KrisBlueNZ
Sadly passed away in 2015
Don't worry about j. The important part of the formulas are:Wikipedia article on impedence says that impedence of capacitor is 1/jwC. I don't know what the j is. Wikipedia article on impedence says that impedence of an inductor is jwL.
XC = 1 / w C
XL = w L
where w is 2 pi f (and f is frequency in hertz).
So those formulas can also be written
XC = 1 / (2 pi f C)
XL = 2 pi f L
where f is frequency in Hz, C is capacitance in farads, and L is inductance in henrys.
No, it's the frequency, in radians per second. The frequency in radians per second is equal to 2 pi multiplied by the frequency in hertz. In your case it's easier to deal with frequencies expressed in hertz (cycles per second), instead of radians per second, so the second pair of formulas above is probably more useful to you.I'm assuming that w is the period in seconds of the sinusoidal AC wave.
Yes. That's because the reactance of a capacitor (Xc) or an inductor (XL) at any particular frequency does not behave like a simple resistance. With a resistor, the current at any moment in time is proportional to the voltage at that moment in time. So if you apply an alternating voltage across a resistor, and measure the current flow in that resistor, you will see a waveform that's identical to the voltage waveform, and is exactly in phase with it. When you're dealing with a capacitance, or an inductance, the current waveform is not in phase with the voltage waveform; the component has reactance, not resistance. I won't go into any more detail at the moment.So those phasor diagrams are designed to allow engineers (and physics students) to use algebra instead of have to solve differential equations, according to Wikipedia.
Perhaps because it's harder to define the impedance of the pickup. It's easy to measure or calculate the DC resistance, but other characteristics depend at least partly on the guitar strings, as well as the pickup. For example the inductance can be measured fairly accurately with no strings on the guitar, but when there's a vibrating string near the pickup, that previously measured inductance does not tell the full story. I'm at the limit of my knowledge here, so don't ask me for any more details, OK?I have no clue why guitar pickup manufactures specify the DC resistance of the pickup, but don't specify the inductance and impedence of the pickup. Maybee its more people like that RadioShack manager who don't have very high technical standards but are very good salesmen so they don't need to have high technical standards because they can fool people.
The DC resistance of the pickup is 7k. The volume pot's resistance is 500k. That's the end-to-end resistance. The resistance you see at the guitar's socket depends on the position of the pot. When the pot is at full volume, the wiper (which connects to the socket) is connected to the top end, which connects to the pickup. So you will measure 7k DC resistance. You can try this measurement if you want. The 500k resistance of the pot track will be in parallel with the 7k but will make little difference. When the pot is around position 7, where there is 250k of track resistance on each side of the wiper, the output resistance will be around 125k. Actually you can measure this with a multimeter on resistance range, connected to the guitar's output socket. You will find a position on the volume control where the resistance is at maximum. This will correspond roughly to the volume setting where the cable capacitance will have the greatest effect and will attenuate high frequencies the most.but why is the impedence of the volume pot 500k if its DC resistance is only 7k?
Right.So this makes me think that 300pF 10 foot long cable and 500k volume pot should already be a low pass filter that attenuates signals at a greater frequency than audio frequencies, and adding further capacitance would just give less of a treble response.
That won't help. My suggestions for avoiding Miley would be:So I could wrap my cables in aluminum foil to block out the RF, since we have established that adding more capacitance to a circuit with high capacitance won't do anything other than kill high end response; but then that would just increase the capacitance and kill the high end response, since aluminum foil wrapped around an insulated conductor is I have built a capacitor.
1. Clean all exposed metal parts of all plugs, and the contact points of all sockets, with isopropyl alcohol and a cotton swab. Press hard. You can use steel wool or emery paper to remove any corrosion but this may cause the metal to corrode again, more quickly. If in doubt, replace the plug or socket.
2. Make sure that cables are tidily and cleanly terminated in all plugs. Make sure solder is bright, and individual strands of wire are not frayed or broken.
