Talk:Differential Manchester encoding
I am fairly certain that the term Conditioned DePhase is incorrect and should be Conditioned Diphase as per FED-STD-1037C. I am changing the article to make this correction.
It has been suggested that Differential Manchester encoding be merged with Biphase Mark Code.
I have come directly to the Differential Manchester encoding section, and I have no knowledge of Biphase Mark Code, nor do I have any desire to read about it. Therefore, it is my opinion that the two pages should be left separate.
Aside, the 'Line coding' section should have links to both Differential Manchester encoding and Biphase Mark Code.
I agree with 'Wickedpygmy'. I only wanted to know the differece between Differential Manchester encoding and Manchester encoding. Keep it serperated ;)
Fair enough. I do agree with you, however the codes are nearly identical. Basically the clock transition is swapped with the data-carrying transition. In BMC, the presence of a transition mid-bit indicates one state, and the omission indicates the other, whereas there is always a transition at the start of the bit. In BMC the data is carried on the period *after* its corresponding clock transition, in DM it's carried on the period *before* its corresponding clock. It becomes clear why a merge (might) be beneficial; the encodings are almost exactly the same.
Re: categorization, I agree and I'll fix this now if I can find it
The codes are different. It must be kept separate. Things can not be "exactly the same" and ""almost" it at same time. We are talking about exact science. The codes are different.
6 and half a dozen are exactly the same. 6 and 5.9999 are NOT the same.
AES and CDs do not care about Diferential Manchester, and DO USE Biphase Mark encoding.
My two drops. Hisatugo.
- No but 6 and 5.99999999999999999999999999999999(recuring) are the same. It is a well known proof. 18.104.22.168 (talk) 14:03, 16 June 2008 (UTC)
- As far as I can tell "differential Manchester" and "biphase mark code" are synonyms for exactly the same thing. So I strongly feel their articles should be merged.
- If you feel that there is some difference between them, please give an example signal where a biphase mark code receiver would decode a different bit sequence than a differential Manchester receiver. --22.214.171.124 19:20, 5 June 2007 (UTC)
I also think, that they are exactly the same. If you look at the two pictures  and  you will notice that they are equal, except for their sign and a negative delay of half a clock cycle (which doesn't make any protocol difference). In the differential manchester encoding seen on the image, a 0 causes a rising or falling edge half of a clock cycle before, while a 1 suppresses an edge at that position. In the BMC encoding seen on the other image, a 1 causes a rising or falling edge, while a 0 suppresses an edge. In both codes there are mandatory edges every full clock cycle, while the presence of an edge between those mandatory edges is directly encoding the data. Given the fact that an inverse encoding can be done in both differential manchester and BMC encoding, the two encodings are two different definitions of the same thing.
Btw. 5.9999999... with an infinite amount of nines following IS in fact 6. As the infinite Sum of 9/10 + 9/100 + 9/1000 + ... equals 1.
126.96.36.199 13:38, 4 December 2007 (UTC)
Currently the article Conditioned Diphase is a stub, vague and says it's the same as Differential Manchester encoding. Differential Manchester encoding says its the same as Conditioned Diphase. If there is no difference, the articles should be merged. If there are differences, these should be properly listed and explained. If both are different conventions over parameters of a common principle, the article should explain the principle, the parameters and the different implementations. This discussion is nearly 4 years old now, time to do something. Past (talk) 10:55, 15 January 2011 (UTC)
- I've done the BMC merge. I'll hold off on the Conditioned Diphase merge until someone else has a chance to massage the one citation I dug up (a page from a US military standard) into a cleaner form. --Damian Yerrick (talk | stalk) 03:35, 6 June 2011 (UTC)
Differential Manchester Encoding and bi-phase mark encoding are NOT the same thing. This article currently implies they are by saying that differential Manchester is also called BMC.
- They sure seem the same to me. I looked at your reference  and that page does not contain the word "Manchester" so I don't get how it's evidence that differential Manchester is not identical to bi-phase mark. Mark.sullivan (talk) 17:31, 11 July 2015 (UTC)
Furthermore I believe some of you are confusing bi-phase mark encoding with bi-phase space encoding. Bi-phase space encoding is indeed very similar to differential Manchester encoding in that it's shifted by a half bit and inverted. Bi-phase mark encoding is completely different (evident by the images in this very article). — Preceding unsigned comment added by Linksmask1 (talk • contribs) 21:23, 12 December 2013 (UTC)
- And bi-phase space is like differential Manchester except the transition is present for a 0 and absent for a 1. Another way to look at it is that you invert the data before the differential Manchester encoder. Mark.sullivan (talk) 17:31, 11 July 2015 (UTC)
Where to start?
