| jneutron |
Here's the new thread for discusion of the difference between bi-wiring and mono wiring.
The basic premise for the analysis is this: When a speaker wire carries an audio signal to a crossover, is there a non linearity in transfer function as a result of the branching of currents at the crossover.
I'll get to definitions on a later post, with pictures or diagrams. For now, I recommend limiting this thread to the technical analysis. Should this analysis have proven, measureable validity, then perhaps listening issues can be discussed.
Any validation testing must be via equipment with resolution at least 10 times the entity being tested.
Oh, almost forgot the most important thing..question ALL assertions, equations, and conclusions. This is a discussion, not a sermon. It is preferable that errors be called when spotted...and we all make errors..:bawling:
Cheers, John |
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| jcx |
1st rule of linear system theory: linear systems do not create nonlinear distortion products of their excitation signals
if wire, crossover and speakers are modeled as linear R,C,L and constant mutual couplings then they are a linear load and do not create distortion no matter how few/many branches you divide the wiring up into
bi-wiring can reduce mutual impedance coupling where current from one driver imposes a voltage on the (linear) mutual impedance and changes the voltage seen by the other driver - but the eq are all linear- - superposition works perfectly
you have to assume a source of nonlinearity in the system for an analysis of distortion to make sense
drivers are of course not linear; voice coil inductance, BL product, surround spring constant are all nonlinear and time varying to some degree - at a much lower level wire tempco causes nonlinear response with power dissipation
as a quick guess if wire is ~100 mOhms vs loads of ~8 Ohms I would expect distortions from one driver's nonlinear current demand to be attenuated by ~.1/8, nearly 40 dB before reaching the other driver and then only distortion products in the pass band would be reproduced
further, nonlinearity in dynamic drivers is mostly related to displacement which is largest at the bottom of its frequency range, assuming several octaves of bandwidth the largest distortion products are in the distorting driver’s own bandwidth and only very small higher order distortion products are present to be coupled to the next higher driver's passband
alright already, I spell checked it this (last) time |
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| jneutron |
| quote: | Originally posted by jcx
1st rule of linear system theory: linear systems do not create nonlinear distortion products of their excitation signals |
Agreed, that is indeed a rule of linear systems. Having had that taught me back in the old days, I have had significant issues with the analysis I have performed as of late.
To that end, I have explained to some of my collegues the dillemma I have "uncovered". They typically will find errors I make, but so far, this issue has not been, um, trashed..
Perhaps you should go peruse the "audio lies thread", to understand what it is we are talking about.
So far, analysis using tried and true concepts still gives the distortion. I am in the middle of configuring a scanner in another room, but will post scans of my "chicken scratches" to give you a better feel for what I am talking about.| quote: | Originally posted by jcx
bi-wiring can reduce mutual impedance coupling where current from one driver imposes a voltage on the the (linear) mutual impedance and chages the voltage seen by the other driver - but the eq are all linear |
That argument is not the same thing as what I am speaking. For my analysis, the amp can have zero impedance.
| quote: | Originally posted by jcx
you have to assume a source of nonlinearity in the system for an analysis of distortion to make sense |
That is what I was also taught. It may be that the "teachers" were incorrect. If so, I will be upgrading their understanding..if not, hey, I've been incorrect before...that's in my job description (and I take my job seriously;) )
| quote: | Originally posted by jcx
drivers are of course not linear; voice coil inductance, BL product, surround spring constant are all nonlinear and time varying to some degree - at a much lower level wire tempco causes nonlinear response with power dissapation |
That is why I asked "phase accurate" to model the system without the non linearities he used for the drivers..to clear the analysis.
Thank you..
Cheers, John |
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| poobah |
Hi Neutron,
I have one point to ponder here:
I think in some ways the initial anaysis was based on the superpositon of power... and I don't recall ever doing that.
Consider the following, we apply a sine voltage to a cable with a resistive termination, a sine current flows, and power is dissipated in the cable (resistor). Now superimpose another sine of equal frequency, and magnitude, but opposite phase on that system... no voltage, no current, no power.
Follow me here?
:xeye: |
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| jneutron |
| quote: | Originally posted by poobah
Hi Neutron,
I have one point to ponder here:
I think in some ways the initial anaysis was based on the superpositon of power... and I don't recall ever doing that.
Consider the following, we apply a sine voltage to a cable with a resistive termination, a sine current flows, and power is dissipated in the cable (resistor). Now superimpose another sine of equal frequency, and magnitude, but opposite phase on that system... no voltage, no current, no power.
Follow me here?
:xeye: |
With only one loop to deal with, there is no branch. This is why a single driver with no crossover will not have this issue; it is only the branching that I'm speaking of.
The whole crux of the branching issue is that the instantaneous power loss in the cable resistance no longer tracks the load dissipations for each driver. Something's gotta give..
Cheers, John |
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| SY |
If there's any posts from the other thread that you'd like me to split out and move over here, I'll be happy to do so. Or you could just post a link to any of the relevant stuff.
Just let me know... |
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| jneutron |
| quote: | Originally posted by SY
If there's any posts from the other thread that you'd like me to split out and move over here, I'll be happy to do so. Or you could just post a link to any of the relevant stuff.
Just let me know... |
Thanks Sy.
Cheers, John |
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| Nordic |
| Unfortunately I lack the education you gentleman posses, but it doesn;t make me any less inquisitive... yesterday I saw a picture which was, if I understand it correctly, trying to convey that low frequency signals pass through the cable almost uniformly except for a small area in the centre.... as the frequency moved up the current would become thiner and thiner layer until its just a skin.... I think the bottomline of the page was the for higher frequencies its better to have two thinner wires than one thick one... does this have any bearing on Biwiring? |
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| jneutron |
Here's a graph showing the dissipation within the cable as a function of the dc current flowing through it. The sine is 1 ampere, and the resistance of the cable is .05 ohms, consistent with a 10 foot run of #12 zip, and neutrik connectors on each end..connector is 3 milliohms per contact, #12 runs 2 milliohms per foot, single conductor...times 2 gives total wire resistance.
The peak power with just the sine is 1 amp squared times .05 ohms, or 50 milliwatts, which happens at 90 degrees.. Since the negative side of the waveform also dissipates, it peaks at 50 mW at 270 degrees. This is shown by the blue line on the graph, the zero dc current line.
With one ampere of dc plus one ampere of ac, look at the 90 degree point...the current there is 2 amperes, and the dissipation in the cable is now 200 mW, while at 270 degrees, the dissipation is.....ZERO..
