3 ms·
It's 'baud', not BAUD/S. Your signal timing is 20 microseconds which equals a signal rate of 50000 baud, or 50 kBd. There are 12 signals in each frame holding
by jonsen 11y ago
It's 'baud', not BAUD/S.
Your signal timing is 20 microseconds which equals a signal rate of 50000 baud, or 50 kBd.
There are 12 signals in each frame holding one byte. So the byte rate is 50000/12 = 4.2kB/s.
- gioscarab 11y agoYou are right, I exluded from the calculation the padding bits, that in the baud measure is not correct, I will correct the readme as soon as possible. Thank you again
- jonsen 11y agoWhat do you mean by 'bandwith'? I would suggest you rather used maybe 'data throughput'? "Channel bandwidth may be confused with useful data throughput...": https://en.m.wikipedia.org/wiki/Bandwidth_(computing) https://en.m.wikipedia.org/wiki/Bandwidth_(computing)
- gioscarab 11y agoThank you so much, I corrected the readme as you suggested :)