CN101047467A - Parallel detection method for double space hour code in multiple input output system - Google Patents

Abstract

A parallel-detecting method of double space hour code in multi-input / multi-output system includes structuring channel matrix H and receiving signal as per space hour packet coding according to received signal, grouping matrix H and confirming position of each group in QR decomposition, carrying out QR decomposition on matrix H to obtain orthogonal matrix Q and top triangular matrix R as well as signal vector, carrying out maximum adaptive likelihood detection in group as per the first branch value M1 to obtain M1 number life path, eliminating interference of the first group on the second one being carried with said detections to obtain M2 number life path.

Description

The parallel detecting method of double space hour code in the multi-input multi-output system

Technical field

The present invention relates to a kind of two (void) Space-Time Block Coding signal detecting methods that relate in multiple-input and multiple-output (MIMO) communication, particularly at two (void) empty time-code, a kind of grouping self adaptation QRM detection algorithm is proposed, can utilize the characteristics of two (void) empty time-code channel matrixes, data are divided into two groups, maximum likelihood detection method in its complexity is significantly less than direct group.

Background technology

Multiple-input and multiple-output (MIMO) technology is the important breakthrough of wireless mobile communications field intelligent antenna technology.The MIMO technology is meant the transmission of data and receives and all adopted many antennas.Studies show that utilize the MIMO technology can improve the capacity of channel, the while also can be improved the reliability of channel, reduces the error rate.The heap(ed) capacity of mimo system or maximum size be linear increasing with the increase of minimum antenna number.And under similarity condition, adopting the common antenna system of many antennas or aerial array at receiving terminal or transmitting terminal, its capacity only increases with the logarithm of antenna number.Comparatively speaking, the MIMO technology has great potentiality for the capacity that improves wireless communication system, is the key technology that the third generation mobile communication system adopts.

From reducing the angle of decoding complexity, Alamouti proposes a kind of transmission method that utilizes two transmit antennas.Fig. 1 shows the theory diagram of Space-Time Block Coding.At first, at transmit leg, data are mapped to planisphere through modulating equipment 101 earlier, are converted into the transmission symbol.Then, will send symbol and be divided into one group of [c of two symbols ₁, c ₂] deliver to space-time block code device 102.Through behind the space-time block code, in two symbol periods, two antennas are launched two symbols simultaneously.The 1st cycle, 1 c of antenna ₁, 2 c of antenna ₂In the 2nd cycle, 1-c of antenna ₂ ^*, 2 c of antenna ₁ ^*(subscript * represents to get complex conjugate).Each row symbol of encoder matrix sends on different antennae simultaneously, and the symbol that sends on the constellation point symbol that sends on the antenna and in addition any antenna is a quadrature.Specifically as shown in table 1

Table 1

	Cycle 1	Cycle 2
			Antenna 1	c 1	-c 2 *
Antenna 2	c 2	c 1 *

In recipient's one side,, can obtain channel by following formula (1) expression as long as there is an antenna just can obtain reliable detection (if channel is constant in the time of these two characters, below the hypothesis reception antenna is 1).

h _i(nT)＝h _i((n+1)T)，i＝1，2 (1)

In the formula (1), T is a symbol period.

If two interior at interval received signals of adjacent characters are r ₁, r ₂Then received signal can with following formula (2a) and (2b) expression.

r ₁＝h ₁c ₁+h ₂c ₂+η ₁ (2a)

$r_2=-h_1{c}_2^{*}+h_2{c}_1^{*}+{\mathrm{\η}}_2---\left(2b\right)$

In formula (2), η ₁, η ₂Represent additive white Gaussian noise.Definition $r={\left[\begin{array}{cc}{r}_1& {\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^{*}\end{array}\right]}^T$ (T represents transposition), $r={\left[\begin{array}{cc}{r}_1& {\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^{*}\end{array}\right]}^T,$ Coded identification vector c=[c ₁c ₂] ^T, noise vector $\mathrm{\&eta;}={\left[\begin{array}{cc}{\mathrm{\&eta;}}_1& {\mathrm{\&eta;}}_2^{*}\end{array}\right]}^T,$ Then formula (2) can be write as shown in following equation (3).

r＝Hc+η (3)

In formula (3), the definition of channel matrix H can be represented with following equation (4).

$H=\left[\begin{array}{cc}{h}_1& h_2\\ h_2^{*}& {-h}_1^{*}\end{array}\right]---\left(4\right)$

H is an orthogonal matrix, the equation (5) below satisfying.

H ^HH＝HH ^H＝pI ₂ (5)

In equation (5), p=|h ₁| ²+ | h ₂| ², I _NThe expression row and column all is the unit matrix (subscript H represents to grip altogether transposition) of N.

Contrast formula (3), definition C is that all possible symbol is to (c ₁, c ₂) set, then Zui Jia maximum likelihood decoder is to use following formula (6) expression.

$\hat{c}=\underset{\hat{c}\&Element;C}{\arg \min }{\left|\right|r-H\hat{c}\left|\right|}^2---\left(6\right)$

Because H is an orthogonal matrix, so the decoding of equation (6) can further be simplified.Definition $\tilde{r}=H^{H}r,$ $\widetilde{\mathrm{\&eta;}}=H^H\mathrm{\&eta;},$ Use H ^HThe both sides of premultiplication equation (3) can obtain equation (7).

$\tilde{r}=\mathrm{pc}+\widetilde{\mathrm{\&eta;}}---\left(7\right)$

The decoding of equation this moment (6) becomes by shown in the following equation (8).

$\hat{c}=\underset{\hat{c}\&Element;C}{\arg \min }{\left|\right|\tilde{r}-p\hat{c}\left|\right|}^2---\left(8\right)$

At this moment, just the maximum likelihood decision problem of two dimension is converted into two one dimension decision problems.

The processing operation of receiving terminal is described below.Antenna 103 received signals, channel estimation unit 104 estimates channel h ₁, h ₂The linear decoder 105 of maximum likelihood obtains the received signal r in adjacent two moment ₁, r ₂Then, the linear decoder 105 of maximum likelihood becomes the signal in two moment $r={\left[\begin{array}{cc}{r}_1& {\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^{*}\end{array}\right]}^T,$ Use H ^HPremultiplication r obtains according to equation (8)

Demodulation output unit 106 will then

Be mapped as bit stream output.

Below with reference to Fig. 2 the transmission and the reception operation of double space hour code are described.Double space hour code uses two code structures when empty to send simultaneously.As shown in Figure 2, use two reception antennas, generally use the method for ZF_SIC to detect the recipient the recipient.

