CN1941759A

CN1941759A - Method for balancing orthogonal frequency division multiplexing signals

Info

Publication number: CN1941759A
Application number: CN 200510107164
Authority: CN
Inventors: 吴强; 李继峰
Original assignee: Matsushita Electric Industrial Co Ltd
Current assignee: Panasonic Holdings Corp
Priority date: 2005-09-28
Filing date: 2005-09-28
Publication date: 2007-04-04

Abstract

The present invention provides a method for equalizing an OFDM signal, which is applied to an OFDM system with M transmitting antennas and M receiving antennas, wherein M is greater than or equal to 1, and the method includes steps : Through the decision feedback method, according to the decision of the previous symbol, the interference between symbols is removed, and the signal after the decision feedback is obtained; the first block column of the inverse matrix of the channel matrix is solved by the method of solving the equation system; the decision feedback The signal takes different parts according to the corresponding antenna, and takes the corresponding part from the first column of the channel matrix inverse matrix for linear convolution; and combines the convolution results obtained for each part to obtain an equalized signal. Compared with other equalization methods, the redundant information provided by the cyclic prefix is used, and the complexity is reduced.

Description

An Equalization Method for OFDM Signals

技术领域technical field

本发明涉及正交频分复用(OFDM)系统中的均衡技术，具体涉及一种能减少均衡时间的低复杂度均衡方法。The invention relates to an equalization technique in an Orthogonal Frequency Division Multiplexing (OFDM) system, in particular to a low-complexity equalization method capable of reducing equalization time.

背景技术Background technique

正交频分复用是一种高效的数据传输方式，其基本思想是在频域内将给定信道分成许多正交子信道，在每个子信道上使用一个子载波进行调制，并且各子载波并行传输。这样，尽管总的信道是非平坦的，具有频率选择性，但是每个子信道是相对平坦的，在每个子信道上进行的是窄带传输，信号带宽小于信道的相应带宽，因此就可以大大消除信号波形间的干扰。Orthogonal frequency division multiplexing is an efficient data transmission method. Its basic idea is to divide a given channel into many orthogonal sub-channels in the frequency domain, use a sub-carrier for modulation on each sub-channel, and each sub-carrier is parallel transmission. In this way, although the overall channel is non-flat and has frequency selectivity, each sub-channel is relatively flat, and narrowband transmission is performed on each sub-channel, and the signal bandwidth is smaller than the corresponding bandwidth of the channel, so the signal waveform can be greatly eliminated. Interference between.

正交频分复用相对于一般的多载波传输的不同之处是它允许子载波频谱部分重叠，只要满足子载波间相互正交，则可以从混叠的子载波上分离出数据信号。由于OFDM允许子载波频谱混叠，其频谱效率大大提高，因而是一种高效的调制方式。OFDM适合在多径传播和多普勒频移的无线移动信道中传输高速数据。它能有效对抗多径效应，消除符号间干扰，对抗频率选择性衰落，而且信道利用率高。OFDM技术先后被欧洲数字音频广播(DAB)、欧洲数字视频广播(DVB)、HIPERLAN和IEEE802.11无线局域网等系统采用。The difference between OFDM and general multi-carrier transmission is that it allows sub-carriers to partially overlap, and as long as the sub-carriers are mutually orthogonal, data signals can be separated from the aliased sub-carriers. Because OFDM allows sub-carrier spectrum to alias, its spectral efficiency is greatly improved, so it is an efficient modulation method. OFDM is suitable for transmitting high-speed data in wireless mobile channels with multipath propagation and Doppler frequency shift. It can effectively resist multipath effects, eliminate intersymbol interference, resist frequency selective fading, and have high channel utilization. OFDM technology has been adopted by systems such as European Digital Audio Broadcasting (DAB), European Digital Video Broadcasting (DVB), HIPERLAN and IEEE802.11 wireless local area network.

通常，在OFDM系统中，要加循环前缀(CP)，循环前缀的长度是固定的。但是，一般希望循环前缀的长度越短越好，因此，在不同的环境多径的长度是不同的。如果CP不是很长，在某些环境中，可能因为循环前缀的长度不够用而造成子载波之间的干扰。Usually, in an OFDM system, a cyclic prefix (CP) is added, and the length of the cyclic prefix is fixed. However, it is generally desirable that the length of the cyclic prefix be as short as possible. Therefore, the length of the multipath is different in different environments. If the CP is not very long, in some environments, interference between subcarriers may be caused due to insufficient length of the cyclic prefix.

文献1(J.Zhu.W.Ser，and A.Nehorai，”channel equalizationfor DMT with insufficient cyclic prefix”，Thirty-FourthAsilomar Conference on Signals.Systems and Computers，vol.2，pp.951-955，Oct-Nov.2000)假设在SISO系统中，OFDM符号的长度为N，信道长度为L+1，其信道为h＝[h₀，h₁，…，h_L]，CP长度为P，且L＞P。根据文献1，s^k为第k个OFDM的频域的符号向量，则时域向量为x^k＝IFFT(s^k)；矩阵表示为x^k＝Q^H×s^k；Q为傅立叶变换矩阵。Document 1 (J.Zhu.W.Ser, and A.Nehorai, "channel equalization for DMT with insufficient cyclic prefix", Thirty-Fourth Asilomar Conference on Signals. Systems and Computers, vol.2, pp.951-955, Oct-Nov .2000) Assume that in the SISO system, the length of the OFDM symbol is N, the channel length is L+1, the channel is h=[h ₀ , h ₁ ,..., h _L ], the CP length is P, and L>P . According to Document 1, s ^k is the symbol vector of the k-th OFDM frequency domain, then the time domain vector is x ^k =IFFT(s ^k ); the matrix is expressed as x ^k =Q ^H ×s ^k ; Q is the Fourier transform matrix.

根据文献1，时域的第k个OFDM符号y^k的接收模型为：According to Document 1, the reception model of the kth OFDM symbol y ^k in the time domain is:

y^k＝H₀x^k-H₁x^k+H₂x^k-1+w^k (1)y ^k ＝H ₀ x ^k -H ₁ x ^k +H ₂ x ^k-1 +w ^k (1)

其中H₂为造成OFDM符号之间干扰的信道矩阵(在OFDM系统中，各个OFDM符号之间的干扰一般称作IBI)，H₀-H₁为造成OFDM符号内或一个OFDM符号各个子载波之间干扰的矩阵(在OFDM系统中，一个OFDM符号内的干扰一般称作ICI)。Where H ₂ is the channel matrix that causes interference between OFDM symbols (in an OFDM system, the interference between each OFDM symbol is generally referred to as IBI), H ₀ -H ₁ is the channel matrix that causes interference within an OFDM symbol or between each subcarrier of an OFDM symbol. A matrix of inter-interference (in an OFDM system, the interference within one OFDM symbol is generally called ICI).

公式(1)中，w^k为加性噪声。H₀、H₁、H₂如下所示：In formula (1), w ^k is additive noise. H ₀ , H ₁ , H ₂ are as follows:

如果令：If order:

A＝H₀-H₁ (3)A＝H ₀ -H ₁ (3)

则在文献1中，s^k的估计为：Then in Document 1, the estimate of s ^k is:

${\overset{~ ~}{s the s}}^{k k} = = Q Q {(({H h}_{00} - - {H h}_{11}))}^{- - 11} (({y the y}^{k k} - - {H h}_{22} {Q Q}^{H h} {\overset{~ ~}{s the s}}^{k k - - 11})) = = {QA QA}^{- - 11} (({y the y}^{k k} - - {H h}_{22} {Q Q}^{H h} {\overset{~ ~}{s the s}}^{k k - - 11})) - - - - - - ((44))$

但是，在信道长度大于CP时，由于现有的均衡方法复杂度很高，其不能应用于多输入多输出(MIMO)系统中，只能用于单输入单输出(SISO)系统。However, when the channel length is greater than the CP, due to the high complexity of the existing equalization method, it cannot be applied to a multiple-input multiple-output (MIMO) system, and can only be used in a single-input single-output (SISO) system.

发明内容Contents of the invention

鉴于上述问题，提出了本发明，以在检测OFDM信号的过程中减少均衡操作的复杂度，减少均衡过程所需的时间，以提高数据检测的速度。In view of the above problems, the present invention is proposed to reduce the complexity of the equalization operation in the process of detecting OFDM signals, reduce the time required for the equalization process, and increase the speed of data detection.

在本发明的一个方面，提出了一种一种正交频分复用信号的均衡方法，其应用于具有M根发送天线和M根接收天线的正交频分复用系统中，其中M大于等于1，所述方法包括步骤：通过判决反馈方法，根据前一个符号的判决，去除符号间的干扰，得到判决反馈后的信号；利用用解方程组的方法来求解信道矩阵的逆矩阵的首块列；将判决反馈后的信号根据相对应的天线取不同的部分，并从信道矩阵逆矩阵的首块列取出相对应的部分做线性卷积；以及合并针对各个部分所得到的卷积结果，以得到均衡后的信号。In one aspect of the present invention, a method for equalizing an OFDM signal is proposed, which is applied in an OFDM system with M transmitting antennas and M receiving antennas, wherein M is greater than Equal to 1, the method includes the steps of: through the decision feedback method, according to the judgment of the previous symbol, the interference between symbols is removed, and the signal after the decision feedback is obtained; the first element of the inverse matrix of the channel matrix is solved by the method of solving equations Block column; take different parts of the signal after the decision feedback according to the corresponding antenna, and take the corresponding part from the first block column of the channel matrix inverse matrix for linear convolution; and merge the convolution results obtained for each part , to get the equalized signal.

采用本发明的均衡方法，与现有的方法相比，大大降低了复杂度，提高了均衡的速度。Compared with the existing method, the equalization method of the present invention greatly reduces the complexity and improves the equalization speed.

附图说明Description of drawings

图1示出了带循环前缀的OFDM符号(CP-OFDM符号)的格式；Figure 1 shows the format of an OFDM symbol with a cyclic prefix (CP-OFDM symbol);

图2是多输入多输出频分多路(MIMO-OFDM)系统的框图；FIG. 2 is a block diagram of a Multiple Input Multiple Output Frequency Division Multiplexing (MIMO-OFDM) system;

图3是通用的信道长度大于循环前缀的OFDM系统的数据检测过程的流程图；Fig. 3 is the flow chart of the data detection process of the general channel length greater than the OFDM system of cyclic prefix;

图4是根据本发明的方法对接收的OFDM符号进行均衡操作的流程图；Fig. 4 is a flow chart of performing an equalization operation on received OFDM symbols according to the method of the present invention;

图5示出了图4所示的步骤S402的详细处理过程；Fig. 5 shows the detailed processing procedure of step S402 shown in Fig. 4;

图6是无循环前缀OFDM信号的信道矩阵的示意图；Fig. 6 is a schematic diagram of a channel matrix of an OFDM signal without a cyclic prefix;

图7示出了在无循环前缀的情况下进行均衡操作的流程图；Figure 7 shows a flow chart of equalization operations without a cyclic prefix;

图8示出了图7所示的步骤S701的详细步骤；Fig. 8 shows the detailed steps of step S701 shown in Fig. 7;

图9示出了图7所示的步骤S702的详细步骤。FIG. 9 shows detailed steps of step S702 shown in FIG. 7 .

