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Reply To: TA


Hi, MKT.

First at all, I learned this topic as a beginner. In fact I never needed this info so I forgot the details.

No metter this I’ll try to explain in breef the main idea. I hope our colleagues will help if something missed or wrong.

MS will send access burst over RACH. The MS Tx is synchronized with BTS Rx ( both know that the beginning of the RACH TS number N is moment “t0” ).
The MS wants access and choose to use RACH TS number N. It waits for the beggining of the RACH TS and in moment “t0” sends the burst.
During the RACH TS number N the BTS receives access burst from unknown MS but in time “t1”.
The BTS knows when beggins RACH TS number N (“t0″). So the BTS computes the delay of the signal due to the propagation distance.
Then ( via the SACCH associated to the AGCH ) BTS informs the MS for its own TA.

TA is useful for SDCCH, voice and data communications. But how exactly?
You know, the DL and UL for one connection is during the same frame TS, but not simultaneous in time. There is a 3 TS period shifting ( this fact could generate some problems with consecutive TS PS data transmission/reception – other topic ).
DL F1 0 1 2 3 4 5 6 7 F2 0 1 2 …
UL F1 X X X 0 1 2 3 4 F2 5 6 7 …

F1 and F2 measure the DL frame.
Its copy down to show the shift for UL

So, lets say, at moment “t0” the BTS transmits burst to MS1 using F1 TS3.
Exactly at same moment the BTS receives burst from MS2 ( which have F1 TS0 – if MS2 present at all).
After 3×0.577ms (“t1”) in DL the BTS will transmit a burst to MS3 and will expect a burst from your MS1 at same time.
MS1 will transmit its burst at moment “t1”-TA. The signal travels some time ( TA ) and arrives exactly at moment “t1”.

That’s what I think about TA…

I don’t know how many clocks have the MS and BTS but I think it’s one… and everything is shifted up or down…

What do you think…?