Ok, this brings me to my next question: The frequency response of a 2.5way system.
Assume you have midrange driver A and woofer B, both with identical sensitivities. Midrange driver A is in a closed box of small volume, whereas woofer B is in a large, ported box. There are one of each of these drivers and they are connected in a 2.5way system; at a given frequency below the low-pass filter of the woofer both are recieving the signal. However, given that driver A is a midrange driver in a small, closed box, it will be rolling off earlier than the woofer. My question is what happens to the output?
Will the output of the system at the frequencies where A is rolling off be the same as if this was a 3-way system and only woofer B was playing? i.e. is frequncy response at the low end of a 2.5way system a sum of driver outputs or is it some other way?
Philip
p.s. what is an OTL amp?
Assume you have midrange driver A and woofer B, both with identical sensitivities. Midrange driver A is in a closed box of small volume, whereas woofer B is in a large, ported box. There are one of each of these drivers and they are connected in a 2.5way system; at a given frequency below the low-pass filter of the woofer both are recieving the signal. However, given that driver A is a midrange driver in a small, closed box, it will be rolling off earlier than the woofer. My question is what happens to the output?
Will the output of the system at the frequencies where A is rolling off be the same as if this was a 3-way system and only woofer B was playing? i.e. is frequncy response at the low end of a 2.5way system a sum of driver outputs or is it some other way?
Philip
p.s. what is an OTL amp?
Sounds like people are mixing and matching efficiency and sensitivity. They aren't the same. Efficiency is SPL at 1 watt power. Sensitivity is SPL at 2.83 volts. If the speaker's impedance is 8 ohms, efficiency equals sensitivity, otherwise they are different.
So, let's say each driver is 8 ohms to keep it simple. Efficiency only depends on the number of drivers, not how they are wired. Double the drivers, efficiency goes up 3 dB because of the increased cone area. Sensitivity is another matter. Parallel the drivers, twice the current flows, +3dB more power at 2.83 volts. Series the drivers, half the current flows, -3dB less power at 2.83 volts.
2 drivers in parallel: +3dB efficiency +3dB power = +6dB sensitivity increase
2 drivers in series: +3dB efficiency -3dB power = 0dB, same sensitivity as one driver
So, let's say each driver is 8 ohms to keep it simple. Efficiency only depends on the number of drivers, not how they are wired. Double the drivers, efficiency goes up 3 dB because of the increased cone area. Sensitivity is another matter. Parallel the drivers, twice the current flows, +3dB more power at 2.83 volts. Series the drivers, half the current flows, -3dB less power at 2.83 volts.
2 drivers in parallel: +3dB efficiency +3dB power = +6dB sensitivity increase
2 drivers in series: +3dB efficiency -3dB power = 0dB, same sensitivity as one driver
catapult said:
This is "common knowledge" and what i've held as true forever (so never thot about) ... a couple recent posts have me questioning it...
I'm still not sure.... if we go back to 1st principles....
! driver has X volts across it and Y amps of current flowing. P = XY which gives N dB out
consider 2 drivers wired in parallel (and assume that doing so does not change the impedance)
They still have X volts but each has Y/2 amps P(one driver) = XY/2 which gives N-3 dB out, add the 2 drivers and you have N dB out with P watts in
consider 2 drivers wired in series (and assume that doing so does not change the impedance)
They still have Y amps but each has X/2 volts P(one driver) = XY/2 which gives N-3 dB out, add the 2 drivers and you have N dB out with P watts in
So any increase in efficiency has to be something acoustic (ie improved coupling to the air, but i'm having trouble visualizing this)
I'd love to have my old beliefs hold out... but ATM i'm either missing something, or my world is starting to crumble.
dave
philip said:
I've never really understood why people would do it this way instead of just making it 3-way (the more normal 2.5 way where the top of one driver is rolled off does make sense)
As the top driver rolls off before the bottom driver it will go out of phase with the bottom driver creating less bass (and confused phase)
dave
I'm not really sure what the relative advantages of 3way vs. 2.5 way are. I thought it might have something to do with fewer crossover components and the lack of a full lower crossover avoiding all the phase problems that come with it. I'm going to look into it more, hopefully that Vance Dickason book will get here today...
Anyways, I just had one quick question back to the baffle step, when you're designing for a 3 or 2.5, do you take baffle step into account for the lower woofer or not? The thing is, chances are, assuming the mid woofer has its own enclosure, its baffle step will have rolled off at several hundred Hz higher than the cabinet of the lower woofer. Therefore, if you were to introduce a bass driver with a 3dB higher sensitivity than than of the the mid woofer to compensate for baffle step, there would be a dip in sensitivity between where baffle step has caused the mid to roll off and where the bass driver comes in. On the other hand, many people seem to just forget about baffle step when they're designing anyways, as 3dB is "just noticeable," and leads to a smoother frequency response that can be somewhat compensated on the amp. Either way, the options seem to be either accept the baffle step roll off and and achieve a smooth frequency response, or compensate for it and have a dip in the frequency response. What do people think?
Philip
Anyways, I just had one quick question back to the baffle step, when you're designing for a 3 or 2.5, do you take baffle step into account for the lower woofer or not? The thing is, chances are, assuming the mid woofer has its own enclosure, its baffle step will have rolled off at several hundred Hz higher than the cabinet of the lower woofer. Therefore, if you were to introduce a bass driver with a 3dB higher sensitivity than than of the the mid woofer to compensate for baffle step, there would be a dip in sensitivity between where baffle step has caused the mid to roll off and where the bass driver comes in. On the other hand, many people seem to just forget about baffle step when they're designing anyways, as 3dB is "just noticeable," and leads to a smoother frequency response that can be somewhat compensated on the amp. Either way, the options seem to be either accept the baffle step roll off and and achieve a smooth frequency response, or compensate for it and have a dip in the frequency response. What do people think?
Philip
You should always consider BS when you are designing
A typical 2.5 way rolls in the 2nd bass driver so that it functions below the baffle-step (the weird 2.5 way mentioned is just weird (and i disgard -- rightly or wrongly -- out of hand).
In a 3-way the woofer can be brought in at the BS frequency so that its level can compensate for BS.
Myself i try to design speakers with no baffle-step. And a 3-way is a small 2-way (or 1-way) with little bass + a pair of separately powered woofers.
dave
A typical 2.5 way rolls in the 2nd bass driver so that it functions below the baffle-step (the weird 2.5 way mentioned is just weird (and i disgard -- rightly or wrongly -- out of hand).
In a 3-way the woofer can be brought in at the BS frequency so that its level can compensate for BS.
Myself i try to design speakers with no baffle-step. And a 3-way is a small 2-way (or 1-way) with little bass + a pair of separately powered woofers.
dave
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.
Francois.
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.
Francois.
DSP_Geek said:
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
planet10 said:
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 said:
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.
audiobomber said:
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.
DSP_Geek said:
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.
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?
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