MOTM

> >That's already a given, and you missed my point anyway. If you sample a
> >sine wave @ 20Khz and a square wave @ 20khz, you will only get a 10khz
> >square wave when you go D to A. The sine wave will lose detail.
> No, it won't. That's the whole point of the Nyquist theorem.
> Everything below the Nyquist frequency is reproduced *exactly* (given
> ideal filters etc.). A 20kHz sine wave is just as detailed when
> sampled at 44.1kHz as when sampled at 96kHz; either way, it contains
> all the information of the original wave.

Misstated example - Replace "sine wave" with "square wave". The square
wave turns in to a sine wave.

> The only thing increasing the sample rate does is allow you to
> represent higher frequencies (and possibly to design a better
> real-world filter).

And represent the original waveshape better provided it's not a sine wave.
;-) Change the input to a sawtooth or a square wave, you'll get a sine
wave out. A higher sample rate will not yield a sine wave as the lower
sample rate will. So again, increasing the sample rate will yield a closer
to the original waveform representation.

-->Neil

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Neil Bradley            In the land of the blind, the one eyed man is not
Synthcom Systems, Inc.  king - he's a prisoner.
ICQ #29402898