CN1941759A - Method for balancing orthogonal frequency division multiplexing signals - Google Patents

Abstract

The method is used in an OFDM system with M sending antennas and M receiving antennas (M>=1) and comprises: getting the signals processed by using decision direct feedback approach based on the decision of previous symbol and removing the interference between signals; solving the first block column of inverse matrix of channel matrix by solving the equation set; picking up different part from the signal processed with decision feedback according to its corresponding antenna, and getting a corresponding part form the first block column of the inverse matrix of channel matrix as the linear convolution; combining the convolution results obtained from each part to get the balanced signals.

Description

The equalization methods of orthogonal frequency-division multiplex singal

Technical field

The present invention relates to the balancing technique in OFDM (OFDM) system, be specifically related to a kind of low complex degree equalization method that can reduce time for balance.

Background technology

OFDM is a kind of data transfer mode efficiently, and its basic thought is in frequency domain given channel to be divided into many orthogonal sub-channels, uses a subcarrier to modulate on each subchannel, and each subcarrier parallel transmission.Like this, although total channel be non-flat forms, have frequency selectivity, but each subchannel is a relatively flat, what carry out on each subchannel is narrow band transmission, and signal bandwidth is less than the respective bandwidth of channel, so the just interference between the erasure signal waveform greatly.

OFDM is that its allows subcarrier spectrum to overlap with respect to the difference of general multi-carrier transmission, as long as satisfy between subcarrier mutually orthogonally, then can isolate data-signal from the subcarrier of aliasing.Because OFDM allows the subcarrier spectrum aliasing, its spectrum efficiency improves greatly, thereby is a kind of modulation system efficiently.OFDM is adapted at transmitting high speed data in the wireless mobile channel of multipath transmisstion and Doppler frequency shift.It can effectively resist multipath effect, eliminates intersymbol interference, the decline of contrary frequency selectivity, and channel utilization height.The OFDM technology is successively adopted by systems such as European digital audio broadcast (DAB), European digital video broadcasting (DVB), HIPERLAN and IEEE802.11 WLAN (wireless local area network).

Usually, in ofdm system, add Cyclic Prefix (CP), the length of Cyclic Prefix is fixed.But, wish that generally the length of Cyclic Prefix is short more good more, therefore, be different in the length of different environment multipaths.If CP is not very long, in some environment, may cause the interference between the subcarrier because the length of Cyclic Prefix is not enough.

Document 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) hypothesis is in the SISO system; the length of OFDM symbol is N, and channel length is L+1, and its channel is h=[h ₀, h ₁..., h _L], CP length is P, and L＞P.According to document 1, s ^kBe the symbolic vector of the frequency domain of k OFDM, then the time domain vector is x ^k=IFFT (s ^k); Matrix notation is x ^k=Q ^H* s ^kQ is a fourier transform matrix.

According to document 1, k OFDM symbol y of time domain ^kThe reception model be:

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

H wherein ₂For causing the channel matrix (in ofdm system, the interference between each OFDM symbol generally is called IBI) that disturbs between the OFDM symbol, H ₀-H ₁For cause in the OFDM symbol or each subcarrier of OFDM symbol between the matrix (in ofdm system, the interference in OFDM symbol generally is called ICI) that disturbs.

In the formula (1), w ^kBe additive noise.H ₀, H ₁, H ₂As follows:

If order:

A＝H ₀-H ₁ (3)

Then in document 1, s ^kBe estimated as:

${\tilde{s}}^k=Q{(H_0-H_1)}^{-1}(y^k-H_2{Q}^H{\tilde{s}}^{k-1})={\mathrm{QA}}^{-1}(y^k-H_2{Q}^H{\tilde{s}}^{k-1})---\left(4\right)$

But during greater than CP, because existing equalization methods complexity is very high, it can not be applied in multiple-input and multiple-output (MIMO) system in channel length, can only be used for the single output of single input (SISO) system.

Summary of the invention

In view of the above problems, proposed the present invention,, reduced the required time of balancing procedure, to improve the speed of Data Detection in the process that detects ofdm signal, to reduce the complexity of equalization operation.

In one aspect of the invention, a kind of equalization methods of orthogonal frequency-division multiplex singal has been proposed, it is applied to have in the ofdm system of M root transmitting antenna and M root reception antenna, wherein M is more than or equal to 1, 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 merge at the resulting convolution results of various piece, to obtain the signal after the equilibrium.

Adopt equalization methods of the present invention, compare, greatly reduce complexity, improved balanced speed with existing method.

Description of drawings

Fig. 1 shows the form of band Cyclic Prefix in OFDM System symbol (CP-OFDM symbol);

Fig. 2 is the block diagram of multiple-input and multiple-output frequency multiplexing (MIMO-OFDM) system;

Fig. 3 is the general channel length flow chart greater than the data detection process of Cyclic Prefix in OFDM System system;

Fig. 4 is the method according to this invention is carried out equalization operation to the OFDM symbol that receives a flow chart;

Fig. 5 shows the detailed process process of step S402 shown in Figure 4;

Fig. 6 is the schematic diagram of the channel matrix of no Cyclic Prefix ofdm signal;

Fig. 7 shows the flow chart that carries out equalization operation under the situation of no Cyclic Prefix;

Fig. 8 shows the detailed step of step S701 shown in Figure 7;

Fig. 9 shows the detailed step of step S702 shown in Figure 7.

Embodiment

1, channel length receives model greater than the signal of circulating prefix-length

Fig. 1 has provided the OFDM sign form that Cyclic Prefix (CP) is arranged.An OFDM symbol is made of Cyclic Prefix and OFDM data usually.