In two Space Time Coding devices among Fig. 2 each all sends according to the rule of empty time-code, satisfies channel constant condition in two symbols.

The recipient, making the 1st signal that the moment receives of first antenna is r ₁₁, the 2nd signal that the moment receives is r ₁₂, making the 1st signal that the moment receives of the 2nd antenna is r ₂₁, the 2nd signal that the moment receives is r ₂₂Make r ₁=r ₁₁, ${\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">r</figure-callout>}}_2={\mathrm{<figure-callout\; id="12"\; label="r"\; filenames="A20061006831100261.png,A20061006831100422.png"\; state="\{\{state\}\}">r</figure-callout>}}_{12}^{*},$ r ₃＝r ₂₁， ${\mathrm{<figure-callout\; id="4"\; label="r"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">r</figure-callout>}}_4={\mathrm{<figure-callout\; id="22"\; label="r"\; filenames="A20061006831100352.png,A20061006831100372.png"\; state="\{\{state\}\}">r</figure-callout>}}_{22}^{*};$ r＝[r ₁，r ₂，r ₃，r ₄] ^T。Then obtain the reception model that following equation (9) provides.

r＝Hs+n (9)

H represents receive channel, and its implication is shown in equation (10a to 10c).N is a noise, s=[s ₁, s ₂, s ₃, s ₄] ^T, s wherein ₁, s ₂Be the symbol that first Space Time Coding device sends, s ₃, s ₄It is the symbol that the 2nd Space Time Coding device sends.

For first user: h ₁₁Channel for first transmitting antenna to the first reception antenna of first sky time-code (two antennas); h ₂₁Be the channel between second transmitting antenna to the first reception antenna of first sky time-code; g ₁₁Be first transmitting antenna of second empty time-code (two antennas) and the channel between first reception antenna; g ₂₁Be second transmitting antenna of second empty time-code and the channel between first reception antenna.Equally, h ₁₂Channel for first transmitting antenna to the second reception antenna of first sky time-code; h ₂₂Be the channel between second transmitting antenna to the second reception antenna of first sky time-code, g ₁₂Be first transmitting antenna of second empty time-code and the channel between second reception antenna; g ₂₂Be second transmitting antenna of second empty time-code and the channel between second reception antenna.Now, what the base station obtained is these eight channel values, and these eight channel values are done as down conversion:

$H_1=\left[\begin{array}{cc}{h}_{11}& h_{21}\\ h_{21}^{*}& -h_{11}^{*}\end{array}\right],$ $G_1=\left[\begin{array}{cc}{g}_{11}& g_{21}\\ {\mathrm{<figure-callout\; id="21"\; label="g"\; filenames="A20061006831100352.png,A20061006831100372.png"\; state="\{\{state\}\}">g</figure-callout>}}_{21}^{*}& -{\mathrm{<figure-callout\; id="11"\; label="g"\; filenames="A20061006831100261.png,A20061006831100421.png"\; state="\{\{state\}\}">g</figure-callout>}}_{11}^{*}\end{array}\right]---\left(10a\right)$

$H_2=\left[\begin{array}{cc}{h}_{12}& h_{22}\\ {\mathrm{<figure-callout\; id="22"\; label="h"\; filenames="A20061006831100352.png,A20061006831100372.png"\; state="\{\{state\}\}">h</figure-callout>}}_{22}^{*}& {-\mathrm{<figure-callout\; id="12"\; label="h"\; filenames="A20061006831100261.png,A20061006831100422.png"\; state="\{\{state\}\}">h</figure-callout>}}_{12}^{*}\end{array}\right],$ ${\mathrm{<figure-callout\; id="2"\; label="G"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">G</figure-callout>}}_2=\left[\begin{array}{cc}{g}_{12}& g_{22}\\ {\mathrm{<figure-callout\; id="22"\; label="g"\; filenames="A20061006831100352.png,A20061006831100372.png"\; state="\{\{state\}\}">g</figure-callout>}}_{22}^{*}& {-\mathrm{<figure-callout\; id="12"\; label="g"\; filenames="A20061006831100261.png,A20061006831100422.png"\; state="\{\{state\}\}">g</figure-callout>}}_{12}^{*}\end{array}\right]---\left(10b\right)$

$H=\left[\begin{array}{cc}{H}_1& G_1\\ H_2& G_2\end{array}\right]---\left(10c\right)$

According to equation (10a to 10c), H ₁, H ₂, G ₁, G ₂Be orthogonal matrix.

In mimo system, transmitting terminal and receiving terminal have a plurality of antennas.Can suppose that transmitting terminal has NT root antenna, receiving terminal has NR root antenna.Receiving terminal utilizes pilot tone to estimate channel information from received signal.The model of received signal can be expressed as following formula (11).

r＝Hs+n (11)

R represents received signal in the equation (11), and H is a channel matrix, and n is a noise signal, and s is the symbol sebolic addressing that transmitting antenna sends.For the NT root transmitting antenna of transmitting terminal, can make sending symbol sebolic addressing s=[s ₁..., s _Nt] ^T, expression transmits NT * 1 dimensional vector of symbol.S wherein _iIt is the symbol of i root antenna transmission.The signal vector of corresponding N R * 1 reception antenna is r=[r ₁..., r _NR] ^TIn formula (11), n=[n ₁..., n _NR] ^TBe illustrated in the zero-mean on the NR root reception antenna, variance is σ ²White Gauss noise.H is NR * NT channel matrix.The purpose of MIMO receiver detector is to recover to send symbol s from receive vectorial r.

In the middle of present detection algorithm, adopt least mean-square error (MMSE) usually, V-BLAST (mode during vertical bell laboratories layered space), maximum likelihood method (MLD).With least mean-square error (MMSE), V-BLAST (mode during vertical bell laboratories layered space) compares, maximum likelihood method (MLD) can obtain the identical error rate or Block Error Rate under lower signal to noise ratio, improved bandwidth availability ratio effectively, satisfies the requirement of high-speed communication.Maximum likelihood method (MLD) is meant all possibilities of traversal s, finds to make | r-Hs| ²Minimum s.Yet well-known, the complexity of maximum likelihood method is counted along with modulation and the increase of antenna amount forms growth exponentially.NTT-Docomo company proposes to carry out QR to channel matrix H and decomposes to combine with the M algorithm and realize MLD, can reduce operand effectively under the prerequisite of not losing performance, becomes the research method that has much future, and this method abbreviates the QRM-MLD algorithm as.

