NR帧结构

1,带宽与采样率

Channel bandwidth (MHz)	Sampling rate (MHz)	Available bandwidth (MHz)	Subcarrier spacing (kHz)	Pass band ratio	Stop band ratio
5	7.68	4.5	15	0.5859	0.6510
5	7.68	3.96	30	0.5156	0.6510
10	15.36	9.36	15	0.6094	0.6510
10	15.36	8.64	30	0.5625	0.6510
15	30.72	14.22	15	0.4629	0.4883
15	30.72	13.68	30	0.4453	0.4883
15	30.72	12.96	60	0.4219	0.4883
20	30.72	19.08	15	0.6211	0.6510
20	30.72	18.72	30	0.6094	0.6510
20	30.72	17.28	60	0.5625	0.6510
25	30.72	24.3	15	0.7910	0.8138
25	30.72	23.4	30	0.7617	0.8138
25	30.72	23.04	60	0.75	0.8138
40	61.44	38.88	15	0.6328	0.6510
40	61.44	38.16	30	0.6211	0.6510
40	61.44	37.44	60	0.6094	0.6510
50	61.44	48.6	15	0.7910	0.8138
50	61.44	47.88	30	0.7793	0.8138
50	61.44	47.22	60	0.7476	0.4883
60	122.88	58.32	30	0.4629	0.4883
60	122.88	56.88	60	0.4629	0.4883
80	122.88	78.12	30	0.6357	0.6510
80	122.88	77.04	60	0.6210	0.6510
100	122.88	98.28	30	0.7998	0.8138
100	122.88	97.2	60	0.7910	0.8138
20	30.72	17.28	60	0.5625	0.6510
25	30.72	23.04	60	0.75	0.8138
40	61.44	37.44	60	0.6094	0.6510
50	61.44	47.22	60	0.7476	0.4883
60	122.88	56.88	60	0.4629	0.4883
80	122.88	77.04	60	0.6210	0.6510
100	122.88	97.2	60	0.7910	0.8138

---

2,基本时间单位

系统中定义最基本的时间单位$T_c=1/(\Delta f_{max}\ast N_f)$,其中$\Delta f_{max}=480\ast 10^3,N_f=4096$,这个个人理解应该是已NR系统中所支持的最大子载波间隔每个样点的间隔所定义。常数$\kappa =\frac{T_s}{T_c}=64$,其中$T_s=1/(\Delta f_{ref}\ast N_{ref}),\Delta f_{ref}=15\ast 10^3,N_{ref}=2048$;

u‌	$\Delta f=2^u.15[kHz]$	Cyclic prefix
0	15	Normal
1	30	Normal
2	60	Normal, Extended
3	120	Normal
4	240	Normal
5	480	Normal
6	960	Normal

---

一个帧的时长为$T_f=(\Delta f_{max}{N}_f/100)=10ms$,一个子帧时常为$T_{sf}=(\Delta f_{max}{N}_f/1000)=1ms$。

3，基本的帧结构

15KHZ 帧结构如下

30KHZ帧结构

Number of OFDM symbols per slot, slots per frame, and slots per subframe for normal cyclic prefix.

μ	$N_{symbol}^{slot}$	$N_{slot}^{frame,\mu }$	$N_{slot}^{subframe,\mu }$
0	14	10	1
1	14	20	2
2	14	40	4
3	14	80	8
4	14	160	16
5	14	320	32
6	14	640	64

采样率与FFT_SIZE 之间的关系

下面我们已$scs=30khz,fftsize=4096$举例。其采样率为$122.88M$即每1ms需要发送$122880$个采样点。根据上面得知一个$subframe$,包含两个slot所以每个slot发送$122880/2=61440$个采样点。14个符号每个符号数据包含4096个样点，符号0 CP 长度为352其余符号CP长度为288.这样：
$$14\ast 4096+352+13\ast 288=61440$$
上面我们介绍了相关 帧结构以及数据样点分布，剩下一个问题就是CP长度的计算($3GPP\mathrm{38.211.5.3.1}$)：
$$\begin{array}{rl}{N}_{CP,l}^u& =\{\begin{array}{ll}512\kappa \cdot 2^{-u},& extendedcyclicprefix\\ 144\kappa \cdot 2^{-u}+16\kappa ,& normalcyclicprefix,l=0orl=7\ast 2^u\\ 144\kappa \cdot 2^{-u},& normalcyclicprefix,l\ne 0andl\ne 7\cdot 2^u\end{array}\end{array}$$

上式子中$\kappa =\frac{T_s}{T_c}=64$,
假设$scs=30khz,即u=1，4096点fft$由上式可得
$$\begin{array}{rl}{N}_{CP,sym0}^{30khz}& =144\ast 64\ast 2^{-1}+16\ast 64=5632T_c\\ N_{CP,symother}^{30khz}& =144\ast 64\cdot 2^{-1}=4608T_c\end{array}$$
一个样点时间单位为
$$\begin{array}{rl}{T}_{sym0,samplingtime}& =\frac{1}{30\cdot 10^3\ast 4096}s=16T_c\\ T_{sym0,sampling,point}& =\frac{N_{CP,sym0}^{30khz}}{T_{sym0,samplingtime}}=\frac{5632T_c}{16Tc}=352\\ T_{symother,sampling,point}& =\frac{N_{CP,symother}^{30khz}}{T_{symother,samplingtime}}=\frac{4608T_c}{16Tc}=288\end{array}$$

【1】 3GPP 38.211
【2】 3GPP 38.101.5.3