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Bi-Wiring, yes, again.

Discussion in 'Audio Hardware' started by Katz, Aug 29, 2019.

  1. timind

    timind phorum rezident

    Westfield, IN USA
    If I had an oscilloscope handy I would post a pic of the signal waveform presented to each speaker binding post. The signal would be identical at both the tweeter and woofer binding posts. I'd love to have someone here prove me wrong.
    Bolster likes this.
  2. Ingenieur

    Ingenieur Forum Resident

    You have been proven wrong.
    By Q Acoustics, they did EXACTLY that.
    But no need, the math already did.


    We have discovered that if your speakers have four binding posts then you may take advantage of the bi-wiring option and that you will need twice as much cable as before. We theorised and proved by measurement, the counter-intuitive notion that the high and low frequencies travel only in their designated cables. We have also shown by measurement that the theory that this would reduce intermodulation distortion caused by non-linearity in the speaker system is proven.
  3. Ingenieur

    Ingenieur Forum Resident

    O-scope current probe measurements
    Signal generator
    Single cable at speaker
    Bi-wire high freq tweeter cable at speaker
    Bi-wire low freq woofer cable at speaker

    based on Kirchhoff and math it is impossible for the bi-wired signals to be the same.
    The measurements confirm it.


    Last edited: Nov 22, 2020
  4. Bolster

    Bolster If it ain't broke try harder..

    Awkward... :rolleyes:
    timind and Ingenieur like this.
  5. timind

    timind phorum rezident

    Westfield, IN USA
    Not really. The images posted by @Ingenieur show current, my post referenced signal. I quoted @Bob_in_OKC's question which I interpreted as whether the signal was the same at each binding post.

    Of course the signal flowing through each driver will be different, the crossover ensures this. But, the signal presented to each crossover is identical.
    Bolster and Gibsonian like this.
  6. Ingenieur

    Ingenieur Forum Resident

    A 'signal' is comprised of V and I (the product of which is power) at a given sum of frequencies.

    The current will be different but the signal the same? That is absurd.
    The signal IS current!
    The reason the current/signal is different is that hi or lo freqs are attenuated by their respective xover.

    in the graphs are different because their 'signal' is different by attenuation of freqs, as is clearly illustrated. It is graphed signal/current amplitude vs freq.

    what parameter would your O-scope measure
  7. Ingenieur

    Ingenieur Forum Resident

    In electronics and telecommunications, (a signal) refers to any time varying voltage, current or electromagnetic wave that carries information.
  8. Ripblade

    Ripblade Forum Resident

    The Six
    If you had an o'scope handy you could prove it to yourself. But since you don't have one, you can carry on believing that the current in each cable somehow magically remains the same despite their differing loads.
    Bolster and Ingenieur like this.
  9. timind

    timind phorum rezident

    Westfield, IN USA
    I'm not talking about the current flowing through each cable, I'm talking about the signal as represented at the output of the source compared to the signal at the output of the amp and also at the input to the speaker. I'm talking about a standard oscope probe which is different from a current probe, and doesn't measure current. With a standard probe, the waveforms would be identical. I get @Ingenieur's point, it's not how I interpreted the question.

    I'd also say that if Q Acoustics wanted to prove their point, they would've provided the results after taking the same measurements at the input to the respective crossovers using single wire with the shorting straps in place. Not difficult to do.
  10. Bolster

    Bolster If it ain't broke try harder..

    Meanwhile, the wag that resurrected this thread yesterday is lying low and about to watch a movie with a side order of scotch.. play nicely chaps :D
    Ingenieur and timind like this.
  11. Ingenieur

    Ingenieur Forum Resident

    the 'signal' IS current!
    You obviously have no education in the subject.
    But that is OK.
    Bear in mind it takes the same knowledge to derive a correct answer as it does to recognize an incorrect one.

    What variable does your 'standard' scope probe measure? V? I? ????
    What are the unit of measurement?
    It must measure a physical, observable, measurable parameter.

    The I of the 'signal' and the V will BOTH
    be different at the end of cable/speaker input terminals. Since Z is different and V = I Z

    And the signal at the amp/beginning of each wire will be the same as at the end/speaker input terminals. They MUST be or violate every principle of physics.

