Klarinet Archive - Posting 000700.txt from 1997/11

From: Roger Shilcock <roger.shilcock@-----.uk>
Subj: RE: Nyquist and analog
Date: Thu, 20 Nov 1997 03:43:29 -0500

No doubt this shows that "greater than" is better than "equal". Ian's
point looks valid, but is it likely, in a real situation, that the
sampling rate is going to be exactly in phase with one (or more)
components? Furthermore, nobody is sampling components - what is being
sampled is a *resultant* waveform.
Roger Shilcock

On Wed, 19 Nov 1997, Ian Dilley wrote:

> Date: Wed, 19 Nov 1997 17:12:44 -0000
> From: Ian Dilley <imd@-----.uk>
> Reply-To: klarinet@-----.us
> To: "'klarinet@-----.us>
> Subject: RE: Nyquist and analog
>
> Here is the disputed definition of Nyquist's theorom
>
> Nyquist's theorem: A theorem, developed by H. Nyquist, which states that
> an analog signal waveform may be uniquely reconstructed, without error,
> from samples taken at equal time intervals. The sampling rate must be
> equal to, or greater than, twice the highest frequency component in the
> analog signal. Synonym sampling theorem.
>
> Well, I'm no expert but I can immediately see a case where this is not
> going to be true. Take as an example a 20 KHz sine wave sampled at 40
> KHz. According to the above definition the samples will contain enough
> information to perfectly reconstruct the 20 KHz sine wave.
>
> Surely you are going to get different results depending on the relative
> phases of the sampling and the sine wave itself. In one extreme case
> you could get a set of samples that are all 0. This occurs because the
> wave crosses the 0 point every 1/2 it's period. From that you aren't
> going to be able to reconstruct very much at all! At the other extreme
> you can get a series of +n, -n, +n, -n ... where n is the amplitude of
> the signal.
>
> Jerry, Jonathon - am I missing something obvious?
>
> Ian Dilley
>