Fig. 3 shows the schematic diagram of QRM_MLD method.Be that the QRM-MLD algorithm utilizes pilot tone that the signal that reception antenna receives is carried out channel estimating,,, arrange again according to from small to large order received signal and channel matrix to each antenna then according to the signal to noise ratio of each antenna channel decline.After this, channel matrix H is carried out QR decompose (S301), obtain equation (12).

H＝QR (12)

In formula (12), R is a upper triangular matrix, and NT * NR matrix Q respectively is listed as mutually orthogonal, and respectively norm of row is 1, promptly is expressed as following formula (13)

Q ^HQ＝I _NT×NT (13)

Utilize conjugate matrices and the received signal r of Q to multiply each other (S302), obtain the vectorial y of following formula (14) expression,

y＝Q ^Hr＝Rs+η (14)

In formula (14), η=Q ^HN, statistical property is the same with noise n.The M algorithm is regarded vectorial y as received signal, and matrix R regards channel as, from last column of R, does the MLD detection algorithm of M reconnaissance step by step.In step after this, duplicate and send symbol s, compute euclidian distances, and select survival node (S303).

In QRM_MLD detects, formula (14) is launched, obtain following formula (15)

Its testing process is from s _NTBeginning is until s ₁In (15), s _NTBe the 1st grade, s ₁It is the NT level.When arriving the first order, obtain the sign estimation of NT root transmitting antenna.In QRM each grade is provided with the number of path of existence, and calculates the Euclidean distance of each grade.

Fig. 4 shows the schematic diagram of each grade of QRM, in Fig. 4, supposes to use the QPSK modulation, and then each symbol has 4 kinds of possibilities.Suppose that number of transmit antennas is 3, the survivor path of each grade is 2,3,4 (survival route (survivor path) is exactly in certain one-level, is retained the path of getting off).Then the sign estimation of last 4 survivor paths that are left is respectively [a, c, f], [a, c, g], [a, d, h], [b, e, i].

The article that is entitled as " Adaptive Selection of SurvivingSymbol Replica Candidates Based on Maximum Reliability in QRM-MLD forOFCDM MIMO Multiplexing " that people such as Kenichi Higuchi deliver (as a reference 1, referring to Globecom 2004,2480-2486) provided based on adaptively selected QRM_MLD algorithm.List of references 1 basic thought is:

According to father node, determine the order (can search) of its child node according to prior sequencing table.

Fig. 5 has provided the relative schematic diagram of father node with child node.The position that can suppose present father node is in the i-1 level, and the reliability of each modulation constellation points sorts among the child node i that will determine now.According to formula (15), can obtain following formula (16)

${\hat{s}}_{N-i+1}=\frac{y_{N-i+1}-\underset{j=N-i+2}{\overset{N}{\mathrm{\&Sigma;}}}{R}_{N-i+1,j}{s}_j}{R_{N-i+1,N-i+1}}---\left(16\right)$

In formula (16), s _N-i+2..., s _NThe above sign estimation of i level under the path for this reason.Then, according to

The ordering of each constellation point is determined in position in two-dimensional space.Can be by tabling look-up to determine the ordering of each constellation point.

Yet M the metric that list of references 1 exists resulting survival route correspondence is not minimum M, thus the problem that performance is incurred loss.

The article that is entitled as " Likelihood function for QRM-MLDsuitable for soft-decision turbo decoding and its performance for OFCDMMIMO multiplexing in multipath fading channel " that people such as Hiroyuki Kawai deliver (as a reference 2, referring to IEICE Trans.Commun., vol.E88-B, No.1, Jan., 2005, pp:47-56) the soft determination methods of a kind of QRM-MLD of being applicable to is disclosed.Yet list of references 2 exists the problems referred to above equally, and the M of a promptly resulting survival route correspondence metric is not minimum M, thereby makes the performance problem that incurs loss.

Summary of the invention

In view of the above problems, the present invention has been proposed.The present invention is directed to two (void) empty time-codes, propose a kind of grouping self adaptation QRM detection method.In being equivalent to organize, the effect of this method carries out the survival route that maximum likelihood obtains M minimum Eustachian distance, the method for Maximum Likelihood Detection in its complexity is significantly less than direct group.

According to an aspect of the present invention, provide a kind of grouping self adaptation QRM detection method, comprise step: a) construct channel matrix H and received signal by space-time block code according to received signal; B) channel matrix H is divided into groups, and determine each group position in QR decomposes; C) channel matrix H is carried out QR and decompose and conversion, obtain orthogonal matrix Q and upper triangular matrix R and received signal vector; D) according to first group branch value M ₁, in group, carry out self-adaptive maximum likelihood and detect, try to achieve M ₁The survival route of individual minimum range; E) according to first group M ₁Individual survival route and sign estimation thereof are removed first group respectively to second group interference; And f) carries out self-adaptive maximum likelihood to second group and detect, obtain M ₂The survival route of individual minimum range and output.

The method according to this invention, the characteristics of two (void) the empty time-code channel matrixes of utilization are divided into two groups with data.In group, count M according to the survival route that is provided with, channel matrix QR decomposes the characteristics of back R matrix, utilizes adaptive method to try to achieve the survival route of M minimum Eustachian distance in the group.

Compare with traditional adaptively selected method, the metric that method of the present invention has overcome the traditional adaptively selected M that a selects survival route may not be M a shortcoming of metric minimum.Compare with QRM completely, reduced the number of times of compute euclidian distances tolerance.And computation complexity is lower than the method for traditional QRM_MLD.

Description of drawings

By reading and understanding the detailed description of the preferred embodiment of the present invention being done below with reference to accompanying drawing, these and other objects of the present invention, feature and advantage will be become apparent.Wherein:

Fig. 1 is the schematic diagram of the Alamouti Space-Time Block Coding of prior art;

Fig. 2 is a double space hour code transmission/reception schematic diagram;

Fig. 3 is the schematic diagram of the method for the traditional QRM_MLD of explanation;

Fig. 4 is the schematic diagram of traditional QRM;

Fig. 5 is the schematic diagram of the relativeness of explanation father node and child node;

Fig. 6 is the schematic diagram that explanation is tabled look-up and sorted according to the position;

Fig. 7 is the method flow diagram that the QR of expression tradition ordering decomposes

Fig. 8 is the schematic diagram of explanation according to the parallel adaptive detection method of the double space hour code of the embodiment of the invention;

Fig. 9 is a schematic diagram of explaining metric in the common group;

Figure 10 is a schematic diagram of explaining the interior metric of group of double space hour code;

Figure 11 is a flow chart of explaining the ordering QR decomposition method of double space hour code;

