# passive crossover basics question

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#### MCPete

Here's my problem/ lack of understanding. Take for example a parallel second order two-way crossover network. As per the usual arrangement, the woofer and tweeter have identical voltage sensitivity. As frequency climbs to the crossover frequency, the intensity of the woofer drops from its mid-band response by -6 dB. At the crossover frequency the tweeter's intensity is also -6 db less than that of its normal intensity. So both drivers have the same intensity which is 6 dB less than the desired level.

For two sound intensities of the same frequency, intensity and phase, the intensity of the summation is + 3 dB relative to the equal intensity of each.

So as the woofer's radiation decreases to -6 dB, the tweeter (connected in reverse) adds to the radiation of the woofer to increase the level to -3 dB relative to the midband response level of the woofer.

However, the above is not what most books/ articles say is the result. According to most, the level at the crossover frequency is flat. One source I read claims that a hump occurs at the crossover frequency.

Rod Elliott in his article on passive crossovers alludes to Q of the filter as the means whereby response is flattened. However, he doesn't explain this. Personally, I'm at a loss at explaining how Q might make a change.

Can anyone provide an explanation? This isn't connected with any project at the moment, I'm just trying to improve my understanding.

Regards,
Pete

#### Jonathan Bright

Hi MCPete, yup this is an easy one. As I understand it the conventional Xovers seek to produce a HALF POWER or -3db result at the cross over frequency and not -6db. In an ideal world if both are -3db (i.e. half power) two halves make a whole. So the result is flat.

(But life is not simple. There IS a popular type of cross over call Linquist-Riley or LR for short (sorry I can't spell and I KNOW that is not quite right!) They are a specialised effort for time aligned speakers and they DO have a -6db response. You need to read up on that one because it is seeking a specialised result and they have their own theoretical approached very well worked out.)

An edit. Re; "Q". I have assumed Butterworth or maximally flat filters in all cases. They have a Q of roughly 0.7.

Hope that helps. If you poke around this topic you'll find a lot of material and it takes a while to get ones head around it.
It's probably fair to say that of all the "black arts" of audio this is one of the more complicated areas. You can have very well reasoned cases for 1st, 2nd, 3rd and 4th order networks (and I have seen 5th being proposed too...) and you will change your mind after reading the case that each proponent puts! (As they say "I used to be indecisive....but now I'm not sure.) In addition some seek time alignment, some flat response, some insist on minimum phase change. Some worry about the off axis response of filters and some (the Full Range guys) despise crossovers in principle!!!!

Good luck, Jonathan

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#### Jonathan Bright

Hi MCPete
I notice sreten has the correct spelling of Linkwitz in his references!!

JB

Mmm, my brain is stuck on the exact explanation, but your numbers ignore the PHASE of the signals being summed.

Two drivers 100% in-phase will sum to +6 dB ON AXIS, by becoming more directional. The POWER is up +3 dB, as you might expect since you have 2 units at the same power.

Only first order crossovers sum truly flat (with "perfect" drivers). All others may be flat on-axis, but will have some compromise in the power response.

#### MCPete

Gave some more thought to this, and I believe that phase is the important factor. In the second order crossovers, the woofer and tweeter are 180 deg. out of phase. So reversing the tweeter makes the drivers in phase, also resulting in the summation being +6 dB with respect to their intensity at the crossover frequency (as Sreten says). Vance Dickason calls this correlated.

The analogy I thought of was adding two electrical sine wave signals of equal frequency, phase and amplitude. The resulting summation has a peak voltage twice that of each of the two components. Electrical power equals the square of the voltage divided by the load resistance, so the result is a power increase of 6 dB.

With the first order crossover, there is a 90 degree phase difference of the two sound intensities which can probably be considered uncorrelated. In that case, the summation is +3 dB with respect to the intensity of the components at the crossover frequency. Not sure exactly about the math behind this, but it sounds reasonable.

I was originally led astray by Backus in his book, The Acoustical Foundations of Music. In his analysis arriving at +3 dB, he makes no mention of phase.

Thanks for the responses,
Pete

#### Jonathan Bright

Hi Pete. I'm not 100% confident of this but I think you'll find acoustic power is proportional to voltage and not voltage squared. But I haven't got all the formulas in front of me at the moment.

#### speaker dave

You are correct that it is all about the phase of the adding sections.

First lets note that most of the textbook crossover examples are ideal cases with no added phase shift or roll offs in the drivers, yet with most real systems a woofer will have depth that gives delay to LF relative to HF. As such most of these simulations are equivalent to adding electrical filters rather than speaker units. They do give some insight into how drivers can add, though.

With the first order examples the units are usually 90 degrees apart. Woofer phase has fallen 45 degress at the crossover point and the tweeter has advanced 45. The 90 degree separation means that units add with +3dB gain rather than +6. This is a simple result of adding two vectors at 90 degrees. Triginometry says the resultant vector is 1.4 times each, rather than 2 times, so +3dB.

The typical 2nd order crossover example gives 180 degree separation between sections so you must phase reverse a driver, usually the tweeter. At this point two drivers 3 dB down would be totally in phase at crossover and give a 3dB hump (6dB higher than -3), so we must either adjust Q (causing the corners to droop) or separate the crossover points a bit. One solution to this, double butterworth targets as used for Linkwitz/Riley crossovers will give units in phase through the crossover region and -6dB down at the corners. Adding units with these properties will give flat response.

If you want to get a glimpse of more real-world crossover design, you might read through this thread I wrote at Classic Speakers on redesigning an AR4x crossover:

Crossover mods for the AR4x - The Classic Speaker Pages Discussion Forums

Regards,
David S.

#### speaker dave

Hi Pete. I'm not 100% confident of this but I think you'll find acoustic power is proportional to voltage and not voltage squared. But I haven't got all the formulas in front of me at the moment.

Voltage is analogous to pressure so acoustic power is pressure squared over acoustic impedance.

David S.

#### speaker dave

An edit. Re; "Q". I have assumed Butterworth or maximally flat filters in all cases. They have a Q of roughly 0.7.

Butterworth equals Q of .7 for 2nd order only!

Every order has a series of required Qs to make a Butterworth total. Usually refered to as the "Butterworth coefficients".

#### MCPete

Thanks to Speaker Dave for the fantastic explanation.

By the way, isn't it very counterintuitive that summing two sounds of equal intensity can have a total intensity four times greater than the original separate intensity of each? Common sense would have you expect a doubling, not four times.

-Pete

#### speaker dave

Thanks to Speaker Dave for the fantastic explanation.

By the way, isn't it very counterintuitive that summing two sounds of equal intensity can have a total intensity four times greater than the original separate intensity of each? Common sense would have you expect a doubling, not four times.

-Pete

I guess I would think in terms of pressure. Two units of equal pressure add to double the pressure, which is intuitively simple.

Intensity is similar to power (proportional to pressure squared), so yes, double the pressure (summing two equals) is four times the intensity or power. This is similar to considering power amplifiers: doubling the output voltage would give 4 time the power. Just the way it is

David S.

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