In the book Computer Networks, the author talks about the maximum data rate of a channel. He presents the Nyquist formula :
C = 2H log\$_2\$ V (bits/sec)
And gives an example for a telephone line :
a noiseless 3-kHz channel cannot transmit binary (i.e., two-level) signals at a rate exceeding 6000 bps.
He then explain the Shannon equation :
C = H log\$_2\$ (1 + S/N) (bits/sec)
And gives (again) an example for a telephone line :
a channel of 3000-Hz bandwidth with a signal to thermal noise ratio of 30 dB (typical parameters of the analog part of the telephone system) can never transmit much more than 30,000 bps
I don't understand why the Nyquist rate is much lower than the Shannon rate, since the Shannon rate takes noise into account. I'm guessing they don't represent the same data rate but the book doesn't explain it.