36.213 随机接入过程中的疑问
时间:04-10
整理:3721RD
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在36.213 随机接入流程中
6.1.1 Timing
For the L1 random access procedure, UE’s uplink transmission timing after a random access preamble transmission is as follows.
If a PDCCH with associated RA-RNTI is detected in subframe n, and the corresponding DL-SCH transport block contains a response to the transmitted preamble sequence, the UE shall, according to the information in the response, transmit an UL-SCH transport block in the first subframe , , if the UL delay field in section 6.2 is set to zero. The UE shall postpone the PUSCH transmission to the next available UL subframe if the field is set to 1.
问题是: 在什么场景下MSG3在UE恻需要延迟发送(UL delay == 1)?下一个有效发送的时刻怎么计算?
6.1.1 Timing
For the L1 random access procedure, UE’s uplink transmission timing after a random access preamble transmission is as follows.
If a PDCCH with associated RA-RNTI is detected in subframe n, and the corresponding DL-SCH transport block contains a response to the transmitted preamble sequence, the UE shall, according to the information in the response, transmit an UL-SCH transport block in the first subframe , , if the UL delay field in section 6.2 is set to zero. The UE shall postpone the PUSCH transmission to the next available UL subframe if the field is set to 1.
问题是: 在什么场景下MSG3在UE恻需要延迟发送(UL delay == 1)?下一个有效发送的时刻怎么计算?
下一个有效发送时刻的计算如下:
如果UE在子帧n成功地接收了自己的RAR,则UE应该在n + k1(其中 k1 ≥ 6)开始的第一个可用上行子帧(对于FDD而言,就是n + 6;对于TDD而言,n + 6可能不是上行子帧,所以 k1 可能≥ 6)发送msg3。RAR所带的UL grant中包含一个1 bit的字段UL delay,如果该值为0,则n + k1 为第一个可用于msg3的上行子帧;如果该值为1,则UE会在n + k1 之后的第一个可用上行子帧来发送msg3(即中间跳过一个上行子帧)。
至于什么时候使用UL delay,个人认为跟负载相关,如果在某个子帧的负载过大,就让其中的某些UE晚点发msg3.
“n + k1 之后的第一个可用上行子帧来发送msg3(即中间跳过一个上行子帧)。”
对于FDD 而言, 下一个可用的上行子帧就是n+7,对吧:victory:
多谢!