具体实施方式Detailed ways

1、信道长度大于循环前缀长度的信号接收模型1. Signal reception model with channel length greater than cyclic prefix length

图1给出了有循环前缀(CP)的OFDM符号格式。一个OFDM符号通常由循环前缀和OFDM数据构成。Figure 1 shows the OFDM symbol format with cyclic prefix (CP). An OFDM symbol usually consists of a cyclic prefix and OFDM data.

MIMO-OFDM系统的示意图见图2，MIMO-OFDM系统是OFDM系统上叠加了多根发送和接收天线。一般使用2×2或4×4的MIMO系统。如图2所示，在发送方，由空时编码模块211进行空时编码，然后在串并转换模块212将串行数据转换成与各个发送天线214相对应的符号，并在IFFT模块213进行IFFT操作，在天线214上发送出去。在接收方，首先由接收天线224将符号接收下来，然后在FFT模块223进行FFT操作。经过FFT运算的符号被传送给信道估计模块222，进行信道估计，例如求出信道矩阵，然后在空时解码模块221进行空时解码操作。The schematic diagram of the MIMO-OFDM system is shown in FIG. 2 . The MIMO-OFDM system is that multiple transmitting and receiving antennas are superimposed on the OFDM system. Generally, 2×2 or 4×4 MIMO systems are used. As shown in Figure 2, on the sending side, the space-time encoding is performed by the space-time encoding module 211, and then the serial data is converted into symbols corresponding to each transmitting antenna 214 in the serial-to-parallel conversion module 212, and then performed in the IFFT module 213 IFFT operation, transmitted on antenna 214. On the receiving side, the symbol is first received by the receiving antenna 224 , and then the FFT operation is performed in the FFT module 223 . The symbol after the FFT operation is sent to the channel estimation module 222 for channel estimation, such as obtaining a channel matrix, and then the space-time decoding module 221 performs space-time decoding operation.

图3是通用的信道长度大于循环前缀的OFDM系统的数据检测过程的流程图。如图3所示，在OFDM系统中，首先要进行判决反馈过程S301，这个步骤与现有的方法是相同的。其次要进行均衡过程S302，本发明的方法对现有技术的改进主要集中在这一步骤。最后，要对均衡后的符号执行解调过程S303，以得到解调数据。Fig. 3 is a flow chart of the data detection process of a general OFDM system whose channel length is greater than the cyclic prefix. As shown in FIG. 3 , in the OFDM system, a decision feedback process S301 is first performed, and this step is the same as the existing method. Next, an equalization process S302 is performed, and the improvement of the method of the present invention on the prior art mainly focuses on this step. Finally, a demodulation process S303 is performed on the equalized symbols to obtain demodulated data.

假设M×M的MIMO系统，一个OFDM符号的长度为N，信道长度为L，第i根发送天线到第j根接收天线的信道为h(i，j)＝[h₀(i，j)，h₁(i，j)，…，h_L(i，j)。假设s^k(i)为第i根发送天线的k个OFDM的频域的符号向量，则时域向量为x^k(i)＝IFFT(s^k)；矩阵表示为x^k(i)＝Q^H×s^k(i)；Q为傅立叶变换矩阵。Assuming an M×M MIMO system, the length of an OFDM symbol is N, the channel length is L, and the channel from the i-th transmitting antenna to the j-th receiving antenna is h(i, j)=[h ₀ (i, j) , h ₁ (i, j), ..., h _L (i, j). Suppose s ^k (i) is the symbol vector of k OFDM frequency domain of the i-th transmitting antenna, then the time domain vector is x ^k (i)=IFFT(s ^k ); the matrix is expressed as x ^k (i)=Q ^H ×s ^k (i); Q is the Fourier transform matrix.

在第k个OFDM符号周期内，时域接收模型为：In the k-th OFDM symbol period, the time-domain reception model is:

y^k＝HX^k+H_IBIX^k-1+w^k (5)y ^k ＝HX ^k +H _IBI X ^k-1 +w ^k (5)

在公式(5)中，y^k为M×(N+P)的向量(P为循环前缀长度)，每根天线都接收了N个数据，令y_j ^k(t)为第j根接收天线的第k个符号在第t个时刻的接收数据。则y^k的第1:M(表示1到M)个元素为y₁ ^k(1)，…y_M ^k(1)；第M+1:2M个元素为y₁ ^k(2)，…y_M ^k(2)；第(i-1)×M+1:i×M个元素为y₁ ^k(i)，…y_M ^k(i)。即，y^k的第(i-1)×M+1:i×M个元素依次为第k个OFDM符号周期内第1:M根接收天线的第i个接收向量，顺序为：第1根天线的第i个接收信号，第2根天线的第i个接收信号，…，第M根天线的第i个接收信号。In formula (5), y ^k is a vector of M×(N+P) (P is the length of the cyclic prefix), each antenna has received N data, let y _j ^k (t) be the jth receiving antenna The received data of the k-th symbol at the t-th time. Then the 1st:M (indicating 1 to M) elements of y ^k are y ₁ ^k (1), ...y _M ^k (1); the M+1:2M elements are y ₁ ^k (2), ...y _M ^k (2); (i-1)×M+1: i×M elements are y ₁ ^k (i), ... y _M ^k (i). That is, the (i-1)×M+1:i×M elements of ^y k are in turn the i-th receiving vectors of the 1st:M receiving antennas in the k-th OFDM symbol period, and the order is: the 1st The i-th receiving signal of the antenna, the i-th receiving signal of the second antenna, ..., the i-th receiving signal of the M-th antenna.

在公式(5)中，矩阵H为OFDM符号的信道矩阵，其通过现有的信道估计方法而求得：In formula (5), the matrix H is the channel matrix of the OFDM symbol, which is obtained by the existing channel estimation method:

在信道矩阵(6)中，H_k为M×M矩阵(0＜＝k＜＝L)，H_k的i，j个元素为第i根发送天线到第j根接收天线的第k径的信道。H是一个M(N+P)行，M(N+P)列的矩阵。In the channel matrix (6), H _k is an M×M matrix (0<=k<=L), and the i and j elements of H _k are the k-th path from the i-th transmitting antenna to the j-th receiving antenna channel. H is a matrix with M(N+P) rows and M(N+P) columns.

在公式(5)中，H_IBI表示前一个符号对当前的符号造成干扰的干扰矩阵：In formula (5), H _IBI represents the interference matrix that the previous symbol interferes with the current symbol:

在矩阵(7)中，H_k(0＜＝k＜＝L)的含义与(6)中的相同。In matrix (7), the meaning of H _k (0<=k<=L) is the same as in (6).

在公式(5)中，X^k为第k个OFDM的符号周期内的时域发送向量，它是M(N+P)×1维的列向量。每根发送天线都发送了N+P个数据。In formula (5), X ^k is the time-domain transmission vector in the k-th OFDM symbol period, which is an M(N+P)×1-dimensional column vector. Each transmit antenna transmits N+P pieces of data.

令z_i ^k(t)为第i根发送天线的第k个OFDM符号周期内的第t个时刻的发送数据，则X^k的第1:M个元素为z₁ ^k(1)，…z_M ^k(1)；第M+1:2M个元素为z₁ ^k(2)，…z_M ^k(2)；第(i-1)×M+1:i×M个元素为z₁ ^k(i)…z_M ^k(i)。即，X^k的第(i-1)×M+1:i×M个元素依次为第k个OFDM符号周期内第1:M根发送天线的第i个发送向量，顺序为：第1根天线的第i个发送信号，第2根天线的第i个接收信号，…，第M根天线的第i个接收信号。Let z _i ^k (t) be the transmission data at the t-th moment in the k-th OFDM symbol period of the i-th transmit antenna, then the 1:M elements of X ^k are z ₁ ^k (1),...z _M ^k (1); the M+1:2M element is z ₁ ^k (2), ...z _M ^k (2); the (i-1)×M+1:i×M element is z ₁ ^k (i)...z _M ^k (i). That is, the (i-1)×M+1:i×M elements of X ^k are the i-th transmit vectors of the 1st:M transmit antennas in the k-th OFDM symbol period, and the order is: the 1st The i-th antenna sends a signal, the i-th antenna receives a signal, ..., the i-th antenna receives a signal.

由于CP段里的内容为循环前缀，所以X^k可以进一步表示(为了简明，以下忽略上标k)，如下：对第i根发送天线来说，待发送的数据为s_i＝[s_i(1)s_i(2)，…，s_i(N)]^T。则时域x_i＝IFFT(s_i)。x_i＝[x_i(1)x_i(2)，…，x_i(N)]^T。然后添加CP，添加CP后的x_i(CP)＝[x_i(N-P+1)，…，x_i(N)，x_i(1)x_i(2)，…，x_i(N)]^T。x_i(CP)是个N+P的向量。一共有M根发送天线。令x_i(j)为x_i的第j个元素，将第j个时刻各个天线的发送信号组成向量，得到x(j)＝[x₁(j)，x₂(j)，…，x_M(j)]^T，x(j)为M×1向量。则X^k＝[x(N-P+1)^T，…，x(N)^T，x(1)^T，x(2)^T，…，x(N)^T]^T。X^k为M×(N+P)的向量。前MP个为M根发送天线的循环前缀。Since the content in the CP segment is a cyclic prefix, X ^k can be further expressed (for simplicity, superscript k is ignored below), as follows: For the i-th transmitting antenna, the data to be transmitted is s _i =[s _i ( 1) s _i (2),..., s _i (N)] ^T . Then the time domain x _i =IFFT(s _i ). x _i =[ _xi (1) x _i (2), . . . , x _i (N)] ^T . Then add CP, after adding CP, x _i (CP)=[ _xi (N-P+1),..., _xi (N), _xi (1) _xi (2),..., _xi (N )] ^T. x _i (CP) is a vector of N+P. There are M transmitting antennas in total. Let x _i (j) be the jth element of x _i , and form a vector of the transmitted signals of each antenna at the jth moment, and obtain x(j)=[x ₁ (j), x ₂ (j),..., x _M (j)] ^T , x(j) is an M×1 vector. Then X ^k =[x(N-P+1) ^T , . . . , x(N) ^T , x(1) ^T , x(2) ^T , . . . , x(N) ^T ] ^T . X ^k is a vector of M×(N+P). The first MP are the cyclic prefixes of the M transmit antennas.

2、均衡方法2. Balance method

图4为根据本发明实施例的均衡方法的流程图。Fig. 4 is a flowchart of an equalization method according to an embodiment of the present invention.