The time varying lossed within the wire resistance are getting larger with the DC current applied.. As the current is increased, the power dissipation has less and less 2x freq profile, and more and more base freq profile..
So, what gives:confused: :confused: :D
Cheers, John
ps..I hate those jpeg artifacts..100 kilobytes is too small..:dead: |
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| jneutron |
| quote: | Originally posted by Nordic
Unfortunately I lack the education you gentleman posses, but it doesn;t make me any less inquisitive... yesterday I saw a picture which was, if I understand it correctly, trying to convey that low frequency signals pass through the cable almost uniformly except for a small area in the centre.... as the frequency moved up the current would become thiner and thiner layer until its just a skin.... I think the bottomline of the page was the for higher frequencies its better to have two thinner wires than one thick one... does this have any bearing on Biwiring? |
No. you refer to skin effect. It happens because all conductors try to push a changing current towards the outside. It has always been calculated incorrectly for audio frequencies, and has not been measured correctly..but that is another thread.
This thread is about the woofer and tweeter current not "playing well together" because they are forced into the same wire.
As jcx pointed out, we were taught that it shouldn't make a difference. But I so far, have shown a mechanism which says it does..and I present it here for all to take potshots at..peer review, so to speak.
Cheers, John |
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| Tweeker |
| quote: | | Unfortunately I lack the education you gentleman posses, but it doesn;t make me any less inquisitive... yesterday I saw a picture which was, if I understand it correctly, trying to convey that low frequency signals pass through the cable almost uniformly except for a small area in the centre.... as the frequency moved up the current would become thiner and thiner layer until its just a skin.... I think the bottomline of the page was the for higher frequencies its better to have two thinner wires than one thick one... does this have any bearing on Biwiring? |
If skin effect is all that matters here and is significant, you really wouldnt need to "biwire" with seperate wires for woof and tweet. Parallel wires for (and connectors?) for both would have the same effect. |
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| Nordic |
| Thanks for the patience John |
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| ingrast |
| quote: | Originally posted by jneutron
Here's a graph showing the dissipation within the cable as a function of the dc current flowing through it. .... |
John:
Please attach your spreadsheet file. You have something wrong, linearity works.
Rodolfo |
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| poobah |
OK John,
Now take your power waveform in post 9 and throw it on a resistor... do you not get a votage sinewave with offset? ... pure?
:confused:
We are being deceived by the shape of the power waveform merely because we are not used to looking at them? |
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| jneutron |
| quote: | Originally posted by ingrast
John:
Please attach your spreadsheet file. You have something wrong, linearity works.
Rodolfo |
Darn* it, I know...something is wrong..
I've been tryin to find it for weeks now..
The full sheet is 947 kilo in size..
Let me see if I can cull it down and still keep it useful. For some reason, copying cells to another worksheet doesn't keep the equations..
Cheers, John |
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| Nordic |
Instead of clicking paste, click paste special, this allows you to copy the formulas too...
Make sure to delete empty datasheets to reduce size... |
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| jneutron |
| quote: | Originally posted by Nordic
Instead of clicking paste, click paste special, this allows you to copy the formulas too...
Make sure to delete empty datasheets to reduce size... |
Thanks. When I tried that,it gave me the choice of text or unicode text. Neither seemed to work, but I'm in a hurry now, carpool and all.
Cheers, John |
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| jneutron |
Got the file down. But, it's not a valid file type for uploading...bummer.
Cheers, John
See ya tomorrow..thanks guys.. |
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| jneutron |
| quote: | Originally posted by poobah
OK John,
Now take your power waveform in post 9 and throw it on a resistor... do you not get a votage sinewave with offset? ... pure?
:confused:
We are being deceived by the shape of the power waveform merely because we are not used to looking at them? |
Looking at the power dissipation is indeed confusing. What nags me is the fact that the power swing gets larger, even though the ac excitation current remains the same..only the dc is changing..
For my premise to be correct, geeze...linearity would hafta be violated, FFT's would hafta miss the distortion...maybe superposition???weird.. too many barriers...that's why I've been bustin my butt trying to find my errors..my assumption is that there is an error somewhere, I just can't find it..:bawling:
Cheers, John |
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| 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. |
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| 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? |
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| jcx |
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 |
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| 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. |
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| jneutron |
| 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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| jneutron |
| 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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| jneutron |
| 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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| jneutron |
| quote: | 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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| jneutron |
| quote: | 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 |
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| 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.
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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| jneutron |
| quote: | 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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| jcx |
if you look closely at Parseval's theorem you will see that it claims there is no net “A*B” power over a cycle of a periodic waveform – the Fourier components are orthogonal and no net “mixed products” enter in the power spectrum
superposition done right:
the power in each branch of the split circuit on the right doesn't add up but the sum of the Ia+b,Va+b in each image when added together do give the correct I*V - superposition works for the linear device I and V, power is not a linear function and doesn't sum when split this way
blue “V(w)*I(Rwoof)” = green “(V(wa)+V(wb))*(I(Rwoofa)+I(Rwoofb))”
the trace is just hidden under green |
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| jneutron |
| quote: | Originally posted by jcx
[B]if you look closely at Parseval's theorem you will see that it claims there is no net “A*B” power over a cycle of a periodic waveform – the Fourier components are orthogonal and no net “mixed products” enter in the power spectrum |
Check out the graph. If you look carefully, the waveform is rotationally symmetric about 180 degrees. the lobes centered at 81 degrees and 284 are equal and opposite, cancelling, the lobes at 129 and 236 are also equal and opposite. (In point of fact, the period of the complex waveforn here is actually 720 degrees on the graph, but every 360 degrees has integral zero.)
So, the theorem has not been violated..the 2AB waveform is net average zero.
Where are the tweeter waveforms? From what I can see, those would be the ones changed by the 2ab power when the other signal is dc.
Re: ""the power in each branch of the split circuit on the right doesn't add up ""
That has been my contention all along..something is going on, as the power delivered by the amp and that delivered to the loads, should be identical for both cases, The monowire case should NOT be capable of causing any difference over the biwire.
Excellent..
Cheers, John |
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| ingrast |
Gentlemen:
Just a couple of comments.
As I understand, this issue was brought about from comparison of separate cabling for tweeter and woofer against a common one.
If this is the case, I understand we are talking either biamplification or having the crossover network upstream of the speaker cables, as compared with the regular single cabling and passive crossover inside the box.
For the first case, the correct simulation model should be at most splitting the high pass and low pass sections, placing each in its proper branch (it doesn't matter whether they are placed before or after the cable). Otherwise, load impedance for this case is not the same as for single wiring and they cannot be compared.
