Abstract:
A method of operating a multiple-input multiple-output (MIMO) receiver includes wirelessly receiving a message over a communication channel using a plurality of antennas. The message includes first data preceded by a plurality of training fields. The method includes generating a first matrix indicative of an estimation of properties of the communication channel, and determining a second matrix and a third matrix by performing a matrix decomposition of the first matrix. The method includes, as each of the plurality of training fields of the message is being received, recursively computing parameters for equalization based on (i) the plurality of training fields, (ii) the second matrix, and (iii) the third matrix. The method includes generating equalizer coefficients based on the parameters for equalization, and applying the equalizer coefficients to the first data of the message to compensate the first data.

Description:
CROSS-REFERENCE TO RELATED APPLICATIONS 
     The present disclosure is a continuation of U.S. patent application Ser. No. 12/902,394, filed on Oct. 12, 2010, which is a continuation of U.S. patent application Ser. No. 11/521,182 (now U.S. Pat. No. 7,813,421), filed on Sept. 14, 2006, which claims the benefit of U.S. Provisional Application No. 60/759,453, filed on Jan. 17, 2006. The entire disclosures of the above referenced applications are incorporated herein by reference. 
    
    
     FIELD 
     The present disclosure relates to channel estimation in a wireless communication system. 
     BACKGROUND 
     The background description provided herein is for the purpose of generally presenting the context of the disclosure. Work of the presently named inventors, to the extent it is described in this background section, as well as aspects of the description which may not otherwise qualify as prior art at the time of filing, are neither expressly or impliedly admitted as prior art against the present disclosure. 
     Some multiple input, multiple output (MIMO) wireless communication systems can estimate channel conditions, or gains, in the communication path between the transmitting and receiving antennas. The channel estimation process can include transmitting known training symbols, receiving the known training symbols, and processing the received symbols to estimate the channel conditions. The estimation is based on differences between the known training symbols and the received symbols. Information regarding the channel conditions can then be used to program coefficients of an equalizer of the receiver. The equalizer then compensates for the channel conditions. 
     Referring now to  FIG. 1 , an example is shown of a MIMO communication system  10  that complies with the Institute of Electrical and Electronics Engineers (IEEE) 802.11n specification, which is hereby incorporated by reference in its entirety. A transmitter module  12  communicates with a receiver module  14  via a wireless communication channel  16 . A matrix H represents signal gains through channel  16 . 
     Transmitter module  12  periodically generates a plurality of long training fields (LTFs)  18 - 1 , . . . ,  18 -j, referred to collectively as LTFs  18 . Each LTF  18  includes a plurality of training symbols  20 - 1 , . . . ,  20 -k, referred to collectively as training symbols  20 . A multiplier module  22  multiplies each training symbol  20  by a corresponding column of a preamble steering matrix P. A number of rows n of matrix P corresponds with a number of transmit antennas  26 - 1 , . . . ,  26 -n, collectively referred to as antennas  26 . The number of columns j of matrix P corresponds with the number of LTFs  18 . Matrix P assures the orthogonality of training symbols  20  as they are transmitted from antennas  26 . Matrix P has a condition number of  1 , i.e. cond(P) = 1 . 
     Receiver module  14  includes receiver antennas  30 - 1 , . . . ,  30 -n, collectively referred to as antennas  30 , that receive the training symbols via channel  16 . After receiving all of the training symbols, receiver module  14  generates matrix H based on known training symbols  20 , matrix P, and the received training symbols. Receiver module  14  can then use matrix H to adjust coefficients of an internal equalization module for signals from antennas  30 . It is generally desirable for receiver module  14  to generate matrix H as quickly as possible. 
     A sample estimation of matrix H will now be described. Assume that n=3 and j=4. Transmitter module  12  then sends 4 LTFs  18  and matrix P is a 3×4 matrix 
     
       
         
           
             P 
             = 
             
               
                 [ 
                 
                   
                     
                       1 
                     
                     
                       
                         - 
                         1 
                       
                     
                     
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                       1 
                     
                   
                   
                     
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                 ] 
               
               . 
             
           
         
       
     
     The effective MIMO channel estimated at receiver module  14  is given by H est =HP           H=H est P −1 , where H est  represents an estimation of matrix H. For the data associated with each LTF  18  the transmitter-to-receiver communication model can be described by y=H est P −1 x+n, where x represents transmitted data symbols. The ZF solution is applied to the matrix H est P −1 .
     If the matrix P were not used, y=Hx+n           {circumflex over (x)}=R −1 Q′y, where y represents received data symbols. With matrix P, y=H est P −l x+n. Receiver module  14  uses each LTF  18  to estimate a column of matrix H est . Matrix H can therefore be estimated by waiting until all columns have been estimated and then estimating the matrix_H est P −1 . Receiver module  14  can then perform orthogonal-triangular decomposition (QR) on H est P −1 , i.e QR(H est P −1 ). However, the computational density increases to the order of n 3 , i.e. O(n 3 ). To meet the processing latency the hardware burden would also increase based on  0 (n 3 ).
     The effect of matrix P will now be described to shed light on the above equations. Let H est =QR. Without matrix P the equalized vector is given by 
                 x   ^     =       1     diag   ⁡     (       R     -   1       ⁢     R     -   *         )         ⁢     R     -   1       ⁢     Q   *     ⁢   y       ,       where   ⁢           ⁢     W   ll       =       1     diag   ⁡     (       R     -   1       ⁢     R     -   *         )         .             
With matrix P the equalized vector is given by:
 
                 x   ^     =       1     diag   ⁡     (         (     RP     -   1       )       -   1       ⁢       (     RP     -   1       )       -   *         )         ⁢     PR     -   1       ⁢     Q   *     ⁢   y       ,     
     ⁢       where   ⁢           ⁢     W   ll       =       1     diag   ⁡     (         (     RP     -   1       )       -   1       ⁢       (     RP     -   1       )       -   *         )         .             
The equivalent matrix RP −1  is a full matrix and it is difficult to compute its inverse for an n×n communication system.
 
