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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..

Rama Siva Subrahmanyam
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2 Answers2

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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.

Neil_UK
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  • There are so many different things called "Ethernet" and likely all your devices are compatible with 10BaseT and it uses Manchester. – Justme Nov 07 '22 at 07:05
  • _"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. – Velvel Nov 07 '22 at 07:05
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    @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. – Justme Nov 07 '22 at 07:08
  • Well what about Rz clock frequency. It is also doubled? – Rama Siva Subrahmanyam Nov 07 '22 at 09:38
  • 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??.... – Rama Siva Subrahmanyam Nov 07 '22 at 09:44
  • @Justme If the data is unconstrained, then a long series of 1s or 0s will shift the mean DC, which causes a problem with data slicing at the receiver. To this end, codes are designed that have zero or constrained DC. Manchester is zero DC at the bit level. An example of a constrained code, that doesn't need to go through a transformer, but does want to avoid wandering baseline, is the CD physical layer, where each 17 bit recorded word has either a +1 or -1 DC shift, and these are tracked in recording to keep the DC within bounds. – Neil_UK Nov 07 '22 at 10:16
  • If clock rate is doubled then why diagram is using same clock rate for every encoding??? – Rama Siva Subrahmanyam Nov 07 '22 at 11:11
  • @RamaSivaSubrahmanyam What they are saying, and what you are inferring, sound to be different. Just look at the pictures to see what's happening, and work out for yourself how they compare. It seems you're trying to double-think what the authors mean by clock doubling. Unfortauntely, after a long career internalising this stuff, I can't really see the problem, as it's obvious what is going on. What the author says has to be read through an understanding of what has to happen to transmit and receive the data to make sense. – Neil_UK Nov 07 '22 at 11:43
  • Why band width is doubled ? .. actually we are sending same data rate bits/second right . – Rama Siva Subrahmanyam Nov 07 '22 at 11:47
  • @RamaSivaSubrahmanyam Just look at the pictures, there's a clock transition in the middle of every data bit, as well as some transitions between bits. More transitions, more bandwidth. – Neil_UK Nov 07 '22 at 13:41
  • Also often used where the clock rate is not standard, like with magnetic-stripe cards. The clock rate is "however fast you swipe it", which can be difficult to standardize. – fectin Nov 07 '22 at 13:52
  • @fectin excellent shout. I've updated my answer, with credit to you. – Neil_UK Nov 07 '22 at 14:10
  • @RamaSivaSubrahmanyam If you take a look at the picture, for each unencoded data bit you send, you basically need to toggle the Manchester encoded data bit twice. – Justme Nov 07 '22 at 17:38
  • Manchester can't support rate changes mid-word, take a closer look. `01010101` has the same pattern of transitions as `0000`. – Ben Voigt Nov 07 '22 at 18:55
  • @BenVoigt I think there's a difference between 'a change', and variable during – Neil_UK Nov 07 '22 at 21:04
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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.)

hotpaw2
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  • 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 – Rama Siva Subrahmanyam Nov 07 '22 at 05:51
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    @RamaSivaSubrahmanyam It does not matter. The Manchster encoded data still requires twice the bandwidth than unencoded NRZ data. – Justme Nov 07 '22 at 06:25