Re: Re: Re: Invoking SY's Second Law again
Well, since we have two of them we don't really have the why,
do we?
Yes, an experiment can show which one, if any of
them, is most consistent with reality. I didn't read the references
in this case, so I don't know if there were only claims or also
arguments. If we only have claims, an experiment may show
one claim to be plausible, but still does not tell us why it it
is true, if it is. If we have two arguments which are contradictory,
something is wrong with at least one of them. An experiment
can tell us which argument is obviously wrong, but not why it
is wrong. That still takes more thinking about why we haven't
realized it is wrong. The experiments can put us on the track
so we know which of the arguments to find the fault in.
Steve Eddy said:
Well, since we have two of them we don't really have the why,
do we?
Yes, an experiment can show which one, if any ofthem, is most consistent with reality. I didn't read the references
in this case, so I don't know if there were only claims or also
arguments. If we only have claims, an experiment may show
one claim to be plausible, but still does not tell us why it it
is true, if it is. If we have two arguments which are contradictory,
something is wrong with at least one of them. An experiment
can tell us which argument is obviously wrong, but not why it
is wrong. That still takes more thinking about why we haven't
realized it is wrong. The experiments can put us on the track
so we know which of the arguments to find the fault in.
Re: ECHOES...WITH A TWIST.
Yes, but the question is whether or not the induced voltage from magnetic fields has a common-mode component. There's no arguing about whether a balanced circuit will reject common-mode noise. It does.
On average they would be. But unless there is a common-mode component to the induced voltage from the magnetic field, it doesn't matter whether they're equidistant.
Yes. Twisting will work toward making each conductor on average equidistant from the source. And as Whitlock says, this will cause the induced voltage to be common-mode. But it still doesn't resolve the issue as to whether there is or can be any common-mode component to the induced voltage.
I don't know that it's proper to think in terms of send and return lines. What we have is a conductive loop. Which according to several references says that the voltage induced in the loop is a function of the amount of flux changing through the loop.
Where are you getting field A and field B from? Are you referring to two separate fields here?
I don't know that they can. One says they are, one says they aren't. If you're talking about two different fields above, I don't feel that the "statement in question" is referring to two different fields, but simply a varying field, as to distinguish it from a static field which wouldn't induce any voltage.
Well, it just seems that most every other reference states that the induced voltage will be differential.
For example, take symetrically round loop of wire and pass a magnet through its center such that each portion of the loop is equidistant from the magnet. Pretty much everything out there I've seen says that this will induce a voltage and subsequently a current that's proportional to the amount of flux changing in the loop and the loop area. The polarity of the induced voltage depending on which way the flux is changing and the polarity of the field.
For example:
That's how I read it. Whitlock says that induced voltages will be equal only if the two wires are equidistant from the source ("...equal voltages will be induced in the two wires of a balanced cable only if they are equidistant from the field source") and that twisting causes the voltages to be equal ("Twisting of the two wires is a first-order technique to make induced voltages identical by averaging the physical positions of the wires").
fdegrove said:
Yes, but the question is whether or not the induced voltage from magnetic fields has a common-mode component. There's no arguing about whether a balanced circuit will reject common-mode noise. It does.
On average they would be. But unless there is a common-mode component to the induced voltage from the magnetic field, it doesn't matter whether they're equidistant.
Yes. Twisting will work toward making each conductor on average equidistant from the source. And as Whitlock says, this will cause the induced voltage to be common-mode. But it still doesn't resolve the issue as to whether there is or can be any common-mode component to the induced voltage.
I don't know that it's proper to think in terms of send and return lines. What we have is a conductive loop. Which according to several references says that the voltage induced in the loop is a function of the amount of flux changing through the loop.
Where are you getting field A and field B from? Are you referring to two separate fields here?
I don't know that they can. One says they are, one says they aren't. If you're talking about two different fields above, I don't feel that the "statement in question" is referring to two different fields, but simply a varying field, as to distinguish it from a static field which wouldn't induce any voltage.
