Xref: utzoo sci.electronics:17795 sci.physics:16817 Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!uunet!wuarchive!emory!gatech!prism!mailer.cc.fsu.edu!uflorida!cs.fau.edu!terryb From: terryb.bbs@shark.cs.fau.edu (terry bohning) Newsgroups: sci.electronics,sci.physics Subject: Re: A question about the Nyquist theorm Message-ID: Date: 16 Feb 91 18:08:48 GMT References: <1751@manta.NOSC.MIL> Sender: bbs@cs.fau.edu (Waffle BBS) Organization: Florida Atlantic University Lines: 17 north@manta.NOSC.MIL (Mark H. North) writes: > > 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. > The catch is that you *know* you're sampling the highest input freq at 2 points per cycle. That is, the input signal is bandlimited. So if someone gives you a set of all zero samples and you know the sample rate is 120 Hz, the only frequency it can be is 60 Hz. The Nyquist theorem is at least, not greater than. Oppenheim & Schafer, "Digital Signal Processing", Prentice-Hall, 1975, pg. 28 bottom. In reality, of course, since ideal filters are unavailable for band-limiting, the rate must be higher.