WO2011110129A2 - Channel estimation method and apparatus - Google Patents

Channel estimation method and apparatus Download PDF

Info

Publication number
WO2011110129A2
WO2011110129A2 PCT/CN2011/073249 CN2011073249W WO2011110129A2 WO 2011110129 A2 WO2011110129 A2 WO 2011110129A2 CN 2011073249 W CN2011073249 W CN 2011073249W WO 2011110129 A2 WO2011110129 A2 WO 2011110129A2
Authority
WO
WIPO (PCT)
Prior art keywords
matrix
inverse
channel estimation
training sequence
sub
Prior art date
Application number
PCT/CN2011/073249
Other languages
French (fr)
Chinese (zh)
Other versions
WO2011110129A3 (en
Inventor
綦睿
袁佳杰
Original Assignee
华为技术有限公司
Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)
Filing date
Publication date
Application filed by 华为技术有限公司 filed Critical 华为技术有限公司
Priority to PCT/CN2011/073249 priority Critical patent/WO2011110129A2/en
Priority to CN201180000278.3A priority patent/CN102246478B/en
Publication of WO2011110129A2 publication Critical patent/WO2011110129A2/en
Publication of WO2011110129A3 publication Critical patent/WO2011110129A3/en

Links

Classifications

    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04LTRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION
    • H04L25/00Baseband systems
    • H04L25/02Details ; arrangements for supplying electrical power along data transmission lines
    • H04L25/0202Channel estimation
    • H04L25/024Channel estimation channel estimation algorithms
    • H04L25/0242Channel estimation channel estimation algorithms using matrix methods
    • H04L25/0244Channel estimation channel estimation algorithms using matrix methods with inversion
    • HELECTRICITY
    • H04ELECTRIC COMMUNICATION TECHNIQUE
    • H04LTRANSMISSION OF DIGITAL INFORMATION, e.g. TELEGRAPHIC COMMUNICATION
    • H04L25/00Baseband systems
    • H04L25/02Details ; arrangements for supplying electrical power along data transmission lines
    • H04L25/0202Channel estimation
    • H04L25/0224Channel estimation using sounding signals
    • H04L25/0228Channel estimation using sounding signals with direct estimation from sounding signals

Abstract

The present invention relates to the field of communication, and discloses a channel estimation method and apparatus. The present invention resolves the problem that amount of data stored in a chip is very large in the prior art. The apparatus includes: a matrix storage unit (101) for storing intermediate inverse matrices; a matrix acquirement unit (102) for acquiring inverse submatrices corresponding to numbers of training sequences of two users from the intermediate inverse matrices stored in the matrix storage unit; a logic calculation unit (103) for implementing multiplication and addition logical operation to the inverse submatrices acquired by the matrix acquirement unit and matrices generated from the training sequences corresponding to the numbers of the training sequences of the two users, and acquiring channel estimation factors; a channel estimation unit (104) for completing channel estimation process according to the channel estimation factors calculated by the logic calculation unit.