Well, it just seems that most every other reference states that the induced voltage will be differential.
For example, take symetrically round loop of wire and pass a magnet through its center such that each portion of the loop is equidistant from the magnet. Pretty much everything out there I've seen says that this will induce a voltage and subsequently a current that's proportional to the amount of flux changing in the loop and the loop area. The polarity of the induced voltage depending on which way the flux is changing and the polarity of the field.
For example:
That's how I read it. Whitlock says that induced voltages will be equal only if the two wires are equidistant from the source ("...equal voltages will be induced in the two wires of a balanced cable only if they are equidistant from the field source") and that twisting causes the voltages to be equal ("Twisting of the two wires is a first-order technique to make induced voltages identical by averaging the physical positions of the wires").
The Capgo reference as well as the page Fred just posted say that twisting works by alternately flipping the loop so that the voltages induced in the loop cancel such that ideally, there's no noise voltage to deal with at all, differential or common-mode
That's how I view it too and as I mentioned before, it works whether we use a balanced or unbalanced receiver which proves the point IMO.Proves which point? To me it seems to prove that there's no common-mode component so equidistance is a non-issue.
But if twisting takes care of the noise locally within the cable itself, it doesn't matter whether it's feeding a balanced or unbalanced input. So what's the point of balance with regard to noise induced into the cable?
According to most references out there, any change of flux within the area of a loop results in a differential voltage.
See, a balanced line only rejects common mode noise, NOT differentail noise, if it did reject that too the net result would be no signal at all.
Which could be a big plus depending on the signal.
se
Re: Re: Re: ECHOES...WITH A TWIST.
It's twisting which does both as far as I'm aware. I don't see how balance would have anything to do with it as whether balanced or not each conductor will be carrying equal and opposite currents producing equal but opposite fields. The twisting flips the polarity of the radiated field so that the field of one twist tends to cancel the field of the next so that on average the radiated field is significantly reduced.
And it seems the same thing is going on in reverse with respect to interference from magnetic fields. The twisting flips the loop around each twist so that the induced voltages are of opposite polarity and also tend to cancel between one twist and the other.
se
purplepeople said:
It's twisting which does both as far as I'm aware. I don't see how balance would have anything to do with it as whether balanced or not each conductor will be carrying equal and opposite currents producing equal but opposite fields. The twisting flips the polarity of the radiated field so that the field of one twist tends to cancel the field of the next so that on average the radiated field is significantly reduced.
And it seems the same thing is going on in reverse with respect to interference from magnetic fields. The twisting flips the loop around each twist so that the induced voltages are of opposite polarity and also tend to cancel between one twist and the other.
se
Re: MOTHER THERESA?
Why? That was never the issue.
The issue was why twisting works. Specifically, whether it's by way of keeping each conductor equidistant to the source and thereby causing the voltage induced in each conductor to be the same, i.e. common-mode as the Whitlock reference states, or is it due to local cancellation in the cable itself as the Capgo and several other references state.
That's the question to be answered.
Right now I'm working on a jig see if that question can be answered by way of some simple measurements. Perhaps I can post some results before the end of the day.
se
fdegrove said:A round of aplause, please...
Why? That was never the issue.
The issue was why twisting works. Specifically, whether it's by way of keeping each conductor equidistant to the source and thereby causing the voltage induced in each conductor to be the same, i.e. common-mode as the Whitlock reference states, or is it due to local cancellation in the cable itself as the Capgo and several other references state.
That's the question to be answered.
Right now I'm working on a jig see if that question can be answered by way of some simple measurements. Perhaps I can post some results before the end of the day.
se
Re: Re:Re:Re:Re:Re: Pffffffffffftttt.
So are you saying that there is no common-mode component to noise induced by magnetic fields? That it's differential and twisting, rather than serving to keep each conductor equidistant from the source so as to make the noise common-mode as Whitlock satates, that the noise reduction from twisting is purely due to local cancellation in the cable and not common-mode rejection at the receiver?
