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Fundamental GSM confusion

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  • #49687 Reply
    Colin
    Guest

    I’m trying to understand a basic thing about GSM RF and timeslots. I’m pretty sure at least one of these statements must be false, because they don’t add up properly.

    a) One pulse of GSM RF can carry 116 bits of data (2 x 58, from ETSI TS 145.002)

    b) A single timeslot of a single frame corresponds to a single pulse of RF.

    c) A control channel in LAPDm carries up to 23 bytes, or 184 bits of data.

    So how can those 184 bits of data fit onto 116 bits in the pulse? What have I got wrong?

    #49688 Reply
    paraHO
    Guest

    I think you are reading the information round the wrong way and mixing your digits.

    A “burst” contains 148 bits, 116 bits of which are real data and the remaining bits are basically for synchronisation.

    #49689 Reply
    Colin
    Guest

    OK, so an RF burst contains 116 bits of data.

    But if a single LAPDm message on the air interface carries 184 bits of data then how can it be carried on a single burst?

    #49690 Reply
    pix
    Guest

    one burst contain 184 bits.

    when the burst is used for TCH, it contains 116 bits of useful data, and 68 bits of other info (training sequence, guard bands, etc).

    a LapDm message might be sent on many timeslots (= many bursts). For instance the measurement results are sent over 4 bursts.

    Regards,

    #49691 Reply
    Colin
    Guest

    OK, I found a bit of information that clarifies the control channels: ‘The 184 bits of raw information are processed to yeild 456 bits. These 456 bits are then transmitted through four time slots’.

    I hadn’t picked up that one LAPDm message is spread over four frames of a timeslot.

    Thanks for your help,
    Colin

    #49692 Reply
    paraHO
    Guest

    Sorry I should have said a “Normal Burst”

    #49693 Reply
    VK
    Guest

    Be informed that a normal burst carries only 114 bits & not 116 bits.

    #49694 Reply
    paraHO
    Guest

    For “Normal Burst” see GSM0502 R99 Cl 5.2.3

    Length of Field 58 – e0-e57
    Length of Field 58 – e58-e115

    Total 116 bits

    #49695 Reply
    VK
    Guest

    3 T + 57 CD + 1 S + 26 TS + 1 S + 57 CD + 3 T + 8.25 GP

    T – Tail bit (3 + 3) = 6
    CD – Coded data (57 + 57) = 114
    S – Stealing Flag (1 + 1) = 2
    TS – Training Sequence (26) = 26
    GP – Guard period (8.25) = 8.25

    #49696 Reply
    VK
    Guest

    Hope everyone knows about the stealing flag.

    #49697 Reply
    pix
    Guest

    everybody seems a bit optimistic … you’re expecting to much from humanity 🙂

    #49698 Reply
    paraHO
    Guest

    The use of stealing flag is optional in the Normal Burst.

Viewing 12 posts - 1 through 12 (of 12 total)
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