Description

信道估计方法和装置 技术领域  Channel estimation method and device
本发明涉及通信领域, 尤其涉及一种信道估计方法和装置。  The present invention relates to the field of communications, and in particular, to a channel estimation method and apparatus.
背景技术 Background technique
随 着 全球移 动 通信 系 统 ( Global System for Mobile Communication, GSM) 用户数量的增加, 语音业务的需求也随之增加, 给现有网络带来了巨大的压力。 为了緩解现有网络的压力,提出了语音 业务多用户正交子信道 ( Voice Service Over Adaptive Mul t i-user Orthogonal Sub Channels, VAMOS) 技术, 它将两个用户复用到同一个 时隙上, 提高了每个基站收发台 (Base Transceiver Station, BTS ) 的语音容量,从而能够更充分的利用现有的网络硬件资源,提高了频谱 资源的利用率。  As the number of Global System for Mobile Communication (GSM) subscribers increases, so does the demand for voice services, putting tremendous pressure on existing networks. In order to alleviate the pressure on the existing network, Voice Service Over Adaptive Mul t i-user Orthogonal Sub Channels (VAMOS) technology is proposed, which multiplexes two users into the same time slot. The voice capacity of each base transceiver station (BTS) is improved, so that the existing network hardware resources can be more fully utilized, and the utilization of spectrum resources is improved.
VAM0S技术中将两个用户复用在同一个时隙上,是通过将新的语音 信道加入到原有语音信道中实现的, 即如果第一个语音信道被占用,一 个与第一个语音信道正交的子信道将会被用来当做另一个语音信道,从 而达到加倍语音容量的目的。 为了最大限度的提高每个 BTS 的语音容 量,新的语音信道的引入有如下原则: 新的语音信道与原有语音信道的 互相关性最低。 现有技术中, 使用 VAM0S 两个用户联合信道估计算法估计新的语 音信道与原有语音信道的相关性, 其具体过程包括: 首先通过可以代表 新的语音信道与原有语音信道的两个用户的训练序列号获取信道估计 因子; 然后将该信道估计因子用于 VAM0S两用户联合信道检测,得到信 道检测结果;最后将该结果用于 VAM0S强弱用户联合信道检测以及后续 其他检测。其中, 通过可以代表新的语音信道与原有语音信道的两个用 户的训练序列号获取信道估计因子的过程包括:从预先存储的信道估计 因子中, 获取两个用户的训练序列号相应的信道估计因子, 该预先存储 的信道估计因子为: y = (ATA)— τ, 其中 Α为 44*20 的矩阵, 具有如下 In the VAM0S technology, two users are multiplexed on the same time slot by adding a new voice channel to the original voice channel, that is, if the first voice channel is occupied, one and the first voice channel are used. The orthogonal subchannel will be used as another voice channel to achieve double the voice capacity. In order to maximize the voice capacity of each BTS, the introduction of the new voice channel has the following principles: The new voice channel has the lowest cross-correlation with the original voice channel. In the prior art, the VAM0S two-user joint channel estimation algorithm is used to estimate the correlation between the new voice channel and the original voice channel, and the specific process includes: first, by using two users that can represent the new voice channel and the original voice channel. The training sequence number obtains the channel estimation factor; then the channel estimation factor is used for the VAM0S two-user joint channel detection to obtain the channel detection result; finally, the result is used for the VAM0S strong and weak user joint channel detection and subsequent detection. The process of obtaining a channel estimation factor by a training sequence number of two users that can represent a new voice channel and an original voice channel includes: estimating from a pre-stored channel In the factor, a channel estimation factor corresponding to the training sequence number of the two users is obtained, and the pre-stored channel estimation factor is: y = (A T A) - τ , where Α is a matrix of 44*20, having the following
0 0 0 0
格式: A , 其中 A为 22*5的矩阵, j = 0、 1; A是  Format: A , where A is a matrix of 22*5, j = 0, 1; A is
0 0 A 由两个用户的训练序列号对应的训练序列生成的;由于两个用户的训练 序列号对应的训练序列均有 16种可能, 因此每一个 有 16种可能。 从上述过程可以看出信道估计过程处于 VAM0S 两个用户联合信道估 计算法的最前端, 其计算精度及实现方式对 VAM0S 两个用户联合信道估 计算法的影响较大。 在实现本发明的过程中, 发明人发现, 由于需要预 先存储信道估计因子, 使得信道估计因子采用 X比特量化时, 芯片的数 据存储量为 2*22*5*16*16*x=56320x 比特, 芯片的数据存储量较大, 提 高了芯片的成本。 0 0 A is generated by the training sequence corresponding to the training sequence numbers of the two users; since there are 16 possibilities for the training sequences corresponding to the training sequence numbers of the two users, there are 16 possibilities for each. It can be seen from the above process that the channel estimation process is at the forefront of the two-user joint channel estimation algorithm of VAM0S, and its computational accuracy and implementation method have a great influence on the two-user joint channel estimation algorithm of VAM0S. In the process of implementing the present invention, the inventors have found that since the channel estimation factor needs to be stored in advance so that the channel estimation factor is quantized by X bits, the data storage amount of the chip is 2*22*5*16*16*x=56320x bits. The chip has a large amount of data storage, which increases the cost of the chip.
发明内容 Summary of the invention
本发明的实施例提供一种信道估计方法和装置, 能够存储少量数 据实现信道估计。  Embodiments of the present invention provide a channel estimation method and apparatus capable of storing a small amount of data to implement channel estimation.
一方面, 提供了一种信道估计装置, 包括:  In one aspect, a channel estimation apparatus is provided, including:
矩阵存储单元, 用于存储中间逆矩阵;  a matrix storage unit, configured to store an intermediate inverse matrix;
矩阵获取单元,  Matrix acquisition unit,
取两个用户的训练序列号相应的子逆矩阵;  Taking the sub-inverse matrix corresponding to the training sequence numbers of the two users;
逻辑计算单元,  Logical computing unit,
两个用户的训练序列号对应的训练序列生成的矩阵进行乘加逻辑运算, 得到信道估计因子;  The matrix generated by the training sequence corresponding to the training sequence numbers of the two users is subjected to multiplication and addition logic operations to obtain a channel estimation factor;
信道估计单元, 用于根据所述逻辑计算单元计算得到的所述信道 估计因子完成信道估计过程。  And a channel estimation unit, configured to complete a channel estimation process according to the channel estimation factor calculated by the logic calculation unit.
另一方面, 提供了一种信道估计方法, 包括: 从预先存储的中间 逆矩阵中获取两个用户的训练序列号相应的子逆矩阵;将所述子逆矩阵 与所述两个用户的训练序列号对应的训练序列生成的矩阵进行乘加逻 辑运算,得到信道估计因子; 根据计算得到的所述信道估计因子完成信 道估计过程。 本发明实施例提供的信道估计方法和装置, 通过逻辑计算单元将矩 阵存储单元存储的中间逆矩阵中两个用户的训练序列号相应的子逆矩 阵, 与训练序列号对应的训练序列生成的矩阵进行乘加逻辑计算, 得到 信道估计因子完成信道估计。 由于只需对中间逆矩阵进行存储, 即采用In another aspect, a channel estimation method is provided, including: acquiring a sub-inverse matrix corresponding to training sequence numbers of two users from a pre-stored intermediate inverse matrix; and training the sub-inverse matrix with the two users The matrix generated by the training sequence corresponding to the serial number is subjected to multiplication and addition logic operations to obtain a channel estimation factor; and the channel estimation process is completed according to the calculated channel estimation factor. The channel estimation method and apparatus provided by the embodiment of the present invention, the matrix generated by the training sequence corresponding to the training sequence number of the two users in the intermediate inverse matrix stored by the matrix storage unit by the logic calculation unit, and the matrix generated by the training sequence corresponding to the training sequence number The multiplication and addition logic calculation is performed to obtain a channel estimation factor to complete channel estimation. Since only the intermediate inverse matrix needs to be stored,
X比特量化时, 最大数据存储量为 X * ( 16 * 25 + 16 * 16 * 25 ) =6800x比特, 使得本发明实施例提供的技术方案能够减少芯片的数据存储量, 从而减 小芯片的面积, 有效降低芯片的成本。 本发明实施例解决了现有技术中 芯片的数据存储量较大的问题。 When the X-bit quantization is performed, the maximum data storage amount is X*(16*25+16*16*25)=6800x bits, so that the technical solution provided by the embodiment of the present invention can reduce the data storage amount of the chip, thereby reducing the area of the chip. , effectively reduce the cost of the chip. The embodiment of the invention solves the problem that the data storage amount of the chip in the prior art is large.
附图说明 DRAWINGS
为了更清楚地说明本发明实施例或现有技术中的技术方案, 下面 将对实施例或现有技术描述中所需要使用的附图作筒单地介绍,显而易 见地, 下面描述中的附图仅仅是本发明的一些实施例,对于本领域普通 技术人员来讲, 在不付出创造性劳动的前提下,还可以根据这些附图获 得其他的附图。  In order to more clearly illustrate the embodiments of the present invention or the technical solutions in the prior art, the drawings to be used in the embodiments or the description of the prior art will be briefly described below. Obviously, the drawings in the following description It is merely some embodiments of the present invention, and those skilled in the art can obtain other drawings according to the drawings without any creative work.
图 1为本发明实施例一提供的信道估计装置的结构示意图; 图 2为本发明实施例二提供的信道估计装置的结构示意图; 图 3 为图 2 所示的信道估计装置中第一矩阵计算模块的结构示意 图;  1 is a schematic structural diagram of a channel estimation apparatus according to Embodiment 1 of the present invention; FIG. 2 is a schematic structural diagram of a channel estimation apparatus according to Embodiment 2 of the present invention; FIG. 3 is a first matrix calculation in the channel estimation apparatus shown in FIG. Schematic diagram of the module;
图 4 为图 2 所示的信道估计装置中第二矩阵计算模块的结构示意 图;  4 is a schematic structural diagram of a second matrix calculation module in the channel estimation apparatus shown in FIG. 2;
图 5为本发明实施例三提供的信道估计方法的流程图;  FIG. 5 is a flowchart of a channel estimation method according to Embodiment 3 of the present invention;
图 6为本发明实施例四提供的信道估计方法的流程图;  6 is a flowchart of a channel estimation method according to Embodiment 4 of the present invention;
图 7为本发明实施例五提供的信道估计方法的流程图。  FIG. 7 is a flowchart of a channel estimation method according to Embodiment 5 of the present invention.
具体实施方式 detailed description
下面将结合本发明实施例中的附图, 对本发明实施例中的技术方 案进行清楚、 完整地描述, 显然, 所描述的实施例仅仅是本发明一部分 实施例, 而不是全部的实施例。 基于本发明中的实施例, 本领域普通技 术人员在没有做出创造性劳动前提下所获得的所有其他实施例,都属于 本发明保护的范围。 为了解决现有技术中芯片内部的数据存储量较大的问题, 提出了 一种信道估计方法和装置。 The technical solutions in the embodiments of the present invention are clearly and completely described in the following with reference to the accompanying drawings in the embodiments of the present invention. It is obvious that the described embodiments are only a part of the embodiments of the present invention, and not all of the embodiments. All other embodiments obtained by a person of ordinary skill in the art based on the embodiments of the present invention without creative efforts are within the scope of the present invention. In order to solve the problem of large amount of data storage inside the chip in the prior art, a channel estimation method and apparatus are proposed.
如图 1所示, 本发明实施例一提供的信道估计装置, 包括: 矩阵存储单元 1 01 , 用于存储中间逆矩阵。  As shown in FIG. 1, a channel estimation apparatus according to Embodiment 1 of the present invention includes: a matrix storage unit 101 for storing an intermediate inverse matrix.
在本实施例中, 矩阵存储单元 1 01 中存储的中间逆矩阵, 可以存 储该中间逆矩阵中各个矩阵的全部元素;也可以根据该中间逆矩阵中各 个矩阵的特性,存储各个矩阵相应的元素, 例如该中间逆矩阵中某个矩 阵为对称矩阵时,存储该矩阵的一半元素等; 还可以通过其他方式存储 该中间逆矩阵, 在此不再——赘述。 两个用户的训练序列号相应的子逆矩阵。  In this embodiment, the intermediate inverse matrix stored in the matrix storage unit 101 may store all the elements of each matrix in the intermediate inverse matrix; and may also store corresponding elements of each matrix according to the characteristics of each matrix in the intermediate inverse matrix. For example, when a matrix in the intermediate inverse matrix is a symmetric matrix, half of the elements of the matrix are stored, etc.; the intermediate inverse matrix may also be stored by other means, which is not repeated here. The sub-inverse matrix corresponding to the training sequence number of the two users.
在本实施例中, 矩阵获取单元 1 02 获取的子逆矩阵, 可以是根据 两个用户的训练序列号与中间逆矩阵的映射关系,从矩阵存储单元 1 01 存储的中间逆矩阵中获取的; 也可以是通过其他方式从矩阵存储单元 In this embodiment, the sub-inverse matrix obtained by the matrix obtaining unit 102 may be obtained from the intermediate inverse matrix stored by the matrix storage unit 101 according to the mapping relationship between the training sequence number of the two users and the intermediate inverse matrix; It can also be from another matrix storage unit
1 01存储的中间逆矩阵中获取的, 在此不再——赘述。 户的训练序列号对应的训练序列生成的矩阵进行乘加逻辑运算,得到信 道估计因子。 1 01 is stored in the intermediate inverse matrix, which is no longer mentioned here. The matrix generated by the training sequence corresponding to the training sequence number of the user performs a multiplication and addition logic operation to obtain a channel estimation factor.
