Why Bi-wiring Makes No Sense.

[jneutron]

You cannot think about power in this way. The speakers do not "receive" power. They receive voltage (or current) which makes them dissipate power. The power they dissipate is dictated by the simple relation Pspeaker = Sq(Ispeaker)*Rspeaker. Simple as that.

As you correctly pointed out, the sum of the power the amp provides must equal the dissipations of the wire, the crossover, and the speaker. (yes, we are being a tad loose with respect to the crossover as a dissipator).

So, lets examine the specific scenario of two identical speakers (loads) being driven by one amplifier, with one speaker biwired the second mono. Assume all wiring is say, 20 guage wire (for lack of a reason to choose another), and the wiring resistance is 5% of the load resistance(russel's max recommendation). And, the breakpoints for each driver is such that there is a big hole in the frequency response....woof breaks at 50 hz, tweet at 10Khz. For this thought experiment, this separation is useful.("you" denotes "not me", for ease of discussion. Do not think I am attributing to you that which you have not stated..)

Both systems receive the exact same voltage. Moot point...one amplifier guarantees this..

Now, lets run a single sine frequency sweep of the systems. At low frequencies, the wire sees a power loss that is an exact scaled replica of the woofer power...at all points in time, the current in the wire is exactly that of the woofer, so the wire dissipates exactly 5% of the power that the woofer does.

At the hf extreme, there is no woof current, and the wire dissipates exactly 5% of the dissipation of the tweeter.

Note that for this portion, there is no significant difference between the two systems. IOW, the fact that there are two wires being used for the biwire does not alter the amount of power loss within the wires at any single sine frequency.

For this condition, the loss within the wire is always I squared R. For continuity from previous discussion, if we used "A" (lf sine) or "B" (hf sine), the wire loss in both cases is either exactly A squared or B squared.

For both cases, the wire loss is always an exact scaled representation of the dissipation of the drivers..

This exactness, this symmetry, breaks down when the amp provides two frequencies which branch to the independent drivers.

In the biwire case, each driver still receives only the signal of interest, so the wire loss is an exact scaling of the driver dissipation.. The sum of the biwire load dissipation is exactly A squared plus B squared, and the sum of the wire loss is an exact scaled version of that, 5% of A squared plus 5% of B squared.

But because the monowire case has both A and B flowing within, the wire dissipation of the monowire is (A + B) squared (or it's algebraic identity A squared plus B squared plus 2AB) ...meanwhile, YOU make the claim that the driver dissipations MUST be identical between the biwire and the monowire scenario... and....the power delivered to each system is identical..so lets run with that... The monowire driver dissipation is A squared plus B squared...

So lets look at the power balance...(I wish I could subscript here, it'd be easier) b is biwire, m is monowire, a is amp, d is drivers, w is wires, c is crossover..

You claim the amp power is identical between the two...Pba = Pma

You claim the drivers dissipate the identical power...Pbd = Pmd

You agree the wires dissipate differently...Pbw <> Pmw

Lets just assume the existance of the crossovers, Pbc and Pmc..

So

Pba = Pbw + Pbc + Pbd for biwire

Pma = Pmw + Pmc + Pmd for monowire

You claim that the amp has to see each system as identical (if not, the discussion is over..)

Pbw + Pbc + Pbd = Pmw + Pmc + Pmd


Your claim that the drivers dissipate equally. Again, if not, the discussion is over.. That means we can subtract the driver dissipation from both equations and maintain equality.

Pbw + Pbc = Pmw + Pmc

We already concur that Pbw <> Pmw, therefore, the "power" of the crossovers must be making up the difference..(again, we are being somewhat loose in jargon in this regard...)

Pbc <> Pmc.

Now, Jim has stated that in the end, the crossovers are doing just this, and that is why the driver dissipations are exact.

But for the crossovers to accomodate this difference, they must have a history which is different. That is not consistent with the drivers dissipating the exact same thing over all time previous, as YOU maintain that the drivers dissipate exactly the same..

That is the conundrum.. The crossovers cannot act differently in the two systems without their two terminal excitations having been different in the past..remember, they both integrate something..

