In article , Chris wrote: :Assume that several packets with a total amount of 100 mbits (not in bytes :to simplify) are send together . Assume that the theoretical speed of 100 :Mbits/sec is a fact.
:Does it mean that they are sent all together (in a kind of parallel :channels) and that the total broadband is busy for a second or are they sent :one after one but the sum of the busy time of all the packets will be one :second?
Neither -- the time required to send 100 Mbits of packets over a 100 Mbit/s link is more than 1 second, due to the mandatory IFG (intra-frame gap) of 96 bit-times. Even if you include the size of the preamble and CRC in your 100 Mbit count, during the IFG nothing is being sent. Including preamble and CRC in the bit counts, you can get slightly over 99 Mbits through in one seconds of 100 Mbit/s.
Different signalling methods are used over different media, some involving discrete bits and some involving phase analysis of particular sample points of analog waveforms. In the analog case, considering the error correction methods and other fine details, bits are not transmitted independantly of each other but rather in small groups. (The interconnection is noticably stronger for gigabit.) Thus the question cannot be resolved without reference to a particular media -- and to a particular definition of what it means to transit bits in parallel.
There are numerous good sources of information on the 'net. One with a useful (but not up-to-date) overview is
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