In LTE, the PDCCH channel conveys all allocation for downlink and the uplink in the subframes. To help manage the overhead in conveying the allocation message, a type field is also included in the PDCCH message. In this blog, we shall look at Type 1 resource allocation and the motivation behind the use of this mode. We shall present a few concrete examples to bring out the many facets of this allocation mode.
Type field in PDCCH:
In LTE, the type field in a PDCCH message can take on one of three values; 0,1, or 2, respectively. Type 0 groups the Resource Blocks (RBs) into groups called as Resource Block Groups (RBGs) whose size can vary with the total number of RBs available. All allocations are in terms of RBGs and is usually conveyed using a bitmap with the number of bits being equal to the number of RBGs. For example, in the 20 MHz case, the number of RBs is 100, the number of RBGs is 25 and thus 25 bits are used to convey the allocation in a PDCCH message. Clearly, the advantage of this method is the reduction in the number of bits used in the bitmap as compared to the case when a bit per RB is used. Typically, consecutive RBGs are assigned for a user and the granularity of the allocation is limited to the size of the RBG.
Type 1 allocation was proposed to address the issues of granularity and frequency diversity which were lacking in the type 0 allocations. In Type 1, the same RBG idea is followed as in Type 0, however, individual RBs inside the RBGs need not be grouped together and assigned to a single user. In addition to the bitmap, 2 extra bits are used to convey the allocation. These extra bits point to the subsets within the RBGs and a bit shift indicator is also used to choose different RBs from the RBGs into a single allocation. Thus, Type 1 allocation can help achieve frequency diversity with the same size of bitmap as in the Type 0 allocation. We shall explain the workings in Type 1 allocation using 2 examples; one for the 5MHz bandwidth and the other for the 10 MHz bandwidth.-
In Type 1 allocation Individual physical RBs (PRBs) can be allocated but this should be within a RBG subset. The resource allocation header using Type 1 contains following fields.
Type 1 Resource Allocation Example
Scenario 1 : Bandwidth = 5 MHz
The size of RBG that is applicable has been obtained from the table below which is given in 3GPP TS 36.213 section 184.108.40.206 table 220.127.116.11-1
Total # of RBGs and RBG Subsets in 5 MHz BW
Number of bits required for ‘Resource allocation header’:
How to allocate RBGs in RBG set ?
A RBG subset k, 0 <=k <P ,consists of every Pth RBG starting from RBG k.(Ex. If P=2, every 2nd RBG in RBG subset 0 with starting RBG 0)
# of PRBs in a selected RBG subset is function of (k,P and NRBDL ) (Ref :3GPP 36.213 section 18.104.22.168)
#. of PRBs in subset k = NRBRBGsubset (k) =
In 5MHz BW, NRBDL = 25 and P = 2 therefore M = floor ((NRBDL-1)/P)mod P = 0
Significance of shift in the bit map
The addressable PRB numbers of a selected RBG subset start from an offset Δshift (k), to the smallest PRB number within the selected RBG subset, which is mapped to the MSB of the bitmap. The offset is in terms of the number of PRBs and is done within the selected RBG subset. The shift is used to utilize the unused PRBs in the subset
If shift bit = 0 then offset for subset k is Δshift (k) = 0
If shift bit =1, then the offset for subset k is Δshift (k) =NRBRBGsubset (k) – C
For the above example (BW-5 MHz), C= 11, NRBRBGsubset(k) =13 for subset0 and 12 for subset 1
PRB numbers in each subset
PRB numbers in a subset 0 (k =0) and subset 1(k=1) is given by
nPRBsubset = ( floor((i + Δshift (k)) /P))*P2 + (k*P) + (i + Δshift (k))mod P
where i = 0,1,2,3,4,5,6,7,8,9,10
Case 1 : for shift bit=0
Δshift (k) = 0 , so PRB numbers in subset 0 (k=0)and subset 1(k=1) is
The above table lists the 11 PRBs usable in each subset for this bit shift. The table below shows the RBGs used in obtaining the PRBs for each subset.
Case 2 : for shift bit=1
Δshift (k) = 2 for k =0 and Δshift (k) = 1 for k =1
PRB numbers in subset 0 (k=0)and subset 1(k=1) is
The above table lists the various PRBs that can be used in the 2 subsets for this bit-shift. It has differences from the earlier table and together they are able to utilize the complete set of 25 PRBs possible for the 5 MHz case. The table below also highlights the difference from the earlier bit-shift as we notice a clear difference in the RBGs used in this case as compared to the earlier case.
Thus it can be seen that type1 offers 2 features which is different from type 0. Consider the entries in the 2 tables which list the PRBs in a subset and the corresponding 11 bits in the bit-map. In the case of type 0, 2 consecutive PRBs will be grouped together for the 5 MHz case. However for type1 it can be seen that for the 4 different cases, the 11 PRBs in the 2 columns of the 2 tables are spread out over the frequency band. Thus, freedom in choosing the number of PRBs at a granularity of 1 PRB is available along with the frequency diversity that can be obtained if a larger number of PRBs are planned to be chosen. The frequency diversity advantage becomes more enhanced when wider bandwidth examples (10/20 MHz, for instance) is considered as each of the subsets now span a much larger bandwidth.
Type 1 Resource Allocation Procedure
Scenario 1: Bandwidth = 10 MHz
* Ref: 3GPP TS 36.213 section 22.214.171.124 table 126.96.36.199-1
Total # of RBGs and RBG Subsets in 10 Mhz BW
How to allocate RBGs in RBG set?
A RBG subset k, 0 <=k <P , consists of every Pth RBG starting from RBG k.(Ex. If P=3, every 3rd RBG in RBG subset 0 with starting RBG 0)
# of PRBs in a selected RBG subset is function of (k,P and NRBDL )
# of PRBs in RBG subset = NRBRBGsubset (k) =
In 10MHz BW, , NRBDL = 50 and P = 3 therefore M = floor((NRBDL-1)/P)mod P = 1
If shift bit =1 then offset for subset k is Δshift (k) =NRBRBGsubset (k) – C
PRB numbers in a subset 0 (k =0), subset 1(k=1) and subset 2(k=2) is given by
nPRBsubset =( floor((i + Δshift (k)) /P))*P2 + (k*P) + (i + Δshift (k))mod P
where i = 0,1,2,3,4,5,6,7,8,9,10,11,12,13
Δshift (k) = 4 for k =0 , Δshift (k) = 3 for k =1 and Δshift (k) = 1
PRB numbers in subset 0 (k=0), subset 1(k=1) and subset 2 are
Type 1 allocation in LTE offers flexibility in choosing the PRBs at a granularity of a single PRB. In addition, using the same number of bits for resource allocation as in Type 0 considerable frequency diversity can be obtained in the allocation. Thus, type 1 can resource allocation can be useful for certain small allocations which need to leverage frequency diversity.
Nice post Prakash... very detailed one.
Do u have similar information for RAT2 (mainly RAT2 distributed)?
Respected Mr. Prakesh,
Very Nice Post Sir, Sir can you please share the detailed view on Random Access Channel (RACH or PRACH).
I am doing my graduation and as a part of my academic project i have taken Implementation and Optimization of RACH. So is there any research work done by you in this UPLINK part. If yes, Please help me out in explaining me sir.
Thanking You in Advance,