Attack of the Clone

[DSP_Geek]

Originally posted by planet10


Yes please... 1st order i knew, i always assummed a roll-off at the -3 dB point, but have never built such a system. I always mount the 0.5 woofer on the back, where a fore-aft cabinet symmetry guarantees perfect BSC (and since i like bipoles always just let the back driver (usually close to FR run all out)

dave

Depending on how thick your cabinet is, you might run into problems with back-to-front delay adding phase shift to your BSC, thus adding ripples to your frequency response. Say your cabinet's 15 inches wide and a foot deep - your BSC -3db point is about 300 Hz, but the frequency for 180 degree phase shift from back to front is 550 Hz, which is about -1.5 dB off half space. However, the backwave signal is around 0.15 of the front signal (about 0.14 of reference, but front signal itself is still diffracting a bit), so you actually get a signal which is 0.7 of the half-space reference, or in other words -3db instead of the 0 db you were looking for. It's not a horrible dip, and you can avoid it by using a wider front baffle and a slimmer cabinet.

In any event, I promised a derivation, and here it is.

Just to keep things clear, s = jw, where j is the root of -1 and w is omega, 2 * pi * frequency. As well, I'll use sqrt(x) for root of x. A(s), B(s), and so on are the representation of gain as a function of s.

A first order low pass transfer function is 1/(s+1), and a first order high pass transfer function is s/(s+1), for w normalised to 1.0. Lowpass functions for denormalised frequencies are w/(s+w) to keep the gain = 1.0 at s = 0. One thing to remember, the absolute value of a complex expression is derived by the square root of the real and imaginary parts. For example, if we have (3 + 4j), then the absolute value is sqrt(3*3 + 4*4) = 5.

Onward!

We can model the baffle step by a constant transfer function added to a first order high pass filter:

Eq.1: BS(s) = 1 + s/(s+1)
Eq.2: BS(s) = (s+1)/(s+1) + s/(s+1)
Eq.3: BS(s) = ((s+1) + s)/(s+1) = (2s+1)/(s+1)

As it turns out, the -3db point is found at w = sqrt(1/2):
Eq.4: BS(0.707j) = (2*sqrt(1/2)j + 1)/(sqrt(1/2)j + 1)

Now we generate the absolute value by squaring the real and imaginary terms, summing them, then taking the root:
Eq.5: abs(BS(0.707j)) = sqrt((2+1)/(1/2+1)) = sqrt(2)

and sqrt(2) happens to be the amplitude which corresponds to +3 dB.

That's all well and good, but how do we fix this? Looking at Eq.3, we can merely flip the fraction to get a good hint as where to go, because we want BS(s)*BSC(s) to be flat with respect to s:

Eq.3: BS(s) = (2s+1)/(s+1)
Eq.6: BSC(s) = (s+1)/(2s+1)

Let's see what happens if we divide the bottom by 2 so we can keep the (s+1) terms intact for cancelling, yet restate the denominator in terms of s:
Eq.7: BSC(s) = (s+1)/(s+1/2)

That implies the correction might happen at w = 1/2, so let's see what the filter looks like, from the LPF form mentioned above:

Eq:8: H(s) = (1/2)/(s + 1/2)

And now we create a baffle step compensator by summing this lowpass filter with a constant gain of 1.0:

Eq.9: BSC(s) = 1 + H(s) = 1 + (1/2)/(s + 1/2) =
(s+1/2)/(s+1/2) + (1/2)/(s + 1/2) =
((s + 1/2) + 1/2) / (s+ 1/2) = (s+1)/(s+1/2)

which is identical to Eq.7 above, so we have a perfect baffle step compensator.

This implies the lowpass filter for one woofer needs a cutoff frequency half that of the equivalent filter formed by the baffle step, but remember from Eq.4 that the baffle -3 dB point is reached at sqrt(1/2), or 0.707, of the baffle filter, so the baffle step filter cutoff is actually (1/2)/sqrt(1/2) = sqrt(1/2) of the baffle -3dB point. The whole system depicted here has a gain of 2, but that's not a problem.

Long story short, if your baffle has a 3 db step at 420 Hz, tune your crossover to first-order lowpass one woofer to 300 Hz, and run the other to the tweeter crossover point.


Cheers,
Francois.