在步骤S401，根据前一个符号的估计，得到

(当前为k)，经过判决反馈，得到当前消除符号间干扰的接收信号的估计。表示为In step S401, according to the estimation of the previous symbol, get

(currently k), through decision feedback, obtain the current estimate of the received signal for eliminating inter-symbol interference. Expressed as

$r r = = {y the y}^{k k} - - {H h}_{IBI IBI} {\overset{^^}{X x}}^{k k - - 11} - - - - - - ((88))$

在步骤S402，利用如后所述的快速算法得到迫零均衡解b，其公式表达为：In step S402, the zero-forcing equilibrium solution b is obtained using a fast algorithm as described later, and its formula is expressed as:

b＝H^-1r (9)b＝H ^- 1r (9)

在步骤S403，在有CP的情况下，利用CP段，合并均衡信号b，其过程如下：In step S403, in the case of CP, use the CP segment to combine the equalized signal b, and the process is as follows:

由于时域的发送向量为X^k＝[x(N-P+1)^T，…，x(N)^T，x(1)^T，x(2)^T，…，x(N)^T]^T，b为X^k的估计，则：Since the sending vector in the time domain is X ^k =[x(N-P+1) ^T ,...,x(N) ^T , x(1) ^T , x(2) ^T ,...,x(N) ^T ] ^T , b is the estimation of X ^k , then:

b＝[(N-P+1)^T，...，(N)^T，(1)^T，(2)^T，...，(N)^T]^T(10)b=[(N-P+1) ^T ,...,(N) ^T ,(1) ^T ,(2) ^T ,...,(N) ^T ] ^T (10)

通过式(10)，可以看出，b的前MP个值和后MP个值都是对[x(N-P+1)^T，…，x(N)^T]的估计，则：From formula (10), it can be seen that the first and last MP values of b are estimates of [x(N-P+1) ^T ,..., x(N) ^T ], then:

[(N-P+1)^T，...，(N)^T]^T＝α×b[1:MP]+β×b[M(N-P)+1:MN](11)[(N-P+1) ^T ,...,(N) ^T ] ^T ＝α×b[1:MP]+β×b[M(NP)+1:MN](11)

在式(11)中，α、β为对两种估计值的加权值，满足α+β＝1，一般情况没有可靠性的先验知识，取α＝β＝0.5。合并后得到：In formula (11), α and β are the weighted values of the two estimated values, satisfying α+β=1, and generally there is no prior knowledge of reliability, so α=β=0.5. After merging we get:

v＝[(1)^T，(2)^T，...，(N)^T]^T (12)v=[(1) ^T ,(2) ^T ,...,(N) ^T ] ^T (12)

在式(12)中，v为MN向量，其后MP个值是经过合并的。In formula (12), v is an MN vector, and then MP values are merged.

在步骤S404，将各个天线的时域合并后的信号，变换到频域，解调得到发送符号。在式(12)中，u为各个天线时域上发送信号的估计。u为MN向量，根据u，得到第一根发送天线的时域发送信号的估计为：In step S404, the combined signals in the time domain of each antenna are transformed into the frequency domain, and demodulated to obtain the transmitted symbols. In formula (12), u is the estimate of the transmitted signal in the time domain of each antenna. u is the MN vector, according to u, the estimate of the time-domain transmitted signal of the first transmitting antenna is obtained as:

₁＝[₁(1)₁(2)，...，₁(N)^T]^T＝[v(1:M:M(N-1)+1)] (13) ₁ ＝[ ₁ (1) ₁ (2),..., ₁ (N) ^T ] ^T ＝[v(1:M:M(N-1)+1)] (13)

在式(13)中，1:M:M(N-1)+1，表示从1开始，每隔M取一个，一直取到M(N-1)+1。即₁＝[v(1)u(M+1)u(2M+1)，…，u(M(N-1)+1)]]^T。In formula (13), 1:M:M(N-1)+1 means starting from 1 and taking every M until M(N-1)+1. That is,  ₁ =[v(1)u(M+1)u(2M+1), . . . , u(M(N-1)+1)]] ^T .

同样，第i根发送天线时域发送信号的估计为Similarly, the estimate of the time-domain transmitted signal of the i-th transmit antenna is

_i＝[v(i:M:M(N-1)+i)] (14) _i = [v(i:M:M(N-1)+i)] (14)

然后，对第_i分别进行FFT运算得到：Then, perform FFT operation on the  _ith respectively to get:

_i＝FFT(_i) (15) _i ＝ FFT( _i ) (15)

接下来，对_i的N个值分别解调得到频域发送的调制符号的估计。在步骤S401-S404中，除了S402外，其他的都可用快速傅立叶变换(FFT)来实现，以降低复杂度。Next, the N values of  _i are respectively demodulated to obtain an estimate of the modulation symbol sent in the frequency domain. In steps S401-S404, except for step S402, other steps can be implemented by fast Fourier transform (FFT) to reduce complexity.

以下对在步骤S402所使用的快速算法进行详细说明。The fast algorithm used in step S402 will be described in detail below.

3、现有计算S402中公式(9)的快速算法3. The existing fast algorithm for calculating formula (9) in S402

根据文献2‘《快速算法》，蒋增荣，国防科技大学出版社，1998年7月’设：According to document 2 "Fast Algorithm", Jiang Zengrong, National University of Defense Technology Press, July 1998':

$T T = = [\begin{matrix} {t t}_{00} & {t t}_{11} & {t t}_{22} & \cdot \cdot \cdot \cdot \cdot \cdot & {t t}_{n no - - 11} \\ {t t}_{- - 11} & {t t}_{00} & {t t}_{11} & \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; & {t t}_{n no - - 22} \\ \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; & \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; & \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; & \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; & \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; \\ {t t}_{- - n no + + 11} & {t t}_{- - n no + + 22} & {t t}_{- - n no + + 11} & \cdot &Center Dot; \cdot &Center Dot; \cdot &Center Dot; & {t t}_{00} \end{matrix}] - - - - - - ((1616))$

其中子块t_i(i＝-n，-n+1，…，n)是m阶方阵，则T为基维为m×m的n阶分块Toeplitz矩阵。分块Toeplitz矩阵的特点是分块对角线上的各个元素彼此相等，平行于分块主对角线上的各分块对角线上的元素也彼此相等。用数学式来表示即为T＝t_ij＝t_j-i(j，i＝1，…n)。即T可以用首行块和首列块来表示。上述文献中给出了分块Toeplitz矩阵求逆的快速算法，复杂度为0(n²m²)。Wherein the sub-blocks t _i (i=-n, -n+1, ..., n) are square matrices of order m, and T is an order-n block Toeplitz matrix with a base dimension of m×m. The characteristic of the block Toeplitz matrix is that each element on the block diagonal is equal to each other, and the elements on each block diagonal parallel to the main diagonal of the block are also equal to each other. Expressed in a mathematical formula, it is T=t _ij =t _ji (j, i=1, . . . n). That is, T can be represented by the first row block and the first column block. A fast algorithm for inverting the block Toeplitz matrix is given in the above literature, and the complexity is 0(n ² m ² ).

从式(6)中可以看出，矩阵H为分块Toeplitz矩阵。在步骤S402中，对应了两个子步骤：It can be seen from formula (6) that the matrix H is a block Toeplitz matrix. In step S402, two sub-steps are corresponding:

●求H^-1，H是M(N+P)阶的方阵，直接求复杂度0(M³(N+P)³)，运用上述文献中的快速算法，复杂度0(M²(N+P)²)。● Find H ^-1 , H is a square matrix of order M(N+P), directly find the complexity 0(M ³ (N+P) ³ ), using the fast algorithm in the above literature, the complexity 0(M ² ( N+P) ² ).

●求H^-1r，直接进行相乘运算，复杂度0(M²(N+P)²)。● Calculate H ^-1 r, directly perform multiplication operation, the complexity is 0(M ² (N+P) ² ).

4、本发明的快速算法4, the fast algorithm of the present invention

根据本说明书的附录1，有定理：分块下三角Toeplitz矩阵T的逆矩阵T^-1，也为分块下三角Toeplitz矩阵。According to Appendix 1 of this specification, there is a theorem: the inverse matrix T ⁻¹ of the block lower triangular Toeplitz matrix T is also the block lower triangular Toeplitz matrix.

由于H是分块下三角Toeplitz矩阵(基维为M×M的N阶分块下三角Toeplitz矩阵)，所以B＝H^-1为分块下三角Toeplitz矩阵，它可由首列块来表示，下面以首列块为例进行说明。Since H is a block lower triangular Toeplitz matrix (the base dimension is an N-order block lower triangular Toeplitz matrix of M×M), so B=H ^-1 is a block lower triangular Toeplitz matrix, which can be represented by the first column block, as follows Take the first block as an example for illustration.

由于HB＝I_(N+P)M(I_(N+P)M为(N+P)M阶单位阵)，I_(N+P)M同样可表示为基维为M×M的(N+P)阶分块下三角Toeplitz矩阵，如下所示：Since HB=I _(N+P)M (I _(N+P)M is the unit matrix of (N+P)M order), I _(N+P)M can also be expressed as (N +P) order block lower triangular Toeplitz matrix, as follows:

在式(17)中，I_M为M×M的单位阵，0_M为M×M的全零矩阵。In formula (17), I _M is an M×M unit matrix, and 0 _M is an M×M all-zero matrix.

上述矩阵I_(N+P)M的首列块V为：The first block V of the above matrix I _(N+P)M is:

$V V = = [\begin{matrix} {I I}_{M m} \\ 00_{M m} \\ . . \\ . . \\ . . \\ 00_{M m} \end{matrix}] - - - - - - ((1818))$

其中，V为(N+P)M×M的矩阵。Wherein, V is a matrix of (N+P)M×M.

下面给出其计算过程：The calculation process is given below:

◆第一步，求B＝H^-1，如下：◆The first step is to find B＝H ^-1 , as follows:

令U为B的第一列块，Let U be the first column block of B,

$V V = = [\begin{matrix} {U u}_{00} \\ {U u}_{11} \\ . . \\ . . \\ . . \\ {U u}_{N N + + P P - - 11} \end{matrix}] - - - - - - ((1919))$

则：but:

HU＝V(20)

这可以通过解方程组的方法来求出。This can be found by solving a system of equations.

由于在OFDM的系统中，N＞＞L+1，矩阵H中零块比较多。根据式(6)、(18)、(19)、(20)，建立方程组：Because in the OFDM system, N>>L+1, there are more zero blocks in the matrix H. According to the formulas (6), (18), (19), (20), the equations are established:

H₀ U₀＝I_M；(21-1)H ₀ U ₀ = I _M ; (21-1)

H₁ U₀+H₀ U₁＝0_M；(21-2)H ₁ U ₀ +H ₀ U ₁ =0 _M ; (21-2)

…...

H₀ U_i+H₁U_i-1+…+H_LU_i-L＝0(12-i)H ₀ U _i +H ₁ U _i-1 +…+H _L U _iL ＝0(12-i)

首先，根据式(12-1)，求出U₀，然后将U₀代入(21-2)，求出U₁，求U_i时，将u_i-1，…，u_i-L代入。最后得到：First, calculate U ₀ according to formula (12-1), then substitute U ₀ into (21-2) to calculate U ₁ , and when calculating U _i , substitute u _i-1 ,..., u _iL . Finally got:

${U u}_{00} = = {H h}_{00}^{- - 11} - - - - - - ((22 twenty two))$

${U u}_{i i} = = - - {H h}_{00}^{- - 11} (({Σ Σ}_{k k = = 11}^{L L} {U u}_{k k} {U u}_{i i - - k k})) 11 < < = = i i < < = = N N + + P P - - 11 - - - - - - ((23 twenty three))$

在式(23)求和项中，所有的下标不小于0，否则下标小于0的项视为0。In the summation items of formula (23), all subscripts are not less than 0, otherwise the items with subscripts less than 0 are regarded as 0.

【复杂度分析】【Complexity Analysis】

●需要计算H₀，H₁，…，H_L的逆矩阵，复杂度为0(L+1)M³)。● It is necessary to calculate the inverse matrix of H ₀ , H ₁ , . . . , H _L , and the complexity is 0(L+1)M ³ ).

●根据H₀，H₁，…，H_L的逆矩阵计算U，总的复杂度为0((N+P)M³)。两者之和的复杂度0((N+P)M³)。● Calculate U according to the inverse matrix of H ₀ , H ₁ , . . . , _HL , and the total complexity is 0 ((N+P)M ³ ). The complexity of the sum of the two is 0((N+P)M ³ ).

