Abstract:
A communication apparatuses, a transmission method, a receiving method of a wireless network system for hybrid automatic repeat request (HARQ) and a tangible machine-readable medium thereof are provided. The wireless network system comprises a base station (BS) and at least one mobile station (MS). The transmission method comprises the following steps of: transmitting at least one first burst having a first symbol and a second symbol to the at least one MS; receiving at least one negative acknowledgement (NAK) from the at least one MS; generating a third symbol by proceeding a linear combination according to the first symbol and the second symbol of the at least one first burst; and transmitting at least one second burst having the third symbol to the at least one MS.

Description:
CROSS-REFERENCES TO RELATED APPLICATIONS 
       [0001]    This application claims the benefit of Provisional Application Ser. No. 61/078,357 filed on Jul. 4, 2008. 
     
    
     BACKGROUND OF THE INVENTION 
       [0002]    1. Field of the Invention 
         [0003]    The present invention relates to a communication apparatuses, a transmission method, a receiving method of a wireless network system for hybrid automatic repeat request (HARQ) and a tangible machine-readable medium thereof. More specifically, the present invention relates to a communication apparatuses, a transmission method, a receiving method of a wireless network system for HARQ and a tangible machine-readable medium thereof which provide a retransmission scheme for saving bandwidth resource. 
         [0004]    2. Descriptions of the Related Art 
         [0005]    Although the IEEE 802.16 standard already provides greater bandwidths, lower building cost, better service quality and expansibility, there still exist some defects of coverage and signal quality of the IEEE 802.16 standard. After the International Telecommunications Union-Radiocommunication Sector (ITU-R) developed the International Mobile Telecommunications-Advanced (IMT-Advanced) air interface standardization, common technical, operational and spectrum-related parameters of systems will maximize the commonality between IMT-Advanced air interfaces. However, no existing IEEE 802 standards or projects can meet the IMT-Advanced target requirements nowadays, such as 100 Mbit/sec in high-speed mobility applications. Therefore, the IEEE 802.16m standard is developed like a raging fire to meet the IMT-Advanced target requirements by the IEEE 802.16 Working Group&#39;s Task Group m. 
         [0006]    The HARQ, adopted in the IEEE 802.16 standard, is an advanced data retransmission strategy, which allows performing possible data retransmissions directly at the physical layer instead of the media access control (MAC) layer and/or higher layers. Since the HARQ is able to achieve data retransmission without involving mechanisms at the higher layers, the delay caused by data retransmission is significantly reduced. However, the original HARQ in the IEEE 802.16 standard fails to meet the IMT-Advanced target requirements. 
         [0007]      FIG. 1A  is a schematic diagram illustrated a wireless network system  1  with a chase combining (CC) HARQ scheme under the IEEE 802.16 standard. The wireless network system  1  comprises a base station (BS)  11  and a mobile station (MS)  13 . 
         [0008]    During a downlink period, the BS transmits a burst  101  (shown in  FIG. 1B ) comprising a cyclic redundancy check (CRC) C 1  and a plurality of symbols (e.g. six symbols S 1 , S 2 , S 3 , S 4 , S 5 , S 6 ). After the MS  13  receives the burst  101 , it will check the burst  101  by the CRC C 1 . More specifically, the MS  13  will first determine whether the burst  101  can be decoded successfully or not by checking the CRC C 1 . If the burst  101  is successfully decoded, the MS  13  will feedback an acknowledgement ACK to the BS  11  and forward the decoded burst to an upper layer. Otherwise, the MS  13  will feedback a negative acknowledgement NAK to the BS  11 . Upon receiving the negative acknowledgement NAK, the BS  11  will retransmit another burst which is the same as the burst  101  to the MS  13 . The retransmission scheme will not stop until the BS  11  receives an acknowledgement ACK from the MS  13 , retransmission time expires, or retry count exhausted, so that the error rate of the wireless network system  1  will drop. 
         [0009]    However, as the number of the negative acknowledgement NAK increases, the wireless network system  1  has to spend more and more bandwidth resource to retransmit the same bursts, such as the burst  101 . In other words, the wireless network system  1  will hold too much resource (e.g. symbols and bandwidth) to increase the throughput so that the spectrum efficiency and capacity of the wireless network system under the IEEE 802.16 standard are both low and thus fail to meet the IMT-Advanced target requirements. 
         [0010]    Accordingly, how to reduce the error probability without over wasting resource of the wireless network system to meet the IMT-Advanced target requirements is still an objective for the industry to endeavor. In view of this, efforts still have to be made in the wireless communication industry to provide a solution to achieve transmission for HARQ under the IEEE 802.16 standard. 
