MOTM

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

Nope. :)

>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):

First of all, you need >2x, not 2x. So say 20.01 kHz. No perfect 
alignment necessary.

>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).

Which is a perfect representation of the <= 10kHz component of each wave.

>The peaks from the
>sawtooth wave are now rounded.

Of course. The high-frequency information is missing.

>Now let's assume a 40khz sample rate with the same 10Khz signals above.
>Each waveform looks quite a bit closer to its original.

They will now be perfect representations of the <= 20kHz component of 
each wave.

>Therefore, a
>higher sample rate == higher detail at the same original input frequency.

Nope. The representation of the <= 10kHz components is identical. 
It's just not the only thing you're representing any more.

>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%.

If you want to use "detail" to mean additional high-frequency 
components, you can, I guess, although I find it counter-intuitive. 
But that's the only way this statement is correct.

Usually when people make this argument, they use "detail" to imply 
that the sampled waveforms are somehow "jagged" or "low-res." I'm not 
quite sure if that's what you're saying, but if it is, 'tain't so.

Is this boring the crap out of everyone but me and Neil?
-- 


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Tim Walters : The Doubtful Palace : http://www.doubtfulpalace.com