randy@uw-june.UUCP (William Randy Day) (11/02/87)
In Message-ID: <1641@uqcspe.OZ> Mark Williams gets on the right track
to answering the question "What is the Max speed possible on
phone lines?" The answer lies in Nyquist's theorem (for noiseless
channels) or Shannon's theorem for channels with random noise.
Shannon's theorem states that:
maximum number of bits/sec = bandwidth * log2(1 + S/N)
where bandwidth is Hertz and S/N is the signal to noise ratio,
expressed as a true ratio of signal power, not as the more
common decibel figure. Mark got all this right.
However, Mark did not properly convert 40 dB to a power ratio.
A decibel is 10*log10(S/N), where S/N is the power ratio. It's logarithmic.
Thus, a S/N ratio of 10/1 is 10 dB, a S/N ratio of 100/1 is 20 dB, etc.
Therefore, a S/N of 40 dB is a power ratio of 10,000/1, not 100/1 as
Mark stated.
Thus, given a bandwidth of 3 khz (the bandwidth of voice phone lines)
and a S/N of 40 dB (probably too generous?), the maximum possible
bit rate is:
= 3000 * log2(1 + 10,000)
= 39,864 bits per second, not 19,931.
This is the maximum possible, regardless of the baud rate or of
the number of bits encoded per baud. As Tannenbaum says, "Counterexamples
should be treated in the same category as perpetual motion machines."
Randy Day.
Internet (ARPA): randy@dbnet.cs.washington.edu
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