Bitrate vs Baudrate
I am really confused how the terms bitrate are baudrate are used in electronics. I never bothered thinking about baudrate being different than bitrate before. But recently, one of my teachers told baudrate being the maximum no. of changes in the bit that can occur during the data transfer.
I really want to visualize how they are different from each other.
3 answers
In short:

The baud rate is the symbol transmission rate.

The bit rate is the data transmission rate.
(A "symbol", in this context, is simply a group of some relatively arbitrary, nonzero number of bits, transmitted as one atomic unit.)
While in simple cases the two can be identical, as soon as one symbol encodes more than one bit, they diverge. And that is incredibly common in modern data transmission systems.
Therefore, in practice, the bit rate of a channel will almost always be higher than its baud rate.
The baud rate matters more when you are interested in the requirements imposed on the data transmission physical link layer (such as a radio link's spectrum bandwidth, or the transition times in a modulator). The bit rate, on the other hand, matters more when you are interested in how much data can actually be transmitted within a given time period. Both are useful metrics, but as they measure different quantities, they answer different questions.
As an example, consider quadrature amplitude modulation; 4096QAM encodes 12 bits (2^{12} = 4096) per symbol. Therefore, the baud rate for a 4096QAM channel is 2^{12} = 1/4096 of the data rate of the same channel.
In practice, some of those bits being transmitted will, in turn, likely be used for various kinds of metadata; things like frame headers, checksums, start and stop bits, onbus addressing, forward error correction, and so on. In places, data is deliberately stretched out, not uncommonly to improve error handling or simplify synchronization. For example, SATA uses 8b/10b encoding on the physical link layer, and CDDA extends data multiple times for various reasons between the raw audio sample data and what's actually on the physical storage medium. Designs like these will reduce the rate at which useful payload data can be transmitted below that of the raw channel bit rate, and that is relevant for the amount of time it takes to usefully transmit some given amount of data, but they do not change the bit rate of the channel itself.
2 comments
In the olden days I liked to use the example of a touchtone telephone. Each tone was a baud, and each tone transmitted 4 bits. Conversely, plain old UARTs have a bit rate that is lower than the baud rate.
@ElliotAlderson At which point you get puzzled looks until you tell people that the DTMF value set is actually 09 plus *, #, A, B, C and D, which adds up nicely to 16 distinct values.
Baudrate is most often used to mean payload data bits/second. Simply picture a manchester encoded protocol, where each data bit corresponds to 2 physical bits in the raw data frame. If you have a bitrate of 9600, you will then have a baudrate of 4800.
There's lots of tutorials out there that describe this in detail, see for example https://www.electronicdesign.com/technologies/communications/article/21802272/whatsthedifferencebetweenbitrateandbaudrate
"one of my teachers told baudrate being the maximum no. of changes in the bit that can occur during the data transfer" No, that's not correct. That sounds more like Hamming distance.
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Baud rate is about the speed of bit slots in the channel or symbols. Bit rate is the rate at which actual data is transmitted.
Put another way, baud rate is what you see when you look at a scope trace of a communication signal. Bit rate is how fast data gets from one end to the other.
The bit rate can be both higher and lower than the baud rate due to encoding and packaging schemes. Modern high speed protocols do various fancy things to encode more than one bit of data into each symbol, so the bit rate is often higher than the baud rate.
However, for most "ordinary" protocols (UART, CAN, IIC, SPI for example), some of the bit slots are used to delimit and package the information, so the bit rate is lower than the baud rate.
Consider the common UART protocol of 115.2 kBaud, 8 data bits, no parity, and 1 stop bit. Each character requires 10 bit slots to send: 1 start bit, 8 actual data bits, and 1 stop bit. Since the baud rate is 115.2 k/s, each bit slot is 8.68 µs long. A whole character therefore takes 86.8 µs to transmit. A total of 8 payload bits are transported during that time, so the bit rate is (8 bits)/(86.8 µs) = 92.2 kBits/s.
@TonyStewart wrote:
Can you give an example where Baud Rate is higher than Bit Rate?
I did. See previous paragraph.
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Dear Olin, I think you are using bit rate efficiency with overhead to define effective data rate, both which have nothing to do with the bit/s baud compression technology Symbols can stuff more bits per symbol such as the Baud symbol in a modem after Émile Baudot who invented the Baudot code (Bodo) for telegraphy in the 1870s e.g. dot dit dot = S which is a character and the dit/dots are binary symbols determined by pulse duration.
There are many types of Baud Symbols that compress bits in time, frequency, phase or time in many combinations by sending many per unit of a bit. such as 9600 modems using 1200+2400 Hz tones with multiple levels to get 8 bits per Baud symbol in a 2400~3000 Hz bandwidth. ( I think, I forget)
... So don't byte me for saying your Byte rate example is not related to bit rate vs baud rate or symbol rate. but does affect character rate. characters are not symbols in this sense yet are symbols in language.
Can you give an example where Baud Rate is higher than Bit Rate?
@TonyStewart Maybe Morse code can be considered one such example, with variablelength symbols? It would probably be tricky to analyze as such, but I suppose technically it fits the definition...
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