in phase/out of phase explanation

[feral]

Hi guys

I did a search of this forum but couldn't find a straight ahead primer on in phase/out of phase. A student of mine asked me what it meant the other day and I said... ummm. Well, if two speakers are out of phase it means that the positive and negative leads are reversed at the inputs and one speaker is travelling away from the magnet while the other is travelling back towards it, thus tending to cancel each other out. But what does it mean for guitar pickups?

The vibrating guitar string causes a disturbance in the magnetic field; this disturbance causes an electrical signal to be transmitted from the pickups. But how would phase function here? The string's oscillation isn't an on/off thing - it is constantly creating a fluctuation in the magnetic field, so I am not sure how that could be seen as '+ve' and '-ve'... unless the string's approach to the +ve pole of the magnet is one 'phase' and the motion toward the -ve pole is the other. Is this what is meant by phase as it relates to pickups? So if two pickups have magnets that are oriented identically, and one pickup is mounted rotated 180 degrees from the other, then the motion of the string out to one end would be towards the +ve pole in one pickup and the -ve pole in the other. Is this what is meant by out of phase? But a phase switch doesn't flip the pickup around in it's mounting bracket, does it...so maybe this can be done electrically by switching the direction of flow through the coils of an active pickup...but what about passive pickups??

Also, in trying to get an answer to this via Google, I saw something about one of the Strat pickup positions being known as 'out of phase' but in fact it isn't really out of phase.

Any help is appreciated guys, thanks.

 

[Artie]

Re: in phase/out of phase explanation

It is pretty much the same thing as in a loudspeaker. Lets say you pull the string back, and let it snap forward. As it moves across the magnetic field, it starts to generate a voltage. If you have two pickups, they'll both generate a positive going voltage. (It could be either polarity. I'm assuming positive to start with.) As the string approaches its opposite excursion, it slows down, stops, reverses, and comes back to its initial position. If you look at this on a o'scope, the waveform will (more or less) follow the string movement.

There's at least three different ways to reverse the phase of one pickup relative to the other. Simply reverse the wires, reverse the magnet, or wind the bobbin the opposite direction. Do two of those things, and you're back to where you started. Thats why a reverse wound/reverse phase (RW/RP) middle pickup will be in-phase with another.

Also, since the actual signal picked up by both pickups won't be identical, due to pickup differences and position on the guitar, you really only have a phase condition with the fundamental frequency. The harmonics won't tend to cancel, and so you'll always get some sound even when they're out-of-phase. Like in the case of loudspeakers, you really only lose the bass.

Hope that helps.

Artie

 

[feral]

Re: in phase/out of phase explanation

It is pretty much the same thing as in a loudspeaker. Lets say you pull the string back, and let it snap forward. As it moves across the magnetic field, it starts to generate a voltage. If you have two pickups, they'll both generate a positive going voltage. (It could be either polarity. I'm assuming positive to start with.) As the string approaches its opposite excursion, it slows down, stops, reverses, and comes back to its initial position. If you look at this on a o'scope, the waveform will (more or less) follow the string movement.

There's at least three different ways to reverse the phase of one pickup relative to the other. Simply reverse the wires, reverse the magnet, or wind the bobbin the opposite direction. Do two of those things, and you're back to where you started. Thats why a reverse wound/reverse phase (RW/RP) middle pickup will be in-phase with another.

Also, since the actual signal picked up by both pickups won't be identical, due to pickup differences and position on the guitar, you really only have a phase condition with the fundamental frequency. The harmonics won't tend to cancel, and so you'll always get some sound even when they're out-of-phase. Like in the case of loudspeakers, you really only lose the bass.

Hope that helps.

Artie

That is great, Artie - thanks for taking the time!

Nik

 

[jdm61]

Re: in phase/out of phase explanation

ok, Artie.....how does it works when the volumes of the 2 pcikups are set differently. What i have read on here is that the bass creeps back in sounding like the louder pickup. I am wanting to have the neck magnet on my new Seth anni's reversed. When Peter Green gets that real thin sound, was he just matching the volume of the two pickups? I always wondered where and how he was picking, not thinking it was also done with the volume knobs.