       SUMMARY OF THE INVENTION 
       [0011]    The first objective of the present invention is to provide a communication apparatus of a wireless network system for HARQ. The wireless network system comprises at least one MS. The communication apparatus comprises a transmitting module, a receiving module and a processing module. The transmitting module is configured to transmit at least one first burst having a first symbol and a second symbol to the at least one MS. The receiving module is configured to receive at least one NAK from the at least one MS. The processing module is configured to generate a third symbol by proceeding a liner combination according to the first symbol and the second symbol of the at least one first burst. Finally, the transmitting module transmits at least one second burst having the third symbol to the at least one MS. 
         [0012]    The second objective of the present invention is to provide another communication apparatus of a wireless network system for HARQ. The wireless network system comprises a BS. The communication apparatus comprises a transmitting module, a receiving module and a processing module. The receiving module is configured to receive at least one first burst having a first symbol and a second symbol from the BS. The processing module is configured to determine that the at least one first burst is incorrect. The transmitting module is configured to transmit at least one NAK to the BS after the processing module determines that the at least one first burst is incorrect. Then the receiving module receives at least one second burst having a third symbol, which is generated by proceeding a liner combination according to the first symbol and the second symbol of the at least one first burst, from the BS. Finally, the processing module estimates the first symbol and the second symbol according to the at least one first burst and the at least one second burst. 
         [0013]    The third objective of the present invention is to provide a transmission method of a wireless network system for HARQ. The wireless network system comprises at least one MS. The transmission method comprises the step of: (a) transmitting at least one first burst having a first symbol and a second symbol to the at least one MS; (b) receiving at least one NAK from the at least one MS; (c) generating a third symbol by proceeding a liner combination according to the first symbol and the second symbol of the at least one first burst; and (d) transmitting at least one second burst having the third symbol to the at least one MS. 
         [0014]    The fourth objective of the present invention is to provide a receiving method of a wireless network system for HARQ. The wireless network system comprises a BS. The receiving method comprises the step of: (a) receiving at least one first burst having a first symbol and a second symbol from the BS; (b) determining that the at least one first burst is incorrect; (c) transmitting at least one NAK to the BS after determining that the at least one first burst is incorrect; (d) receiving at least one second burst having a third symbol from the BS, wherein the third symbol is generated by proceeding a liner combination according to the first symbol and the second symbol of the at least one first burst via the BS; and (e) estimating the first symbol and the second symbol according to the at least one first burst and the at least one second burst. 
         [0015]    This invention provides a tangible machine-readable medium storing a program which, when being executed, enables a communication apparatus to execute the transmission/receiving method of the wireless sensor network for HARQ described above. 
         [0016]    Accordingly, the present invention can provide a retransmission scheme of saving bandwidth resource by transmitting half symbol number than that of the prior art. The spectrum efficient and system capacity therefore will be approved with low gain loss by appropriate decoding methods. 
         [0017]    The detailed technology and preferred embodiments implemented for the subject invention are described in the following paragraphs accompanying the appended drawings for people skilled in this field to well appreciate the features of the claimed invention. 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0018]      FIG. 1A  is a schematic diagram illustrating a conventional wireless network system with a chase combining HARQ scheme; 
           [0019]      FIG. 1B  is a schematic diagram illustrating a burst of the conventional wireless network system; 
           [0020]      FIG. 2A  is a schematic diagram illustrating the first embodiment in accordance with the present invention; 
           [0021]      FIG. 2B  is a schematic diagram illustrating bursts of the first embodiment; 
           [0022]      FIG. 2C  is a schematic diagram illustrating maximum gain combining decoding scheme of the first embodiment; 
           [0023]      FIG. 3A  is a flow chart of the second embodiment of the present invention; and 
           [0024]      FIG. 3B  is a flow chart of the third embodiment of the present invention. 
       