Figure 12 is the schematic diagram of the example (first group of group is interior) of explanation double space hour code metric;

Figure 13 is first group testing process figure according to present embodiment;

Figure 14 a to Figure 14 i is a schematic diagram of explaining first group of example that detects;

Figure 15 is the schematic diagram of the tree graph of explanation group 2, wherein M ₁=3;

Figure 16 explains second group testing process figure

Figure 17 a to Figure 171 is a schematic diagram of explaining second group of example that detects;

Figure 18 is the schematic diagram of measure of explanation;

Figure 19 explains the transmission of empty DSTTD and the schematic diagram of reception;

Figure 20 a explains the transmission of empty DSTTD and the isoboles of reception;

Figure 20 b explains the transmission of empty DSTTD and the isoboles (mixed structure) of reception;

Embodiment

With reference to the accompanying drawings the grouping self adaptation QRM detection method as the embodiment of the invention is elaborated, having omitted in the description process is unnecessary details and function for the present invention, obscures to prevent that the understanding of the present invention from causing.

The front illustrates that in effective maximum likelihood detection method, as ball decoding, QRM_MLD etc. need to calculate QR and decompose.According to above-mentioned formula (10a), (10b), (10c) and classical Gram-Schmidt algorithm, the QR of the H in the formula (15) decomposed can obtain the feature that following formula (17) provides:

H＝QR (17a)

$R=\left[\begin{array}{cccc}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">R</figure-callout>}}_1& 0& a& c\\ 0& R_1& \mathrm{<figure-callout\; id="0"\; label="b\; d"\; filenames="A20061006831100381.png,A20061006831100431.png"\; state="\{\{state\}\}">b</figure-callout>}& d\\ 0& 0& {\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">R</figure-callout>}}_2& 0\\ 0& 0& 0& R_2\end{array}\right]---\left(17b\right)$

$Q=\left[\begin{array}{cccc}{q}_1& q_2& q_5& q_6\\ {\mathrm{<figure-callout\; id="2"\; label="q"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">q</figure-callout>}}_2^{*}& {-\mathrm{<figure-callout\; id="1"\; label="q"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">q</figure-callout>}}_1^{*}& {\mathrm{<figure-callout\; id="6"\; label="q"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">q</figure-callout>}}_6^{*}& {-\mathrm{<figure-callout\; id="5"\; label="q"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">q</figure-callout>}}_5^{*}\\ q_3& q_4& q_7& q_8\\ {\mathrm{<figure-callout\; id="4"\; label="q"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">q</figure-callout>}}_4^{*}& {-\mathrm{<figure-callout\; id="3"\; label="q"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">q</figure-callout>}}_3^{*}& {\mathrm{<figure-callout\; id="8"\; label="q"\; filenames="A20061006831100261.png,A20061006831100311.png"\; state="\{\{state\}\}">q</figure-callout>}}_8^{*}& {-\mathrm{<figure-callout\; id="7"\; label="q"\; filenames="A20061006831100261.png,A20061006831100341.png"\; state="\{\{state\}\}">q</figure-callout>}}_7^{*}\end{array}\right]---\left(17c\right)$

The particularity that the present invention decomposes according to empty time-code QR is improved QR decomposition and detection algorithm.

General decomposition is first row that calculate Q and R earlier, calculates secondary series again, and the rest may be inferred.Because the particularity of the channel matrix of double space hour code can reduce amount of calculation.

As can be seen, the Q matrix has similar structure to H from formula (17c), is listed as the secondary series that can determine the Q matrix from first of Q matrix, can determine the 4th row of Q matrix from the 3rd row of Q.

Because in (DSTTD), channel has special structure when two sky, process that its ordering QR decomposes becomes simple.The channel matrix H of double space hour code and the character that QR decomposes thereof have been described in the front.

The character of H matrix is: the H matrix is the piecemeal orthogonal matrix.Its first row and secondary series quadrature, the 3rd row and the 4th row quadrature, and also the norm of first row is with the and the norm that is listed as is the same, and tertial norm is the same with the 4th norm that is listed as.The structure of H is shown in formula (16a), (16b), (16c).Like this, caused the R matrix as shown in Equation (17).

The character of R matrix is: r ₁₁=r ₂₂, r ₃₃=r ₄₄

Like this, as long as each row of matrix H are divided into two groups, first group is the 1st, 2 row, and second group is the 3rd, 4 row.In addition, because concerning any matrix A, first row, the first row (r of the R matrix that its QR decomposes ₁₁) be the norm value of matrix A first row, i.e. R (1,1)=r ₁₁=‖ A (:, 1) ‖ ₂So concerning double space hour code, ordering QR decomposes as follows:

A. calculate the norm of first (or two) row and the 3rd (or four) row of H.

If B. the norm of first row is less than tertial norm, then the order of each row of ordering back channel matrix is 1,2,3,4 (order is constant).

If C. the norm of first row is greater than tertial norm, then the order of each row of ordering back channel matrix is 3,4,1,2.

From top description as can be known, the ordering QR of double space hour code decomposes much simple, and decomposes the ordering that the back is exactly an optimum.

The QRM-MLD that double space hour code is described below detects.For the Mathematical Modeling shown in formula (11) and (9), existing MIMO detection method can be used.But because the R matrix has its particularity, the method for detection can obtain simplifying.

Below with reference to the characteristics of Fig. 7, a kind of grouping parallel adaptive QRM algorithm is proposed at the R matrix.As shown in Figure 7, QR decomposes and ordering owing to mentioned in front, below in the main explanation group parallel adaptive Maximum Likelihood Detection how to realize.

The character of R matrix is for the R matrix being written as the combination of 2 * 2 matrix in block form, shown in following formula (18).

$R=\left[\begin{array}{cc}{D}_1& x\\ 0_2& D_2\end{array}\right]---\left(18\right)$

In formula (18), D ₁And D ₂Be 2 * 2 diagonal matrix.And D ₁Diagonal element identical; D ₂Diagonal element identical.According to (17) and (18), s is described ₁With s ₂Detection separate, s ₃With s ₄Detection separate.During then according to the method computing metric of QRM, can utilize above autonomous behavior.

Fig. 9 has provided the tree graph (in the group) of general QRM.In Fig. 9, the metric of partial each branch is relevant with the father node of the first order.Figure 10 shows the tree graph (in the group) of double space hour code.In Figure 10, the metric of partial each branch and the father node of the first order are irrelevant.

Detection according to the double space hour code of first embodiment of the invention is described below.

