LTE PUCCH 1 资源计算
2011-11-24 09:50
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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)
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)
所以计算PUCCH 1 资源的时候需要将subframe 对应的n_cce进行调整
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 |
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 | - |
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