With regards to the power composition issue, this frequency discrimination must also be considered. In the crossover region, i.e. at the frequency where both branches receive equal current, it is obvious the actual comparison should be with a single cable **with half the series resistance** (since for biwiring the cables are in paralell). Of course this also holds at any other frequency, I only wanted to highlight the fact that considering the same series resistance for the interconnect in both situations is wrong and will give misleading results when computing power (de)composition.
With respect to jcx simulation, it is correct nonwithstanding said above, in the sense of showinghow superposition holds for linear systems.
Rodolfo |
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| jneutron |
| quote: | Originally posted by ingrast
As I understand, this issue was brought about from comparison of separate cabling for tweeter and woofer against a common one. |
Yes| quote: | Originally posted by ingrast
If this is the case, I understand we are talking either biamplification or having the crossover network upstream of the speaker cables, as compared with the regular single cabling and passive crossover inside the box. |
Not biamp, just the difference between a single cable feeding the crossover and two cables, each dedicated to it's portion of the crossover.
| quote: | Originally posted by ingrast
For the first case, the correct simulation model should be at most splitting the high pass and low pass sections, placing each in its proper branch (it doesn't matter whether they are placed before or after the cable). Otherwise, load impedance for this case is not the same as for single wiring and they cannot be compared. |
Agreed.
| quote: | Originally posted by ingrast
With regards to the power composition issue, this frequency discrimination must also be considered. In the crossover region, i.e. at the frequency where both branches receive equal current, it is obvious the actual comparison should be with a single cable **with half the series resistance** (since for biwiring the cables are in paralell). Of course this also holds at any other frequency, I only wanted to highlight the fact that considering the same series resistance for the interconnect in both situations is wrong and will give misleading results when computing power (de)composition. |
Agreed. to keep the analysis simple, I used a DC component and an AC one, to remove the crossover region interaction out of the mix, so to speak.
| quote: | Originally posted by ingrast
With respect to jcx simulation, it is correct nonwithstanding said above, in the sense of showinghow superposition holds for linear systems.Rodolfo |
The comment about the power not adding up is the inconsistency I keep coming back to. Using one cable with the appropriate resistance should be equivalent to the two. This includes all power transferred and lost.
Cheers, John
ps..welcome to the thread, your name is unfamiliar to me.. |
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| ingrast |
| quote: | Originally posted by jneutron
...The comment about the power not adding up is the inconsistency I keep coming back to. Using one cable with the appropriate resistance should be equivalent to the two. This includes all power transferred and lost... |
Please sketch the circuit where you find power inconsistencies.
Rodolfo |
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| jneutron |
| quote: | Originally posted by ingrast
Please sketch the circuit where you find power inconsistencies.
Rodolfo |
Here are the two circuits. sorry it'll take two posts to explain.
Top is single cable, both currents flow through the one wire.
Bottom is bi-wire. Each current flows through it's respective cable.
John |
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| jneutron |
This is the graph of the dissipations.
The dark blue is the hf dissipation within the biwire setup.
The red? is the lf dissipation within the biwire setup. DC for simplicity.
the Light blue is the sum of the two power dissipations in the biwire case.
The Yellow is the dissipation within the cable for both the low and hf signals present within the resistor.
The difference is the 2AB term of the equation, this I stated in post #28.
(A + B)[sup2[/sup] = A2 + 2AB + B2
There should be no difference.
This model used .05 ohms cable, 1 amp dc, 1 amp peak ac, 8 ohm loads..each load gets 8 watts of power.
The scale is in watts.
The difference in peak wattage loss is 100 milliwatts, .1watt/8watts is 1.25 % difference. That is the difference between the yellow and the light blue waveform.
Cheers, John |
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| AndrewT |
Hi Jn,
can you confirm that the graphs shown in post37 are not sinusoidal?
It appears that the rms power loss in the hi only graph is about 0.015 to 0.02 watts.
If you add the low loss =0.05W and the high loss then the total= 0.065W to 0.07W based on visual rms estimation.
If the total loss of the combined cable were also at this range then your equivalence is established. The total graph losses are substantially below 0.1W. It appears to me that the numbers are very close.
Now considering superposition, the instantaneous currents in the combined cable will predict accurately the voltage losses in the cable. This instantaneous addition will also show the voltage intermodulation that the two speakers suffer from when run on a combined cable with actual resistance. (i don't know how to examine the impedance case).
Could the differences you are finding be phase differences where the two loads have leading and lagging currents with respect to the drive voltage? |
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| phase_accurate |
I don't see why John is interested in how INSTANTANEOUS power adds at all.
Instantaneous power carries no energy because it's duarion time is zero. I therefore doubt the validity of any conlcusions taken from it.
In order to represent energy instanataneous power has to be integrated over time. And it has to be done over an integer amount of signal periods (making it a little inconvenient for non-integer frequency ratios).
As John already found out the 2AB term is changeing between positive and negative and is therefore averaging out to zero.
Regards
Charles |
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| AndrewT |
Hi Phase,
why should power be related to time?
P=IV and this holds true for any time period, however short or long. Peak power is just such a short term case
I said| quote: | | instantaneous currents |
andI did not refer to instantaneous power.
My comment| quote: | | instantaneous addition will also show the voltage intermodulation |
was an alternative approach to understanding the problem since the rest of you are getting tied in knots as you all go down the same plug hole. I'm not too good at lateral thinking but just trying to help. |
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| phase_accurate |
| quote: | | why should power be related to time? |
Simply because we look at time-variant signals. And losses should be referred to energy and not instantaneous power IMO.
Regards
Charles |
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| Christer |
Power is the time derivative of energy. If P is the power and E the energy, we have
P = dE/dt
but I think you already knew that Charles, so what am I missing in your argument? |
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| phase_accurate |
I simply doubt that instantaneous power is a valid parameter to check for losses in time-variant signals. This accounts for voltages and currents as well IMO.
One has to look at energy, average power and RMS -voltages and -currents.
Regards
Charles |
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| pinkmouse |
Hi folks
I have been lurking here, mostly 'cos I don't think I can make a meaningful contribution. However, as a self taught, practical kind of guy, who's first response to a problem is to fire up the table saw, I just wanted to make a quick comment. Rather than delving deep into theory or simulation, which as I said, I'm not qualified to comment on, wouldn't be worth testing a real system to see if this kind of behaviour exists? I'd do it myself only I don't have the test gear.
al/the guy who dusts the mirrors at the Mount Wilson of diyAudio. |
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| jneutron |
| quote: | Originally posted by AndrewT
Hi Jn,
can you confirm that the graphs shown in post37 are not sinusoidal? |
Graph 37 is the output of an excel spreadsheet. All of the waveforms shown are the squares of the currents time the resistance of the cables.| quote: | Originally posted by AndrewT
It appears that the rms power loss in the hi only graph is about 0.015 to 0.02 watts.