    
    
     DESCRIPTION OF THE PREFERRED EMBODIMENT 
       [0025]    In the following description, the present invention will be explained with reference to embodiments thereof. However, the descriptions of these embodiments are only for purposes of illustration rather than limitations. It should be appreciated that in the following embodiments and the attached drawings, the elements not related directly to this invention are omitted from depiction and dimensional relationships among individual elements in the attached drawings are illustrated only for ease of understanding, and not limitation. 
         [0026]      FIG. 2A  illustrates a wireless network system  2  of a first embodiment under the IEEE 802.16 standard in accordance with the present invention, wherein the wireless network system  2  is with a symbol-level multiplexing HARQ retransmission scheme. The IEEE 802.16 standard can be any standard which will be developed in the future, such as the IEEE 802.16m standard. 
         [0027]    The wireless network system  2  comprises a first communication apparatus, such as a BS  21  and at least one second communication apparatus, such as a MS  23 . The BS  21  comprises a transmitting module  211 , a processing module  212  and a receiving module  213 . Similarly, the MS  23  comprises a transmitting module  231 , a processing module  232  and a receiving module  233 . 
         [0028]    During a downlink period, the transmitting module  211  of the BS  21  initially transmits a first burst  20  (shown in  FIG. 2B ) to the MS  23 , wherein the first burst  20  comprises a plurality of symbols and a CRC C 20 . For the sake of convenience, there are two symbols described in the first burst  20  (e.g. a first symbol S 201  and a second symbol S 202 ). After the receiving module  233  of the MS  23  receives the first burst  20 , the processing module  232  of the MS  23  will determine whether the first burst  20  is incorrect or not. In other words, the processing module  232  will decode the first burst  20  according to the CRC C 20 . If the first burst  20  can be decoded successfully, the MS  23  will feedback an acknowledgement ACK to the BS  21  and forward the decoded burst to an upper layer. Otherwise, the transmitting module  231  of the MS  23  will transmit a negative acknowledgement NAK to the BS  21  after the processing module  232  determines that the first  20  is incorrect. 
         [0029]    In the first embodiment, the receiving module  213  receives the negative acknowledgement NAK from the MS  23 , and the processing module  212  of the BS  21  will generate a third symbol S 221  (as shown in  FIG. 2B ) by proceeding a liner combination according to the first symbol S 201  and the second symbol  202  of the first burst  20  (shown in  FIG. 2B ). Then the transmitting module  211  of the BS  21  transmits a second burst  22  having the third symbol S 221  to the MS  23 . 
         [0030]    More specifically, the BS  21  will perform the linear combination function according to the first and the second symbols S 201 , S 202  to generate the third symbol S 221  before transmitting the second burst  22 . The first and the second symbols S 201 , S 202  may be adjacent or not adjacent, and in the embedment, the first and the second symbols S 201 , S 202  are adjacent. 
         [0031]    For example, the processing module  212  of the BS  21  may generate the third symbol S 221  by performing a subtraction between the first symbol S 201  and the second symbol S 202  of the first burst  20 . By the linear combination function, the symbol number of the second burst  22  is half than that of the first burst  20 . In this embedment, the third symbol S 221  is a subtraction between the first symbol S 201  and the second symbol S 202  of the first burst  20 . 
         [0032]    It should be noted that the symbol number of the second burst  22  ought to depend on that of the first burst  20 . For example, if the first burst  20  comprises six symbols, i.e. the 1 st , 2 nd , 3 rd , 4 th , 5 th , and 6 th  symbol in order, the second burst  22  will then comprise three symbols by the BS  21  performing the linear combination function of the six symbols of the first burst  20 . More specifically, the three symbols of the second burst  22  may be generated by the subtraction or an addition between the 1 st  symbol and the 2 nd  symbol of the first burst  20 , the subtraction or the addition between the 3 rd  symbol and the 4 th  symbol of the first burst  20 , and the subtraction or the addition between the 5 th  symbol and the 6 th  symbol of the first burst  20 , respectively. Those skilled in the art can understand the corresponding operations of linear combination function of the BS  21  by the explanation of the above description, and thus more detailed explanation is unnecessary. 
         [0033]    After the receiving module  233  of the MS  23  receives the second burst  22  having the third symbol S 221 , processing module  232  of the MS  23  will start to estimate the first and the second symbols S 201 , S 202  according to the first burst  20  and the second burst  22 . More specifically, although the processing module  232  can not decode the first burst  20  to retrieval the first and the second symbols S 201 , S 202  successfully, it can still estimate the first and the second symbols S 201 , S 202  according to the second burst  22 . The estimation will be described as the following description. 
         [0034]    When starting to estimate the first and the second symbols S 201  and S 202 , the processing module  232  first obtains h 1 S 1 +h 1 ′S 2  by adding the first and the second symbols S 201 , S 202  in the first burst  20 , wherein S 1  represents the first symbol S 201 , S 2  represents the second symbol S 202 , and h 1  represents the channel response of S 1  and h 1  represents the channel response of S 2  in the transmission of the first burst  20 . The processing module  232  further obtains h 2 S 1 −h 2 S 2  according to the third symbol S 221  in the second burst  22 , wherein h 2  represents the channel response of the third symbol S 221  in the transmission of the second burst  22 . Note that channel responses h 1 , h 1 ′ and h 2  may be obtained by arbitrary channel estimation techniques. Since the first and the second symbols S 201  and S 202  are two adjacent symbols, for simplified the derivation, it is reasonable to regard that their channel responses are approximately the same, which means h 1 =h 1 ′. However, the decoding methods described below will also applicable without this constraint. 
         [0035]    The equation (1) shows the combining for the h 1 S 1 +h 1 S 2  and h 2 S 1 −h 2 S 2 : 
         [0000]    
       
