Why In Manchester encoding, the bit rate is half of the baud rate?

Question

I think baud rate is the rate of the symbols, and if each symbol contains n bit, then the bit rate should be n x baud rate

In Ethernet( Manchester encoding) ,if bit rate is half of the baud rate, then a symbol contains 1/2 bit ? As far as I know, bit rate should at least not less than symbol rate (baud rate).

About the relationship of baud rate and bit rate, my understanding have no problems, yet when it comes to the Manchester code, it's totally counterintuitive, could anyone explain about these?

Fpc

Answer 1

Bit rate is related to the speed of the transmission of the digital bit, while baudrate is related to the speed of change of symbols, which are significancies in analog signal. These can be either in amplitude, frequency or phase or more complex modulation methods. In manchester encoding, one bit is reprsented by two different levels of voltage. Therefore, lets say if you want to transfer 1Mbit digital data in one second, then you will need to make ~ 2 million changes in the level of the analogous signal. That is why, your bit rate will be 1Mbs, while your baud rate will be 2M bauds.

In NRZ encoding, one bit is represented by one symbol. Therfore rates will be equal.

Romanov

I like this explanation. It's intuitive. But it's confusing when a specific example such as 10101010 is imagined. The transitions in this case as seen here are equivalent in both NRZ and Manchester encoding. – Faculty

Answer 2

The Wikipedia article for Baud says that it can be defined as pulses per second. In the case of Manchester Encoding, this results in the baud rate being defined as "clock transitions".

A transition is what occurs when the signaling voltage goes from a low voltage to a high voltage, or vice versa. If you look at this diagram:

You will notice that the Manchester wave always makes a transition from either low to high or high to low when the clock transitions from high to low. The bits are encoded in that transition; a transition from low to high indicates a 1, and a transition from high to low indicates a 0. The low-to-high clock transitions are used to get the Manchester wave in a position where it can make the correct transition for the next bit. As you can see, there are never more than two clock transitions between one Manchester transition and the next; the clock is effectively encoded in the Manchester wave itself.

If the bits were encoded in a single clock transition (i.e. high being 1 and low being 0), then the clock (baud) rate and the bit rate would be the same, but then you would have to run a separate line for the clock. Because Manchester guarantees a transition every

Adrenaline

I did read the Wikipedia article , and I could understand why they said Manchester code, the bit rate is half the baud rate, what I puzzled is that these saying is incompatible with other sayings about bit rate and baud rate, for example :#20534917 – Fpc

You're being confused by your assumption that, in Manchester Encoding, a symbol must contain a 1/2 bit. It would be more accurate to say that it takes two symbols (clock transitions, in this case) to encode one bit. – Adrenaline

I'm comfused by what symbol means. In my opinion, symbols is what I want to transfer by the channel let's say 10110 and bits are what actually get transferred in the channel is 10.01.10.10.01 ( I add the period to make it more readable) , assume these is all transfer in a second, so we can say the baud rate is 5 baud/s and the bit rate is 10 bps(ten bits in a second). – Fpc

A symbol is the smallest possible thing in a system which can encode information. In an older modem, a symbol is a tone, and each tone corresponds to 1 bit. It's not a transition, because a modem only understands tones. Some symbols correspond to multiple bits. For example, a voltage with four possible levels can record two bits. In Manchester Encoding, it takes two symbols (clock transitions) to record a single bit. – Adrenaline

that's a point, I am still quite comfused, I will think about it for a while, thank you for answering. – Fpc

I think this sentence in the article is incorrect:"Common communication links such as 10 Mbit/s Ethernet (10Base-T), USB, and FireWire typically have a symbol rate slightly lower than the data bit rate, due to the overhead of extra non-data symbols used for self-synchronizing code and error detection.", at least in the Ethernet case. – Fpc

Answer 3

You can think of Manchester encoding not only transmitting the actual data, but also the clock (meta data) due to its self clocking characteristic.

http://en.wikipedia.org/wiki/Self-clocking_signal

Brawn

Answer 4

All you need to understand is that WITHIN any ONE state in Manchester encoding ( i.e either 1 or 0 ) there would be a transition . . as depicted in DIAGRAM above. . .the sole reason for transition being for reciever to synchronize

This being said, it means if we compare this encoding scheme to others. . Like NRZ. there would be double the transitions in manchester encoding as compared to other techniques ( for a sequence of 10101 manchester will have 10 transitions while NRZ would have 5 ). . there may be exceptions. This means baud rate for manchester would be 10 while for NRZ would be 5. .

In designing we use to say that if any recvr is capable of syncing to baud rate of 10 . . . that means with manchester it transmts five Bits while with NRZ it would transmit 10 bis

Aronarondel

Answer 5

You are mixing up the meaning of bit and baud. The baud rate is the rate of symbols on the wire and the bit rate is the rate of bits seen by the application. In the example below the application sends 11 bits and the wire transfers 22 symbols.

Waksman