Given a baseband channel with 3-kHz bandwidth and a 40-dB signal-to-noise ratio, what is the theoretical capacity of this channel in bits per second?
The impulse response of an ideal channel with exactly 3 kHz of bandwidth is a sinc (i.e., sin(x)/x) pulse in the time domain that has nulls that are 1/6,000 s apart. This means you could send a series of impulses through this channel at a 6,000 pulses per second rate. And, if you sampled at exactly the correct instants on the receiving end, you could recover the amplitudes of each of those pulses with no interference from the other pulses on either side of it.
However, a 40-dB signal-to-noise ratio implies that the noise power is 1/10,000 of the maximum signal power. In terms of distinguishing voltage or current levels, this means you can send at most sqrt(10,000) = 100 distinct levels through the channel before they start to overlap, making it impossible to separate one from another at the receiving end.
100 levels translates to log2100 = 6.64 binary bits of information. This means the total channel capacity is 3,9840 bits/s (i.e., 6,000 pulses/s × 6.64 bits/pulse).
This is the basis for the Shannon-Hartley channel capacity theorem.