         
           
             
               
                 
                   X 
                   = 
                   
                     
                       [ 
                       
                         
                           
                             
                               x 
                               1 
                             
                           
                         
                         
                           
                             
                               x 
                               2 
                             
                           
                         
                       
                       ] 
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                     = 
                     
                       
                         [ 
                         
                           
                             
                               
                                 
                                   
                                     h 
                                     1 
                                   
                                    
                                   
                                     S 
                                     1 
                                   
                                 
                                 + 
                                 
                                   
                                     h 
                                     1 
                                   
                                    
                                   
                                     S 
                                     2 
                                   
                                 
                               
                             
                           
                           
                             
                               
                                 
                                   
                                     h 
                                     2 
                                   
                                    
                                   
                                     S 
                                     1 
                                   
                                 
                                 - 
                                 
                                   
                                     h 
                                     2 
                                   
                                    
                                   
                                     S 
                                     2 
                                   
                                 
                               
                             
                           
                         
                         ] 
                       
                        
                       
                         
 
                       
                        
                       
                           
                       
                       = 
                       
                         
                           
                             
                               [ 
                               
                                 
                                   
                                     
                                       h 
                                       1 
                                     
                                   
                                   
                                     
                                       h 
                                       1 
                                     
                                   
                                 
                                 
                                   
