Bi wiring/single wire distortion thread

[jneutron]

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 2ABR_c 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.

 

[jneutron]

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

 

[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...

 

[jneutron]

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.

Cheers, John

 

[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.

 

[jneutron]

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..

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..

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

 

[poobah]

John you started on paper... finish on paper. If it is there... it will show up on paper.

 

[jneutron]

John you started on paper... finish on paper. If it is there... it will show up on paper.

Umm, I did.

The component of error voltage that is predicted is of the form:

V_error = (2*A_current * B_current )^1/2 * R_cable

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.

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.

 

[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

 

[jneutron]

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.

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.

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.

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.

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.

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.

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

 

[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

 

[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..

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

 

[jneutron]

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.

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.

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..

I 'm not holding my breath however

Me either.

I am doing this for the knowledge and understanding..

Cheers, John

 

[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

 

[jneutron]

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 definition, will have additional power dissipated.

To state otherwise is indeed a violation of the law of conservation of energy. THAT cannot be allowed..

Think of a dc signal, and a signal that is .1 hz, they branch as a result of one honkin inductor and capacitor..

If you look at the 2ab component here, you will see that for 10 seconds, the product 2ab is either positive or negative..yes, integral zero, but...instantaneously not.. From your assertion (actually, everybody except me), that is no different from two independent wires..

But yet, for several seconds at a time, the single wire will have up to 2 times the power dissipation as the two separate wires..

There is a penalty for doubling the current in a wire, and it is not double the dissipation, it is four times..

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).
Best regards,
Kurt

As I said, spectra was not a good word to choose, as it is leading you in the wrong direction.

Look at the additional power that is being forced to dissipate as a result of the difference in configuration.

Account for that power..at all instants in time, that power must be accounted for. The amp delivers an exact amount, the loads dissipate an exact amount...for the mono and biwire cases to be identical, they MUST dissipate the exact same amount of power in the wires...they do not. They differ by 2ABR_c..

As P = V²_error / R_c

V²_error =PR_c

V²_error = (2AB) R²_c)

V_error = (2AB)^1/2 R_c

With one ampere dc and one ampere peak ac, this predicts an additional drop of 1.414 R_c.

For a 50 milliohm wire, that is an additional 70 millivolt drop, and if you examine the dissipation envelope, it is a reduction in the ac waveform..the fundamental frequency. This will show up at the load, instead of 8 volts peak minus the 50 millivolts, or 7.95 volts peak, there should be 7.88 volts.

The dc load will remain unchanged..but the AC load will go from 7.95 volts to 7.88 volts (both peak) when the DC current is turned on..

To expect equality at the instantaneous level is to disregard conservation of energy. All you are stating is quite accurate, but you are neglecting the power and conservation of it. I have a funny feeling superposition neglects this for orthogonal waveforms that branch.

Superposition works at the RMS level, and for non orthogonality.

Positive testing results for this issue will require modification of the theory of superposition..cool..I like being on the leading edge..kinda hope I'm dead nuts on..

Cheers, John

Oh, happy new year...

 

[phase_accurate]

To expect equality at the instantaneous level is to disregard conservation of energy. All you are stating is quite accurate, but you are neglecting the power and conservation of it.

I still question the validity of power conservation (is there any physical law that is talking about POWER conservation at all) rather one must look at ENERGY conservation which has to be done over completed signal periods.

Regards

Charles

 

[jneutron]

I still question the validity of power conservation (is there any physical law that is talking about POWER conservation at all) rather one must look at ENERGY conservation which has to be done over completed signal periods.

Regards

Charles

Energy conservation is only power conservation over a specific timeframe. For rms, that is a one cycle timeframe.

There always has to be power conservation. The delivery loss and load dissipation ALWAYS have to be equal to the delivered power...at all instants. Inclusion of course, of storage elements.

I have also considered the possibility that the mechanism responsible for the branching, the inductor and capacitor, are capable of storing energy to account for the instantaneous power component 2ABR_c, however a doubling of the cable resistance does not double the storage of the L and C, whereas it would double the 2ABR component. Therefore, the L and C cannot be responsible for the power difference.


Cheers, John

 

[phase_accurate]

John

Don't forget that your amp doesn't deliver two voltages

V1 = vpeak1 * sin (2 * pi * f1 * t) and
V2 = vpeak2 * sin (2 * pi * f2 * t)

but ONE voltage that is

Vtot = (vpeak1 * sin (2 * pi * f1 * t) + vpeak2 * sin (2 * pi * f2 * t))

So there is not two instantaneous powers but one at a time.

Regards

Charles

 

[jneutron]

John

Don't forget that your amp doesn't deliver two voltages

V1 = vpeak1 * sin (2 * pi * f1 * t) and
V2 = vpeak2 * sin (2 * pi * f2 * t)

but ONE voltage that is

Vtot = (vpeak1 * sin (2 * pi * f1 * t) + vpeak2 * sin (2 * pi * f2 * t))

So there is not two instantaneous powers but one at a time.

Regards

Charles

Yup..I do realize that.

But when you square that voltage to get power, the sum has the added component that is not there for individual signals.

Conferred with a friend..

We're questioning the validity of the error voltage as being of the form (2 AB)^1/2 R. The basis for questioning unfortunately resting on a circuit analysis with no orthogonal branches, which doesn't settle easy with me..however, it seems reasonable that we have to normalize the total power, as opposed to simply using the incremental power error.

However, we are unable to eliminate the modulation of each branch by the other's signal.

This is gonna require some testing..

Cheers, John

 

[Omicron]

Hi John,

Happy new year to you as well!

I have a feeling we are running around in circles in our discussion. Please reread my previous posts. The bi-wire case and the mono-wire case are NOT equivalent in respect to power dissipated. It is utterly incorrect to assume that both cases would draw an equal amount of power from the amplifier, dissipate an equal amount of power in the load or dissipate an equal amount of power in the wires. Not power averaged and not instantaneous power. Makes not one iota difference.

You can arrange things so that ONE of these dissipations is equal in both cases but not ALL THREE. It simply isn't the same circuit you are looking at! And yes, losses in the cable are higher in one case than the other. That's perfectly normal and exactly as one would expect as one circuit has a different topology to the other and an extra Rc in it! Hence I don't really see what point you are trying to make.

Conservation of power is not violated in any of my statements, it IS in yours if you blindly assume that the power in the load AND the power delivered by the amp AND the power lost in the cable(s) is the same in both cases. It simply isn't and never can be! And sure the voltages won't be the same either. That's perfectly normal too as the both circuits look different to the amplifier: they don't have the same resistance as seen from the amp! So if your are going to keep the currents constant then the voltages are going to differ. Pretty basic stuff it seems to me.

Nowhere in your experiment, and i really mean NOWHERE, is there ANY voltage that is not EXACTLY the same shape as the currents that are flowing. Hence there is NO DISTORTION! To claim otherwise is to reject ohm's law. It's as simple as that.

Best regards,
Kurt