 

[hamerfan]

Re: in phase/out of phase explanation

Yes you have to match the volume of two pup to get the out of phase sound in full. The if you go down to 7 on one pup the influence of this pups coil goes down so that the frequency chancellation ability is minimal.

 

[75lespaul]

Re: in phase/out of phase explanation

I was just messing with that last night with my Les Paul. I take it we are talking in part about the type of sound Jimmy Page made famous by using the middle position on his selector switch. When I lowered the volumes to get a cleaner tone, I had to play with one knob's volume just to get it right. a tiny bit louder or softer would eliminate the OOP sound and it would sound like either the bridge or neck PU alone. Cool stuff.

 

[feral]

Re: in phase/out of phase explanation

It is pretty much the same thing as in a loudspeaker. Lets say you pull the string back, and let it snap forward. As it moves across the magnetic field, it starts to generate a voltage. If you have two pickups, they'll both generate a positive going voltage. (It could be either polarity. I'm assuming positive to start with.) As the string approaches its opposite excursion, it slows down, stops, reverses, and comes back to its initial position. If you look at this on a o'scope, the waveform will (more or less) follow the string movement.

There's at least three different ways to reverse the phase of one pickup relative to the other. Simply reverse the wires, reverse the magnet, or wind the bobbin the opposite direction. Do two of those things, and you're back to where you started. Thats why a reverse wound/reverse phase (RW/RP) middle pickup will be in-phase with another.

Also, since the actual signal picked up by both pickups won't be identical, due to pickup differences and position on the guitar, you really only have a phase condition with the fundamental frequency. The harmonics won't tend to cancel, and so you'll always get some sound even when they're out-of-phase. Like in the case of loudspeakers, you really only lose the bass.

Hope that helps.

Artie

Hi again

Well you guys, i didn't want to look like an idiot, so I decided to wait a day or two before posting again. Unfortunately, I still don't understand what in phase/out of phase means.

Ok.. so when you strike a string it oscillates - when you say it generates a 'positive going voltage'... you mean a consistently positive voltage? Or does the voltage itself oscillate with the string? I am not an electrical engineer as you can tell.

When you say 'it could be either polarity'... what could be either polarity?

When you say it's easy to reverse the polarity'... assuming I understand what you mean by 'the polarity' (which I don't), you say that you can reverse the wires. Which wires? Do you mean the leads from the pickup? So the pickup leads run into the volume pots, then through any active electronics, then out the output plug, right?

Are two pickups which are out of phase the same as two pickups whose polarity has been reversed? Since the magnets are fixed in the pickups, can polarity be reversed by manipualting the way the current flows thorugh them? Is this what you mean when you say that one way to put two pickups out of phase is to reverse the direction of the windings?

Ok, so the output of two in-phase pickups, when put on an o'scope, will mirror the motion of the string. The you reverse the polarity of one of the pickups. Since the motion of the string does not provide an 'on-off' type signal, I am not sure I understand how, in this case, one signal would cancel another. There is no qualitative difference between the signal being generated at the 'north' end of it's excursion and the 'south' end.

I guess I'm hopeless but I still don't get it.

Any further help wold be great.

 

[zionstrat]

Re: in phase/out of phase explanation

I'm always nervous follwing Artie in as he's so good at this, but I'll give it a try..


Ok.. so when you strike a string it oscillates - when you say it generates a 'positive going voltage'... you mean a consistently positive voltage? Or does the voltage itself oscillate with the string? I am not an electrical engineer as you can tell.

The point is you create a waveform that we will call positive in this example. Of course it will really bounce around both sides of 0 db, but let's just call this 'positive'.

When you say 'it could be either polarity'... what could be either polarity?
Doesnt matter- Call one positive and the opposite negitive.

So for the fun of it, let's take the output from a single pup and devide it in 2- Then let's combine them back together down the signal chain, but invert the 'positive' and 'negitive' leads of one of the 2 signal chains. In this case the pup signal is being combined with exactly the opposite signal and everything will be canceled out- Imagine the wave form-

Where the wave is going up in the graph for this pup, the copy that is inverted will be going down exactly the same amount so they cancel each other out.


