Why Bi-wiring Makes No Sense.

[EJ1]

I enjoy reading all of your posts, jneutron and Omicron, so please continue. Just letting you know you have at least 1 person in the audience.

 

[jneutron]

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.

We are discussing.

We certainly agree on many many things.

We disagree on some.

We are discussing in on a forum entitled "why biwiring makes no sense"

If you do not care to read, then don't. Simple..nobody is forcing your nose to the screen.

Cheers, John

 

[jneutron]

I enjoy reading all of your posts, jneutron and Omicron, so please continue. Just letting you know you have at least 1 person in the audience.

Oh, thanks.

Keep in mind, however. What Kurt is saying is consistent with all the electrical engineering we have all learned over the years..

What I am providing in terms of analysis, seems to fly in the face of all that we were taught..

His way, until proven otherwise, is the correct one...linearity of circuits has been gospel now for a century or so, and serves us well.

My position cannot be adopted unless many others have duplicated my test results...

After all, engineering is not anarchy..

Cheers, John

 

[Seth=L]

nvm then, I guess someone follows it

 

[mouettus]

nvm then, I guess someone follows it

I thought nobody cared either.

For me, it's a yawn. I think I'll remove it from my subscriptions. Bummer... cuz it was a great subject.

 

[stratman]

I think it's great, all the techno-speak and all, it's like an eminent train wreck that you can't stop looking at! I'm telling you, one will draw a light saber, then pfffft! It'll be all over and either Jedi J or Lord O will win this fascinating duel of wits, semantics and what have you, also it's a bit like the Hatfieds and McCoys, I hope you guys are married or at least have girlfriends, because really, you're both very intelligent, but fellas, I gotta say, no disrespect now, but it's all a little weird, not that there's anything wrong with that.

 

[Omicron]

So really, what you're here for...the entire reason you have registered here

The reason I am here is that we had this discussion before and that I stumbled into it is pure coincidence.

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..

I'll try again. I am nothing if not persistent.

Let's consider the case of mono-wire. We will concentrate on the tweeter. A is the tweeter current, B is the woofer current.

The voltage at the end of the mono wires can be seen as (A+B)*Kw with Kw some constant incorporating the resistance of the wires etc.

The voltage over the tweeter is A*Rtweet.

Now also consider the characteristics for the ideal filter you are using. Such a filter has 0 impedance for signal A (it passes it unattenuated) meaning that no voltage component due to A can exist over this filter. Also consider that it has infinite impedance for signal B (it blocks it completely, see where the infinities pop up?). We can then calculate the voltage of the filter as follows:

Vfilter = Vend-wires - Vtweet = (A+B)*Kw - A*Rtweet

and since the component in A must be 0, this reduces to:

Vfilter = B*Kw

In other words the voltage over the filter is such that it exactly compensates for the B voltage to the left of it so that the tweeter only sees an A voltage. This is a VERY important observation as we will see next.

We can now calculate the power contribution of the filter (yes, we can I was wrong earlier...I'm not always as smart as I think I am ).

Pfilter = Vfilter * Ifilter = B*Kw*A

Well now...what about that? We see a factor AB popping up in the filter power contribution as well! Also notice that this power contribution contains the Kw term which incorporates the wire resistance. The bigger the wire resistance, the LOWER Kw. This means that if more of the AB term occurs in the wire, less will occur in the filter and vice versa.

Let us think about that a little deeper. If the wire resistance is 0 then we have the largest possible value for Kw. Let us call this value simply K. K is such that the voltage at the amp output is:

Vamp = K(A+B)

We can now write the voltage drop over a wire that does have resistance as follows:

Vwire = K(A+B) - (A+B)Kw

From this follows the dissipation in the wire:

Pwire = Vwire(A+B) = sqr(A+B)(K - Kw)

When considering only the 2AB component of this we get:

Pabw=2AB(K - Kw)

Combining the 2AB terms of both wire and filter gives us this:

Pab = Pabw + 2Pabf = 2AB(K - Kw) + 2ABKw

The factor 2 comes about because off course we have 2 filters and not one. The reasoning for the woofer is identical and the result is the same as for the tweeter so I do not include it.

