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Everyone seems to have different definitions everywhere I look.

According to my lecturer:

$R_{bit} = \frac{bits}{time} $

$R_{baud} = \frac{data}{time} $

According to manufacturers :

$R_{bit} = \frac{data}{time} $

$R_{baud} = \frac{bits}{time} $

Which is the correct one?

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  • \$\begingroup\$ I considered it but saw that there were signal questions on this SE so seems to be within the acceptable range (especially since it is about a fundamental physical definition)... \$\endgroup\$ – Psi Jan 25 '17 at 11:54
  • \$\begingroup\$ I'm not saying the question is necessarily inappropriate here, but you're more likely to get an answer on the Electronics SE. \$\endgroup\$ – John Rennie Jan 25 '17 at 11:55
  • \$\begingroup\$ Thanks! I'll go there if it is not answered here! I am simply worried people in the Electronics SE will go with the definition that is most common or that they use. I am really searching for the absolute physically correct definition. \$\endgroup\$ – Psi Jan 25 '17 at 11:56
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    \$\begingroup\$ I agree. This is electronics or computer science. @ThomasHollis , what in the world is a "physically correct" definition? I teach computer science and I'm not entirely certain what you mean by that. \$\endgroup\$ – Trixie Wolf Jan 25 '17 at 12:01
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Baud rate is the number of symbols per second. Bit rate is the number of the information bits per second.
To see the difference, consider a transmission line capable of having 4 different states, say 0, 1, 2 and 3V. And the end equipment is able to detect each one of them. In this case each symbol would be one of these four states. And each symbol will carry two bits of information as opposed to the simple binary (2-state) line carrying a single bit at a time. Now consider both, the 4-state line and the 2-state line are capable to switch the states in the same rate, which is making both of them to have the same baud rate. But the 4-state line will transmit twice as many bits as the 2-state one, having the bit-rate higher two times.

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  • \$\begingroup\$ This is a better answer because it does explain what is meant by symbols. Otherwise I'd think symbol refers to a text character. \$\endgroup\$ – Ruslan Jan 25 '17 at 15:28
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This really belongs somewhere else.

  • Baud is the symbol rate: the number of symbols per second.
  • Bit rate is the number of bits per second.

If your symbols are bits then baud and bit rate are the same. If your symbols are equivalent to more than one bit of information (which they often are for later models of modem for instance, and do in many other cases) then the bit rate is higher than the baud.

A search on Wikipedia for 'baud' would have found this.

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  • \$\begingroup\$ The point is wiki searches directly conflict with what my lecturer says. Due to conflicting definitions I was after the physically-correct method of definition and why it is defined as such. But thanks for contributing. \$\endgroup\$ – Psi Jan 25 '17 at 12:35
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    \$\begingroup\$ Thomas Hollis, If I were you I would print out with large letters what @tfb wrote and carry it in my wallet to look at it when in doubt and thereby save myself much grief and misunderstanding in the future. Note that "symbol" in this context is the symbol that is mapped directly and one-to-one into waveforms. Just one more thing: "baud" is the "symbol rate", the expression "baud rate" is redundant. \$\endgroup\$ – hyportnex Jan 25 '17 at 13:29
  • \$\begingroup\$ @hyportnex: you are right, I have edited my answer to remove the spurions. I always get this wrong. \$\endgroup\$ – tfb Jan 25 '17 at 13:58
  • \$\begingroup\$ @ThomasHollis: there is no 'physically-correct' method of definition. This is, however, the standard definition of the two terms. If your lecturer is using some other definition then their definition is non-standard. \$\endgroup\$ – tfb Jan 25 '17 at 14:01
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Bit rate describes the information transmitted. Baud rate describes the number of "signal transitions" (symbols) used to transmit that information.

Depending on the transmission scheme used, your baud rate can be lower or higher than the bit rate. Things are further complicated by the fact that most data transmission includes (significant) overhead - frame information, start/stop bits, error correction etc.

A nice tutorial covering much of this in detail can be found at http://m.electronicdesign.com/communications/what-s-difference-between-bit-rate-and-baud-rate

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    \$\begingroup\$ Well, it's completely different now. And correct. The overhead might or might not be taken in account though.. \$\endgroup\$ – Eugene Sh. Jan 25 '17 at 15:30

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