在本实施例中, 逻辑计算单元 1 0 3 中将子逆矩阵与两个用户的训 练序列号对应的训练序列生成的矩阵进行乘加逻辑运算,可以包括对该 子逆矩阵和训练序列生成的矩阵进行乘法逻辑运算、加法逻辑运算和转 置运算等, 在此不再——赘述。  In this embodiment, the logical computing unit 1 0 3 performs a multiplication and addition logic operation on the matrix generated by the training matrix corresponding to the training sequence number of the two users, and may include the matrix inverse matrix and the training sequence generated. The matrix performs multiplication logic operations, addition logic operations, and transpose operations, etc., and is not repeated here.
信道估计单元 1 04 ,用于根据逻辑计算单元计算得到的信道估计因 子完成信道估计过程。  The channel estimation unit 104 is configured to complete the channel estimation process according to the channel estimation factor calculated by the logic calculation unit.
在本实施例中, 通过信道估计单元 1 04 完成信道估计过程, 可以 包括将逻辑计算单元 103计算得到的信道估计因子用于 VAMOS两个用户 联合信道检测,得到检测结果; 并将该检测结果用于 VAM0S强弱用户联 合信道检测及其他后续检测等, 在此不再——赘述。 In this embodiment, the channel estimation process is completed by the channel estimation unit 104, and The channel estimation factor calculated by the logic calculation unit 103 is used for VAMOS two-user joint channel detection, and the detection result is obtained; and the detection result is used for the VAM0S strong and weak user joint channel detection and other subsequent detections, etc., - Narration.
本发明实施例提供的信道估计装置, 通过逻辑计算单元将矩阵存 储单元存储的中间逆矩阵中两个用户的训练序列号相应的子逆矩阵,与 训练序列号对应的训练序列生成的矩阵进行乘加逻辑计算,得到信道估 计因子实现信道估计。 由于只需对中间逆矩阵进行存储, 即采用 X比特 量化时, 最大的数据存储量为 X* ( 16*25 + 16*16*25 ) =6800x比特, 使 得本发明实施例提供的技术方案能够减少芯片的数据存储量,从而减小 芯片的面积,有效降低芯片的成本。本发明实施例解决了现有技术中芯 片的数据存储量较大的问题。  The channel estimation apparatus provided by the embodiment of the present invention multiplies the sub-inverse matrix corresponding to the training sequence number of the two users in the intermediate inverse matrix stored in the matrix storage unit by the logical calculation unit, and multiplies the matrix generated by the training sequence corresponding to the training sequence number. Add logic calculation to get the channel estimation factor to achieve channel estimation. The maximum technical data storage capacity is X* (16*25 + 16*16*25)=6800x bits, so that the technical solution provided by the embodiment of the present invention can be used. Reduce the amount of data storage of the chip, thereby reducing the area of the chip and effectively reducing the cost of the chip. The embodiment of the invention solves the problem of large data storage capacity of the chip in the prior art.
如图 2所示, 本发明实施例二提供的信道估计装置, 包括: d矩阵存储模块 201, 用于存储中间逆矩阵中的 d矩阵, 该 d矩阵 为( - ^( 4)— 1矩阵。 As shown in FIG. 2, the channel estimation apparatus provided in Embodiment 2 of the present invention includes: a d matrix storage module 201, configured to store a d matrix in an intermediate inverse matrix, where the d matrix is a (-^(4) -1 matrix.
在本实施例中, d矩阵存储模块 201存储的 «Α-Α^ ^Α^^ΑΓ1 矩阵中, A、 Α是由所述两个用户的训练序列号对应的训练序列生成的 22*5的矩阵;由于两个用户的训练序列号对应的训练序列有 16种可能, 因此每一个 A、 A都有 16种可能。 In the present embodiment, d matrix storage module 201 stores the «Α-Α ^ ^ Α ^^ ΑΓ 1 matrix, A, Α training sequence generated by the training sequence numbers corresponding to the two users 22 * 5 The matrix; since there are 16 possibilities for the training sequence corresponding to the training sequence number of the two users, there are 16 possibilities for each A and A.
在本实施例中, d矩阵存储模块 201可以通过第一存储子模块存储 d矩阵的全部元素; 由于^^^ -^ ^^ )—1^^)-1) ^/, 即 d矩阵为 对称矩阵, 因此 d矩阵存储模块 201还可以通过第二存储子模块存储 d
Figure imgf000007_0001
0010
In this embodiment, the d matrix storage module 201 can store all elements of the d matrix through the first storage submodule; since ^^^ -^ ^^ ) - 1 ^^) - 1 ) ^/, that is, the d matrix is symmetric a matrix, so the d matrix storage module 201 can also store d through the second storage submodule
Figure imgf000007_0001
0010
A 0 0 0001  A 0 0 0001
B = P 时, τ = ΡΑΊ , 即 Α和 Β可以线性转 O A1 0 A 1000 When B = P, τ = ΡΑ Ί, i.e. Α and Β be converted linearly OA 1 0 A 1000
0100 化, 其对应的脉沖响应也可以线性转化, 因此 d矩阵存储模块 201也可 以通过第三存储子模块存储两个用户中用户 1 的训练序列号小于两个 用户中用户 2的训练序列号的 d矩阵的元素;由于 d矩阵既为对称矩阵, 并且 A和 B的脉沖响应可以线性转化,因此 d矩阵存储模块 201可以通 过第四存储子模块存储 d矩阵的一半元素中,两个用户中用户 1的训练 序列号小于两个用户中用户 2的训练序列号的元素; d矩阵存储模块 201 还可以通过其他方式存储 d矩阵的元素, 在此不再——赘述。  0100, its corresponding impulse response can also be linearly transformed, so the d matrix storage module 201 can also store the training sequence number of the user 1 in the two users through the third storage submodule is smaller than the training serial number of the user 2 in the two users. The elements of the d matrix; since the d matrix is both a symmetric matrix, and the impulse responses of A and B can be linearly transformed, the d matrix storage module 201 can store half of the elements of the d matrix through the fourth storage submodule, the users of the two users The training sequence number of 1 is smaller than the element of the training sequence number of the user 2 of the two users; the d matrix storage module 201 can also store the elements of the d matrix by other means, which are not repeated here.
在本实施例中, d矩阵存储模块 201通过第一存储子模块存储 d矩 阵的元素, 采用 X比特量化时,数据存储量为 16*16*25*x=6656x比特; d矩阵存储模块 201通过第二存储子模块存储 d矩阵的元素, 采用 X比 特量化时,数据存储量为 16*16*15*x=3840x比特; d矩阵存储模块 201 通过第三存储子模块存储 d矩阵的元素, 采用 X比特量化时,数据存储 量为 120*25*x=3000x比特; d矩阵存储模块 201通过第四存储子模块 存储 d矩阵的元素,采用 X比特量化时,数据存储量为 120*15*x=1800x 比特。 其中, d矩阵存储模块 201通过第一存储子模块存储 d矩阵的元 素时, 由于每一个 、 A都有 16种可能, 且 d矩阵为 5*5的矩阵, 因 此采用 X比特量化时, 数据存储量为 16*16*25*x=6656x比特; d矩阵 存储模块 201通过第二存储子模块存储 d矩阵的元素时,由于每一个 4、  In this embodiment, the d matrix storage module 201 stores the elements of the d matrix through the first storage submodule, and when the X bit quantization is used, the data storage amount is 16*16*25*x=6656x bits; the d matrix storage module 201 passes The second storage submodule stores the elements of the d matrix. When the X bit quantization is used, the data storage amount is 16*16*15*x=3840x bits; the d matrix storage module 201 stores the elements of the d matrix through the third storage submodule, In the case of X-bit quantization, the data storage amount is 120*25*x=3000x bits; the d-matrix storage module 201 stores the elements of the d-matrix through the fourth storage sub-module, and when the X-bit quantization is used, the data storage amount is 120*15*x. =1800x bits. Wherein, when the d matrix storage module 201 stores the elements of the d matrix through the first storage submodule, since each and A has 16 possibilities, and the d matrix is a matrix of 5*5, when using X-bit quantization, data storage The quantity is 16*16*25*x=6656x bits; the d matrix storage module 201 stores the elements of the d matrix through the second storage submodule, since each 4
A都有 16种可能, 且 d矩阵为 5*5的矩阵, 存储一半元素时, 每个需 要存储 15 个元素, 因此采用 X 比特量化时, 数据存储量为 16*16*15*x=3840x比特; d矩阵存储模块 201通过第三存储子模块存储 d矩阵的元素时, 由于只需存储两个用户中用户 1的训练序列号小于两 个用户中用户 2的训练序列号的 d矩阵的元素,且用户 1和用户 2的训 练序列号都有 16种可能, 因此当用户 1的训练序列号为 0时, 需要存 储 15个 d矩阵;当用户 1的训练序列号为 1时,需要存储 14个 d矩阵; 当用户 1的训练序列号为 2时, 需要存储 14个 d矩阵; ......; 当用户A has 16 possibilities, and the d matrix is a matrix of 5*5. When storing half of the elements, each needs to store 15 elements. Therefore, when using X-bit quantization, the data storage capacity is 16*16*15*x=3840x. The d matrix storage module 201 stores the elements of the d matrix through the third storage submodule, since it is only necessary to store the elements of the d matrix of the training sequence number of the user 1 of the two users that are smaller than the training sequence number of the user 2 of the two users. And the training of user 1 and user 2 There are 16 possibilities for the training serial number. Therefore, when the training sequence number of user 1 is 0, 15 d matrices need to be stored; when the training sequence number of user 1 is 1, 14 d matrices need to be stored; When the training serial number is 2, 14 d matrices need to be stored; ......; when the user
1 的训练序列号为 15 时, 需要存储 0 个 d 矩阵, 因此 d 矩阵共有 15 + 14+13+…… +1 + 0=120 个, 因此采用 x 比特量化时, 数据存储量为 120*25*x=3000x比特; d矩阵存储模块 201通过第四存储子模块存储 d 矩阵的元素的数据存储量, 与上述计算过程相似, 在此不再——赘述。 When the training sequence number of 1 is 15, it needs to store 0 d matrices, so the d matrix has 15 + 14+13+... +1 + 0=120, so when using x-bit quantization, the data storage capacity is 120*25. *x=3000x bits; The d matrix storage module 201 stores the data storage amount of the elements of the d matrix through the fourth storage submodule, which is similar to the above calculation process, and is not repeated here.
逆矩阵存储模块 202, 用于存储中间逆矩阵中的逆矩阵, 该逆矩阵 为 矩阵。  The inverse matrix storage module 202 is configured to store an inverse matrix in the intermediate inverse matrix, where the inverse matrix is a matrix.
在本实施例中, 逆矩阵存储模块 202存储的( 矩阵, 、 是 由所述两个用户的训练序列号对应的训练序列生成的 22*5的矩阵; 由 于两个用户的训练序列号对应的训练序列有 16种可能,因此每一个 、 A都有 16种可能。 在本实施例中, 逆矩阵存储模块 202 可以通过第五存储子模块存 储逆矩阵的全部元素; 由于( =(( ) 即逆矩阵为对称矩阵, 因此逆矩阵存储模块 202 也可以通过第六存储子模块存储逆矩阵的一 半元素; 逆矩阵存储模块 202还可以通过其他方式存储逆矩阵的元素, 在此不再一一赘述。  In this embodiment, the inverse matrix storage module 202 stores (a matrix, which is a 22*5 matrix generated by the training sequence corresponding to the training sequence numbers of the two users; corresponding to the training sequence numbers of the two users. There are 16 possibilities for the training sequence, so there are 16 possibilities for each and A. In this embodiment, the inverse matrix storage module 202 can store all elements of the inverse matrix through the fifth storage submodule; since (=(() The inverse matrix is a symmetric matrix, so the inverse matrix storage module 202 can also store half of the elements of the inverse matrix through the sixth storage sub-module; the inverse matrix storage module 202 can also store the elements of the inverse matrix by other means, and will not be described herein. .
在本实施例中, 逆矩阵存储模块 202 通过第五存储子模块存储逆 矩阵的元素, 采用 X比特量化时, 数据存储量为 15*25*x=400x比特; 逆矩阵存储模块 202 通过第六存储子模块存储逆矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*15*x=240x比特。 其中, 逆矩阵存储模 块 202通过第五存储子模块存储逆矩阵的元素时, 由于由于每一个 、 In this embodiment, the inverse matrix storage module 202 stores the elements of the inverse matrix through the fifth storage submodule, and when using X-bit quantization, the data storage amount is 15*25*x=400x bits; the inverse matrix storage module 202 passes the sixth The storage sub-module stores the elements of the inverse matrix. When X-bit quantization is used, the data storage amount is 16*15*x=240x bits. Wherein, the inverse matrix storage module 202 stores the elements of the inverse matrix through the fifth storage submodule, due to each
A都有 16种可能, 且逆矩阵为 5*5的矩阵, 因此采用 X比特量化时, 数据存储量为 15*25*x=400x比特; 逆矩阵存储模块 202通过第六存储 子模块存储逆矩阵的元素时, 由于由于每一个 A、 A都有 16种可能, 且逆矩阵为 5*5的矩阵, 存储一般元素时, 每个需要存储 15个元素, 因此采用 X比特量化时, 数据存储量为 16*15*x=240x比特。 A has 16 possibilities, and the inverse matrix is a matrix of 5*5, so when using X-bit quantization, the data storage amount is 15*25*x=400x bits; the inverse matrix storage module 202 passes the sixth storage. When the submodule stores the elements of the inverse matrix, since there are 16 possibilities for each A and A, and the inverse matrix is a matrix of 5*5, when storing the general elements, each needs to store 15 elements, so X-bit quantization is used. At the time, the amount of data storage is 16*15*x=240x bits.
子逆矩阵获取模块 203, 用于从 d矩阵存储模块和逆矩阵存储模块 存储的 d 矩阵和逆矩阵中获取两个用户的训练序列号相应的子逆矩阵 dj和 m1The sub-inverse matrix obtaining module 203 is configured to obtain the sub-inverse matrices dj and m 1 corresponding to the training sequence numbers of the two users from the d matrix and the inverse matrix stored by the d matrix storage module and the inverse matrix storage module.
在本实施例中, 子逆矩阵获取模块 203获取的逆矩阵 d pn 可以 是根据两个用户的训练序列号与 d矩阵存储模块 201和逆矩阵存储模块 202存储的 d矩阵和逆矩阵的映射关系,从 d矩阵存储模块 201和逆矩 阵存储模块 202存储的 d矩阵和逆矩阵中获取的;也可以是通过其他方 式从 d矩阵存储模块 201和逆矩阵存储模块 202存储的 d矩阵和逆矩阵 中获取的, 在此不再——赘述。  In this embodiment, the inverse matrix d pn obtained by the sub-inverse matrix obtaining module 203 may be a mapping relationship between the d matrix and the inverse matrix stored by the d matrix storage module 201 and the inverse matrix storage module 202 according to the training sequence numbers of the two users. Obtained from the d matrix and the inverse matrix stored by the d matrix storage module 201 and the inverse matrix storage module 202; or may be the d matrix and the inverse matrix stored by the d matrix storage module 201 and the inverse matrix storage module 202 by other means. Obtained, no longer here - repeat.
