LTE downlink Physical layer transport mechanism (6)-Downlink resource allocation method (Resource Allocation Type)

The resource allocation of downlink RB (Resource Allocation) has three ways, namely resource allocation mode 0, resource allocation Method 1 and resource allocation Method 2. In the previous post, "LTE downlink Physical layer transport mechanism (5)-DCI format selection and dci1a" mentioned in Dci1a, referring to DCI1A can only allocate continuous RB, and this way RIV(Resource Indication Value ), then this allocation is actually a resource allocation method of 2. The DCI2 and DCI2A formats use another 2 different allocation methods, namely, resource allocation mode 0 and resource allocation Method 1. Therefore, before talking about DCI2 and dci2a, it is necessary to introduce these 2 kinds of resource allocation methods first.

1. Resource allocation Method 0

DCI with resource allocation Method 0:DCI1,DCI2,dci2a,dci2b. Each DCI has a resource allocation field that is used to indicate which RB is assigned to the UE.

In RB resource allocation Mode 0 (Resource Allocation Type 0), all RB resources are composed of different RBG(RB Group), so the allocation method 0 is based on the RBG as the basic unit allocation . With this allocation, the Resource Allocation field in the DCI uses a bitmap table to allocate the RB resource, and each bit of the bitmap table represents a RBG. Each RBG corresponds to a P - continuous virtual RB (i.e., VRB, where the VRB and the physical RB are one by one corresponding, hereinafter referred to as RB). Each RBG specifically corresponds to how many RB is associated with downlink bandwidth, as shown in the table below.

For example, if the downlink is 20MHz bandwidth (n_dl_rb=100), then if you allocate RB resources according to resource allocation mode 0, each RBG will consist of 4 RB (P=4). If the bandwidth is 1.4MHz (n_dl_rb=6), then each RBG will consist of only 1 RB.

different bandwidth, resource allocation mode 0 can use the number of RBG is also fixed. If you use the variable N_RBG to represent this value, n_rbg = ( n_dl_rb  /  P ) rounding up . N_RBG Bit-length bitmap resource Allocation table, this bitmap allocation table is encoded into the DCI Stream, the UE from this allocation table can be pushed to export the Pdsch used by the RB resource .

For example, the downlink bandwidth of 3MHz, its RB number n_dl_rb=15,P= 2, so n_rbg=(n_dl_rb/p) rounding up = (15/2) rounding up = 8. However, it is important to note that if it is (n_dl_rb mod P)! = 0 bandwidth, its last RBG size is not the same as other RBG sizes: The Last RBG contains the number of RB =N_dl_rb-p * (n_dl_rb/p), not the last RBG of the other RBG, each RBG contains the number of RB =P . For example, 3MHz downlink bandwidth, in addition to the last RBG, the other (n_dl_rb/p=15/2=) 7 RBG, each RBG has 2 RB, at this time the 7 RBG a total of 7*2=14 a RB, less than the entire bandwidth of 15 RB, So the last RBG contains the number of RB =n_dl_rb-p *(n_dl_rb/p) = 15-2 * (15/2) = 1. As shown in.

In addition, since it is a bitmap table, there are high and low level problems. The agreement also specifies that therbg_0 corresponds to the bitmap's high position, the MSB, and Rbg_ (n_rbg-1) corresponds to the low-level, or LSB, of the bitmap table. If the UE decodes to a bit bit = 1, the corresponding RBG is assigned to the UE. For example, 3M bandwidth, the bitmap table occupies the bit number n_rbg= (15/2) rounded up =8,bitmap stream = Binary bin (00001011), then the UE used by the RBG resource group is RBG4, RBG6 and RBG7 , so the Rb-id used were: Rb8-rb9,rb12-rb14. So, allocation mode 0 can allocate discrete RB resources, but the greater the bandwidth, the distribution of the RB grain size P is thicker.

2. Resource Allocation Method 1

The DCI with resource allocation 1 is:DCI1,DCI2,dci2a,dci2b, same as resource allocation mode 0. Each DCI has a resource allocation field that is used to indicate which RB is assigned to the UE. Just as with allocation mode 0, the Resource Allocation field for resource Allocation 1 has three domains, not just one bitmap domain, as specified below.

In resource allocation Mode 1, all RB resources comprise different RBG (this is the same as resource allocation mode 0), and different RBG make up a different subset of RBG (RBG subset), so allocation mode 1 is based on RBG subset allocation . All RBG are divided into P-rbg subsets, or the RBG subset itself is a collection, each of which consists of several RBG, and when ENB assigns a resource to a UE, it selects a subset of RBG and then sets the corresponding RB in that sub-set to 1. In summary, each bandwidth contains a subset of P RBG, each RBG subset contains several RBG, each RBG also contains p consecutive RB (the last RBG contains the same number of RB as the allocation method 0), The composition is shown below.

Different bandwidths, the number of RBG subsets it contains, and the number of RB contained in each RBG, are equal to the P -values in table 7.1.6.1-1 above. The signal is a bandwidth of 10MHz when the composition of each RBG subset, at this time n_dl_rb=50,P= 3, so there are 3 rbg subsets, each RBG contains 3 A continuous RB.

