe. We haven't set a specification for the impedance of the pickup. But I expect it might be similar to the 7.7k reactance of the 47nF cap. If equal, this would result in a 50% voltage division. Half voltage is not half volume. Half voltage would be 6dB. Another strange question is this: So suppose we have a volume pot, also 500K, and the tone pot is at 0%, so the signal will have a significant easier path to ground through the tone pot. I mean, just roll off the volume pot to say 70%, the signal will face 150K resistance at this pot. But only 7,6K in the tone pot. Shouldn't the signal be affected? I mean, shouldn't the signal volume drop significantly? It all depends on where the tone control is located. If it's before the volume control, your math doesn't apply. But if the tone pot is located after the volume control, you are correct. The signal at 440Hz will be affected drastically. Much more than the 6dB mentioned above. In this case, the impedance of the pickup and the resistance (350k) of the lower portion of the volume control (in parallel with the tone cap 7.7k) are minor factors. Since they aren't as significant, we'll ignore them for now and just look at the series resistance (150k) of the upper portion of the volume control and the reactance (7.7k) of the tone cap. 7.7k / (150k + 7.7k) = 0.049 or roughly 5%. This is a very serious voltage division, resulting in a 26.2dB attenuation. At this setting of the volume control, the tone control would attenuate the fundamental so it would appear only 1/4 the volume, compared with having the tone control before the volume control. This interaction, where the position of the volume knob drastically affects the effectiveness of the tone control is annoying and (imho) avoided at all costs. If you only have three holes for controls and want two volumes, you might think you're stuck with having a single tone control that by definition must come after the volume controls. But you aren't. The tone control could be either a ganged (two elements controlled by a single knob) or a concentric (two elements each with a separate knob, upper and lower). That way, the tone control(s) can be connected before the volume controls. So Zed, +1 for a couple of 'stupid' questions that turned out to be not at all stupid. And led to a useful discussion.



Post by JohnH on Jul 13, 2012 16:35:01 GMT 5
Complex numbers are fun aren't they?
I understood them about 20% when i was 18, forgot them entirely for the next 30 years, then got back up to about 25%, enough go make the GuitarFreak spreadsheet (see reference section). They give you a graphical way to get your head around things such as voltage drops across two points in LRC circuits.
Series combinations of impedances become easily represented on a graph, and solved by geometry,adding the i and j terms. The thing i had to hunt for and think about harder was how to do parallel combinations. Once you have that, you can solve most networks, such as in the spreadsheet. If you open it up and unhide the maths, you can follow it as It goes from left to right,from pickup to output
John