     SUMMARY 
     A receiver module includes an input that receives a data message from a wireless communication channel. The data message has a plurality of training fields and data. A channel estimator module recursively estimates a matrix H that represents the channel based on the plurality of training fields. The recursive estimation is performed as the plurality of training fields are being received. An equalizer module applies coefficients to the data based on the matrix H. 
     Further areas of applicability of the present disclosure will become apparent from the detailed description provided hereinafter. It should be understood that the detailed description and specific examples, while indicating the preferred embodiment of the disclosure, are intended for purposes of illustration only and are not intended to limit the scope of the disclosure. 
    
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
       The present disclosure will become more fully understood from the detailed description and the accompanying drawings, wherein: 
         FIG. 1  is a functional block diagram of a multiple input, multiple output (MIMO) communication system according to the prior art; 
         FIG. 2  is a functional block diagram of a MIMO communication system that includes a receiver that employs a recursive channel estimation method; 
         FIG. 3  is a functional block diagram of a MIMO transceiver that includes the receiver of  FIG. 2 ; 
         FIG. 4  is a data diagram of a prior art data message that is transmitted by a transmitter module of the communication system of  FIG. 2 ; 
         FIG. 5  is a flowchart of the recursive channel estimation method; 
         FIG. 6A  is a functional block diagram of a high definition television; 
         FIG. 6B  is a functional block diagram of a vehicle control system; 
         FIG. 6C  is a functional block diagram of a cellular phone; 
         FIG. 6D  is a functional block diagram of a set top box; and 
         FIG. 6E  is a functional block diagram of a media player. 
     
    
    
     DETAILED DESCRIPTION 
     The following description is merely exemplary in nature and is in no way intended to limit the disclosure, its application, or uses. For purposes of clarity, the same reference numbers will be used in the drawings to identify similar elements. As used herein, the term module, circuit and/or device refers to an Application Specific Integrated Circuit (ASIC), an electronic circuit, a processor (shared, dedicated, or group) and memory that execute one or more software or firmware programs, a combinational logic circuit, and/or other suitable components that provide the described functionality. As used herein, the phrase at least one of A, B, and C should be construed to mean a logical (A or B or C), using a non-exclusive logical or. It should be understood that steps within a method may be executed in different order without altering the principles of the present disclosure. 
     Referring now to  FIG. 2 , a functional block diagram is shown of a MIMO communication system  50 . Communication system  50  includes a receiver module  52  that employs a recursive channel estimation method (shown in  FIG. 5 ). The recursive channel estimation method estimates conditions in a wireless communication channel  54 . The recursive estimation method begins with a first long training sequence that is sent by a transmitter module  56 . The recursive estimation method ends by generating a channel estimation matrix H when the last long training sequence has been received. Since the recursive estimation method develops a basis for generating matrix H while the long training sequences are being received, instead of starting after the long training sequences have been received, the recursive estimation method can generate matrix H faster than previously known methods. The matrix H can then be used to perform channel equalization in receiver module  52  to compensate for the effects of communication channel  54 . 
     Communication system  50  will now be described in pertinent part. A baseband module  58  generates data messages based on m streams of incoming data. Baseband module  58  communicates the data messages (shown in  FIG. 4 ) to an encoder module  60 . The data messages include respective long training fields that are compliant with IEEE 802.11n. Encoder module  60  encodes the long training fields into n data streams based on the matrix P, which is shown in  FIG. 1 . Encoder module  60  communicates the n data streams to n respective transmit channels  62 - 1 , . . . ,  62 -n, which are referred to collectively as transmit channels  62 . Each transmit channel  62  includes a respective modulation module  64  that modulates its respective data stream, such as with quadrature-amplitude modulation (QAM), and communicates the modulated data stream to a respective inverse fast-Fourier transform (IFFT) module  66 . IFFT modules  66  convert their respective data streams from a frequency domain signal to a time domain signal. IFFT modules  66  communicate the time domain signals to respective radio frequency (RF) transmitters that are represented by antennas  68 . 
     The transmitted data streams propagate through communication channel  54 . Communication channel  54  perturbs the transmitted data streams due to phenomena such as reflections, signal attenuation, and so forth. The perturbations can be represented by matrix H. 
     Receiver module  52  includes n RF receivers that are represented by antennas  70 - 1 , . . . ,  70 -n. The RF receivers receive the transmitted data streams and communicate the perturbed time domain signals to a channel estimator module  72 . Channel estimator module  72  estimates matrix H based on matrix P and the long training fields that are included in the received data streams. In some embodiments channel estimator module  72  includes a processor  73  and associated memory  75  for storing and/or executing the recursive channel estimation methods that are described below. 
     Channel estimator module  72  communicates the n received data streams to n respective fast-Fourier transform (FFT) modules  74  and adjusts gains of an equalizer module  76 . FFT modules  74  convert the time-domain data streams to frequency-domain data streams and communicate them to equalizer module  76 . Equalizer module  76  compensates the respective data streams based on the gains and communicates the compensated gains to a Viterbi decoder module  78 . Viterbi decoder module  78  decodes the n data streams to generate received data streams y m . 
     Referring now to  FIG. 3 , a functional block diagram is shown of a transceiver  80  that includes transmitter module  56  and receiver module  52 . Transceiver  80  can communicate with other transceivers  80  via antennas  82 - 1 , . . . ,  82 -n. An antennas switch module  84  selectively connects antennas  82  to transmitter module  56  or receiver module  52  based on whether transceiver  80  is transmitting or receiving. 
     Referring now to  FIG. 4 , a data diagram is shown of an IEEE 802.11n data message  90 . Data message  90  includes data  92  and a preamble  94  that contains a plurality of training fields. Preamble  94  is divided into a first portion  96  and a second portion  98 . First portion  96  may be used by legacy systems, e.g. non-MIMO, IEEE 802.11 communication systems. Second portion  98  includes a signal filed field  100 , a short training field  102 , and x long training fields (LTFs)  104 - 1 , . . . ,  104 -x, where x is an integer. Each LTF  104  includes k training symbols  106  or tones. Short training field  102  is generally used by receiver module  52  to establish symbol timing of data message  90 . Channel estimator module  72  uses LTFs  104  and their k respective training symbols  106  to estimate matrix H based on methods that are described below. 
     Referring now to  FIG. 5 , a method  120  is shown for estimating matrix H. Method  120  can be executed by channel estimator module  72 . In some embodiments method  120  can be implemented as a computer program or firmware that is stored in memory  75  and executed by processor  73 . 
     Control enters at a block  122  and proceeds to decision block  124 . In decision block  124  control determines whether an LTF  104  is being received. If not then control returns to block  122 . If an LTF  104  is being received then control branches from decision block  124  to block  126 . In block  126  control receives a training symbol  106  that is associated with the current LTF  104 . Control then proceeds to block  128  and updates a matrix H est , which is described below in more detail, based on the current training symbol  106 . Control then proceeds to decision block  130  and determines whether the current training symbol  106  was the last training symbol  106  of the present LTF  104 . If not then control branches to block  132  and waits for the next training symbol  106  of the current LTF  104 . When the next training symbol  106  is received control returns to block  126  and repeats the aforementioned steps for the new training symbol  106 . On the other hand, if the training symbol  106  in decision block  130  was the last training symbol  106  of the present LTF  104  then control branches to decision block  134 . 
     In decision block  134  control determines whether the current LTF  104  was the last LTF  104  (i.e. LTF  104 -x) of the current group of LTFs  104 . If not then control branches to block  136  and waits for the next LTF  104  to begin before returning to block  126 . On the other hand, if the current LTF  104  was the last LTF  104 -x then control branches from decision block  134  to block  138 . In block  138  control generates matrix H based on matrix H est  and matrix P. Control then proceeds to block  140  and adjusts the gains of equalizer module  76  based on matrix H. Control then returns to other processes via termination block  142 . 
     As channel estimator module  72  executes method  120  it performs distributed QR across LTFs  104 . That is, QR(H est ). Computational density therefore increases as O(n 2 ) and the processing latency of associated hardware and/or processor  73  would increase by about O(n 2 ). This represents an improvement, e.g. reduced need for processing power, over the prior art. Example estimations of matrix H est  will now be provided for various MIMO dimensions of communications system  50 . 
     2×2 and 2×3 MIMO Cases 
     In 2×2 and 2×3 MIMO cases the equalized vector is given by 
               x   ^     =       1     diag   ⁡     (       P     2   ×   2       ⁢     R     2   ×   2       -   1       ⁢     R     2   ×   2       -   *       ⁢     P     2   ×   2     T       )         ⁢     P     2   ×   2       ⁢     R     2   ×   2       -   1       ⁢     Q   *     ⁢   y           
It can be seen that
 
