Bi wiring/single wire distortion thread - Page 3 - diyAudio
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 28th December 2005, 10:34 PM #21 diyAudio Member   Join Date: Sep 2002 Location: Sweden Something that is bothering me is that the filters are ignored. They are assumed to exist, so the the signal splits between the branches, but neither the filters nor their effects are modelled in the equations. I have been thinking quite a lot about this during the past days (at least there is one good thing with insomnia ),but I still haven't figured out quite what this means and how it affects the whole thing. Have you considered the phase of voltages and currents in the branches, for instance?
 29th December 2005, 06:02 AM #22 diyAudio Member   Join Date: Feb 2003 Location: .. is Parseval's theorem related to your question? also Christer is on to another possible issue - the biwire situation is not an exact equivalent to "mono" circuit by some "inverse superpostion " - they describe different linear systems with (slightly) dfferent linear responses
 29th December 2005, 07:43 AM #23 diyAudio Member   Join Date: Dec 2005 Location: Kuala Lumpur Two more possible causes: 1) A speaker cable carrying current I and with a spacing d has a repulsion u.I.I/2.pi.d N/m. At 1A and d=2mm this is 10-4 N/m, which might be enough to cause the wires to move apart in the insulation. If they move then work has been done and energy absorbed. 2) A speaker cable with current flowing is a moving coil microphone. The sound from the speakers may be picked up. Both of these effects depend on the current flowing and will be reduced by biwiring. Their significance is another matter.
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Quote:
 Originally posted by poobah John, Look at your equation: A^2 + 2 AB + B^2 It ain't legal to pull out the A^2 term and look at it by itself... throw in a sinewave for A and a constant for B... boil it all down and then look at A.
I have been thinking of it this way. (currents) A being AC, B being DC.

The A2 term is a result of signal A. Alone, the cable will dissipate A2Rc watts, and the load will dissipate A2RloadA watts.

The B signal will do the same by itself, producing B2Rc watts, and the load will dissipate B2RloadB watts.

When the signals are combined, each load should see the exact same dissipation it saw independently. However, the cable resistance now sees BOTH currents flowing in it at the same time. That means, the equation for the cable dissipation now has to be Rc * (A + B)2 watts. This produces the expected A2Rc term associated with the A signal and A load, the B signal term B2Rc, and as a result of the math, a 2AB Rc term..

The 2AB term of dissipation within the cable is a different time dependent function than the squared terms in the loads. Where did that time dependent power come from? The entire system power dissipation cannot change as a result of the one or two wire case, they should be equivalent as far as the amplifier can tell, the loads should dissipate the exact same thing, but yet, that 2AB term is there, dissipating at the cable..

Therein lies the issue..

Cheers, John
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Quote:
 Originally posted by Christer Something that is bothering me is that the filters are ignored. They are assumed to exist, so the the signal splits between the branches, but neither the filters nor their effects are modelled in the equations. I have been thinking quite a lot about this during the past days (at least there is one good thing with insomnia ),but I still haven't figured out quite what this means and how it affects the whole thing. Have you considered the phase of voltages and currents in the branches, for instance?
The reactances are completely ignored for this thought experiment. By using DC for one signal, and a high frequency for the AC, the L and C should be sufficiently low in reactance to be considered as shorts for their respective signals.

A real analysis using real components of course will be needed for testing this hypo (as phase accurate is doing), but we only get there if this infernal 2AB cannot be explained using current understanding...if it is explained away, then there's nuttin to do...is there?

However, if harmonics are generated, we have some thinking to do, don't we?

I love this stuff.

Cheers, John
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Quote:
 Originally posted by jcx is Parseval's theorem related to your question?
Good question.

Related only in that it cannot be violated. The crux of my concern is because of the conservation of energy. Since the sum dissipates more energy in the cable than the separate dissipations, the 2AB term must be taking power from elsewhere..

(Parseval's theorem states that integration of power in the time domain equals integration of power in the frequency domain)..I believe my analysis integrates to equivalence, it's the instantaneous power I am considering, which is not what Parseval's theorem states, it is integral from -infinity to +infinity.

Quote:
 Originally posted by jcx also Christer is on to another possible issue - the biwire situation is not an exact equivalent to "mono" circuit by some "inverse superpostion " - they describe different linear systems with (slightly) dfferent linear responses
While I do not concur, if that is indeed the case, then one has proven the case for bi-wiring, no?.

Either result is equally acceptable, as is of course the result that they are exact equivalents.. I sleep soundly regardless of the outcome of this discussion..I enjoy the discussion..thank you.

It's the journey, not the destination..

Cheers, John
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 Originally posted by davidsrsb Two more possible causes: 1) A speaker cable carrying current I and with a spacing d has a repulsion u.I.I/2.pi.d N/m. At 1A and d=2mm this is 10-4 N/m, which might be enough to cause the wires to move apart in the insulation. If they move then work has been done and energy absorbed. 2) A speaker cable with current flowing is a moving coil microphone. The sound from the speakers may be picked up. Both of these effects depend on the current flowing and will be reduced by biwiring. Their significance is another matter.