Or

Pab = 2ABK

Isn't that quite the surprise? It means the 2AB term is INDEPENDENT of the wire resistance (which is incorporated in Kw only).

It means the 2AB component is always present in the mono wire case and increasing the wire resistance simply shifts where it occurs (more in the wire and less in the filters or vice versa). The SUM is always the same.

Now we can consider the special case again where Rwire is zero. As shown even in this case the 2AB term is STILL there. Yet, as I assume you would agree there is no distortion in this case (it is identical to the bi-wire case with zero wire resistance after all).

Since increasing Rwire does not alter the total 2AB component I now assert that no AB related distortion can pop up either. You can also easily see that the bi-wire case suffers from the same 2AB ills, although the 2AB factor is always inside the filters there.

Please critique my analysis and point me to any errors.

 

[Omicron]

Pab = 2ABK

Isn't that quite the surprise? It means the 2AB term is INDEPENDENT of the wire resistance (which is incorporated in Kw only).

Ooops, that part of my analysis is incorrect I see now. In fact K incorporates the wire resistance, not Kw:

K = Rw + Kw

If you would keep K (total system impedance) constant then the AB term would remain constant also. But thats not very helpful.

What does remain true however is that you cannot make the AB term disappear. Even without wire resistance it is there due to the action of the filters. So both mono and bi-wire setups suffer from it. The magnitude depends on Rw and Kw for mono and Kw alone for bi-wire situations. If even the distortion free case has this component, how can there be any distortion mechanism?

 

[jneutron]

I think it's great, all the techno-speak and all, it's like an eminent train wreck that you can't stop looking at! I'm telling you, one will draw a light saber, then pfffft! It'll be all over and either Jedi J or Lord O will win this fascinating duel of wits, semantics and what have you, also it's a bit like the Hatfieds and McCoys.....

Wits? I make no personal claim in that regard..

but fellas, I gotta say, no disrespect now, but it's all a little weird, not that there's anything wrong with that.

No, that is incorrect.

As you can see, we both are passionate about the discipline of EE.

As such, I am presenting a concept and analysis that defies several of the tenets we learned as EE's...the common knowledge we share..

Kurt is passionately definding what we (yes, we) learned to become EE's.

I am presenting a seemingly contradictory analysis to a problem that is thought to be entirely understood, that analysis different as I approach it from the balance of power equations..

Consider this more as a debate, a point/counterpoint session, where one side has been tasked with defending an assertion, the other, prosecuting it.

Emotions...we are both human, so that is to be expected on occasion..

Cheers, John

 

[jneutron]

The reason I am here is that we had this discussion before and that I stumbled into it is pure coincidence.

Ah, ok..

I'll try again. I am nothing if not persistent.

You say that as if it's a bad thing??

Let's consider the case of mono-wire. We will concentrate on the tweeter. A is the tweeter current, B is the woofer current.

The voltage at the end of the mono wires can be seen as (A+B)*Kw with Kw some constant incorporating the resistance of the wires etc.

The voltage over the tweeter is A*Rtweet.

BZZZZZTTTT.

Stop right there.

You just made an assumption. You assumed here that the crossover network devices compensated exactly for the wireloss components...you therefore have built in the conclusion you are looking to support..

My analysis presents the 2AB aspect of the wireloss dissipation as an added "perk" so to speak.

As I stated earlier, you have just made a foregone conclusion...

Since we are discussing the effect of an entity on the output, we cannot make the assumption that the output is the same.

Without that assumption, the rest of your analysis falls apart like a cheap suit..(I had ta say that, I love that phrase.)

You have to perform the analysis without making the foregone conclusion.

