Practice Examples for Shannon’s Law: Source: http://www.fast-product-development.com/maximum_data_rate.html

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

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.