The Panning Problem

An interesting illustration of the difference between coherent and noncoherent signal addition comes up in the problem of stereo panning between two loudspeakers. Let $ x_L(t)$ and $ x_R(t)$ denote the signals going to the left and right loudspeakers, respectively, and let $ g_L$ and $ g_R$ denote their respective gain factors (the ``panning'' gains, between 0 and 1). When $ (g_L,g_R)=(1,0)$ , sound comes only from the left speaker, and when $ (g_L,g_R)=(0,1)$ , sound comes only from the right speaker. These are the easy cases. The harder question is what should the gains be for a sound directly in front? It turns out that the answer depends upon the listening geometry and the signal frequency content.

If the listener is sitting exactly between the speakers, the ideal ``front image'' channel gains are $ g_L = g_R = 1/2$ , provided that the shortest wavelength in the signal is much larger than the ear-to-ear separation of the listener. This restriction is necessary because only those frequencies (below a few kHz, say), will combine coherently from both speakers at each ear. At higher frequencies, the signals from the two speakers decorrelate at each ear because the propagation path lengths differs significantly in units of wavelengths. (In addition, ``head shadowing'' becomes a factor at frequencies this high.) In the perfectly uncorrelated case (e.g., independent white noise coming from each speaker), the energy-preserving gains are $ g_L = g_R =
1/\sqrt{2}$ . (This value is typically used in practice since the listener may be anywhere in relation to the speakers.)

To summarize, in ordinary stereo panning, decorrelated high frequencies are attenuated by about 3dB, on average, when using gains $ g_L = g_R = 1/2 = -6$ dB. At any particular high frequency, the actual gain at each ear can be anywhere between 0 and 1, but on average, they combine on a power basis to provide a 3 dB boost on top of the $ -6$ dB cut, leaving an overall $ -3$ dB change in the level at high frequencies.


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