Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site petrus.UUCP Path: utzoo!watmath!clyde!burl!ulysses!allegra!bellcore!petrus!karn From: karn@petrus.UUCP Newsgroups: net.audio Subject: Re: CD Reflections - 44.1k? Message-ID: <267@petrus.UUCP> Date: Mon, 21-Jan-85 12:19:55 EST Article-I.D.: petrus.267 Posted: Mon Jan 21 12:19:55 1985 Date-Received: Tue, 22-Jan-85 05:28:26 EST References: <15100001@hpfcmp.UUCP> <3411@mit-eddie.UUCP> <1420@hplabs.UUCP> <755@clyde.UUCP> <258@petrus.UUCP> <272@mtxinu.UUCP> Organization: Bell Communications Research, Inc Lines: 42 > > The Nyquist theorem is valid for ANY bandlimited signal, it doesn't matter > > whether it is periodic or not. ... > > ... Check out your communications > > theory textbook on this one. > > > > Phil > > When I studied signal theory briefly about 12 years ago, there was > a theorem stating that it was *impossible* to push more information > through a signal than the bandwidth of the signal, e.g., one can't > send more than k bits per second through a k Hz bandlimnited channel. > > Telephone voice-grade channels are 2700 Hz limited, filtering to allow > signals only from 300 Hz to 3000 Hz. So how do 4800 and 9600 bps > modems work over dialup circuits? (The telco carriers, by the way, > are strict about bandlimiting their signals, since they frequency- > multiplex them onto higher-bandwidth channels.) > > The answer seems to be that the theory that generated that theorem > wasn't completely correct. Maybe the Nyquist theorem shouldn't be > regarded as gospel, either. > > -- > Ed Gould mt Xinu, 739 Allston Way, Berkeley, CA 94710 USA > {ucbvax,decvax}!mtxinu!ed +1 415 644 0146 Wrongo! The answer seems to be that you don't remember the theory correctly. The Nyquist limit still holds. High speed modems work by sending more than one bit per transition on the line, e.g., the Bell 212 sends two bits per signal transition, which can therefore take one of 2^2 or 4 possible states. The theoretical limit on the bit rate through a given channel also depends on the signal-to-noise ratio as well as the bandwidth, and is given by C = B * log2(1 + S/(N0 * B)) where C = channel capacity in bits/sec B = bandwidth, hertz S = signal power, watts N0 = noise spectral density, watts/hertz This is the famous Shannon limit.