The baud rate is

Concept

A digital link is described by two different rates that must not be merged. The baud rate, also called the signalling rate or symbol rate, counts the signal elements placed on the line each second, where a signal element is one discrete state of the carrier; its unit is the baud. The bit rate counts the information bits carried each second, in bits per second.

If a modulation scheme uses M distinguishable signal levels, one signal element carries log2M bits, so bit rate = baud rate × log2M. The two rates are numerically equal only in the special case M = 2, where exactly one bit rides on each signal element.

How fast signal elements may be pushed through a channel is set by its bandwidth, not by the number of levels. For an ideal (noiseless) channel of bandwidth B hertz, the Nyquist result fixes the maximum number of signal elements the receiver can still tell apart at 2B per second: maximum baud rate = 2B.

Application

1. Fix the definition. Baud rate is the number of signal elements sent per second, and signalling rate is the standard synonym for that same count, so the two names denote one quantity.

2. Separate it from the bit rate. With M signal levels, bit rate = baud rate × log2 M, so the two coincide only when M = 2. In a four-level scheme such as QPSK each signal element carries two bits, and the bit rate is double the baud rate.

3. Bound it by bandwidth. Nyquist gives an ideal noiseless channel of bandwidth B hertz a ceiling of 2B distinguishable signal elements per second, that is two signal elements for every hertz.

4. Combine the three. The rate the question asks about is the signalling rate, it is not tied to the bit rate in general, and on an ideal channel it is governed by the bandwidth through the Nyquist relation, which allows at most 2B baud. Among the four statements offered, only "equal to twice the bandwidth of an ideal channel" expresses that standard bandwidth relation; the other three contradict either the definition of the baud or the Nyquist relation itself.

Cross-check

Take an ideal channel of bandwidth B = 3000 Hz, the classic telephone-grade figure. Nyquist gives a maximum of 2 × 3000 = 6000 baud. With two-level signalling (M = 2) that is 6000 bits per second; with four-level signalling (M = 4, so two bits per element) the same 6000 baud now carries 12000 bits per second. The baud rate did not move while the bit rate doubled, which confirms that the bandwidth relation constrains the symbol rate and not the bit rate.

• Reading the baud rate as always equal to the bit transfer rate requires one bit per signal element, which is only the M = 2 case and fails for every multi-level scheme.

• Reading the baud rate as not equal to the signalling rate contradicts the definition, since signalling rate is another name for the same symbols-per-second count.

• Reading it as half of the bandwidth gives one signal element for every two hertz, a quarter of the two-per-hertz density that Nyquist allows on an ideal channel.

Result

The relation being tested is the Nyquist one: an ideal channel of bandwidth B hertz supports at most 2B signal elements per second, which is the relation the answer "equal to twice the bandwidth of an ideal channel" states. Read strictly, 2B is a ceiling rather than a value every link must take, and a real link may of course be operated below it; textbooks and examination keys state the relation in this abbreviated form, and it is this bandwidth relation, not the bit rate, that the baud rate of an ideal channel obeys.

The other two properties of the baud settle the remaining statements. Baud rate is the signalling rate by definition, so the two can never differ; and it equals the bit rate only in the special case where each signal element carries a single bit, which is why no unconditional equality with the bit transfer rate holds.