     
       
         
           
             
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     4×4 Spatial-multiplexing (SM) MIMO Case 
     For a general n×n MIMO communication system  50  the equalized vector is given by 
     
       
         
           
             
               
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     Methods are known in the art for recursively solving the Q n *y term of the above equation. Recursive computation of the P n R n   −1  and  w ll,n   =1./diag(P n R n   −1 R n   − *P n   T ) terms of the above equation will now be described. 
     4×4 SM MIMO Case—Substream Signal—to Noise Ratio (SNR) Recursion 
     Let  w   ll,n =1./diag(P n×n R n×n   −1 R n×n   − *P n×n   T ). It can be seen that for 1≦j&lt;n the j th  element of the w 11  vector for n streams can be recursively computed as follows: 
     
       
         
           
             
               
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                 For   ⁢           ⁢   j     =   n     ,       w     ll   ,   n     n     =     1       λ   n     +     k   n                 
where λ n =P(n,1:n−1)R n−1   −1 R n−1   − *P(n,1:n−1) T .
 
     A proof of the immediately preceding equations will now be provided. 
     
       
         
           
             
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                                 n 
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               R 
                               
                                 n 
                                 - 
                                 1 
                               
                               
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               R 
                               
                                 n 
                                 - 
                                 1 
                               
                               
                                 - 
                                 * 
                               
                             
                             ⁢ 
                             
                               P 
                               
                                 n 
                                 - 
                                 1 
                               
                               T 
                             
                           
                           , 
                           
                             
                               P 
                               
                                 ( 
                                 
                                   n 
                                   , 
                                   
                                     1 
                                     : 
                                     
                                       n 
                                       - 
                                       1 
                                     
                                   
                                 
                                 ) 
                               
                             
                             ⁢ 
                             
                               R 
                               
                                 n 
                                 - 
                                 1 
                               
                               
                                 - 
                                 1 
                               
                             
                             ⁢ 
                             
                               R 
                               
                                 n 
                                 - 
                                 1 
                               
                               
                                 - 
                                 * 
                               
                             
                             ⁢ 
                             
                               P 
                               
                                 ( 
                                 
                                   n 
                                   , 
                                   
                                     1 
                                     : 
                                     
                                       n 
                                       - 
                                       1 
                                     
                                   
                                 
                                 ) 
                               
                               T 
                             
                           
                         
                         ) 
                       
                     
                     + 
                     
                       diag 
                       ⁡ 
                       
                         ( 
                         
                           
                             P 
                             n 
                           
                           ⁢ 
                           
                             
                               vv 
                               _ 
                             
                             * 
                           
                           ⁢ 
                           
                             P 
                             n 
                             T 
                           
                         
                         ) 
                       
                     
                   
                 
               
             
           
         
       
     
     4×4 SM MIMO Case—Processing the 2 ND  LTF  104   
     Compute R 2×2 . 
     
