The PBCH carries the MIB, the first piece of system information the UE reads after synchronization. The MIB has only 24 bits, but the UE must decode it at the cell edge and without any prior configuration. So those 24 bits go through heavy coding and are spread over 40 ms. This page follows them through lteMIB, lteBCH, ltePBCH and ltePBCHIndices.

If you don't know what PBCH(Physical Broadcast Channel) is, refer to Physical Layer Channel : Downlink : PBCH (Physical Broadcast Channel) page first.

Followings are the topics to be covered in this page.

• Generating BCH bit sequence
• Generating PBCH Symbol Data
• PBCH RE Mapping
• Appendix - A : Numerical Output for Encoding Process
• Reference

Generating BCH bit sequence

The first step turns the MIB into the BCH codeword. The code below builds the 24 MIB bits with lteMIB and encodes them with lteBCH, and the three plots after it show how the codeword reacts to a change in the bandwidth and in the frame number.

% First you have to define properites of a eNodeB.  

% NDLRB indicate System Bandwith in the unit of RBs.

% NDLRB 6 = 1.4 Mhz, NDLRB 15 = 3.0 Mhz, NDLRB 25 = 5.0 Mhz,

% NDLRB 50 = 10 Mhz, NDLRB 75 = 15 Mhz, NDLRB 100 = 20 Mhz

% CellRefP indicate number of downlink Antenna. CellRefP = 1 means 1 transmission antenna (SISO)

% NCellID indicate PCI (Physical Cell Identity) of the Cell

% NSubframe indicate the subframe number.

enb.CyclicPrefix = 'Normal';

enb.PHICHDuration = 'Normal';

enb.Ng = 'Sixth';

enb.NDLRB = 6;

enb.CellRefP = 1;

enb.DuplexMode = 'FDD';

 

enb.NCellID = 1;

enb.NSubframe = 0;

enb.CFI = 1;

enb.NFrame = 0;

 

% Now you have to generate MIB bits. This is basically the bit string that you can see from ASN Decoder.

% If you pass eNodeB information into lteMIB() function, it will generate MIB bits based on the information

% configured in enb.

mib_bits = lteMIB(enb);

 

% Next step is to apply the channel coding process to the MIB bits (a kind of user data).

% This is done by lteBCH()

% as shown below. Basically this performs step (1)~(3) of procedures descringed in

% PBCH (Physical Broadcast Channel)

bch_cw = lteBCH(enb,mib_bits);

 

% Now I want to show you the two sets of data generated above into a plot.

% this plot does not have any physical meaning since it is just user data and channel coding process

% but it would be easiler to see overall meaning rather than looking at all the ones and zeros.

 

% this is the plot to show the MIB message itself. It is just layer 3 bits.

subplot(2,1,1);

stem(mib_bits); xlim([0 length(mib_bits)]); ylim([0 1]);

title('MIB bits');

 

% this is the plot to show the result of channel coding (the result of lteBCH()).

subplot(2,1,2);

stem(bch_cw); xlim([0 length(bch_cw)]); ylim([0 1]);

title('BCH codeword');

 

The first example show the case of MIB message with System BW = 1.4 Mhz and Frame Number = 0. As you see, all zeros in layer 3 message and all 0 even after the channel coding.

One possible motivation from this example would be just to have 'question' (no need to have answer yet) how the small number of bits can get converted into such a long/huge number of bits (over 1800 bits).

enb.PHICHDuration = 'Normal'; enb.Ng = 'Sixth'; enb.NDLRB = 6; enb.CellRefP = 1;   enb.NCellID = 1; enb.NSubframe = 0; enb.NFrame = 0;

 

This example is almost same as previous example. the only difference is the channel bandwidth. I used 3.0 Mhz Bandwidth and it sets only one bit to be '1'. As you see in the second graph (the output of channel coding), you would see much more complicated patterns of ones and zeros.  Why ? Just have a question is enough.

enb.PHICHDuration = 'Normal'; enb.Ng = 'Sixth'; enb.NDLRB = 15; enb.CellRefP = 1;   enb.NCellID = 1; enb.NSubframe = 0; enb.NFrame = 0;

 

Following is for the case which is same as the first example except NFrame (SFN). I guess you would see the difference.

If you have LteToolbox, Try enb.NFrame = 1,2,3 and see what you get ? Are they all the same ? or different ?