The schematic diagram of MIMO-OFDM system is seen Fig. 2, and the MIMO-OFDM system has superposeed on the ofdm system many to send and reception antenna.General 2 * 2 or 4 * 4 the mimo system that uses.As shown in Figure 2, at transmit leg, carry out Space Time Coding by Space Time Coding module 211, at string and modular converter 212 serial data is converted to and each transmitting antenna 214 corresponding symbols then, and carry out IFFT in IFFT module 213 and operate, on antenna 214, send.The recipient, at first symbol is received by reception antenna 224, carry out the FFT operation in FFT module 223 then.Symbol through the FFT computing is transmitted to channel estimation module 222, carries out channel estimating, for example obtains channel matrix, decode operation when decoder module 221 carries out sky when sky then.

Fig. 3 is the general channel length flow chart greater than the data detection process of Cyclic Prefix in OFDM System system.As shown in Figure 3, in ofdm system, at first will carry out decision-feedback process S301, this step is identical with existing method.Its less important balancing procedure S302 that carries out, method improvements over the prior art of the present invention mainly concentrate on this step.At last, carry out demodulating process S303 to the symbol after the equilibrium, to obtain demodulating data.

Suppose the mimo system of M * M, the length of an OFDM symbol is N, and channel length is L, i root transmitting antenna to the channel of j root reception antenna be h (i, j)=[h ₀(i, j), h ₁(i, j) ..., h _L(i, j).Suppose s ^k(i) be the symbolic vector of frequency domain of k OFDM of i root transmitting antenna, then the time domain vector is x ^k(i)=IFFT (s ^k); Matrix notation is x ^k(i)=Q ^H* s ^k(i); Q is a fourier transform matrix.

In k OFDM symbol period, time domain receives model and is:

y ^k＝HX ^k+H _IBIX ^k-1+w ^k (5)

In formula (5), y ^kBe the vector (P is a circulating prefix-length) of M * (N+P), every antenna has all received N data, makes y _j ^k(t) be the reception data of k symbol of j root reception antenna t the moment.Y then ^kThe individual element of 1:M (expression 1 to M) be y ₁ ^k(1) ... y _M ^k(1); M+1:2M element is y ₁ ^k(2) ... y _M ^k(2); (i-1) * M+1:i * M element is y ₁ ^k(i) ... y _M ^k(i).That is y, ^k(i-1) * M+1:i * M element be followed successively by i of 1:M root reception antenna in k the OFDM symbol period and receive vectorial, order is: i received signal of the 1st antenna, i received signal of the 2nd antenna ..., i received signal of M root antenna.

In formula (5), matrix H is the channel matrix of OFDM symbol, and it is tried to achieve by existing channel estimation methods:

In channel matrix (6), H _kBe M * Metzler matrix (0＜=k＜=L), H _kI, j element is the channel of i root transmitting antenna to the k footpath of j root reception antenna.H be a M (N+P) OK, the matrix of M (N+P) row.

In formula (5), H _IBIRepresent that previous symbol causes the interference matrix of interference to current symbol:

In matrix (7), H _k(identical in the implication of 0＜=k＜=L) and (6).

In formula (5), X ^kThe interior time domain of symbol period that is k OFDM sends vector, and it is the column vector of M (N+P) * 1 dimension.Every transmitting antenna has all sent N+P data.

Make z _i ^k(t) be t interior transmission data constantly, the then X of k OFDM symbol period of i root transmitting antenna ^k1:M element be z ₁ ^k(1) ... z _M ^k(1); M+1:2M element is z ₁ ^k(2) ... z _M ^k(2); (i-1) * M+1:i * M element is z ₁ ^k(i) ... z _M ^k(i).That is X, ^k(i-1) * M+1:i * M element be followed successively by i of 1:M root transmitting antenna in k the OFDM symbol period and send vectorial, order is: i of the 1st antenna sends signal, i received signal of the 2nd antenna ..., i received signal of M root antenna.

Because the content in the CP section is a Cyclic Prefix, so X ^kCan further represent (for the sake of simplicity, below ignoring subscript k), as follows: concerning i root transmitting antenna, data to be sent are s _i=[s _i(1) s _i(2) ..., s _i(N)] ^TTime domain x then _i=IFFT (s _i).x _i＝[x _i(1)x _i(2)，…，x _i(N)] ^T。Add CP then, the x behind the interpolation CP _i(CP)=[x _i(N-P+1) ..., x _i(N), x _i(1) x _i(2) ..., x _i(N)] ^Tx _i(CP) be the vector of a N+P.One total M root transmitting antenna.Make x _i(j) be x _iJ element, with j the transmission signal composition of vector of each antenna constantly, obtain x (j)=[x ₁(j), x ₂(j) ..., x _M(j)] ^T, x (j) is M * 1 vector.X then ^k=[x (N-P+1) ^T..., x (N) ^T, x (1) ^T, x (2) ^T..., x (N) ^T] ^TX ^kVector for M * (N+P).Preceding MP is the Cyclic Prefix of M root transmitting antenna.

2, equalization methods

Fig. 4 is the flow chart according to the equalization methods of the embodiment of the invention.

At step S401, the estimation according to previous symbol obtains

(current is k) through decision-feedback, obtains the estimation of the received signal of current elimination intersymbol interference.Be expressed as

$r=y^k-H_{\mathrm{IBI}}{\hat{X}}^{k-1}---\left(8\right)$

At step S402, utilization fast algorithm as described later obtains zero forcing equalization and separates b, and its equation expression is:

b＝H ^-1r (9)

At step S403, having under the situation of CP, utilize the CP section, merge equalizing signal b, its process is as follows:

Because the transmission vector of 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 X ^kEstimation, then:

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

Through type (10), as can be seen, MP value of preceding MP value and back of b all is to [x (N-P+1) ^T..., x (N) ^T] estimation, then:

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

In formula (11), α, β are the weighted value to two kinds of estimated values, satisfy alpha+beta=1, and ordinary circumstance does not have the priori of reliability, get α=β=0.5.Obtain after the merging:

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

In formula (12), v is the MN vector, and MP value is through merging thereafter.

