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In Reply to: RE: Without knowledge it won't make sense posted by Craiger56 on July 16, 2015 at 10:22:04
Try this..
Rig up a high pass filter that you can temporarily place at the terminals of one of your speakers, or on a spare woofer if you have one.
Use two individual wires as your speaker cable for this test.
Place a clamp-on ammeter on one of the wires (+ or - but not both) to the speaker.
Play a test CD with a steady low bass tone.
Measure the milliamperes (AC) with, and without the high pass filter over the entire length of the wires.
You will see that with the high pass filter in place the milliamperes are lower, throughout the length of the wires than without the filter.
No shit.
(I assume you meant for this experiment to be done on the LF terminals of the speaker)Do it, and you will see that your statement "however this is an entirely moot point because the high frequency speaker cables are already being driven with a full range signal containing the same bass frequencies by the amp" Is incorrect.
With bi-wiring, the high pass cable is not subjected to the "measurable field" generated by bass frequencies, over the entire length, of the high pass cable.
This is where you are wrong. I assume you understand Ohm's Law, V=IR.
Do you agree that at the amplifier end, the voltage signal applied to both cables is the same? And do you also agree that it is full range, so that the voltage applied to the HF cable at the amp end includes the bass frequencies?
Now do you also agree that the current in the HF cable is NOT full range, because the high pass filter in the crossover is blocking current flow in its stop band?
If we're good so far, then apply Ohm's Law to calculate the voltage drop across the length of the HF speaker cable at a bass frequency, e.g. 100 Hz. Since the current is essentially zero at this frequency, the voltage drop is also essentially zero, and therefore the voltage at the speaker terminals is the same as at the amp end. Ergo, a full range signal is present at the HF speaker terminals.
Running separate wires from the amp to the LF and HF terminals of the speaker does not make the low frequencies magically disappear from the signal at the input to the high pass portion of the crossover, or the high frequencies disappear from the input of the low pass portion.
Follow Ups:
Do I agree that at the amp, the voltage signal applied to both cables is the same? Yep, absolutely.
Do I also agree the current in the HF cable is NOT full range, because the high pass filter in the crossover is blocking current flow in its stop band? That's for damn sure.
Don't need to go any further because we've already proved that CURRENT is blocked by the high pass filter, and it's CURRENT that creates "measurable fields" that we want to avoid throughout the length of our cables.
In reply to your last paragraph, the frequencies don't disappear, they are attenuated, and not just at the terminals, but as you wrote "Now do you also agree that the current in the HF cable is NOT full range, because the high pass filter in the crossover is blocking current flow in its stop band?" So... How would low frequencies make it to the HF terminals, or the high frequencies to the LF terminals without current?
So... How would low frequencies make it to the HF terminals, or the high frequencies to the LF terminals without current?OK, since we're going to be pedantic and precice, what exactly do you mean by "low frequencies make it to the HF terminals"? Specifically "make it"?
Thanks.
Edits: 07/16/15
"Make it" was my poor substitute for conducted, in reply to your less than ideal term "disappear".
As in, when bi-wiring low frequencies are attenuated over the entire length of high frequency cable, and high frequencies are attenuated over the entire length of low frequency cable.
Good evening
Right. It has to be over the entire length. It's the current that is being attenuated. The voltage is not attenuated until post filter. But the current is attenuated all the way back to the amplifier terminals where the bi-wire split is made.I think Dave is adding discussion about interaction between cables where the individual currents can modulate one another when separated into adjacent cables. This is complex, because we need to know about more than just the distance separating the cables, but the cable geometry including insulation thickness and turns per unit length.
If we're going that far, there are lots of details that would need to be added to the theoretical model if that specific detail were to be added, such as the individual voltage drops *across* each bi-wired cable, which would be different, since the currents going through them are also different, despite each cable having the same impedance. In that case the voltages presented to each filter input are NOT identical, albeit close.
The, introduce two different cables into the model and ALL kinds of mathematical hell breaks loose! ;)
Cheers,
Presto
Edits: 07/17/15
It does not make the full range voltage go away.
It does make the portion of the filtered current go away.
You can't speak of "signal" without confusing the issue.
When you say "signal" are you referring to the full range voltage at the filter terminals? Or the filtered current?
You can't have "unfiltered current" going into the filter unless it is ONLY a shunt filter. If there are series components, then the filter is actually blocking the current below the Fc of the crossover point, with attenuation (impedance) increasing as frequency goes away from Fc.
You most definitely have unfiltered voltage present. Voltage does not get filtered. Current does.
Replace the word "signal" in all of your posts with either the word "current" or "voltage" and the argument would end immediately with either you two agreeing or one of you being wrong. ;)
Cheers,
Presto
You most definitely have unfiltered voltage present *at the input to each filter*.
That clarifies what I said...
It does not make the full range voltage go away.
It does make the portion of the filtered current go away.You can't speak of "signal" without confusing the issue.
When you say "signal" are you referring to the full range voltage at the filter terminals? Or the filtered current?
