🚀 go-pugleaf

In article <12122@ucrmath.ucr.edu> stebbins@musial.ucr.edu (john stebbins) writes:

>Suppose I was sampleing a 20khz sine wave at 44khz and my first sample
>just happened to occur at the positive peak of the sine wave.  My next
>sample would occur a little before ( and thus above ) the negative peak.
>And the next would occur a little more before ( and a little more below )
>the next positive peak.  This continues until zero crossing at which point
>my samples start growing instead of decreasing.  Its pretty easy to see why
>filtering the sample back down to 20khz will reproduce a 20khz signal, but
>how does the filtering recover the original amplitude of my sign wave. 
>It appears that what I'll get is an am modulated signal that is some 
>combination of the 20khz signal and the 44khz sample rate.

	Your last comment is the key to the solution.  The combination
is the difference frequency, 24 kHz.  A sum of equal-amplitude 20 kHz
and 24 kHz pure sine waves gives exactly the AM-modulated signal that
you describe.
	So, it is the responsibility of the CD playback unit (because
these numbers are appropriate for high-frequency audio in a CD player)
to correctly erase the spurious 24 kHz tone.  To the best of my knowledge,
no CD players actually work at 1x sampling, but ALL (even the oldest
and cheapest) first calculate a bandwidth-limited intermediate sample
from the recorded samples (with a FIR filter, comprised of simple
arithmetic operations).  The simplest mechanism, 2x oversampling,
gets rid of that 24 kHz signal and introduces instead (because
the new Nyquist limit is 44 kHz) some junk at 46 kHz or so.
The analog filter then can be trusted with the task of getting
rid of the 44 kHz-and-up junk, while passing (with low distortion)
all of the 22 kHz-and-under signal.

	John Whitmore