Re: [Phys-l] high frequency sounds

["John Clement" <clement@hal-pc.org>]

But of course you do not have just one cycle of a 20 kHz wave.  So with a
more time extended wave you do have a better representation.  There are a
number of programs which can slow down playback, and even some that can
change the pitch of a wave without changing the tempo.

Winamp is a free audio player which comes with some slow down software, and
it is possible to download pitch/temp shifting plugins.

Changing the digital to analog requires a DAC which can be conveniently
purchased in chip form.  But supposedly the best conversion is done by a 1
bit processor which puts out a number of pulses equal to the size of the
number.

Most CDs have been encoded using the normal 44 kHz ADCs, but some have been
encoded using higher sample rates and are then digitally downshifted to the
CD sample rate.  Some golden eared people claim that encoding at 96kHz is
better.  But many DVDs are encoded at 48kHz.  DVDs actually have a range of
encoding frequencies.

John M. Clement
Houston, TX

> > With a 44kHz sampling rate,  one cycle of a 20 kHz sound wave will be
> > represented by only 2 numbers.   So above 15kHz, the digital version of
> the
> > sound is not a very good 'analog'.
>
> Why not? As per the sampling theorem, a 44kHz sampling rate allows
> one to "perfectly" recreate a waveform with frequency components up
> to 22kHz, well above most everyone's hearing range. Even with just 2
> numbers.
>
> But I admit to not knowing exactly how the playback electronics
> works. I'm assuming the CD is encoded at 44kHz regardless, and it is
> up to the electronics to create the analog interpolation. I think
> most recording is done at roughly 4x oversampling, but is
> downsampled/decimated for the CD.
>
> Nuances (aka filter shapes/bandwidths), economics of chipsets, and
> audiophile arguments abound wrt these issues, and I admit I'm not up
> on all that, but are they enough to thwart the sampling theorem in
> its "first approximation?"