At step S404, the signal with after the time domain merging of each antenna transforms to frequency domain, and demodulation obtains sending symbol.In formula (12), u is the estimation that sends signal on each antenna time domain.U is the MN vector, and according to u, the time domain that obtains first transmitting antenna sends being estimated as of signal:

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

In formula (13), 1:M:M (N-1)+1, expression is got one since 1 every M, gets M (N-1)+1 always.Be  ₁=[v (1) u (M+1) u (2M+1) ..., u (M (N-1)+1)]] ^T

Equally, i root transmitting antenna time domain sends being estimated as of signal

 _i＝[v(i:M:M(N-1)+i)] (14)

Then, to  _iCarrying out the FFT computing respectively obtains:

 _i＝FFT( _i) (15)

Next, to  _iN be worth the estimation that demodulation respectively obtains the modulation symbol that frequency domain sends.In step S401-S404, except S402, other all available fast Fourier transform (FFT) is realized, to reduce complexity.

Below to being elaborated at the employed fast algorithm of step S402.

3, the existing fast algorithm that calculates formula (9) among the S402

Establish according to document 2 ' " fast algorithm ", Jiang Zengrong, publishing house of the National University of Defense technology, in July, 1998 ':

$T=\left[\begin{array}{ccccc}{t}_0& t_1& t_2& \·\·\·& t_{n-1}\\ t_{-1}& t_0& t_1& \·\·\·& t_{n-2}\\ \·\·\·& \·\·\·& \·\·\·& \·\·\·& \·\·\·\\ t_{-n+1}& t_{-n+2}& t_{-n+1}& \·\·\·& t_0\end{array}\right]---\left(16\right)$

Its neutron piece t _i(i=-n ,-n+1 ..., n) be m rank square formations, then T is that the base dimension is the n rank piecemeal Toeplitz matrix of m * m.The characteristics of piecemeal Toeplitz matrix are that each element on the piecemeal diagonal is equal to each other, and the element that is parallel on each the piecemeal diagonal on the piecemeal leading diagonal also is equal to each other.Represent to be T=t with mathematical expression _Ij=t _J-i(j, i=1 ... n).Be that T can represent with first trip piece and first piece.Provided the fast algorithm of piecemeal Toeplitz matrix inversion in the above-mentioned document, complexity is 0 (n ²m ²).

As can be seen, matrix H is a piecemeal Toeplitz matrix from formula (6).In step S402, corresponding two sub-steps:

● ask H ^-1, H is the square formation on M (N+P) rank, directly asks complexity 0 (M ³(N+P) ³), use the fast algorithm in the above-mentioned document, complexity 0 (M ²(N+P) ²).

● ask H ^-1R directly carries out the phase multiplication, complexity 0 (M ²(N+P) ²).

4, fast algorithm of the present invention

Appendix 1 according to this specification has theorem: the inverse matrix T of triangle Toeplitz matrix T under the piecemeal ^-1, also be triangle Toeplitz matrix under the piecemeal.

Because H is triangle Toeplitz matrix under the piecemeal (the base dimension is triangle Toeplitz matrix under the N rank piecemeal of M * M), so B=H ^-1Be triangle Toeplitz matrix under the piecemeal, it can be represented by first piece, be that example describes with first piece below.

Because HB=I _{(N+P) M}(I _{(N+P) M}Be (N+P) M rank unit matrix), I _{(N+P) M}Can be expressed as the base dimension equally and be triangle Toeplitz matrix under (N+P) rank piecemeal of M * M, as follows:

In formula (17), I _MBe the unit matrix of M * M, 0 _MFull null matrix for M * M.

Above-mentioned matrix I _{(N+P) M}First piece V be:

$V=\left[\begin{array}{c}{I}_M\\ 0_M\\ .\\ .\\ .\\ 0_M\end{array}\right]---\left(18\right)$

Wherein, V is the matrix of (N+P) M * M.

Provide its computational process below:

◆ the first step, ask B=H ^-1, as follows:

Make that U is the first row piece of B,

$V=\left[\begin{array}{c}{U}_0\\ U_1\\ .\\ .\\ .\\ U_{N+P-1}\end{array}\right]---\left(19\right)$

Then:

HU＝V(20)

This can obtain by the method for the group of solving an equation.

Since in the system of OFDM, N＞＞L+1, zero piece is many in the matrix H.According to formula (6), (18), (19), (20), set up equation group:

H ₀ U ₀＝I _M；(21-1)

H ₁ U ₀+H ₀ U ₁＝0 _M；(21-2)

…

H ₀ U _i+H ₁U _i-1+…+H _LU _i-L＝0(12-i)

At first, according to formula (12-1), obtain U ₀, then with U ₀U is obtained in substitution (21-2) ₁, ask U _iThe time, with u _I-1..., u _I-LSubstitution.Obtain at last:

$U_0=H_0^{-1}---\left(22\right)$

$U_i=-H_0^{-1}\left(\underset{k=1}{\overset{L}{\mathrm{\&Sigma;}}}{U}_k{U}_{i-k}\right)1<=i<=N+P-1---\left(23\right)$

In formula (23) sum term, all subscripts are not less than 0, otherwise subscript less than 0 be considered as 0.

[analysis of complexity]

● need to calculate H ₀, H ₁..., H _LInverse matrix, complexity is 0 (L+1) M ³).

● according to H ₀, H ₁..., H _LInverse matrix calculate U, total complexity is 0 ((N+P) M ³).The complexity 0 of both sums ((N+P) M ³).

◆ in second step, calculate b=H ^-1R, as follows:

Computing formula (9), if directly find the solution, complexity is 0 ((N+P) ²(M) ²).

Below, utilize H ^-1Be triangle Toeplitz matrix under the piecemeal, reduce the complexity of calculating, wherein, can be with H ^-1Be divided into M * M matrix sum, each submatrix all has the similar feature of triangle Toeplitz matrix down.

