The theoretical calculation for the maximum data rate is based on Shannon's law also known as Shannon Hartley law. The law is a cornerstone of communications theory and is in all text books. However it is usually surrounded by so much maths that we miss the significance of it. Here I will try and correlate it to common communications example. Here is Shannon's Law
Maximum Capacity in Bits/Second = Bandwidth * Log2(1 + Signal/Noise)
The maximum data rate of any channel is determined by its bandwidth and signal to noise ratio. Wireless channels may have signal to noise ratios (SNR) of 7 to 100 or so, while wired channels are much cleaner and have SNRs in 1000s.
The signal to noise ratio also gives an indication of which modulation schemes to use. For example if the SNR is around 7 or 8 it makes no sense to 64 QAM which gives 4 bits per symbol. However 8PSK which gives 3 bits per symbol is ideally suited. In fact the next generation of GSM Enhanced Data rates for GSM Evolution (EDGE) uses 8PSK modulation.
Looking closer at the above equation reveals that it is possible to transmit data even if the signal is weaker than the background nose level. For example if the Signal to Noise Ratio (SNR) is 0.5 and the channel is 1 MHz wide the
Maximum data rate = 1 * log2(1 + 0.5) = 1 * log2(1.5) = 0.58 Mbps
This is the basis of spread spectrum communications.
For 802.11B, the allotted bandwidth is 16 MHz. If we assume a conservative Signal to Noise ratio of 7, the maximum data rate works out to be
16 * log2(1 +7)= 16 * log2(8)= 48 Mbits/per second
The maximum data rate for 802.11 B is 11 Mbits per second so we see that it is using only 1/3 of its maximum theoretical capacity.
For a narrow band 12.5 kHz channel, similarly assuming a conservative signal to noise ratio of 7 the maximum capacity works out to be
12.5 * 3 = 37.5 kbits per second
Well in reality there will be some guard bands to comply with FCC so we will not be able to utilize 12.5 kHz fully. Even then we see the performance of current narrowband data radio modems is much below what is possible. Also we can see that there is no point in using a modulation scheme that gives more than 3 bits per symbol.
For a cable modem that uses 6 MHz of bandwidth, the signal noise ratio is substantially better about 2000. So its maximum data capacity is, (assuming SNR of 2047 for simplicity)
6 * log2(1 + 2047) = 6 * 11 Mbps = 66 Mbps
The actual rate achieved by a cable modem is about 30 Mbps.
Since the noise characteristics of any channel change, it is advantageous to adjust the modulation schemes to adapt to differing noise levels in the channel.
Shannon's law applies to any communication channel. For example it can be applied to acoustic modems used for underwater communications.