       
         
           
             
               Compute 
               ⁢ 
               
                   
               
               ⁢ 
               
                 R 
                 
                   2 
                   ⁣ 
                   
                     × 
                     2 
                   
                 
                 
                   - 
                   1 
                 
               
             
             = 
             
               
                 [ 
                 
                   
                     
                       
                         1 
                         / 
                       
                     
                     
                       
                         
                           
                             - 
                             
                               r 
                               12 
                             
                           
                           / 
                           
                             r 
                             11 
                           
                         
                         ⁢ 
                         
                           r 
                           22 
                         
                       
                     
                   
                   
                     
                       0 
                     
                     
                       
                         1 
                         / 
                         
                           r 
                           22 
                         
                       
                     
                   
                 
                 ] 
               
               . 
             
           
         
       
     
     Compute 1/w ll,1 =(r 22   2 +∥r 11 +r 12 ∥ 2 )/r 11   2 r 22   2 . 
     Compute 1/w ll,2 =(r 22   2 +∥r 11 −r 12 ∥ 2 )/r 11   2 r 22   2 . 
     Let λ=1/w ll,2 . 
     4×4 SM MIMO Case—Processing the 3 RD  LTF  104   
     Update the inverse of the triangular matrix based on 
     
       
         
           
             
               R 
               
                 3 
                 × 
                 3 
               
             
             = 
             
               
                 
                   [ 
                   
                     
                       
                         
                           
                             
                               R 
                               
                                 2 
                                 × 
                                 2 
                               
                             
                           
                         
                         
                           
                             
                               0 
                               - 
                             
                           
                         
                       
                       ⁡ 
                       
                         [ 
                         
                           
                             
                               
                                 r 
                                 13 
                               
                             
                             
                               
                                 r 
                                 23 
                               
                             
                             
                               
                                 r 
                                 33 
                               
                             
                           
                         
                         ] 
                       
                     
                     T 
                   
                   ] 
                 
                 ⁢ 
                 
                   R 
                   
                     3 
                     × 
                     3 
                   
                   
                     - 
                     1 
                   
                 
               
               = 
               
                 [ 
                 
                   
                     
                       
                         
                           
                             R 
                             
                               2 
                               × 
                               2 
                             
                             
                               - 
                               1 
                             
                           
                         
                       
                       
                         
                           
                             0 
                             - 
                           
                         
                       
                     
                     ⁡ 
                     
                       [ 
                       
                         
                           
                             
                               ρ 
                               1 
                             
                           
                           
                             
                               ρ 
                               2 
                             
                           
                           
                             
                               ρ 
                               3 
                             
                           
                         
                       
                       ] 
                     
                   
                   T 
                 
                 ] 
               
             
           
         
       
       
         
           
             
               where 
               ⁢ 
               
                   
               
               ⁢ 
               
                 ρ 
                 1 
               
             
             = 
             
               
                 
                   ( 
                   
                     
                       
                         r 
                         12 
                       
                       ⁢ 
                       
                         r 
                         23 
                       
                     
                     - 
                     
                       
                         r 
                         13 
                       
                       ⁢ 
                       
                         r 
                         22 
                       
                     
                   
                   ) 
                 
                 / 
                 
                   r 
                   11 
                 
               
               ⁢ 
               
                 r 
                 22 
               
               ⁢ 
               
                 r 
                 33 
               
             
           
         
       
       
         
           
             
               ρ 
               2 
             
             = 
             
               
                 
                   - 
                   
                     r 
                     23 
                   
                 
                 / 
                 
                   r 
                   22 
                 
               
               ⁢ 
               
                 r 
                 33 
               
             
           
         
       
       
         
           
             
               ρ 
               3 
             
             = 
             
               1 
               / 
               
                 
                   r 
                   33 
                 
                 . 
               
             
           
         
       
     
     Update the substream SNRs based on
 
1/ w   ll,1 →1/ w   ll,1 +∥ρ 1 −ρ 2 +ρ 3 ∥ 2 .
 
1/ w   ll,2 →1/ w   ll,2 +∥ρ 1 +ρ 2 +ρ 3 ∥ 2 .
 
1/ w   ll,3 →λ+∥ρ 1 +ρ 2 +ρ 3 ∥ 2 .
 
     Update the lamda factor based on
 
λ=1/ w   ll,1 +1/ w   ll,1 −1/ w   ll,3 +8 real (ρ 2 )ρ 3 .
 
     4×4 SM MIMO Case—Processing the 4 TH  LTF  104   
     Update the inverse of the triangular matrix based on 
     
       
         
           
             
               R 
               
                 4 
                 × 
                 4 
               
             
             = 
             
               
                 
                   [ 
                   
                     
                       
                         
                           
                             
                               R 
                               
                                 3 
                                 × 
                                 3 
                               
                             
                           
                         
                         
                           
                             
                               0 
                               - 
                             
                           
                         
                       
                       ⁡ 
                       
                         [ 
                         
                           
                             
                               
                                 r 
                                 14 
                               
                             
                             
                               
                                 r 
                                 24 
                               
                             
                             
                               
                                 r 
                                 34 
                               
                             
                             
                               
                                 r 
                                 44 
                               
                             
                           
                         
                         ] 
                       
                     
                     T 
                   
                   ] 
                 
                 ⁢ 
                 
                     
                 
                 ⁢ 
                 
                   R 
                   
                     4 
                     × 
                     4 
                   
                 
               
               = 
               
                 [ 
                 
                   
                     
                       
                         
                           
                             R 
                             
                               3 
                               × 
                               3 
                             
                           
                         
                       
                       
                         
                           
                             0 
                             - 
                           
                         
                       
                     
                     ⁡ 
                     
                       [ 
                       
                         
                           
                             
                               r 
                               14 
                             
                           
                           
                             
                               r 
                               24 
                             
                           
                           
                             
                               r 
                               34 
                             
                           
                           
                             
                               r 
                               44 
                             
                           
                         
                       
                       ] 
                     
                   
                   T 
                 
                 ] 
               
             
           
         
       
     
     where
 
ρ 1 =( r   33 ( r   12   r   24   −r   14   r   22 )−( r   12   r   23   −r   13   r   22 ) r   34 )/ r   11   r   22   r   33   r   44 ,
 
ρ 2 =( r   34   r   23   −r   24   r   332 )/ r   22   r   33   r   44 , ρ 3   =−r   33   /r   33   r   44 , ρ 4 =1/ r   44 .
 