I had calculated the forces on a #18 guage wire a while back. With 10 amperes, no insulation (1 mm spacing), the force was .02 N/m. That is .0018 ounces per inch of wire. awfully low, and with the modulus of the insulation considered, the movement will be very small. The change in spacing which would alter the inductance of the wire needs to be very large in order to shed or store energy from the internal currents. I also did those calcs somewhere, just don't remember where..

For case 2, again, the amount of wire to wire movement must very large to generate voltage of any significance.. I suspect humans cannot survive those levels of vibration.

In any case, you have presented very good possibilities to be tested if we were trying to explain why bi-wiring changes the sound.. My analysis is using ideal components to determine how the overall system should work.

Thanks.
Cheers, John
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 Originally posted by phase accurate [B][ A^2 + B^2 unequal (A+B)^2 Post #198 I had a deeper look at John's posting regarding the missing 2AB. I now see what is wrong with it. He makes the wrong assumption that the audio signal follows the sum of the acoustic powers and therefore at every point down the audio-chain we would have a scaled version of the sum of acoustic power again which can be split back into seperate powers by simple linear operations. It is not that simple though: The signal-voltage represents the sum of sound-PRESSURE levels. There are the same relationships between sound-power and SPL as there are between electrical power and voltage. That means for the same circumstances power raises with the square of the RMS -voltage/-SPL ! So far so good. What happens if we now add two sinusoids of different frequency ? Is the RMS voltage of the two added signals of RMS value A and B now A+B ??? No it isn't !!!! Since they sum UNCORRELATED their new RMS value is: SQRT(A^2 + B^2) Now check by yourself if you would still get a missing 2AB ! BTW: If John's observations were true they would indeed have nothing to do with cables as such but they would occur in every single little part of the whole chain ! Regards Charles
Here's a graph of the instantaneous power calculated from two sines. One at freq 1, the other at freq 3.5. both have peak value of 1.

The power of freq 1 is the dark blue line. The power of the higher freq is the pink?? line. The sum of those two is the yellow one. This represents the power dissipated in both loads, and the power that would be dissipated in separate resistors.

The purple line is the 2ab line. the bright blue one is sum of all three, Asq plus Bsq plus 2AB.

What I find really fascinating is the fact that 2AB swings plus AND minus, this represents negative power?????? Also note that the sum of all three terms NEVER goes below zero power into negative territory.

The fact that 2AB goes negative, while seemingly nonsense, remember that it in itself is an artifact of the product of the A and B, so it can never exceed in negative value, that of the two squares summed does in the positive value..In other words, it is impossible mathematically for the power to go negative..

I'd be sellin refrigerators if that were the case...and, to the king of sweden, no less..

Cheers, John
Attached Images
 2ab picture.jpg (84.6 KB, 94 views)
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 29th December 2005, 03:57 PM #29 diyAudio Member   Join Date: Dec 2005 Location: Kuala Lumpur Motor effect can be strong, if you have ever connected a jump lead between car batteries with one battery having a shorted cell so a large current flows - the cable will kick. The microphone effect can be measurable, I tried it with 1A dc flowing and tapping the cable was visible on a scope. The magnetic field around a close pair is strong enough. Another thought - normal strands of multistrand wire with current flowing will attract each other and this may have resistance effects. This would explain why cat5 based cables using a few conductors of individually insulated wire seem to work well for me.
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 Originally posted by davidsrsb Motor effect can be strong, if you have ever connected a jump lead between car batteries with one battery having a shorted cell so a large current flows - the cable will kick.
Agreed. The force goes as current squared. For current levels within a speaker wire, I think ten amps was sufficient to show the level of forces involved, and how little the wires will move.

I need this equation daily, as I typically run 1500 amps in wires that are about 16 guage, and the 50 mil spacing raises lots of forces. If I do not keep the conductors still to the tune of 50 microinches in a 2 to 5 tesla background field, they have "problems", ala "red october".

Quote:
 Originally posted by davidsrsb The microphone effect can be measurable, I tried it with 1A dc flowing and tapping the cable was visible on a scope. The magnetic field around a close pair is strong enough..
Again, absolutely agree. The effect of course, depends on the rate of change of the flux within the current loop, this in turn dependent on the inductance of the loop and how fast the energy sheds into the voltage produced.

What you should have done is push the amp dc into a zip cord, hook up the scope, and put it into a room where a stereo is playing loudly. That would have given you a better indication of level of effect.

Quote:
 Originally posted by davidsrsb Another thought - normal strands of multistrand wire with current flowing will attract each other and this may have resistance effects. This would explain why cat5 based cables using a few conductors of individually insulated wire seem to work well for me.
You are again correct...and, most people are not aware of the pinch effect. It is an issue for my work, as well as for those who use molten copper conductors in furnaces. (weird, huh?).

I suspect cat 5 issues are significantly different from what is being discussed here. Cat 5 pairs are orthogonal so inductance goes down as parallel inductors would..capacitance goes up with number of conductors..the characteristic impedance is also parallel calculated with a start at 100 ohms, 8 pair gives 12.5 ohms Zline.

Cool stuff, but not applicable to the instantaneous power issue we are discussing..

Excellent points, however..thank you.

Cheers, John
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