The Shannon channel capacity is the maximum rate at which bits can be transmitted reliably through a communication channel.

It is given by the following formula:

C = W * log2(1 + SNR) bits/second

Where:

• C is the Shannon channel capacity
• W is the bandwidth of the channel
• SNR is the signal-to-noise ratio

For example, a modern telephone channel has a maximum useful bandwidth of about 3400 Hz. Assuming an SNR of 40 dB (slightly over the maximum possible in a phone line), the formula above gives us a channel capacity of about 44.8 kbps (rounding down since we assumed a high bandwidth).

  40 = 10 * log10(SNR)
   4 = log10(SNR)
 104 = SNR
 SNR = 10000

C = (3400) * log2(1 + 10000) ~= 3400 * 13.2 ~= 44.8 kbps

What's interesting is how 56 kbps modems seem to exceed this fundamental limit. In reality, the bit rate inbound to the telephone network is limited to 33.6 kbps and the modem signal must undergo an analog-to-digital conversion since it needs to converted to PCM for digital transmission (in a modern network). This step introduces quantization noise and so a higher bit rate is unattainable. Ah, but you see, from the ISP to the user, the signal is already digital and does not need to undergo A/D conversion! Hence, no quantization noise is introduced and a higher SNR can be achieved. Therefore, downstream speeds of up to 56 kbps (theoretically) are possible.