That makes it a tad more difficult, doesn't it..

It means the 2AB component is always present in the mono wire case and increasing the wire resistance simply shifts where it occurs (more in the wire and less in the filters or vice versa). The SUM is always the same.

You made that assumption up front when you assigned the load power.

You have used circular logic...the output is identical....because the output is identical..therefore, the crossover has to compensate exactly.

Please critique my analysis and point me to any errors.

Done. Again, as I said earlier, incorrect but well thought out. The difficulty you are having is you are assuming the output, not deriving it..

What does remain true however is that you cannot make the AB term disappear. Even without wire resistance it is there due to the action of the filters. So both mono and bi-wire setups suffer from it. The magnitude depends on Rw and Kw for mono and Kw alone for bi-wire situations. If even the distortion free case has this component, how can there be any distortion mechanism?

It is worse than what you think..

You kept "harping" (for lack of a more intellectual word on my part) on "measure the 2AB, it must be obvious, anybody can see..

But, what is 2AB? How exactly would you propose measuring it??

Go through the math.. The only place the 2AB dissipation modulation loss exists is in the monowire. 2AB is a power, not a voltage. To view it as a voltage, what do you look at? The wire end to end voltage??pffft.

The wire end to end voltage is sqr(A squared plus B squared plus 2AB).

So what is viewed will be exactly A plus B. It's in the math..

What you need to look at is the difference. Can't view it by looking directly at the wire.

Look for the end item response..

This is what I told you, um, about a thousand posts ago.

You fell into a logic trap.

Hey, I tried to upload the schematic, but I can't..Mattt...heeeellllpp...

Cheers, John

 

[gene]

Here we go again with made up electrical theory that only some desk jockey working on a Govt superconductor can come up with

Here is the true math of biwiring:
http://www.audioholics.com/education/cables/bi-wiring-part-2-the-cable-conundrum

 

[stratman]

Here we go again with made up electrical theory that only some desk jockey working on a Govt superconductor can come up with

Here is the true math of biwiring:
http://www.audioholics.com/education/cables/bi-wiring-part-2-the-cable-conundrum

Pray tell boss, what took you soooo long??

And Yoda exits, leaving the two combatants in a daze.

 

[jneutron]

Here we go again with made up electrical theory that only some desk jockey working on a Govt superconductor can come up with

LOOK HERE BUDDY BOY...it's not my fault they have safety rules here (sumptin about giving an engineer tools or objects we could hurt ourselves with)

Here is the true math of biwiring:
http://www.audioholics.com/education/cables/bi-wiring-part-2-the-cable-conundrum

Hey there Gene, hows it going?

"True math"?

I guess you weren't following along with the discussion, eh?

Jim made the assumption that the outputs were exactly the same, so he forced the crossovers to do exactly what was necessary to make the output what he said it was.

Jim is using circular logic..he assumed the output was the same, then used that assumption to prove that the output was the same..


That's not a scientifically correct process..

Thank goodness Kurt injected himself into the discussion. He's the one who pointed out the flaw in Jim's analysis...

MY definition of "true math" does not include stating what the outcome is, then adjusting everything else around it to prove I'm right..


BTW, why can I not put a schematic here? Is there something wrong with the software, or have I reached a limit or sumptin?

Also, it seems that when I look at the last post in the thread, the page numbers string always has the next page as being there, even though there isn't one..

Cheers, John

 

[Omicron]

Thank goodness Kurt injected himself into the discussion. He's the one who pointed out the flaw in Jim's analysis...

I admit the ideal filters got me confused but there is no flaw in Jim's analysis. Sorry if I should have implied that, I was wrong.

If you insist I can do the maths another way. I'll take care not to assume anything about the output. I'm sure you'll agree that:

Pspeakers = Pamp - Pwire - Pfilters.