第一矩阵计算模块 204, 用于将子逆矩阵获取模块获取的子逆矩阵 d pn 与训练序列生成的矩阵进行乘加逻辑运算, 得到 bT +4 矩 阵。 The first matrix calculation module 204 is configured to perform multiplication and logical operations on the sub-inverse matrix d pn obtained by the sub-inverse matrix acquisition module and the matrix generated by the training sequence to obtain a b T +4 matrix.
在本实施例中, 第一矩阵计算模块 204得到的 bT^+A 矩阵中, b 矩阵为
Figure imgf000010_0001
, 该训练序列生成的矩阵可以为 A和 A。
In this embodiment, in the bT^+A matrix obtained by the first matrix calculation module 204, the b matrix is
Figure imgf000010_0001
The matrix generated by the training sequence may be A and A.
进一步的, 如图 3所示, 本实施例中第一矩阵计算模块 204, 还可 以包括:  Further, as shown in FIG. 3, the first matrix calculation module 204 in this embodiment may further include:
第一计算子模块 2041, 用于将 A、 A进行乘法逻辑运算, 得到 A 矩阵。  The first calculation sub-module 2041 is configured to perform multiplication logic operations on A and A to obtain an A matrix.
第二计算子模块 2042, 用于将 A矩阵与子逆矩阵^进行乘法逻 辑运算, 得到 t矩阵, 该 t矩阵为 ^( )。  The second calculation sub-module 2042 is configured to perform a multiplication logic operation on the A matrix and the sub-inverse matrix ^ to obtain a t matrix, where the t matrix is ^( ).
第三计算子模块 2043, 用于将 t矩阵与 矩阵进行乘法逻辑运算, 得到 b矩阵, 该 b矩阵为 - ^。 第四计算子模块 2044, 用于将 b矩阵与 A进行乘法逻辑运算, 得 到 矩阵。 The third calculation sub-module 2043 is configured to perform a multiplication logic operation on the t matrix and the matrix to obtain a b matrix, where the b matrix is -^. The fourth calculation sub-module 2044 is configured to perform a multiplication logic operation on the b matrix and A to obtain a matrix.
第五计算子模块 2045, 用于将 ^矩阵与 A进行乘法逻辑运算, 得 到^ 4f矩阵。 法逻辑运算, 得到所述 矩阵。  The fifth calculation sub-module 2045 is configured to perform a multiplication logic operation on the ^ matrix and A to obtain a ^4f matrix. The logical operation is performed to obtain the matrix.
第二矩阵计算模块 205, 用于将子逆矩阵获取模块获取的子逆矩阵 d pn 与训练序列生成的矩阵进行乘加逻辑计算, 得到 +^矩阵。  The second matrix calculation module 205 is configured to perform multiplication and logical calculation on the sub-inverse matrix d pn obtained by the sub-inverse matrix acquisition module and the matrix generated by the training sequence to obtain a matrix of +^.
在本实施例中, 第二矩阵计算模块 205得到的^+b 矩阵中, 《矩 阵为 mi
Figure imgf000011_0001
, 该训练序列生成的矩阵可以为 Α和 。
In this embodiment, in the ^+b matrix obtained by the second matrix calculation module 205, the matrix is mi
Figure imgf000011_0001
The matrix generated by the training sequence may be a sum.
进一步的, 如图 4所示, 本实施例中第二矩阵计算模块 205, 还可 以包括:  Further, as shown in FIG. 4, the second matrix calculation module 205 in this embodiment may further include:
第一计算子模块 2051, 用于将 Α、 Α进行乘法逻辑运算, 得到 A 矩阵。  The first calculation sub-module 2051 is configured to perform multiplication logic operations on Α and , to obtain an A matrix.
第二计算子模块 2052, 用于将 A矩阵与子逆矩阵^进行乘法逻 辑运算, 得到 t矩阵, 该 t矩阵为 ^( )。  The second calculation sub-module 2052 is configured to perform a multiplication logic operation on the A matrix and the sub-inverse matrix to obtain a t matrix, where the t matrix is ^( ).
第三计算子模块 2053, 用于将 t矩阵与 矩阵进行乘法逻辑运算, 得到 b矩阵, 该 b矩阵为 - ^。  The third calculation sub-module 2053 is configured to perform a multiplication logic operation on the t matrix and the matrix to obtain a b matrix, where the b matrix is -^.
第七计算子模块 2054, 用于将 b矩阵与 t矩阵进行乘法逻辑运算, 得到 b*[]T矩阵。 The seventh calculation sub-module 2054 is configured to perform a multiplication logic operation on the b matrix and the t matrix to obtain a b*[] T matrix.
第八计算子模块 2055, 用于将子逆矩阵 ^与 *[^矩阵进行乘法逻 辑运算, 得到《矩阵, 该《矩阵为!¾+ [^。  The eighth calculation sub-module 2055 is configured to perform a multiplication logic operation on the sub-inverse matrix ^ and the *[^ matrix to obtain a matrix, the matrix is! 3⁄4+ [^.
第九计算子模块 2056, 用于将所述《矩阵与 4进行乘法逻辑运算, 得到^ 矩阵。 第十计算子模块 2057, 用于将所述 b矩阵与 A进行乘法逻辑运算, 得到 bAf矩阵。 The ninth calculation sub-module 2056 is configured to perform the multiplication logic operation on the matrix and the 4 to obtain a ^ matrix. The tenth calculation sub-module 2057 is configured to perform multiplication logic operation on the b matrix and A to obtain a bAf matrix.
第十一计算子模块 2058, 用于将所述^ 矩阵与所述 b 矩阵进行 加法逻辑运算, 得到所 矩阵。  The eleventh calculation sub-module 2058 is configured to perform an addition logic operation on the ^ matrix and the b matrix to obtain a matrix.
拼接模块 206,用于将第一矩阵计算模块得到的 bT +«矩阵和第 二矩阵计算模块得到的 α^τ+Μ 矩阵拼接, 得到信道估计因子 The splicing module 206 is configured to splicing the b T +« matrix obtained by the first matrix calculation module and the α^τ+Μ matrix obtained by the second matrix calculation module to obtain a channel estimation factor.
.
Figure imgf000012_0001
信道估计模块 207,用于根据逻辑计算单元计算得到的信道估计因 子完成信道估计过程。
Figure imgf000012_0001
The channel estimation module 207 is configured to complete the channel estimation process according to the channel estimation factor calculated by the logical computing unit.
在本实施例中, 通过信道估计模块 207 完成信道估计过程, 可以 包括将拼接模块 206拼接得到的信道估计因子用于 VAM0S两个用户联合 信道检测,得到检测结果; 并将该检测结果用于 VAM0S强弱用户联合信 道检测及其他后续检测等, 在此不再——赘述。  In this embodiment, the channel estimation process is performed by the channel estimation module 207, and the channel estimation factor obtained by splicing the splicing module 206 is used for the joint channel detection of two users of the VAM0S, and the detection result is obtained; and the detection result is used for the VAM0S. The joint channel detection and other subsequent detection of strong and weak users are not repeated here.
本发明实施例提供的信道估计装置, 采用 X 比特量化时, 最大数 据存储量为 X* ( 16*25 + 16*16*25 ) =6800x 比特, 仅为原有算法的 6800x/56320x=12. 07%, 可以节省大量的存储空间。  In the channel estimation apparatus provided by the embodiment of the present invention, when X-bit quantization is used, the maximum data storage amount is X* (16*25 + 16*16*25) = 6800x bits, which is only 6800x/56320x=12 of the original algorithm. 07%, can save a lot of storage space.
本发明实施例提供的信道估计装置,通过逻辑计算单元将矩阵存储 单元存储的中间逆矩阵中两个用户的训练序列号相应的子逆矩阵,与训 练序列号对应的训练序列生成的矩阵进行乘加逻辑计算,得到信道估计 因子实现信道估计。 由于只需对中间逆矩阵进行存储, 即采用 X比特量 化时, 最大数据存储量为 X* ( 16*25 + 16*16*25 ) =6800x比特, 使得本 发明实施例提供的技术方案能够减少芯片的数据存储量,从而减小芯片 的面积,有效降低芯片的成本。本发明实施例解决了现有技术中芯片的 数据存储量较大的问题。 The channel estimation apparatus provided by the embodiment of the present invention multiplies the sub-inverse matrix corresponding to the training sequence number of two users in the intermediate inverse matrix stored in the matrix storage unit by the logic calculation unit, and multiplies the matrix generated by the training sequence corresponding to the training sequence number. Add logic calculation to get the channel estimation factor to achieve channel estimation. Since only the intermediate inverse matrix is stored, that is, when the X-bit quantization is used, the maximum data storage amount is X*(16*25+16*16*25)=6800x bits, so that the technical solution provided by the embodiment of the present invention can be reduced. The amount of data stored in the chip, thereby reducing the area of the chip, effectively reducing the cost of the chip. The embodiment of the invention solves the chip in the prior art The problem of large data storage.
如图 5所示, 本发明实施例三提供的信道估计方法, 包括: 相应的子逆矩阵。  As shown in FIG. 5, the channel estimation method provided in Embodiment 3 of the present invention includes: a corresponding sub-inverse matrix.
在本实施例中, 所述预先存储的中间逆矩阵, 可以存储该中间逆矩 阵中各个矩阵的全部元素; 也可以根据该中间逆矩阵中各个矩阵的特 性,存储各个矩阵相应的元素, 例如该中间逆矩阵中某个矩阵为对称矩 阵时,存储该矩阵的一半元素等; 还可以通过其他方式存储该中间逆矩 阵, 在此不再——赘述。  In this embodiment, the pre-stored intermediate inverse matrix may store all elements of each matrix in the intermediate inverse matrix; and may also store corresponding elements of each matrix according to characteristics of each matrix in the intermediate inverse matrix, for example, When a matrix in the intermediate inverse matrix is a symmetric matrix, half of the elements of the matrix are stored, etc.; the intermediate inverse matrix can also be stored by other means, which is not repeated here.
在本实施例中, 通过步骤 501 获取的子逆矩阵, 可以是根据两个 用户的训练序列号与中间逆矩阵的映射关系,从预先存储的中间逆矩阵 在此不再一一赘述。  In this embodiment, the sub-inverse matrix obtained by step 501 may be based on the mapping relationship between the training sequence numbers of the two users and the intermediate inverse matrix, and the intermediate inverse matrix stored in advance will not be further described herein.
步骤 502 ,将子逆矩阵与两个用户的训练序列号对应的训练序列生 成的矩阵进行乘加逻辑运算, 得到信道估计因子。  Step 502: Perform a multiplication and logical operation on the matrix generated by the sub-inverse matrix and the training sequence corresponding to the training sequence numbers of the two users, to obtain a channel estimation factor.
在本实施例中, 将子逆矩阵与两个用户的训练序列号对应的训练 序列生成的矩阵进行乘加逻辑运算,可以包括对该子逆矩阵和训练序列 生成的矩阵进行乘法逻辑运算、加法逻辑运算和转置运算等, 在此不再 In this embodiment, the matrix generated by the training sequence corresponding to the training sequence number of the two users of the sub-inverse matrix is multiplied and logically operated, and may include multiplication logic operations and additions on the matrix generated by the sub-inverse matrix and the training sequence. Logical operations and transposition operations, etc., no longer here
"""" ' "赞述。 """" ' "Like.
步骤 503 , 根据计算得到的信道估计因子完成信道估计过程。 在本实施例中, 通过步骤 503 完成信道估计过程, 可以包括将通 过步骤 502计算得到的信道估计因子用于 VAM0S 两个用户联合信道检 测,得到检测结果; 并将该检测结果用于 VAM0S强弱用户联合信道检测 及其他后续检测等, 在此不再——赘述。  Step 503: Complete a channel estimation process according to the calculated channel estimation factor. In this embodiment, the channel estimation process is completed in step 503, which may include using the channel estimation factor calculated in step 502 for the two-user joint channel detection of the VAM0S, and obtaining the detection result; and using the detection result for the VAM0S strength User joint channel detection and other subsequent detections are not repeated here.
本发明实施例提供的信道估计方法,通过将预先存储的中间逆矩阵 中两个用户的训练序列号相应的子逆矩阵,与训练序列号对应的训练序 列生成的矩阵进行乘加逻辑计算, 从而得到信道估计因子实现信道估 计。 由于只需对中间逆矩阵进行预先存储, 即采用 X比特量化时, 最大 的数据存储量为 X* ( 16*25 + 16*16*25 ) =6800x比特, 使得本发明实施 例提供的技术方案能够减少芯片的数据存储量, 从而减小芯片的面积, 有效降低芯片的成本。本发明实施例解决了现有技术中芯片的数据存储 量较大的问题。 A channel estimation method provided by an embodiment of the present invention, by using a pre-stored intermediate inverse matrix The sub-inverse matrix corresponding to the training sequence number of the two users is multiplied and logically calculated with the matrix generated by the training sequence corresponding to the training sequence number, thereby obtaining a channel estimation factor to implement channel estimation. The maximum amount of data storage is X* (16*25 + 16*16*25)=6800x bits, so that the technical solution provided by the embodiment of the present invention is provided. It can reduce the amount of data storage of the chip, thereby reducing the area of the chip and effectively reducing the cost of the chip. The embodiment of the invention solves the problem that the data storage amount of the chip in the prior art is large.
如图 6所示, 本发明实施例四提供的信道估计方法, 包括: 步骤 601 , 存储中间逆矩阵中的 d 矩阵和逆矩阵, d 矩阵为 (ATA - ΑΤ ( Τ )— 1 A )- 1矩阵, 逆矩阵为( A )— 1矩阵。 As shown in FIG. 6, the channel estimation method provided in Embodiment 4 of the present invention includes: Step 601: Store a d matrix and an inverse matrix in an intermediate inverse matrix, where the d matrix is (A T A - Α Τ ( Τ ) - 1 A ) - 1 matrix, the inverse matrix is the ( A ) - 1 matrix.
在本实施例中, 通过步骤 601存储的 d矩阵和逆矩阵中, 、 是 由所述两个用户的训练序列号对应的训练序列生成的 22*5的矩阵; 由 于两个用户的训练序列号对应的训练序列有 16种可能,因此每一个 、 A都有 16种可能。  In this embodiment, in the d matrix and the inverse matrix stored in step 601, a 22*5 matrix generated by the training sequence corresponding to the training sequence numbers of the two users; due to the training sequence numbers of the two users There are 16 possibilities for the corresponding training sequence, so each and A has 16 possibilities.