Embodiment 1

Under the situation of the detection of double space hour code, transmit leg is protected polished introduction in " signal of communication processing " book of National Defense Industry Press's publication in 2000 and is sent processing according to opening a prominent personage.The recipient's who the present invention relates to detection.Below with reference to Fig. 8 recipient's operating process is described.

At step S801, construct channel H and received signal r according to received signal by empty time-code, obtain the Mathematical Modeling of aforementioned formula (9).After this, at step S802,, and determine each group position in QR decomposes with the H grouping.Its process is as follows: the norm of calculating first (or two) row and the 3rd (or four) row of H; If the norm of first row is less than tertial norm, then the order of each row of ordering back channel matrix is 1,2,3,4 (order is constant); If the norm of first row is greater than tertial norm, then the order of each row of ordering back channel matrix is 3,4,1,2.The order of the group that the QR that obtains thus sorting decomposes.Next,, carry out QR and decompose and conversion, obtain QR matrix as shown in Equation (17) at step S803, and the received signal shown in the formula (14).At step S804, according to first group M ₁, in group, carry out self-adaptive maximum likelihood and detect, try to achieve M ₁The survival route of individual minimum range.At step S805, according to first group M ₁Individual survival route according to its sign estimation, removes the 1st group of interference to the 2nd group respectively.

Suppose in first group detection, to have obtained M ₁Individual survival route so, has just obtained M ₁Individual s ₃, s ₄Estimation, be designated as [s ₃ ¹s ₄ ¹], [s ₃ ²s ₄ ²] ..., [s ₃ ^M1s ₄ ^M1].According to formula (14) and (17).First group of interference to second group deducted, can obtain formula (19)

$\left[\begin{array}{c}{z}_1^i\\ z_2^i\end{array}\right]=\left[\begin{array}{c}{r}_1\\ r_2\end{array}\right]-\left[\begin{array}{cc}a& c\\ b& d\end{array}\right]\left[\begin{array}{c}{s}_3^i\\ s_4^i\end{array}\right]i=1,...M_1---\left(19\right)$

During detection, according to detecting as following equation (20).

$\left[\begin{array}{c}{z}_1^i\\ z_2^i\end{array}\right]=\left[\begin{array}{cc}{\mathrm{<figure-callout\; id="1"\; label="R"\; filenames="A20061006831100261.png,A20061006831100281.png"\; state="\{\{state\}\}">R</figure-callout>}}_1& 0\\ 0& R_1\end{array}\right]\left[\begin{array}{c}{s}_1\\ s_2\end{array}\right]i=1,...M_1---\left(20\right)$

Suppose M ₁=3, there are two child nodes then can get as shown in figure 15 tree graph under each father node.

At step S806, carry out self-adaptive maximum likelihood to second group and detect, obtain M ₂The survival route of individual minimum range (calculate apart from time, the 1st group distance be added).

Figure 11 has provided the flow chart of S802 and S803.As shown in figure 11, at step S1101, to the channel matrix H of double space hour code, matrix H is divided into two groups, wherein the 1st and 2 classifies first group as, the 3rd and four classify second group as.At step S1102, the norm of first (or two) of calculating channel matrix H row and the 3rd (or four) row.After this at step S1103, determine the position of group according to the size of norm.Then, at step S1104, channel matrix H is carried out QR decompose.

In aforesaid step S804 according to first group M ₁, in group, carry out self-adaptive maximum likelihood and detect, try to achieve M ₁In the process of the survival route of individual minimum range, can suppose has 3 child nodes under each father node.The branch metric as shown in figure 12, the first order has 3, each father node has 3 child nodes.Altogether be combined as 3 * 3=9 kind, suppose to get wherein M as survival route.Present problem is how to obtain M minimum branch tolerance and survival route and metric thereof.

Figure 13 has provided first group testing process figure.When detecting for first group, the initial value of the branch of root node 0 (example as shown in figure 12) tolerance is 0.Testing process following (supposing M=3).At step s1301, calculate the metric of independent branch respectively.Obtain 13,17 in the first order, 8.The second level obtains 15,4,8 (the measuring independent with partial branch of the first order).At step S1302, by ordering from small to large, obtain 8,13,17 in the first order, by ordering from small to large, obtain 4,8,15 (step S1302 ') in the second level.At step S1303, the pointer of the sensing of initialization sequence 1 oneself, and the pointer of sensing sequence 2; The pointer of the sensing of initialization sequence 2 oneself, and the pointer of sensing sequence 1 (step S1303 ').During beginning, the pointed 8 of sequence 1 itself, sequence 1 is pointed to the sensing 4 of sequence 2; The pointed 4 of sequence 2 itself, sequence 2 is pointed to the sensing 8 of sequences 1, shown in Figure 14 a.At step S1304, calculate branch tolerance and, obtain 4+8=12.At step S1305, the metric of the branch 1 that relatively calculates is got less value output with regard to the metric of branch 2, is 12 (having only a value) in this example.At step S1306, judge whether to obtain M survival route.Because M=3 has only exported minimum one now, so enter S1307.At step S1307, the pointer that sequence 1 is pointed to oneself is constant, and sequence 1 is pointed to pointer+1 of sequence 2, points to 15; Simultaneously, the pointer that sequence 2 is pointed to oneself is constant, and sequence 2 is pointed to the pointed 13 of sequence 1.Shown in Figure 14 a.Forward S1304 then again to.At step S1304, calculate its branch tolerance in sequence 1, its computational process is the pointed 8 that sequence 1 is pointed to oneself, sequence 1 is pointed to the pointed 15 of sequence 2,15+8=23; Calculate its branch tolerance in sequence 2, its computational process is the pointed 4 that sequence 2 is pointed to oneself, and sequence 2 is pointed to the pointed 13 of sequence 1,4+13=17.Shown in Figure 14-2.Enter S1305 then.At step S1305, because 17＜23, output 17.At step S1306, obtain the 2nd output valve now, 2＜M=3 enters step S1307.At step S1307, the position of moving hand.Because in step S1305,17 is little,, point to 17 so the pointer that sequence 2 is pointed to sequence 1 adds 1.Shown in Figure 14 b.Enter S1304 then.At step S1304, calculate its branch tolerance in sequence 1, be 23 last once having obtained, need not calculate; Calculate its branch tolerance (having upgraded pointer) in sequence 2, its computational process is the pointed 4 that sequence 2 is pointed to oneself, and sequence 2 is pointed to the pointed 17 of sequence 1,4+17=21.Shown in Figure 14 c.Enter step S1305 then.At step S1305,21＜23, output 21.At step S1306, obtained 3 values, 3=M enters step S1308.At step S1308, output branch tolerance and 12,17,21 and corresponding survival route, this flow process end.