I mean, if it's local cancellation, period, then the common-mode rejection of the receiver is of no value with regard to noise induced by magnetic fields. Rejection is a function of the cable.
I have. And there are some seeming contradictions with regard to the Whitlock reference.
se
fdegrove said:
So are you saying that there is no common-mode component to noise induced by magnetic fields? That it's differential and twisting, rather than serving to keep each conductor equidistant from the source so as to make the noise common-mode as Whitlock satates, that the noise reduction from twisting is purely due to local cancellation in the cable and not common-mode rejection at the receiver?
I mean, if it's local cancellation, period, then the common-mode rejection of the receiver is of no value with regard to noise induced by magnetic fields. Rejection is a function of the cable.
I have. And there are some seeming contradictions with regard to the Whitlock reference.
se
Hi,
No, I never said that.
You really CAN NOT make induced EMI fully common mode when the source wasn't in the first place.( Please read this carefully)
Remember "varying" magnetic fields?
No amount of twisting is going to make that differential signal common mode and no common mode receiver is going to treat it any different.
I feel some people read Whitlock wrong amongst other things.
The receiver only treats differential signals, whatever is left common mode on the loop is rejected, be that RFI or EMI the balanced receiver doesn't care one iota about it.
The twisted cable does not entirely reject all common mode signals but it highly reduces pickup up of EMI and whatever is picked up is reduced to mostly insignificant levels and when it's differential not even the best balanced receiver is going to be able to reject it, the twisted wire can only do so much to reduce it but it's no miracle worker.
Although it does come close.
Comprende?
No, I never said that.
You really CAN NOT make induced EMI fully common mode when the source wasn't in the first place.( Please read this carefully)
Remember "varying" magnetic fields?
No amount of twisting is going to make that differential signal common mode and no common mode receiver is going to treat it any different.
I feel some people read Whitlock wrong amongst other things.
The receiver only treats differential signals, whatever is left common mode on the loop is rejected, be that RFI or EMI the balanced receiver doesn't care one iota about it.
The twisted cable does not entirely reject all common mode signals but it highly reduces pickup up of EMI and whatever is picked up is reduced to mostly insignificant levels and when it's differential not even the best balanced receiver is going to be able to reject it, the twisted wire can only do so much to reduce it but it's no miracle worker.
Although it does come close.
Comprende?
I don't have much in the way of formal training on these matters (too little, too long ago), but I'd say there are two cases of EMI cancellation:
- If the input impedance (or output impedance, for that matter) is infinite, we have no loop, and the wires work independently. Twisting serves by averaging the field lines through the wires, thereby creating common mode voltages which cancel at the inputs.
- If the input impedance (and output impedance) is zero, we have a perfect loop with the induced current depending on the flux density inside the loop and the angle of the loop, which twisting keeps alternating between 0 degreees and 180 degrees, thereby cancelling locally.
[/list=1]
The practical case would of course be somewhere in between. For loudspeaker connections the loop element will probably be dominant, whereas for interconnects the loopless element will be dominant, unless we're dealing with classical 600 ohm (transformer) interfaces.
In most audio applications - loudspeakers aside - there will be an additional common ground connection, so for a balanced connection we will effectively have three loops, two of which are made up of ground and each signal connector. In these loops there will be no alternating loop angle, but the induced voltages in the signal wires will be approximately the same, once they are twisted. Thus - a common mode voltage.
Rune
Sorry guys, I just couldn't resist since Fred has mysteriously
failed to post this.