The filters are a necessary item to set up the conditions of your experiment (i.e. split A and B apart perfectly). Note also that in real life no filter can behave the way you assumed (being infinitely steep). So the filters you use aren't exactly your regular basic reactances. Why are they able to exactly compensate? Simply because the circuit you created doesn't leave them any other choice courtesy of the law of conservation of energy.

Think about the statement "Why are they able to exactly compensate". You are making the claim that one crossover is reacting in a different fashion with respect to the other, to compensate..as it were..

The crossovers are only able to use previous history and present excitation to achieve a state. To say that the crossovers are reacting differently absolutely states that either their historical excitation or their present exitation (or both) are somehow different...and given the fact that they are elements in series with the drivers, you are claiming that the drivers had different currents within...this is in conflict with your statement that they are the same..

Off course there is if you're talking about instantaneous power in the time domain (which you insist on doing). All reactive elements will show positive (storing energy) and negative (releasing energy) instantaneous power. Because they cannot dissipate energy the power has to average out to zero over time for any purely reactive element.

Ummm...that was one of the first statements I made, remember?? If you examine a plot of the 2AB waveform, you will notice that the lobe areas above zero dissipation equal those of the lobe areas below the zero dissipation line.. Integrated as an average, the 2AB power envelope represents zero energy, zero power..

A term 2AB doesn't sound like a subtle distortion but quite a big one. Don't you think we should easily be able to see it in the time domain using a simple analogue oscilloscope?

H##L YES!!! a 5% deviation should be slappin everybody in the face.

Why nobody has seen this beats the heck outta me..this is why I have taken my sweet time with this analysis, as it flies in the face of what we were taught in school..

Given the assumption that my analysis is dead nuts on, let's look at the confounders that might prevent seeing such a gross entity.

1. The 2AB component is a zero integral power envelope. Is a math package that is designed to find the power of a signal over a specific time capable of seeing a modulation signal which does not alter the average power.

2. The 2AB power envelope is a fictional entity that cannot exist on it's own.. The negative power regions guarantee that. It can only exist as part of a power envelope comprised of A squared plus B squared, (it can easily be shown that 2AB is always <= A squared + B squared.

3. Measurement across the speaker wire itself gets confounded by the existance of the two signals..IOW, how does one differentiate between the two wiring cases when the terminals at the speakers do not have the same IR drops.

4. Measurement across actual drivers is certainly wrought with issues with respect to accuracy.

Other methods must be used. Hence my test schematics posted previously.

I've also come up with another test topology, a much more novel approach... This one I came up as a direct result of the discussions with you, which is WHY these discussions are so valuable..

I'm gonna work on that new test topology, it's gonna be fun as well as being consistent with the work I am currently involved in...danfysik current transducers will be involved..I have to see what the bandwidth of the ones available to me are..

Cheers, John


ps..don't get hung up on the real world vs idealized components...if the analysis doesn't hold for ideals, it cannot for real world..

 

[Omicron]

H##L YES!!! a 5% deviation should be slappin everybody in the face.

Why nobody has seen this beats the heck outta me..this is why I have taken my sweet time with this analysis, as it flies in the face of what we were taught in school..

It certainly should slap everybody in the face if it was there. You don't need an FFT to see it, you can make it as large as you want by controlling the wire resistance and the speaker resistance. Hell you can make it SO large you could easily HEAR it or see it on any analogue oscilloscope! An experiment that is trivially easy to do. So why haven't you?

We already concur that Pbw <> Pmw, therefore, the "power" of the crossovers must be making up the difference..(again, we are being somewhat loose in jargon in this regard...)

Pbc <> Pmc.

Now, Jim has stated that in the end, the crossovers are doing just this, and that is why the driver dissipations are exact.

All very true.

But for the crossovers to accomodate this difference, they must have a history which is different. That is not consistent with the drivers dissipating the exact same thing over all time previous, as YOU maintain that the drivers dissipate exactly the same..

That is the conundrum.. The crossovers cannot act differently in the two systems without their two terminal excitations having been different in the past..remember, they both integrate something..