                                     
                                       h 
                                       2 
                                     
                                   
                                   
                                     
                                       - 
                                       
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                                         2 
                                       
                                     
                                   
                                 
                               
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                              
                             
                               [ 
                               
                                 
                                   
                                     
                                       S 
                                       1 
                                     
                                   
                                 
                                 
                                   
                                     
                                       S 
                                       2 
                                     
                                   
                                 
                               
                               ] 
                             
                           
                           + 
                           
                             [ 
                             
                               
                                 
                                   
                                     
                                       n 
                                       1 
                                     
                                     + 
                                     
                                       n 
                                       2 
                                     
                                   
                                 
                               
                               
                                 
                                   
                                     n 
                                     3 
                                   
                                 
                               
                             
                             ] 
                           
                         
                          
                         
                           
 
                         
                          
                         
                             
                         
                         = 
                         
                           HS 
                           + 
                           N 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
         [0036]    Wherein n 1 +n 2  represents a noise part of the transmission of the first burst  20 , n 3  represents a noise part of the transmission of the second burst  22 , H represents a matrix related to the gain, S represents a matrix related to the first and the second bursts  20 ,  22 , and N represents a matrix related to noise. 
         [0037]    There are some approaches, which are described as the following description, for the processing module  232  to estimate S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) according to the equation (1). 
         [0038]    Approach 1: Maximum Likelihood (ML) 
         [0039]    One alternative approach is using Maximum Likelihood according to the following equation: 
         [0000]    
       
         
           
             
               S 
               ^ 
             
             = 
             
               
                 arg 
                 S 
               
                
               min 
                
               
                  
                 
                   X 
                   - 
                   HS 
                 
                  
               
             
           
         
       
     
         [0040]    S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) can be estimated such that the term of ∥X−HS∥ has a minimum value. The ML is a well-known method in the prior art, and the details will not be mentions here. 
         [0041]    Approach 2: QR Decomposition 
         [0042]    By using QR decomposition method, the matrix H will be decomposed into the multiplication of the matrix Q and the matrix R according to the equation (2): 
         [0000]    
       
         
           
             
               
                 
                   X 
                   = 
                   
                     
                       [ 
                       
                         
                           
                             
                               x 
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                      
                     
                       
 
                     
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                           + 
                           N 
                         
                          
                         
                           
 
                         
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                         = 
                         
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                           + 
                           N 
                           + 
                           QRS 
                           + 
                           N 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   2 
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         [0043]    The matrix Q is a unitary matrix as shown in the following equation: 
         [0000]    
       
         
           
             Q 
             = 
             
               [ 
               
                 
                   
                     
                       
                         h 
                         1 
                       
                       
                         
                           
                             
                                
                               
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                                
                             
                             2 
                           
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                                   2 
                                 
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                         h 
                         2 
                       
                       
                         
                           
                             
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                           + 
                           
                             
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                                 2 
                               
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                                   2 
                                 
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               ] 
             
           
         
       
     
         [0044]    The matrix R is an upper triangular matrix as shown in the following equation: 
         [0000]    
       
         
           
             R 
             = 
             
               [ 
               
                 
                   
                     
                       
                         
                           
                              
                             
                               h 
                               1 
                             
                              
                           
                           2 
                         
                         + 
                         
                           
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               ] 
             
           
         
       
     
         [0045]    After canceling the matrix Q by multiplying Q H  at the front of H, the equation (2) is modified as the following equation: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       Q 
                       H 
                     
                      
                     X 
                   
                   = 
                     
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                     RS 
                     + 
                     
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                       ′ 
                     
                   
                 
               
             
             
               
                 
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                                        
                                       
                                         h 
                                         2 
                                       
                                        
                                     
                                     2 
                                   
                                 
                               
                             
                           
                         
                       
                     
                     ] 
                   
                 
               
             
             
               
                 
                     
                    
                   
                     