If you add the low loss =0.05W and the high loss then the total= 0.065W to 0.07W based on visual rms estimation.
If the total loss of the combined cable were also at this range then your equivalence is established. The total graph losses are substantially below 0.1W. It appears to me that the numbers are very close. |
My assumption is that the rms powers add up. If they didn't, we'd be breaking a law of thermodynamics..| quote: | Originally posted by AndrewT
Now considering superposition, the instantaneous currents in the combined cable will predict accurately the voltage losses in the cable. This instantaneous addition will also show the voltage intermodulation that the two speakers suffer from when run on a combined cable with actual resistance. (i don't know how to examine the impedance case). |
Yes. It is the voltage intermodulation that must be there in order for the instantaneous power dissipated to ad up to the power delivered by the source.
The biggest problem I can see is the lack of reporting of this effect. So, either my analysis has some basic flaw, or it has been there all along but nobody has cared look for it, or the tools used to look for it have a hole in them big enough for a 2AB truck to drive through..
I note this right now: under no circumstances can a report of audibility of biwiring be used to justify this analysis. This analysis has been spurred BY anecdotal reports of audibility...it is not valid to use those anecdotes as support of my hypothesis..only measurement is valid..
| quote: | Originally posted by AndrewT
Could the differences you are finding be phase differences where the two loads have leading and lagging currents with respect to the drive voltage? |
By using DC and an arbitrary high frequency, I believe I have eliminated the need to consider the L and C reactances from the equation. It is certainly possible that this is an incorrect assumption, and that energy storage needs consideration, however, that may not explain the 2AB component.
Cheers, John |
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| jneutron |
| quote: | Originally posted by phase_accurate
I don't see why John is interested in how INSTANTANEOUS power adds at all.
Instantaneous power carries no energy because it's duarion time is zero. I therefore doubt the validity of any conlcusions taken from it.
In order to represent energy instanataneous power has to be integrated over time. And it has to be done over an integer amount of signal periods (making it a little inconvenient for non-integer frequency ratios).
As John already found out the 2AB term is changeing between positive and negative and is therefore averaging out to zero.
Regards
Charles |
We concur in all but the statement ""I therefore doubt the validity of any conlcusions taken from it. ""
Only because, the value of instantaneous power is defined by the instantaneous current and the instantaneous voltage. So if the instantaneous power is not what is predicted, then both I and V are also not as predicted. Hence, the dillemma.
Cheers, John |
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| jneutron |
| quote: | Originally posted by phase_accurate
I simply doubt that instantaneous power is a valid parameter to check for losses in time-variant signals. This accounts for voltages and currents as well IMO.
One has to look at energy, average power and RMS -voltages and -currents.
Regards
Charles |
Since 2AB integrates to zero, the rms value will not be any different. Conservation of energy.
Do not consider rms power.
Cheers, John |
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| jneutron |
| quote: | Originally posted by pinkmouse
Hi folks
I have been lurking here, mostly 'cos I don't think I can make a meaningful contribution. However, as a self taught, practical kind of guy, who's first response to a problem is to fire up the table saw, I just wanted to make a quick comment. Rather than delving deep into theory or simulation, which as I said, I'm not qualified to comment on, wouldn't be worth testing a real system to see if this kind of behaviour exists? I'd do it myself only I don't have the test gear.
al/the guy who dusts the mirrors at the Mount Wilson of diyAudio. |
""who's first response to a problem is to fire up the table saw,""
Remind me never to hire you for heart surgery, or to fix a flat..:D
Yes, that is a very good thing. As I stated within the initial post..| quote: | Originally posted by jneutron
Should this analysis have proven, measureable validity, then perhaps listening issues can be discussed.
Any validation testing must be via equipment with resolution at least 10 times the entity being tested. |
What we are doing is modelling what the system should do. I have asserted that all in not as it seems in Kansas, and we are discussing the validity of the model.
Measurement of the prediction of the model I am proposing is an absolute must for closure. The level of the effect should be on the order of the sqr of (2ABRc)1/2. Failure to find this error voltage means there is something wrong with the model..sending me back to the blackboard.
Finding this error voltage sends us all back to the blackboard, as, this has never been reported previously..
It would also tend to affirm the validity of bi-wiring based solely on resistive effects of branch circuitry.
Either conclusion is a data point, so I win either way...;) HA..
Cheers, John |
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| Omicron |
Hi John,
As I understand it your argument hinges on the term 2AB. I should perhaps point out that power is not a linear function of current (or voltage) and hence you can't just add to currents and expect the powers to add as well. Put more plainly: if you increase the current trough a resistance just a little bit then the power will increase a lot more. Superposition does NOT APPLY TO POWER!
Let us simplify the case to 2 DC current sources driving a simple resistor. If I switch on only one of the sources (say A) then the power dissipated in the resistance is obviously sqr(A)*R. If I switch on only source B then this power is sqr(B)*R. Now, if we swith on both current sources is the power then equal to sqr(A)*R + sqr(B)*R? No, it is in fact equal to sqr(A+B)*R = sqr(A)*R + sqr(B)*R + 2*A*B!
So where did this mysterious 2AB come from? Simply from both current sources. They are now together supplying more power than the sum of either one alone acting on the resistance. This is because power does not increase linearly with current (or voltage).
In the case you give indeed more power is lost in the cable with both signals present. But also MORE power is dissipated in the load at this point and MORE power is being delivered by the amplifier. ALL off these powers go up with an equivalent '2*A*B' term. Hence there is no distortion...
Kind regards,
Kurt |
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| Omicron |
Oops.
obviously where I said '2*A*B' I should have said '2*A*B*R' in my previous post... |
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| jneutron |
| quote: | Originally posted by Omicron
Hi John,
As I understand it your argument hinges on the term 2AB. I should perhaps point out that power is not a linear function of current (or voltage) and hence you can't just add to currents and expect the powers to add as well. Put more plainly: if you increase the current trough a resistance just a little bit then the power will increase a lot more. Superposition does NOT APPLY TO POWER!
Let us simplify the case to 2 DC current sources driving a simple resistor. If I switch on only one of the sources (say A) then the power dissipated in the resistance is obviously sqr(A)*R. If I switch on only source B then this power is sqr(B)*R. Now, if we swith on both current sources is the power then equal to sqr(A)*R + sqr(B)*R? No, it is in fact equal to sqr(A+B)*R = sqr(A)*R + sqr(B)*R + 2*A*B!
