the baud rate
/ baud -> 'bawd' /
Imagine flashing a torch to a friend across a valley. If you only flash bright or dark, each flash tells them one bit. But if you can flash four distinct brightness levels, each single flash now carries two bits' worth of choice (dark, dim, medium, bright = 00, 01, 10, 11). The baud rate is how many flashes per second you make; the bit rate is how much information all those flashes deliver per second.
More precisely, a symbol is one such discrete signal state held for one time-slot, and the baud rate (or symbol rate) is the number of symbols sent per second. The bit rate equals the baud rate times the number of bits each symbol encodes: bit rate = baud rate times log2(M), where M is the number of distinct symbol states. With two states (M = 2), one symbol carries log2(2) = 1 bit, so baud and bit rate match. With M = 16 states, each symbol carries log2(16) = 4 bits, so 1,000 baud delivers 4,000 bps.
This distinction matters because the physical medium limits how many symbols per second you can cram in (that is the Nyquist ceiling, tied to bandwidth), but it does not directly limit how many bits each symbol holds — that is limited instead by noise, because the more levels you pack in, the closer they sit and the easier noise makes them collide. So engineers raise the bit rate above the baud rate by using richer symbols (as QAM does), right up to the point where noise starts causing errors. People loosely say 'baud' to mean 'bit rate', but they are equal only when each symbol is a single bit.
A modem runs at 2,400 baud but advertises 9,600 bps. The trick: each symbol uses one of 16 distinct states, so each symbol carries log2(16) = 4 bits, and 2,400 symbols/second times 4 bits = 9,600 bps.
Bit rate = baud rate times bits-per-symbol; they coincide only for two-level signalling.
'Baud' and 'bps' are not synonyms. Calling a 56k modem '56k baud' is wrong: it sent far fewer symbols per second, each carrying many bits via dense modulation.