I've applied the De-mud mod to one of my guitars and was thinking about the how/why of it recently. Reading Wikipedia, I'm getting the impression it's a simple first order low-cut/high-pass filter, though I don't understand what the "first order" bit means.
Here's the diagram of a such a filter from Wikipedia, and ArtieToo's diagram for comparison, noting that the volume pot acts at the resistor to ground, even though it's not shown in Artie's diagram. (edit, found a new image of a high pass filter. Ignore the cap and resistor values.)
First thing is that since the volume pot acts as the resistor to ground, the amount of resistance is going to change as the volume is turned up and down, thus effecting the cuttoff frequency. Per Wikpedia, the formula cutoff frequencey of a High pass filter is f = 1/2*Pi*t, or f = 1/2*Pi*R*C, where f is the cutoff frequency in hertz, R is the resistance in Ohms and C is the capacitance in Farads. If my math is right (and if it isn't, please point out what I did wrong), the cutoff frequency of a De-mud mod with a 500K volume pot all the way up should be a 318.31hz. Now if the pot is turned down halfway, assuming it's a pot with a perfect linear taper, that gives us 250k resistance to ground, giving us a cutoff frequency of 636.62hz. Now, I realize pots aren't going going to measure perfectly their rated value in Ohms, nor is it going to have a perfect taper, liner or audio(logarithmic). But the general idea remains the same: the De-mud mod is going to cut out more and more bass as the volume pot is turned down. I don't know if that's an issue or not, but I guess it could cause the bass to get trimmed off too much making the lower volume setting sound thin if you used the wrong capacitor value.
Looking at Artie's diagram, what I'm trying to figure out is what the resistor parallel to the capacitor, but in series to the signal is supposed to do, if anything. Does that some how change the resistance to ground, even though it's wired in series with the pickup? If so, does the pickup's own resistance count for anything as far as the filter is concerned, since the other end of the pickup is connected to ground, and if so, that messes up my math above, doesn't it?
Here's the diagram of a such a filter from Wikipedia, and ArtieToo's diagram for comparison, noting that the volume pot acts at the resistor to ground, even though it's not shown in Artie's diagram. (edit, found a new image of a high pass filter. Ignore the cap and resistor values.)
First thing is that since the volume pot acts as the resistor to ground, the amount of resistance is going to change as the volume is turned up and down, thus effecting the cuttoff frequency. Per Wikpedia, the formula cutoff frequencey of a High pass filter is f = 1/2*Pi*t, or f = 1/2*Pi*R*C, where f is the cutoff frequency in hertz, R is the resistance in Ohms and C is the capacitance in Farads. If my math is right (and if it isn't, please point out what I did wrong), the cutoff frequency of a De-mud mod with a 500K volume pot all the way up should be a 318.31hz. Now if the pot is turned down halfway, assuming it's a pot with a perfect linear taper, that gives us 250k resistance to ground, giving us a cutoff frequency of 636.62hz. Now, I realize pots aren't going going to measure perfectly their rated value in Ohms, nor is it going to have a perfect taper, liner or audio(logarithmic). But the general idea remains the same: the De-mud mod is going to cut out more and more bass as the volume pot is turned down. I don't know if that's an issue or not, but I guess it could cause the bass to get trimmed off too much making the lower volume setting sound thin if you used the wrong capacitor value.
Looking at Artie's diagram, what I'm trying to figure out is what the resistor parallel to the capacitor, but in series to the signal is supposed to do, if anything. Does that some how change the resistance to ground, even though it's wired in series with the pickup? If so, does the pickup's own resistance count for anything as far as the filter is concerned, since the other end of the pickup is connected to ground, and if so, that messes up my math above, doesn't it?
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