                       [ 
                       
                           
                       
                        
                       
                         
                           
                             
                               S 
                               1 
                             
                           
                         
                         
                           
                             
                               S 
                               2 
                             
                           
                         
                       
                       ] 
                     
                     + 
                     
                         
                     
                      
                     
                       N 
                       ′ 
                     
                   
                 
               
             
           
         
       
     
         [0046]    wherein N′ is the result of N multiplied by Q H . 
         [0047]    According to the property of the upper triangular matrix R, there will be no interference term for S 2  (i.e. the second symbol S 202 ). Thus S 2  (i.e. the second symbol S 202 ) can be estimated, and the gain of S 2  equals to 
         [0000]    
       
         
           
             
               
                 
                   
                      
                     
                       h 
                       1 
                     
                      
                   
                   2 
                 
                 + 
                 
                   
                      
                     
                       h 
                       2 
                     
                      
                   
                   2 
                 
                 - 
                 
                   
                     
                       ( 
                       
                         
                           
                              
                             
                               h 
                               1 
                             
                              
                           
                           2 
                         
                         - 
                         
                           
                              
                             
                               h 
                               2 
                             
                              
                           
                           2 
                         
                       
                       ) 
                     
                     2 
                   
                   
                     
                       
                          
                         
                           h 
                           1 
                         
                          
                       
                       2 
                     
                     + 
                     
                       
                          
                         
                           h 
                           2 
                         
                          
                       
                       2 
                     
                   
                 
               
             
             . 
           
         
       
     
         [0048]    After S 2  (i.e. the second symbol S 202 ) is decoded, S 1  (i.e. the first symbol S 201 ) can be also estimated by applying the same QR decomposition approach. Those skilled in the art can understand the corresponding approach of decoding S 1  (i.e. the first symbol S 201 ) by the explanation of the above description, and thus more detailed explanation is unnecessary. 
         [0049]    Approach 3: Sphere Decoding Algorithm 
         [0050]    Sphere detection algorithm (SDA) is another common approach, which is near-optimal by adjusting radius of search, to estimate S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) with lower calculation complexity. The SDA is also a well-known method used in the prior art, so the details will not be mentioned here. 
         [0051]    Approach 4: Maximum Gain Combining (MGC) 
         [0052]    Maximum Gain Combining approach can also provide near optimum performance with considerably lower complexity by linearly combining S 1 , S 2 , and S 1 -S 2  (i.e. the first, the second and the third symbols S 201 , S 202 , S 221 ) with different coefficients in order to achieve a maximum gain. Essentially, it is a generalized approach since linear decoding methods mentioned above could be modified to special cases of linear combining. Please refer to  FIG. 2C , which is a schematic diagram illustrated a model of the maximum ratio combining decoding scheme of the first embodiment. 
         [0053]    For the MGC approach, the processing module  232  will estimate one symbol and eliminate other symbols. For example, the processing module  232  may estimate S 1  (i.e. the first symbol S 201 ) or S 2  (i.e. the second symbol S 202 ) by maxing a gain of S 1  (i.e. the first symbol S 201 ) and eliminating S 2  (i.e. the second symbol S 202 ) of the first burst  20 , and estimate S 2  (i.e. the second symbol S 202 ) according to S 1  (i.e. the first symbol S 201 ). For illustration of the decoding procedure, the following description takes S 1  (i.e. the first symbol  201 ) as an example. As shown in  FIG. 2C , the S 1 , S 2 , S 1 -S 2  (i.e. the first, second, third symbols S 201 , S 202 , S 221 ) are linearly combined by different coefficients α 1 , α 2 , and α 3 . Note that the coefficients can be real number or complex numbers. After the combination, the output of the combiner  25  is (α 1 h 1 +α 3 h 2 )S 1 +(α 1 h 1 −α 3 h 2 )S 2 +√{square root over ((|α 1 | 2 +|α 2 | 2 +|α 3 | 2 ))}n′, wherein n′ has a normal distribution whose mean is zero and variance is σ. 
         [0054]    More specifically, the SINR of S 1  (denoted as γ S     1   ) can be represented as the equation (3): 
         [0000]    
       