So where did this mysterious 2AB come from? Simply from both current sources. They are now together supplying more power than the sum of either one alone acting on the resistance. This is because power does not increase linearly with current (or voltage).
In the case you give indeed more power is lost in the cable with both signals present. But also MORE power is dissipated in the load at this point and MORE power is being delivered by the amplifier. ALL off these powers go up with an equivalent '2*A*B' term. Hence there is no distortion...
Kind regards,
Kurt |
Excellent points.
In fact, it is exactly the issue of power dissipation going up as the square that I am speaking of. Where your thought model differs from mine, is the use of two loads and a branch. If only one load were dissipating power, then what you have stated is exactly correct. However, consider the two branch model I speak of:
If you push only 1 ampere dc through, the cable will dissipate 50 mW. The load in series with the inductor will dissipate 8 watts.
If you push only 1 ampere peak sine through this circuit, the cable will dissipate 50 mW peak, and the load in series with the capacitor will dissipate 8 watts peak.
If you push both signals at the same time, then the cable will see at some points in time, 2 amperes peak, and the resistor will dissipate 200 mW at that instant in time. During the negative peak of the ac signal, the resistor will have no net current, and therefore, zero dissipation.
So, the change from a single cable to two cables changes the amount of power that is being dissipated by the entity which delivers the power to the load.
If the amplifier is sending the same power to the circuit, and the wires to the loads dissipate power differently, how is it the loads can see the exact same energy dissipation regardless of the wiring scheme?
The term 2ABR appears to be re-distributing how the power is being dissipated within the circuit on a time varying basis, but not on an integral one.
Cheers, John
btw, the editing capability of this forum allows changes for 30 minutes I believe. You could have easily fixed your post instead of a second post. |
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| poobah |
John,
You are closer... The VOLTAGE amplifier must supply the "missing" 2AB power term... and this term is small. Power amplifier is perhaps a poor term... these are voltage amplifiers with (hopefully) low ouput impedance.
You are starting with currents... no-no? You are presuming a transconductance amp.
Start with a voltage (the instaneous sum of sines). Derive currents based on system or branch impedance and the mysteries go away. Take the current(s) and apply them to their respective loads and see what comes out... pure signs.
The only thing I can see is that biwiring simplifies the impedance when modeling the individual crossover networks. This interaction between being cable resistance and and effective crossover frequency was handled well in the link posted earlier: "BI-WIRING". Mutual impedance, in the single wire case, can shift the HP filter lower and the LP filter higher creating an overlap and an apparent lump in output power. This effect diminishes when we size cable such that its resistance is small in comparison load impedance(s)... square one. |
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| Omicron |
I my view your last premise is wrong. I.e. for a total current of A+B to flow (no matter if they are AC or DC) the amplifier must deliver MORE power to the circuit than simply the sum of the powers needed to let A or B flow alone.
So, yes the losses caused by the resistance of the wire are greater when you apply both signals together. However, this difference is not taken from the power at the load (the power there can't have changed since the same currents are flowing there). Instead the amplifier is required to deliver this extra power. If it can't then the total current will not be A+B but it wil be less.
In this perspective all you do by adding another wire is lowering the losses in the wire (the resistance of the entire thing halves) and thus let the amplifier do a very tiny amount less work to deliver the same amount of power at the load.
PS: this was the very first time I posted on this forum and I've yet to figure out how to edit a post after it was submitted :-) |
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| Cal Weldon |
| There is an edit button at the bottom of your reply window. It is active for about 1/2 hour after you first post. |
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| jneutron |
| quote: | Originally posted by poobah
John,
You are closer... The VOLTAGE amplifier must supply the "missing" 2AB power term... and this term is small. Power amplifier is perhaps a poor term... these are voltage amplifiers with (hopefully) low ouput impedance.
You are starting with currents... no-no? You are presuming a transconductance amp.
Start with a voltage (the instaneous sum of sines). Derive currents based on system or branch impedance and the mysteries go away. Take the current(s) and apply them to their respective loads and see what comes out... pure signs.
The only thing I can see is that biwiring simplifies the impedance when modeling the individual crossover networks. This interaction between being cable resistance and and effective crossover frequency was handled well in the link posted earlier: "BI-WIRING". This effect diminishes when we size cable such that its resistance is small in comparison load impedance... square one. |
Your answer has not addressed the difference.
I used a current source to simplify the discussion. By using current, the cable dissipation is fixed and invariant. I chose that so that everybody would find understanding a bit easier.
Had I chosen voltage as the locked entitiy, the currents would have to be derived.
By choosing current for the though experiment, it is simple to see the reaction of the circuit to both configurations by viewing the input node...the 2ABRc should make itself apparent there, added to the load's reaction to the stimulus.
Had I chosen voltage, then it would be necessary to view the current out of the input node. A bit more difficult to do without disrupting the circuit loop impedance.
Previous threads did not address this 2ABRc power loss resulting from branch analysis. There is no use in re-hashing that analysis here.
Cheers, John |
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| jneutron |
| quote: | Originally posted by Omicron
I my view your last premise is wrong. I.e. for a total current of A+B to flow (no matter if they are AC or DC) the amplifier must deliver MORE power to the circuit than simply the sum of the powers needed to let A or B flow alone.
So, yes the losses caused by the resistance of the wire are greater when you apply both signals together. However, this difference is not taken from the power at the load (the power there can't have changed since the same currents are flowing there). Instead the amplifier is required to deliver this extra power. If it can't then the total current will not be A+B but it wil be less.
In this perspective all you do by adding another wire is lowering the losses in the wire (the resistance of the entire thing halves) and thus let the amplifier do a very tiny amount less work to deliver the same amount of power at the load.
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Good..
So, then what you are saying is this..when the switch from a bi-wire configuration is made to a monowire, that the resultant increase in power dissipation in the resistor is made up by the amplifier simply providing more power..
Let's examine the consequences of that statement..
1. The resistor losses change from that of A2Rc plus B2Rc, to A2Rc plus B2Rc plus 2ABRc.
2. The load losses have remained the same..A2RloadA plus B2RloadB.
3. The amplifier now provides more power to the system, to add up to A2RloadA plus B2RloadB and the cable loss of A2Rc plus B2Rc plus 2ABRc.
Re-organized: The amp power out becomes:A2(RloadA + Rc) plus B2(RloadB + Rc) and the strange term 2ABRc.
How does the amplifier know to provide the additional power component, a mix of the two signals? It knows only current and voltage.