◆第二步，计算b＝H^-1r，如下：◆The second step, calculate b=H ^-1 r, as follows:

计算公式(9)，如果直接求解，复杂度为0((N+P)²(M)²)。Calculation formula (9), if solved directly, the complexity is 0 ((N+P) ² (M) ² ).

以下，利用H^-1是分块下三角Toeplitz矩阵，来降低计算的复杂度，其中，可以将H^-1分成M×M个矩阵之和，每个子矩阵都具有类似下三角Toeplitz矩阵的特征。In the following, H ^-1 is a block lower triangular Toeplitz matrix to reduce the computational complexity, wherein H ^-1 can be divided into the sum of M×M matrices, and each sub-matrix has characteristics similar to the lower triangular Toeplitz matrix.

第i，j个矩阵C_i，j(维数与H^-1一样)与H^-1的关系为：The relationship between the i, j-th matrix C _{i, j} (the dimension is the same as H ^-1 ) and H ^-1 is:

C_ij的(p-1)×M+i行(1＜＝p＜＝N+P)，(q-1)×M+j列(1＜＝q＜＝N+P)的元素为H^-1的(p-1)×M+i行，(q-1)×M+j列的元素，其他为零。举例如下：The (p-1)×M+i row (1<=p<=N+P) of C _ij , the element of (q-1)×M+j column (1<=q<=N+P) is H Elements of (p-1)×M+i rows, (q-1)×M+ ^j columns of -1, and zero for others. Examples are as follows:

假设B的H^-1是一个基维是2×2的3阶分块下三角矩阵，则Assuming that H ^-1 of B is a 3-order block lower triangular matrix whose base dimension is 2×2, then

$B B = = [\begin{matrix} {B B}_{11} \\ {B B}_{22} & {B B}_{11} \\ {B B}_{33} & {B B}_{22} & {B B}_{11} \end{matrix}] = = [\begin{matrix} {a a}_{11} & {a a}_{22} & 00 & 00 & 00 & 00 \\ {a a}_{33} & {a a}_{44} & 00 & 00 & 00 & 00 \\ {a a}_{55} & {a a}_{66} & {a a}_{11} & {a a}_{22} & 00 & 00 \\ {a a}_{77} & {a a}_{88} & {a a}_{33} & {a a}_{44} & 00 & 00 \\ {a a}_{99} & {a a}_{1010} & {a a}_{55} & {a a}_{66} & {a a}_{11} & {a a}_{22} \\ {a a}_{1111} & {a a}_{1212} & {a a}_{77} & {a a}_{88} & {a a}_{33} & {a a}_{44} \end{matrix}] - - - - - - ((24 twenty four))$

则：but:

${C C}_{1111} = = [\begin{matrix} {a a}_{11} & 00 & 00 & 00 & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ {a a}_{55} & 00 & {a a}_{11} & 00 & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ {a a}_{99} & 00 & {a a}_{55} & 00 & {a a}_{11} & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \end{matrix}] - - - - - - ((2525 - - 11))$

${C C}_{21 twenty one} = = [\begin{matrix} 00 & 00 & 00 & 00 & 00 & 00 \\ {a a}_{33} & 00 & 00 & 00 & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ {a a}_{77} & 00 & {a a}_{33} & 00 & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ {a a}_{1111} & 00 & {a a}_{77} & 00 & {a a}_{33} & 00 \end{matrix}] - - - - - - ((2525 - - 22))$

${C C}_{1212} = = [\begin{matrix} 00 & {a a}_{22} & 00 & 00 & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ 00 & {a a}_{66} & 00 & {a a}_{22} & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ 00 & {a a}_{1010} & 00 & {a a}_{66} & 00 & {a a}_{22} \\ 00 & 00 & 00 & 00 & 00 & 00 \end{matrix}] - - - - - - ((2525 - - 33))$

${C C}_{22 twenty two} = = [\begin{matrix} 00 & 00 & 00 & 00 & 00 & 00 \\ 00 & {a a}_{44} & 00 & 00 & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ 00 & {a a}_{88} & 00 & {a a}_{44} & 00 & 00 \\ 00 & 00 & 00 & 00 & 00 & 00 \\ 00 & {a a}_{1212} & 00 & {a a}_{88} & 00 & {a a}_{44} \end{matrix}] - - - - - - ((2525 - - 44))$

其中：in:

$B B = = {Σ Σ}_{i i = = 11}^{M m} {Σ Σ}_{j j = = 11}^{M m} {C C}_{ij ij} - - - - - - ((2626))$

则B与一个向量r的乘积可表示如下：Then the product of B and a vector r can be expressed as follows:

$d d = = Br Br = = {Σ Σ}_{i i = = 11}^{M m} {Σ Σ}_{j j = = 11}^{M m} {C C}_{ij ij} r r - - - - - - ((2727))$

接上例，令e＝C₁₁×r，e＝[e₁，e₂，e₃，e₄，e₅，e₆]^T，则e₂＝e₄＝e₆＝0，且：Continuing from the above example, let e=C ₁₁ ×r, e=[e ₁ , e ₂ , e ₃ , e ₄ , e ₅ , e ₆ ] ^T , then e ₂ =e ₄ =e ₆ =0, and:

$[\begin{matrix} {e e}_{11} \\ {e e}_{33} \\ {e e}_{55} \end{matrix}] = = [\begin{matrix} {a a}_{11} \\ {a a}_{55} & {a a}_{11} \\ {a a}_{99} & {a a}_{55} & {a a}_{11} \end{matrix}] [\begin{matrix} {r r}_{11} \\ {r r}_{33} \\ {r r}_{55} \end{matrix}] = = {T T}_{1111} [\begin{matrix} {r r}_{11} \\ {r r}_{33} \\ {r r}_{55} \end{matrix}] - - - - - - ((2828))$

在式(28)中， $T_{11} = [\begin{matrix} a_{1} \\ a_{5} & a_{1} \\ a_{9} & a_{5} & a_{1} \end{matrix}]$ 为下三角Toeplitz矩阵。由于T₁₁是下三角Toeplitz矩阵，所以式(28)中T₁₁与[r₁，r₃，r₅]^T的乘积可以看成是T₁₁的第1列与[r₁，r₃，r₅]^T做线性卷积，得到结果后取前3个值，而线性卷积可以用FFT来计算。其他的C_ij有同样的特性。In formula (28), $T_{11} = [\begin{matrix} a_{1} \\ a_{5} & a_{1} \\ a_{9} & a_{5} & a_{1} \end{matrix}]$ is a lower triangular Toeplitz matrix. Since T ₁₁ is a lower triangular Toeplitz matrix, the product of T ₁₁ and [r ₁ , r ₃ , r ₅ ] ^T in formula (28) can be regarded as the first column of T ₁₁ and [r ₁ , r ₃ , r ₅ ] ^T does linear convolution, and takes the first 3 values after getting the result, and linear convolution can be calculated by FFT. Other C _ij have the same characteristics.

综上所述，计算H^-1r的快速算法如下：To sum up, the fast algorithm for computing H ^-1 r is as follows:

1)对C_ij来说(1＜＝i，j＜＝M)，得到一个(N+P)×1的向量vec_ij，此向量是(28)中，T_ij的第一列。vec_ij的取法如下：1) For C _ij (1<=i, j<=M), get a (N+P)×1 vector vec _ij , which is the first column of T _ij in (28). The method of vec _ij is as follows:

vec_ij＝H^-1(i:M:i+(N+P-1)*M，j) (29)vec _ij = H ^-1 (i:M:i+(N+P-1)*M, j) (29)

2)从r中M(N+P)维向量，中取N+P个作为r_vec_ij，取法如下：2) From the M(N+P) dimensional vector in r, take N+P as r_vec _ij , the method is as follows:

r_vec_ij＝r(j:M:j+(N+P-1)*M) (30)r_vec _ij = r(j:M:j+(N+P-1)*M) (30)

3)即r_vec_ij是从r的第j个元素开始，以间隔M来取值。3) That is, r_vec _ij starts from the jth element of r and takes values at an interval of M.

4)将r_vec_ij补N+P个零得到r_e_ij，将vec_ij补N+P个零得到vec_e_ij；4) Complement r_vec _ij with N+P zeros to obtain r_e _ij , and supplement vec _ij with N+P zeros to obtain vec_e _ij ;

f_ij＝ifft(fft(r_vec_ij).*fft(vec_ij)) (31)f _ij = ifft(fft(r_vec _ij ).*fft(vec _ij )) (31)

5)然后令u_ij为f_ij的前(N+P)个值。5) Then let u _ij be the first (N+P) values of f _ij .

6)初始化part_ij为M(N+P)×1的全零列向量。赋值：6) Initialize part _ij as an all-zero column vector of M(N+P)×1. assignment:

part_ij(i:M:i+(N+P-1)*M，1)＝u_ij (32)part_ij从i个元素开始，每隔M个，被赋予u_ij对应的值，即part_ij(i，1)＝u_ij(1)，part_ij(i+M，1)＝u_ij(2)，part_ij(i+2M，1)＝u_ij(3)，…。part _ij (i:M:i+(N+P-1)*M, 1)=u _ij (32)part _ij starts from i elements, every M, is given the corresponding value of u _ij , that is, part _ij (i, 1) = u _ij (1), part _ij (i+M, 1) = u _ij (2), part _ij (i+2M, 1) = u _ij (3), . . .

7)得到所有的part_ij之后，计算：7) After getting all part _ij , calculate:

$b b = = {H h}^{- - 11} r r = = {B B}^{- - 11} r r = = {Σ Σ}_{i i = = 11}^{M m} {Σ Σ}_{j j = = 11}^{M m} {part part}_{ij ij} - - - - - - ((3333))$

【复杂度分析】【Complexity Analysis】

每计算一次part_ij，复杂度为0(2(N+P)log₂(2(N+P))，一共要计算M²次，总的复杂度为0(M²2(N+P)log₂(2(N+P))。Every time part _ij is calculated, the complexity is 0(2(N+P)log ₂ (2(N+P)), a total of M ² calculations are required, and the total complexity is 0(M ² 2(N+P) log ₂ (2(N+P)).

5、本发明的方法与现有技术的复杂度比较5. The method of the present invention is compared with the complexity of the prior art

当CP长度大于(信道长度-1)时，无符号间干扰和子载波之间的干扰，其信道在频域为单径信道，MIMO检测在频域进行，迫零算法的复杂度为0(M³N)。When the CP length is greater than (channel length-1), there is no inter-symbol interference and interference between subcarriers, the channel is a single-path channel in the frequency domain, MIMO detection is performed in the frequency domain, and the complexity of the zero-forcing algorithm is 0(M ³ N).

表1给出了CP不足时迫零算法的比较。本发明现有快速算法未使用快速算法复杂度 Max{0(M²2(N+P)log₂(2(N+P))，0((N+P)M³)) 0((N+P)²M²) 0((N+P)³M³) Table 1 shows the comparison of zero-forcing algorithms when CP is insufficient. this invention Existing Fast Algorithms fast algorithm not used the complexity Max{0(M ² 2(N+P)log ₂ (2(N+P)), 0((N+P)M ³ )) 0((N+P) ² M ² ) 0((N+P) ³ M ³ )

第一实例(M×M正交频分复用系统)The first example (M×M OFDM system)

以下描述CP不为0的情况下的具体均衡过程。The specific equalization process in the case where CP is not 0 is described below.

