公式： n_pucch_(1) (ref. 213)
假設 DL RB = UL RB = 50, PHICH 資源 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 |
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)
當 Delta_shift_pucch=1， 1RB容納36個PUCCH 1資源； Delta_shift_pucch=2, 18個PUCCH 1; Delta_shift_pucch=3, 12個PUCCH
CFI (1, 2, 3) => for (2, 3, 5)
=> for (3, 6, 10)
=> for (4, 9, 15)
注：下行cce資源的計算，是按照subframe 的順序進行，但是Downlink association set index 則如下圖
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 |
- |
所以計算PUCCH 1 資源的時候需要將subframe 對應的n_cce進行調整