LTE PUCCH 1 资源计算

 

公式：  n_pucch_(1) (ref. 213)

 

假设 DL RB = UL RB = 50, PHICH 资源 1/6

 

Number of CCE depends on CFI。To calculate the CCE resources,

 

1) Cal. no. of REG in a subframe using CFI.

 

2) Cal. no. of REG to be used by PHICH group

 

TDD, the number of PHICH groups may vary between downlink subframes and is given by  m_i * Phich_group(213)

Uplink-downlink configuration	Subframe number
	0	1	2	3	4	5	6	7	8	9
0	2	1	-	-	-	2	1	-	-	-
1	0	1	-	-	1	0	1	-	-	1
2	0	0	-	1	0	0	0	-	1	0
3	1	0	-	-	-	0	0	0	1	1
4	0	0	-	-	0	0	0	0	1	1
5	0	0	-	0	0	0	0	0	1	0
6	1	1	-	-	-	1	1	-	-	1

 

Note: m = 2 for TDD0 SF 0 & 5, as two UL SF could be ACK/NACKed in these two SFs.

 

1  N_Phich_group = 3 REG (HI in PHICH is coded with 3 bits (212 5.3.5), each bit is spreaded to 4 (211 6.9.1); as BPSK in use, 12 RE needed to send 12 bits, i.e. 3 REGs)；if Ng = 1/6,  N_Phich_group = 2 (normal CF, non Cfg0, Sf 0)

so, the CCE used by Phich = 6 REG (normal CF, non Cfg0, Sf 0)

 

3) No. of CCE = floor( (Total no. of REG - REG for PHICH - REG for CFI) / 9)

According to 212. 5.3.4.1 CFI = 32bits, QPSK in use; Thus 1 RE carries 2 bits, and 16 RE needed to carry CFIx; As 1 REG = 4 RE, thus CFI needs 4 REG

 

The number of CCE is calculated by following (1 antenna, TDD0 Ng = 1/6, sf = 1),

 

CFI = 1.

2 REG per RB, thus 50 RB = 100 REG, 1 CCE request 9 REG

Thus, DL CCE at most (100 – 6 – 4)/9 = 10

 

CFI = 2,

5 REG per RB, thus 50 RB = 250 REG, 1 CCE request 9 REG

Thus, DL CCE at most (250 – 6 – 4)/9 = 26

 

 

CFI = 3,

8 REG per RB, thus 50 RB = 400 REG; 1 CCE request 9 REG

thus DL CCE at most (400 – 6 – 4)/9 = 43

 

According to the definition

 = {0, 11, 27, 44}

CFI (1, 2, 3) =>  (42, 106, 174)

 

当 Delta_shift_pucch=1， 1RB容纳36个PUCCH 1资源； Delta_shift_pucch=2, 18个PUCCH 1; Delta_shift_pucch=3, 12个PUCCH

 

CFI (1, 2, 3) =>  for  (2, 3, 5)

                        =>  for  (3, 6, 10)

                        =>  for  (4, 9, 15)

 

 

注：下行cce资源的计算，是按照subframe 的顺序进行，但是Downlink association set index 则如下图

 

Table 10.1-1: Downlink association set index :  for TDD (36.213)

UL-DL Configuration	Subframe n
	0	1	2	3	4	5	6	7	8	9
0	-	-	6	-	4	-	-	6	-	4
1	-	-	7, 6	4	-	-	-	7, 6	4	-
2	-	-	8, 7, 4, 6	-	-	-	-	8, 7, 4, 6	-	-
3	-	-	7, 6, 11	6, 5	5, 4	-	-	-	-	-
4	-	-	12, 8, 7, 11	6, 5, 4, 7	-	-	-	-	-	-
5	-	-	13, 12, 9, 8, 7, 5, 4, 11, 6	-	-	-	-	-	-	-
6	-	-	7	7	5	-	-	7	7	-

 

所以计算PUCCH 1 资源的时候需要将subframe 对应的n_cce进行调整