在本实施例中,存储中间逆矩阵中的 d矩阵时, 步骤 601可以通过 方法一存储 d矩阵的全部元素; 由于 ^^( ^「 ^( )—1^^)—1) ^ 即 d矩阵为对称矩阵,因此步骤 601还可以通过方法二存储 d矩阵的一 半元素; 由于 AIn this embodiment, when the d matrix in the intermediate inverse matrix is stored, step 601 can store all the elements of the d matrix by method one; since ^^( ^ " ^( ) - 1 ^^) - 1 ) ^ is the d matrix Is a symmetric matrix, so step 601 can also store half of the elements of the d matrix by method two;
Figure imgf000014_0001
Figure imgf000014_0001
0010  0010
0001  0001
Ρ 即 Α和 Β可以线性转化, 其对应的脉沖响应也 1000  Ρ Α Α and Β can be linearly transformed, and the corresponding impulse response is also 1000
0100 可以线性转化, 因此步骤 601也可以通过方法三存储两个用户中用户 1 的训练序列号小于两个用户中用户 2的训练序列号的 d矩阵的元素;由 于 d矩阵既为对称矩阵, 并且 A和 B的脉沖响应可以线性转化, 因此步 骤 601 可以通过方法四存储 d矩阵的一半元素中, 两个用户中用户 1 的训练序列号小于两个用户中用户 2 的训练序列号的元素; 步骤 601 还可以通过其他方法存储 d矩阵的元素, 在此不再——赘述。 0100 can be linearly transformed, so step 601 can also store, by method three, the elements of the d matrix of the user 1 whose training sequence number is smaller than the training sequence number of the user 2 of the two users; since the d matrix is both a symmetric matrix, and The impulse responses of A and B can be linearly transformed, so the step Step 601 may store half of the elements of the d matrix by method four, where the training sequence number of the user 1 of the two users is smaller than the element of the training sequence number of the user 2 of the two users; Step 601 may also store the elements of the d matrix by other methods. , no longer here - repeat.
在本实施例中, 步骤 601 通过方法一存储 d矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*16*25*x=6656x比特; 步骤 601通过方 法二存储 d 矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*16*15*x=3840x比特; 步骤 601通过方法三存储 d矩阵的元素, 采 用 X比特量化时, 数据存储量为 120*25*x=3000x比特; 步骤 601通过 方法四存储 d矩阵的元素, 采用 X比特量化时,数据方法一第一存储子 模块存储 d矩阵的元素时, 由于每一个 A、 A都有 16种可能, 且 d矩 阵为 5*5 的矩阵, 因此采用 X 比特量化时, 数据存储量为 16*16*25*x=6656x比特; 步骤 601通过方法二存储 d矩阵的元素时, 由于每一个 、 A都有 16种可能, 且 d矩阵为 5*5的矩阵, 存储一半 元素时, 每个需要存储 15个元素, 因此采用 X比特量化时, 数据存储 量为 16*16*15*x=3840x比特; 步骤 601通过方法三存储 d矩阵的元素 时,由于只需存储两个用户中用户 1的训练序列号小于两个用户中用户 2的训练序列号的 d矩阵的元素, 且用户 1和用户 2的训练序列号都有 16种可能,因此当用户 1的训练序列号为 0时,需要存储 15个 d矩阵; 当用户 1的训练序列号为 1时, 需要存储 14个 d矩阵; 当用户 1的训 练序列号为 2时, 需要存储 14个 d矩阵; ......; 当用户 1的训练序列 号为 15 时, 需要存储 0 个 d矩阵, 因此 d矩阵共有 15 + 14+13+…… + 1 + 0=120个, 因此采用 X比特量化时, 数据存储量为 120*25*x=3000x 比特; 步骤 601通过方法四存储 d矩阵的元素的数据存储量, 与上述计 算过程相似, 在此不再——赘述。  In this embodiment, step 601 stores the elements of the d matrix by method one, and when the X bit is used for quantization, the data storage amount is 16*16*25*x=6656x bits; step 601 stores the elements of the d matrix by method two, adopting In the case of X-bit quantization, the data storage amount is 16*16*15*x=3840x bits; Step 601 stores the elements of the d matrix by Method 3, and when X-bit quantization is used, the data storage amount is 120*25*x=3000x bits; Step 601 stores the elements of the d matrix by method four, and when the X-bit quantization is used, when the first storage sub-module stores the elements of the d-matrix, there are 16 possibilities for each A and A, and the d matrix is 5*. 5 matrix, so when using X-bit quantization, the data storage amount is 16*16*25*x=6656x bits; when step 601 stores the elements of the d matrix by method two, since each and A has 16 possibilities, and The d matrix is a 5*5 matrix. When storing half of the elements, each of the 15 elements needs to be stored. Therefore, when using X-bit quantization, the data storage amount is 16*16*15*x=3840x bits; Step 601 is stored by Method Three.For the elements of the d matrix, since only the elements of the d matrix of the user 1 whose training sequence number is smaller than the training sequence number of the user 2 of the two users are stored, and the training sequence numbers of the user 1 and the user 2 are 16 It is possible, therefore, when the training sequence number of user 1 is 0, 15 d matrices need to be stored; when the training sequence number of user 1 is 1, 14 d matrices need to be stored; when the training sequence number of user 1 is 2 , need to store 14 d matrices; ...; When user 1's training serial number is 15, you need to store 0 d matrices, so the d matrix has a total of 15 + 14 + 13 + ... + 1 + 0 = 120, so when using X-bit quantization, the data storage amount is 120*25*x=3000x bits; Step 601 stores the data storage amount of the elements of the d matrix by Method 4, which is similar to the above calculation process, and no longer - Narration.
在本实施例中, 步骤 601可以通过方法五存储逆矩阵的全部元素; 由于
Figure imgf000016_0001
, 即逆矩阵为对称矩阵, 因此步骤 601也可以通 过方法六存储逆矩阵的一半元素;步骤 601还可以通过其他方法存储逆 矩阵的元素, 在此不再——赘述。
In this embodiment, step 601 may store all elements of the inverse matrix by method five; due to
Figure imgf000016_0001
That is, the inverse matrix is a symmetric matrix, so step 601 can also store half of the elements of the inverse matrix by method six; step 601 can also store the elements of the inverse matrix by other methods, which are not repeated here.
在本实施例中, 步骤 601 通过方法五存储逆矩阵的元素, 采用 X 比特量化时, 数据存储量为 15*25*χ=400χ比特; 步骤 601通过方法六 存储逆矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*15*χ=240χ 比特。 其中, 步骤 601通过方法五存储逆矩阵的元素时, 由于由于每一 个 Α、 Α都有 16种可能, 且逆矩阵为 5*5的矩阵, 因此采用 X步骤 601 通过方法六存储逆矩阵的元素时, 由于由于每一个 、 A都有 16种可 能, 且逆矩阵为 5*5的矩阵, 存储一般元素时, 每个需要存储 15个元 素, 因此采用 X比特量化时, 数据存储量为 16*15*x=240x比特。 相应的子逆矩阵。  In this embodiment, step 601 stores the elements of the inverse matrix by method five, and when the X-bit quantization is used, the data storage amount is 15*25*χ=400χ bits; step 601 stores the elements of the inverse matrix by method six, adopts X bits. When quantizing, the amount of data stored is 16*15*χ=240χ bits. Wherein step 601 stores the elements of the inverse matrix by method five, since there are 16 possibilities for each Α and Α, and the inverse matrix is a matrix of 5*5, the X step 601 is used to store the elements of the inverse matrix by method six. Since, since each and A have 16 possibilities, and the inverse matrix is a matrix of 5*5, when storing general elements, each needs to store 15 elements, so when using X-bit quantization, the data storage amount is 16*. 15*x=240x bits. The corresponding sub-inverse matrix.
在本实施例中, 步骤 602 中的子逆矩阵, 可以是根据两个用户的 训练序列号与中间逆矩阵的映射关系,从预先存储的中间逆矩阵中获取 再一一赞述。  In this embodiment, the sub-inverse matrix in step 602 may be obtained from the pre-stored intermediate inverse matrix according to the mapping relationship between the training sequence numbers of the two users and the intermediate inverse matrix.
步骤 603, 将子逆矩阵与两个用户的训练序列号对应的训练序列生 成的矩阵进行乘加逻辑运算, 得到信道估计因子。 对应的训练序列生成的矩阵进行乘加逻辑运算,可以包括对该子逆矩阵 和训练序列生成的矩阵进行乘法逻辑运算、 加法逻辑运算和转置运算 等, 在此不再一一赘述 o  Step 603: Perform multiplication and addition logic on the matrix generated by the training matrix of the two inverse training matrices corresponding to the training sequence numbers of the two users to obtain a channel estimation factor. The matrix generated by the corresponding training sequence performs multiplication and addition logic operations, and may include multiplication logic operations, addition logic operations, and transposition operations on the matrix generated by the sub-inverse matrix and the training sequence, and will not be repeated here.
步骤 604, 根据计算得到的所述信道估计因子完成信道估计过程。 在本实施例中, 通过步骤 604 完成信道估计过程, 可以包括将通 过步骤 603计算得到的信道估计因子用于 VAM0S 两个用户联合信道检 测,得到检测结果; 并将该检测结果用于 VAM0S强弱用户联合信道检测 及其他后续检测等, 在此不再——赘述。 Step 604: Complete a channel estimation process according to the calculated channel estimation factor. In this embodiment, the channel estimation process is completed by step 604, which may include The channel estimation factor calculated in step 603 is used for two user joint channel detection of VAM0S, and the detection result is obtained; and the detection result is used for joint channel detection of VAM0S strong and weak users and other subsequent detections, etc., .
本发明实施例提供的信道估计方法, 采用 X比特量化时, 最大数据 存储量为 X* ( 16*25 + 16*16*25 ) =6800x 比特, 仅为原有算法的 6800x/56320x=12. 07%, 可以节省大量的存储空间。  In the channel estimation method provided by the embodiment of the present invention, when X-bit quantization is used, the maximum data storage amount is X* (16*25 + 16*16*25) = 6800x bits, which is only 6800x/56320x=12 of the original algorithm. 07%, can save a lot of storage space.
本发明实施例提供的信道估计方法,通过将预先存储的中间逆矩阵 中两个用户的训练序列号相应的子逆矩阵,与训练序列号对应的训练序 列生成的矩阵进行乘加逻辑计算, 从而得到信道估计因子实现信道估 计。 由于只需对中间逆矩阵进行预先存储, 即采用 X比特量化时, 最大 的数据存储量为 X* ( 16*25 + 16*16*25 ) =6800x比特, 使得本发明实施 例提供的技术方案能够减少芯片的数据存储量, 从而减小芯片的面积, 有效降低芯片的成本。本发明实施例解决了现有技术中芯片的数据存储 量较大的问题。  The channel estimation method provided by the embodiment of the present invention performs multiplication and addition logic calculation on a matrix generated by a training sequence corresponding to a training sequence number by using a sub-inverse matrix corresponding to training sequence numbers of two users in a pre-stored intermediate inverse matrix, thereby A channel estimation factor is obtained to achieve channel estimation. The maximum amount of data storage is X* (16*25 + 16*16*25)=6800x bits, so that the technical solution provided by the embodiment of the present invention is provided. It can reduce the amount of data storage of the chip, thereby reducing the area of the chip and effectively reducing the cost of the chip. The embodiment of the invention solves the problem that the data storage amount of the chip in the prior art is large.
如图 7所示, 本发明实施例五提供的信道估计方法, 包括: 步骤 701 , 存储中间逆矩阵中的 d 矩阵和逆矩阵, d 矩阵为 (ATA - ΑΤ ( Τ )— 1 A )- 1矩阵, 逆矩阵为( A )— 1矩阵。 As shown in FIG. 7, the channel estimation method provided in Embodiment 5 of the present invention includes: Step 701: Store a d matrix and an inverse matrix in an intermediate inverse matrix, where the d matrix is (A T A - Α Τ ( Τ ) - 1 A ) - 1 matrix, the inverse matrix is the ( A ) - 1 matrix.
在本实施例中, 通过步骤 701存储的 d矩阵和逆矩阵中, 4、 是 由所述两个用户的训练序列号对应的训练序列生成的 22*5的矩阵; 由 于两个用户的训练序列号对应的训练序列有 16种可能,因此每一个 、 A都有 16种可能。  In this embodiment, in the d matrix and the inverse matrix stored in step 701, 4 is a 22*5 matrix generated by the training sequence corresponding to the training sequence numbers of the two users; due to the training sequence of the two users There are 16 possibilities for the training sequence corresponding to the number, so there are 16 possibilities for each and A.
在本实施例中,存储中间逆矩阵中的 d矩阵时, 步骤 701可以通过 方法一存储 d矩阵的全部元素; 由于 ^^( ^「 ^( )—1^^)—1) ^ 即 d矩阵为对称矩阵,因此步骤 701还可以通过方法二存储 d矩阵的一 半元素; 由于 AIn this embodiment, when the d matrix in the intermediate inverse matrix is stored, step 701 may store all the elements of the d matrix by method one; since ^^( ^ " ^( ) - 1 ^^) - 1 ) ^ is the d matrix Is a symmetric matrix, so step 701 can also store one of the d matrices by method two. Semi-element; due to A
Figure imgf000018_0001
Figure imgf000018_0001
0010  0010
0001  0001
P 即 A和 B可以线性转化, 其对应的脉沖响应也 1000  P ie A and B can be linearly transformed, and the corresponding impulse response is also 1000
0100 可以线性转化, 因此步骤 701也可以通过方法三存储两个用户中用户 1 的训练序列号小于两个用户中用户 2的训练序列号的 d矩阵的元素;由 于 d矩阵既为对称矩阵, 并且 A和 B的脉沖响应可以线性转化, 因此步 骤 701 可以通过方法四存储 d矩阵的一半元素中, 两个用户中用户 1 的训练序列号小于两个用户中用户 2 的训练序列号的元素; 步骤 701 还可以通过其他方法存储 d矩阵的元素, 在此不再——赘述。  0100 can be linearly transformed, so step 701 can also store, by method three, the elements of the d matrix of the user 1 whose training sequence number is smaller than the training sequence number of the user 2 of the two users; since the d matrix is both a symmetric matrix, and The impulse responses of A and B can be linearly transformed, so step 701 can store half of the elements of the d matrix by method four, wherein the training sequence number of user 1 of the two users is smaller than the element of the training sequence number of user 2 of the two users; 701 can also store the elements of the d matrix by other methods, no longer here - repeat.