3. Make sure that the screening in the cable is tightly woven. If it isn't, buy another cable. Also, borrow a good quality cable from a friend and see if that makes any difference. Also, use the shortest possible cable.
4. You may be able to modify your amplifier to reject RF on its input. But I don't want to encourage you to poke around in there. Let me know if those other fixes don't work.
Why does the problem only occur at night?So then my "project" is a waste of time, and there's nothing I can do to remove radio interference if I don't want my treble response to be affected other than (in order of lowest cost):
(1) don't record at night if don't want to record WBBM.
That's always a good idea.(2) instead of building a low pass filter to attenuate RF, build a noise gate that only allows the strong signals generated by the guitar strings vibrating in the magnetic field of the pickups to pass and attenuates weak noises such as RF noises.
What are you plugging your guitar into when you get this interference? Is it a guitar amp? Or are you going direct into a mixer or sound card?
Yeah, probably not LOL(3) install electric outlets into my basement and record in the basement (probably not a good idea for me to mess around with mains since I'm new to this).
You could try that.(4) build a one transistor short range super regenerative FM guitar transmitter and receiver tuned to a frequency not used for broadcasting and add in the damn low pass filter to attenuate RF in the radio receiver.
Good idea.(5) buy a noise gate.
Good idea.(6) buy a wireless guitar system to eliminate the capacitance of the cable altogether and make sure that the wireless receiver has a low pass filter to attenuate RF and use a 6 inch guitar cable to connect the wireless receiver to the amp.
Good idea.(7) buy a better amp that was designed to remove RF.
LOL(8) buy out Star 105.5 and WBBM and turn off their transmitters whenever I want to record.
Also... if there's room in the body of the guitar, add active electronics to it.
Also, make sure the metal body of the pickup(s), the strings, and any other metal components on the guitar are grounded to the socket metal. Check with a multimeter.
No, a decade is a ratio of 10 times. But a single pole filter attenuates at a rate of 3 dB per octave, not per decade. An octave is a frequency ratio of 2 times. So for every doubling in frequency, the signal is attenuated 3 dB more.Okay, I searched for that search term and I found that what your saying agrees with Wikipedia. The cutoff frequency does not literally stop all frequencies above or below it, it attenuates the frequencies at 3db per decade (I think decade means 1khz).
Dieter, you're one of a kind. Thank goodness!I was curious ... dipole antenna ... 3.4 MHZ ... FM stations ... green oxidized penny ... Led Zeppelin ... classic rock station ... it turns out you don't even need a guitar to make a radio ... I think I heard air traffic control too ... cleared to land on runway 9 left ...
OK, interesting.the inductance of guitar pickups ranges from 1 Henry for single coil pick ups to 4 Henry for high output humbucker pickups. More inductance increases the output of the pickup.
So is everyone else!I like being right. I'm used to everything I post on here is complete nonsense.
You mean a 1 henry pickup? And what is the actual resonant frequency?Your right! A 1 farad pickup ensures that the LC resonant circuit is in the audio frequency range.
Yes, Q is unitless. It stands for "quality factor" and is an indication of the sharpness of the peak in the response of a tuned circuit. A higher Q corresponds to a sharper peak, with steeper sides, and a narrower bandwidth.Okay, I Googled Q and bandwidth and I found that Google says that Q is a unit less concept that is the ratio of (2*pi*f*L ) / R .
Almost right. You've got the general idea.Oh...so at that position 7, the wiper is connecting the two resistances (the resistance of the pickup and the resistance of the pot) in parallel instead of in series, and in parallel will actually lower the resistance since its the reciprocal.
OK. That will be interesting to know.I'll tell you what the inductance of my pickups are when I find out.
When you know the pickup inductance, you will be able to calculate the resonant frequency for various cable capacitances, such as 300 pF.
The formula is
f = 1 / (2 pi sqrt(L C))
where f is the resonant frequency in Hz
sqrt() means the square root of the bit in parentheses
L and C are in henrys and farads respectively.