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LTE: How many minimum RBs are required to transmit master information block ?

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What things are considered to calculate the minimum RBs that need to broadcast master information block information ? I am expecting a detailed information to strengthen my knowledge. Thanks in advance.

posted Aug 24, 2016 by Harshita

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2 Answers

+1 vote

If you see the content of master information block, you can easily calculate the size of master information block. Total size of content of MIB is 24 bits. Calculation details is as following:
3 bits for downlink bandwidth
3 bits for phich-config
8 bits for System Frame Number
10 bits as spare
These 24 bits are not sent directly over PBCH. There are lot of things done to send over PBCH.
16 bits long CRC is generated and scrambled with antenna specific mask.
Now total 40 bits (24 +16) is passed to convolution encoding which generates 3 streams of 40 bits. Now 120 bits are passed to rate matching function which just repeats these 120 bits 16 times and information becomes 1920 bits long.
QPSK is used to transmit the MIB information that's why 1920 bits information becomes 960 QPSK symbols.

These 940 QPSK symbols should be transmitted within 40ms since a new MIB information can come to broadcast. 960 / 4 = 240 QPSK symbols should be transmitted in each radio frame. For that purpose 240 Resource Elements are required in subframe zero of each radio frame since MIB is transmitted in slot 1 of subframe zero. Since in each Radio block, 8 REs are used for reference signals, so only 40 REs will be available per RB to carry MIBs information. Because there are 240 bits so you need 240/40 = 6 RBs to transmit MIB information. However total 960 QPSK symbols are transmitted by using 4 radio frames.

Hope, this detailed information will help.

answer Aug 25, 2016 by Vimal Kumar Mishra
72 RE or 84 RE per RB based on cyclic prefix but from your answer 40 RE per RB taken and 8RE fixed for ref signals what about remaining RE's per RB.
0 votes

Hi harshita,

MIB transmission is doesn't require RB's at Scheduling side, It mainly processed at PHY layer. PDCCH and other control information is not needed RB's to transmit like wise MIB also processed at PHY layer and will be transmitted in cell specfic position.

Actually In 20 MHZ there will be 200 RB's in two slot but Scheduling will be dont only on 100 Rb's , remaining RB's will be utilized implicitely by PHY.

answer Aug 24, 2016 by Jaganathan
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