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Johannes Schmitz jsemail at gmx.de> I understand that one TDMA frame has a period of 4.615 ms and consists > of 8 Time slots, each of a period of 576.9 us. When a MS registers to > the BTS, it gets one Time-slot, but how is speech transfered? Normally > the Fs (samplefrequency) is 8 KHz, this means every 128 us one byte > sample. If I have one timeslot of let's say 577 us , than there is a gap > of 7 other timeslots, which makes a 4 ms gap. Also the speech data is > compressed. How is that damn speech data multiplexed with other MSs and > finally received on the other side as 8 bit samplevalue at a rate of 128 > us (8 KHz). I am not sure if this is what you want to know but I suggest you have to take a look at GSM fullrate and halfrate speech codecs. I am currently doing the speech processing lecture in here in Aachen so you might have a look at the following books: "Digital Speech Transmisson" by Peter Vary and Rainer Martin and "Wireless Communications Second Edition" by Theodore S. Rappaport. Or you read it from the standard... You will find that speech is not transmitted by sending just the 8kHz samples. There is a complex prediction filter structure. In principle the filter coefficients and the error are send and the receiver does a reconstruction of the signal. Additionally everything is quantized heavily to achieve the low datarates. If you use fullrate, every 8th timeslot is given to you, if you use halfrate you get every 16th timeslot. You also need to now about multiplexing of the logical GSM channels. Sometimes a timeslot is used to send signaling or control information instead of speech data. The Speechcodec is able to hide this missing data so you will not notice that while talking. regards, Johannes