FDM allocates portions of bandwidth to each circuit in the link. According to the Nquist equation, this would propotionally divide the data rate among the circuits.
Now with TDM time slots are allocated to each circuit, but for each time slot allocated, a circuits gets the whole bandwidth of the link. Then is the data rate divided or a circuit is able to utilise the whole data rate of the link?
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1 Answer
The bandwidth is measured in how many bits-per-second you can send on the link. FDM will allow simultaneously sending that many bits in aggregate by dividing the frequency into sub-frequencies, each with a smaller bandwidth (think cable television that simultaneously sends all the channels across a link on different frequencies, each only taking a small part of the total link bandwidth).
TDM, on the other hand, can only send one virtual circuit at a time, so it allocates a certain time period to each virtual circuit. For example, a 10 Mbps TDM circuit with 10 virtual circuits would allocate 1/10 of a second to each virtual circuit, allowing each virtual circuit to send 1 Mb each second. The hosts could serialize the bits onto the wire at the full 10 Mbps, but only for 1/10 of a second during each second.
answered Jul 22, 2018 at 23:59
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Thus, link data rate is divided among circuits in both FDM and TDM? Commented Jul 23, 2018 at 0:05
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Either way only gives each virtual circuit a part of the total available physical circuit bandwidth. If the answer answered your question, please accept it so that the question doesn't keep popping up forever, looking for an answer.– Ron Maupin ♦Commented Jul 23, 2018 at 0:08