It should be measurable.
Cheers, John |
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| jneutron |
| quote: | Originally posted by poobah
Huh? |
I am sorry. You will have to be a little more specific, as I cannot discern your question.
Cheers, John |
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| poobah |
| The amplifier knows only voltage... |
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| Omicron |
John, the amplifier only "sees" the complete impedance of the entire circuit. This impedance is not going to be the same for the mono and bi-wire cases. Therefore, you will find that for the same currents to flow in both cases you will have to fiddle a tiny bit with the volume knob of the driving amplifier in our thought experiment. The adjustment needed will be exactly that which will give rise to the 2ABR term.
Obviously if you use current sources then the "fiddling with the volume knob" is done automatically. The current source will adjust its output voltage so that the same currents will flow in both cases. Again, this will exactly compensate for the 2ABR term. |
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| jneutron |
| quote: | Originally posted by poobah
The amplifier knows only voltage... |
Ah, ok.
What I meant is that the amplifier should not be able to determine that the cable is configured as either a biwire or as a monowire setup, so it does not have enough information to determine that the wire will be dissipating additional power of the term 2ABRc in order that it may keep the loads purring the same.
For it to do so, it must be able to detect the changes in the circuit that require that oddball component of the power.
Cheers, John
I must apologize for incorrectly stating "additional power", as that can be confusing in light of the fact that the "additional power" varies from positive to negative. |
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| jneutron |
| quote: | Originally posted by Omicron
John, the amplifier only "sees" the complete impedance of the entire circuit. This impedance is not going to be the same for the mono and bi-wire cases. Therefore, you will find that for the same currents to flow in both cases you will have to fiddle a tiny bit with the volume knob of the driving amplifier in our thought experiment. The adjustment needed will be exactly that which will give rise to the 2ABR term. |
Ah, but consider the fact that the spectra of the 2ABR term is not the same as that of the two signals, it is a multiplication. Of note is that the voltage error from the 2AB power term is actually proportional to the square root of the error power term.
If the hf signal is going to both right and left load at the same level, and then a low frequency is applied to only the left channel, thereby introducing the additional loss component of 2ABR, we do not have the option of increasing that channel gain to compensate for the losses. Indeed, if you recall, the 2ABR term is in fact integral zero, it has an equal amount of postive power and negative power (better stated as more and less power). So in fact, the gain required to compensate will vary as the 2ABR term itself. Obviously, this is not possible.
If one uses a low frequency sine and a high frequency sine, then the modulation of the hf waveform will be rapidly changing based on the modulation of the 2ABR term. It is unknown if this type of modulation is visible through FFT algorithms.
This hypothesis is very easily tested. Create a test setup with one input node, a biwire feed to separate crossover/loads, and another monowire feed to another crossover/loads. Four loads. Use 50 mohm cables for the monowire and two 50 mohm cables for the biwire.
Force the hf voltage into the system, look at the difference between the hf loads...there should be no difference.
Then, force only dc voltage into the system. There should also be no difference.
Then, with the hf only and the scope verifying zero hf load voltage difference, add the dc voltage to the input node.
If there is a 2ABR dissipation within the monowire resistance, there will be harmonics generated at the monowire hf load as well as amplitude variation, whereas there should be none generated at the biwire hf load..the difference should be evident. If the scope is simply triggering on the hf wave, the abberation may simply look like a fuzzy scope trace. I will have to consider how to trigger the scope, as it must be locked to both waveforms. This will require an ARB setup.
Since I do not know if FFT's can see what we are looking for, I recommend the use of a differential probe setup. (or, maybe a table saw..;) )
Cheers, John |
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| poobah |
All I'm saying here is that starting with currents and heading for power is clouding the analysis. The numbers are right... the conclusions are???
Start with a voltage source... apply it to the impedance... find the current(s)... the resultant voltages, powers, etc... |
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| jneutron |
| quote: | Originally posted by poobah
All I'm saying here is that starting with currents and heading for power is clouding the analysis. The numbers are right... the conclusions are??? |
The analysis is in no way cloudy to me. It is very easy for me to understand..it should be since I am the one who originated it.
That of course, makes it neither correct or incorrect.
It was easier to use currents for the initial methodology of understanding..to arrive at the 2ABR power term that is in question.
From that use of currents, it is now possible to predict the error voltage which should be generated, as well as a methodology to look for it, the limitations in measurement technique which must be overcome to see what is predicted, and if found, the possible consequences and further work.
If nothing rears it's ugly head, I will persue the analysis further to discover what in my conceptual understanding is incorrect..
If however, the error term is visible, it calls to question, why has it not been seen before.:confused:
Cheers, John |
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| Omicron |
John, I think you are muddying the waters here. Best to go back to the basics before we start talking spectra. Consider this:
In your thought experiment the current trough each load stays the same: A in one load and B in the other. If we may assume that ohm's law is correct then it follows that the voltages over the loads are also the same. And indeed the power in the loads. So, I ask you now: where then is this distortion you speak of if NOTHING AT ALL has changed for load A or B?
Conclusion: What happens because of Rc is quite irrelevant in regards to load A and load B. It only affects the voltage as measured at the output of the current sources you are using. |
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| jneutron |
| quote: | Originally posted by Omicron
John, I think you are muddying the waters here. Best to go back to the basics before we start talking spectra. |
Sorry, I have worked my model considerably further along than what we are discussing at present, so I get ahead of the game..my fault..
| quote: | Originally posted by Omicron
Consider this:
In your thought experiment the current trough each load stays the same: A in one load and B in the other. If we may assume that ohm's law is correct then it follows that the voltages over the loads are also the same. And indeed the power in the loads. So, I ask you now: where then is this distortion you speak of if NOTHING AT ALL has changed for load A or B? |
I must apologize for my style of writing.
My statement is this. If the two wiring configurations make no difference in the resultant voltage and current that is delivered to both loads, then how do we account for the change in the power dissipation within the cable resistance?
I believe we all agree that the monowire dissipation has the included 2ABR dissipation, but yet, how can the load still have the same dissipation in light of that 2ABR for audio amplifiers of low output impedance, which is of course, the final configuration to which we wish to apply the results of this experiment..
| quote: | Originally posted by Omicron
Conclusion: What happens because of Rc is quite irrelevant in regards to load A and load B. It only affects the voltage as measured at the output of the current sources you are using. |
By using current sources, we guarantee that the load V and I are exactly what we desire them to be. The (2ABR)1/2 will appear across the cable, and it will be added to the load voltages and appear on the current source terminal.
