MOTM

> >>  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.
> Not in any way except by representing higher frequencies.
> Weren't we just here?

Yes, and what you're stating is incorrect. ;-)

If I have a 20khz sample rate, and I have the following waveforms being
sampled (assuming PERFECT alignment of the sample point and the peaks of
each cycle of each waveform):

10Khz Sine wave
10Khz Square wave
10Khz Sawtooth wave
10Khz Pulse wave

When played back at the same 20khz sample rate, they are *ALL* going to be
sine waves (assuming an ideal filter, of course). The peaks from the
sawtooth wave are now rounded.

Now let's assume a 40khz sample rate with the same 10Khz signals above.
Each waveform looks quite a bit closer to its original. Therefore, a
higher sample rate == higher detail at the same original input frequency.

If you double the sample rate, you double the significant samples within a
waveform, making it closer to the original. Hopefully this clears it up
100%.

-->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