Acoustic filter of the Sealed Enclosure

[amo]

I am trying to understand the ABCs of speaker design, and stumbled on the following question: If we take a mid or small mid-bass driver, we can theoretically increase its power handling AND decrease its distortion by cutting off lower frequencies, which substantially increase the excursion. We also know that a sealed enclosure forms a 12 dB/Oct acoustic high-pass filter at some frequency based on specific driver characteristics. So my question is, will this filter actually decrease cone motion, just like an electric filter with the same slope would if attached to the driver suspended in free air? Can this "filter" be used for decreasing distortion? Or is this filter closer related to the 6 dB/Oct "filter" of the dipole woofer, which is more often considered the unfortunate side-effect, and is often equalized, and where the cone motion does not decrease although output does? I hope this makes since.

[kelticwizard]

The filter is exactly the same as an electrical filter. It decreases cone motion as well as output. It is one of the reasons closed box aficionados cite as an advantage of the closed box over the ported box.

Actualy, since the output rolls off at 12 dB/octave, and going an octave lower requires four times the cone motion to maintain the same SPL, what it amounts to is the cone motion will remain equal no matter how low you go, once you hit the point of rolloff. Only the output will decrease-cone motion remains the same.

Example:
A closed box speaker begins rolling off 12 dB/octave at 50 Hz. At 50 Hz, it is playing at 100 dB SPL.

By 25 Hz, it is 12 dB down. However, to play 25 Hz at 88 dB requires the same air displacement as it takes to play 50 Hz at 100 dB SPL.

And to play 12.5 Hz at 76 dB SPL requires the same displacement as is required to play 25 Hz @ 88 dB and 50 Hz @ 100 dB.

So the cone excursion stays equal no matter how far down the scale you go, unless you raise the SPL requirements.

[kelticwizard]

These two charts, in normal and metric measurements, give the displacement needed to produce a certain SPL at a certain frequency. As ou can see, to maintain the same SPL while moving an octave lower, fourt times as much air must be moved. Which translates into four times the excursion.

Normal measurments:
http://www.diyaudio.com/forums/attac...p?postid=55867

Metric measurements:
http://www.diyaudio.com/forums/attac...p?postid=55869

[Svante]

The roll-off at low frequencies of the closed box speaker will have exactly the same effect on cone amplitude as an electrical filter would. It is not related to dipole roll-off, which is an effect of the back and front volume flow cancelling each other. In the closed box such cancelling is impossible (because of the closed box...).

The distortion might be higher with the speaker roll-off, though. This is due to that the motion below the resonance is controlled by the suspension of the driver and box. This impedance is often more non-linear than the mass impedance (which dominates above the resonance).

[BillFitzmaurice]

While the previous posts are correct they don't mention the fact that you still have to electrically limit the frequency input to the speaker in question. While a small sealed box can drive upward the Fb with resultant acoustic high-pass filtering it will still try to output frequencies below Fb if you don't electrically prevent them from entering the voice coil. A good example is a sealed-back midrange, which often shares the same motor structure with a mid-bass driver but by dint of the small rear chamber pushes the driver Fs up by two octaves or so; it still requires an electrical crossover to perform properly.

[amo]

Makes sense..... but the bottom line is that you do not want to send anything below Fb to the driver. On the bright side one can use a first order xo and still end up with 18 db/oct rolloff.

Thanks!

[BillFitzmaurice]

Yes, but my personal opinion is that you can't have too much crossover slope. In this case while the total rolloff will be 18dB the driver will still be looking at constant excursion below Fb as the 6dB electrical roll off is only compensatory to the 6dB increase in excursion with each lowering of an octave. A moot point perhaps in home use with moderate power input, but potentially very significant in pro-sound usage with high power input.

[Few]

Perhaps this goes without saying, but I guess I'm saying it anyway: Another factor to keep in mind, in addition to the asymptotic slope of the rolloff (12 dB/8ve for sealed box) is the sharpness of the knee at the cutoff point. A sharper cutoff in the frequency response (sharp kink in the SPL vs frequency plot) tends to correlate with poorer time domain behavior. When adjusting box size to play with its highpass filtering effect on the woofer, you affect not just the cutoff frequency, but also the Q (just as with electrical filters). Excessively high Q's can lead to undesirable ringing in the time domain.

[sreten]

This thread is verging towards garbage.

Rolling off a driver reduces distortion end
of story, as asked in the original question.

sreten.

[kelticwizard]

Originally posted by BillFitzmaurice
...the 6dB electrical roll off is only compensatory to the 6dB increase in excursion with each lowering of an octave.

I believe that is 12 dB.

At 100 Hz and 100 dB SPL:

If you want to raise your SPL level 12 dB, you increase excursion four times

OR:

If you want to maintain 100 dB SPL but move down one octave, you also increase your excursion four times.



Four times the excursion = One octave lower = 12 dB higher SPL.

Please consult the displacement charts in post #3.