-----Original Message----- From: Edward Lewis [mailto:edlewis@arin.net] /---Snip...snip...snip--- | | The 8000 times a second comes from the Shannon's (or was it | Nyquist's?) research that found that sampling at 2 x the bandwidth | (bandpass - the range of frequencies passing) was optimal. I.e., | the local loop is assumed to have a 4k hertz range (by the sampling | systems). | \---Snip...snip...snip--- It was Nyquist. But he didn't take into account the signal-to-noise ratio of the communication channel. His formula took into account a perfect, noiseless medium, which - as we all know - is not of this world. Nyquist's sampling theorem states that for a given bandwidth with maximum frequency Fm, the sampling frequency Fs must be at least equal (or greater) to twice the maximum frequency Fm : Fs >= 2 x Fm in order to have the signal be reconstructed without aliasing. The frequency "2 x Fm" is sometimes called the Nyquist sampling rate. It means that in order to sample a voice signal up to 4000Hz, we need to sample at 8000Hz. It also means that to sample music up to 22050Hz, we need to sample at 44100Hz. Shannon came with an additional factor - the signal-to-noise ratio. He predicted that on an ordinary telephone line filtered at around 3400 Hertz with an average S/N, we would once achieve a bitrate of around 30000 bps - which we actually achieved with the V.34 standard. Shannon's Theorem calculates the capacity of a channel, in bits per second, as a function of the bandwidth AND the signal-to-noise ratio of the given channel. The Theorem can be stated as : C = B x log2(1 + S/N) where C is the achievable channel capacity in bits per second, B is the bandwidth of the channel in Hertz and S/N is the signal-to-noise ratio. Outside the scope of this formula, the signal-to-noise ratio (S/N) is usually expressed in decibels (dB) according to the formula : 10 x log10(S/N) so for example a signal-to-noise ratio of 1000 is commonly expressed as : 10 x log10(1000) = 30 dB So, if a telephone line has a signal-to-noise ratio of 30dB, it means that the signal-to-noise ratio is 1000. Here is an example of the use of Shannon's Theorem : for a typical telephone line with a signal-to-noise ratio of 30dB and an audio bandwidth of 3kHz, we get a maximum data rate of : C = 3000 x log2(1 + 1000) which is a little less than 30 kbps. To go beyond that maximum bitrate, we need workarounds, like data compression techniques or complex V.90/V.92 techniques (which use digital to analog conversion at high downstream rates and analog to digital conversion at low upstream rates). E.T.