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-   -   At which frequency does "ear canal gain" start for headphones? (http://www.diyaudio.com/forums/multi-way/37829-frequency-does-ear-canal-gain-start-headphones.html)

454Casull 14th July 2004 09:49 PM

At which frequency does "ear canal gain" start for headphones?
 
Assuming a typical circumaural headphone, when would the 12dB/octave gain start?

Svante 14th July 2004 10:42 PM

Re: At which frequency does "ear canal gain" start for headphones?
 
Quote:

Originally posted by 454Casull
Assuming a typical circumaural headphone, when would the 12dB/octave gain start?
I'm not really an expert on this, but I could possibly make an educated guess. I would suggest that it would occur at the same frequency as any "room gain", just that the "room" is pretty much smaller. :D If the diameter of the cavity is some 4 cm, half a wavelength would fit into that at some 4 kHz. That would be a rough guesstimate. Hmm, this means that headphones should be compensated for this. Maybe they are, I don't know.

Edit: But wait, you call it "ear canal gain" in the subject line. The ear canal would have nothing to do with what I described. The ear canal acts like a pipe resonator, also around those frequencies. That resonance is responsible for the increased sensitivity of the ear in the range 1-5 kHz.

454Casull 14th July 2004 10:59 PM

Re: Re: At which frequency does "ear canal gain" start for headphones?
 
Quote:

Originally posted by Svante


I'm not really an expert on this, but I could possibly make an educated guess. I would suggest that it would occur at the same frequency as any "room gain", just that the "room" is pretty much smaller. :D If the diameter of the cavity is some 4 cm, half a wavelength would fit into that at some 4 kHz. That would be a rough guesstimate. Hmm, this means that headphones should be compensated for this. Maybe they are, I don't know.

Edit: But wait, you call it "ear canal gain" in the subject line. The ear canal would have nothing to do with what I described. The ear canal acts like a pipe resonator, also around those frequencies. That resonance is responsible for the increased sensitivity of the ear in the range 1-5 kHz.

I thought the gain started at the frequency whose wavelength was longer than 1/2 the longest dimension of the volume of the listening enclosure. Wouldn't the length of the ear canal be the longest dimension?

Svante 14th July 2004 11:46 PM

454: Again, with little practical experience from this: I think that this 1/2-wavelength-of-the-biggest-dimension rule is appropriate for box shaped rooms. With headphones, we have h cavity outside the head, and then the earcanal. This is not a box-shaped room, so I guess that that rule does not nessecarily hold. OTOH the resonance in the earcanal is maybe 3kHz, so the number turns out about the same. ;)

Why are you asking in the first place?

454Casull 14th July 2004 11:51 PM

I was just curious.

Actually, the question stemmed from my wondering how much excursion a headphone driver would have to undergo to produce the entire audible spectrum.

BillFitzmaurice 15th July 2004 02:46 AM

There's no such thing in an ear, any more than there is in a car. In small spaces pure air pressure achieved via compression of the air within the space is the medium by which frequencies are transmitted. This, by the way, is the origin of the term SPL- Sound Pressure Level.

Svante 15th July 2004 08:49 AM

Quote:

Originally posted by BillFitzmaurice
There's no such thing in an ear, any more than there is in a car. In small spaces pure air pressure achieved via compression of the air within the space is the medium by which frequencies are transmitted.
The main question here is "when is the space small?". It is certainly small for low frequencies, but for high frequencies (eg 20 kHz) it is definitely large. Somewhere in between those frequencies the "room gain" starts. This frequency would be roughly 4 kHz for the headphones, roughly 100 Hz for a normal living room.

"A tweeter is big, and a woofer is small, acoustically."

Svante 15th July 2004 09:04 AM

Quote:

Originally posted by 454Casull
I was just curious.

Actually, the question stemmed from my wondering how much excursion a headphone driver would have to undergo to produce the entire audible spectrum.

Ok, so the pressure inside a small cavity is the volume flow multiplied by the acoustic impedance of the cavity. In absolute terms:

p=U*1/(wC)

Since the velocity (v) is the derivative of displacement (x), and flow (U) is the velocity multiplied with the surface (S) of the membrane:

p=wxS/(wC)=xS/C

C is the acoustic compliance

C=V/(rho0*c^2)

so p=xS*rho0*c^2/V, or

x=pV/(S*rho0*c^2)

Assuming the p=1 Pa (94 dB), V=10e-6 m3 (10 ml), S=0.0005 m2 (5 cm2), rho0=1.2kg/m3 and c=345 m/s, the displacement would be 0.14 um. (I Hope I got it right...)


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