1)在步骤S401，根据前一个符号的估计，得到

(当前为k)，根据式(8)，经过判决反馈，得到当前消除符号间干扰的接收信号的估计。1) In step S401, according to the estimation of the previous symbol, get

(currently k), according to formula (8), through decision feedback, the current estimate of the received signal for eliminating inter-symbol interference is obtained.

2)在步骤S402，按照式(9)，利用快速算法得到迫零均衡解b。由于计算步骤S402的快速算法说明起来比较复杂，将在后面单独给出。2) In step S402, according to formula (9), a zero-forcing equilibrium solution b is obtained using a fast algorithm. Since the description of the fast algorithm of the calculation step S402 is complicated, it will be given separately later.

3)在步骤S403，利用CP段，合并均衡信号b，其详细过程如下：3) In step S403, use the CP segment to combine the equalized signal b, the detailed process is as follows:

b为M(N+P)×1的列向量，令v为MN×1的列向量，则有：b is a column vector of M(N+P)×1, let v be a column vector of MN×1, then:

v(1:M(N-P))＝b(1+MP:MP+M(N-P)) (34-1)v(1:M(N-P))＝b(1+MP:MP+M(N-P)) (34-1)

v(M(N-P)+1:MN)＝0.5×(b(1:MP)+b(MN+1:M(N+P))) (34-2)v(M(N-P)+1:MN)＝0.5×(b(1:MP)+b(MN+1:M(N+P))) (34-2)

4)在步骤S404，将各个天线的时域合并后的信号，变换到频域，解调得到发送符号。对第i根发送天线时域发送信号的估计为_i＝[v(i:M:M(N-1)+i)]，(i:M:M(N-1)+i表示从第i个开始，每隔M取一个，一直取到M(N-1)+I。然后对第_i分别进行FFT运算得到_i＝FFT(_i)。然后对_i的N个值分别解调得到频域发送的调制符号的估计。4) In step S404, transform the time-domain combined signals of each antenna into the frequency domain, and demodulate to obtain the transmitted symbols. The estimation of the time-domain transmission signal of the i-th transmitting antenna is  _i =[v(i:M:M(N-1)+i)], (i:M:M(N-1)+i means from the i start, get one every M, always get to M(N-1)+I. Then carry out FFT operation to get  _i =FFT( _i ) to  _i respectively. Then to the N values of  _i respectively Demodulation yields an estimate of the modulation symbols sent in the frequency domain.

【步骤S402的实现】[Implementation of Step S402]

根据式(9)，在步骤S02计算b＝H^-1r。其计算步骤如图7所示。According to formula (9), b=H ⁻¹ r is calculated in step S02. Its calculation steps are shown in Figure 7.

在步骤S501，首先要输入：In step S501, first input:

◆信道矩阵首列块的非零子块，根据式(6)，得到首列块的非零子块为H₀，H₁，…，H_L。◆The non-zero sub-blocks of the first block of the channel matrix, according to the formula (6), the non-zero sub-blocks of the first block are H ₀ , H ₁ ,...,H _L .

◆判决反馈后的信号r，天线数M，子载波数N，循环前缀长度P。◆The signal r after decision feedback, the number of antennas M, the number of subcarriers N, and the length of the cyclic prefix P.

步骤S402的具体步骤如下：The specific steps of step S402 are as follows:

在步骤S501，根据信道矩阵首列块的非零子块，通过解方程组的方法得到，信道矩阵逆矩阵的首列块U，其计算公式见公式(22)、(23)。进入循环1＜＝i＜+M，1＜＝j＜+M。In step S501, according to the non-zero sub-blocks in the first block of the channel matrix, the first block U of the inverse channel matrix is obtained by solving equations, and its calculation formula is shown in formulas (22) and (23). Enter the loop 1<=i<+M, 1<=j<+M.

在步骤S502，在U的j列中，从第i个开始，以M为间隔，取N+P个值，得到向量vec_ij，从向量r中，以第j个开始，以M为间隔，取N+P个值，得到向量r_vec_ij。vec_ij的计算公式见(29)，r_vec_ij的计算公式见(30)。In step S502, in the column j of U, starting from the i-th column, taking M as an interval, taking N+P values to obtain a vector vec _ij , starting from the j-th column in the vector r, and taking M as an interval, Take N+P values to get the vector r_vec _ij . See (29) for the calculation formula of vec _ij , and see (30) for the calculation formula of r_vec _ij .

在步骤S703，通过FFT计算vec_ij与r_vec_ij的线性卷积，并取线性卷积结果的前N+P个，得到u_ij。其具体过程为将r_vec_ij补N+P个零得到r_e_ij，将vec_ij补N+P个零得到vec_e_ij。然后按照式(31)，得到f_ij，然后令u_ij为f_ij的前(N+P)个值。In step S703, the linear convolution of vec _ij and r_vec _ij is calculated by FFT, and the first N+P linear convolution results are taken to obtain u _ij . The specific process is to add N+P zeros to r_vec _ij to get r_e _ij , and add N+P zeros to vec _ij to get vec_e _ij . Then, f _ij is obtained according to formula (31), and then u _ij is set as the first (N+P) values of f _ij .

在步骤S504中，初始化part_ij为M(N+P)×1的全零列向量，并将u_ij赋值给part_ij以第i个元素开始，间隔为M的元素(N+P个)，part_ij赋值的公式见式(32)。In step S504, initialize part _ij to be an all-zero column vector of M(N+P)×1, and assign u _ij to part _ij starting with the i-th element, with an interval of M elements (N+P), The formula for assigning part _ij is shown in formula (32).

得到所有的part_ij时，进入步骤S505。在步骤S505，将所有的part_ij求和，得到b，参见式(33)。When all part _ij are obtained, go to step S505. In step S505, sum all part _ij to obtain b, see formula (33).

举例：2×2系统，多径长度为L+1＝3，N＝4，P＝1，其信道矩阵为：Example: In a 2×2 system, the multipath length is L+1=3, N=4, P=1, and its channel matrix is:

$H h = = [\begin{matrix} {H h}_{00} & 00_{22} & 00_{22} & 00_{22} & 00_{22} \\ {H h}_{11} & {H h}_{00} & 00_{22} & 00_{22} & 00_{22} \\ {H h}_{22} & {H h}_{11} & {H h}_{00} & 00_{22} & 00_{22} \\ 00_{22} & {H h}_{22} & {H h}_{11} & {H h}_{00} & 00_{22} \\ 00_{22} & 00_{22} & {H h}_{22} & {H h}_{11} & {H h}_{00} \end{matrix}] - - - - - - ((3535))$

在式(35)中，0₂为2×2的零矩阵。H₀、H₁、H₂为2×2矩阵。In formula (35), 0 ₂ is a 2×2 zero matrix. H ₀ , H ₁ , and H ₂ are 2×2 matrices.

1)根据前一个符号的估计，得到

(当前为k)，根据式(8)，经过判决反馈，得到当前消除符号间干扰的接收信号的估计r＝[r₁ r₂r₃ r₄ r₅ r₆ r₇ r₈ r₉ r₁₀]^T。1) According to the estimation of the previous symbol, we get

(currently k), according to formula (8), after decision feedback, the current estimated received signal r=[r ₁ r ₂ r ₃ r ₄ r ₅ r ₆ r ₇ r ₈ r ₉ r ₁₀ ] ^T.

2)根据式(19)，H^-1的首列块U为：2) According to formula (19), the first block U of H ^-1 is:

$U u = = [\begin{matrix} {U u}_{00} \\ {U u}_{11} \\ {U u}_{22} \\ {U u}_{33} \\ {U u}_{44} \end{matrix}] - - - - - - ((3636))$

则：but:

$[\begin{matrix} {H h}_{00} & 00_{22} & 00_{22} & 00_{22} & 00_{22} \\ {H h}_{11} & {H h}_{00} & 00_{22} & 00_{22} & 00_{22} \\ {H h}_{22} & {H h}_{11} & {H h}_{00} & 00_{22} & 00_{22} \\ 00_{22} & {H h}_{22} & {H h}_{11} & {H h}_{00} & 00_{22} \\ 00_{22} & 00_{22} & {H h}_{22} & {H h}_{11} & {H h}_{00} \end{matrix}] [\begin{matrix} {U u}_{00} \\ {U u}_{11} \\ {U u}_{22} \\ {U u}_{33} \\ {U u}_{44} \end{matrix}] = = [\begin{matrix} {I I}_{22} \\ 00_{22} \\ 00_{22} \\ 00_{22} \\ 00_{22} \end{matrix}] - - - - - - ((3737))$

式(37)中，I₂为2×2的单位阵，O₂为2×2的零矩阵。则In formula (37), I ₂ is a 2×2 unit matrix, and O ₂ is a 2×2 zero matrix. but

H₀ U₀＝I₂ (38-1)H ₀ U ₀ =I ₂ (38-1)

H₁ U₀+H₀ U₁＝O₂ (38-2)H ₁ U ₀ +H ₀ U ₁ =O ₂ (38-2)

H₂ U₀+H₁ U₁+H₀ U₂＝O₂ (38-3)H ₂ U ₀ +H ₁ U ₁ +H ₀ U ₂ =O ₂ (38-3)

H₂ U₁+H₁ U₂+H₀ U₃＝O₂ (38-4)H ₂ U ₁ +H ₁ U ₂ +H ₀ U ₃ ＝O ₂ (38-4)

H₂ U₂+H₁ U₃+H₀ U₄＝O₂ (38-5)H ₂ U ₂ +H ₁ U ₃ +H ₀ U ₄ ＝O ₂ (38-5)

从(38-1)开始解，得到：Starting from (38-1), we get:

${U u}_{00} = = {H h}_{00}^{- - 11} - - - - - - ((3939 - - 11))$

${U u}_{11} = = - - {H h}_{00}^{- - 11} (({H h}_{11} {U u}_{00})) - - - - - - ((3939 - - 22))$

${U u}_{22} = = - - {H h}_{00}^{- - 11} (({H h}_{22} {U u}_{00} + + {H h}_{11} {U u}_{11})) - - - - - - ((3939 - - 33))$

${U u}_{33} = = - - {H h}_{00}^{- - 11} (({H h}_{22} {U u}_{11} + + {H h}_{11} {U u}_{22})) - - - - - - ((3939 - - 44))$

${U u}_{44} = = - - {H h}_{00}^{- - 11} (({H h}_{22} {U u}_{22} + + {H h}_{11} {U u}_{33})) - - - - - - ((3939 - - 55))$

得到H^-1的首块列U为，如表2所示：The first column U of H ^-1 is obtained, as shown in Table 2:

表2 U 第一列第二列 U0 a₁ a₂ a₃ a₄ U₁ a₅ a₆ a₇ a₈ U₂ a₉ a₁₀ a₁₁ a₁₂ U₃ a₁₃ a₁₄ a₁₅ a₁₆ U₄ a₁₇ a₁₈ a₁₉ a₂₀ Table 2 u first row The second column U0 a ₁ a ₂ a ₃ a ₄ U ₁ a ₅ a ₆ a ₇ a ₈ U ₂ a ₉ a ₁₀ a ₁₁ a ₁₂ U ₃ a ₁₃ a ₁₄ a ₁₅ a ₁₆ U ₄ a ₁₇ a ₁₈ a ₁₉ a ₂₀

3)根据首列块U，(29)，得到：3) According to the first block U, (29), get:

vec₁₁＝[a₁ a₅ a₉ a₁₃ a₁₇]^T；vec ₁₁ = [a ₁ a ₅ a ₉ a ₁₃ a ₁₇ ] ^T ;

vec₁₂＝[a₂ a₆ a₁₀ a₁₄ a₁₈]^T；vec ₁₂ = [a ₂ a ₆ a ₁₀ a ₁₄ a ₁₈ ] ^T ;

vec₂₁＝[a₃ a₇ a₁₁ a₁₅ a₁₉]^T；vec ₂₁ = [a ₃ a ₇ a ₁₁ a ₁₅ a ₁₉ ] ^T ;

vec₂₂＝[a₄ a₈ a₁₂ a₁₆ a₂₀]^T；vec ₂₂ = [a ₄ a ₈ a ₁₂ a ₁₆ a ₂₀ ] ^T ;

然后，将r分为两部分：Then, split r into two parts:

R1＝[r₁ r₃ r₅ r₇ r₉]^T。R1=[r ₁ r ₃ r ₅ r ₇ r ₉ ] ^T .