As you can see from the above, if you use allocation Mode 1, then each DCI must carry the currently selected subset index and the bitmap used by that subset. However, it is important to note that each bit in the bitmap table represents the corresponding RB-id, not the RBG, which is different from the allocation mode 0 .

Also, take bandwidth 10MHz as an example to explain how the RB allocation information in DCI is understood. As shown, it can be seen that the DCI uses a subset of RBG to indicate resource allocations, whereas in a specific bitmap table, Bit2, bit3, bit5 are equal to 1, and the 3rd RB, 4th RB, and 6th RB for subset 1 are assigned to the UE. It is also possible to see that each UE uses the RB allocation indicator information is given in the form of Rb-bitmap, this rb-bitmap (rb0/rb1/rb2/... ) is an index that belongs to the same RBG subset, the rb0/rb1/rb2/of a subset of different RBG ... The Rb-id of the reference is different.

It can also be seen that not all RBG subsets contain the same number of RBG and RB , for example, subset 0 contains the number of RBG is 6, a total of 18 RB, and although subset 1 contains 6 of the RBG, but only 17 RB While subset 2 contains only 5 RBG, the number of RB is only 15.

The above describes a practical use of resource allocation Mode 1, where there are several places not mentioned: the meaning of the shift field in the DCI, the number of bits that are allocated in the bitmap table, and so on. The following is a detailed description of the resource allocation Method 1 in the DCI, how its resources allocated content to fill in.

The protocol stipulates that the resource Allocation field of resource Allocation 1 takes the number of bits (n_dl_rb/p) up , including 3 fields, which are 3 fields:

(1)RBG subset indicates the domain . This field indicates which subset of the current resource allocation is selected in the P -RBG Sub-set, the number of bit bits used by the field n = log2 (P) rounding up . such as 10MHz bandwidth, p=3, then n=log2 (P) rounding up =log2 (3) Rounding up = (1.58) Rounding up = 2, which is the first 2 bits of the RB Resource Allocation field in the DCI stream, is used to indicate which RBG subset is currently in use. SUBSET=01, which means that a subset of the DCI allocations is subset 1.

(2)Shift resource Offset field . The field is fixed to occupy 1 bit, if the value of the field = 0, the selected RBG subset of RB is not offset, the value =1 indicates that the selected RBG subset of RB needs to perform an offset, the protocol uses variable delta_shift (p) to represent the the RB offset within the P subset, which is offset from the RB low (that is, the RB0 end instead of the RB99 end), or from the MSB bit of the bitmap. Specific how to offset see below.

(3)bitmap table field . the number of bit bits occupied by this field n_type1_rb = (n_dl_rb/p) rounding up-log2 (P) rounding up-1, each bit value represents a RB in the current RBG sub-set, The MSB bit of the bitmap table corresponds to a low frequency rb-id, while the LSB bit corresponds to a rb-id with a high RB frequency (this rule is the same as the allocation mode 0).

The above describes the use of resource allocation 1 o'clock the resource Allocation field in the DCI contains the contents of each field, you can see that the shift field determines the understanding of the bitmap table field, the following 10M bandwidth (50 RB) as an example to explain the specific.

(A) If each subset is not offset by a shift=0 of three subsets, the offset of each subset Delta_shift (p)=delta_shift (0)=Delta_shift (1) =Delta_shift (2)= 0, then:

RBG subset 0 consists of RBG0 (RB0/RB1/RB2), RBG3 (RB3/4/5), RBG6 (RB6/7/8), RBG9 (RB9/10/11), RBG12 (RB12/13/14), RBG15 (RB15/16/17 ) composition.

RBG subset 1 consists of RBG1 (RB0/RB1/RB2), RBG4 (RB3/4/5), RBG7 (RB6/7/8), RBG10 (RB9/10/11), RBG13 (RB12/13/14), RBG16 (rb15/16 Composition

RBG subset 2 consists of RBG2 (RB0/RB1/RB2), RBG5 (RB3/4/5), RBG8 (RB6/7/8), RBG11 (RB9/10/11), RBG14 (RB12/13/14).

However, it should be noted that the number of RB contained in each subset is actually different (subset 0 has 18 RB, subset 1 has 17 RB, subset 2 has 15 RB), and according to the calculation formula of DCI's Bitmap table field, each subset can contain only the number of bits N_TYPE1_RB = (50/3) rounding Up-log2 (3) rounding up-1 = 17-2-1 = 14, that is, each time the dispatch, the ENB can only allocate 14 RB to the UE, so under shift=0 conditions, in fact, only the first 14 RB per subset can be used (from RB G0 or RB low-frequency end), and RB high-frequency end there are 4 RB can not be allocated to use, as shown in.

Because there is a limit on the number of RB per schedule, it is not appropriate to use resource allocation Mode 1 in some cases, such as when testing throughput.

