:: > This is a little bit problematic, because the number of distorted
:: > bands does not tell you the weight of distortions you will get.
:: > What do you think sounds uglier:
:: > 20 distorted bands, each 0.1 dB
:: > or
:: > 1 distorted band by 2 dB
:: > ???
::
:: I would say that 20 distorted bands at 0.1dB is preferable, and assume
:: that this is the choice of lame's algorithms. Is this assumption wrong ?
::
Yes and no.
That depends on the distortion of the undistorted bands. Also undistorted
bands have an distortion, it is negative.
First you need a table of the probability to hear distortions:
distortion probability error noise
[d=dB] [p=%] [i]
-10 50.0 0.0000
-3 50.8 0.0004
-2.5 52.0 0.0025
-2 53.2 0.0064
-1.5 55.6 0.020
-1 57.9 0.040
-0.5 63.3 0.116
-0.25 66.3 0.176
0 70.2 0.281
+0.1 72.2 0.336
+0.2 74.2 0.422
+0.3 76.1 0.504
+0.4 78.0 0.593
+0.5 80.0 0.706
+1 87.1 1.28
+1.5 93.7 2.34
+2 97.2 3.65
+2.5 99.0 5.43
+3 99.8 8.4
+3.5 99.9 9.5
probability is taken from an experiment (propability to recognize
distortion depending on the level of the distortion).
It is used to compute the steepness of the penalty function.
[19*-0.25 dB + 2 dB] gives a penalty of 3.65+19*0.176 = 6.994
[19*0 dB + 2 dB] gives a penalty of 3.65+19*0.281 = 8.989
[20*0.1 dB] gives a penalty of 0.336*20 = 6.720
So 20*0.1 dB is the best of the three examples. Such *fuzzy* error
calculations working with probability distributions and are much more
realistic than the black sheep model.
A pleasant side effect is the also the numeric stability.
Operators like min() and max() are speckling the search space (bad
english?).
Tables may be taken from JAES papers.
--
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Frank Klemm
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