     Update the substream SNR based on
 
1/ w   ll,2 →1/ w   ll,2 +∥ρ 1 +ρ 2 −ρ 3 +ρ 4 ∥ 2  
 
1/ w   ll,3 →1/ w   ll,3 +∥ρ 1 +ρ 2 +ρ 3 −ρ 4 ∥ 2  
 
1/ w   ll,4   →λ+∥−ρ   1 +ρ 2 +ρ 3 +ρ 4 ∥ 2  
 
1/ w   ll,4   →λ+∥−ρ   1 +ρ 2 +ρ 3 −ρ 4 ∥ 2 .
 
     Compute their inverses and store them in a memory that can be included in channel estimator module  72 . 
     3×3 SM MIMO Case—Processing the 4 TH  LTF  104   
     The 3×3 MIMO case employs a non-square matrix P. For 3 streams transmitter module  56  sends 4 LTFs  104  and employs P of P 3    
     
       
         
           
             
               P 
               3 
             
             = 
             
               
                 [ 
                 
                   
                     
                       1 
                     
                     
                       
                         - 
                         1 
                       
                     
                     
                       1 
                     
                     
                       1 
                     
                   
                   
                     
                       1 
                     
                     
                       1 
                     
                     
                       
                         - 
                         1 
                       
                     
                     
                       1 
                     
                   
                   
                     
                       1 
                     
                     
                       1 
                     
                     
                       1 
                     
                     
                       
                         - 
                         1 
                       
                     
                   
                 
                 ] 
               
               . 
             
           
         
       
     
     Channel estimator module  72  estimates channel matrix H est3×4  and the real matrix is H=H est3×4 P 3×4   ⊥ . The received vector can be based on y=H est3×4 P 3×4   ⊥ x=Hx. 
     A distributed solution may also be employed by working directly with H est3×4  without forming H. In the distributed solution let
 
 H   est3×4   =QR   3×4  and
 
 H   est3×4   =[H   est3×3   h   4 ].
 
Then  Q*H   est3×4   =[R Q*h   4 ]where  QR=H   est3×3 .
 
     In some embodiments the equalized vector can be based on
 
{circumflex over ( x )}=( H   est,3×4   P   3×4   ⊥ ) ⊥   y.  
 
It can be seen that
 
{circumflex over (   x   )}= P   1   −1     z   −μ   v    where    u   = R   −1   Q*h   4 , and μ=[1−1 1]   z   .
 
The vector v is based on
 
   v   = kP   1   −1     u    where    u   = R   −1   Q*h   4 .
 
The scalar k is based on
 
 k= 1/(1+[1−1 1]   u   ).
 
The matrix P 1   −1  is based on
 
     
       
         
           
             
               P 
               1 
               
                 - 
                 1 
               
             
             = 
             
               
                 2 
                 ⁡ 
                 
                   [ 
                   
                     
                       
                         1 
                       
                       
                         
                           - 
                           1 
                         
                       
                       
                         0 
                       
                     
                     
                       
                         1 
                       
                       
                         0 
                       
                       
                         
                           - 
                           1 
                         
                       
                     
                     
                       
                         0 
                       
                       
                         1 
                       
                       
                         1 
                       
                     
                   
                   ] 
                 
               
               . 
             
           
         
       
     
     A proof of the immediately preceding equations will now be provided. 
     A solution is given by {circumflex over (x)}=(H est,3×4 P 3×4   ⊥ ) −1 y. 
     Let 
               P     3   ×   4     ⊥     ⁡     [           P   1               p   1           ]           
and write the matrix as
 
                 (       H     est   ,     3   ×   4         ⁢     P     3   ×   4     ⊥       )       -   1       =         [         H     est   ,     3   ×   3         ⁢     P   1       +       h   4     ⁢     p   1         ]       -   1       =         [       QRP   1     +       h   4     ⁢     p   1         ]       -   1       =           P   1     -   1       ⁢     R     -   1       ⁢     Q   *       -         P   1     -   1       ⁢     R     -   1       ⁢     Q   *     ⁢     h   4     ⁢     p   1     ⁢     P   1     -   1       ⁢     R     -   1       ⁢     Q   *         1   +       p   1     ⁢     P   1     -   1       ⁢     R     -   1       ⁢     Q   *     ⁢     h   4             =       [     I   -       vp   1       1   +       p   1     ⁢   v           ]     ⁢     P   1     -   1       ⁢     R     -   1       ⁢     Q   *                   
where v=P 1   −1 R −1 Q*h 4 .
 
     
       
         
           
             
               
                 P 
                 1 
                 
                   - 
                   1 
                 
               
               = 
               
                 2 
                 ⁡ 
                 
                   [ 
                   
                     
                       
                         1 
                       
                       
                         
                           - 
                           1 
                         
                       
                       
                         0 
                       
                     
                     
                       
                         1 
                       
                       
                         0 
                       
                       
                         
                           - 
                           1 
                         
                       
                     
                     
                       
                         0 
                       
                       
                         1 
                       
                       
                         1 
                       
                     
                   
                   ] 
                 
               
             
             , 
             
               
                 
                   p 
                   1 
                 
                 ⁢ 
                 
                   P 
                   1 
                   
                     - 
                     1 
                   
                 
               
               = 
               
                 
                   [ 
                   
                     
                       
                         1 
                       
                       
                         
                           - 
                           1 
                         
                       
                       
                         
                           - 
                           1 
                         
                       
                     
                   
                   ] 
                 
                 . 
               
