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Implemented Model for ATH

From [2] I take the approximation of the curve in Fig. 2 as
\begin{displaymath}
ATH(f) = 3.64(f/1000)^{-0.8}-6.5e^{-0.6(f/1000-3.3)^2}+10^{-3}(f/1000)^4,
\end{displaymath} (6)

where $ATH(f)$ denotes the ATH in dB, and $f$ is the frequency in Hz. The problem in digital audio coding, is that one cannot know what absolute level the sound will be played at. One common solution is to set the lowest point on the curve in eq. 6 to be equal to the sound pressure level of a sine with amplitude $\pm 1$ LSB. In the coder, 16-bit samples with normalized amplitude to $\pm 1$ are used. Thus, the smallest possible sine has amplitude $A= 1/2^{15}$, and has a power of $A^2/2 = 1/2^{2\cdot15+1} = -93.32 {\rm dB}
\approx -90\ {\rm dB}$. The model used in the coder will thus be $ATH'(f) = ATH(f)-90$.


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Download bosse.pdf

``An Experimental High Fidelity Perceptual Audio Coder'', by Bosse Lincoln<bosse@ccrma.stanford.edu>, (Final Project, Music 420, Winter '97-'98).
Copyright © 2006-01-03 by Bosse Lincoln<bosse@ccrma.stanford.edu>
Center for Computer Research in Music and Acoustics (CCRMA),   Stanford University
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