If they are all same, would you be able to explain why ?

enb.PHICHDuration = 'Normal'; enb.Ng = 'Sixth'; enb.NDLRB = 6; enb.CellRefP = 1;   enb.NCellID = 1; enb.NSubframe = 0; enb.NFrame = 4;

NFrame = 4 sets MIB bit 14, the last bit of the SFN field. MIB bit 3 is 0, so the bandwidth in this example is NDLRB = 6.

The 24 MIB bits follow the MasterInformationBlock of 36.331 v19.3.0. The table below lists the fields in the order they are sent. Only the first three fields matter to the toolbox, which fills the rest with zeros.

The table answers both questions on the page. With NDLRB = 6, every MIB bit is zero. The CRC of an all-zero block is zero, the antenna mask for CellRefP = 1 is all zeros, and the convolutional code is linear, so the whole codeword stays zero. With NDLRB = 15, dl-Bandwidth becomes n15 = 001, and that single 1 spreads through the CRC, the coding and the rate matching into the pattern above.

The MIB carries only the 8 most significant bits of the 10-bit SFN, which is SFN divided by 4. NFrame = 0, 1, 2 and 3 therefore give the same MIB and the same codeword, and NFrame = 4 is the first value that changes it. The UE finds the two missing bits from where the PBCH sits inside its 40 ms period, as the next section shows.

• 24 MIB bits : 3 for the bandwidth, 3 for the PHICH, 8 for the SFN and 10 for later additions.
• All zeros in, all zeros out : the CRC, the mask for one port and the coding are all linear.
• NFrame 0 to 3 give the same MIB : the MIB carries SFN divided by 4.

Generating PBCH Symbol Data

The BCH codeword is still a string of bits. The next step turns it into QPSK symbols with ltePBCH, and the plot below shows why the 960 symbols are best read as four blocks of 240 rather than one long sequence.

In this section, we will convert the encoded BCH data into physical layer symbols and plot them into constellation and sequence plot.

 

% First you have to define properites of a eNodeB.  

% NDLRB indicate System Bandwith in the unit of RBs.

% NDLRB 6 = 1.4 Mhz, NDLRB 15 = 3.0 Mhz, NDLRB 25 = 5.0 Mhz,

% NDLRB 50 = 10 Mhz, NDLRB 75 = 15 Mhz, NDLRB 100 = 20 Mhz

% CellRefP indicate number of downlink Antenna. CellRefP = 1 means 1 transmission antenna (SISO)

% NCellID indicate PCI (Physical Cell Identity) of the Cell

% NSubframe indicate the subframe number.

 

enb.CyclicPrefix = 'Normal';

enb.PHICHDuration = 'Normal';

enb.Ng = 'Sixth';

enb.NDLRB = 6;

enb.CellRefP = 1;

enb.DuplexMode = 'FDD';

 

enb.NCellID = 0;

enb.NSubframe = 0;

enb.CFI = 1;

enb.NFrame = 0;

 

% Now you have to generate MIB bits. This is basically the bit string that you can see from ASN Decoder.

% If you pass eNodeB information into lteMIB() function, it will generate MIB bits based on the information

% configured in enb.

mib_bits = lteMIB(enb);

 

% Next step is to apply the channel coding process to the MIB bits (a kind of user data).

% This is done by lteBCH()

% as shown below. Basically this performs step (1)~(3) of procedures descringed in

% PBCH (Physical Broadcast Channel)

bch_cw = lteBCH(enb,mib_bits);

 

% Now if you  pass the encoded BCH bit stream into ltePBCH, it will generate physical layer symbols for BCH

%,which  is PBCH symbol.

 

pbch_sym = ltePBCH(enb,bch_cw);

pbch_sym_arrayIndex = 0:length(pbch_sym)-1;

pbch_sym_ind = ltePBCHIndices(enb,{'1based','re'});

 

% Now I will create a bunch of graphs showing different aspects of the PBCH symbols.

% The first two plot shows whole PBCH symbols in a single plot. The plot on top left will show you the

% constellation of the whole PBCH symbols generated above.

% The plot on the top right show you the whole PBCH symbols in sequence number (just the index of the array)

subplot(5,3,1);

plot(real(pbch_sym),imag(pbch_sym),'ro','MarkerFaceColor',[1 0 0]);

axis([-1 1 -1 1]);

subplot(5,3,[2 3]);

plot(pbch_sym_arrayIndex,real(pbch_sym),'ro-',pbch_sym_arrayIndex,imag(pbch_sym),'bo-');

xlim([0 max(pbch_sym_arrayIndex)]);ylim([-1 1]);

 

 

% Then, I split the generated symbols into 4 clusters, each of them has 240 symbols and plot the four clusters

% onto separate graph. Why did I split into 4 clusters ? What would be the meaning of the each cluster ?