             
           
         
       
     
     3×3 SM MIMO Case—Solution for W ll    
     Channel estimator module  72  computes
 
 W   ll =1/(( H   est,3×4   P   3×4   ⊥ )*( H   est,3×4   P   3×4   ⊥ )) −1  
 
based on terms of v=kP 1   −1 R −1 Q*h 4  and R −1 R 31 *.
 
Let this matrix be
 
                 R     -   1       ⁢     R     -   *         =     [             I   11     ⁢     jQ   11               I   21     ⁢     jQ   21               I   31     ⁢     jQ   31                   I   21     ⁢     jQ   21               I   22     ⁢     jQ   22               I   32     ⁢     jQ   32                   I   31     ⁢     jQ   31               I   32     ⁢     jQ   32               I   33     ⁢     jQ   33             ]           
and during LTFs  104  compute
 
 S   1 =4( I   11 −2 I   21   +I   22 )
 
 S   2 =4( I   11 −2 I   31   +I   33 )
 
 S   3 =4( I   22 +2 I   32   +I   33 )
 
 S= ( I   11   +I   22   +I   33 )−2( I   21   +I   33   I   32  )
 
 S   4   =I   11 −21 I   21   +I   22   −I   31   +I   32   j ( Q   11   +Q   22   +Q   31    −Q   32 )
 
 S   5   =I   11   −I   21   +I   32  −2 I   31    I   33    +j ( Q   11   +   21    Q   32    Q   33 )
 
 S   6   =I   21   −I   22   +I   31 −2 I   32   −I   33    +j ( Q   21   −Q   22   +Q   31   −Q   33 )
 
It can be seen that
 
 hd ll 1   =1/( S   1   +S∥v   1 ∥ 2 −4real( S   4   v   1 *))
 
 W   ll2 =1/( S   2   +S∥V   2 ∥ 2 −4real( S   5   v   2 *))
 
 W   ll3 =1/( S   3   +S∥v   3 ∥ 2 −4real( S   6   v   3 *))
 
and  v =kP 1   −1   u .
 
     Proof of the above solution for W ll  will now be provided. Let
 
 H=[QR h   4   ]P   3×4   ⊥               Q*H=[R Q*h   4   ]P   3×4   ⊥ =
 
 R ( P   1 +   u     p   1 ),    u   = R   1   Q*h   4  

     Then 
                 (       H   *     ⁢   H     )       -   1       =           (       P   1     +       u   _     ⁢     p   1         )       -   1       ⁢     R     -   1       ⁢         R     -   *       ⁡     (       P   1   T     +       p   1   T     ⁢       u   _     *         )         -   1         =         (     I   -         P   1     -   1       ⁢     u   _     ⁢     p   1         1   +       [         1         -   1           -   1           ]     ⁢     u   _             )     ⁢     P   1     -   1       ⁢     R     -   1       ⁢     R     -   *       ⁢       R   1     -   T       ⁡     (     I   -         p   1   T     ⁢       u   _     *     ⁢     P   1     -   T           1   +       [         1         -   1           -   1           ]     ⁢     u   _             )         =         (       P   1     -   1       -     v   ⁡     [         1         -   1           -   1           ]         )     ⁢     R     -   1       ⁢       R     -   *       ⁡     (       P   1     -   T       -         [         1         -   1           -   1           ]     T     ⁢     v   *         )         =         P   1     -   1       ⁢     R     -   1       ⁢     R     -   *       ⁢     P   1     -   T         +       v   ⁡     [         1         -   1           -   1           ]       ⁢     R     -   1       ⁢         R     -   *       ⁡     [         1         -   1           -   1           ]       T     ⁢     v   *       -     2   ⁢           ⁢     real   ⁡     (       P   1     -   1       ⁢     R     -   1       ⁢       R     -   *       ⁡     [         1         -   1           -   1           ]       ⁢     v   *       )                         v=kP   1   −1   R   −1   Q*h   4 . 
     Channel estimator module  72  still needs to recursively update R −1 R − * after determining  v =kP 1   −1   u  as described above. At a time n let
 
 R   n   −1   =[R   n−1   −1 ρ]
 
and apply the following identity
 
 R   n   −1   R   n   −   *=R   n−1   −* +ρρ*.
 
For example,
 
     
       
         
           
             
               R 
               
                 3 
                 × 
                 3 
               
               
                 - 
                 1 
               
             
             = 
             
               [ 
               
                 
                   
                     
                       
                         
                           R 
                           
                             2 
                             × 
                             2 
                           
                           
                             - 
                             1 
                           
                         
                       
                     
                     
                       
                         
                           0 
                           - 
                         
                       
                     
                   
                   ⁡ 
                   
                     [ 
                     
                       
                         
                           
                             ρ 
                             1 
                           
                         
                         
                           
                             ρ 
                             2 
                           
                         
                         
                           
                             ρ 
                             3 
                           
                         
                       
                     
                     ] 
                   
                 
                 T 
               
               ] 
             
           
         
       
       
         
           
             
               
                 ρ 
                 1 
               
               = 
               
                 
                   
                     ( 
                     
                       
                         
                           r 
                           12 
                         
                         ⁢ 
                         
                           r 
                           23 
                         
                       
                       - 
                       
                         
                           r 
                           13 
                         
                         ⁢ 
                         
                           r 
                           22 
                         
                       
                     
                     ) 
                   
                   / 
                   
                     r 
                     11 
                   
                 
                 ⁢ 
                 
                   r 
                   22 
                 
                 ⁢ 
                 
                   r 
                   33 
                 
               
             
             ⁢ 
             
                 
             
           
         
       
       
         
           
             
               ρ 
               2 
             
             = 
             
               
                 
                   - 
                   
                     r 
                     23 
                   
                 
                 / 
                 
                   r 
                   22 
                 
               
               ⁢ 
               
                 r 
                 33 
               
             
           
         
       
       
         
           
             
               ρ 
               3 
             
             = 
             
               1 
               / 
               
                 
                   r 
                   33 
                 
                 . 
               