Next, provide the selection course of M=1～9 respectively, shown in Figure 14 a to 14i.

In step 1: select minimum value from the ordering of the first order and obtain 8, obtain 4 from partial ordering being selected minimum value, both summations are obtained 12, corresponding path is selected.Because 8 is minimum in the sequence 1, the 4th, minimum in the sequence 2.Then respectively get a minimum in two sequences, then and must be minimum, process such as Figure 14 a.

In step 2: shown in Figure 14 b, the pointer that points to another sequence is added 1, as shown below.Obtain 4+13=17,8+15=23.Since 17＜23, output 17.Because 17 4 point to 13 in the sequences 1 and obtain,, point to 17 with 4 pointing to pointer position+1 in the sequences 1 in the sequence 2 in sequence 2.

In step 3: shown in Figure 14 c, calculate 4+17=21.Since 21＜23, output 21.Because 4 all values that traveled through in the sequence 1 in sequence 2, thus in sequence 2 with pointer position+1 of itself, obtain 15.Because in sequence 1, pointer position is to 8, with the position in 15 the sensing sequence 1, for position+1 of pointer in the sequence 1 own obtains position 13.Because the current pointer position of sequence 2 points to 15, the position of the current pointer of sequence 1 points to 8.So concerning sequence 1, pointer position points to the sequence that all travels through 2 before 8, all selected coming out.For sequence 2, pointer position had all traveled through sequence 1 before 15, so the current location of pointer is 8 in sequence 1, its pointer in sequence 2 is certain to point to 15.So current location points to 15 in sequence 2.Point to pointer in the sequence 1 and just needn't point to 8, and will point to 13.

In step 4: shown in Figure 14 d, calculate 15+13=28.Since 23＜28, output 23.And the pointed 18 of sequence 2 will be pointed in the sequence 1.

In step 5: shown in Figure 14 e, calculate 8+18=26.Since 26＜28, therefore export 26.And, point to the pointed 18 of sequence 2 with self pointed 13 in the sequence 1.

In step 6: shown in Figure 14 f, calculate 13+18=31.Since 28＜31, therefore export 28.And the pointed 17 of sequence 1 will be pointed in the sequence 2.

In step 7: shown in Figure 14 g, calculate 15+17=32.Since 31＜32, output 31.And, point to the pointed 18 of sequence 2 with self pointed 17 in the sequence 1.

In step 8: shown in Figure 14 h, calculate 17+18=35.Since 32＜35, output 32.Because only remaining last two, self pointer of sequence 2 does not need to have moved, and directly removes to get final product.

In step 9: shown in Figure 14 i, export last 35, this flow process finishes.

Need to prove that the metric of selecting by this kind method all is minimum in candidate's combination.In addition, can will select the path of M minimum degree value to reduce following problem: promptly, sequence 1 has the K number, and sequence 2 has the K number.Select a number for a kind from sequence, from sequence 2, select a number, then with this two numbers addition.In all may combination, select M minimum.

According to above example, respectively sequence 1 and sequence 2 are sorted.Then minimum value in the sequence 1 and the minimum value addition in the sequence 2 must be minimum values, see the step 1 among Figure 14 a.When calculating the 2nd, (see Figure 14-2, step 2), because 8 is that selectable value in the current sequence 1 (value that current pointer points in the sequence 1) is (to the step 6 shown in Figure 14 f, selectable minimum value has just become 13 in the sequence 1) in minimum value, then value in the sequence 2 and the value addition in the sequence 1 must be to have selected minimum value 8 in the sequence 1.In like manner, in the sequence 1 with sequence 2 in addition, must be to have selected minimum value 4 (as the step 2 of Figure 14 b) in the sequence 2.And whole minimum value one is decided to be one in this possibility, so second also is minimum.In like manner, when selecting n, necessarily also be the minimum value in optionally worthwhile.So when selecting M, be to elect according to metric order from small to large.

Figure 16 has gone out second group testing process figure.As an example, can suppose first order M ₁=3 as initial value, obtain as shown in figure 15 tree graph and the value of each branch.In addition, can suppose M ₂=3, obtained M at first group ₁Individual branch metric also is assumed to be a ₁, a ₂, a ₃

At step S1601, to M ₁The metric of individual survival route carries out demodulation, tables look-up, and obtains M ₁The ordering of each branch under the individual survivor path.Provided the method for tabling look-up among Fig. 6 of list of references 1,, omitted its description, obscured with content of the present invention avoiding at this in view of this look-up method is not a content of the present invention.In step S1602 (comprising S1602 '), obtain the ordering of modulation constellation points at these two branches.For example, as shown in figure 15, first survivor path of the 1st group has the two-stage branch, is designated as the third level and the fourth stage respectively.In the branch of the third level, the branch value be 5 come first, the branch value is 10 come second (at this moment, and do not know the value of branch, but by tabling look-up, can know the ordering of branch size).After this, in step S1603 (comprising S1603 '), the pointer of initialization sequence 1, sequence 2.Shown in Figure 17 a, under first survival route of the 1st group, sequence 1 is pointed to the pointed 5 of oneself, points to the pointed 18 of sequence 2; Sequence 2 is pointed to the pointed 18 of oneself, points to the pointed 5 (these two actual is) of sequence 1.Under the 1st group the 2nd survival route, sequence 1 is pointed to the pointed 15 of oneself, points to the sensing 1 of sequence 2; Sequence 2 is pointed to the pointed 1 of oneself, points to the pointed 15 of sequence 1.Under the 1st group the 2nd survival route, sequence 1 is pointed to the pointed 18 of oneself, points to the pointed 3 of sequence 2; Sequence 2 is pointed to the pointed 3 of oneself, points to the pointed 18 of sequence 1.