I can see your point. But you are basing all this on crossovers that cannot (even in theory) exist. No reactive element could split A and B apart perfectly. The simplification you make introduces an infinity and then assuming that these filters are simple beasts that behave like a capacitor and an inductor is flawed! You incorrectly assume that simply moving the breakpoints far enough apart is the same thing as an infinitely steep filter. It isn't.

When you make simplifications like that and start getting absurd results, then the first thing to question are the simplifications. Not basic circuit theory or physics.

The only thing we CAN know for sure is that the currents in the speakers did not change. This is the boundary condition for your circuit. All else is based on it. Where would any other current components come from? Do they appear out of thin air after the filters?

Now, if you're arguing that the power put out by the speakers did change even though the currents trough them remained the same then you are invalidating the very premise your argument is built upon, i.e. that P = sqr(I)*R.

Assume you make the resistance of the speakers ZERO (i.e. you short out the tweeter and woofer but keep the filters in place). You can increase the wire resistances so that the same currents keep flowing as before. Now there is a HUGE 2AB component yet no speakers who can vary their power output to compensate for it. How do you explain that one?

And I have yet another thought experiment for you. Assume we would use a perfectly steep perfectly narrow filter. So these filters would pass ONLY A and ONLY B. It doesn't change anything to the power arguments made. Yet it is clear that this circuit will not allow the speakers to distort in any way because the filters will allow only a perfect sine wave to pass through them...

See how easy it is to get weirdness?

 

[jneutron]

It certainly should slap everybody in the face if it was there. You don't need an FFT to see it, you can make it as large as you want by controlling the wire resistance and the speaker resistance. Hell you can make it SO large you could easily HEAR it or see it on any analogue oscilloscope! An experiment that is trivially easy to do. So why haven't you?

Um...I've done both audibility as well as ele test.. Received positives on both.

The problem is one of test setup and confounders. Till the last coupla days, I had no serious contender for alternative test setup...So have not been able to verify by independent means, the accuracy of the intial test results. As I previously noted, my initial test setup required matched LR and C's...L and R were easy enough, R with extremely low inductance ta boot..C's are not as easy to sort, and the dielectric subtleties are not easily discounted as confounders. Positive results on a single test BEGS for an independent and entirely alternative methodology. A million people can confirm the same result using the same test setup, but if there is a fundamental flaw in the test setup, a million people will be incorrect. An alternative test setup reduces the possibility of a fundamental flaw..

As for the positive audibility test....expectation bias plays a big role, so anecdotal accounts of a test is fine and dandy, it certainly is not rigorous. And certainly not worthy of publication...

So as you can see, it is not so much an issue changing the world, or flying in the face of physics, and more an issue of a different test regime which does not share common possible confounders.

But you are basing all this on crossovers that cannot (even in theory) exist. No reactive element could split A and B apart perfectly.

So what???? Who cares if it's split perfectly or not? THAT DOESN"T matter, does it. You are diverting away from the mathematical analysis aren't you....but you already knew that..

Who cares if there is 1,2,or 5% bleedthrough intercomponent. You are guilty of introducing a red herring..

The simplification you make introduces an infinity and then assuming that these filters are simple beasts that behave like a capacitor and an inductor is flawed!

No..assuming that some trivially real component of the ideal components throws the entire thing off is flawed. Nice try.. You are attempting to divert.


The effect is independent of the frequencies... I could just as easily chosen 100 Megahertz as the hf and DC as the low. My resistors are good to about 5 Ghz, but I fear the caps wouldn't be.. Plus, I can only sort the caps to 1 Mhz...and the amps I've used certainly can't do 100 Meg..

For example, I would very much like to see you try to calculate the impedance as seen by the amplifier using these infinitely steep filters. You incorrectly assume that simply moving the breakpoints far enough apart is the same thing. It isn't.

What does an infinitely sloped crossover have to do with this discussion???Answer, nothing. Another red herring..

Stick to the basics...explain why the two crossovers will react differently, yet be provided the same historical and present stimulus.

This is the crux of the argument you are avoiding like the plague...

