Linear sweeps don't work, they have to be log in order for the harmonics to line up in a sweep plot. There are no mathematics that can do this with a linear sweep.
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.
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
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.....
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Btw, I don't really see why a linear sweep wouldn't work?
You can even get a distortion number from a single frequency.
It's just a matter of FFTing and crosscorelation.
I can only see that a log would be much more efficient way.
You can even get a distortion number from a single frequency.
It's just a matter of FFTing and crosscorelation.
I can only see that a log would be much more efficient way.
You forget the first picture, which shows the amount of boost.
Even says in the description EQ curve.
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.
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??
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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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.
Don't know about the data, but it lowers Qt' = higher roll-off = less LF eff..
(Qt'): (Qt) + any added series resistance (Rs)
The rest can also be found in that paper (which I assume you read)
as well as;
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:
as well as;
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:
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Yep, we're always 'robbing Peter to pay Paul' or in speaker design lingo: 'trading efficiency for bandwidth'.
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.
If you compare to the the other graph (I traced them just for good comparison)
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! 🙂
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.
If you compare to the the other graph (I traced them just for good comparison)
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! 🙂