- This topic has 11 replies, 1 voice, and was last updated 13 years, 3 months ago by paraHO.
14th November 2007 at 17:29 #49687ColinGuest
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?14th November 2007 at 19:15 #49688paraHOGuest
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.15th November 2007 at 09:31 #49689ColinGuest
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?15th November 2007 at 09:48 #49690pixGuest
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,15th November 2007 at 10:52 #49691ColinGuest
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,
Colin15th November 2007 at 12:53 #49692paraHOGuest
Sorry I should have said a “Normal Burst”15th November 2007 at 15:14 #49693VKGuest
Be informed that a normal burst carries only 114 bits & not 116 bits.15th November 2007 at 17:30 #49694paraHOGuest
For “Normal Burst” see GSM0502 R99 Cl 5.2.3
Length of Field 58 – e0-e57
Length of Field 58 – e58-e115
Total 116 bits17th November 2007 at 11:05 #49695VKGuest
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) = 8.2517th November 2007 at 11:07 #49696VKGuest
Hope everyone knows about the stealing flag.17th November 2007 at 17:41 #49697pixGuest
everybody seems a bit optimistic … you’re expecting to much from humanity 🙂18th November 2007 at 11:50 #49698paraHOGuest
The use of stealing flag is optional in the Normal Burst.