AFDM_tutorial

Date: 2024.10.25 17:16
Author: Joffrey LC

AFDM: A Full Diversity Next Generation Waveform for High Mobility Communications. Ali Bemani et.al. 2021 IEEE International Conference on Communications Workshops (ICC Workshops), June 2021 (pdf) (Citations 26)

The corresponding video can be found in Bilibili.

Quick Overview

AFDM: Affline Frequency Division Multiplexing

chirp-based multicarrier transceiver
based on discrete affline Fourier transform (DAFT)
Adjust the ==two parameters== to better cope with the doubly dispersive channels. And avoid the time domain channel paths with distinct delays or Doppler frequency shifts overlap in DAFT domain.
Achieved full diversity of linear time-varying (LTV) channel.

OFDM/single-carrier FDMA (SC-FDMA) achieved even optimal performance in ==time-invariant== frequency selective channels.

AFDM for communication

OFDM is designed for LTI channel, and we can use the orthogonal to modulate/demodulate

general AFT: 调整abcd以适用于各种变换

discrete form：2 parameters and chirp periodic prefix (CPP) to avoid ISI like CP in OFDM

相比于传统的OFDM 仅 IFFT 多了左右两个matrix, 主要特性来自于C1 matrix（frequency变化斜率）

CP-CPP

OFDM子载波是固定频率
AFDM子载波是chrip信号线性调频

信道模型：双色散信道， multiple paths, each of which has its own delay and Doppler shift

大频率间隔或者大持续时间下，这两个现象可以被忽略（观察分母）

==时延和多普勒被分为整数部分和分数部分？？==

分数部分的delay处理起来比较困难，而且如上图所示有个取模的现象（增大到边界后下降到0再增加），这点很多文章都没注意。

只有整数时，如左侧图所示

而分数情况如右侧所示

有了分数的Doppler shift之后，他们可能会重叠，如左侧图所示；或者如右图所示不重叠

==峰值的位置取决于C1==

想要让整个delay-Doppler 表示能够有区分，需要满足以下公式：

k_max denotes the maximal Doppler frequency shift, k_v is driven from the fractional Doppler shift.

AFDM with Zak transform

diversity

diversity, rank=P (number of channel paths)

如果不满足2*k_max+l_max+2*k_max*l_max的条件，会由于前面聊到的取模操作而产生重叠。

Only one pilot for AFDM channel estimation

P_GI depends on the max Doppler frequency shift and delay
可以把zeros位的功率全部给pilot，这样使得一个frame的功率不会发生变化
for the channel estimation, we should only focus on the blue parts.

整个公式都是聚焦于blue part， only one pilot symbol.

同时，SNR也被称为overhead，因为如果我们想增大SNR，就意味着要给pilot增加更大的功率，就需要更多的guard（zeros）

Number of guard: AFDM需要更==少==的Guards，几乎是OTFS的一半

粉色和紫色是OTFS和AFDM的==共同的/重叠的==guard位
所有颜色的和是OTFS的guards
红色的等于是OTFS的额外开销（减小了channel capacity）

如果有稀疏性，C1的值就不需要取太大

Situation a: We have a distribution of delay, and the Doppler shift is random. 多普勒random但是在不同delay中几乎都一样？？，
Situation b: Everything is random.
Situation c: Delay is random and we have the distribution of Doppler shift.每个谱图delay的多普勒几乎都一样？？？？