R2＝[r₁ r₃ r₅ r₇ r₉]^T。R2 = [r ₁ r ₃ r ₅ r ₇ r ₉ ] ^T .

然后计算：Then calculate:

f₁₁＝vec₁₁*R1 (40-1)f ₁₁ =vec ₁₁ *R1 (40-1)

f₁₂＝vec₁₂*R2 (40-2)f ₁₂ =vec ₁₂ *R2 (40-2)

f₂₁＝vec₂₁*R1 (40-3)f ₂₁ =vec ₂₁ *R1 (40-3)

f₂₂＝vec₂₂*R2 (40-3)f ₂₂ =vec ₂₂ *R2 (40-3)

在式(40)中，*表示卷积，可以用FFT来实现，见式(30)、(31)。在式(40)中，卷积得到的点是10个。然后令u₁₁为f₁₁的前5个点，u₁₂为f₁₂的前5个点，u₁₃为f₁₃的前5个点，u₁₄为f₁₄的前5个点。In formula (40), * means convolution, which can be realized by FFT, see formula (30), (31). In formula (40), the number of points obtained by convolution is 10. Then let u ₁₁ be the first 5 points of f ₁₁ , u ₁₂ be the first 5 points of f ₁₂ , u ₁₃ be the first 5 points of f ₁₃ , u ₁₄ be the first 5 points of f ₁₄ .

然后，令part₁₁为初始化为零的10×1向量，按照(32)将u₁₁的5个点赋给part₁₁的1、3、5、7、9个点上，即：Then, let part ₁₁ be a 10×1 vector initialized to zero, and assign the 5 points of u ₁₁ to the 1, 3, 5, 7, and 9 points of part ₁₁ according to (32), that is:

part₁₁＝[u₁₁(1) 0 u₁₁(2) 0 u₁₁(3) 0 u₁₁(4) 0 u₁₁(5)0]^T；part ₁₁ = [u ₁₁ (1) 0 u ₁₁ (2) 0 u ₁₁ (3) 0 u ₁₁ (4) 0 u ₁₁ (5) 0] ^T ;

part₁₂＝[u₁₂(1) 0 u₁₂(2) 0 u₁₂(3) 0 u₁₂(4) 0 u₁₂(5)0]^T；part ₁₂ = [u ₁₂ (1) 0 u ₁₂ (2) 0 u ₁₂ (3) 0 u ₁₂ (4) 0 u ₁₂ (5) 0] ^T ;

part₂₁＝[0 u₂₁(1) 0 u₂₁(2) 0 u₂₁(3) 0 u₂₁(4) 0 u₂₁(5)]^T；part ₂₁ = [ _0u21 (1) _0u21 (2) _0u21 (3) _0u21 (4) _0u21 (5)] ^T ;

part₂₂＝[0 u₂₂(1) 0 u₂₂(2) 0 u₂₂(3) 0 u₂₂(4) 0 u₂₂(5)]^T。part ₂₂ = [0 u ₂₂ (1) 0 u ₂₂ (2) 0 u ₂₂ (3) 0 u ₂₂ (4) 0 u ₂₂ (5)] ^T .

4)然后，按照式(33)得到b＝part₁₁+part₁₂+part₂₁+part₂₂。得到：4) Then, b=part ₁₁ +part ₁₂ +part ₂₁ +part ₂₂ is obtained according to formula (33). get:

b＝[b₁ b₂ b₃ b₄ b₅ b₆ b₇ b₈ b₉ b₁₀]^T。b=[b ₁ b ₂ b ₃ b ₄ b ₅ b ₆ b ₇ b ₈ b ₉ b ₁₀ ] ^T .

按照式(34)，得到v＝[b₃ b₄ b₅ b₆ b₇ b₈ 0.5×(b₁+b₉)0.5×(b₂+b₁₀)]^T＝[v₁ v₂ v₃ v₄ v₅ v₆ v₇ v₈]^T。According to formula (34), v=[b ₃ b ₄ b ₅ b ₆ b ₇ b ₈ 0.5×(b ₁ +b ₉ )0.5×(b ₂ +b ₁₀ )] ^T =[v ₁ v ₂ v ₃ v ₄ v ₅ v ₆ v ₇ v ₈ ] ^T .

5)按照式(14)，得到₁＝[v₁ v₃ v₅ v₇]^T，₂＝[v₂ v₄ v₆ v₈]^T。然后，将₁、₂按式(15)变换到频域并解调，得到两根发送天线上发送的调制符号的估计。5) According to formula (14),  ₁ =[v ₁ v ₃ v ₅ v ₇ ] ^T ,  ₂ =[v ₂ v ₄ v ₆ v ₈ ] ^T . Then,  ₁ ,  ₂ are transformed into frequency domain according to formula (15) and demodulated to obtain the estimates of the modulation symbols sent on the two sending antennas.

第二实例(1×1正交频分复用系统)The second example (1×1 OFDM system)

以下描述无CP的情况下的均衡方法。The equalization method in the case of no CP is described below.

在无CP的情况下，处理模型与文献1中的一样。In the case of no CP, the processing model is the same as in Reference 1.

同样，如图3所示，本例中需要进行如下的三个过程。Likewise, as shown in FIG. 3 , the following three processes need to be performed in this example.

1)判决反馈过程1) Judgment feedback process

从式(4)看，有：From formula (4), we have:

$d d = = {y the y}^{k k} - - {H h}_{22} {Q Q}^{H h} {\overset{~ ~}{s the s}}^{k k - - 11} - - - - - - ((4141))$

在式(5)中，y^k为第k个OFDM符号的接收向量。

为对第k-1个OFDM符号上发送符号的估计向量(频域上的所有的子载波上发送符号的估计，N维向量)，

为

对应的时域信号，Q为FFT变换矩阵。In formula (5), y ^k is the receiving vector of the kth OFDM symbol.

For the estimation vector of the transmitted symbol on the k-1th OFDM symbol (the estimation of the transmitted symbol on all subcarriers in the frequency domain, N-dimensional vector),

for

Corresponding to the time domain signal, Q is the FFT transformation matrix.

2)均衡过程2) Equalization process

从式(4)看，有：From formula (4), we have:

$b b = = {(({H h}_{00} - - {H h}_{11}))}^{- - 11} (({y the y}^{k k} - - {H h}_{22} {Q Q}^{H h} {\overset{~ ~}{s the s}}^{k k - - 11})) = = {(({H h}_{00} - - {H h}_{11}))}^{- - 11} d d - - - - - - ((4242))$

在式(6)中，d为时域判决反馈后的信号。A＝H₀-H₁为造成OFDM的ICI干扰的矩阵，b为经过判决反馈均衡后的信号。In formula (6), d is the signal after time domain decision feedback. A=H ₀ -H ₁ is a matrix of ICI interference causing OFDM, and b is a signal after decision feedback equalization.

3)解调过程3) Demodulation process

将时域均衡后的信号b变换到频域，然后解调，得到发送符号的估计，有：Transform the time-domain equalized signal b into the frequency domain, and then demodulate to obtain the estimate of the transmitted symbol, which is:

${\overset{~ ~}{s the s}}^{k k} = = Qb Qb - - - - - - ((4343))$

然后，对进行解调，得到发送符号的估计。then, yes Demodulation is performed to obtain an estimate of the transmitted symbol.

在式(7)中，b为时域均衡后的信号，Qb是将时域均衡后的信号变换到频域，然后解调，得到发送符号的估计。In formula (7), b is the time-domain equalized signal, and Qb is to transform the time-domain equalized signal into the frequency domain, and then demodulate it to obtain the estimate of the transmitted symbol.

现有方法的计算复杂度：Computational complexity of existing methods:

1)判决反馈过程：可以用FFT来计算，复杂度Nlog₂(N)；1) Decision feedback process: it can be calculated by FFT, and the complexity is Nlog ₂ (N);

2)均衡过程：矩阵求逆，复杂度0(N³)；2) Equalization process: matrix inversion, complexity 0(N ³ );

3)解调过程：复杂度Nlog₂(N)。3) Demodulation process: complexity Nlog ₂ (N).

可以看出，主要是均衡过程复杂度很高，阻碍了实际的应用。It can be seen that the main reason is that the equalization process is very complicated, which hinders the actual application.

以下详细说明均衡过程，根据式(1)、(2)，得到：The following is a detailed description of the equalization process. According to formulas (1) and (2), we can get:

式(44)中矩阵A的表达式如图8所示。在本例中，目的是降低均衡的复杂度，更改图3中均衡过程，降低求解式(42)的复杂度。其他部分不变。The expression of matrix A in formula (44) is shown in Figure 8. In this example, the purpose is to reduce the complexity of equalization, change the equalization process in Figure 3, and reduce the complexity of solving formula (42). Other parts remain unchanged.

第一步，求A^-1 The first step is to find A ^-1

根据附录1的证明，可得对角不为零的下三角T矩阵的逆矩阵T^-1，为下三角T矩阵。According to the proof in Appendix 1, the inverse matrix T ^-1 of the lower triangular T matrix whose diagonal is not zero can be obtained, which is the lower triangular T matrix.

用解方程的方法(利用A的零值)求A^-1的首列：Find the first column of A ^-1 by solving the equation (using the zero value of A):

令u＝[u₀，u₂，...，u_N-1]^T为A^-1的首列。Let u=[u ₀ , u ₂ , . . . , u _N-1 ] ^T is the first column of A ^-1 .

由于在OFDM的系统中，N＞＞L+1，类似于式(17)-(21)：Since in the OFDM system, N>>L+1, similar to formula (17)-(21):

h₀ u₀＝1；(45-1)h ₀ u ₀ = 1; (45-1)

h₁ u₀+h₀ u₁＝0；(45-2)h ₁ u ₀ +h ₀ u ₁ =0; (45-2)

…...

h₀ u_i+h₁u_i-1+…+h_Lu_i-L＝0(45-i+1)h ₀ u _i +h ₁ u _i-1 +…+h _L u _iL ＝0(45-i+1)

根据(45-1)，求出u₀，然后将u₀代入(45-2)，求出u₁，求u_i时，将u_i-1，…，u_i-L代入。Calculate u ₀ according to (45-1), and then substitute u ₀ into (45-2) to calculate u ₁ , and when calculating u _i , substitute u _i-1 ,..., u _iL .