If we're going to discuss things with pendantry, I guess you have a fair point. But really, which did you think I meant?
You can't have "unfiltered current" going into the filter unless it is ONLY a shunt filter. If there are series components, then the filter is actually blocking the current below the Fc of the crossover point, with attenuation (impedance) increasing as frequency goes away from Fc.Very true, and this is exactly what I stated earlier. We're talking about actual bi-wire capable speakers in this thread, which means parallel LPF and HPF crossover networks. And I don't believe there are any which implement the HPF via shunt inductors only without a series cap. So when I said earlier that the HPF blocks low frequencies, I meant that in the literal sense.
You most definitely have unfiltered voltage present. Voltage does not get filtered. Current does.Not true! Since we're being pedantic now, please reconsider your statement. Voltage most definitely DOES get filtered. If not, the filter wouldn't work. If the full range voltage is present at the output of the high pass filter, bass frequencies and all, what would stop it from producing full range currents through the tweeter? Think about it.
Edits: 07/16/15
Don't move the goalposts. We're talking CABLES. Cables are pre-filter.
Voltage does not get attenuated pre-filter. Only post filter.
Current is attenuated in the whole branch, i.e., in the individual cables - with one feeding each filter.Let's make it even simpler.
In a pair of bi-wire cables, the current to the low pass filter is only low pass current. The current to the high-pass filter is only high-pass current.
The sum of both currents electrically would be equal to the highpass current and the lowpass current combined, which is precisely what you see at the amplifier terminals IF only one amp is used.
If two amps are used, in a passive bi-wire configuration, each amp (like each cable) sees respective filter current only. (High sees high, low sees low).
Cheers,
Presto
Edits: 07/17/15
This should be entirely obvious if you read back through the thread where I've stated it several times, but I'll clarify once again:
I'm talking about the EMF induced on the HF cable from the magnetic field produced by the bass currents in the LF cable.
If you measure the voltage at the HF speaker terminals, it will come from two sources: the amp and induced EMF. Both sources will contain bass frequency content, and the high pass crossover section will filter out the bass frequency content regardless of source.
If the HF speaker cable is exposed to low frequency EMI from any source, it's not going to bother the tweeter as long as the frequencies are well into the stop band of the HF crossover section.
Oh, I am referring to an ideal model where there is only current driven by a voltage source and not induced.
In any case, a case is made for separating high and low pass filters from each other. Some 'philes have heard differences simply by splitting the board in half and moving the respective filters apart. By virtue of this, the cable in a bi-wire config would also move apart. All this said, the theoretical field around two conductors (send/return) in parallel and close proximity is zero. There may be interaction within a single cable jacket with two pairs cables, but if the cables are inches apart, I don't think it matter nearly as much if at all.
Cheers,
Presto
Oh, I am referring to an ideal model where there is only current driven by a voltage source and not induced.
The whole point of this sub-thread was RV's hypothesis regarding the magnetic field interaction between the two cables causing low frequencies to modulate high frequencies.
Oh, okay. To me it sounded like at certain points that things became fuzzy about current versus voltage at various points in the circuit.One thing that came out of this thread that is of interest to me though, ponder this:
Why would having low and high-pass filter currents in the same cable be less deleterious than having two separate, but in proximity, cables for low and high-pass filter currents? Aka, could not the low frequency content also modulate the high frequency content if in the same cable? In that case, the field associated with the lower frequencies is located immediately to the high frequency field.
In the case for separate cables, they can be moved further apart - reducing or even eliminating any possible effects.
Cheers,
Presto
Edits: 07/20/15
Why would having low and high-pass filter currents in the same cable be less deleterious than having two separate, but in proximity, cables for low and high-pass filter currents? Aka, could not the low frequency content also modulate the high frequency content if in the same cable? In that case, the field associated with the lower frequencies is located immediately to the high frequency field.I don't believe the low frequency content in the cable can modulate the high frequency content in the cable. Modulation is a non-linear process. What non-linear property of the cable is there that could produce this modulation?
EDIT: Let me expand on this a bit. Suppose we could replace regular wire with some kind of non-linear semi-conducting material that did cause one frequency band to be modulated by another. Given that a full range voltage is applied to the cable by the amp, and thus a full range electric field is present in the cable, the modulation would occur regardless of whether or not there is induced EMF from the magnetic field of a nearby cable.
Regarding the basic question of how can single wiring be less deleterious, I would expect it's system dependent. The single and bi-wire circuits are different and not equivalent to each other, so it's plausible that with some systems and cables bi-wiring might make things sound worse.
Edits: 07/22/15
The more I think about it the only thing I can up with is that the low frequency content in a full range signal is *supposed* to be there along with the high frequency content where with bi-wire any low frequency current induced in the high frequency cable is *not* supposed to be there.In a moving speaker cone, multiple frequencies being reproduced at once is another matter...
I wonder if cable self inductance can be a possible source of modulation within the cable for a full range signal? (In this case I thought we were dealing with relatively tiny inductance values that had very little effect...)
Cheers,
Presto
Edits: 07/22/15
Self-inductance is still linear
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