# [LAD] Ardour session archiver (Was: new lossless/lossy audio compressor)

Gregory Alan Hildstrom hildstromoo at yahoo.com
Wed Aug 1 21:18:14 UTC 2007

```Yes, I understand that any single 32-bit floating point number occupies 32 bits. I am specifically
talking about sets of numbers and the number of bits actually used to represent the set. My basic
thought is summarized in this example:

If you limit your range of possible values to positive numbers, the sign bit is useless to you
because it is always 0. This is true because of how you are using the variable; not because of the
variable's implementation. The sign bit will still be stored in the float variable, but it is
wasted space and precision if you do not use it. Your data could be represented with 31 bits
instead of 32. Obviously wav data is signed, but this illustrates my point.

I saved a 32-bit 44.1kHz 2-channel wav file using Audacity. The left channel was a full-amplitude
sine wave sweep from 20Hz to 20kHz. The right channel was a full-amplitude square wave sweep from
20kHz to 20Hz. The exponent portion of a float variable is 8 bits, which can represent 0-255 if
treated as an unsigned integer. I added a line of code to j2kaudio to print out the exponents of
each float sample while reading the wav file, which I redirected to a file and plotted with
gnuplot. The exponents ranged from 106 to 127, which is a range of 21. The 21 exponent values can
be represented with 5 bits (2^5=32).

I would argue that my wav file, saved with Audacity and ranging from -1 to +1 could be saved with
a 5-bit exponent instead of an 8-bit exponent. Each 32-bit float value occupies 32 bits, but at
most only 32-8+5=29 bits needed to be flipped in my wav file; I did not examine the mantissa.

I then tried to create another 32-bit wav file with a greater dynamic range. I created a 1Hz sine
wave at full amplitude. I then used the amplify tool (-50dB) to reduce its peak amplitude to
-600dB, which is an incredibly tiny signal. I then added a second channel with a full amplitude
sine wav. The small signal would be totally inaudible next to the normal signal. The exponent
range was 0-10 for the incredibly small sine wave values near 0. The exponent range was 106-128
for the full amplitude sine wave. The total range I was able to create with Audacity was 0-127,
still 1-bit shy of an 8-bit exponent, but those additional exponent values I manufactured were
only used to define incredibly tiny numbers near the zero crossing.

I guess I do not understand why I would use such a high-dynamic-range variable (32-bit float) and
artificially limit its range to +/-1 (31 bits). I also do not understand why such a
high-dynamic-range variable is necessary for audio reproduction given the limits of human hearing.
I do understand why you might want to use it for various computations. I do not know if this is an
Audacity-specific limitation or standard industry practice.

These are just a couple of my hangups that I did not do a very good job of explaining before.
Please clue me in if I am missing something.

Thanks. -Greg

--- Paul Winkler <pw_lists at slinkp.com> wrote:

> On Wed, Aug 01, 2007 at 10:45:59AM -0700, Gregory Alan Hildstrom wrote:
> > Are there any wav/audio programs out there that use the full float range of 3.4e+/-38? It
> seems to
> > me that this larger range would use more of the 32 bits.
>
> That's not how float works.  *Any* floating point number uses the full
> precision of all 32 bits (or 64, or whatever). This is because the
> radix point (like decimal point, but binary) can be anywhere.
>
> What you're saying would be true of fixed-point numbers.
>
> http://en.wikipedia.org/wiki/Floating_point
>
> --
>
> Paul Winkler