【复杂度分析】【Complexity Analysis】

由于有N个等式，每个等式最多有L个数乘加，复杂度0(LN)。这样，就求得了A^-1的第一列。Since there are N equations, each equation has at most L multiplication and addition, and the complexity is 0(LN). In this way, the first column of A ^-1 is obtained.

第二步，求A^-1dThe second step is to find A ^-1 d

由于A^-1为下三角T矩阵，求：Since A ^-1 is a lower triangular T matrix, find:

r＝A^-1d (46)r = A ^-1 d (46)

如果直接求，复杂度0(N²)。If it is calculated directly, the complexity is 0(N ² ).

A^-1是下三角T阵，假设其第一列为u＝u₀，u₁₁，...，u_N-1，则：A ^-1 is a lower triangular T matrix, assuming that its first column is u=u ₀ , u ₁₁ ,..., u _N-1 , then:

r₀＝d₀u₀(47-1)r ₀ =d ₀ u ₀ (47-1)

r₁＝d₁u₀+d₀u₁(47-2)r ₁ ＝d ₁ u ₀ +d ₀ u ₁ (47-2)

r₂＝d₂u₀+d₁u₁+d₀u₂(47-3)r ₂ ＝d ₂ u ₀ +d ₁ u ₁ +d ₀ u ₂ (47-3)

……...

把A^-1d与线性卷积比较，r的值相当与a与v线性卷积的前N个值。然后利用FFT求线性卷积的方法，可得到r。Comparing A ^-1 d with linear convolution, the value of r is equivalent to the first N values of a linear convolution with v. Then use the method of FFT to find linear convolution to get r.

令u_e表示u补N个0，d_e表示d补N个0，然后计算：Let u_e mean that u is filled with N zeros, and d_e means that d is filled with N zeros, and then calculate:

r_e＝ifft(fft(u_e).*fft(a_e)) (48)r_e＝ifft(fft(u_e).*fft(a_e)) (48)

在式(48)中，.*表示点乘，然后令r等于r_e的前N的值。In formula (48), .* represents dot multiplication, and then let r be equal to the value of the first N of r_e.

【复杂度分析】【Complexity Analysis】

第一步，求A^-1，复杂度(L+1)N；The first step is to find A ^-1 , complexity (L+1)N;

第二步，求A^-1d，复杂度0(2Nlog₂(2N))。The second step is to find A ^-1 d, the complexity is 0(2Nlog ₂ (2N)).

因此，总的复杂度0(2Nlog₂(2N))+(L+1)N。一般情况下第二步的复杂度要大于第一步，总的复杂度认为是0(2Nlog₂(2N))。表3给出与其它方法的复杂度比较。Therefore, the total complexity is 0(2Nlog ₂ (2N))+(L+1)N. In general, the complexity of the second step is greater than that of the first step, and the total complexity is considered to be 0(2Nlog ₂ (2N)). Table 3 gives the complexity comparison with other methods.

表3在1×1系统的情况下与其它方法的比较本发明 Levinson快速算法未使用快速算法复杂度 0(2Nlog₂(2N)) 0(N²) 0(N³) Table 3 Comparison with other methods in the case of 1×1 system this invention Levinson Fast Algorithm fast algorithm not used the complexity 0(2Nlog ₂ (2N)) 0(N ² ) 0(N ³ )

以下以具体的实例来说明如何在无CP的情况下进行本发明的均衡方法。The following uses specific examples to illustrate how to implement the equalization method of the present invention without a CP.

图7示出了在无CP的情况下进行本发明的均衡方法的流程图。Fig. 7 shows a flow chart of the equalization method of the present invention without CP.

在步骤S701，得到S401得到判决反馈后的接收信号d，信道的估计值h₀，h₂，…，h_L。从而得到信道矩阵的第一列a＝[h₀，h₂，…，h_L，0，…0]^T。计算A^-1的第一列，记为向量u。图8示出了步骤S701的详细过程。In step S701, the received signal d and channel estimated values h ₀ , h ₂ , . . . , h _L after the decision feedback obtained in step S401 are obtained. Thus, the first column a=[h ₀ , h ₂ , . . . , h _L , 0, . . . 0] ^T of the channel matrix is obtained. Compute the first column of A ^-1 , denoted as vector u. FIG. 8 shows the detailed process of step S701.

如图8所示，计算A^-1的第一列。A^-1的第一列有N个元素，为u＝[u₀，u₂，...，u_N-1]^T。As shown in Figure 8, calculate the first column of A ^-1 . The first column of A ^-1 has N elements, which is u=[u ₀ , u ₂ , . . . , u _N-1 ] ^T .

在步骤801，计算首先得到信道值h₀，h₂，…，h_L，计算u₀＝1/h₀。In step 801, the calculation firstly obtains the channel values h ₀ , h ₂ , ..., h _L , and calculates u ₀ =1/h ₀ .

在步骤802，计算u₁到u_L。根据h₁到h_k，u₁到u_k-1，计算u_k(2＜＝k＜＝L)。In step 802, u ₁ to u _L are calculated. From h ₁ to h _k , u ₁ to u _k-1 , u _k is calculated (2<=k<=L).

${u u}_{k k} = = - - (({Σ Σ}_{t t = = 11}^{k k} {h h}_{t t} {u u}_{k k - - t t})) / / {h h}_{00} - - - - - - ((4949))$

在步骤803，计算u_L+1到u_N-1。根据h₀到h_L，u_k-L到u_k-1，计算u_k(L＜k＜＝N)。In step 803, u _L+1 to u _N-1 are calculated. From h ₀ to h _L , u _kL to u _k-1 , u _k is calculated (L<k<=N).

${u u}_{k k} = = - - (({Σ Σ}_{t t = = 11}^{L L} {h h}_{t t} {u u}_{k k - - t t})) / / {h h}_{00} - - - - - - ((5050))$

这样，在步骤S803后，得到了全部的u＝[u₀，u₁，...，u_N-1]^T。In this way, after step S803, all u=[u ₀ , u ₁ , . . . , u _N-1 ] ^T are obtained.

例如，假设N＝5，L+1＝3。则For example, assuming N=5, L+1=3. but

$A A = = [\begin{matrix} {h h}_{00} & 00 & 00 & 00 & 00 \\ {h h}_{11} & {h h}_{00} & 00 & 00 & 00 \\ {h h}_{22} & {h h}_{11} & {h h}_{00} & 00 & 00 \\ 00 & {h h}_{22} & {h h}_{11} & {h h}_{00} & 00 \\ 00 & 00 & {h h}_{22} & {h h}_{11} & {h h}_{00} \end{matrix}] - - - - - - ((5151))$

则A^-1的第一列可如下式表示：Then the first column of A ^-1 can be expressed as follows:

$[\begin{matrix} {h h}_{00} & 00 & 00 & 00 & 00 \\ {h h}_{11} & {h h}_{00} & 00 & 00 & 00 \\ {h h}_{22} & {h h}_{11} & {h h}_{00} & 00 & 00 \\ 00 & {h h}_{22} & {h h}_{11} & {h h}_{00} & 00 \\ 00 & 00 & {h h}_{22} & {h h}_{11} & {h h}_{00} \end{matrix}] [\begin{matrix} {u u}_{00} \\ {u u}_{11} \\ {u u}_{22} \\ {u u}_{33} \\ {u u}_{44} \end{matrix}] = = [\begin{matrix} 11 \\ 00 \\ 00 \\ 00 \\ 00 \end{matrix}] - - - - - - ((5252))$

根据式(19)，有如下5个等式：According to formula (19), there are five equations as follows:

h₀ u₀＝1 (53-1)h ₀ u ₀ =1 (53-1)

h₁ u₀+h₀ u₁＝0 (53-2)h ₁ u ₀ +h ₀ u ₁ ＝0 (53-2)

h₂ u₀+h₁ u₁+h₀ u₂＝0 (53-3)h ₂ u ₀ +h ₁ u ₁ +h ₀ u ₂ =0 (53-3)

h₂ u₁+h₁ u₂+h₀ u₃＝0 (53-4)h ₂ u ₁ +h ₁ u ₂ +h ₀ u ₃ =0 (53-4)

h₂ u₂+h₁ u₃+h₀ u₄＝0 (53-5)h ₂ u ₂ +h ₁ u ₃ +h ₀ u ₄ ＝0 (53-5)

从式(53-1)开始解，得到：Starting from equation (53-1), we get:

u₀＝1/h₀ (54-1)u ₀ =1/h ₀ (54-1)

u₁＝-(h₁ u₀)/h₀ (54-2)u ₁ =-(h ₁ u ₀ )/h ₀ (54-2)

u₂＝-(h₂ u₀+h₁ u₁)/h₀ (54-3)u ₂ ＝-(h ₂ u ₀ +h ₁ u ₁ )/h ₀ (54-3)

u₃＝-(h₂ u₁+h₁ u₂)/h₀ (54-4)u ₃ ＝-(h ₂ u ₁ +h ₁ u ₂ )/h ₀ (54-4)

u₄＝-(h₂ u₂+h₁ u₃)/h₀ (54-5)u ₄ ＝-(h ₂ u ₂ +h ₁ u ₃ )/h ₀ (54-5)

最后，得到u＝[u₀，u₂，...，u_N-1]^T。Finally, u=[u ₀ , u ₂ , . . . , u _N-1 ] ^T is obtained.

在步骤S702，根据得到的u，和判决反馈后的d，利用FFT的方法求A^-1d，计算见式(48)。图9示出了步骤S702的详细过程。In step S702, according to the obtained u and the d after decision feedback, use the method of FFT to calculate A ^-1 d, and the calculation is shown in formula (48). FIG. 9 shows the detailed process of step S702.

如图9所示，计算A^-1d的第一列。A^-1的第一列有N个元素，为u＝[u₀，u₁，...，u_N-1]^T。Compute the first column of A ^-1 d as shown in Figure 9. The first column of A ^-1 has N elements, which is u=[u ₀ , u ₁ , . . . , u _N-1 ] ^T .

根据式(5)输入判决反馈后的接收信号d(N×1向量)，从S701得到信道逆矩阵的第一列u(N×1向量)，在步骤S901，令d补N个0得到d_e(2N×1向量)，u补N个0得到u_e(2N×1向量)。Input the received signal d (N×1 vector) after decision feedback according to formula (5), and obtain the first column u (N×1 vector) of the channel inverse matrix from S701, and in step S901, let d complement N zeros to obtain d_e (2N×1 vector), u complements N 0s to get u_e (2N×1 vector).

在步骤S902，将d_e做2N点的FFT变换得到d_f(2N×1向量)，将u_e做2N点的FFT变换得到u_f(2N×1向量)。In step S902, perform 2N-point FFT transformation on d_e to obtain d_f (2N×1 vector), and perform 2N-point FFT transformation on u_e to obtain u_f (2N×1 vector).

在步骤S903，计算u_f.*d_f，记对应的元素相除，得到r_f(2N×1向量)。In step S903, calculate u_f.*d_f, and divide corresponding elements to obtain r_f (2N×1 vector).

在步骤S904，将r_f进行2N点IFFT变换，得到b_e，然后将b_e取前N个点得到均衡后的时域值b。In step S904, r_f is subjected to 2N-point IFFT transformation to obtain b_e, and then the first N points of b_e are taken to obtain an equalized time-domain value b.

