Is it possible to cover the whole spectrum, high SPL, low distortion with a 2-way?

Farina's method was quite clever and rather revolutionary. What happens specifically with a log sweep (and not with a linear sweep or other broadband stimulus signals) is that a series of "distortion impulse responses" appear individually separated in the IFFT calculated IR at negative time points ahead of the larger "linear" IR. That isolates the linear part from the distortion parts avoiding one source of contamination, and also allows easy access, via windowing and FFT, to distortion curves over frequency of each individual harmonic distortion order (2nd. 3rd. 4th, etc). You get the separated harmonic distortion orders over all frequencies (at least where product frequencies are below the nyquist frequencies) in one quick take, for almost free while you get the linear frequency response. An added benefit of log sweep over linear sweep is that it applies more test energy as frequencies get lower (roughly a "pink" spectrum), which matches well with where most environmental room noise tends to. You can certainly calculate IR and FR by deconvolving lots of different test signals (even music), but none have the unique advantages of the log sweep.
 
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To be perfectly honest, I actually don't know.
Mostly because I don't care to much about the why in this case, haha :D

Fact is just that BL shifts that line up, very similar to like an amplifier and its gain/sensitivity vs low end frequency response.
So I guess if we take the equivalent circuit again, we can find our answer in there, since the BL is nothing more than a ratio.

I don't know how that will end up in math, but I bet you will find your answer there.

But again, for me this is something like I know it's there, but the question why, isn't really that important for any practical sense.
If that makes any sense? :)

For subwoofers, I don't really look at parameters that much anymore.
Just needs to have plenty of cone excursion, low Fs, power and a Qt that is low so we don't need an enormous cabinet.
Fact is that you have to just simulate things and within 30 seconds you know how it will perform in a specific cabinet with a specific target curve and SPL etc.
Makes total sense to me. I'm all about staying as pragmatic as possible.

My ongoing DIY speaker goal is to always make looking for the greatest avenue of marginal improvement,
a heck of a lot more important than striving for complete technical understanding, or trying to perfect any particular aspect of the design/build.
 
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Makes total sense to me. I'm all about staying as pragmatic as possible.

My ongoing DIY speaker goal is to always make looking for the greatest avenue of marginal improvement,
a heck of a lot more important than striving for complete technical understanding, or trying to perfect any particular aspect of the design/build.
So just off the cuff, it basically acts the same way as a transformer.

So with an higher turn ratio, you also push the low end of the frequency side up.

You can also see that in the paper, where the calculated values will get smaller.

I think that's called reflected impedance?

Anyway, maybe not a very satisfying and very abstract answer. Lol
 
We ended up lowering the BL actually, because otherwise a boost of more than 6dB was needed to get a certain target response.
I am confused... your data doesn't support the logic of your statement. You data showed higher BL having more sensitivity than the lower BL driver. No where did you need to boost.... IF this is frequency matched and green is higher BL the lowest chart shows less energy being used by the BL driver.....
1705013387929.png
1705013408015.png

 
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I am confused... your data doesn't support the logic of your statement. You data showed higher BL having more sensitivity than the lower BL driver. No where did you need to boost.... IF this is frequency matched and green is higher BL the lowest chart shows less energy being used by the BL driver.....
View attachment 1258898 View attachment 1258901
You forget the first picture, which shows the amount of boost.

Even says in the description EQ curve.
 
An higher BL also shifts up the frequency response, as can be seen in the paper as well.
Meaning that for the same acoustic response, we have to boost more in the low-end.
You didnt have to boost the low end, you chose to, you could just as easily cut the high end. This chart shows Green using less energy regardless of what eq you used to frequency match.
1705013642404.png
 
The filter used vs net gain or loss of input voltage. You adjusted the transfer function...

High Bl is more efficient. Raising BL does not lower LF efficiency according to that data...

Now that I reread what you posted I see that you're not even talking about that you're talking about the issue of the gain stage I guess??
 
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The filter used vs net gain or loss of inout voltagr You adjusted the transfer function...

High Bl is more efficient. Raising BL does not lower LF efficiency
For low frequency efficiency, see the post after;
https://www.diyaudio.com/community/...ion-with-a-2-way.334757/page-638#post-7564666

There is clearly a difference.

I don't know what else there is to say sorry.

The only thing I did, is to show a practical implementation.
We actually started that same driver with even an higher BL.
Resulting in having to boost 11.5dB for that same target response.
Which is just not doable for your pre-amp stage unless you want to clip the entire thing.

As I also said in the post, the difference in power is only a little bit.
The difference in total amplifier voltage was also only 1.5dB.
 
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The rest can also be found in that paper (which I assume you read)

Fig. 1. Theoretical Thiele-Small nominal power efficiency frequency response of the systems described in Section 3.2 for two values of Bl (Bl = 8N/A and Bl = 40 N/A). The nominal power efficiency model predicts the frequency response of the system when driven by a constant-voltage source. Note that according to this model, the high Bl value raises the upper-frequency efficiency of the system to 60% which is 14 dB above the Bl = 8 N/A efficiency, but severely rolls off the low -frequency response of the system.

as well as;

When the true efficiency of the driver is considered, it is clear that increasing the Bl factor will directly result in higher efficiency values at all frequencies. Unfortunately, the constant-voltage-drive low-frequency response may suffer, but this only means equalization must be used to flatten the frequency response.

It's about that last part.
Which sounds like amazing magic!

Unfortunately, there is always a catch in the world of physics.
The catch here is that EQ'ing is all great, be we can't boost an infinite amount.
And this is what I was just showing you.

Which btw, can also be found in the conclusion of the paper:
The downside of increasing the Bl product is the requirement that the amplifier must provide much greater voltage swing and significant bass equalization is required to drive the speaker to flat response as compared to the moderate-Bl driver
system.
 
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Btw, there is even an additional catch that even that paper glossed over.

A subwoofer only has a very limited bandwidth.

So while the total system efficiency seems to be much greater, if we focus and optimize on just a smal portion, the difference becomes actually very small.

1705017522190.png


If you compare to the the other graph (I traced them just for good comparison)

1705017972700.png



You will see that the efficiency is actually extremely similar in that part of the freq resp.
This is where the Fs is of course! :)
 
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