# [music-dsp] stereo-wide pan law?

Jerry lanceboyle at qwest.net
Mon Feb 20 23:26:21 EST 2012

```On Feb 11, 2012, at 1:00 AM, Ross Bencina wrote:

>
>
> On 11/02/2012 2:27 PM, Jerry wrote:
>> Glad to help. With your set-up, if you try to put a loud low frequency signal well outside the loudspeaker array, you will notice that your speakers and/or amplifiers will have melted. To the extent that sin(theta_A) = theta_A (small-angle approximation), for every halving that you reduce the loudspeaker spacing, there is a doubling of low frequency amplitude requirements for images at 90 degrees. There is a cure for this--let me know if this situation bites you. (Also, for every halving of loudspeaker angle, there is an octave added to the frequency range over which the widening works.)
>
> Hi Jerry,
>
> I'm not sure I follow you here.
>
> I'm not using ITD, only amplitude (a sum of in-phase and 180-degree phase signals). I don't see how this can impact the frequency response. Can you give me a clue?
>
> Thanks
>
> Ross.

Sure I can give you a clue. (And sorry for the retarded response. ;-)

Here's an answer for your ITD question, although it doesn't seem related to the problem at hand.

You _are_ using ITD (Interaural Time Difference). Don't confuse loudspeaker time difference with ITD. Any sound which seems to be at any angle but 0 or 180 degrees has a non-zero ITD and you've just told me that Bauer is working, so therefore....

Let's constrain our thinking to low frequencies, where the interaural spacing is much less than a wavelength. (The ear spacing is equal to a wavelength at approximately 1 KHz.) With this assumption, we can ignore the head as such and consider only the ear locations as points. If the acoustical spacing of the ears is D and we use the same notation as we used for Bauer's Law of Sines (which has been deleted at this point), then with c the speed of sound, the ITD, using the left ear as the phase reference, is

(S_l - S_r)   D * sin(theta_A)
ITD = ----------- * ----------------
(S_l + S_r)          c

I just derived this from scratch (I'm sure its been published somewhere). I don't think I've made a mistake because it is easy to show that when using the Bauer relation the ITD is the same as for a real sound emanating from theta_I.

But I'm not sure why you ask about ITD. Your better question is about frequency response. For the moment, retain the low frequency scenario. Then the sound from one speaker will be delayed relative to the other speaker causing a severe comb filter as a function of frequency. This happens at both ears. If we look at what happens at higher frequencies (and with an actual head in place), the frequency dependence is all over the place, varying by easily 15-20 dB depending on the setting of your pan pot, relative to a real sound source at the angle that you have panned to. Amplitude panners fall apart in this respect above about 600 Hz. To fix this problem, one must use HRTF-based panners. Alternately, I am told that mixing people sometimes dial in an ad hoc EQ which can ameliorate the problem.

However, this still doesn't (quite) get to what I think you want to ask about, which is why the increased amplitude requirements when panning a low frequency sound far to the side. Consider that the speakers are acoustically close together, as they are at low frequencies. To make the phantom image far to the side, you are instructing the woofer cones to move in opposite directions at all times. (Look at the panning law.) Sources operating in these conditions are nearly cancelling each other out, with first one cone pushing air out while at the same time the other is sucking it in. Only a little bit of the sound actually radiates outwards. This is all in a valiant effort to make the air in the vicinity of your head wiggle left-and-right. The closer the speakers, the more cancellation. Looked at this way, it all makes sense, doesn't it?

HTH,
Jerry
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