(B) If subset 0 uses an offset of shift=1, it means that the RB that makes up RBG subset 0 is offset adjusted, offset delta_shift (p)= N_RBGSUBSET_RB (p)-N_ TYPE1_RB= N_RBGSUBSET_RB (0)-N_TYPE1_RB. wherein, the parameter N_RBGSUBSET_RB (p) represents the number of RB in the RBG subset P (if it is a subset of 0, note that the value is equal to 18 instead of equal to), N_TYPE1_RB is the bitmap domain bit length, such as schematic.

As you can see from the formula above, the offset delta_shift (p) = n_rbgsubset_rb (p)-N_TYPE1_RB Also represents the number of RB that cannot be allocated in each subset , such as 4 RB in subset 0 cannot be used, 3 RB in subset 1 is not available, and 2 RB in subset 1 is not available. So how to decide which RB can not be used, that is, the real purpose of the shift domain: if shift=0, it means that the RBG high-frequency end of a few RB is not used, and if shift=1, it means that RBG low band of a few RBG can not be used .

Still take 10MHz bandwidth for example, at this time n_dl_rb=50,p=3. So shift=1, judging condition [(n_dl_rb-1)/P down rounding] mod P = [(49/3) rounding down] MoD 3 = mod 3 =1. So:

For subset 0, p=0<1, so RB offset delta_shift (0)= N_RBGSUBSET_RB (0)-N_TYPE1_RB = [(50-1)/9] rounded down * 3 + 3-(17-2-1 ) = 5 * 3 + 3-14 =4.

For subset 1, p=1=1, so RB offset delta_shift (1)= N_RBGSUBSET_RB (1)-N_TYPE1_RB = [49/9] Rounding down * 3 + (mod 3) + 1-(17-2- 1) = + 2-14 =3.

For subset 2, p=2>1, so RB offset delta_shift (2)= N_RBGSUBSET_RB (2)-N_TYPE1_RB = [(50-1)/9] rounded down * 3-(17-2-1) = 5 * 3-14 =1.

The offset of three subsets is calculated according to the formula Delta_shift (p) is 4, 3, 1, which is consistent with the number of RB numbers that cannot be assigned in each subset described in the previous article. The actual allocation is as shown in this case.

As can be seen from the figure, the subset 0 contains RB starting from RB10 until RB47, a total of 14 RB, in the DCI bitmap domain, the first RB corresponds to RB10, and so on. The protocol also gives a sequence of each RB in the case of Shift=1, as shown in each sub-set.

In the case of 10MHz bandwidth, the P=3,n_type1_rb=14,i range is 0~13, so:

For a subset of 0,delta_shift (0) =4,i=0 (first RB position), N_RBGSUBSET_VRB (0) = (0+4)/3 * 9 + 0*3 + (0+4) mod3 = 9 + 0 + 1 =; i=13 (last A RB position), N_RBGSUBSET_VRB (0) =(13+4)/3 * 9 + 0*3 + (13+4) mod3 = 45+0+2=.

for a subset of 1,delta_shift (1) =3,i=0 (first RB position), N_RBGSUBSET_VRB (1) = (0+3)/3 * 9 + 1*3 + (0+3) mod3 = 9 + 3 + 0 = ; i= (last RB position) , N_RBGSUBSET_VRB (1) =(13+3)/3 * 9 + 1*3 + (13+3) mod3 = 45+3+1=.

For a subset of 2,delta_shift (2) =1,i=0 (first RB position), N_RBGSUBSET_VRB (2) = (0+1)/3 * 9 + 2*3 + (0+1) mod3 = 0 + 6 + 1 = 7; i=13 (last RB position) , N_RBGSUBSET_VRB (2) = (13+1)/3 * 9 + 2*3 + (13+1) mod3 = 36+6+2=.

The results are calculated according to the formula and are consistent with the above displacement results.

In summary, the allocation of resources allocation of 1 of the granularity is RB, the relative allocation of 0 of the RBG distribution granularity, more flexible and free, but because only one subset of the RB resources can be allocated at a time, so the allocation of the maximum number of RB can be limited by 1 o'clock, the maximum flow will be limited. if you want to maximize the downlink traffic of the LTE system, ENB can not adopt resource allocation Mode 1, but need to adopt the dual-stream mode of resource allocation mode 0 .

3. Resource Allocation Method 2

DCI with resource allocation 0 has:dci1a,dci1b, Dci1c,dci1d, and no overlap with resource allocation methods 0 and 1. This approach has been mentioned in the blog "LTE downlink physical layer transport mechanism (5)-DCI format selection and dci1a", which is no longer described here.

Reference documents:

(1) 3GPP TS 36.212 V9.4.0 (2011-09) multiplexing and channel coding

(2) 3GPP TS 36.213 V9.3.0 (2010-09) Physical layer procedures

(3) 3GPP TS 36.321 V9.6.0 (2012-03) Medium Access Control (MAC) protocol specification

(4) http://www.sharetechnote.com/

LTE downlink Physical layer transport mechanism (6)-Downlink resource allocation method (Resource Allocation Type)