在本实施例中, 步骤 701 通过方法一存储 d矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*16*25*x=6656x比特; 步骤 701通过方 法二存储 d 矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*16*15*x=3840x比特; 步骤 701通过方法三存储 d矩阵的元素, 采 用 X比特量化时, 数据存储量为 120*25*x=3000x比特; 步骤 701通过 方法四存储 d矩阵的元素, 采用 X比特量化时,数据方法一第一存储子 模块存储 d矩阵的元素时, 由于每一个 、 A都有 16种可能, 且 d矩 阵为 5*5 的矩阵, 因此采用 X 比特量化时, 数据存储量为 16*16*25*x=6656x比特; 步骤 701通过方法二存储 d矩阵的元素时, 由于每一个 、 A都有 16种可能, 且 d矩阵为 5*5的矩阵, 存储一半 元素时, 每个需要存储 15个元素, 因此采用 X比特量化时, 数据存储 量为 16*16*15*x=3840x比特; 步骤 701通过方法三存储 d矩阵的元素 时,由于只需存储两个用户中用户 1的训练序列号小于两个用户中用户 2的训练序列号的 d矩阵的元素, 且用户 1和用户 2的训练序列号都有 16种可能,因此当用户 1的训练序列号为 0时,需要存储 15个 d矩阵; 当用户 1的训练序列号为 1时, 需要存储 14个 d矩阵; 当用户 1的训 练序列号为 2时, 需要存储 14个 d矩阵; ......; 当用户 1的训练序列 号为 15 时, 需要存储 0 个 d矩阵, 因此 d矩阵共有 15 + 14+13+…… + 1 + 0=120个, 因此采用 X比特量化时, 数据存储量为 120*25*x=3000x 比特; 步骤 701通过方法四存储 d矩阵的元素的数据存储量, 与上述计 算过程相似, 在此不再——赘述。 In this embodiment, step 701 stores the elements of the d matrix by method 1, and when the X bit is used for quantization, the data storage amount is 16*16*25*x=6656x bits; step 701 stores the elements of the d matrix by method two, adopting In the case of X-bit quantization, the data storage amount is 16*16*15*x=3840x bits; Step 701 stores the elements of the d matrix by Method 3, and when X-bit quantization is used, the data storage amount is 120*25*x=3000x bits; Step 701 stores the elements of the d matrix by method four, and when the X-bit quantization is used, the data method-first storage sub-module stores the elements of the d-matrix, since each and A has 16 possibilities, and the d matrix is 5*5. The matrix, therefore, when X-bit quantization is used, the data storage amount is 16*16*25*x=6656x bits; when step 701 stores the elements of the d-matrix by method two, since each and A has 16 possibilities, and d The matrix is a 5*5 matrix. When storing half of the elements, each of the 15 elements needs to be stored. Therefore, when X-bit quantization is used, the data storage amount is 16*16*15*x=3840x bits; Step 701 is stored by method three. The elements of the matrix, since only storage The training sequence number of user 1 is smaller than the element of the d matrix of the training sequence number of user 2 of the two users, and the training sequence numbers of user 1 and user 2 have 16 possibilities, so when the training sequence number of user 1 is When 0, it is necessary to store 15 d matrices; When the training sequence number of the user 1 is 1, 14 d matrices need to be stored; when the training sequence number of the user 1 is 2, 14 d matrices need to be stored; ...; when the training serial number of the user 1 is When it is 15, it needs to store 0 d matrices, so the d matrix has 15 + 14+13+... + 1 + 0=120, so when using X-bit quantization, the data storage is 120*25*x=3000x bits. Step 701 stores the data storage amount of the elements of the d matrix by method four, which is similar to the above calculation process, and is not repeated here.
在本实施例中, 步骤 701可以通过方法五存储逆矩阵的全部元素; 由于 ¾ =(( 4 ) 即逆矩阵为对称矩阵, 因此步骤 701也可以通 过方法六存储逆矩阵的一半元素;步骤 701还可以通过其他方法存储逆 矩阵的元素, 在此不再——赘述。  In this embodiment, step 701 may store all elements of the inverse matrix by method five; since 3⁄4 = ((4), ie, the inverse matrix is a symmetric matrix, step 701 may also store half of the elements of the inverse matrix by method six; step 701 It is also possible to store the elements of the inverse matrix by other methods, which are not repeated here.
在本实施例中, 步骤 701 通过方法五存储逆矩阵的元素, 采用 X 比特量化时, 数据存储量为 15*25*x=4QQx比特; 步骤 701通过方法六 存储逆矩阵的元素, 采用 X 比特量化时, 数据存储量为 16*15*x=240x 比特。 其中, 步骤 701通过方法五存储逆矩阵的元素时, 由于由于每一 个 A、 A都有 16种可能, 且逆矩阵为 5*5的矩阵, 因此采用 X步骤 701 通过方法六存储逆矩阵的元素时, 由于由于每一个 、 A都有 16种可 能, 且逆矩阵为 5*5的矩阵, 存储一般元素时, 每个需要存储 15个元 素, 因此采用 X比特量化时, 数据存储量为 16*15*x=240x比特。  In this embodiment, step 701 stores the elements of the inverse matrix by method five, and when the X-bit quantization is used, the data storage amount is 15*25*x=4QQx bits; step 701 stores the elements of the inverse matrix by method six, adopts X bits. When quantized, the amount of data stored is 16*15*x=240x bits. Wherein step 701 stores the elements of the inverse matrix by method five, since there are 16 possibilities for each A and A, and the inverse matrix is a matrix of 5*5, the X step 701 is used to store the elements of the inverse matrix by method six. Since, since each and A have 16 possibilities, and the inverse matrix is a matrix of 5*5, when storing general elements, each needs to store 15 elements, so when using X-bit quantization, the data storage amount is 16*. 15*x=240x bits.
步骤 702,从 d矩阵和逆矩阵中获取两个用户的训练序列号相应的 子逆矩阵 ^和 m,。  Step 702: Obtain sub-inverse matrices ^ and m corresponding to training sequence numbers of two users from the d matrix and the inverse matrix.
在本实施例中, 步骤 702获取的逆矩阵 d pn 可以是根据两个用 户的训练序列号与通过步骤 701存储的 d矩阵和逆矩阵的映射关系,从 步骤 701存储的 d矩阵和逆矩阵中获取的;也可以是通过其他方式从步 骤 701存储的 d矩阵和逆矩阵中获取的, 在此不再——赘述。  In this embodiment, the inverse matrix d pn obtained in step 702 may be based on the mapping relationship between the training sequence numbers of the two users and the d matrix and the inverse matrix stored in step 701, from the d matrix and the inverse matrix stored in step 701. The obtained data may also be obtained from the d matrix and the inverse matrix stored in step 701 by other means, and is not described here again.
步骤 703, 将子逆矩阵 d pn 与训练序列生成的矩阵进行乘加逻 辑运算, 得到 bT^+A 矩阵。 Step 703, multiplying and adding the sub-inverse matrix d pn and the matrix generated by the training sequence The operation is calculated to obtain the bT^+A matrix.
在本实施例中, 步骤 703 得到的 矩阵中, b 矩阵为 -mx ( A )*d, , 该训练序列生成的矩阵可以为 A和 A。 In this embodiment, in the matrix obtained in step 703, the b matrix is -m x ( A )*d, and the matrix generated by the training sequence may be A and A.
在本实施例中, 通过步骤 703 得到 矩阵的过程, 具体可 以包括:  In this embodiment, the process of obtaining the matrix by step 703 may specifically include:
一、 将 A、 A进行乘法逻辑运算, 得到 A矩阵。  First, A and A are multiplied by logical operations to obtain an A matrix.
二、 将 矩阵与子逆矩阵^进行乘法逻辑运算, 得到 t矩阵, t 矩阵为!!^( 4)。  Second, multiply the matrix and the sub-inverse matrix ^ to obtain the t matrix, and the t matrix is! !^( 4).
三、 将 t矩阵与 ^矩阵进行乘法逻辑运算, 得到 b矩阵, b矩阵为 Third, multiply the t matrix and the ^ matrix to obtain a b matrix, and the b matrix is
-t*^。 -t*^.
四、 将 b矩阵与矩阵 进行乘法逻辑运算, 得到 矩阵。  4. Perform a multiplication logic operation on the b matrix and the matrix to obtain a matrix.
五、 将^矩阵与矩阵 A进行乘法逻辑运算, 得到 4 矩阵。  5. Multiply the logic between the ^ matrix and the matrix A to obtain a 4 matrix.
六、将 矩阵与 4 矩阵进行加法逻辑运算,得到 矩阵。 步骤 704, 将子逆矩阵 d pn 与训练序列生成的矩阵进行乘加逻 辑运算, 得到^ + 矩阵。  Sixth, the matrix and the 4 matrix are added to the logical operation to obtain the matrix. Step 704: Multiply and add the sub-inverse matrix d pn to the matrix generated by the training sequence to obtain a ^ + matrix.
在本实施例中, 通过步骤 704 得到的 ^+b^矩阵中, 《矩阵为 m1 +b* [m, ( A )]τ , 该训练序列生成的矩阵可以为 Α和 A。 In this embodiment, in the ^+b^ matrix obtained by step 704, the matrix is m 1 + b* [m, ( A )] τ , and the matrix generated by the training sequence may be Α and A.
在本实施例中, 通过步骤 704 得到的 ^+Μ矩阵的过程, 具体可 以包括:  In this embodiment, the process of the ^+Μ matrix obtained in step 704 may specifically include:
一、 将 Α、 Α进行乘法逻辑运算, 得到 A矩阵。  1. Perform multiplication logic operations on Α and , to obtain an A matrix.
二、 将 A矩阵与子逆矩阵^进行乘法逻辑运算, 得到 t矩阵, t 矩阵为!!^( 4)。  Second, the A matrix and the sub-inverse matrix ^ are multiplied by logical operations to obtain a t matrix, and the t matrix is! !^( 4).
三、 将 t矩阵与 ^矩阵进行乘法逻辑运算, 得到 b矩阵, b矩阵为 Third, multiply the t matrix and the ^ matrix to obtain a b matrix, and the b matrix is
-t*^。 四、 将 b矩阵与 t矩阵进行乘法逻辑运算, 得到 b*wT矩阵。 -t*^. 4. Perform a multiplication logic operation on the b matrix and the t matrix to obtain a b*w T matrix.
五、 将子逆矩阵 ^与^^^矩阵进行乘法逻辑运算, 得到"矩阵, " 矩阵为 n^+b ir。  5. Multiply the sub-inverse matrix ^ with the ^^^ matrix to obtain a "matrix," and the matrix is n^+b ir.
六、 将"矩阵与 A进行乘法逻辑运算, 得到 α 矩阵。  6. The matrix and A are multiplied by logical operations to obtain an α matrix.
七、 将 b矩阵与 Α进行乘法逻辑运算, 得到 Mf矩阵。  7. Perform a multiplication logic operation on the b matrix and Α to obtain an Mf matrix.
八、 将^ 矩阵与 fe 矩阵进行加法逻辑运算, 得 矩阵。 步骤 705, 将 矩阵与^ +^矩阵拼接, 得到信道估计因 子
Figure imgf000021_0001
Eight, the ^ matrix and the fe matrix are added to the logical operation, resulting in a matrix. Step 705, splicing the matrix with the ^^^ matrix to obtain a channel estimation factor.
Figure imgf000021_0001
步骤 706, 根据计算得到的信道估计因子完成信道估计过程。  Step 706: Complete a channel estimation process according to the calculated channel estimation factor.
在本实施例中, 通过步骤 706 完成信道估计过程, 可以包括将通 过步骤 705 计算得到的信道估计因子用于 VAM0S 两个用户联合信道检 测,得到检测结果; 并将该检测结果用于 VAM0S强弱用户联合信道检测 及其他后续检测等, 在此不再——赘述。  In this embodiment, the channel estimation process is completed in step 706, which may include using the channel estimation factor calculated in step 705 for two user joint channel detection of VAM0S, and obtaining a detection result; and using the detection result for the VAM0S strength User joint channel detection and other subsequent detections are not repeated here.
本发明实施例提供的信道估计方法, 采用 X比特量化时, 最大数据 存储量为 X* ( 16*25 + 16*16*25 ) =6800x 比特, 仅为原有算法的 6800x/56320x=12. 07%, 可以节省大量的存储空间。  In the channel estimation method provided by the embodiment of the present invention, when X-bit quantization is used, the maximum data storage amount is X* (16*25 + 16*16*25) = 6800x bits, which is only 6800x/56320x=12 of the original algorithm. 07%, can save a lot of storage space.
本发明实施例提供的信道估计方法,通过将预先存储的中间逆矩阵 中两个用户的训练序列号相应的子逆矩阵,与训练序列号对应的训练序 列生成的矩阵进行乘加逻辑计算, 从而得到信道估计因子实现信道估 计。 由于只需对中间逆矩阵进行预先存储, 即采用 X比特量化时, 最大 的数据存储量为 X* ( 16*25 + 16*16*25 ) =6800x比特, 使得本发明实施 例提供的技术方案能够减少芯片的数据存储量, 从而减小芯片的面积, 有效降低芯片的成本。本发明实施例解决了现有技术中芯片的数据存储 量较大的问题。 The channel estimation method provided by the embodiment of the present invention performs multiplication and addition logic calculation on a matrix generated by a training sequence corresponding to a training sequence number by using a sub-inverse matrix corresponding to training sequence numbers of two users in a pre-stored intermediate inverse matrix, thereby A channel estimation factor is obtained to achieve channel estimation. The maximum amount of data storage is X* (16*25 + 16*16*25)=6800x bits, so that the technical solution provided by the embodiment of the present invention is provided. It can reduce the amount of data storage of the chip, thereby reducing the area of the chip and effectively reducing the cost of the chip. The embodiment of the invention solves the data storage of the chip in the prior art A large amount of problems.
本发明实施例提供的信道估计方法和装置,可以应用在 VAMO S技术 中, 检测新的语音信道与原有语音信道的相关性。  The channel estimation method and apparatus provided by the embodiments of the present invention can be applied to the VAMO S technology to detect the correlation between a new voice channel and an original voice channel.
结合本文中所公开的实施例描述的方法或算法的步骤可以直接用 硬件、 处理器执行的软件模块, 或者二者的结合来实施。 软件模块可以 置于随机存储器 (RAM ) 、 内存、 只读存储器 (ROM ) 、 电可编程 R0M、 电可擦除可编程 R0M、 寄存器、 硬盘、 可移动磁盘、 CD-R0M、 或技术领 域内所公知的任意其它形式的存储介质中。  The steps of a method or algorithm described in connection with the embodiments disclosed herein can be implemented directly in hardware, a software module executed by a processor, or a combination of both. Software modules can be placed in random access memory (RAM), memory, read only memory (ROM), electrically programmable ROM, electrically erasable programmable ROM, registers, hard disk, removable disk, CD-ROM, or technical field Any other form of storage medium known.
以上所述,仅为本发明的具体实施方式,但本发明的保护范围并不 局限于此, 任何熟悉本技术领域的技术人员在本发明揭露的技术范围 内,可轻易想到变化或替换,都应涵盖在本发明的保护范围之内。 因此, 本发明的保护范围应所述以权利要求的保护范围为准。  The above is only the specific embodiment of the present invention, but the scope of the present invention is not limited thereto, and any person skilled in the art can easily think of changes or substitutions within the technical scope of the present invention. It should be covered by the scope of the present invention. Therefore, the scope of the invention should be determined by the scope of the claims.