Let's Twist Again
Come on everybody clap your hands
Now you're looking good
I'm gonna sing my song and you won't take long
We gotta do the twist and it goes like this
Come on let's twist again like we did last summer
Yea, let's twist again like we did last year
Do you remember when things were really hummin'
Yea, let's twist again, twistin' time is here
Yeah round 'n around 'n up 'n down we go again
Oh baby make me know you love me so then
Come on let's twist again like we did last summer
Yea, let's twist again, twistin' time is here
Come on let's twist again like we did last summer
Yea, let's twist again like we did last year
Do you remember when things were really hummin'
Yea, let's twist again, twistin' time is here
Yeah round 'n around 'n up 'n down we go again
Oh baby make me know you love me so then
Come on let's twist again like we did last summer
Yea, let's twist again, twistin' time is here
failed to post this.
Let's Twist Again
Come on everybody clap your hands
Now you're looking good
I'm gonna sing my song and you won't take long
We gotta do the twist and it goes like this
Come on let's twist again like we did last summer
Yea, let's twist again like we did last year
Do you remember when things were really hummin'
Yea, let's twist again, twistin' time is here
Yeah round 'n around 'n up 'n down we go again
Oh baby make me know you love me so then
Come on let's twist again like we did last summer
Yea, let's twist again, twistin' time is here
Come on let's twist again like we did last summer
Yea, let's twist again like we did last year
Do you remember when things were really hummin'
Yea, let's twist again, twistin' time is here
Yeah round 'n around 'n up 'n down we go again
Oh baby make me know you love me so then
Come on let's twist again like we did last summer
Yea, let's twist again, twistin' time is here
An old poem found in a Baltimore church
(This one is mainly for SY.)
Go placidly amid the noise and waste,
And remember what comfort
there may be in owning a piece thereof.
Avoid quiet and passive persons
unless you are in need of sleep.
Rotate your tires.
Speak glowingly of those greater than yourself,
And heed well their advice,
even though they be turkeys.
Know what to kiss and when.
Consider that two wrongs never make a right,
But that three lefts do.
Wherever possible put people on "HOLD".
Be comforted that in the face of all aridity and disillusionment,
And despite the changing fortunes of time,
There is always a big future in computer maintenance.
Remember the Pueblo.
Strive at all times to bend, fold, spindle and mutilate.
Know yourself.
If you need help, call the FBI.
Exercise caution in your daily affairs,
Especially with those persons closest to you;
That lemon on your left for instance.
Be assured that a walk through the ocean of most souls,
Would scarcely get your feet wet.
Fall not in love therefore;
It will stick to your face.
Carefully surrender the things of youth:
Birds, clean air, tuna, Taiwan,
And let not the sands of time get in your lunch.
For a good time, call 606-4311.
Take heart amid the deepening gloom that your dog
Is finally getting enough cheese;
And reflect that whatever fortunes may be your lot,
It could only be worse in Milwaukee.
You are a fluke of the Universe.
You have no right to be here,
And whether you can hear it or not,
The Universe is laughing behind your back.
Therefore make peace with your God whatever you conceive him to be,
Hairy Thunderer or Cosmic Muffin.
With all its hopes, dreams, promises, and urban renewal,
The world continues to deteriorate.
Give up.
(This one is mainly for SY.)
Go placidly amid the noise and waste,
And remember what comfort
there may be in owning a piece thereof.
Avoid quiet and passive persons
unless you are in need of sleep.
Rotate your tires.
Speak glowingly of those greater than yourself,
And heed well their advice,
even though they be turkeys.
Know what to kiss and when.
Consider that two wrongs never make a right,
But that three lefts do.
Wherever possible put people on "HOLD".
Be comforted that in the face of all aridity and disillusionment,
And despite the changing fortunes of time,
There is always a big future in computer maintenance.
Remember the Pueblo.
Strive at all times to bend, fold, spindle and mutilate.
Know yourself.
If you need help, call the FBI.
Exercise caution in your daily affairs,
Especially with those persons closest to you;
That lemon on your left for instance.
Be assured that a walk through the ocean of most souls,
Would scarcely get your feet wet.
Fall not in love therefore;
It will stick to your face.
Carefully surrender the things of youth:
Birds, clean air, tuna, Taiwan,
And let not the sands of time get in your lunch.