Next, in step S1604 (comprising S1604 '), calculate the branch value.According to the order of ordering, two branch values that obtain under first survival route of the 1st group are 5,18; Two branch values under second survival route are 15,1; Under second survival route is 15,1; Be 18,3 (a) under the 3rd survival route referring to Figure 15 and Figure 17.At this, the branch value is to have considered first group of resulting M ₁Individual branch tolerance and a ₁, a ₂, a ₃Owing to will calculate total metric sum at last, as shown in figure 18, then which grade branch initial value is added on and can change last result.Therefore, in Figure 15, the branch value of the third level hypothesis is by original branch value being added the nodal value (a of initial survivor path ₁, a ₂, a ₃) obtain.In step S1605 (comprising S1605 '), calculate branch tolerance and, under each survival node of first group, get minimum metric.First is 5+18=23 (having only a value), and second is 15+1=16, and the 3rd is 18+3=21.In step S1606, because in these three metrics 23,16,21 that obtain, 16 minimums, (Figure 17 is a) in output 16.After this, in step S1607, judge whether to obtain M ₂Individual survival route.In this example, owing to do not obtain M ₂Individual survival route, flow process enter step S1608.In step S1608, the change pointer position is shown in Figure 17 a.After this, flow process is returned step S1604.In step S1604, calculate the branch value, because 15 and 1 branch value obtains, obtain 22,25 respectively now.As Figure 17 b.In step S1605, calculate 15+22=37,1+25=26 is shown in Figure 17 b.Since 26＜27, so under the 1st group the 2nd survival route, select 26.In step S1606, original metric 23,21 and 26 ratios that obtain now, 21 minimums, so output degree value 21 are shown in Figure 17 b.In step S1607, owing to do not obtain M ₂Individual survival route, flow process still enter step S1608.In step S1608, the change pointer position is shown in Figure 17 b.Enter step S1604 then.In step S1604, calculate the branch value, because 15 and 1 branch value obtains, obtain branch value 29,25 now respectively, shown in Figure 17 c.In step S1605, calculate branch value 18+25=43,3+29=32 is as Figure 17 c.Since 32＜43, so under the 1st group the 3rd survival route, select 32.After this, in step S1606, from branch value 23,26,32, choosing 23 of minimum, and exporting 23, as Figure 17 c.In step S1607, obtained M ₂=3 survival routes, so flow process enters step S1609.In step S1609, the M that output obtains ₂Individual survival route, flow process so far finishes.

Provide M=M below respectively ₂=1～12 selection course, as Figure 17 a to shown in the 17l.

Can adopt example shown in Figure 15, first group survival route M ₂=3.

Step 1 (Figure 17 a, M=1): the symbol order that obtains according to ordering, calculate the branch value respectively, under first survival node of first group, obtain (5,18); Under second node, obtain (15,1); Under the 3rd node, obtain (18,3).Calculate 5+18=23; 15+1=16; 18+3=21.Wherein Zui Xiao branch value is 16, output 16 and corresponding survival route thereof.Because branch value 16 is under first group the 2nd survival node, under this node, mobile sequence 1 points to the pointer of sequence 2, and sequence 2 is pointed to the pointer of sequence 1, shown in Figure 17 a.

Step 2 (Figure 17 b, M=2): calculate branch value 25,22.Calculate 15+22=37,1+25=26.Since 26＜37, the output of branch value 26 therefore chosen as first group of the 2nd survival node.Then branch value 26 and last computation go out 23,21 in choose minimum value, obtain 21 and export 21.Moving hand is shown in Figure 17 b then.

Step 3 (Figure 17 c, M=3): calculate branch value 29,25.Calculate 18+25=43,3+29=32.Since 32＜43, the output of branch value 32 therefore chosen as first group of the 3rd survival node.Then 32 with last time remaining 23,26 in select minimum value, obtain 23 and export 23.Moving hand is shown in Figure 17 c then.

Step 4 (Figure 17 d, M=4): calculate branch value 10,19.Calculate 5+19=24,18+10=28.Since 24＜28,24 outputs therefore chosen as first group of the 1st survival node.Then 24 with last time remaining 32,26 in select minimum value, obtain 24 and export 24.Moving hand is shown in Figure 17 d then.

Step 5 (Figure 17 e, M=5): calculate 10+19=29, with 29 and 28 comparisons.Since 28＜29,28 outputs therefore chosen as first group of the 1st survival node.Then 28 with last time remaining 32,26 in select minimum value, obtain 26 and export 26.Moving hand is shown in Figure 17 e then.

Step 6 (Figure 17 f, M=6): calculate 22+25=47, with 47 and 37 relatively, because 37＜47, so choose 37 outputs as first group of the 2nd survival node.Then 37 with last time remaining 32,28 in select minimum value, obtain 28 and export 28.Moving hand shown in Figure 17 f then.At this moment, minimum value is 29 under first group of the 1st survival node.

Step 7 (Figure 17 g, M=7): in 29,37 and 32, choose minimum value, obtain 29.Output 29.Moving hand is shown in Figure 17 g then.

Step 8 (Figure 17 h, M=8): in 37 and 32, select minimum value, obtain 32 and export 32.Moving hand is shown in Figure 17 f then.

Step 9 (Figure 17 i, M=9): calculate 29+25=54.Since 43＜54, under first group of the 3rd survival route, output 43.Compare 37 and 43, because 37＜43, therefore export 37.Moving hand is shown in Figure 17 i then.

Step 10 (Figure 17 j, M=10): relatively 47 and 43.Since 43＜47, therefore export 43.Moving hand is shown in Figure 17 j then.

Step 11 (Figure 17 k, M=11): relatively 47 and 54.Since 47＜54, therefore export 47.Moving hand is shown in Figure 17 k then.

Step 12 (Figure 17 l, M=12): export last 54.

Embodiment 2

Under the situation of Figure 19 explanation as the detection of (void) double space hour code of second embodiment of the invention.Referring to Figure 19, (void) double space hour code is meant that 1901 for example adopt that two antennas carry out space-time block code, and 1902 have only an effective antenna to transmit data (thinking that equally the channel in two moment of 1902 is the same).Therefore, can think that 1902 have only an antenna, can think that also 1902 formed empty time-code by two transmitting antennas originally, but wherein one because of former such as fault thereby can't send.In addition, the recipient still has for example two antennas.Empty DSTTD can be regarded as the special case of DSTTD.It receives handles with double space hour code is identical substantially.

Example 1

At transmit leg, (as Figure 20 a), antenna 1 sends according to the rule of space-time block code to regard empty space-time block code structure as with 1902.

The recipient, think that the channel gain of 1902 antenna 2 is 0, can constitute the double space hour code structure equally.Testing process is identical with the process of description among the embodiment 1.

Example 2

At transmit leg, regard that with 1902 common single antenna transmits as.Like this, just constituted the hybrid system of a space-time block code 1901 and spatial reuse 1902.

The recipient, can think that the channel gain of 1902 empty antenna 2 is 0, can constitute Mathematical Modeling as shown in Equation (9) equally.It is identical with the process of description among the embodiment 1 that process detects.

For the symbol of 1902 detected transmissions, odd number symbols is calculated according to resulting symbol, gets to grip altogether by the dual numbers symbol to obtain sending symbol.This conjugation that requires modulation constellation points is also in modulation constellation points.Usually, this situation can be met.

