> As the Whittaker-Nyquist-Kotelnikov-Shannon sampling theorem
> When sampling a band-limited signal (e.g., converting from an
> analogue signal to digital), the sampling frequency must be
> greater than twice the input signal bandwidth in order to be able
> to reconstruct the original perfectly from the sampled version.
> I'm not sure I understand this here Nyquist limit. Wouldn't the
> frequency have to be _much_ greater than twice the highest
> recorded? F'rinstance, say you want to record a 22.05KHz sine
wave. The best
> you could do with a 44.1KHz sample rate is a 22.05KHz square wave,
> regardless of bit depth. The quantitative frequency would be
present, but in
> a considerably altered shape. Seems like the minimum sample rate
> to be, oh, say, 5,644.8KHz.
I agree that it is difficult to understand these things. However, I
believe that it is important to note that our ears cannot be
interpreted as a sampling system that simply analyzes the shape of
the waveforms (you are correct that a sampled signal that is close
to 22.05 kHz looks like a square wave in the digital domain).
As far as I understand, our ears function like a filter bank
(comparable to a frequency-domain spectrogram analysis) that
exhibits an upper limit of less than 20 kHz. Therefore, we cannot
distinguish a high-frequency (lets say 10 kHz) sine wave signal from
a square wave signal of the same fundamental frequency. The upper
harmonis of such a square wave (starting at 30 kHz) would just be
inaudible to the human ear.
Also note that the original shape of the audible part of such high-
frequency signals is usually beeing reconstructed during playback.
This reconstruction is performed by a low-pass filter at the output
of the D/A converter. High-quality audio gear uses additional
oversampling techniques that indeed use D/A converter clock rates
that are much higher than the original sample rate. As a result, a
high-frequency sine wave appears at the analog output of a CD player
in a perfect sinusoidal shape (you might check that by using an