I, j Matrix C _{I, j}(dimension and H ^-1The same) and H ^-1The pass be:

C _Ij(p-1) * M+i capable (1＜=p＜=N+P), (q-1) * (element of 1＜=q＜=N+P) is H to the M+j row ^-1(p-1) * M+i capable, (q-1) * element of M+j row, other are zero.Be exemplified below:

Suppose the H of B ^-1Be that a base dimension is 2 * 23 rank piecemeal lower triangular matrixs, then

$B=\left[\begin{array}{ccc}{B}_1& & \\ B_2& B_1& \\ B_3& B_2& B_1\end{array}\right]=\left[\begin{array}{cccccc}{a}_1& a_2& 0& 0& 0& 0\\ a_3& a_4& 0& 0& 0& 0\\ a_5& a_6& a_1& a_2& 0& 0\\ a_7& a_8& a_3& a_4& 0& 0\\ a_9& a_{10}& a_5& a_6& a_1& a_2\\ a_{11}& a_{12}& a_7& a_8& a_3& a_4\end{array}\right]---\left(24\right)$

Then:

$C_{11}=\left[\begin{array}{cccccc}{a}_1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ a_5& 0& a_1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ a_9& 0& a_5& 0& a_1& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right]---(25-1)$

$C_{21}=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ a_3& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ a_7& 0& a_3& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ a_{11}& 0& a_7& 0& a_3& 0\end{array}\right]---(25-2)$

$C_{12}=\left[\begin{array}{cccccc}0& a_2& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& a_6& 0& a_2& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& a_{10}& 0& a_6& 0& a_2\\ 0& 0& 0& 0& 0& 0\end{array}\right]---(25-3)$

$C_{22}=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& a_4& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& a_8& 0& a_4& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& a_{12}& 0& a_8& 0& a_4\end{array}\right]---(25-4)$

Wherein:

$B=\underset{i=1}{\overset{M}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{M}{\mathrm{\&Sigma;}}}{C}_{\mathrm{ij}}---\left(26\right)$

Then the product of a B and a vectorial r can be expressed as follows:

$d=\mathrm{Br}=\underset{i=1}{\overset{M}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{M}{\mathrm{\&Sigma;}}}{C}_{\mathrm{ij}}r---\left(27\right)$

Connect example, make e=C ₁₁* r, e=[e ₁, e ₂, e ₃, e ₄, e ₅, e ₆] ^T, e then ₂=e ₄=e ₆=0, and:

$\left[\begin{array}{c}{e}_1\\ e_3\\ e_5\end{array}\right]=\left[\begin{array}{ccc}{a}_1& & \\ a_5& a_1& \\ a_9& a_5& a_1\end{array}\right]\left[\begin{array}{c}{r}_1\\ r_3\\ r_5\end{array}\right]=T_{11}\left[\begin{array}{c}{r}_1\\ r_3\\ r_5\end{array}\right]---\left(28\right)$

In formula (28), $T_{11}=\left[\begin{array}{ccc}{a}_1& & \\ a_5& a_1& \\ a_9& a_5& a_1\end{array}\right]$ Be following triangle Toeplitz matrix.Because T ₁₁So triangle Toeplitz matrix under being is T in the formula (28) ₁₁With [r ₁, r ₃, r ₅] ^TProduct can regard T as ₁₁The 1st row with [r ₁, r ₃, r ₅] ^TDo linear convolution, obtain getting preceding 3 values behind the result, and linear convolution can calculate with FFT.Other C _IjSame characteristic is arranged.

In sum, calculate H ^-1The fast algorithm of r is as follows:

1) to C _Ij(1＜=i, j＜=M), obtain the vector v ec of (N+P) * 1 _Ij, this vector is in (28), T _IjFirst row.Vec _IjFollow the example of as follows:

vec _ij＝H ^-1(i:M:i+(N+P-1)*M，j) (29)

2) M (N+P) dimensional vector from r, in get N+P as r_vec _Ij, follow the example of as follows:

r_vec _ij＝r(j:M:j+(N+P-1)*M) (30)

3) be r_vec _IjBe j element from r, M comes value with the interval.

4) with r_vec _IjMend N+P the zero r_e that obtains _Ij, with vec _IjMend N+P the zero vec_e that obtains _Ij

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

5) make u then _IjBe f _IjPreceding (N+P) individual value.

6) initialization part _IjComplete zero column vector for M (N+P) * 1.Assignment:

Part _Ij(i:M:i+ (N+P-1) * M, 1)=u _Ij(32) part _IjFrom i element,, be endowed u every M _IjCorresponding value, i.e. part _Ij(i, 1)=u _Ij(1), part _Ij(i+M, 1)=u _Ij(2), part _Ij(i+2M, 1)=u _Ij(3) ...

7) obtain all part _IjAfterwards, calculate:

$b=H^{-1}r=B^{-1}r=\underset{i=1}{\overset{M}{\mathrm{\&Sigma;}}}\underset{j=1}{\overset{M}{\mathrm{\&Sigma;}}}{\mathrm{part}}_{\mathrm{ij}}---\left(33\right)$

[analysis of complexity]

Part of every calculating _Ij, complexity is 0 (2 (N+P) log ₂(2 (N+P)) will calculate M altogether ²Inferior, total complexity is 0 (M ²2 (N+P) log ₂(2 (N+P)).

5, the complexity of method of the present invention and prior art relatively

When CP length during greater than (channel length-1), the interference between no intersymbol interference and the subcarrier, its channel is single footpath channel at frequency domain, and MIMO detects and carries out at frequency domain, and the complexity of zero forcing algorithm is 0 (M ³N).

Table 1 has provided the comparison of zero forcing algorithm when CP is not enough.

The present invention

Existing fast algorithm

Do not use fast algorithm

Complexity

Max{0(M 22(N+P) log 2(2(N+P))，0((N+P) M 3))

0((N+P) 2M 2)

0((N+P) 3M 3)

First example (M * M ofdm system)

CP below described be not the concrete balancing procedure under 0 the situation.