    BUT the signal (V, I, P) will be different in each wire.

    if I is different, and Z is different, V must be different, hence power is different.
    The high cable will have attenuated lows
    The low cable attenuated highs

    Q Acoustics proved their point (reality).
    It was the PERFECT test set-up.
    What you propose is pointless.
    Last edited: Nov 22, 2020
    vinnn and Ripblade like this.
  12. timind

    timind phorum rezident

    Westfield, IN USA
    My last go at this. Q Acoustics proved the current flowing through a cable is dependent on the load provided by the circuit. They proved the obvious. If Q Acoustics wanted to prove their thesis, they would've measured the input to to the crossover when wired with single wire and straps in place. It would've taken all of 30 minutes to setup. If they had a dual trace scope, they could've run the test simultaneously and shown the differences in real time. How does this not make sense?

    Also, when I say the signal doesn't change, maybe I should've used the term, waveform. That said, I'm talking about measuring the signal with a standard scope probe which measures the signal's amplitude and frequency. It won't measure the current. If we remove the biwire cables from the speaker binding posts and measure the waveform, both cables will measure the same. When we hook the cables to the binding posts, the amplitude and frequency of the waveform will still be the same. That is how I answered the question originally. If you don't agree with this...well I don't know what to say.

    I will admit it's been 30 years since I've worked with any of this, but I'm confident everything I wrote above is correct. Especially the fact the only thing Q Acoustics provided proof of is current flowing through the tweeter's circuit and the woofer's circuit is different. Thanks, and have a good holiday week.
  13. Ingenieur

    Ingenieur Forum Resident

    wrong on every count:
    Q Acoustics measured the signal at the end of the wire/at speaker input, not AFTER the filter, but BEFORE.
    Single and bi-wired and the difference was attenuation of freq. Simple and elegant, proved their hypothesis. Period

    The waveform changes, that is what they measured.
    The hi cable has attenuated lows
    And vice versa
    Current (hence V since there is Z)

    the signal (I, V, etc.) is different in each cable

    What you need to say is you are wrong but ego or lack of knowledge precludes it.

    You never answer questions:
    -what is your scope probe and what variable does it measure, the ones I posted were magnitude vs freq. what is yours?

    -What accounts for the difference in current in their test? And why were the low and hi freqs attenuated?

    I have a MSEE and design filters.
    One application is for VFD's. The inverter section produces harmonics that travel into the power source and load. These are currents, they have a waveform, you could call them a signal but since they convey no information not appropriate.
    We filter to attenuate the harmonics the free up line capacity for useful power, not harmonic garbage.

    It is the EXACT same principle, science and engineering.
    Ripblade likes this.
  14. Ingenieur

    Ingenieur Forum Resident

    The reason I am insistent is that I do not want misinformation spread for those considering this. Can you hear the difference? That is up to the listener.

    But if the signal is the same in each wire there is no reason to do it. You would achieve the same by upsizing wire and using the jumper.

    And all the engineers at KEP, Focal, etc. wasted there time and $ building speakers that can be bi-wired. But I'm guessing they did it for a proven technical reason.
  15. vinnn

    vinnn Forum Resident

    You've made the classic mistake of thinking an amp pushes current out to a speaker, it doesn't. Current flows towards the negative terminal like water flowing downhill to ground. The current or signal on the wire is determined by the characteristics and load of the speaker driver so in a bi-wire scenario each wire ends up carrying a different signal, doesn't matter at what end of the wire you're looking at.
    MGW and Ingenieur like this.
  16. Ingenieur

    Ingenieur Forum Resident

    An amp is a variable voltage source.
    The current is determined by the load I = V/Z
    The load varies with frequency

    You can look at a speakers Z magnitude and phase chart to learn about it.

    The Z at low freq is different than at high end
    This due to the filters
    Low being Z ~ freq, as f increases so does Z
    Hi being Z ~ 1/freq as f increases Z decreases

    the magnitude is in Ohms
    The phase in degrees, the ratio of L and C reactance to R (arctan of)
    When 0 degree all R
    At these points C and L reactance are equal and cancel.
    As degrees go negative L is in play, lo pass
    As phase goes positive C is in play, hi pass

  17. George P

    George P Notable Member

    Can you explain what L, C and R stand for here? Is it length. capacitance and resistance?
  18. Ingenieur

    Ingenieur Forum Resident

    L inductance, related to the magnetic field in Henries
    C capacitance, electric field in Fareds
    R resistance in Ohms

    L and C have to be converted to get Ohms
    That value is reactance
    Xc = 1/(2 Pi f C)
    Xl = 2 Pi f L
    Pi = 3.141596..,,
    f = frequency in Hz
    you can see these are freq dependent

    Z impedance magnitude in Ohms =
    sqrt(R^2 + (Xl - Xc)^2)
    Ripblade and George P like this.

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