             
           
         
       
     
     3×3 SM MIMO Case—Procesing the 1 ST  LTF  104   
     Channel estimator module  72  performs first column nulling by computing
 
 R   1   −1 =1/ r   11 .
 
     3×3 SM MIMO Case—Processing the 2 ND  LTF  104   
     Channel estimator module  72  performs second column QR processing based on 
                 R     2   ×   2       -   1       =     [               R   1     -   1                 0   -           ⁡     [           ρ   1           ρ   2           ]       T     ]       ⁢                           where   ⁢           ⁢     ρ   1       =         -     r   12       /     r   11       ⁢     r   22         ,       and   ⁢           ⁢     ρ   2       =     1   /       r   22     .               
Channel estimator module  72  then computes ∥ρ 1 ∥ 2 ,∥ρ 2 ∥ 2 , ρ 1 ρ 2 * and updates R −1 R − * based on
 
     
       
         
           
             
               
                 R 
                 
                   2 
                   × 
                   2 
                 
                 
                   - 
                   1 
                 
               
               ⁢ 
               
                 R 
                 
                   2 
                   × 
                   2 
                 
                 
                   - 
                   * 
                 
               
             
             = 
             
               
                 [ 
                 
                   
                     
                       
                         
                           1 
                           / 
                           
                             r 
                             11 
                           
                         
                         + 
                         
                           
                              
                             
                               ρ 
                               1 
                             
                              
                           
                           2 
                         
                       
                     
                     
                       
                         
                           ρ 
                           1 
                         
                         ⁢ 
                         
                           ρ 
                           2 
                           * 
                         
                       
                     
                   
                   
                     
                       
                         
                           ρ 
                           1 
                           * 
                         
                         ⁢ 
                         
                           ρ 
                           2 
                         
                       
                     
                     
                       
                         
                            
                           
                             ρ 
                             2 
                           
                            
                         
                         2 
                       
                     
                   
                 
                 ] 
               
               . 
             
           
         
       
     
     3−3 SM MIMO Case—Procesing the 3 RD  LTF  104   
     Channel estimator module  72  performs third column QR processing by recursively updating 
               R     3   ×   3       -   1       =     [               R     2   ×   2       -   1                 0   -           ⁡     [           ρ   1           ρ   2           ρ   3           ]       T     ]           
where ρ 1 =(r 12 r 23 −r 13 r 22  )/r 11 r 22 r 33 ρ 2 =−r 23 /r 22 r 33 ρ 3 =1/r 33 .
 
Channel estimator module  72  can then compute
 
∥ρ 1 ∥ 2 ,∥ρ 2 ∥ 2 ,∥ρ 2 ∥ 2 , ρ 1 ρ 2 *, ρ 1 ρ 3 *, ρ 2ρ   3 *
 
and update R −1 R − * based on
 
 R   3×3   −1   R   3×3   −   *=R   2×2   −1   R   2×2   − *+ρρ*.
 
     Channel estimator module  72  can then compute the sums
 
 S   1 =4( I   11 −2 I   21   +I   22 )
 
 S   2 =4( I   11 −2 I   31   +I   33 )
 
 S   3 =4( I   22 +2 I   32   +I   33 )
 
 S= ( I   11   +I   22   +I   33 )−2( I   21   +I   33   I   32  )
 
 S   4   =I   11 −2 I   21   +I   22   −I   31   +I   32   j ( Q   11   +Q   22   +Q   31    −Q   32 )
 
 S   5   =I   11   −I   21   +I   32  −2 I   31    I   33    +j ( Q   11   +   21    Q   32    Q   33 )
 
 S   6   =I   21   −I   22   +I   31 −2 I   32   −I   33    +j ( Q   21   −Q   22   +Q   31   −Q   33 )
 
     3×3 SM MIMO Case—Procesing the 4 TH  LTF  104   
     Channel estimator module  72  computes
 
   u   = R   −1   Q*h   4 ,
 
 k= 1/(1+[1−1 1]   u   ), and
 
   v   = kP   1   −1     u   .
 
Channel estimator module can store  v  in memory  75 .
 
     Channel estimator module  72  computes substream SNR based on
 
 W   ll1 =1/( S   1   +S∥v   1 ∥ 2 −4real( S   4   v   1 *))
 
 W   ll2 =1/( S   2   +S∥v   2 ∥ 2 −4real( S   5   v   2 *))
 
 W   ll3 =1/( S   3   +S∥v   3 ∥ 2 −4real( S   6   v   3 *))
 
     and store the substream SNR in memory  75 . 
     3×3 SM MIMO Case—Procesing Data  92   
     Channel estimator module  72  can compute
 
   z   = R   −1   Q*y  and μ=[1−1 1]   z   
 
and read  v  from memory  75 . Channel estimator module  72  can then compute.
 
 {circumflex over ( x )} = P   1   −1     z   −μ   v   
 
and read the substream SNR from memory  75 . Equalizer module  76  can scale the equalized vector based on the SNRs.
 