Claims

权利要求 书 Claim
1、 一种信道估计装置, 其特征在于, 包括:  A channel estimation apparatus, comprising:
矩阵存储单元, 用于存储中间逆矩阵;  a matrix storage unit, configured to store an intermediate inverse matrix;
矩阵获取单元,  Matrix acquisition unit,
两个用户的训练序列号相应的子逆矩阵 Sub-inverse matrix corresponding to the training sequence number of two users
逻辑计算单元,  Logical computing unit,
个用户的训练序列号对应的训练序列生成的矩阵进行乘加逻辑运算, 得 到信道估计因子; The matrix generated by the training sequence corresponding to the training sequence number of the user performs multiplication and addition logic operations to obtain a channel estimation factor;
信道估计单元, 用于根据所述逻辑计算单元计算得到的所述信道估 计因子完成信道估计过程。  And a channel estimation unit, configured to complete a channel estimation process according to the channel estimation factor calculated by the logic calculation unit.
2、 根据权利要求 1所述的信道估计装置, 其特征在于, 所述矩阵存 储单元, 包括:  2. The channel estimation apparatus according to claim 1, wherein the matrix storage unit comprises:
d矩阵存储模块, 用于存储所述中间逆矩阵中的 d矩阵, 所述 d矩阵 为( - ^( )—1^^)-1矩阵; a d matrix storage module, configured to store a d matrix in the intermediate inverse matrix, where the d matrix is a matrix of (-^()- 1 ^^)- 1 ;
逆矩阵存储模块, 用于存储所述中间逆矩阵中的逆矩阵, 所述逆矩 阵为 矩阵;  An inverse matrix storage module, configured to store an inverse matrix in the intermediate inverse matrix, wherein the inverse matrix is a matrix;
其中, 所述 A、 A是由所述训练序列生成的 22 * 5的矩阵。  Wherein A, A are a matrix of 22 * 5 generated by the training sequence.
3、 根据权利要求 2所述的信道估计装置, 其特征在于, 所述 d矩阵 存储模块, 包括:  The channel estimation apparatus according to claim 2, wherein the d matrix storage module comprises:
第一存储子模块, 用于存储所述 d矩阵的全部元素; 或者, 第二存储子模块, 用于存储所述 d矩阵的一半元素; 或者, 第三存储子模块, 用于存储所述两个用户中用户 1 的训练序列号小 于所述两个用户中用户 2的训练序列号的 d矩阵的元素; 或者,  a first storage submodule for storing all the elements of the d matrix; or a second storage submodule for storing half of the elements of the d matrix; or a third storage submodule for storing the two The training sequence number of user 1 among the users is smaller than the element of the d matrix of the training sequence number of user 2 of the two users; or
第四存储子模块, 用于存储所述 d 矩阵的一半元素中, 所述两个用 户中用户 1 的训练序列号小于所述两个用户中用户 2 的训练序列号的元 素。 And a fourth storage submodule, configured to store one of the two elements of the d matrix, wherein the training sequence number of the user 1 of the two users is smaller than the training sequence number of the user 2 of the two users.
4、 根据权利要求 2所述的信道估计装置, 其特征在于, 所述逆矩阵 存储模块, 包括: The channel estimation apparatus according to claim 2, wherein the inverse matrix storage module comprises:
第五存储子模块, 用于存储所述逆矩阵的全部元素; 或者,  a fifth storage submodule, configured to store all elements of the inverse matrix; or
第六存储子模块, 用于存储所述逆矩阵的一半元素。  And a sixth storage submodule, configured to store half of the elements of the inverse matrix.
5、 根据权利要求 2所述的信道估计装置, 其特征在于, 所述矩阵获 取单元, 包括:  The channel estimation apparatus according to claim 2, wherein the matrix obtaining unit comprises:
子逆矩阵获取模块, 用于从所述 d 矩阵存储模块和逆矩阵存储模块 存储的 d 矩阵和逆矩阵中获取所述两个用户的训练序列号相应的子逆矩 阵 ^和 miAnd a sub-inverse matrix obtaining module, configured to obtain the sub-inverse matrices ^ and mi corresponding to the training sequence numbers of the two users from the d matrix and the inverse matrix stored by the d matrix storage module and the inverse matrix storage module.
6、 根据权利要求 5所述的信道估计装置, 其特征在于, 所述逻辑计 算单元, 包括:  The channel estimation apparatus according to claim 5, wherein the logic calculation unit comprises:
第一矩阵计算模块,用于将所述子逆矩阵获取模块获取的子逆矩阵 ^ 和 mi , 与所述训练序列生成的矩阵进行乘加逻辑运算, 得到 ^ +^^矩 阵; a first matrix calculation module, configured to multiply and add a sub-inverse matrix ^ and mi obtained by the sub-inverse matrix acquisition module, and a matrix generated by the training sequence to obtain a ^^^^ matrix;
第二矩阵计算模块,用于将所述子逆矩阵获取模块获取的子逆矩阵 ^ 和 n 与所述训练序列生成的矩阵进行乘加逻辑计算,得到
Figure imgf000024_0001
+Μ矩阵; 拼接模块, 用于将所述第一矩阵计算模块得到的 bT + 矩阵和所 述第二矩阵计算模块得到的 α + Μτ矩阵拼接, 得到所述信道估计因子
a second matrix calculation module, configured to perform multiplication and logical calculation on the sub-inverse matrices ^ and n obtained by the sub-inverse matrix acquisition module and the matrix generated by the training sequence, to obtain
Figure imgf000024_0001
a Μ splicing module, configured to splicing the b T + matrix obtained by the first matrix calculation module and the α + Μ τ matrix obtained by the second matrix calculation module to obtain the channel estimation factor
Figure imgf000024_0002
Figure imgf000024_0002
其中, 所述 b矩阵为
Figure imgf000024_0003
Wherein the b matrix is
Figure imgf000024_0003
.
7、 根据权利要求 6所述的信道估计装置, 其特征在于, 所述第一矩 阵计算模块, 包括:  The channel estimation apparatus according to claim 6, wherein the first matrix calculation module comprises:
第一计算子模块, 用于将所述 A、 A进行乘法逻辑运算, 得到 A矩 阵; a first calculation submodule, configured to perform multiplication logic operations on the A and A to obtain an A moment Array
第二计算子模块, 用于将所述^ ^矩阵与所述子逆矩阵 mi进行乘法逻 辑运算, 得到 t矩阵, 所述 t矩阵为 ^( ); a second calculation submodule, configured to perform a multiplication logic operation on the ^^ matrix and the sub-inverse matrix mi to obtain a t matrix, where the t matrix is ^( );
第三计算子模块, 用于将所述 t 矩阵与所述 矩阵进行乘法逻辑运 算, 得到 b矩阵, 所述 b矩阵为  a third calculation submodule, configured to perform multiplication logic operation on the t matrix and the matrix to obtain a b matrix, where the b matrix is
第四计算子模块, 用于将所述 b 矩阵与所述 A进行乘法逻辑运算, 得到 矩阵;  a fourth calculation submodule, configured to perform a multiplication logic operation on the b matrix and the A to obtain a matrix;
第五计算子模块, 用于将所述 ^矩阵与所述 A进行乘法逻辑运算, 得 到 矩阵;  a fifth calculation submodule, configured to perform a multiplication logic operation on the ^ matrix and the A to obtain a matrix;
第六计算子模块, 用于将所述 矩阵与所述 矩阵进行加法逻辑 运算, 得到所述 矩阵。  And a sixth calculation submodule, configured to perform an addition logic operation on the matrix and the matrix to obtain the matrix.
8、 根据权利要求 6所述的信道估计装置, 其特征在于, 所述第三矩 阵计算模块, 包括:  The channel estimation apparatus according to claim 6, wherein the third matrix calculation module comprises:
第一计算子模块, 用于将所述 、 A进行乘法逻辑运算, 得到 A矩 阵;  a first calculation sub-module, configured to perform multiplication logic operations on the A and the A to obtain an A matrix;
第二计算子模块,用于将所述 A矩阵与所述子逆矩阵 进行乘法逻 辑运算, 得到 t矩阵, 所述 t矩阵为 ^( );  a second calculation submodule, configured to perform a multiplication logic operation on the A matrix and the sub inverse matrix to obtain a t matrix, where the t matrix is ^( );
第三计算子模块, 用于将所述 t 矩阵与所述 矩阵进行乘法逻辑运 算, 得到 b矩阵, 所述 b矩阵为 - *^ ;  a third calculation submodule, configured to perform multiplication logic operation on the t matrix and the matrix to obtain a b matrix, where the b matrix is - *^ ;
第七计算子模块, 用于将所述 b矩阵与所述 t 矩阵进行乘法逻辑运 算, 得到 b*[ ]T矩阵; a seventh calculation submodule, configured to perform a multiplication logic operation on the b matrix and the t matrix to obtain a b*[ ] T matrix;
第八计算子模块, 用于将所述子逆矩阵 mi与所述 b*wT矩阵进行乘法 逻辑运算, 得到"矩阵, 所述《矩阵为 mi +b *[i]T ; An eighth calculation submodule, configured to perform a multiplication logic operation on the sub-inverse matrix mi and the b*w T matrix to obtain a matrix, wherein the matrix is mi +b *[i] T ;
第九计算子模块, 用于将所述《矩阵与所述 进行乘法逻辑运算, 得 到^ 矩阵; a ninth calculation sub-module, configured to perform the multiplication logic operation on the matrix To ^ matrix;
第十计算子模块, 用于将所述 b矩阵与所述 A进行乘法逻辑运算,得 到 b 矩阵;  a tenth calculation submodule, configured to perform a multiplication logic operation on the b matrix and the A to obtain a b matrix;
第十一计算子模块,用于将所述 α 矩阵与所述 Mf矩阵进行加法逻辑 运算, 得到所 矩阵。 The eleventh calculation submodule is configured to perform an addition logic operation on the alpha matrix and the Mf matrix to obtain a matrix.
9、 一种信道估计方法, 其特征在于, 包括: 矩阵;  9. A channel estimation method, comprising: a matrix;
将所述子逆矩阵与所述两个用户的训练序列号对应的训练序列生成 的矩阵进行乘加逻辑运算, 得到信道估计因子;  Performing multiplication and logical operations on the matrix generated by the training matrix corresponding to the training sequence number of the two users, to obtain a channel estimation factor;
根据计算得到的所述信道估计因子完成信道估计过程。  The channel estimation process is completed according to the calculated channel estimation factor.
1 0、 根据权利要求 9 所述的信道估计方法, 其特征在于, 所述从预 先存储的中间逆矩阵中获取两个用户的训练序列号相应的子逆矩阵之 前, 所述方法还包括:  The channel estimation method according to claim 9, wherein the method further comprises: before acquiring the sub-inverse matrix corresponding to the training sequence number of the two users from the pre-stored intermediate inverse matrix, the method further includes:
存储所述中间逆矩阵中的 d 矩阵和逆矩阵, 所述 d 矩阵为 Storing a d matrix and an inverse matrix in the intermediate inverse matrix, wherein the d matrix is
)- 1矩阵, 所述逆矩阵为 1矩阵; ) - 1 matrix, the inverse matrix is a matrix;
其中, 所述 A、 A是由所述训练序列生成的 22 * 5的矩阵。  Wherein A, A are a matrix of 22 * 5 generated by the training sequence.
1 1、 根据权利要求 1 0所述的信道估计方法, 其特征在于, 所述存储 所述中间逆矩阵中的 d矩阵, 包括:  The channel estimation method according to claim 10, wherein the storing the d matrix in the intermediate inverse matrix comprises:
存储所述 d矩阵的全部元素; 或者,  Storing all elements of the d matrix; or
存储所述 d矩阵的一半元素; 或者,  Storing half of the elements of the d matrix; or
存储所述两个用户中用户 1的训练序列号小于所述两个用户中用户 2 的训练序列号的 d矩阵的元素; 或者,  Storing an element of the d matrix of the user 1 whose training sequence number is smaller than the training sequence number of the user 2 of the two users; or
存储所述 d矩阵的一半元素中, 所述两个用户中用户 1 的训练序列 号小于所述两个用户中用户 2的训练序列号的元素。 In storing half of the elements of the d matrix, the training sequence number of the user 1 of the two users is smaller than the element of the training sequence number of the user 2 of the two users.
12、 根据权利要求 10所述的信道估计方法, 其特征在于, 所述存储 所述中间逆矩阵中的逆矩阵, 包括: The channel estimation method according to claim 10, wherein the storing the inverse matrix in the intermediate inverse matrix comprises:
存储所述逆矩阵的全部元素; 或者,  Storing all elements of the inverse matrix; or
存储所述逆矩阵的一半元素。  Stores half of the elements of the inverse matrix.
1 3、 根据权利要求 10所述的信道估计方法, 其特征在于, 所述从预 先存储的中间逆矩阵中获取两个用户的训练序列号相应的子逆矩阵, 包 括:  The channel estimation method according to claim 10, wherein the sub-inverse matrix corresponding to the training sequence numbers of the two users is obtained from the pre-stored intermediate inverse matrix, and includes:
从所述 d 矩阵和逆矩阵中获取所述两个用户的训练序列号相应的子 逆矩阵 ^和  Obtaining a sub-inverse matrix corresponding to the training sequence numbers of the two users from the d matrix and the inverse matrix
14、 根据权利要求 1 3所述的信道估计方法, 其特征在于, 所述将所 述子逆矩阵与所述两个用户的训练序列号对应的训练序列生成的矩阵进 行乘加逻辑运算, 包括:  The channel estimation method according to claim 13, wherein the matrix generated by the training sequence corresponding to the training sequence number of the two users is multiplied and logically operated, including :
将所述子逆矩阵 ^和 mi,与所述训练序列生成的矩阵进行乘加逻辑运 算, 得到 矩阵; And multiplying and adding the sub-inverse matrix ^ and mi to the matrix generated by the training sequence to obtain a matrix;
将所述子逆矩阵 ^和 mi ,与所述训练序列生成的矩阵进行乘加逻辑运 算, 得到 T +^T矩阵; And multiplying and adding the sub-inverse matrix ^ and mi to the matrix generated by the training sequence to obtain a T + ^ T matrix;
将所述 bT + 矩阵与所述 + bAT矩阵拼接, 得到所述信道估计因 子 . And splicing the b T + matrix with the + bA T matrix to obtain the channel estimation factor.
 ,
Figure imgf000027_0001
Figure imgf000027_0001
其中, 所述 b矩阵为
Figure imgf000027_0002
Wherein the b matrix is
Figure imgf000027_0002
.
15、 根据权利要求 14所述的信道估计方法, 其特征在于, 所述将所 述子逆矩阵 d P n 与所述训练序列生成的矩阵进行乘加逻辑运算, 得到 矩阵, 包括:  The channel estimation method according to claim 14, wherein the multiply-adding logical operation of the sub-inverse matrix d P n and the matrix generated by the training sequence to obtain a matrix comprises:
将所述 A、 A进行乘法逻辑运算, 得到 A矩阵; 将所述 A矩阵与所述子逆矩阵 mi进行乘法逻辑运算, 得到 t矩阵, 所述 t矩阵为 ^( ); Performing multiplication logic operations on the A and A to obtain an A matrix; Performing multiplication logic operation on the A matrix and the sub-inverse matrix mi to obtain a t matrix, where the t matrix is ^( );
将所述 t矩阵与所述 ^矩阵进行乘法逻辑运算, 得到 b矩阵, 所述 b 矩阵为 - *^;  Performing multiplication logic operation on the t matrix and the ^ matrix to obtain a b matrix, where the b matrix is - *^;
将所述 b矩阵与所述矩阵 A进行乘法逻辑运算, 得到 矩阵; 将所述 ^矩阵与所述矩阵 A进行乘法逻辑运算, 得到 ^ 矩阵; 将所述 bT 矩阵与所述 ^ 矩阵进行加法逻辑运算, 得到所述 br +4 矩阵。 Performing multiplication logic operation on the b matrix and the matrix A to obtain a matrix; performing multiplication logic operation on the matrix and the matrix A to obtain a ^ matrix; adding the b T matrix and the ^ matrix Logically, the b r +4 matrix is obtained.
16、 根据权利要求 14所述的信道估计方法, 其特征在于, 所述将所 述子逆矩阵 d pn 与所述训练序列生成的矩阵进行乘加逻辑运算, 得到 a +b 矩阵, 包括:  The channel estimation method according to claim 14, wherein the multiply-adding logical operation of the sub-inverse matrix d pn and the matrix generated by the training sequence to obtain an a +b matrix comprises:
将所述 A、 A进行乘法逻辑运算, 得到 A矩阵;  Performing multiplication logic operations on the A and A to obtain an A matrix;
将所述 A矩阵与所述子逆矩阵 mi进行乘法逻辑运算, 得到 t矩阵, 所述 t矩阵为 !^( 4); Performing a multiplication logic operation on the A matrix and the sub-inverse matrix mi to obtain a t matrix, the t matrix being !^(4);
将所述 t矩阵与所述 ^矩阵进行乘法逻辑运算, 得到 b矩阵, 所述 b 矩阵为 - *^;  Performing multiplication logic operation on the t matrix and the ^ matrix to obtain a b matrix, where the b matrix is - *^;
将所述 b矩阵与所述 t矩阵进行乘法逻辑运算, 得到 b*wT矩阵; 将所述子逆矩阵^与所述 b*WT矩阵进行乘法逻辑运算, 得到"矩阵, 所述 "矩阵为 mi+b*[i]T; Performing a multiplication logic operation on the b matrix and the t matrix to obtain a b*w T matrix; performing a multiplication logic operation on the sub inverse matrix ^ and the b*W T matrix to obtain a "matrix, the" matrix For mi +b*[i] T ;
将所述 "矩阵与所述 A进行乘法逻辑运算, 得到 矩阵;  Performing a multiplication logic operation on the matrix and the A to obtain a matrix;
将所述 b矩阵与所述 A进行乘法逻辑运算, 得到 Mf矩阵;  Performing a multiplication logic operation on the b matrix and the A to obtain an Mf matrix;
将所述 a 矩阵与所述 矩阵进行加法逻辑运算, 得到所述 a + bA 矩阵。  Adding the a matrix to the matrix to perform an additive logic operation to obtain the a + bA matrix.
PCT/CN2011/073249 2011-04-25 2011-04-25 Channel estimation method and apparatus WO2011110129A2 (en)