例如，假设N＝5，就得到了u₀，u₁，u₂，u₃，u₄。从S401得到了判决反馈后的信号d₀，d₁，d₂，d₃，d₄。For example, assuming N=5, u ₀ , u ₁ , u ₂ , u ₃ , u ₄ are obtained. Signals d ₀ , d ₁ , d ₂ , d ₃ , and d ₄ after decision feedback are obtained from S401.

根据上述的方法，d_e＝[d₀ d₁ d₂ d₃ d₄ 0 0 0 0 0]^T，u_e＝[u₀u₁ u₂ u₃ u₄ 0 0 0 0 0]^T。According to the above method, d_e=[d ₀ d ₁ d ₂ d ₃ d ₄ 0 0 0 0 0] ^T , u_e=[u ₀ u ₁ u ₂ u ₃ u ₄ 0 0 0 0 0] ^T .

然后，将d_e做10点的FFT，得到d_f＝FFT(d_e)，d_f＝[df₀ df₁df₂ df₃ df₄ df₅ df₆ df₇ df₈ df₉]^T。将u_e做10点的FFT得到u_f＝FFT(u_e)，u_f＝[uf₀ uf₁ uf₂ uf₃ uf₄ uf₅ uf₆ uf₇ uf₈ uf₉]^T。Then, do a 10-point FFT on d_e, and get d_f=FFT(d_e), d_f=[df ₀ df ₁ df ₂ df 3 df ₄ df ₅ df ₆ df ₇ _df ₈ df ₉ ] ^T . Do 10-point FFT of u_e to get u_f=FFT(u_e), u_f=[uf ₀ uf ₁ uf ₂ uf ₃ uf ₄ uf ₅ uf ₆ uf ₇ uf ₈ uf ₉ ] ^T .

接下来，令r_f＝[rf₀＝uf₀×df₀ rf₁＝uf₁×df₁ rf₂＝uf₂×df₂rf₃＝uf₃×df₃ rf₄＝uf₄×df₄ rf₅＝uf₅×df₅ rf₆＝uf₆×df₆ rf₇＝uf₇×df₇ rf₈＝uf₈×df₈ rf₉＝uf₉×df₉]^T。Next, let r_f=[rf ₀ =uf ₀ ×df ₀ rf ₁ =uf ₁ ×df ₁ rf ₂ =uf ₂ ×df ₂ rf ₃ =uf ₃ ×df ₃ rf ₄ =uf ₄ ×df ₄ rf ₅ = uf ₅ ×df ₅ rf ₆ =uf ₆ ×df ₆ rf ₇ =uf ₇ ×df ₇ rf ₈ =uf ₈ ×df ₈ rf ₉ =uf ₉ ×df ₉ ] ^T .

将r_f做10点的IFFT变换得到，b_e＝IFFT(r_f)。b_e＝[be₀ be₁be₂ be₃ be₄ be₅ be₆ be₇ be₈ be₉]^T。Perform 10-point IFFT transformation on r_f to get, b_e=IFFT(r_f). b_e＝[be ₀ be ₁ be ₂ be ₃ be ₄ be ₅ be ₆ be ₇ be ₈ be ₉ ] ^T .

最后，将b_e取前5个点得到均衡后的时域值b＝[be₀ be₁ be₂ be₃be₄]^T。Finally, take the first 5 points of b_e to obtain the equalized time domain value b=[be ₀ be ₁ be ₂ be ₃ be ₄ ] ^T .

附录1Appendix 1

定理(发明)：分块下三角Toeplitz矩阵A的逆矩阵A^-1，也为分块下三角Toeplitz矩阵。Theorem (invention): The inverse matrix A ^-1 of the block lower triangular Toeplitz matrix A is also the block lower triangular Toeplitz matrix.

在以下运算中X_ij表示的是m×m的矩阵，X为任一nm×nm的矩阵，分块如式(16)所示。In the following operations, X _ij represents an m×m matrix, X is any nm×nm matrix, and the blocks are shown in formula (16).

证明：prove:

假设A为基维为m×m的n阶分块下三角Toeplitz矩阵。则A的三个性质：Assume that A is a block lower triangular Toeplitz matrix of order n with base dimension m×m. Then there are three properties of A:

1、A^-1唯一。1. A ^-1 is unique.

2、由于A是下三角结构，所以A^-1也为下三角结构。2. Since A is a lower triangular structure, A ^-1 is also a lower triangular structure.

3、由于A是分块下三角Toeplitz矩阵，所以A可由A的首列块来表示。令A的第一列块的元素为A₁₁＝a₁，A₂₁＝a₂，...，A_n1＝a_n。由于A是T矩阵，所以A_ij＝A_i+P，j+P。以上ai为m×m矩阵，A_ij也为m×m矩阵。3. Since A is a block lower triangular Toeplitz matrix, A can be represented by the first block of A. Let the elements of the first column block of A be A ₁₁ =a ₁ , A ₂₁ =a ₂ , . . . , A _n1 =a _n . Since A is a T matrix, A _ij =A _{i+P, j+P} . Above ai is an m×m matrix, and A _ij is also an m×m matrix.

4、令B为下三角矩阵，令下三角矩阵C＝AB则4. Let B be the lower triangular matrix, let the lower triangular matrix C=AB then

$C_{ij} = Σ_{k = 1}^{n} A_{ik} B_{kj}$ (附录1-1) $C_{ij} = Σ_{k = 1}^{no} A_{ik} B_{kj}$ (Appendix 1-1)

在(附录1-1)中，C_ij，A_ik，B_kj都为m×m矩阵。In (Appendix 1-1), C _ij , A _ik , and B _kj are all m×m matrices.

由于A，B为下三角矩阵，则Since A and B are lower triangular matrices, then

当k＞i时，A_ik＝0。When k>i, A _ik =0.

当j＞k时，B_kj＝0。When j>k, B _kj =0.

所以，有：F:

$C_{ij} = Σ_{k = j}^{i} A_{ik} B_{kj}$ (附录1-2) $C_{ij} = Σ_{k = j}^{i} A_{ik} B_{kj}$ (Appendix 1-2)

令B＝A^-1，则，则C＝AB＝I_nm(I_nm为n×M阶单位阵)。此时C等于单位阵，由于单位阵也是分块下三角T矩阵，所以C也是分块下三角T矩阵。满足：Let B=A ^-1 , then C=AB=I _nm (I _nm is an n×M order unit matrix). At this time, C is equal to the unit matrix. Since the unit matrix is also a block lower triangular T matrix, C is also a block lower triangular T matrix. satisfy:

C_ij＝C_i+p，j+p(附录1-3)C _ij ＝C _{i+p, j+p} (Appendix 1-3)

在(附录1-3)中，C_ij为m×m的矩阵。p为整数，且i、j、i+p、j+p∈1，…，n。由于C可由C的首列块来表示，则可令(附录1-3)中j＝1。得到In (Appendix 1-3), C _ij is an m×m matrix. p is an integer, and i, j, i+p, j+p∈1,...,n. Since C can be represented by the block in the first column of C, j=1 in (Appendix 1-3). get

C_i+p，l+p＝C_i，l(附录1-4)C _{i+p, l+p} = C _{i, l} (Appendix 1-4)

根据(附录1-2)，有：According to (Appendix 1-2), there are:

$C_{1 + p, 1 + p} = Σ_{k = 1 + p}^{i + p} A_{i + p, k} B_{k, 1 + p}$ (附录1-5) $C_{1 + p, 1 + p} = Σ_{k = 1 + p}^{i + p} A_{i + p, k} B_{k, 1 + p}$ (Appendix 1-5)

根据(附录1-3)、(附录1-4)、(附录1-5)得到：According to (Appendix 1-3), (Appendix 1-4), (Appendix 1-5):

$Σ_{k = 1}^{i} A_{ik} B_{k 1} = Σ_{m = 1 + p}^{i + p} A_{i + p, m} B_{m, 1 + p}$ (附录1-6) $Σ_{k = 1}^{i} A_{ik} B_{k 1} = Σ_{m = 1 + p}^{i + p} A_{i + p, m} B_{m, 1 + p}$ (Appendix 1-6)

令(附录1-6)中，q＝m-p，则有：In order (appendix 1-6), q=m-p, then have:

$Σ_{k = 1}^{i} A_{ik} B_{k 1} = Σ_{q = 1}^{i} A_{i + p, q + p} B_{q + p, 1 + p}$ (附录1-7) $Σ_{k = 1}^{i} A_{ik} B_{k 1} = Σ_{q = 1}^{i} A_{i + p, q + p} B_{q + p, 1 + p}$ (Appendix 1-7)

并令(附录1-7)中等式右边，令q＝k，则有：And let (appendix 1-7) the right side of the equation, let q=k, then have:

${Σ Σ}_{k k = = 11}^{i i} {A A}_{ik ik} {B B}_{k k 11} - - {Σ Σ}_{k k = = 11}^{i i} {A A}_{i i + + p p,, k k + + p p} {B B}_{k k + + p p,, 11 + + p p} = = 00$

$Σ_{k = 1}^{i} A_{ik} (B_{k 1} - B_{k + p, 1 + p}) = 0$ (附录1-8) $Σ_{k = 1}^{i} A_{ik} (B_{k 1} - B_{k + p, 1 + p}) = 0$ (Appendix 1-8)

显然，如果B_k+p，l+p＝B_k1，是(附录1-8)的一个解。由于B为A的逆矩阵根据性质(1)，A^-1唯一，所以，即B为分块下三角Toeplitz矩阵。证毕。Obviously, if B _{k+p, l+p} = B _k1 , it is a solution of (Appendix 1-8). Since B is the inverse matrix of A, according to property (1), A ^-1 is unique, so B is a block lower triangular Toeplitz matrix. Certificate completed.

同理，假设A为基维为1×1的n阶分块下三角Toeplitz矩阵，即A为下三角Toeplitz矩阵，则A^-1为下三角Toeplitz矩阵。Similarly, assuming that A is an n-order block lower triangular Toeplitz matrix with a base dimension of 1×1, that is, A is a lower triangular Toeplitz matrix, then A ^-1 is a lower triangular Toeplitz matrix.

Claims

1, a kind of equalization methods of orthogonal frequency-division multiplex singal, it is applied to have in the ofdm system of M root transmitting antenna and M root reception antenna, and wherein M is more than or equal to 1, and described method comprises step:

By the decision-feedback method, according to the judgement of previous symbol, remove intersymbol interference, obtain the signal after the decision-feedback;

Utilize the first row of finding the solution the inverse matrix of channel matrix with the method for the group of solving an equation;

Signal after the decision-feedback is got different parts according to corresponding antenna, and do linear convolution from the first corresponding part of row taking-up of channel matrix inverse matrix; And

Merging is at the resulting convolution results of various piece, to obtain the signal after the equilibrium.

2, the method for claim 1 is characterized in that, also comprises step:

Signal after utilizing cyclic prefix information to equilibrium merges, and obtains sending the time domain estimation of signal; And

The time domain that sends signal is estimated that transforming to frequency domain carries out demodulation.

3, the method for claim 1 is characterized in that, also comprises step:

Choosing the first half or the latter half of the signal after the equilibrium estimates as the time domain that sends signal; And

As the described method of one of claim 1-3, it is characterized in that 4, described linear convolution is undertaken by fast fourier transform.

5, method as claimed in claim 4 is characterized in that, utilizing the channel matrix inverse matrix is the step that the characteristic of piecemeal Toeplitz lower triangular matrix is carried out first row of the described inverse matrix of finding the solution channel matrix.