# Why is the Manchester encoded bit rate half the baud rate?

I think the baud rate is the symbol rate, and if each character contains n bits, then the baud rate should be `n x baud rate`

In Ethernet (Manchester encoding), if the bit rate is half the bit rate, then the character contains 1/2 bit? As far as I know, the bit rate should be no less than the symbol rate (baud rate).

Regarding the relation of baud rate to baud rate, my understanding has no problem, but when it comes to Manchester code, it is completely illogical, can anyone explain about it?

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The Wikipedia article for Bode says it can be defined as pulses per second. In the case of Manchester coding, this results in the bit rate being defined as "clock hops".

A transition occurs when the signal voltage goes from low voltage to high voltage or vice versa. If you look at this diagram:

You will notice that the Manchester wave always makes a low to high or high to low transition as the watch goes from high to low. Bits are encoded in this transition; the low to high transition indicates 1, and the high to low transition indicates 0. The low and high clock transitions are used to get the Manchester waveform in a position where it can make the correct transition for the next bit, as you you can see that there will be no more than two clock transitions between one passage in Manchester and the next; the clock is effectively encoded in the Manchester wave itself.

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

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You can think of a Manchester encoding that not only conveys actual data, but also clocks (metadata) due to its self-timing characteristic.

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The data rate is related to the digital bit rate, while the transfer rate is related to the rate of change of symbols that are significant in an analog signal. These can be either amplitude or frequency, or phase or more complex modulation methods. In manchester, one bit is represented by two different voltage levels. Therefore, say, if you want to transfer 1Mbit digital data in one second, you would need to make ~ 2 million changes in the level of the same signal. Therefore, your baud rate will be 1 MB and your baud rate will be 2M baud.

In NRZ encoding, one bit is represented by one character. These rates will be equal.

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All you need to understand is that WITHIN any ONE state in Manchester encoding (i.e. 1 or 0) will transition. As shown in the DIAGRAM above. The only reason for the transition is for the receiver to sync

That being said, it means that we are comparing this coding scheme with others, like NRZ. it would be a double hop in manchester coding compared to other methods (for sequence 10101 manchester will have 10 hops, while NRZ will have 5). there might be exceptions. This means that the baud rate for manchester will be 10, and for NRZ it will be 5 ..

In design, we say that if any recvr is capable of synchronizing with a baud rate of 10. this means that with manchester it will transmit five bits, while with NRZ it will transmit 10 bis

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