[planet10]

Originally posted by DSP_Geek
In any event, I promised a derivation,

Thanx Francois

I'll persuse that when i'm a bit less rushed... can i add that to the series of BS articles i have on my website?

dave

[audiobomber]

Originally posted by DSP_Geek
Actually, if you want to do a 2.5 way with one driver rolling off early to provide baffle step compensation, that first rolloff must be first order since the baffle step is first order, and the corner frequency (-3 dB point) must be 0.707 of the baffle step 3 dB point. I have a derivation lying around if anyone's interested.

Audiobomber crossed over too high and at too high a slope for BSC to work well; given the system description there might have been a midrange peak around 400 Hz.

It wasn't a 2.5-way system. Both the mid and woofer were rolled off at 480 Hz.

[audiobomber]

Originally posted by planet10
three-ways work best if the XO is below 250-300 Hz... which means you need a baffle on the order of 13-15" wide (or more).

I wouldn't try crossing at 500 Hz again, but not all systems follow that rule. I know there are 3-way designs that cross at 500 Hz and that's the recommended XO for the JX53, which I was considering until I tried a similar crossover point with the WR125S.

[dwk123]

Originally posted by planet10



So any increase in efficiency has to be something acoustic (ie improved coupling to the air, but i'm having trouble visualizing this)

dave

My understanding is that it is basically exactly this - the drivers couple to increase efficiency. I believe the analogy is that power goes as amplitude^2. Two drivers couple together to produce double the amplitude with double the electrical power input. However, that amplitude causes 4x the acoustic power, which gives you your 3dB increase. Only works if the drivers are close enough to couple, though - about 1/4 wavelength.

A bit of hand-waving there, but I think it holds.

[planet10]

Originally posted by audiobomber
I know there are 3-way designs that cross at 500 Hz

Unfortunately true... none i'd consider

dave

[DSP_Geek]

Originally posted by planet10


Thanx Francois

I'll persuse that when i'm a bit less rushed... can i add that to the series of BS articles i have on my website?

dave

Attribute that bad boy and we're golden. Mods, how do we get this under a title where people could see/use it?


Thanks in advance,
Francois.

[DSP_Geek]

Originally posted by audiobomber


It wasn't a 2.5-way system. Both the mid and woofer were rolled off at 480 Hz.

I didn't express myself all that well; the second paragraph was more or less independent of the first. Still, if you have a second order rolloff on the woofers, they'll be coming in too steeply below crossover to properly compensate for baffle step, ergo my guess at a 400 Hz peak.


Francois.

[audiobomber]

Originally posted by DSP_Geek
Still, if you have a second order rolloff on the woofers, they'll be coming in too steeply below crossover to properly compensate for baffle step, ergo my guess at a 400 Hz peak.

I didn't measure, so I can't say if your speculation is correct. But I'm reasonably certain that the sound I was so displeased with was not related to frequency response. I'm quite sensitive to coherence and that's what was lacking. Voices did not sound natural, they sounded kind of fractured or imprecise. I'm not sure how to describe it. It just wasn't a convincing illusion of a real person singing.

[catapult]

So any increase in efficiency has to be something acoustic (ie improved coupling to the air, but i'm having trouble visualizing this)

Dave, I don't think your belief system and the "common knowledge" are in any danger.

First we know the 3dB efficiency gain when you double the drivers thing is true because it's verified experimentally by measuring power and SPL.

Let me take a crack at explaining it. I've never really thought it through before so lets see if if works. Let's work with a voltage source amp. At a given frequency voltage is proportional to cone displacement (X). Pressure in Pascals is proportional to the volume of air moved (Vd) which is X times the cone area (Sd).

Single 8 ohm driver
1 watt = 2.83 volts at 8 ohms
X = 1 mm
Sd = 100 cm^2
Vd = 10 cm^3

Two 8 ohm drivers in parallel ---> 4 ohm load
1 watt = 2.0 volts at 4 ohms
X = 2/2.83 = .7 mm
Sd = 200 cm^2
Vd = 14 cm^3

Two 8 ohm drivers in series ---> 16 ohm load
1 watt = 4.0 volts at 16 ohms
Each series driver sees 2 volts
X = 2/2.83 = .7 mm
Sd = 200 cm^2
Vd = 14 cm^3

So at 1 watt, 2 drivers move 1.4 times more air than 1 driver, no matter how they are wired. The sound pressure will also increase 1.4x. Convert to relative SPL:

SPL increase = 20*log(1.4) = 3dB.

Make sense?