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In Ethernet when Manchester encoding is used, the bit rate is:

1. Half the baud rate

2. Twice the baud rate

3. Same as the baud rate

4. None of the above

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Is this topic present in the current syllabus?
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You can think of Manchester encoding not only transmitting the actual data, but also the clock (meta data) due to its self clocking characteristic.

Bit rate is half the baud rate in Manchester encoding as bits are transferred only during a positive transition of the clock.

http://stackoverflow.com/questions/25834577/why-in-manchester-encoding-the-bit-rate-is-half-of-the-baud-rate

Correct Answer: $A$

edited
Basically there is a relation between bit rate and baud rate which is:

BaudRate = Bitrate * c * (1/r)

c = Case factor usually = 1 (if nothing mentioned in question)

r = data element per signal element

In manchester signal or differential manchester a data bit is encoded in two signal element so r = 1/2 because for half duration level is up and for half of bit duration its below so two signal element per data element

hence if we substitue the data of r in above we get Baud rate = 2* Bit rate
ans b)
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how?
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Baud rate = 2* Bit Rate.