When you make simplifications like that and start getting absurd results, then the first thing to question are the simplifications. Not basic circuit theory or physics.

Who is questioning the physics??? Red herring again. And I've taken the analysis and self criticism far beyond anything either you or Jim has done. Haven't you wondered why it's been trivially easy to discount all your statements and red herrings?? I've already argued every point you have made years ago in my attempts to knockdown my analysis. In other words, I've been far more capable as an opponent of my own analysis than either of you have been to date.

I am applying physics..you are ignoring the crux..explain why the two crossovers will react differently even though you said the drivers see the exact same thing..

The only thing we CAN know for sure is that the currents in the speakers did not change. This is the boundary condition for your circuit.

Ummm, no. Keep track of the model please. The currents cannot change because YOU said they must be identical....(you again being "not me", but a construct who is attacking my analysis.. I have also taken the position of "you" as the devil's advocate)

All else is based on it. Where would any other current components come from? Do they appear out of thin air after the filters? No...so the power in the speakers is the same. THAT is the only thing you can be sure of in your experiment. That and the losses in the wires, off course.

Again, please stop diverting from the crux of the issue. You make the claim that the crossovers fix the problem by compensating exactly for the wire loss difference..

So the speakers receive the exact same stimulus, even though the power being presented to the cabinet is different (according to me, Jim, and yourself? Again, that infernal crossover has been given intelligence to compensate exactly it's energy storage and release independent of the conditions of it's terminals..

Let me suggest one other experiment to you. Assume you make the resistance of the speakers ZERO (i.e. you short them out). You can increase the wire resistances so that the same currents keep flowing as before. Now there is a HUGE 2AB component yet no speakers who can vary their power output to compensate for it. How do you explain that one?

First, your gonna have to rephrase the question to make sense..

The 2AB component is a result of kirchoffs law and the algebraic equality of (A+B) squared = A squared + B squared + 2AB.

Kirchoffs law applies to a node, where that node appears alters the dissipation of the delivery system (wire) from that of Asq + Bsq to (A + B) squared.

Now explain how the crossover in the mono cabinet knows to store or release the exact amount of power known as 2AB, in light of the fact that the reactive components are orthogonal (90 degrees out).

And I have yet another thought experiment for you. Assume we would use a perfectly steep perfectly narrow filter. So these filters would pass ONLY A and ONLY B. It doesn't change anything in the power arguments made. Yet it is clear that this circuit will not allow the speakers to distort in any way because the filters will allow only a perfect sine wave to pass through them...

You keep producing ridiculous red herring arguments..

Stick to the question at hand. How do the monowire crossover components "know" what energy to store or release, which is different for the biwire case, even though their terminals are "supposed to have the exact same history and present" stimulus.?

See how easy it is to get weirdness?

Actually, I see someone diverting from the crux with red herring arguments.

Stick to the program, eh??


Are you in a position to perform any ele tests at all, or will your participation be limited to tossing red herrings about?

You are welcome as a devil's advocate, but so far, you've not approached a credible argument against my analysis, and you've not approached the level of the arguments I have attempted to use to shoot it down either..

Cheers, John

 

[Omicron]

Um...I've done both audibility as well as ele test.. Received positives on both.

Then show us the results and how to reproduce them. Don't go blabbering on about "special setups" for a distortion that can be made arbitrarily large so anyone with half a mind could see it immediately on an oscilloscope picture.

Actually, I see someone diverting from the crux with red herring arguments.

Stick to the program, eh??

Let me try again then. You base your arguments on power correct? Your test circuit contains resistances and you can calculate instantaneous power in the by using P=sqr(I)*R. Correct?

Now unfortunately you have simplified your filters so that you are now unable to calculate the instantaneous power in them based on the current. Correct me if I'm wrong (but do show me the equation you use). Don't tell me there is no power in the filters, I'm talking instantaneous power here. And no Jim only calculated the power in them indirectly by making the same assumptions about the speaker power as I do. I want to see a power relation in terms of A and B if you want to convince me.

Now if YOU cannot calculate the power contributions of ALL elements in your experiment based in the currents A and B ONLY then YOU do NOT have an argument let alone you would have a basis to claim distortion occurs.