% see SIB Scheduling page.

 

subplot(5,3,[4 6]);

plot_index = 1:240;

plot(plot_index,real(pbch_sym(plot_index)),'ro-',plot_index,imag(pbch_sym(plot_index)),'bo-');

xlim([min(plot_index) max(plot_index)]);ylim([-1 1]);

 

subplot(5,3,[7 9]);

plot_index = 241:480;

plot(plot_index,real(pbch_sym(plot_index)),'ro-',plot_index,imag(pbch_sym(plot_index)),'bo-');

xlim([min(plot_index) max(plot_index)]);ylim([-1 1]);

 

subplot(5,3,[10 12]);

plot_index = 481:720;

plot(plot_index,real(pbch_sym(plot_index)),'ro-',plot_index,imag(pbch_sym(plot_index)),'bo-');

xlim([min(plot_index) max(plot_index)]);ylim([-1 1]);

 

subplot(5,3,[13 15]);

plot_index = 721:960;

plot(plot_index,real(pbch_sym(plot_index)),'ro-',plot_index,imag(pbch_sym(plot_index)),'bo-');

xlim([min(plot_index) max(plot_index)]);ylim([-1 1]);

 

enb.PHICHDuration = 'Normal'; enb.Ng = 'Sixth'; enb.NDLRB = 6; enb.CellRefP = 1;   enb.NCellID = 1; enb.NSubframe = 0; enb.NFrame = 0;

960 QPSK symbols in total. The four lower plots split them into blocks of 240, one for each radio frame of the 40 ms PBCH period.

The 1920 bits of the BCH codeword become 960 QPSK symbols. Before modulation, 36.211 v19.3.0 clause 6.6.1 scrambles them with a cell-specific sequence, initialized with N_ID^cell in each radio frame where the frame number is a multiple of 4. The scrambling runs across all 1920 bits, so each of the four blocks of 240 symbols is scrambled differently.

That is the meaning of the four blocks. Each radio frame carries one block of 240 symbols, and each block holds a full copy of the coded MIB in a different scrambling phase. A UE with a good signal can decode one block alone. The scrambling phase that works tells the UE which frame of the 40 ms period it is in, and that gives the 2 SFN bits the MIB leaves out.

• 1920 bits become 960 QPSK symbols : 4 blocks of 240.
• One block per radio frame : the PBCH period is 40 ms.
• The scrambling phase gives the 2 lowest SFN bits : the MIB carries the other 8.

PBCH RE Mapping

The UE has to find the PBCH before it knows the bandwidth or the number of antenna ports, because the MIB carries the first and the PBCH decoding reveals the second. So the PBCH position must not depend on either. The grids below show where ltePBCHIndices places it.

In this step, I will show the Resource Elements that will carry the PBCH symbols generated at previous step.

 

% First you have to define properites of a eNodeB.  

% NDLRB indicate System Bandwith in the unit of RBs.

% NDLRB 6 = 1.4 Mhz, NDLRB 15 = 3.0 Mhz, NDLRB 25 = 5.0 Mhz,

% NDLRB 50 = 10 Mhz, NDLRB 75 = 15 Mhz, NDLRB 100 = 20 Mhz

% CellRefP indicate number of downlink Antenna. CellRefP = 1 means 1 transmission antenna (SISO)

% NCellID indicate PCI (Physical Cell Identity) of the Cell

% NSubframe indicate the subframe number.

enb.CyclicPrefix = 'Normal';

enb.PHICHDuration = 'Normal';

enb.Ng = 'Sixth';

enb.NDLRB = 6;

enb.CellRefP = 1;

enb.DuplexMode = 'FDD';

 

enb.NCellID = 0;

enb.NSubframe = 0;

enb.CFI = 1;

enb.NFrame = 0;

 

% following is PHICH parameters that will be used ltePHICH() function in later step.

 

PHICH_Group_Index = 0;

PHICH_Sequence_Index = 1;

HARQ_Indicator_Value = 0; % 0 = NACK, 1 = ACK

 

% Following is to create an empty resource grid for one subframe.