Btw. I defined the test as looking for the differential across the hf loads, because the measurement of a voltage across a low impedance resistance of 50 milliohms is very very difficult for high frequencies. The B dot error component inherent in this measurement is never considered properly, leading to much error.
To measure it correctly requires a coaxial voltage pickup technique, and I would recommend using the outer braid of a coax as the 50 milliohm cable, and using the center conductor of the coax as the voltage tap to the far end of the "resistor". (there is no magnetic field within a cylinder of current, so there will be no voltage pickup due to the time varying magnetic field.). Measuring the voltage between the center and the outer will show the voltage drop on the shield accurately out to in excess of several hundred megahertz, more than sufficient for our needs. Using a simple resistor will not suffice, I have experienced that.
Cheers, John |
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| poobah |
John you started on paper... finish on paper. If it is there... it will show up on paper.
:D |
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| jneutron |
| quote: | Originally posted by poobah
John you started on paper... finish on paper. If it is there... it will show up on paper.
:D |
Umm, I did.
The component of error voltage that is predicted is of the form:
Verror = (2*Acurrent * Bcurrent )1/2 * Rcable
For a current driven system, this error voltage will show up at the input terminal as a result of this error voltage across the cable resistance.
For a voltage driven system, this error voltage will result in a drop in the level of the voltage at the loads.
For a voltage system where one of the drive currents is DC, all of the error component will end up across the hf load.
I believe that the spectra of the error will have the majority of it's power at one half of the frequency of the hf signal. (but I admit that this is a gut feeling, unconfirmed by math as of yet.)
There. It's on paper...well, ok, pixels...same thing..;)
I will start to develop the test setup upon my return from bejiing.
Unless of course, someone else is able to confirm my predictions before that. I am unsure of my ability to access the net over there.:bawling:
Cheers, John
Oh, forgot...I also predict that it will not be very easy to see using normal measurement techniques, as low impedance measurements are very difficult to do correctly. |
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| Omicron |
Well, obviously if you're using a normal amplifier instead of current sources then the voltage drop over the cables will affect the voltage over the loads. But the point of your experiment was, I thought, to see if this causes DISTORTION. And that is where the argument goes astray.
Let's recap for a moment:
1) mono wire: Ptot = sqr(A)Ra + sqr(B)Rb + 2ABRc
2) bi-wire: Ptot = sqr(A)Ra + sqr(B)Rb
How to explain away the 2ABRc term?
My explanation:
The two cases are NOT equivalent. We are measuring a small effect here and hence we must be careful in what kind of approximations we make. In this case we may not assume that the total impedance as seen from the side of the amplifier is the same in both cases. Because, obviously the bi-wire case has an extra Rc in it's circuit in a new leg. The effect we are studying here is also in the order of Rc so we cannot neglect this.
In case 1 we have a total impedance of Rc + ( (Xa + Ra) || (Xb + Rb) )
In case 2 we have a total impedance of (Rc + Xa + Ra) || (Rc +Xb + Rb)
Where Xa and Wb represent the impedances of the capacitor and the coil.
Clearly these two are different from each-other.
So yes, the power dissipated in both circuits is not the same when driven with the same voltages. Well, it can't be because the total impedance (and total resistance) is not the same either. Nothing mysterious here. Nothing there to cause any distortion either.
Since the impedances are not the same you will only get the same currents in the loads A and B in both cases if you adjust the output level of the amplifier to compensate for the different impedances seen by the amplifier. The difference in power that the amplifier delivers will be EXACTLY equal to the difference in wire losses between the two cases.
And also yes, when driven by current sources the power in load A and B will be the same in both cases but the power delivered by the current sources will NOT. Again the difference being exactly the difference in the wire losses between the two cases.
I really can't see any relevance to distortion in the test you propose. All you are going to measure is cable loss...pure and simple the ohmic voltage drop over the cable. And this quite simply CANNOT produce distortion if Rc is linear. And if it can, not only is our basic knowledge of physics wrong but our basic mathematics as well! So...good luck to you :-) |
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| jneutron |
| quote: | Originally posted by Omicron
Well, obviously if you're using a normal amplifier instead of current sources then the voltage drop over the cables will affect the voltage over the loads. But the point of your experiment was, I thought, to see if this causes DISTORTION. And that is where the argument goes astray. |
If the dc is capable of changing in any way, the hf signal at the hf load, I consider that to be distortion. If it is even a simple amplitude modulation, that is still distortion. So in all cases, I am looking for a change.| quote: | Originally posted by Omicron
Let's recap for a moment:
1) mono wire: Ptot = sqr(A)Ra + sqr(B)Rb + 2ABRc
2) bi-wire: Ptot = sqr(A)Ra + sqr(B)Rb
How to explain away the 2ABRc term?
My explanation:
The two cases are NOT equivalent. We are measuring a small effect here and hence we must be careful in what kind of approximations we make. In this case we may not assume that the total impedance as seen from the side of the amplifier is the same in both cases. Because, obviously the bi-wire case has an extra Rc in it's circuit in a new leg. The effect we are studying here is also in the order of Rc so we cannot neglect this. |
Agreed.| quote: | Originally posted by Omicron
In case 1 we have a total impedance of Rc + ( (Xa + Ra) || (Xb + Rb) )
In case 2 we have a total impedance of (Rc + Xa + Ra) || (Rc +Xb + Rb)
Where Xa and Wb represent the impedances of the capacitor and the coil.
Clearly these two are different from each-other. |
Agreed.| quote: | Originally posted by Omicron
So yes, the power dissipated in both circuits is not the same when driven with the same voltages. Well, it can't be because the total impedance (and total resistance) is not the same either. Nothing mysterious here. Nothing there to cause any distortion either.. |
Disagree. the 2ABR term of dissipation occurs with a spectra unlike that of either of the load dissipations.
| quote: | Originally posted by Omicron
Since the impedances are not the same you will only get the same currents in the loads A and B in both cases if you adjust the output level of the amplifier to compensate for the different impedances seen by the amplifier. The difference in power that the amplifier delivers will be EXACTLY equal to the difference in wire losses between the two cases... |
RMS yes, spectral content, I disagree.
| quote: | Originally posted by Omicron
I really can't see any relevance to distortion in the test you propose. All you are going to measure is cable loss...pure and simple the ohmic voltage drop over the cable. |
If the dc component alters the hf component, there is distortion.
| quote: | Originally posted by Omicron
And this quite simply CANNOT produce distortion if Rc is linear. And if it can, not only is our basic knowledge of physics wrong but our basic mathematics as well! |
Perhaps not. I do not believe I am attempting to re-write physics (although I am not averse to doing so if I find it needs to be re-written).
