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- Date: Fri, 22 Jan 1993 13:51:42 +0000
- Sender: Discussions on all forms of Music <ALLMUSIC@AUVM.BITNET>
- From: "Out,
- out - you'll not feel the fall-out..." <OLIVOTTO@ITNCISCA.BITNET>
- Subject: Re: Technonerdophilia
- Lines: 33
-
- Paul Heroy says:
-
- >I don't remember the exact math, but every bit basically adds 6 dB to the
- >range, giving a nominal 96 dB range for 16 bits. How then do most CD players
- >claim a dynamic range of over 100 dB??? (Actually this is legitimate and
- >depends on the way you measure noise, or something like that.)
-
- The decibel is defined as 20*log(Aout/Ain), where Aout is the out-
- put signal amplitude, Ain the input signal amplitude and the log is taken
- in base 10. In general, Aout and Ain need only be amplitudes of comparable
- signals -- you're not always dealing with the *same* signal processed.
- In binary, if you shift a signal one bit left, you actually double
- its amplitude (ex: 1000 --> 16, 10000 --> 32), thus Aout/Ain = 2 and
-
- 20*log(2) = 20*0.301 = 6.02 dB.
-
- If you have 16 bits, the maximum amplitude is 2^16 (minus 1, to be
- precise, but let's forget), and the minimum *significant* amplitude is 1.
- So:
-
- 20*log(2^16) = 20*4.816 = 96.33 dB.
-
- Sticking to the original definition of dB, there's little one can do
- to push this further. Oversampling may be the key to this...?
-
- Ciao, Marconerd
-
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- Like a shot, like a paparazzo picture gone to pot...
- ---------------------------------------------------------------------------
- Marco Olivotto, C.I.S.C.A. - Univ. degli Studi di Trento, POVO (TN) - ITALY
- olivotto@itnvax.cineca.it OLIVOTTO@ITNCISCA
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-