If you would bother to do the maths with a real capacitor and a real indictor (real as in having a reactance you can actually calculate with) you would have figured this out years ago. Heck, two minutes of simulation time in spice would give you all the power graphs for every single element in the circuit (and no these tools are not based on FFTs so quit that story already, they use kirchoff's laws just like you). A time domain plot will show anything you want to see.

You are welcome as a devil's advocate, but so far, you've not approached a credible argument against my analysis, and you've not approached the level of the arguments I have attempted to use to shoot it down either..

I'm getting a bit tired of your condescending tone John. If you want to be taken serious then you'll have to do a heck of a lot better than this.

 

[Matt34]

Holy geek-speak central!

If there was every a more fitting pic for what about 98% of this forum probably thinks of the last couple of pages....

 

[Matt34]

My bad, didn't mean to lock the thread, continue to speak which I do not understand.

 

[stratman]

MATT, NOOO MATT!!!! PLEASE NOOOOOO!!!!

Or in a "Janice-from-friends" voice:


OH, MY, CRON!!!! please give it a rest!!!

 

[jneutron]

First things first...

I'm getting a bit tired of your condescending tone John. If you want to be taken serious then you'll have to do a heck of a lot better than this.

I am not being condescending, calling it such doesn't make it so.

My point is, you are presenting arguments that I have already used against my analysis.. You are years behind me in attempting to trash my assertions...I always play devils advocate to my own work..

Then show us the results and how to reproduce them.

I posted the schematic already. Serves you right for coming in a year late and restarting an old thread.

Don't go blabbering on about "special setups" for a distortion that can be made arbitrarily large so anyone with half a mind could see it immediately on an oscilloscope picture.

Now now, let's not get emotional...stick with professionalism please..

Let me try again then. You base your arguments on power correct? Your test circuit contains resistances and you can calculate instantaneous power in the by using P=sqr(I)*R. Correct?

so far, so good..

Now unfortunately you have simplified your filters so that you are now unable to calculate the instantaneous power in them based on the current.

How does using an ideal capacitor or inductor prevent one from using simple math to do this??? Jim did it easy enough..where are you getting this??

Don't tell me there is no power in the filters, I'm talking instantaneous power here.

Where are you getting this??? Who said that? Is this another diversion??

And no Jim only calculated the power in them indirectly by making the same assumptions about the speaker power as I do.

WAITTAMINUTE???....the "fact" of identical speaker power was in fact, an assumption??? That's not good at all... That is what is known as a "foregone colclusion"...why would you do that???

Are you sure Jim did that??

WOW...egg on my face...I can't believe I missed an error of that magnitude..

Thanks for pointing that error out..(well, I assume you read him correctly, I'll have to go back to it and look again..)

Cheers, John

 

[Omicron]

WAITTAMINUTE???....the "fact" of identical speaker power was in fact, an assumption??? That's not good at all... That is what is known as a "foregone colclusion"...why would you do that???

It's not a forgone conclusion because the currents are known. And hence the power in the speakers follows from it.

You cannot on the one hand calculate the power in the wires using P = sqr(I)*R and then ignore that same relation for the speakers. That's an absurd argument.

If you argue that the currents through the speakers are not the same, then where did the additional current components come from? I haven't heard you argue any such thing.

 

[jneutron]

It's not a forgone conclusion because the currents are known. And hence the power in the speakers follows from it.

You are confusing the models.

The constant current model which I believe you are remembering from two or three years ago, was useful to easily depict the differences between the systems..

Use a constant voltage source to drive both the biwire and monowire systems at the same time..then you will see the disparity of wire dissipation.

You cannot on the one hand calculate the power in the wires using P = sqr(I)*R and then ignore that same relation for the speakers. That's an absurd argument.

Get the models straight..

If you argue that the currents through the speakers are not the same, then where did the additional current components come from? I haven't heard you argue any such thing.

You've mixed the models up sufficiently enough, that I don't know how you can say anything about anything..

I'll post a model, as well as a test schematic say, tomorrow or the next day....then we will be able to discuss a tad more rationally.