With relatively comparing of QRM-MLD, QRM_MLD is that one-level one-level ground is chosen, and the method for invention is that two-stage is chosen together.What method of the present invention obtained is the effect of grouping maximum likelihood.With first group be example (detect s3, s4), if detect s4, the survivor path M that QRM_MLD chooses ₁Insufficient, M for example ₁=2, and that wilfully survive is not s4, does not separate on the path at place at maximum likelihood, thereby mistake has taken place.And just might avoid these mistakes according to the method for invention, it is branch value minimum in the first order and the second level that method of the present invention obtains.In addition, if with first group be one group, second group is one group, carries out Maximum Likelihood Detection in group, then obtain be the grouping maximum likelihood the result.The M that occurs in the table below ₁, M ₂, M ₃, M ₄Being meant according to QRM_MLD to be divided into 4 grades, for purposes of the invention, is M ₂, M ₄Useful.In table 2, the number of times of the 1st group of calculative branch tolerance makes modulation constellation count and is C

Table 2

The grouping maximum likelihood	QRM_MLD	Method of the present invention
			C 2	C+M 1C	C+C

Table 3 is QRM_MLD and method of the present invention in the situation of the number of times of the 1st group of calculative cumulative metrics (computation measure and), wherein makes modulation constellation count and is C.

Table 3

QRM_MLD	Method of the present invention
		M 1C	M 2+1

Table 4 is in the contrast of the number of times of the 2nd group of calculative branch tolerance, wherein makes modulation constellation count and is C.

Table 4

The grouping maximum likelihood	QRM_MLD	The present invention's (usage space is changed complexity)
			M 2C 2	M 2C+M 3C	M 2+M 4+1

Last row in the table 4, " usage space is changed complexity " refer to by in the list of references 1, table look-up according to the position of the constellation point of demodulation, obtain the ordering of each constellation point, as shown in Figure 6.

Table 5 is in the number of times of the 2nd group of calculative cumulative metrics (computation measure and) contrast, wherein makes modulation constellation count and is C.

Table 5

QRM_MLD	Method of the present invention
		M 3C	M 2+M 4+1

The result who provides from top table 2 to table 5 as can be seen, method of the present invention is compared with QRM_MLD with the grouping maximum likelihood, has reduced amount of calculation.

Following table 6 has provided method of the present invention and has compared with grouping maximum likelihood and the resulting performance of QRM_MLD.

Table 6

The grouping maximum likelihood	QRM_MLD	Invention
			In group, obtain M minimum survival route	Each grade maximum likelihood, in the group not necessarily	With the grouping maximum likelihood

Data surface in the table 6, identical with the survival route that method of the present invention obtains with the grouping maximum likelihood method, be better than QRM_MLD.

In addition, according to the model of formula (15) and (1), the structure of double space hour code need be provided with 4 grades survival route according to the method for general QRM, is respectively M ₁, M ₂, M ₃, M ₄And data are divided into two groups according to method of the present invention, and need a survival route be set in each group, like this, the survival route that needs is two M ₁, M ₂M in the present invention ₁The M that is equivalent to traditional Q RM ₂, the M among the present invention ₂The M that is equivalent to traditional Q RM ₄Therefore, compare, obviously reduced the number of times of compute euclidian distances tolerance with QRM_MLD.

So far invention has been described in conjunction with the preferred embodiments.Should be appreciated that those skilled in the art can carry out various other change, replacement and interpolations under the situation that does not break away from the spirit and scope of the present invention.Therefore, scope of the present invention is not limited to above-mentioned specific embodiment, and should be limited by claims.

Claims (14)

1. grouping self adaptation QRM detection method comprises step:

A) construct channel matrix H and received signal according to received signal by space-time block code;

B) channel matrix H is divided into groups, and determine each group position in QR decomposes;

C) channel matrix H is carried out QR and decompose and conversion, obtain orthogonal matrix Q and upper triangular matrix R and received signal vector;

D) according to first group branch value M ₁, in group, carry out self-adaptive maximum likelihood and detect, try to achieve M ₁The survival route of individual minimum range;

E) according to first group M ₁Individual survival route and sign estimation thereof are removed first group respectively to second group interference; With

F) carry out self-adaptive maximum likelihood to second group and detect, obtain M ₂The survival route of individual minimum range and output.

2. method according to claim 1, wherein the step that channel matrix H is divided into groups comprises first row from channel matrix H, per two row are divided into one group step.

3. method according to claim 2 wherein further comprises any one norm that is listed as in two row that are divided into a group in the calculating channel matrix H, and the step of determining the position of group according to the size of the norm of calculating.

4. method according to claim 1, wherein said steps d comprises the metric that calculates independent branch respectively, and the metric of each independent branch is formed the step of corresponding sequence according to rank order from small to large.

5. method according to claim 4 further comprises the pointer of each sequence sensing of initialization oneself, and points to the step of the pointer of other sequence.

6. method according to claim 5, further comprise the branch that calculates each sequence tolerance and step.

7. method according to claim 6 further comprises the metric of each branch that relatively calculates, and chooses the step of minimum value output.

8. method according to claim 7 further comprises according to the minimum value of being exported judging whether to obtain M ₁The step of individual survival route is if obtained M ₁Individual survival route is then exported this M ₁The step of individual survival route.

9. method according to claim 7 further comprises if do not obtain M yet ₁Individual survival route is then changed position indicator pointer, continues to calculate the metric sum of each branch, and the final election of laying equal stress on is got minimum value output and judged whether to obtain M ₁The step of individual survival route is up to obtaining M ₁Individual survival route and output.

10. method according to claim 1, wherein said step f comprises the M with first group ₁The metric of individual survival route carries out demodulation, tables look-up as initial value, to obtain M ₁The step that puts in order of each sequence under the individual survival route.

11. method according to claim 10 further comprises the pointer of each sequence of initialization, and calculates the step of the branch value of each sequence.

12. method according to claim 11, further comprise the branch that calculates each sequence tolerance and and the metric of each branch of relatively calculating and, and choose the step of minimum value output.

13. method according to claim 12 further comprises according to the minimum value of being exported judging whether to obtain second group M ₂If the step of individual survival route and obtained M ₂Individual survival route is then exported this M ₂The step of individual survival route.

14. method according to claim 12 further comprises if do not obtain M yet ₂Individual survival route is then changed position indicator pointer, continues to calculate the metric and the metric sum of each branch, and the final election of laying equal stress on is got minimum value output and judged whether to obtain M ₂The step of individual survival route is up to output M ₂Individual survival route.

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