The statement given is
Bi-phase Encoding
The signal level is checked twice for every bit time, both initially and in the middle. Hence, the clock rate is double the data transfer rate and thus the modulation rate is also doubled ..
How the clock rate is doubled??
In the given picture clock rate is same for every encoding?? How they are saying clock rate ia doubled in Manchester encoding..
The signal level is checked twice for every bit time, both initially and in the middle.
The signal is changing at those times, so we actually need to check 1/4 into the bit time, and 3/4 into the bit time.
Compared to a NRZ code, which only needs to be checked half way through the bit time, that's twice as often.
Conventionally, we talk about a clock being used to either generate the edges, or check the levels. In practice, actual hardware will often use a clock signal that is many multiples of that, certainly on the receive side, to handle its offset with the bit clock on the data line.
The two main reasons for using Manchester Encoding, which embeds the clock signal in the data like that, are that without any further line coding -
The tradeoff is twice as many transitions on the line.
Other modulations can be made to have zero DC, but at the expense of coding the data somehow. For clock recovery, NRZ data might limit the maximum run length of 1's or 0's.
Manchester tends to be used where transitions are cheap (low bit rate), reliability is important (aircraft busses like 1553) and the technology is old (and simple). Ethernet will go through transformers, but is a rather newer standard.
edit - thanks to fectin in comments below. Another advantage of Manchester coding is where the data rate is not standardised, and can even be variable during the word, like a manually-swiped card in a reader. Embedding the clock transition in every single bit makes it possible to extract the clock, even with a rapidly varying clock rate that could defeat RZ modulation types.
In typical basic digital hardware, every output signal edge is generated by a (rising) clock edge. In Manchester encoding, an edge can happen either at a bit cell boundary or in the middle of a bit cell. Thus there are two possible signal edge locations per bit cell. Thus two (rising) clock edges are required per bit cell, to provide those two possible output signal edge locations. Thus, the clock has to be (at least) twice as fast as the bit cell rate. (4x or 8x (etc.) faster would also work, but isn't needed.)