 

resourceGrid = lteDLResourceGrid(enb);

 

% Following is to create symbols for Cell Specific Reference Signal and make a list of resource index for the

% reference signal.

 

rsAnt0 = lteCellRS(enb,0);

indAnt0 = lteCellRSIndices(enb,0);

resourceGrid(indAnt0) = rsAnt0;

 

% Following is to create symbols for PCFICH and make a list of resource index for the signal

 

cfi_cw = lteCFI(enb);

pcfich_sym = ltePCFICH(enb,cfi_cw);

pcfich_sym_arrayIndex = 0:length(pcfich_sym)-1;

pcfich_sym_ind = ltePCFICHIndices(enb,{'1based','re'});

 

% Following is to create symbols for PBCH and make a list of resource index for the signal (channel)

 

mib_bits = lteMIB(enb);

bch_cw = lteBCH(enb,mib_bits);

pbch_sym = ltePBCH(enb,bch_cw);

pbch_sym_arrayIndex = 0:length(pbch_sym)-1;

pbch_sym_ind = ltePBCHIndices(enb,{'1based','re'});

 

% Following is to create symbols for PSS and make a list of resource index for the signal

 

pss = ltePSS(enb);

pss_arrayIndex = 0:length(pss)-1;

pss_sym_ind = ltePSSIndices(enb,0,{'1based','re'});

 

% Following is to create symbols for SSS and make a list of resource index for the signal

 

sss = lteSSS(enb);

sss_arrayIndex = 0:length(sss)-1;

sss_sym_ind = lteSSSIndices(enb,0,{'1based','re'});

 

% Following is to create symbols for PHICH and make a list of resource index for the signal

 

phich_sym = ltePHICH(enb,[PHICH_Group_Index,PHICH_Sequence_Index,HARQ_Indicator_Value]);

phich_sym_arrayIndex = 0:length(phich_sym)-1;

phich_sym_ind = ltePHICHIndices(enb,{'1based','re'});

 

% Following part is filling the resource grid with each of the signal.. but if you see carefully I didn't fill this

% with real symbol number, I just filled it with a constant that I arbitrarily set. This is just for visualization..

% just to allocate constant/outstanding color for each signal. When you  use this resource grid for real

% transmission (not for visualization), fill the resourceGrid with real symbol value you generated above.

 

pss_scale = 0.2;

sss_scale = 0.4;

phich_scale = 0.7;

pcfich_scale = 0.5;

pbch_scale = 0.7;

 

resourceGrid(pss_sym_ind) = pss_scale;

resourceGrid(sss_sym_ind) = sss_scale;

resourceGrid(pcfich_sym_ind) = pcfich_scale;

resourceGrid(phich_sym_ind) = phich_scale;

resourceGrid(pbch_sym_ind) = pbch_scale;

 

% Following is to display the resource grid. I didn't find any proper functions in the toolbox to display

% one subframe grid as I like. So I used a little bit of tricks. First I plot 3D surface graph with the grid and

% move the view point right on top of the plot so that it looks like plane 2D grid.

 

xStep = 0:13;

yStep = 0:(enb.NDLRB*12-1);

surface(xStep,yStep,abs(resourceGrid));

axis([0 13 0 (enb.NDLRB*12-1) 0 1]);

view([0,90]);

set(gca,'xtick',[0 6 7 13]);

set(gca,'ytick',[[0:12:enb.NDLRB*12-1] [enb.NDLRB*12-1]]);

 

The first set of examples shows the location of PBCH Resource Elements in 1.4 Mhz BW. The light green REs in the symbol 7~10 represents PBCH. Can you count how many Resource Elements are for PBCH ? Maybe a little confusing to count, but it should be 240 REs, meaning it should carry 240 symbols.

1.4 MHz. The PBCH fills symbols 7 to 10 across the whole 72 subcarriers, apart from the RS positions in symbols 7 and 8.

The second set of examples shows the location of PBCH Resource Elements in 5 Mhz BW. The light green REs in the symbol 7~10 represents PBCH. Can you count how many Resource Elements are for PBCH ? Maybe a little confusing to count, but it should be 240 REs, meaning it should carry 240 symbols. Number of REs stay same regardless of system bandwidth.