1) at step S401, the estimation according to previous symbol obtains

(current is k) according to formula (8), through decision-feedback, obtains the estimation of the received signal of current elimination intersymbol interference.

2),, utilize fast algorithm to obtain zero forcing equalization and separate b according to formula (9) at step S402.Because the fast algorithm of calculation procedure S402 illustrates more complicated, will provide separately in the back.

3) at step S403, utilize the CP section, merge equalizing signal b, its detailed process is as follows:

B is the column vector of M (N+P) * 1, makes that v is the column vector of MN * 1, then has:

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)

4) at step S404, the signal with after the time domain merging of each antenna transforms to frequency domain, and demodulation obtains sending symbol.I root transmitting antenna time domain is sent the  that is estimated as of signal _i=[v (i:M:M (N-1)+i)], (i:M:M (N-1)+i represents to get one since i every M, gets M (N-1)+I always.Then to  _iCarry out the FFT computing respectively and obtain  _i=FFT ( _i).Then to  _iN be worth the estimation that demodulation respectively obtains the modulation symbol that frequency domain sends.

[realization of step S402]

According to formula (9), calculate b=H at step S02 ^-1R.Its calculation procedure as shown in Figure 7.

At step S501, at first to import:

◆ the sub-piece of the non-zero of first piece of channel matrix, according to formula (6), the sub-piece of non-zero that obtains first piece is H ₀, H ₁..., H _L

◆ the signal r after the decision-feedback, antenna number M, sub-carrier number N, circulating prefix-length P.

The concrete steps of step S402 are as follows:

At step S501, according to the sub-piece of the non-zero of first piece of channel matrix, obtain by the method for organizing of solving an equation, first piece U of channel matrix inverse matrix, its computing formula is seen formula (22), (23).Enter the circulation 1＜=i＜+M, 1＜=j＜+M.

At step S502, in the j of U row,, be at interval with M since i, get N+P value, obtain vector v ec _Ij, from vectorial r,, be at interval with M with j beginning, get N+P value, obtain vectorial r_vec _IjVec _IjComputing formula see (29), r_vec _IjComputing formula see (30).

At step S703, calculate vec by FFT _IjWith r_vec _IjLinear convolution, and the preceding N+P of line taking convolution results obtains u _IjIts detailed process is with r_vec _IjMend N+P the zero r_e that obtains _Ij, with vec _IjMend N+P the zero vec_e that obtains _IjAccording to formula (31), obtain f then _Ij, make u then _IjBe f _IjPreceding (N+P) individual value.

In step S504, initialization part _IjBe complete zero column vector of M (N+P) * 1, and with u _IjAssignment is given part _IjBegin with i element, be spaced apart the element (N+P) of M, part _IjThe formula of assignment is seen formula (32).

Obtain all part _IjThe time, enter step S505.At step S505, with all part _IjSummation obtains b, referring to formula (33).

For example: 2 * 2 systems, multipath length are L+1=3, N=4, and P=1, its channel matrix is:

$H=\left[\begin{array}{ccccc}{H}_0& 0_2& 0_2& 0_2& 0_2\\ H_1& H_0& 0_2& 0_2& 0_2\\ H_2& H_1& H_0& 0_2& 0_2\\ 0_2& H_2& H_1& H_0& 0_2\\ 0_2& 0_2& H_2& H_1& H_0\end{array}\right]---\left(35\right)$

In formula (35), 0 ₂It is 2 * 2 null matrix.H ₀, H ₁, H ₂Be 2 * 2 matrixes.

1) according to the estimation of previous symbol, obtains

(current is k) according to formula (8), through decision-feedback, obtains the estimation r=[r of the received signal of current elimination intersymbol interference ₁r ₂r ₃r ₄r ₅r ₆r ₇r ₈r ₉r ₁₀] ^T

2) according to formula (19), H ^-1First piece U be:

$U=\left[\begin{array}{c}{U}_0\\ U_1\\ U_2\\ U_3\\ U_4\end{array}\right]---\left(36\right)$

Then:

$\left[\begin{array}{ccccc}{H}_0& 0_2& 0_2& 0_2& 0_2\\ H_1& H_0& 0_2& 0_2& 0_2\\ H_2& H_1& H_0& 0_2& 0_2\\ 0_2& H_2& H_1& H_0& 0_2\\ 0_2& 0_2& H_2& H_1& H_0\end{array}\right]\left[\begin{array}{c}{U}_0\\ U_1\\ U_2\\ U_3\\ U_4\end{array}\right]=\left[\begin{array}{c}{I}_2\\ 0_2\\ 0_2\\ 0_2\\ 0_2\end{array}\right]---\left(37\right)$

In the formula (37), I ₂Be 2 * 2 unit matrix, O ₂It is 2 * 2 null matrix.Then

H ₀ U ₀＝I ₂ (38-1)

H ₁ U ₀+H ₀ U ₁＝O ₂ (38-2)

H ₂ U ₀+H ₁ U ₁+H ₀ U ₂＝O ₂ (38-3)

H ₂ U ₁+H ₁ U ₂+H ₀ U ₃＝O ₂ (38-4)

H ₂ U ₂+H ₁ U ₃+H ₀ U ₄＝O ₂ (38-5)

Begin to separate from (38-1), obtain:

$U_0=H_0^{-1}---(39-1)$

$U_1=-H_0^{-1}\left(H_1{U}_0\right)---(39-2)$

$U_2=-H_0^{-1}(H_2{U}_0+H_1{U}_1)---(39-3)$

$U_3=-H_0^{-1}(H_2{U}_1+H_1{U}_2)---(39-4)$

$U_4=-H_0^{-1}(H_2{U}_2+H_1{U}_3)---(39-5)$

Obtain H ^-1First row U be, as shown in table 2:

Table 2

U	First row	Secondary series
			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) according to first piece U, (29) obtain:

vec ₁₁＝[a ₁ a ₅ a ₉ a ₁₃ a ₁₇] ^T；

vec ₁₂＝[a ₂ a ₆ a ₁₀ a ₁₄ a ₁₈] ^T；

vec ₂₁＝[a ₃ a ₇ a ₁₁ a ₁₅ a ₁₉] ^T；

vec ₂₂＝[a ₄ a ₈ a ₁₂ a ₁₆ a ₂₀] ^T；

Then, r is divided into two parts:

R1＝[r ₁ r ₃ r ₅ r ₇ r ₉] ^T。

R2＝[r ₁ r ₃ r ₅ r ₇ r ₉] ^T。

Calculate then:

f ₁₁＝vec ₁₁*R1 (40-1)

f ₁₂＝vec ₁₂*R2 (40-2)

f ₂₁＝vec ₂₁*R1 (40-3)

f ₂₂＝vec ₂₂*R2 (40-3)

In formula (40), * represents convolution, can realize with FFT, sees formula (30), (31).In formula (40), the point that convolution obtains is 10.Make u then ₁₁Be f ₁₁Preceding 5 points, u ₁₂Be f ₁₂Preceding 5 points, u ₁₃Be f ₁₃Preceding 5 points, u ₁₄Be f ₁₄Preceding 5 points.

Then, make part ₁₁For being initialized as zero 10 * 1 vectors, according to (32) with u ₁₁5 points compose to part ₁₁1,3,5,7,9 point on, that is:

part ₁₁＝[u ₁₁(1) 0 u ₁₁(2) 0 u ₁₁(3) 0 u ₁₁(4) 0 u ₁₁(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；

part ₂₁＝[0 u ₂₁(1) 0 u ₂₁(2) 0 u ₂₁(3) 0 u ₂₁(4) 0 u ₂₁(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) then, obtain b=part according to formula (33) ₁₁+ part ₁₂+ part ₂₁+ part ₂₂Obtain:

b＝[b ₁ b ₂ b ₃ b ₄ b ₅ b ₆ b ₇ b ₈ b ₉ b ₁₀] ^T。

According to formula (34), obtain v=[b ₃b ₄b ₅b ₆b ₇b ₈0.5 * (b ₁+ b ₉) 0.5 * (b ₂+ b ₁₀)] ^T=[v ₁v ₂v ₃v ₄v ₅v ₆v ₇v ₈] ^T

5) according to formula (14), obtain  ₁=[v ₁v ₃v ₅v ₇] ^T,  ₂=[v ₂v ₄v ₆v ₈] ^TThen, with  ₁,  ₂Transform to frequency domain and demodulation by formula (15), obtain the estimation of the modulation symbol that sends on two transmitting antennas.

Second example (1 * 1 ofdm system)

Equalization methods under the situation of no CP is below described.

Under the situation of no CP, the same in transaction module and the document 1.

Equally, as shown in Figure 3, need to carry out three following processes in this example.

1) decision-feedback process

From formula (4), have:

$d=y^k-H_2{Q}^H{\tilde{s}}^{k-1}---\left(41\right)$

In formula (5), y ^kIt is the reception vector of k OFDM symbol.

Be estimate vector (sending the estimation of symbol on all subcarriers on the frequency domain, the N dimensional vector) to sending symbol on k-1 the OFDM symbol,

For

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

2) balancing procedure

From formula (4), have:

$b={(H_0-H_1)}^{-1}(y^k-H_2{Q}^H{\tilde{s}}^{k-1})={(H_0-H_1)}^{-1}d---\left(42\right)$

In formula (6), d is the signal after the time domain decision-feedback.A=H ₀-H ₁Be the matrix that causes the ICI interference of OFDM, b is through the signal behind the decision feedback equalization.

3) demodulating process

Signal b behind the time domain equalization is transformed to frequency domain, and demodulation then obtains sending the estimation of symbol, has:

${\tilde{s}}^k=\mathrm{Qb}---\left(43\right)$

Then, right Carry out demodulation, obtain sending the estimation of symbol.

In formula (7), b is the signal behind the time domain equalization, and Qb arrives frequency domain with the signal transformation behind the time domain equalization, and demodulation then obtains sending the estimation of symbol.

The computation complexity of existing method:

1) decision-feedback process: can calculate complexity Nlog with FFT ₂(N);

2) balancing procedure: matrix inversion, complexity 0 (N ³);

3) demodulating process: complexity Nlog ₂(N).

As can be seen, mainly be that the balancing procedure complexity is very high, hindered actual application.

Below describe balancing procedure in detail,, obtain according to formula (1), (2):

The expression formula of matrix A as shown in Figure 8 in the formula (44).In this example, purpose is to reduce balanced complexity, and balancing procedure among change Fig. 3 reduces and finds the solution the complexity of formula (42).Other parts are constant.

The first step is asked A ^-1

According to the proof of appendix 1, can get the non-vanishing following triangle T inverse of a matrix matrix T in diagonal angle ^-1, be following triangle T matrix.

Ask A with the method for solving an equation (utilizing the null value of A) ^-1First:

Make u=[u ₀, u ₂..., u _N-1] ^TBe A ^-1First.

Since in the system of OFDM, N＞＞L+1, be similar to formula (17)-(21):

h ₀ u ₀＝1；(45-1)

h ₁ u ₀+h ₀ u ₁＝0；(45-2)

…

h ₀ u _i+h ₁u _i-1+…+h _Lu _i-L＝0(45-i+1)

According to (45-1), obtain u ₀, then with u ₀U is obtained in substitution (45-2) ₁, ask u _iThe time, with u _I-1..., u _I-LSubstitution.

[analysis of complexity]

Add complexity 0 (LN) owing to have N equation, each equation to have the L number to take advantage of at most.Like this, just tried to achieve A ^-1First row.

