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7th December 2007 at 06:27 #49967MKTGuest
I have one doubt regarding RF hopping.
A cell with 4TRX. TRX1 is BCCH TRX so no hopping for it.TRX1 has BCCH on TS0 and SDCCH/8 on TS1.
Further due to half rate implemented on remaining 3 TRX’s one more SDCCH/8 is added and this time we have choosen TS0 of TRX4.Doubt is TRX2 TRX3 TRX4 are put to hopping mode(RF) then how MS will be able to use the SDCCH/8 of TRX4. I think the there should not be hopping on SDCCH. hopping is only for TCH.??Pl correct7th December 2007 at 09:55 #49968pixGuest
hopping on SDCCH is done the same way as on TCH. So on TRX TS0, the SDCCH/8 will be hopping.
Just one remark : in synthetized hopping the BCCH TRX is not hopping, but in BBH, it will be hopping (except TS0).
pix8th December 2007 at 06:53 #49969MKTGuest
thanks PIX for your reply. one more question it is said that one SDCCH/8 logical channel can serve 8 MS. while one SDCCH/4 can do it for 4 MS. How is it calculated? I mean at any instance (live time) the SDCCH will be serving one MS only and after that it will become free and can be used by other MS and so on(pl correct if not so). So does that means if we consider a 51-TDMA frame structure plus assuming that none of the frame will go idle 50/8=6 frames are required by a MS for completing the signalling work.8th December 2007 at 23:36 #49970pixGuest
in sdcch/8 51-multiframe :
8 SDCCH blocks of 4 bursts (0 … 7)
4 SACCH blocks of 4 bursts (for sdcch 0…3)
next multiframe, the 4 SACCH blocks will be for SDCCH 4…7
there are 3 blocks which are unused.