Priority Applications (2)

Application Number Priority Date Filing Date Title
PCT/CN2011/073249 WO2011110129A2 (en) 2011-04-25 2011-04-25 Channel estimation method and apparatus
CN201180000278.3A CN102246478B (en) 2011-04-25 2011-04-25 Channel estimation method and apparatus

Applications Claiming Priority (1)

Application Number Priority Date Filing Date Title
PCT/CN2011/073249 WO2011110129A2 (en) 2011-04-25 2011-04-25 Channel estimation method and apparatus

Publications (2)

Publication Number Publication Date
WO2011110129A2 true WO2011110129A2 (en) 2011-09-15
WO2011110129A3 WO2011110129A3 (en) 2012-04-05

Family

ID=44563908

Family Applications (1)

Application Number Title Priority Date Filing Date
PCT/CN2011/073249 WO2011110129A2 (en) 2011-04-25 2011-04-25 Channel estimation method and apparatus

Country Status (2)

Country Link
CN (1) CN102246478B (en)
WO (1) WO2011110129A2 (en)

Families Citing this family (1)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
CN103427933B (en) * 2012-05-15 2018-06-08 深圳市中兴微电子技术有限公司 Vector processor and its method for generating scrambler sequence

Citations (5)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
CN1351427A (en) * 2000-10-26 2002-05-29 华为技术有限公司 Method and equipment for fast channel estimation with training sequence
US20040234009A1 (en) * 2003-05-23 2004-11-25 Mark Fimoff Best linear unbiased channel estimation for frequency selective multipath channels with long delay spreads
CN1905432A (en) * 2006-07-31 2007-01-31 华为技术有限公司 Method for testing signal in multi-antenna digital wireless communication system
CN101292481A (en) * 2005-09-06 2008-10-22 皇家飞利浦电子股份有限公司 Method and apparatus for estimating channel based on implicit training sequence
WO2009124568A1 (en) * 2008-04-11 2009-10-15 Trident Microsystems (Far East) Ltd. Method and circuit device for correlation and reshaping based channel estimation in a dtmb receiver

Patent Citations (5)

* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
CN1351427A (en) * 2000-10-26 2002-05-29 华为技术有限公司 Method and equipment for fast channel estimation with training sequence
US20040234009A1 (en) * 2003-05-23 2004-11-25 Mark Fimoff Best linear unbiased channel estimation for frequency selective multipath channels with long delay spreads
CN101292481A (en) * 2005-09-06 2008-10-22 皇家飞利浦电子股份有限公司 Method and apparatus for estimating channel based on implicit training sequence
CN1905432A (en) * 2006-07-31 2007-01-31 华为技术有限公司 Method for testing signal in multi-antenna digital wireless communication system
WO2009124568A1 (en) * 2008-04-11 2009-10-15 Trident Microsystems (Far East) Ltd. Method and circuit device for correlation and reshaping based channel estimation in a dtmb receiver

Also Published As

Publication number Publication date
CN102246478B (en) 2013-10-09
WO2011110129A3 (en) 2012-04-05
CN102246478A (en) 2011-11-16

Similar Documents

Publication Publication Date Title
WO2011082681A1 (en) Method and device for resource mapping and code division multiplexing
JP2008512951A (en) Closed loop feedback in MIMO systems
JP2009507400A (en) Wireless communication device
EP3861811A1 (en) Communication apparatus and communication method for channel estimation
WO2011079429A1 (en) Method and device for generating precoding matrix codebook
WO2022007932A1 (en) Signal transmission method, channel estimation method, transmitting end device, and receiving end device
WO2017097269A1 (en) Interference estimation method and device
WO2016090587A1 (en) Data processing method, apparatus and device
WO2016029482A1 (en) Data transmitting method, channel estimation method and device
WO2010124456A1 (en) Data transmitting processing method and apparatus, data receiving processing method and apparatus
CN114285523B (en) Multi-service-demand-oriented large-scale MTC (machine type communication) unlicensed multi-user detection method and system
CN103916340B (en) A kind of noise power estimation method and network equipment
US20190312602A1 (en) Channel coding method, channel coding apparatus, chip system, and storage medium
WO2013117032A1 (en) Method and device for transmitting and receiving channel state information
WO2017059719A1 (en) Data transmission method and device
CN106685625B (en) User channel estimation method
WO2011110129A2 (en) Channel estimation method and apparatus
JP2008278474A (en) Wireless communications apparatus
WO2012103716A1 (en) Detecting method and system for multiple-input multiple-output system
WO2017185935A1 (en) Message transmitting or receiving method and device, and base station
WO2014187356A1 (en) Multiple-input multiple-output (mimo) detection method, apparatus and system for transmitting signal
TWI609577B (en) Method of performing uplink channel estimation and base station using the same
WO2015143602A1 (en) Method for transmitting physical layer data and data transmission device
WO2010121508A1 (en) Grouped decoding method and receiver based on partial interference cancellation
WO2022042259A1 (en) Wireless communication method and apparatus

Legal Events

Date Code Title Description
WWE Wipo information: entry into national phase

Ref document number: 201180000278.3

Country of ref document: CN

121 Ep: the epo has been informed by wipo that ep was designated in this application

Ref document number: 11752872

Country of ref document: EP

Kind code of ref document: A2

NENP Non-entry into the national phase in:

Ref country code: DE

122 Ep: pct application non-entry in european phase

Ref document number: 11752872

Country of ref document: EP

Kind code of ref document: A2