In second step, ask A ^-1d

Because A ^-1Be following triangle T matrix, ask:

r＝A ^-1d (46)

If directly ask, complexity 0 (N ²).

A ^-1Be following triangle T battle array, suppose that it first classifies u=u as ₀, u ₁₁..., u _N-1, then:

r ₀＝d ₀u ₀(47-1)

r ₁＝d ₁u ₀+d ₀u ₁(47-2)

r ₂＝d ₂u ₀+d ₁u ₁+d ₀u ₂(47-3)

……

A ^-1D and linear convolution relatively, the value of r is quite and the top n value of a and v linear convolution.Utilize FFT to ask the method for linear convolution then, can obtain r.

Make u_e represent that u mends N 0, d_e represents that d mends N 0, calculates then:

r_e＝ifft(fft(u_e).*fft(a_e)) (48)

In formula (48) .* represents dot product, makes r equal the value of the preceding N of r_e then.

[analysis of complexity]

The first step is asked A ^-1, complexity (L+1) N;

In second step, ask A ^-1D, complexity 0 (2Nlog ₂(2N)).

Therefore, total complexity 0 (2Nlog ₂(2N))+(L+1) N.The complexity in second step is greater than the first step generally speaking, and total complexity is thought 0 (2Nlog ₂(2N)).Table 3 provides the complexity comparison with other method.

Table 3 under the situation of 1 * 1 system with the comparison of other method

	The present invention	The Levinson fast algorithm	Do not use fast algorithm
				Complexity	0(2Nlog 2(2N))	0(N 2)	0(N 3)

Below illustrate how under the situation of no CP, to carry out equalization methods of the present invention with concrete example.

Fig. 7 shows the flow chart that carries out equalization methods of the present invention under the situation of no CP.

At step S701, obtain the received signal d after S401 obtains decision-feedback, the estimated value h of channel ₀, h ₂..., h _LThereby obtain the first row a=[h of channel matrix ₀, h ₂..., h _L, 0 ... 0] ^TCalculate A ^-1First row, be designated as vectorial u.Fig. 8 shows the detailed process of step S701.

As shown in Figure 8, calculate A ^-1First row.A ^-1First show N element, be u=[u ₀, u ₂..., u _N-1] ^T

In step 801, calculate and at first obtain channel value h ₀, h ₂..., h _L, calculate u ₀=1/h ₀

In step 802, calculate u ₁To u _LAccording to h ₁To h _k, u ₁To u _K-1, calculate u _k(2＜=k＜=L).

$u_k=-\left(\underset{t=1}{\overset{k}{\mathrm{\&Sigma;}}}{h}_t{u}_{k-t}\right)/h_0---\left(49\right)$

In step 803, calculate u _L+1To u _N-1According to h ₀To h _L, u _K-LTo u _K-1, calculate u _k(L＜k＜=N).

$u_k=-\left(\underset{t=1}{\overset{L}{\mathrm{\&Sigma;}}}{h}_t{u}_{k-t}\right)/h_0---\left(50\right)$

Like this, behind step S803, obtained whole u=[u ₀, u ₁..., u _N-1] ^T

For example, suppose N=5, L+1=3.Then

$A=\left[\begin{array}{ccccc}{h}_0& 0& 0& 0& 0\\ h_1& h_0& 0& 0& 0\\ h_2& h_1& h_0& 0& 0\\ 0& h_2& h_1& h_0& 0\\ 0& 0& h_2& h_1& h_0\end{array}\right]---\left(51\right)$

A then ^-1First row can be as shown in the formula expression:

$\left[\begin{array}{ccccc}{h}_0& 0& 0& 0& 0\\ h_1& h_0& 0& 0& 0\\ h_2& h_1& h_0& 0& 0\\ 0& h_2& h_1& h_0& 0\\ 0& 0& h_2& h_1& h_0\end{array}\right]\left[\begin{array}{c}{u}_0\\ u_1\\ u_2\\ u_3\\ u_4\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\\ 0\end{array}\right]---\left(52\right)$

According to formula (19), following 5 equatioies are arranged:

h ₀ u ₀＝1 (53-1)

h ₁ u ₀+h ₀ u ₁＝0 (53-2)

h ₂ u ₀+h ₁ u ₁+h ₀ u ₂＝0 (53-3)

h ₂ u ₁+h ₁ u ₂+h ₀ u ₃＝0 (53-4)

h ₂ u ₂+h ₁ u ₃+h ₀ u ₄＝0 (53-5)

(53-1) begins to separate from formula, obtains:

u ₀＝1/h ₀ (54-1)

u ₁＝-(h ₁ u ₀)/h ₀ (54-2)

u ₂＝-(h ₂ u ₀+h ₁ u ₁)/h ₀ (54-3)

u ₃＝-(h ₂ u ₁+h ₁ u ₂)/h ₀ (54-4)

u ₄＝-(h ₂ u ₂+h ₁ u ₃)/h ₀ (54-5)

At last, obtain u=[u ₀, u ₂..., u _N-1] ^T

At step S702,, utilize the method for FFT to ask A according to the d after u that obtains and the decision-feedback ^-1D calculates and sees formula (48).Fig. 9 shows the detailed process of step S702.

As shown in Figure 9, calculate A ^-1First row of d.A ^-1First show N element, be u=[u ₀, u ₁..., u _N-1] ^T

According to the received signal d (N * 1 vector) after formula (5) the input decision-feedback, obtain the first row u (N * 1 vector) of channel reverse matrix from S701, at step S901, make d mend N 0 and obtain d_e (2N * 1 vector), u mends N 0 and obtains u_e (2N * 1 vector).

At step S902, d_e is done the FFT conversion that 2N orders obtain d_f (2N * 1 vector), u_e is done the FFT conversion that 2N orders obtain u_f (2N * 1 vector).

At step S903, calculate u_f.*d_f, the corresponding element of note is divided by, and obtains r_f (2N * 1 vector).

