Xref: utzoo sci.electronics:17856 sci.physics:16866
Path: utzoo!utgpu!cunews!software.mitel.com!grayt
From: grayt@Software.Mitel.COM (Tom Gray)
Newsgroups: sci.electronics,sci.physics
Subject: Re: A question about the Nyquist theorm
Message-ID: <6607@healey>
Date: 19 Feb 91 14:00:08 GMT
References: <91046.095459F0O@psuvm.psu.edu> <1751@manta.NOSC.MIL>
Organization: Mitel. Kanata (Ontario). Canada.
Lines: 71

In article <1751@manta.NOSC.MIL> north@manta.NOSC.MIL (Mark H. North) writes:
}In article <91046.095459F0O@psuvm.psu.edu> F0O@psuvm.psu.edu writes:
}>
}>     I was reading an article that states the Nyquist theorm as:
}>     "The sample frequency must be at least twice the highest frequency
}>component within the analog signal for an accurate representation of the
}>analog signal".
}
}This is an incorrect statement of the Nyquist theorem. The sample freq
}must be *greater* than twice the highest freq component...
}

This is an incorrect correction. The original satement is accurate.
Sampling at twice or greater than the highest freqeuncy n a
band limited signal is all that is required for Nyquist sampling. 

}>     I'd guess here he is talking about complex signals.  But what do you
}>do with a pure sine wave?  There is only one frequency component in a sine
}>wave(the fundamental), and if you sample at twice that, you're not going
}>to get a good representation of the signal.
}
}A pure sine wave is fine. As long as you sample at greater than twice its
}freq. Even though it may appear that you are not getting a good represen-
}tation of the signal it can be shown with Fourier analysis that the
}sample set is unique to this component and hence the exact signal can
}be recovered from the sample set.
}
}>     i.e. If you have a 60HZ sine wave, and you sample at 120HZ, you're
}>only going to get two points per cycle.
}
}And imagine that those two points are phased such that they land at the
}zero crossing of the 60Hz signal. All your samples are zero! This is
}why you must sample at greater than 2nu.
}
This is a common misconception. The sampling pulses are of finite widht.
The shape of the wave is preserved within the sampling pulse. This
information allows representation of a signal at exactly 1/2
the Nyquist freqency.

The origin of this misconcetion is a confusion about the sampling
methods assumed for the Nyquist theorem. Nyquist assumed
natural sampling in which the shape of the signal is preseved
by multiplication with the sampling pulse. This is a simple
multilication of the two signals in the time domain.
Digital sample storage cannot do this, Only one value
of the signal can be obtained per sample (not the
continuous representation through out the sampling
period which is obtained for natural). The digital
method of sampling is called commonly flat top sampling.
Flat top sampling cannot represent signals at the
half sampling frequency. it is a limitation of flat top
sampling and not of sampling in general ( including
Nyquist sampling) which makes this limitation.

If you have text books proving Nyquist by multyplying
with instabtaneous pulses and referring the Dirac
delat functions, I have text books which properly prove
Nyquist with pulses of any width. The instantaneous
pulse case is only a special case and is not true
in general since it implies limitations which do not
occur for pulses of finite width (ie all REAL sampling
pulses).
 
} 
} A good reference is "Digital Signal Analysis" by Samuel D Stearns. It is
} no longer in print but is available in most engr. libraries. Also there
} is a new edition of this book published by Printice Hall.
}

Most text books play fast and lose with Nyquist.