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Speaker Asylum General speaker questions for audio and home theater. |
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Speaker Asylum General speaker questions for audio and home theater. |
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In Reply to: Re: The most important question may be... posted by Jim Austin on February 24, 2007 at 05:59:56:
"Thanks Jim. I should started by saying "there is a constant phase difference between the two drivers, which means there is no relative phase shift between the two drivers at all frequencies. Thus, the first-order circuit does not disturb the original phase differences of the fundamental and harmonics".
To think of this as time differences: For any given frequency, the complete first-order circuit injects a constant 1/4-wave-period of time difference BETWEEN the two drivers. More on this below... Thus, with that relative time difference between mid and tweeter remaining constant, a simple first-order circuit does not disturb the original time relationships of the fundamentals and harmonics.
What keeps a simple simple first-order crossover from working as it should?
Non-linear drivers (those with distortion and resonances)
Drivers with limited bandwidths
Drivers with tilted responses
Drivers with varying impedance curves (varying at each frequency, and with stroke)
Cabinet resonances, reflections and diffractions
Imperfect inductors and capacitors in the crossover circuit.
As with any man-made device, there is no perfect driver or cabinet or crossover part, only 'better'.
What should be made clear about a simple first-order crossover circuit used on a perfect mid and perfect tweeter is:
The upper range of the mid driver gets a varying offset (away from you = delay) as we go up the scale.
This changing amount of delay begins at zero down in the bass, and reaches +45 degrees at the crossover point, for a single inductor in series with the mid. We can also represent that 45 degrees as 1/8th of that wave's period, if we want to calculate the actual 'time delay' at that crossover frequency.
The bottom end of the tweeter gets a varying offset (towards you = advance) as we go down the scale.
This changing amount of advance begins at zero in the ultra-high frequencies, and reaches -45 degrees (or 1/8th of the wave's period) at the crossover point, for a single capacitor in series with the tweeter.
But how can we 'advance' time? We can't, so for this analysis, we are simply setting t=0 at say, 50kHz, and not worrying about anything else except what the plus and minus signs tell us relative to that arbitrary zero point up at 50kHz. What is really happening is that as we go down the scale from 50kHz, that tweeter's capacitor is injecting less and less delay, which can then be thought of as more and more of an 'advance'... I hope that's clear!
At the actual crossover point, the mid is delayed by 1/8-period of that crossover frequency, and the tweeter is advanced by that same 1/8-period amount, for a difference of 1/4 period.
As we go down the scale, the mid's delay goes back to zero, while the tweeter's advance levels off at 1/4-wave period advance, for a difference of 1/4 period.
As we go up the scale, the tweeter's advance goes to zero, while the mid's delay levels off at 1/4-wave-period delay, for a difference of 1/4 period.
At all other frequencies, the total difference between the two is always a constant 1/4-wave period.
Thus no RELATIVE timing change occurs between the two.
What then happens is that the outputs from this mid and tweeter sum together at your ear as if they could be replaced by ONE single driver that you and the measurement mic would believe was doing all the work at all the frequencies, with no time delays at any frequency.
In contrast, high-order crossovers inject advance/delay differences (varying phase differences) between these same two drivers, which means that there are 'frequency-dependant phase shifts' between the two drivers. Your imaginary 'one-driver source' is moving farther away or closer to you, depending on the frequency.
Any sine-wave's period is 1/(its frequency) = time
Example for a 40Hz sine wave: period = 1/40Hz = 1/40th of a second. 1/4 period of 40hz = 1/160th second or '90 degrees'.
Thanks again.
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Follow Ups
- Re: The most important question may be... - mauimusicman 02:05:45 02/25/07 (1)
- Good description - dlr 06:53:20 02/25/07 (0)
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