At step S904, r_f is carried out 2N point IFFT conversion, obtain b_e, then b_e is got the top n point and obtain time thresholding b after the equilibrium.

For example, suppose N=5, just obtained u ₀, u ₁, u ₂, u ₃, u ₄Obtained the signal d after the decision-feedback from S401 ₀, d ₁, d ₂, d ₃, d ₄

According to above-mentioned method, d_e=[d ₀d ₁d ₂d ₃d ₄0000 0] ^T, u_e=[u ₀u ₁u ₂u ₃u ₄0000 0] ^T

Then, d_e is 10 FFT, obtains d_f=FFT (d_e), d_f=[df ₀Df ₁Df ₂Df ₃Df ₄Df ₅Df ₆Df ₇Df ₈Df ₉] ^TThe FFT that u_e is done at 10 obtains u_f=FFT (u_e), u_f=[uf ₀Uf ₁Uf ₂Uf ₃Uf ₄Uf ₅Uf ₆Uf ₇Uf ₈Uf ₉] ^T

Next, make r_f=[rf ₀=uf ₀* df ₀Rf ₁=uf ₁* df ₁Rf ₂=uf ₂* df ₂Rf ₃=uf ₃* df ₃Rf ₄=uf ₄* df ₄Rf ₅=uf ₅* df ₅Rf ₆=uf ₆* df ₆Rf ₇=uf ₇* df ₇Rf ₈=uf ₈* df ₈Rf ₉=uf ₉* df ₉] ^T

The IFFT conversion of r_f being done at 10 obtains b_e=IFFT (r_f).b_e＝[be ₀ be ₁be ₂ be ₃ be ₄ be ₅ be ₆ be ₇ be ₈ be ₉] ^T。

At last, b_e is got preceding 5 points and obtain time thresholding b=[be after the equilibrium ₀Be ₁Be ₂Be ₃Be ₄] ^T

Appendix 1

Theorem (invention): the inverse matrix A of triangle Toeplitz matrix A under the piecemeal ^-1, also be triangle Toeplitz matrix under the piecemeal.

X in following computing _IjWhat represent is the matrix of m * m, and X is the matrix of arbitrary nm * nm, and piecemeal as the formula (16).

Proof:

Suppose that A is triangle Toeplitz matrix under the n rank piecemeal of m * m for the base dimension.Three of A character then:

1, A ^-1Unique.

2, because A is a lower triangular structure, so A ^-1It also is lower triangular structure.

3, because A is a triangle Toeplitz matrix under the piecemeal, so A can be represented by first piece of A.The element that makes the first row piece of A is A ₁₁=a ₁, A ₂₁=a ₂..., A _N1=a _nBecause A is the T matrix, so A _Ij=A _{I+P, j+P}Above ai is m * m matrix, A _IjIt also is m * m matrix.

4, make that B is a lower triangular matrix, make lower triangular matrix C=AB then

$C_{\mathrm{ij}}=\underset{k=1}{\overset{n}{\mathrm{\&Sigma;}}}{A}_{\mathrm{ik}}{B}_{\mathrm{kj}}$ (appendix 1-1)

In (appendix 1-1), C _Ij, A _Ik, B _KjIt all is m * m matrix.

Because A, B is a lower triangular matrix, then

When k＞i, A _Ik=0.

When j＞k, B _Kj=0.

So, have:

$C_{\mathrm{ij}}=\underset{k=j}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{\mathrm{ik}}{B}_{\mathrm{kj}}$ (appendix 1-2)

Make B=A ^-1, then, C=AB=I then _Nm(I _NmBe n * M rank unit matrix).This moment, C equaled unit matrix, because unit matrix also is a triangle T matrix under the piecemeal, so C also is a triangle T matrix under the piecemeal.Satisfy:

C _Ij=C _{I+p, j+p}(appendix 1-3)

In (appendix 1-3), C _IjMatrix for m * m.P is an integer, and i, j, i+p, j+p ∈ 1 ..., n.Because C can be represented by first piece of C, then can make j=1 in (appendix 1-3).Obtain

C _{I+p, l+p}=C _{I, l}(appendix 1-4)

According to (appendix 1-2), have:

$C_{1+p,1+p}=\underset{k=1+p}{\overset{i+p}{\mathrm{\&Sigma;}}}{A}_{i+p,k}{B}_{k,1+p}$ (appendix 1-5)

Obtain according to (appendix 1-3), (appendix 1-4), (appendix 1-5):

$\underset{k=1}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{\mathrm{ik}}{B}_{k1}=\underset{m=1+p}{\overset{i+p}{\mathrm{\&Sigma;}}}{A}_{i+p,m}{B}_{m,1+p}$ (appendix 1-6)

Make in (appendix 1-6), q=m-p then has:

$\underset{k=1}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{\mathrm{ik}}{B}_{k1}=\underset{q=1}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{i+p,q+p}{B}_{q+p,1+p}$ (appendix 1-7)

And make equation the right in (appendix 1-7), and make q=k, then have:

$\underset{k=1}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{\mathrm{ik}}{B}_{k1}-\underset{k=1}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{i+p,k+p}{B}_{k+p,1+p}=0$

$\underset{k=1}{\overset{i}{\mathrm{\&Sigma;}}}{A}_{\mathrm{ik}}(B_{k1}-B_{k+p,1+p})=0$ (appendix 1-8)

Obviously, if B _{K+p, l+p}=B _K1, be that (appendix 1-8) one separates.Because B is that the inverse matrix of A is according to character (1), A ^-1Unique, so promptly B is a triangle Toeplitz matrix under the piecemeal.Card is finished.

In like manner, suppose that A is triangle Toeplitz matrix under 1 * 1 the n rank piecemeal for base dimension, promptly A is triangle Toeplitz matrix, then A down ^-1Be following triangle Toeplitz matrix.

Claims (5)

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

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

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.

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