For a good time, call 606-4311.
Take heart amid the deepening gloom that your dog
Is finally getting enough cheese;
And reflect that whatever fortunes may be your lot,
It could only be worse in Milwaukee.
You are a fluke of the Universe.
You have no right to be here,
And whether you can hear it or not,
The Universe is laughing behind your back.
Therefore make peace with your God whatever you conceive him to be,
Hairy Thunderer or Cosmic Muffin.
With all its hopes, dreams, promises, and urban renewal,
The world continues to deteriorate.
Give up.
Ok, did some playing around today.
First, I wanted a nice big loop with which to better produce a strong enough field to get some measurements. So I took a scrap of some MDF I had laying around and using some insulated terminal posts, staked out the four corners of a rectangle, 10-1/5" x 18".
I then took some insulated, 24 gauge solid core copper wire and wrapped it once around each post forming a rectangular loop. I'd put two additional posts close together in the middle of one of the 18" sides and from there I twisted the two ends of the loop together to reduce any interference that might be caused by the leads to the power source.
Then I cut some strips of walnut, 7/8" wide, 10-1/5" long and 3/8" thick. Around the edge I taped some of the same 24 gauge wire I used for the loop to create another, smaller loop that could be moved about while measuring.
To drive current through the loop, I initially used a 9V AC, 800mA wall wart with an 8 ohm resistor in series with each leg.
I connected the 'scope to the test loop and placed it perpendicular to one of the 10-1/5" ends of the current loop. Even with the 'scope at 2mV/div and the probe at 1:1, I couldn't see much other than noise.
Plan B.
Since the induced voltage is a function of frequency, I took my little JVC FS-2000 "executive" system from my desktop, generated a 20kHz full scale sinewave with GoldWave and burned it onto a CD.
I snipped off some of the cable that was running from the wall wart to the resistors and current loop, stripped them, and plugged them into the speaker outputs of the FS-2000.
I fired up the CD and with the somewhat higher output voltage from the FS-2000 and moreso the 20kHz sinewave as opposed to the 60Hz wave from the wall wart, placing one side of the movable loop cose to the current loop, I got a good 12mV peak sinewave.
Positioning the movable loop such that each half of the loop was the same distance from the one leg of the current loop I found that I could null the voltage. Since this could be the case either for there simply being no induced voltage or each half of the loop having a common-mode voltage, I modified the movable loop.
I took two pair of 47.5 ohm resistors and put two in series at each end of the movable loop.
First I measured across both resistors at one end and found that I could still null the output.
With the output nulled across both resistors, I moved the 'scope probe to the center tap between the two resistors and the other end of one of them. I still got no signal. I swithed the probe tip to the opposite resistor and got the same thing. Nothing.
This would indicate that there is no common-mode voltage across the two conductors. That there's either some differential voltage or there's no voltage. If there were a common-move voltage across the two conductors, I'd have expected to see signal measured across each resistor.
I then repositioned the movable loop to get the maximum signal across both resistors. This being the case mentioned above with one side of the movable loop close to the one leg of the current loop.
Then I measured across each resistor. Across each resistor I measured half the voltage I'd measured across both resistors. Again, indicating that there is no common-mode component, only differential.
Finally, I ran a wire down the center of the movable loop, connecting the center taps of the two resistor pairs. It was a bit more fussy getting it to null because there were now two loops, but ultimately it behaved as the others had.
So there you have it. Make of it what you will. Or better yet, try it for yourself and see what sort of results you get.
As it stands, I'd have to say that the Whitlock reference is incorrect.
se
First, I wanted a nice big loop with which to better produce a strong enough field to get some measurements. So I took a scrap of some MDF I had laying around and using some insulated terminal posts, staked out the four corners of a rectangle, 10-1/5" x 18".
I then took some insulated, 24 gauge solid core copper wire and wrapped it once around each post forming a rectangular loop. I'd put two additional posts close together in the middle of one of the 18" sides and from there I twisted the two ends of the loop together to reduce any interference that might be caused by the leads to the power source.