         
           
             
               
                 
                   
                     γ 
                     
                       s 
                       1 
                     
                   
                   = 
                   
                     
                       
                         
                           
                             ( 
                             
                               
                                 
                                   α 
                                   1 
                                 
                                  
                                 
                                   h 
                                   1 
                                 
                               
                               + 
                               
                                 
                                   α 
                                   3 
                                 
                                  
                                 
                                   h 
                                   2 
                                 
                               
                             
                             ) 
                           
                           2 
                         
                          
                         
                           A 
                           2 
                         
                       
                       
                         
                           
                             
                               ( 
                               
                                 
                                   
                                     α 
                                     2 
                                   
                                    
                                   
                                     h 
                                     1 
                                   
                                 
                                 - 
                                 
                                   
                                     α 
                                     3 
                                   
                                    
                                   
                                     h 
                                     2 
                                   
                                 
                               
                               ) 
                             
                             2 
                           
                            
                           
                             A 
                             2 
                           
                         
                         + 
                         
                           
                             ( 
                             
                               
                                 
                                    
                                   
                                     α 
                                     1 
                                   
                                    
                                 
                                 2 
                               
                               + 
                               
                                 
                                    
                                   
                                     α 
                                     2 
                                   
                                    
                                 
                                 2 
                               
                               + 
                               
                                 
                                    
                                   
                                     α 
                                     3 
                                   
                                    
                                 
                                 2 
                               
                             
                             ) 
                           
                            
                           σ 
                         
                       
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                     = 
                     
                       
                         
                           
                             ( 
                             
                               
                                 
                                   α 
                                   1 
                                 
                                  
                                 
                                   h 
                                   1 
                                 
                               
                               + 
                               
                                 
                                   α 
                                   3 
                                 
                                  
                                 
                                   h 
                                   2 
                                 
                               
                             
                             ) 
                           
                           2 
                         
                          
                         SNR 
                       
                       
                         
                           
                             
                               ( 
                               
                                 
                                   
                                     α 
                                     2 
                                   
                                    
                                   
                                     h 
                                     1 
                                   
                                 
                                 - 
                                 
                                   
                                     α 
                                     3 
                                   
                                    
                                   
                                     h 
                                     2 
                                   
                                 
                               
                               ) 
                             
                             2 
                           
                            
                           SNR 
                         
                         + 
                         
                           ( 
                           
                             
                               
                                  
                                 
                                   α 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   α 
                                   2 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   α 
                                   3 
                                 
                                  
                               
                               2 
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   3 
                   ) 
                 
               
             
           
         
       
     
         [0055]    Wherein A represents the amplitude of S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) which are assumed to be the same, SNR is defined as 
         [0000]    
       
         
           
             
               
                 A 
                 2 
               
               σ 
             
             . 
           
         
       
     
         [0000]    There are two observations on the equation (3) to find the set of coefficients α 1 , α 2  and α 3 . 
         [0056]    Observation 1: 
         [0000]    
       
         
           
             
               α 
               1 
             
             = 
             
               
                 
                   
                     h 
                     1 
                     * 
                   
                    
                   
                     α 
                     3 
                   
                 
                 
                   h 
                   2 
                   * 
                 
               
               . 
             