If my assertions and predictions are correct, it may be necessary to modify certain assumptions a bit..
Cheers, John |
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| Omicron |
So, I think we can summarise your statement simply as follows:
Linear superposition can cause non linear distortion.
This is so basic and fundamendal a claim that it really needs some pretty extraordinary proof.
No need to make complex setups with sensitive equipement. You can build a circuit with discrete resistors, a capacitor and a coil that is fully equivalent to your bi-wire thought experiment. You can make Rc very very large so the effect is magnified beyond proportion. If you can show ANY non linear distortion this way you will have won yourself one (or more) nobel prizes.
I 'm not holding my breath however :-) |
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| jneutron |
Fellow "fanatics".
I would like to thank all of you for your contributions to this thread. It has been a most wonderful discussion, I have looked forward to all of your responses.
Despite the fact that I am presenting an analysis with consequences which are contrary to all that we have learned by education, you have all been very nice and pleasurable to have this discussion with.
I completely understand the ramifications of my assertions being correct, and yet you have all been nice.
I also am aware of the ramifications of my being wrong. I have been so many times, so that is familiar territory..:bawling:
This forum has the distinction of being able to discuss this type of far reaching analysis without the breakdown typical of other forums. I attribute that to the level of the participants, as well as to the moderators responsible for policing this forum..(yes, even that really, really, old guy from Napa..;) )
Thank you all.
I wish all of you a very happy new year, and look forward to my return.
Cheers, John |
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| jneutron |
| quote: | Originally posted by Omicron
So, I think we can summarise your statement simply as follows:
Linear superposition can cause non linear distortion.
This is so basic and fundamendal a claim that it really needs some pretty extraordinary proof. |
Agreed.| quote: | Originally posted by Omicron
No need to make complex setups with sensitive equipement. You can build a circuit with discrete resistors, a capacitor and a coil that is fully equivalent to your bi-wire thought experiment. You can make Rc very very large so the effect is magnified beyond proportion. |
It must remain low impedance, it is a current driven entity. So care is needed.
| quote: | Originally posted by Omicron
If you can show ANY non linear distortion this way you will have won yourself one (or more) nobel prizes. |
Honestly, I don't think this is that big a deal..however, I certainly would not turn one down...I have three children and college costs to contend with..
| quote: | Originally posted by Omicron
I 'm not holding my breath however :-) |
Me either.
I am doing this for the knowledge and understanding..
Cheers, John |
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| Omicron |
Hi John,
I got to think a little bit more about the remark you made about the spectrum of the 2AB term not seeming to "fit" with the rest of the spectra. So here's my take on this and some more thoughts for you to ponder over:
Let us consider a simple resistor R driven by 2 sources A and B (current or voltage, it doesn't really matter...if you want to think about voltages instead of currents just replace R by 1/R int he following equations). The total power is then given by:
Ptot = sqr(A)R + sqr(B)R + 2ABR
Lets assume A and B are sine waves with A having a high frequency Fa and B a low frequency Fb.
If you examine the spectrum of this waveform you will find components with the following frequencies:
2*Fa due to the term sqr(A)R
2*Fb due to the term sqr(B)R
Fa - Fb and Fa + Fb due to the term 2ABR
As you can see NONE of the frequencies of the original current (or voltage) signals are present in the spectrum of the power waveform!
So, what does this tell us? It seems to tell us that we can't just look at the spectrum of the power waveform and then draw conclusions about the current or voltage waveforms. I.e. we can't say that there must be an Fa - Fb or Fa + Fb frequency in the current somewhere just like we can't say from this that there has to be a frequency component 2*Fa in the current waveform. Because clearly, there isn't (the current waveform spectrum only containing Fa and Fb).
The reason why the spectra are not related is again due to the fact that the relationship between P and I or P and V is not linear.
We can contrive some examples that illustrate this even more clearly:
Assume we drive a resistor with a symmetric square wave with amplitude X (so it will be +X half of the time and -X half of the time). The spectrum of the current and voltage waveforms contains a fundamental and an infinite number of uneven harmonics. But now look at the power waveform: it's simply a flat line. It's spectrum contains only a "DC" component! Actually from this waveform we can't tell what kind of current or voltage waveform caused this power dissipation (it could have been dc or any symmetrical square wave with amplitude X).
Another example: consider that both a pure sine wave and it's doubly rectified equivalent produce the same power waveform and the same power waveform spectrum. Nevertheless the spectra of a sine wave and a doubly rectified sine wave are vastly different!
So all in all I think it is invalid to look at the spectral content of the power waveform and use that to draw any conclusions about how the corresponding current or voltage waveform should look like. It is perfectly legal for the power waveform spectrum to contain "weird" frequencies WITHOUT the current or voltage spectra having to show any "harmonics" or "intermodulation" frequencies whatsoever.
Best regards,
Kurt |
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| jneutron |
| quote: | Originally posted by Omicron
Hi John,
I got to think a little bit more about the remark you made about the spectrum of the 2AB term not seeming to "fit" with the rest of the spectra. So here's my take on this and some more thoughts for you to ponder over:
Let us consider a simple resistor R driven by 2 sources A and B (current or voltage, it doesn't really matter...if you want to think about voltages instead of currents just replace R by 1/R int he following equations). The total power is then given by:
Ptot = sqr(A)R + sqr(B)R + 2ABR
Lets assume A and B are sine waves with A having a high frequency Fa and B a low frequency Fb.
If you examine the spectrum of this waveform you will find components with the following frequencies:
2*Fa due to the term sqr(A)R
2*Fb due to the term sqr(B)R
Fa - Fb and Fa + Fb due to the term 2ABR
As you can see NONE of the frequencies of the original current (or voltage) signals are present in the spectrum of the power waveform! |
In thinking over the vaca, I realize that the word "spectrum" is an insufficient descriptor for the dissipation...I concur..
Consider the problem from the viewpoint I first understood the issue from.
With one wire carrying one frequency, the dissipation waveform will have a frequency component of 2*Fa, as you correctly state.
Regardless of the resistance, there will be only the component 2*Fa in the dissipation. This is exactly the same component that will dissipate within the load...the only diff between the wire and the load will be the scaling factor...but same 2*Fa component.
So consider two wires, two loads..one wire dissipates 2*Fa, the other, 2*Fb. Again, that is the entire system delivery loss, the sum of 2*Fa and 2*Fb.
Now, put both signals into one wire. Now, 2*Fa, 2*Fb, and that infernal 2AB, which as you show, is sum and difference frequency.
That is a difference.. one wire carrying the two currents by | | | |