We are talking instantaneous power here so yes, the filters do contribute. Pfilter can be positive (filters storing energy) or negative (filters releasing energy) and must average out to 0. So, all we have to do is calculate these powers separately. I will consider the mono wire case here. The schematic looks like this:

Amplifier -> Wire resistance -> parallel a couple of filters and speakers

I'm sure we know it all by heart now so I do not need to draw it.

Let us start with Pamp:

Pamp = Iamp * Vamp = (Ia + Ib)*(Va + Vb)

We can find the relation between Va and Ia by noting that the filter of the tweeter will have 0 impedance for the "a" (HF) signal and the impedance of the woofer filter will be infinite. So:

Va = Ia * (Rwire + Rtweeter) and similarly:
Vb = Ib * (Rwire +Rwoofer)

Let us call Rwire + Rtweeter = Kt and Rwire + Rwoofer = Kw

So:

Pamp = (Ia + Ib)(Ia * Kt + Ib * Kw)

Pamp = sqr(Ia)Kt + sqr(Ib)Kw + IaIb(Kw + Kt)

Note that the power delivered by the amp ALSO has an AB component, namely:

IaIb(Kw + Kt) = IaIb(2Rwire + Rtweeter + Rwoofer)

We already know that the part (sqr(Ia) + sqr(Ib) + IaIb*2)*Rwire corresponds to the wire losses. In fact we can now write:

Pamp - Pwire = sqr(Ia)Rtweeter + sqr(Ib)Rwoofer + IaIb(Rtweeter + Rwoofer)

and this obviously must be equal to Pfilters + Pspeakers

Let us now calculate Pfilters. We consider the filter for the tweeter:

Pfilter1 = Vfilter1 * Ifilter1

We know filter 1 has infinite impedance for current Ib, so it cannot pass it. We also know it has zero impedance for current Ia so it cannot drop any voltage due to that. Also consider that the voltage at the end of the wire is:

Vend-wire = Vb - Ib*Rwire + Va - Ia*Rwire

This voltage drops over the filter and the speaker. Since the filter cannot have a voltage component due to Ia over it it's voltage must be:

Vfilter1 = Vb - Ib*Rwire

and

Vtweeter = Va - Ia*Rwire

Ifilter1 = Itweeter = Ia (remember the impedance of the filter is infinite for signal b)

So we can calculate the power in the tweeter filter as:

Pfilter1 = (Vb - Ib*Rwire) * Ia = (Ib * (Rwire +Rwoofer) - Ib*Rwire)*Ia

Or:

Pfilter1 = Ib*Ia*Rwoofer

In the same way the power in the other filter can be calculated as:

Pfilter2 = Ib*Ia*Rtweeter

So the total power in the filters is: Ib*Ia*(Rwoofer + Rtweeter)

As a sanity check we note that this power term indeed averages out to 0 over time so the filters do not dissipate energy, which is what we expect.

Since we already established that Pamp - Pwire = sqr(Ia)Rtweeter + sqr(Ib)Rwoofer + IaIb(Rtweeter + Rwoofer)

We can now write that:

Pamp - Pwire - Pfilter = sqr(Ia)Rtweeter + sqr(Ib)Rwoofer

As you can see...no terms in this expression that include a product of Ia and Ib!

Now I have given you an equation for the power in all parts of the system and shown that the AB term occurs in:

The power output by the amp
The wire losses
The power exchanged by the filters

And these 3 add up to 0 so in the speakers we see only the terms we expect to see in the undistorted case.

The powers all add up. The voltages all add up. And the current are all such that they obey Kirchoff.

The ball is in your court now John to prove this analysis is incorrect. If you think any assumption of mine was incorrect then please show so by an equation.

 

[jneutron]

I admit the ideal filters got me confused but there is no flaw in Jim's analysis. Sorry if I should have implied that, I was wrong.

No, you were very correct. The assumption that the output is identical to force the expected result ...that is indeed incorrect. You were absolutely right to mention that. Scientists always try to eliminate that predisposition bias from their experiments..

