Date: Wed, 22 Aug 2001 16:04:00 -0400 (EDT) From: Stuart Barkley <stuartb@4gh.net> To: The Anarcat <anarcat@anarcat.dyndns.org> Cc: <freebsd-multimedia@FreeBSD.ORG> Subject: Re: precisions of pcm driver problems and update of rec program Message-ID: <20010822154117.S12745-100000@precipice.4gh.net> In-Reply-To: <20010822135453.W8939-100000@precipice.4gh.net>
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On Wed, 22 Aug 2001, Stuart Barkley wrote:
> The main thing is to find minimum and maximum values over an
> interval of interest. You will be clipping if you are hitting the
> maximum or minimum values. My initial code just displayed these
> values over each 1 second interval so I could visually see what
> was happening.
>
> To convert these numbers to dB use something like:
>
> #include <limits.h>
> #include <math.h>
>
> float dB = log10((double)SHRT_MAX / max) * 10.0;
>
> This can let you see the range of your data (how many dB below the
> maximum signal). For recording you do not want to reach 0dB or
> you will be clipping.
Opps. That formula I gave above is wrong. It should really be:
float dB = log10((double)SHRT_MAX / max) * 20.0;
The extra factor of two is necessary because dB is log10(x^2) and
pulling the square outside of the log10 is more efficient
computationally.
In addition, sometimes you will see systems showing negative dB.
This comes from inverting the SHRT_MAX/max component or equivalently
multiplying log10 by -1. dB can also be shown relative to other base
values which can effectively add or subtract constants when converting
between dB levels.
You might also want to use the difference between min and max to get
the full power range. However, if the signal is offset from zero it
will be the the actual max or min which causes clipping.
log10( (2^16)^2 ) * 10 = log10(2^16) * 20 = 96.3 which is where the 96
dB range for 16 bit sound cards comes from.
log10(2^1) * 20 = 6.02 dB per bit.
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