Then I cut some strips of walnut, 7/8" wide, 10-1/5" long and 3/8" thick. Around the edge I taped some of the same 24 gauge wire I used for the loop to create another, smaller loop that could be moved about while measuring.
To drive current through the loop, I initially used a 9V AC, 800mA wall wart with an 8 ohm resistor in series with each leg.
I connected the 'scope to the test loop and placed it perpendicular to one of the 10-1/5" ends of the current loop. Even with the 'scope at 2mV/div and the probe at 1:1, I couldn't see much other than noise.
Plan B.
Since the induced voltage is a function of frequency, I took my little JVC FS-2000 "executive" system from my desktop, generated a 20kHz full scale sinewave with GoldWave and burned it onto a CD.
I snipped off some of the cable that was running from the wall wart to the resistors and current loop, stripped them, and plugged them into the speaker outputs of the FS-2000.
I fired up the CD and with the somewhat higher output voltage from the FS-2000 and moreso the 20kHz sinewave as opposed to the 60Hz wave from the wall wart, placing one side of the movable loop cose to the current loop, I got a good 12mV peak sinewave.
Positioning the movable loop such that each half of the loop was the same distance from the one leg of the current loop I found that I could null the voltage. Since this could be the case either for there simply being no induced voltage or each half of the loop having a common-mode voltage, I modified the movable loop.
I took two pair of 47.5 ohm resistors and put two in series at each end of the movable loop.
First I measured across both resistors at one end and found that I could still null the output.
With the output nulled across both resistors, I moved the 'scope probe to the center tap between the two resistors and the other end of one of them. I still got no signal. I swithed the probe tip to the opposite resistor and got the same thing. Nothing.
This would indicate that there is no common-mode voltage across the two conductors. That there's either some differential voltage or there's no voltage. If there were a common-move voltage across the two conductors, I'd have expected to see signal measured across each resistor.
I then repositioned the movable loop to get the maximum signal across both resistors. This being the case mentioned above with one side of the movable loop close to the one leg of the current loop.
Then I measured across each resistor. Across each resistor I measured half the voltage I'd measured across both resistors. Again, indicating that there is no common-mode component, only differential.
Finally, I ran a wire down the center of the movable loop, connecting the center taps of the two resistor pairs. It was a bit more fussy getting it to null because there were now two loops, but ultimately it behaved as the others had.
So there you have it. Make of it what you will. Or better yet, try it for yourself and see what sort of results you get.
As it stands, I'd have to say that the Whitlock reference is incorrect.
se
fdegrove said:
Ok, I've read it carefully. And I've no idea what you mean when you speak of the source of the magnetic field being "common-mode." Explain.
Yes, it just means a magnetic field which changes over time, as opposed to a static field which doesn't. It's only a changing field which induces voltage, not a static field.
What differential signal? Again, I don't understand what you mean about the source of a magnetic field being common-mode or differential.
Yes.
The Capgo and other references don't say that the cable rejects common-mode signals. What they say the cable does is cancel differential signals. The differential voltage at one twist will be of opposite polarity of the differential voltage at the next twist. It's only the differential input which cancels common-mode signals.
And by the way, are you using "signals" to mean noise and not the signals that are being transmitted over the line?
By the way, I forgot to mention that I also did a comparison between a parallel pair and a twisted pair.
I laid two lengths of wire side-by-side as close together as their insulation would allow. The insulation's about 0.01" thick and the conductors are 0.02" in diameter, so their center-to-center spacing was about 0.04".
I took the same length of wire and put about 70 twists in it which comes to about 7 twists per inch.
I laid the parallel pair down such that one wire was closest to the source and the other was farthest. Laying down the twisted pair along the same axis as the parallel pair, I measured about a 6dB difference between the two, i.e. the voltage induced into the twisted pair was about 6dB lower than the parallel pair.
se
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