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TA

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  • #50407 Reply
    MKT
    Guest

    Hi,
    In a live situation how can a MS take
    a decision to advance its TX time. I mean either there should be certain sort of buffering make TA possible in MS but that will give delay. if not then how TA is possible.
    I think max TA should be 3+7 max. how
    64 values are possible.

    pl correct..i know i am wrong.

    #50408 Reply
    Pix
    Guest

    MKT,

    1. The BTS sends the value of the TA to the MS. The MS doesn’t decide anything !

    2. The timing advance is computed in symbols, not in timeslots. 1 symbol is extremely short ! (don’t have the value in mind).

    3. The MS is sending 1 burst of compressed voice every 4.615 ms. So there is obviously a buffer in the mobile. More than that, there is a whole encoding, modulating, ciphering, etc… chain within the MS. 1 burst contains the equivalent of 4.6ms of voice (approx).
    The TA will just tell him exactly when to send it, compared to the “time reference” provided by the BTS.

    #50409 Reply
    NK
    Guest

    Hi! MKT,

    6 bits are assigned for TA. Therefore, value is from 0 to 63. What MS does is it multplied TA from BTS by 3.692 microsec(duration of 1 bit) and divided by 2 as timing advance recieve from BTS is for round trip delay. Hope you understand.

    Regards,
    NK

    #50410 Reply
    MKT
    Guest

    Thanks for reply.
    i got it. It is all there in the name itself. Its “Timing Advance”.

    So…
    I think BTS should also used a TA on per TS basis for individual MS….Is it correct?

    Second…when does BTS tells a MS to use this TA…….i mean is it done on AGCH initially…?

    In dedicated mode it must be on SACCH .
    Thanks MKT.

    #50411 Reply
    pix
    Guest

    MKT,

    you got it all right 🙂 it’s per TS (per MS actually, so if you have a HR timeslots with 2 calls, each call will have a independant T.A.)

    it’s on the AGCH initially (along with the IMMEDIATE ASSIGNMENT COMMAND) and it is updated every .5s on the SACCH (SACCH of the SDCCH and/or SACCH of the TCH)

    #50412 Reply
    MKT
    Guest

    Let us say it like this:

    A MS will use SCH initially to know the time(TDMA) of network/BTS first.

    But before entering into the dedicated mode the MS will be told by the BTS to correct its watch(the one which is to be used for TX)using the given TA.

    Also for RX the MS need not to do anything as BTS is taking care of adjusting its watch attached to the particular TS of MS under question.

    So, does this mean MS has to use two watches in dedicated mode.

    #50413 Reply
    TNS
    Guest

    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.
    TA=”t1”-“t0”.
    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…?

    BR,
    TNS

    #50414 Reply
    MKT
    Guest

    Hi TNS,
    The concept of difference of 3 TS between UL/DL is altogether a different theory.
    As far I know it was done to make the circuitry inside MS somewhat simpler (As MS need not have to send and receive simultaneously).

    But yes indeed it is one of basis for my two clocks theory:
    Just have a look at my logic:
    MS adjusts its clock as per the clock(1) of BTS. Under hypothetical condition(assuming no delay)….MS will use clock(1) for listening(RX) and speaking(TX)…….MS will apply a difference of 3TS between TX and RX. Everything is fine here and happy ending.

    But in reality..a delay is there….BTS senses this….and he(BTS) tells the MS to advance its timing for speaking a little bit(TA)….MS will now start speaking early(TA)….and BTS starts listening the MS well at the schedule time…But same should be applicable to BTS also….he knows that if is getting the speech of MS delivered late to him so do the MS also….hence BTS also applies a TA on the TS allocated to the particular MS…..

    But MS has to apply the concept of speaking(TX) 3 TS later after RX…..Now he is listening to BTS as the clock(1)……..MS cant do any changes wrt to clock(1)…hence MS has to use a separate clock(2) for TX. However the clock(2) will no doubt be derived from clock(1).

    Sequence may be
    a) get clock(1) from BTS through SCH
    b) use this clock(1) for RX
    c) apply a difference of 3 TS to this clock(1) for TX . let us say the new time is clock(2) (can say that TX watch is running 3 seconds slower then the RX watch.)
    d) But as per the wish of BTS apply a correction of TA to clock(2) and finally use this resulting clock time for transmitting.

    May be taken as , a person is wearing wrist watches in both wrists. Let us say he will use the wrist watch of left hand to RX and wrist watch of right hand to speak. The wrist watch of right hand is running slow by a 3 seconds(3TS). But as per the order of BTS the person has to correct the time of right hand TX watch by an amount of TA.

    Sorry friends … I think above discussion was not needed indeed. But I have just expressed thoughts of my mind which may not be true.

    So, should I conclude that TA will be,
    1. calculated by BTS
    2. used by MS in UL
    3. used by BTS in downlink

    #50415 Reply
    Pix
    Guest

    TNS, i think you’re right 🙂

    MKT, there is no downlink adjustment. No need : the delay of the path in DL will make the DL signal arrives exactly with a diff of 3 TS.

    #50416 Reply
    TNS
    Guest

    Hi, MKT.

    If you are interested in details it’s Ok.
    I think you should check “GSM 05.10 – Radio Synchronization” (if you are not already ).

    BR,
    TNS

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