     Referring now to  FIGS. 6A-6E , various exemplary implementations of the receiver module are shown. Referring now to  FIG. 6A , the receiver module can be implemented in a high definition television (HDTV)  420 . The receiver module may implement and/or be implemented in a WLAN interface  429 . The HDTV  420  receives HDTV input signals in either a wired or wireless format and generates HDTV output signals for a display  426 . In some implementations, signal processing circuit and/or control circuit  422  and/or other circuits (not shown) of the HDTV  420  may process data, perform coding and/or encryption, perform calculations, format data and/or perform any other type of HDTV processing that may be required. 
     The HDTV  420  may communicate with mass data storage  427  that stores data in a nonvolatile manner such as optical and/or magnetic storage devices. Mass data storage  427  may include at least one hard disk drive (HDD) and/or at least one digital versatile disk (DVD) drive. The HDD may be a mini HDD that includes one or more platters having a diameter that is smaller than approximately 1.8″. The HDTV  420  may be connected to memory  428  such as RAM, ROM, low latency nonvolatile memory such as flash memory and/or other suitable electronic data storage. The HDTV  420  also may support connections with a WLAN via WLAN network interface  429 . The HDTV  420  also includes a power supply  423 . 
     Referring now to  FIG. 6B , the receiver module may implement and/or be implemented in a WLAN interface  448  of a vehicle  430 . In some implementations WLAN interface  448  communicates with a powertrain control system  432  that receives inputs from one or more sensors. Examples of sensors includes temperature sensors, pressure sensors, rotational sensors, airflow sensors and/or any other suitable sensors and/or that generates one or more output control signals such as engine operating parameters, transmission operating parameters, and/or other control signals. 
     The receiver module may also be implemented in other control systems  440  of the vehicle  430 . The control system  440  may likewise receive signals from input sensors  442  and/or output control signals to one or more output devices  444 . In some implementations, the control system  440  may be part of an anti-lock braking system (ABS), a navigation system, a telematics system, a vehicle telematics system, a lane departure system, an adaptive cruise control system, a vehicle entertainment system such as a stereo, DVD, compact disc and the like. Still other implementations are contemplated. 
     The powertrain control system  432  may communicate with mass data storage  446  that stores data in a nonvolatile manner. Mass data storage  446  may include at least one HDD and/or at least one DVD drive. The HDD may be a mini HDD that includes one or more platters having a diameter that is smaller than approximately 1.8″. The powertrain control system  432  may be connected to memory  447  such as RAM, ROM, low latency nonvolatile memory such as flash memory and/or other suitable electronic data storage. The powertrain control system  432  also may support connections with a WLAN via a WLAN network interface  448 . The control system  440  may also include mass data storage, memory and/or a WLAN interface (all not shown). Vehicle  430  may also include a power supply  433 . 
     Referring now to  FIG. 6C , the receiver module can be implemented in a cellular phone  450  that may include a cellular antenna  451 . The receiver module may implement and/or be implemented in a WLAN interface  468 . In some implementations, the cellular phone  450  includes a microphone  456 , an audio output  458  such as a speaker and/or audio output jack, a display  460  and/or an input device  462  such as a keypad, pointing device, voice actuation and/or other input device. The signal processing and/or control circuits  452  and/or other circuits (not shown) in the cellular phone  450  may process data, perform coding and/or encryption, perform calculations, format data and/or perform other cellular phone functions. 
     The cellular phone  450  may communicate with mass data storage  464  that stores data in a nonvolatile manner. Mass data storage  450  may include at least one HDD and/or at least one DVD drive. The HDD may be a mini HDD that includes one or more platters having a diameter that is smaller than approximately 1.8″. The cellular phone  450  may be connected to memory  466  such as RAM, ROM, low latency nonvolatile memory such as flash memory and/or other suitable electronic data storage. The cellular phone  450  also may support connections with a WLAN via the WLAN network interface  468 . The cellular phone  450  may also include a power supply  453 . 
     Referring now to  FIG. 6D , the receiver module can be implemented in a set top box  480 . The receiver module may implement and/or be implemented in a WLAN interface  496 . The set top box  480  receives signals from a source such as a broadband source and outputs standard and/or high definition audio/video signals suitable for a display  488  such as a television and/or monitor and/or other video and/or audio output devices. The signal processing and/or control circuits  484  and/or other circuits (not shown) of the set top box  480  may process data, perform coding and/or encryption, perform calculations, format data and/or perform any other set top box function. 
     The set top box  480  may communicate with mass data storage  490  that stores data in a nonvolatile manner. Mass data storage  490  may include at least one HDD and/or at least one DVD drive. The HDD may be a mini HDD that includes one or more platters having a diameter that is smaller than approximately 1.8″. The set top box  480  may be connected to memory  494  such as RAM, ROM, low latency nonvolatile memory such as flash memory and/or other suitable electronic data storage. The set top box  480  also may support connections with a WLAN via the WLAN network interface  496 . The set top box  480  may include a power supply  483 . 
     Referring now to  FIG. 6E , the receiver module can be implemented in a media player  500 . The receiver module may implement and/or be implemented in a WLAN interface  516 . In some implementations, the media player  500  includes a display  507  and/or a user input  508  such as a keypad, touchpad and the like. In some implementations, the media player  500  may employ a graphical user interface (GUI) that typically employs menus, drop down menus, icons and/or a point-and-click interface via the display  507  and/or user input  508 . The media player  500  further includes an audio output  509  such as a speaker and/or audio output jack. The signal processing and/or control circuits  504  and/or other circuits (not shown) of the media player  500  may process data, perform coding and/or encryption, perform calculations, format data and/or perform any other media player function. 
     The media player  500  may communicate with mass data storage  510  that stores data such as compressed audio and/or video content in a nonvolatile manner. In some implementations, the compressed audio files include files that are compliant with MP3 format or other suitable compressed audio and/or video formats. Mass data storage  510  may include at least one HDD and/or at least one DVD drive. The HDD may be a mini HDD that includes one or more platters having a diameter that is smaller than approximately 1.8″. The media player  500  may be connected to memory  514  such as 
     RAM, ROM, low latency nonvolatile memory such as flash memory and/or other suitable electronic data storage. The media player  500  also may support connections with a WLAN via the WLAN network interface  516 . The media player  500  may also include a power supply  513 . Still other implementations in addition to those described above are contemplated. 
     Those skilled in the art can now appreciate from the foregoing description that the broad teachings of the disclosure can be implemented in a variety of forms. Therefore, while this disclosure includes particular examples, the true scope of the disclosure should not be so limited since other modifications will become apparent to the skilled practitioner upon a study of the drawings, the specification and the following claims.