5 MHz. The PBCH stays in the central 72 subcarriers of symbols 7 to 10.

The count of 240 follows from 36.211 clause 6.6.4. The PBCH uses the first four OFDM symbols of slot 1 in subframe 0, which are symbols 7 to 10 of the subframe, on the central 72 subcarriers. That gives 288 REs. The mapping always skips the RS positions of four antenna ports, whatever CellRefP is, because the UE does not know the number of ports before it decodes the PBCH. Those positions take 24 REs in symbol 7 and 24 in symbol 8, which leaves 240.

The code writes pbch_scale = 0.7 and phich_scale = 0.7, so the PBCH and the PHICH appear in the same colour in these grids. The PHICH is the part of that colour in symbol 0, as the PHICH page shows.

• Symbols 7 to 10 of subframe 0 : slot 1, the central 72 subcarriers.
• 240 REs every radio frame : 288 minus 48 RS positions for four ports.
• Same position for every bandwidth and cell : the UE can find the PBCH before it knows either.

Appendix - A : Numerical Output for Encoding Process

The appendix repeats lteBCH step by step with lower-level toolbox functions, for NFrame = 4, and checks that the result is the same. Each output below can be counted against 36.212 v19.3.0 clause 5.3.1.

enb.CyclicPrefix = 'Normal';

enb.PHICHDuration = 'Normal';

enb.Ng = 'Sixth';

enb.NDLRB = 6;

enb.CellRefP = 1;

enb.DuplexMode = 'FDD';

 

enb.NCellID = 0;

enb.NSubframe = 0;

enb.CFI = 1;

enb.NFrame = 4;

 

mib_bits = lteMIB(enb);

bch_cw = lteBCH(enb,mib_bits);

 

mib_bits_crc = lteCRCEncode(mib_bits,'16',0);

bch_Econde = lteConvolutionalEncode(mib_bits_crc);

 

bch_Econde_RM = lteRateMatchConvolutional(bch_Econde,length(bch_cw));

 

isequal(bch_cw,bch_Econde_RM)

 

enb.PHICHDuration = 'Normal'; enb.Ng = 'Sixth'; enb.NDLRB = 6; enb.CellRefP = 1;   enb.NCellID = 0; enb.NSubframe = 0; enb.CFI = 1; enb.NFrame = 4;

mib_bits' (24 bits)  0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 mib_bits_crc' (40 bits = 24 + 16)  0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0 1 0 0 bch_Econde' (120 bits = ( 24 + 16 ) x 3)    1 0 1 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 0 0 0 0 1 1 1 0 0 0 1 1 0 0 1 1 1 0 0 0  1 0 0 1 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0  0 1 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 0 1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 0   bch_Econde_RM' (1920 bits)    0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 1 0 0 0 1 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 1 1 1 1 0 1 0 0 0  0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0  0 0 1 1 1 1 1 0 0 0 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0

 

The lengths match the clause. The function lteCRCEncode appends a 16-bit CRC to the 24 MIB bits, giving 40 bits, and the mask 0 corresponds to one antenna port. The function lteConvolutionalEncode applies the rate 1/3 tail-biting convolutional code, which gives 120 bits. The function lteRateMatchConvolutional first reorders the 120 bits with a sub-block interleaver, and then repeats the reordered block to fill 1920 bits for the normal cyclic prefix. The last output therefore repeats with a period of exactly 120 bits, 16 times, but in a different order from bch_Econde.

• 24 + 16 = 40 bits : MIB plus CRC, masked by the number of antenna ports.
• 40 x 3 = 120 bits : the tail-biting convolutional code.
• 120 x 16 = 1920 bits : rate matching for the normal cyclic prefix.

Disclaimer !

This page is only to show you the overall logics and visualization for various LTE physical layer channels. I haven't investigated much about verifying about the accuracy.

If you think the code is not so efficient, it is 100% my fault. I haven't made any effort for effiecient code. I just tried to create code as simple as possible for the readers. As you know, easy-to-read code is not always efficient for a specific chipset.

If you find any mistake in terms of accuracy, it is also very highly likely be my fault. Not the problem of Matlab tool box itself.

Any comment and corrections if you find any mistake will be welcome and appreciated.

Reference

[1] 3GPP TS 36.211 v19.3.0 - clause 6.6, Physical broadcast channel

[2] 3GPP TS 36.212 v19.3.0 - clause 5.3.1, Broadcast channel

[3] 3GPP TS 36.331 v19.3.0 - MasterInformationBlock

[4] Physical Layer Channel : Downlink : PBCH