I have an exam tomorrow and I'm confused on the following question, that may be asked in the exam:

One byte data with parity bit by using asynchronous serial transmission with two stop bits. For 2K-byte/second transfer rate, what is the baud rate?


closed as off-topic by Chris Stratton, pipe, RoyC, Finbarr, Dmitry Grigoryev Nov 22 '18 at 14:11

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    \$\begingroup\$ Welcome, Gökay! We don't want to ruin your learning experience by doing your exam for you, but many of us here will be more than happy to guide you through it. So, where exactly are you stuck? Can you make any progress at all? Can you show us that you made a decent attempt at solving the problem? \$\endgroup\$ – Ricardo Jun 30 '14 at 19:37
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    \$\begingroup\$ I'm voting to close this question as off-topic because it is an utterly unattempted problem in education. \$\endgroup\$ – Chris Stratton Nov 15 '18 at 3:20

2 kbytes = 2048 bytes.

Transmission of a byte requires:

  • 1 start bit
  • 8 bits = actual byte
  • 1 bit parity
  • 2 bits stop

Total of 12 bits.

The baud is 2048 * 12 = 24576 baud

Have a look at this Wikipedia article.

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    \$\begingroup\$ I'm not sure you can assume the OP means 2,048 bytes per second - he's probably not understanding that concept yet. \$\endgroup\$ – Andy aka Jun 30 '14 at 18:59
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    \$\begingroup\$ 2 kB = 2000 bytes, 2 KiB = 2048 bytes. en.wikipedia.org/wiki/Kibibyte \$\endgroup\$ – Splitlocked Apr 21 '17 at 1:21

Baud rate is the rate at which highs and lows must be sampled to decode the signal. The signal includes:

  1. protocol overhead (e.g. start bits, stop bits, and parity bits) and
  2. actual application specific data.

The "2K-byte/second transfer rate" almost certainly refers to the effective rate at which actual application data is transferred, where 2K-byte probably means 210 = 2048 bytes per second (although this is admittedly ambiguous with respect to storage size vs. transfer rates).

With the above settled, it comes down to an algebra problem. Every byte of application data you send needs protocol overhead bits associated with it.

  • Define baud rate B with units of "total bits per second"
  • Define data rate D with units of "application data bits per second"
  • Define b to be the total number of parity, start, and stop bits with units of "overhead bits per application byte"
  • Define n to be the total number of "application data bits per application data byte", in common serial protocols, n tends to be 8, 9, or 10.

$$D = B \cdot \frac{n}{b + n}$$


$$B = D \cdot \frac{b + n}{n}$$

Convert D from "bits per second" to "K-bytes per second" in the appropriate manner.

  • \$\begingroup\$ +1 That's what I call a decent answer to a homework question. \$\endgroup\$ – Ricardo Jun 30 '14 at 19:42
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    \$\begingroup\$ To be correct, the baud rate is actually defined to be the number of symbols per second rather than the number of bits. A symbol could represent many bits of information. \$\endgroup\$ – sherrellbc Jun 30 '14 at 19:43
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    \$\begingroup\$ @sherrellbc yea that's pedantically accurate, but in the context of common serial protocols over wires between microcontrollers and devices without fancy encoders/modulators and decoders/demodulators between them, symbol is synonymous with bits \$\endgroup\$ – vicatcu Jun 30 '14 at 19:45

Your question is confusing, but if it you're sending a start bit, 8 bits of data, a parity bit, and two stop bits all encoded as one symbol, 2048 times a second, then that's 2 kilobaud where "kilo" = 2048.

Then, since there's 12 bits per baud, the bit rate would be 12 bits * 2048 baud = 24576 bits per second.

An easy-to-understand example:

Assume I'm sending asynchronous data using two different analog tones which I can turn on or off independently and simultaneously, and that the presence of a tone corresponds to a "1" while its absence corresponds to a "0", so that with the two tones I can encode "00", "01", "10", or "11".

Now assume that I'm sending the tone pairs at a rate of 1000 pairs per second, and that the receiver is receiving - and decoding - them at the same rate.

So, since every "package" of tone pairs is being sent at 1000 packages per second and contains two bits of data which are decoded once per millisecond, the data rate is 2000 bits per second while the package - or "symbol" - rate is 1000 baud.


On the basis that you may not know that 2048 is sometimes called "2k", I think the answer you are looking for is 24,000 bits per second. Hey it's either 24000 or what Cornelius wrote.

  • \$\begingroup\$ You are assuming that 1 byte is an octet, 8 bits. ;o) According to Wikipedia: "The size of the byte has historically been hardware dependent and no definitive standards existed that mandated the size." \$\endgroup\$ – jippie Jun 30 '14 at 19:07

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