hi bigdogguru,
thanks for the reply, I think I found the part realted to my question. in the first link you have recommended it says:
"Clock synchronization
Another intrinsic value to Manchester encoding is the fact that the synchronizing clock is embedded within the signal. This fact is exploited in Ethernet, which uses on-board circuitry to maintain clock synchronization. A Digital Phase Locked Loop (DPLL) circuit monitors the incoming Manchester-encoded signal and makes adjustments to its internal oscillator to keep it in constant synchronization with the transmitter's clock frequency.
The DPLL functions by sampling the incoming Manchester-encoded data with its own local clock. A simple shift register, driven by the local clock, accumulates all the shifted bits.
If the local oscillator is in synchronization with the transmitter's clock, there will be an equal number of binary 1's and 0's across the shift register. If an imbalance occurs between binary 1's and 0's, the local clock is adjusted based on the number of binary bits off center. This is why you will find a preamble at the beginning of each packet transmitted via Ethernet. "
but why do the numbers of 1's and 0's have to be equal?