As the shape of a bit depends on its preceding bit, I guess there is a need for a definition of how the first bit has to look like. --Abdull 18:51, 9 July 2007 (UTC)
- This is usually implementation dependent. It is the presence or absence of a transition that encodes data, which only implicitly relies on the previous state of the line. In a real system, there will be a natural 'idle' state of the line when the system is started up, and the first transition will deviate from this state. There may be other implementation and physical layer dependencies that crop up in certain cases as well, but in general there is an obvious and/or specified state that the line will be in at startup and when no data is present. Does that answer your question?
- I suppose it may be worthy of mention in the article, but I'm really on the fence. The actual conditions depend heavily on the physical medium in use, and vary from case to case.
- --Ktims (talk) 13:32, 20 November 2007 (UTC)
- Physical systems that are point-to-point or point-to-multipoint will generally send either a continuous stream of 0s or a continuous stream of 1s when the line is idle. This way the clock recovery means can stay synchronized and you can use transformers. In a multipoint-to-multipoint system, there is a zero voltage on the line between transmissions but each message (packet) has to start with a preamble that gives the clock recovery an opportunity to get synchronized.
- Indded in the diagram that accompanies the article, although the first bit has been encoded, there is no way for the decoder to determine that it was a '1' as the signal is previously undefined. 188.8.131.52 (talk) 14:02, 16 June 2008 (UTC)
yet another slight variation
- There seem to be several slightly-different line codes, all of which would work with a decoder that decoded "transition in the middle of the bit-time" (hi-lo or lo-hi) to 0, and "no transition for the entire bit-time" ( hi-hi or lo-lo ) to 1.
- This category includes coded mark inversion, the 2-level version of "Return-to-zero in optical communication", and differential Manchester encoding.
- The difference between them is in the details of how the transmitter chooses whether or not to add another transition between bits.
- When the transmitter *always* adds a transition between bits, it generates differential Manchester encoding.
- When the transmitter *never* adds a transition between bits (always maintains the previous level), except it always adds a transition between 2 consecutive 1 data bits -- is there a name for this kind of line code?
- --184.108.40.206 (talk) 06:59, 27 July 2008 (UTC)
isn't this just...
...a high speed variation on the Kansas City Standard, or to be more precise, 1200 baud CUTS? Or maybe they're a slower variant on this, depending which came first.
If you see it in terms of an audio wave, a 0 is being encoded as two cycles of a high frequency master clock, and a 1 as a single cycle of a wave at half that frequency. Which is exactly how CUTS works (KCS quadruples up the number of cycles for a more reliable 300-baud). Presumably there may also be provision for a carrier signal at 4x the 1-code rate or similar... ;) 220.127.116.11 (talk) 13:40, 25 September 2010 (UTC)
Differential coding examples.
Problem with images
The first image is wrong: if "data" row is still not differential-encoded, then the result should be:
(we cannot know the status of the first bit because we cannot know the state of the previous one)
I point out that "differential manchester" row would be wrong even if we assumed "data" as already differentiated.
The second image (BMC) could be correct. But "DATA" refers to already differentiated data, where "0" means change between bit k and bit k-1 (usually 1 means change between one bit and previous one). Additionally, the whole line should be shifted half a bit to the right. So I suggest to leave only one image and add the "differential encoded" line, in order to make it more clear. I can do this as far as I have time to do it, but I'd like to listen to some opinions before doing it.
Problem with the example graph of differential Manchester
In the example graph, the voltage level starts out at 0V then begins transitioning between a positive and a negative voltage. The voltage at the end is 0V. I believe that the part of the signal where it is pictured at 0V should not be shown. When the signal transitions between a positive and a negative voltage, it would never stop at 0V. It would just pass through 0V on its way to the opposite extreme. — Preceding unsigned comment added by 18.104.22.168 (talk) 23:39, 11 April 2012 (UTC)
Hi, are we sure that the first image is correct? In my opinion it should be different when data are 1 continuously (as in the image, it loses clock!). Thanks,
As others have indicated, the graph is incorrect. I've edited it. Of course, inverted polarity would also be correct and it depends on the previous state of the line.