Why SFN size is 8 bit in MIB?

Specs & Protocol Testing
1

Hi Experts.
Any one know why SFN size is 8 bit in MIB?
Any specific reason?

2

2 bits are implicitly used by UE at reception to figure out exact SFN.
00 01 10 11

This is what i know, any better explanation is welcome.

1 Like
3

MIB has a periodicity of 40 ms and repetitions are made within 40 ms at 10 ms interval.
These 4 repetitions of MIB will help to derive the remaining 2 bits of SFN.

4

Let us discuss more on this since i have always doubted this part.

Initially while Channel encoding from enb end, SFN 10 bits are encoded at higher layers but from phy layer only 8 bits are CRC coded and send towards UE.
UE at receiving at randomly attempts with 00, 01,10, 11 and with any of bit if CRC decoding gets successful that particular bit no. is considered to as SFN no. out of 4.

E.g. if 01 helped decoding CRC at UE end then UE figures out that SFN no is 2 within that 8 bit SFN which are 256 total sending same SFN No for next 40ms.

Let me know if others too know the same thing.
I was unable to find its explanation anywhere.

Like below -
SFN = 4 (received inside MIB)
00 = SFN-0
01 = SFN-1
10 = SFN-2
11 = SFN-3

SFN = 5 (received inside MIB)
00 = SFN-0
01 = SFN-1
10 = SFN-2
11 = SFN-3

5

I think MIB will not contain full 10bits of SFN, but only 8bits. Rest 2bits can be decoded by UE by:

Best Case: UE decodes first MIB then it will need decode rest 3 (all 4) MIBs successfully to figure the rest 2bits(full SFN).

Worst Case: UE decodes second MIB then it will need decode rest 7 (3 RV MIBs for previous SFN and all 4 MIBs for current SFN) to be in sync with ongoing SFN (from eNB) and also to get sync on 40ms boundary.

This is my understanding.

6

Best Case: UE decodes first MIB then it will need decode rest 3 (all 4) MIBs successfully to figure the rest 2bits(full SFN). ----------->>>>

What i know here, if UE decode 1st MIB successfully it won;t be decoding rest 3 MIB’s since all RV versions of MIB are independent of each other where same set of 240 symbols are copied 4 times.

Overall, UE will read more than 1 RV only in case when UE won;t be able to decode 240 symbols successfully in that case UE can read copied RV’s to get complete info.

Correct me if sounds wrong.

In Logs, if we read MIB, it will show same SFN for all 4 RV’s e.g. 137 137 137 137 which is actually 137 138 139 140. So, to figure out what is the exact SFN no. UE use 00 01 10 11 value to received 8 bit SFN CRC from MIB, so if 11 helps CRC decoding then SFN no which UE will understand is 140.

UE will use these bits (00 01 10 11) one by one in randomly manner.

7

Are you sure about this?

I thought LTE PBCH CRC corresponds to no of antennas.
UE does blind decodes to figure out the antenna info.

8

That is PBCH masking which is 16 bit info denoting 1, 2, 4 port Antenna.

CRC is 16 bit Convolution encoding.

9

Ok.
If CRC is taken care at channel level, why is there one more CRC at MIB?
Doesn’t sound right…

10

Both are different.

CRC is done in multiple steps, first CRC is done before segmentation then 2nd CRC is done after segmentation.

11

You are talking about code book segmentation and concatenation.
Is this applicable to MIB too?
I thought this is only for data.
Do you have a reference for this?

12

You are right that segmentation is for turbo coding only, for convolution coding segmentation not required.

13

So, do you agree that there is no CRC for MIB?
Only CRC is for PBCH?

14

CRC for data channels and control channels are different.
Control channels will have 16 bit CRC.
Data channels will have 24 bit CRC.
16 bit CRC used here for MIB since MIB is control channel.
CRC (16 bit) in MIB is different and CRC for PBCH is 16 bit masking (XOR) operation.

15

That is fine.
You use this CRC to perform validation on PBCH payload and to derive the antenna info implicitly.
Once the PBCH payload is validated, MIB is derived from it.

Now, why is CRC coming into picture again to derive the SFN again?
That’s what doesn’t seem right.

16

With this we are back to the basic question. SFN is 8 bit transmitted in MIB.
That means single values of SFN is used for all copied RV’s.

Question is – Suppose enb sending 137 137 137 137 towards UE, so out of all 4 137’s UE decoded which SFN? How UE will come to know?

17

Let me check the spec and revert.

18

What is your confusion here?

8 bits are transmitted in MIB and remaining 2 bits are derived by UE.
This together gives unique 10 bit SFN.

19

Question is – (Taking an Example)

Suppose enb sending 137 137 137 137 towards UE (as 4 MIB’s per 10ms), so out of all 4 137’s UE decoded which SFN?

How UE will come to know?

The explanation of this statement, can you break it down how UE using those 2 bits and how is it figuring out the exact SFN?

You can read the description I have given for this in previous texts… but those are just a mouth words, I was unable to find proper source any where…

20

Remaining 2 bits derived as follows

First 137 corresponding to 2 bits as 00, next 137 to 01, third 137 to 10 and last 137 as 11

For your 137 example of 4 transmissions of MIB, SFN is derived as follows

First 137 (0b10001001) + 00 (2 bits derived by UE) = 0b1000100100 (SFN: 548)

Second 137 (0b10001001) + 01 (2 bits derived by UE) = 0b1000100101 (SFN: 549)

Third 137 (0b10001001) + 10 (2 bits derived by UE) = 0b1000100110 (SFN: 550)

Fourth 137 (0b10001001) + 11 (2 bits derived by UE) = 0b1000100111 (SFN: 551)

So, UE gets 548, 549, 550 and 551 as SFN numbers.