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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?? enter image description here

In the given picture clock rate is same for every encoding?? How they are saying clock rate ia doubled in Manchester encoding..

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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 -

  • it has zero DC, so can be sent through a transformer
  • the receiver can use a robust clock recovery scheme

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.

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  • \$\begingroup\$ There are so many different things called "Ethernet" and likely all your devices are compatible with 10BaseT and it uses Manchester. \$\endgroup\$
    – Justme
    Nov 7, 2022 at 7:05
  • \$\begingroup\$ "it has zero DC, so can be sent through a transformer" This is true when the signal is bipolar (which you would use when it's your goal to send it through a transformer). I don't know why, but the question only displays the unipolar variant. \$\endgroup\$
    – Velvet
    Nov 7, 2022 at 7:05
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    \$\begingroup\$ @VelVel You can remove DC with a single capacitor. For the sake of example, it does not matter. And optical interfaces must use on/off as there is no negative light. \$\endgroup\$
    – Justme
    Nov 7, 2022 at 7:08
  • \$\begingroup\$ Well what about Rz clock frequency. It is also doubled? \$\endgroup\$ Nov 7, 2022 at 9:38
  • \$\begingroup\$ Another question is , generally clock cycle time is equal to bit interval time... but if we use 2 clock cycles in bit interval time as you said it needs to check two times in single bit interval and if we read the signal data at every clock positive edge we may get double the data right??.... \$\endgroup\$ Nov 7, 2022 at 9:44
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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.)

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  • \$\begingroup\$ But we don't need to check what's happening at the beginning of the bit interval because we read only at the middle of the bit interval ...as both clock and data present at the middle of the bit interval \$\endgroup\$ Nov 7, 2022 at 5:51
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    \$\begingroup\$ @RamaSivaSubrahmanyam It does not matter. The Manchster encoded data still requires twice the bandwidth than unencoded NRZ data. \$\endgroup\$
    – Justme
    Nov 7, 2022 at 6:25

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