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Old 17th April 2006, 07:22 AM   #1
real is offline real  Australia
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Default Crossover Summed Response

Hi All,

I am in the process of building some speakers, specifically in designing the crossover. I am having trouble assessing the summed response. All sources and websites I have visited so far have simply used the sum of the output voltages from each leg of the crossover - this doesn't seem right to me.

Should it not be the sum of the power output from each leg of the crossover, with respect to the phase that it is output at? Assuming a resistive load (complex loads are much too complicated for this purpose), the power output will be proportional to the square of the voltage magnitude, at the voltage angle (assuming similar driver sensitivites and impedances/resistances):

Output = |Vwoofer|^2 * exp(i.angle(Vwoofer)) + |Vmid|^2 * exp(i.angle(Vmid)) + |Vtweeter|^2 * exp(i.angle(Vtweeter))

I have used this formula, in conjunction with many different filter types (Bessel, Butterworth, Chebychev and others) to get a 3 way 3rd order crossover with the flattest frequency response (~ 2dB variation).

I must use at least a 2nd order crossover for the tweeter. I could not get a flat enough response with the second order crossover though.

What are peoples thoughts?

Thanks, regards,

Chris.
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Old 17th April 2006, 10:40 AM   #2
lndm is offline lndm  Australia
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You are right in that it is more complex than that. You need four plots per driver to properly construct a crossover. It can be complicated but is drastically improved with a crossover simulator.

You need an impedance magnitude plot and that impedance's phase plot to calculate the effect of the crossover components on the driver. The filter transfer function translates to the effect on the driver's measured acoustic frequency response plot. The driver's acoustic phase plot is required to sum multiple drivers, and to see group delay issues.

All of the above can be measured with simple equipment and will take an initial investment of time.
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Old 17th April 2006, 02:34 PM   #3
sreten is offline sreten  United Kingdom
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Hi,

for what its worth Linkwitz/Riley croosovers are the only filters
that theorectically sum to a flat power response, it is this that
defines this filter type - which is only relevant to crossovers.

I say theorectically because the response required is acoustic,
not electrical, and the units also have to be time aligned.

/sreten.
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Old 17th April 2006, 04:56 PM   #4
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Quote:
for what its worth Linkwitz/Riley croosovers are the only filters that theorectically sum to a flat power response, it is this that defines this filter type - which is only relevant to crossovers.
NO - others definitely sum flat as well.

Quote:
I say theorectically because the response required is acoustic,
That part however is true but again not only for LR.

Regards

Charles
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Old 17th April 2006, 06:03 PM   #5
real is offline real  Australia
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lndm, I'm not sure I have the equipment or time to measure the acoustic phase plot of my drivers! However, I do have a signal generator and oscilloscope, and I have measured the |impedance| of the drivers, and will use zobels/resonant circuits to make it flat.

Sreten, I am currently looking into Linkwitz-Riley circuits, thanks for that.

Charles, which others sum flat? It should be the |voltage|^2 at voltage angles which sum flat, correct?

Thanks for the help,

Chris.

PS, is there any problem using high (3rd to 4th) order crossovers in speakers apart from cost? I have been reading that their time/transient response is not that good.
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Old 17th April 2006, 09:35 PM   #6
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If you need some testing or simulations done Real, drop me line, I might be able to help you (I'm also in Melbourne).
David
www.gattiweb.com
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Old 18th April 2006, 01:02 AM   #7
lndm is offline lndm  Australia
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You'd be surprised what you can do with a cro configured for lissajous (X-Y) patterns, a frequency table on paper, a little common sense, time and determination, for measuring phase.

These days I plug my mic into my soundcard and use FFT software to do it in seconds.

Having spent many years designing crossovers by hand, I believe that no matter how much theory you absorb, or calcutations you make, you will virtually always need to tweak by hand/ear afterward, with an open mind.
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Old 18th April 2006, 01:46 PM   #8
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Quote:
Charles, which others sum flat? It should be the |voltage|^2 at voltage angles which sum flat, correct?
Third-order Butterworth for example and all crossovers that behave like:

H(s) = (1+sT+s^2*T^2+ ... + s^n*T^n)/(1+sT+s^2*T^2+ ... + s^n*T^n)

Regards

Charles
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Old 18th April 2006, 04:02 PM   #9
real is offline real  Australia
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David, thanks for the offer, I will pass my design by you, see what you think. It won't be for a few weeks though as I am currently overseas! In fact, I'll just post on the forum when I get back - I have done plots on MATLAB too of the summed responses.

lndm, I am not sure what you are talking about wrt the X-Y patterns, could you explain to me please?


Quote:
Originally posted by phase_accurate

H(s) = (1+sT+s^2*T^2+ ... + s^n*T^n)/(1+sT+s^2*T^2+ ... + s^n*T^n)

Charles,

H(s) = 1 in this case! Third order butterworth is what I have designed thus far - I am mainly checking to see that I have used the right formulas for getting the summed response - since no one has said otherwise, I assume that I am correct!

Additionally, I mentioned that I did impedance measurements of the drivers, though this was in free space, no enclosure. Is this an issue? It only really affects the woofer (mid is dome type), which only needs compensation for the impedance rise at high frequencies - will this be different when in the enclosure? I will check it anyway when I have time to do all this stuff!

Thanks all,

Chris.
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Old 18th April 2006, 09:04 PM   #10
Svante is offline Svante  Sweden
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Quote:
Originally posted by sreten
Hi,

for what its worth Linkwitz/Riley croosovers are the only filters
that theorectically sum to a flat power response, it is this that
defines this filter type - which is only relevant to crossovers.
I don't agree. Odd order filters sum up to a flat response if the branches are butterworth. The easiest case would be the first order filter, which cannot be LR (they are always even ordered):

Hlp=1/(1+s/s0)

Hhp=s/s0 / (1+s/s0)

The sum of the two is (1+s/s0)/(1+s/s0)=1
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