Tomorrow also allows both of us time off to persue other endeavors...

Cheers, John

 

[Omicron]

Get the models straight..

I am simply referring to the system schematic I posted yesterday and I copied that from the website to which YOU directed me.

Simple case currents A + B flowing trough the mono wire. A flowing trough the tweeter after the filter. B through the woofer after its filter. No other currents present. Both pure sines.

Ergo power dissipation in the speakers is known:

Rtweet*sqr(A) + Rwoof*sqr(B)

What's so hard to understand about that?

YOU keep skirting around the issue by refusing to answer me how in gods name the power in the speakers could be anything else than that. All your arguments are based on dissipation in the wires which could only lead you to say anything about the power dissipated in the speakers if you would also calculate the power contribution of the filters.

 

[Omicron]

MATT, NOOO MATT!!!! PLEASE NOOOOOO!!!!

Or in a "Janice-from-friends" voice:


OH, MY, CRON!!!! please give it a rest!!!

If I'm breaking any rules of the forum then off course I will stop this discussion. If not, well John has been wasting his time on this issue for far too long. It's time someone was kind enough to let him see the light

 

[jneutron]

I am simply referring to the system schematic I posted yesterday and I copied that from the website to which YOU directed me.

Ah, no wonder you're confused...that's not my schematic. I told you about that site because he provides an argument which was consistent with yours....I was helping you, guy..by providing some arguments against my analysis..sheesh...your welcome..

How the heck can you argue coherently with that schematic and my verbage based on my model and schematic...sheesh.

Coupla days friend..coupla days..

Cheers, John

 

[Seth=L]

Why can't this just stay locked? Hasn't it run it's course?

 

[jneutron]

If I'm breaking any rules of the forum then off course I will stop this discussion. If not, well John has been wasting his time on this issue for far too long. It's time someone was kind enough to let him see the light

I don't think your breaking any rules here..nor I.

When you are arguing with the correct model in hand, maybe you will provide something that is credible and coherent to kill the analysis.....course, maybe ya wont..but arguing with the wrong model in mind, is just silly and non productive.

John

 

[jneutron]

Why can't this just stay locked? Hasn't it run it's course?

No.

BTW, you do realize that you don't have to read it..nobody is forcing you.

John

 

[Omicron]

Ah, no wonder you're confused...that's not my schematic. I told you about that site because he provides an argument which was consistent with yours....I was helping you, guy..by providing some arguments against my analysis..sheesh...your welcome..

How the heck can you argue coherently with that schematic and my verbage based on my model and schematic...sheesh.

Coupla days friend..coupla days..

Cheers, John

You are welcome. But unless you can show a schematic and accompanying equations that give us either the power OR the current OR the voltage in any of the SPEAKERS, then don't bother because there will be nothing to discuss.

 

[jneutron]

You are welcome. But unless you can show a schematic and accompanying equations that give us either the power OR the current OR the voltage in any of the SPEAKERS, then don't bother because there will be nothing to discuss.

Ah, I see..It's not merely enough to point out the flawed assumption where the speakers get the same power, therefore the crossover has to do what I tell it...foregone conclusions...

Or provide a test schematic which YOU could actually put together to duplicate the test results of another. You know, the part of the scientific process where others have to be able to reproduce the results?

You won't have anything to do with actual tests, eh?

Nor the discussion of an actual schematic and the disparity of dissipation..

So really, what you're here for...the entire reason you have registered here, is to toss as much floobydust around as possible, never answer the question I've asked you (you know, how can the crossover know that the wires have dissipated that additional 2AB so that they can make up the difference for the speakers..), conjure up red herring scenarios that have nothing to do with the analysis....all that by searching the archives for a thread that was sleeping peacefully for over a year...remember the "oh I didn't know it was that old??? Got news for ya, we all know ya gotta trek back into the archives to go a year back, yet you feign surprise??

A shame..

Cheers, John

 

[Rickster71]

I can think of two guys that need to discover women.

 

[Seth=L]

I am simply saying that this public bickering is pointless, get each other's numbers, go have lunch and beat this dead horse somewhere more private.