# Concept of Time Division Multiplexing [duplicate]

I have been studying the concept of Time Division Multiplexing, but I have not been able to understand some stuffs, for instance, I have not been able to understand why the sampling rate for audio is 8Kbps and then they say the frame length is 128ms, I saw on another documents that the frame length would be 1/8000 which gives 128µs, I am really getting confused, as a matter of fact I am confused, on another documents I was also asked to Explain Time Division in terms of sources and their bit rates. I would like some help on all these please, maybe a little explanation of the concept would be also good, so that I can view it from another persons point of view, thanks...

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## marked as duplicate by Dave Tweed♦, placeholder, PeterJ, Camil Staps, Anindo GhoshMay 16 '13 at 10:58

1/8000 gives you the number of seconds per bit, so you can transmit 1 bit in 128us and hence you can transfer 1000 bits in 128ms. – nonsensickle May 15 '13 at 22:17
If you have additional points that were not addressed in your original question, you should bring them up there, rather than starting another question. – Dave Tweed May 15 '13 at 22:25

"Explain Time Division in terms of sources and their bit rates" is one of those badly phrased questions that appear in modern textbooks; it's supposed to trigger recall of some previous paragraph, not make sense in terms of basic concepts.

Time division multiplexing is any technique for transmitting multiple logical streams of data over a single physical channel by allocating timeslots to transmitters.

Consider a very simple example where there are two 8kbps streams you want to time division multiplex onto a 16kpbs channel. (Channel total bit capacity obviously has to be greater than the sum of things to multiplex onto it).

One 8kbps stream generates a bit every 125us (not 128us). The 16kbit channel transmits one bit every 62.5us.

Let's say we declare our timeslots to be 5ms long, which is 80 bits in the channel, assuming all data no padding. The multiplexer accumulates input bits from both channels for 10ms, after which it has 80 bits from channel 1 and 80 bits from channel 20. It then spends 5ms transmitting a frame of the first 80 bits, then 5ms transmitting another frame of channel 2. During that time it's accumulated another frame from each channel to transmit, and so on.

Edit: this is entirely separate from any consideration of audio sampling rates which might be going on - I'm explaining it in pure bit terms.

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