If you insist I can do the maths another way. I'll take care not to assume anything about the output.

You have made the assumption that current and voltage are scalars.

They are vectors. For some aspects of the problem, the assumption of scalars is good enough...but not for the reactive components.

Consider the tweeter with it's current at the highest positive..at this time, the capacitor is seeing the most charging current, so is therefore absorbing power.

Now consider two cases where the tweeter is at that point...woof at most positive, and woof at most negative.

When the woofer is at the most negative, the wireloss is zero. And the cap is at the highest charging current, absorbing energy..(power).

When the woofer is at the most positive, the wireloss is at the peak...and the cap is at the highest charging current...absorbing energy.

EXCEPT...for biwiring, the first case doesn't exist...for the second, the resistive loss is 1/4..

hmmm.

I have been the first to admit..it is rather difficult to solve this problem if the rule of superposition are not allowed (as using that rule is circular logic.)

Darn frustrating when that tool is disallowed...eh?

The powers all add up. The voltages all add up. And the current are all such that they obey Kirchoff.

The average power always adds up..I've been saying that all along...

Phase is important.

The ball is in your court now John to prove this analysis is incorrect. If you think any assumption of mine was incorrect then please show so by an equation.

The assumption that phase is unimportant messes up the equations for the crossover elements. For a cap/resistor series string, the time relationship between the resistors dissipation and the capacitors energy storage are not scalar entities.... If the cap is being discharged by the resistor's current, then the case for the cap equalling and opposing the resistors heatloss is certainly a possibility...but if the capacitor is charging, it is absorbing while the resistor is STILL dissipating, and that is the opposite condition..

Vector analysis would probably be more fitting...I haven't used phasors since '74..major rust..

Cheers, John

ps...isn't it great to exercise the mind? I love it..

As long as we can be civil, that is....

 

[bandphan]

you guys ever play pong?

 

[jneutron]

you guys ever play pong?

Where does everbody get these smiley things?? Cool..

Nah, don't answer..

Coupla months ago, I was involved in refurbing a duplicate of the first "pong" made.. Used a systron donner 10 channel analog computer...TUBES..tubes, tubes, and more tubes..

And 30 year old electrolytic capacitors that were unhappy with life..at random times, of course...

Higgenbothen...Willy Higgenbothen..he invented the first video game.."table tennis for two"..

Cheers, John

 

[Omicron]

You have made the assumption that current and voltage are scalars.

They are vectors. For some aspects of the problem, the assumption of scalars is good enough...but not for the reactive components.

Vb, Va, Ia and Ib are all functions in time. This is perfectly valid. If not, show us why.

I've given you a perfectly self consistent set of expressions for the INSTANTANEOUS powers of all elements in the circuit. They prove there is nothing strange going on if you care to look at them.

Now I have put a lot of my time in this. If you want further input from me you'll have to start showing equations of your own.

 

[stratman]

You gotta love these guys, Pong! Carry on then!

 

[jneutron]

Oh, really? Where did I make that assumption? Do I really need to draw little arrows over Va, Vb, Ia and Ib? Off course they are vectors!

Well then, where's the phase??

And far more important is the differential at the drivers when both the bi and mono systems are being driven by the same amplifier.

Now I have put a lot of my time in this. If you want further input from me you'll have to start showing equations of your own.

Let's examine this assumption:

Let us call Rwire + Rtweeter = Kt and Rwire + Rwoofer = Kw

So:

Pamp = (Ia + Ib)(Ia * Kt + Ib * Kw)

Can't assume that. You are adding the woofer loop voltage and the tweeter loop voltage, but yet both of them are derived with the wire resistance as part. Your messing up the 2AB, both in magnitude and phase.

And...Circular logic..

Actually, I'm trying to plop the schematic up to further the discussion..you are losing the forest for the trees.

Gene, why can't I upload a jpeg????

Cheers, John