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LTE: In ideal condition, how many random access preambles can be decoded successfully at the eNodeB ?

+1 vote

RACH procedure can be done by any number of UEs in a cell. But I want to know how many maximum RACH preambles can be decoded successfully at eNodeB and what factors derive this ?

posted Aug 27, 2016 by Ganesh

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2 Answers

0 votes

Hi ganesh,

At ENB side in one PRACH SF it can succesfully detect only one preamble. Because PRACH resource is 6 RB which is common to all UE, then If all UE tries to send PRACH to ENB then it will collide in RF and will be discarded at ENB. Based on UL delay some UE's preamble reach ENB then ENB will process it. I hope you know CRM procedure, this will let other(failed UE) UE to do rach again.

answer Aug 29, 2016 by Jaganathan
0 votes

Total number of rach preambles in LTE are 64. Rach preamble information is broadcasted in sib2 for contention based rach. The number of rach preambles available in sib2 can be used by different UEs who wants to perform contention based rach with enode. All the rach preambles those are used by ue uniquely in single sub frame in a cell will be decoded by eNodeB successfully if they are reaching to eNodeB. Also if any ue performing rach using dedicated preamble will also be decoded by eNodeB. So eNodeB can decode unique all max 64 preambles in single subframe.

answer Sep 2, 2016 by Veer Pal Singh Yadav
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