When you say it's easy to reverse the polarity'... assuming I understand what you mean by 'the polarity' (which I don't), you say that you can reverse the wires. Which wires? Do you mean the leads from the pickup? So the pickup leads run into the volume pots, then through any active electronics, then out the output plug, right?

A single coil pup has 2 wires- Simply invert them- A humbucker can have up to 4 conductors, but you would simply swap the "in and out" leads.

Are two pickups which are out of phase the same as two pickups whose polarity has been reversed? Since the magnets are fixed in the pickups, can polarity be reversed by manipualting the way the current flows thorugh them? Is this what you mean when you say that one way to put two pickups out of phase is to reverse the direction of the windings?

yep think you got it-

Ok, so the output of two in-phase pickups, when put on an o'scope, will mirror the motion of the string. The you reverse the polarity of one of the pickups. Since the motion of the string does not provide an 'on-off' type signal, I am not sure I understand how, in this case, one signal would cancel another. There is no qualitative difference between the signal being generated at the 'north' end of it's excursion and the 'south' end.

This is the example I gave above- Excluding the difference because 2 physical pups will actually be 'looking' at differnt parts of the string, two out of phase pups (by any of the 3 methods that ARtie listed) will register exactly oppositve voltages at any given time so the average signal is 0-

THis help?


Any further help wold be great

 

[feral]

Re: in phase/out of phase explanation

I'm always nervous follwing Artie in as he's so good at this, but I'll give it a try..

Thanks - there is no doubt in my mind that I'm just not smart enough to learn from Artie's explanation

The point is you create a waveform that we will call positive in this example. Of course it will really bounce around both sides of 0 db, but let's just call this 'positive'.

So for the fun of it, let's take the output from a single pup and devide it in 2- Then let's combine them back together down the signal chain, but invert the 'positive' and 'negitive' leads of one of the 2 signal chains. In this case the pup signal is being combined with exactly the opposite signal and everything will be canceled out- Imagine the wave form-

Where the wave is going up in the graph for this pup, the copy that is inverted will be going down exactly the same amount so they cancel each other out.

Ok.. thanks for trying. I guess I am going to have to go with what I have, Maybe I am trying to make this too compcated. I understand the analogy between the peaks and troughs on the o'scope cancelling themselves out and i.e. two out of phase speakers cancelling each other out - it's easy to visualize. So it's as simple as saying that the signals from two out of phase pickups are mirror images of each other - the fact that the voltage is in a 'negative' phase or a 'positive' phase coming from one or the other pup doesn't effect the sound being produced by either one - either one would produce identical sounds if they were running on their own. So at the exact moment when the first signal is transmitted down the wire as the string starts to vibrate, the oscillation from one pup starts on one 'side' of the peak-trough cycle and in the othe out of phase pup, the signal start sout on the opposite 'side' of the peak-trough cycle.

If this is correct, I can at least start to talk about it with my student, who is a bright spark and will sense any weakness in my explanation and ask me a pointed question about it.

I wonder if that is close.

wait a minute... ok, the oscillation of the string isn't 'on' or 'off' - it is an oscillation. Could it be that the oscillation in the voltage is the same thing? i.e. that the oscillation of the voltage doesn't.... AHA!!! Ok, I think I have figured it out - the oscillation in the voltage doesn't produce a qualitative difference in the sound, it is just an exact analog of the oscillation of the string, and will cause a commensurate oscillation in whatever the output of the pickup is hooked up to!!

AHA!

 

[zionstrat]

Re: in phase/out of phase explanation

yep, that's right- Very analog as opposed to digital- The voltage represents the current 'position' of the string (in very loose terms) and the number of times the voltage cycles through 0 in a second gives you the fundamental frequency or a harmonic if you sample at a node-
cheers

 

[feral]

Re: in phase/out of phase explanation

yep, that's right- Very analog as opposed to digital- The voltage represents the current 'position' of the string (in very loose terms) and the number of times the voltage cycles through 0 in a second gives you the fundamental frequency or a harmonic if you sample at a node-
cheers

That's great - sorry for the delay in getting back to you, and thanks for taking the time.

My student now asked me -'What is an example of a solo or song which is clearly played with out of phase pickups?' So I am off to search the forums for an example. If anyone wants to point me in the right direction, that would be great.