           
         
       
     
         [0000]    This relationship can be proved by applying Cauchy-Schwartz inequality to the numerator of the equation (3). The Cauchy-Schwartz inequality is a well-known to the people skilled in this art, the details will not be mentioned here. 
         [0057]    Observation 2: 
         [0058]    To make the term (α 2 h 1 −α 3 h 2 ) 2  in denominator of the equation (3) as small as possible, such as ∠α 2 h 1 =∠α 3 h 2 →∠α 2 =∠α 3 +∠h 2 −∠h 1 . Thus the relationship of phases of α 2  and α 3  are known, then |α 2 | can further represents as x|α 3 |, where xε         . 
         [0059]    From observation 1 and 2, the equation (3) can be modified as the equation (4): 
         [0000]    
       
         
           
             
               
                 
                   
                     γ 
                     
                       s 
                       1 
                     
                   
                   = 
                   
                     
                       
                         
                           
                             ( 
                             
                               
                                 
                                   
                                     
                                       h 
                                       1 
                                       * 
                                     
                                      
                                     
                                       h 
                                       1 
                                     
                                   
                                   
                                     h 
                                     2 
                                     * 
                                   
                                 
                                  
                                 
                                   α 
                                   3 
                                 
                               
                               + 
                               
                                 
                                   h 
                                   2 
                                 
                                  
                                 
                                   α 
                                   3 
                                 
                               
                             
                             ) 
                           
                           2 
                         
                          
                         SNR 
                       
                       
                         
                           
                             
                               
                                 
                                   ( 
                                   
                                     
                                       
                                         x 
                                         2 
                                       
                                        
                                       
                                         
                                            
                                           
                                             h 
                                             1 
                                           
                                            
                                         
                                         2 
                                       
                                     
                                     - 
                                     
                                       2 
                                        
                                       x 
                                        
                                       
                                          
                                         
                                           h 
                                           1 
                                         
                                          
                                       
                                        
                                       
                                          
                                         
                                           h 
                                           2 
                                         
                                          
                                       
                                     
                                     + 
                                     
                                       
                                          
                                         
                                           h 
                                           2 
                                         
                                          
                                       
                                       2 
                                     
                                   
                                   ) 
                                 
                                  
                                 SNR 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               ( 
                               
                                 
                                   
                                      
                                     
                                       α 
                                       1 
                                     
                                      
                                   
                                   2 
                                 
                                 + 
                                 
                                   
                                      
                                     
                                       α 
                                       2 
                                     
                                      
                                   
                                   2 
                                 
                                 + 
                                 
                                   
                                      
                                     
                                       α 
                                       3 
                                     
                                      
                                   
                                   2 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                     = 
                     
                       
                         
                           
                             
                               ( 
                               
                                 
                                   
                                      
                                     
                                       h 
                                       1 
                                     
                                      
                                   
                                   2 
                                 
                                 + 
                                 
                                   
                                      
                                     
                                       h 
                                       2 
                                     
                                      
                                   
                                   2 
                                 
                               
                               ) 
                             
                             2 
                           
                           
                             
                                
                               
                                 h 
                                 2 
                               
                                
                             
                             2 
                           
                         
                          
                         
                           
                              
                             
                               α 
                               3 
                             
                              
                           
                           2 
                         
                          
                         SNR 
                       
                       
                         
                           
                             
                               
                                 
                                   ( 
                                   
                                     
                                       
                                         x 
                                         2 
                                       
                                        
                                       
                                         
                                            
                                           
                                             h 
                                             1 
                                           
                                            
                                         
                                         2 
                                       
                                     
                                     - 
                                     
                                       2 
                                        
                                       x 
                                        
                                       
                                          
                                         
                                           h 
                                           1 
                                         
                                          
                                       
                                        
                                       
                                          
                                         
                                           h 
                                           2 
                                         
                                          
                                       
                                     
                                     + 
                                     
                                       
                                          
                                         
                                           h 
                                           2 
                                         
                                          
                                       
                                       2 
                                     
                                   
                                   ) 
                                 
                                  
                                 
                                   
                                      
                                     
                                       α 
                                       3 
                                     
                                      
                                   
                                   2 
                                 
                                  
                                 SNR 
                               
                               + 
                             
                           
                         
                         
                           
                             
                               ( 
                               
                                 
                                   
                                      
                                     
                                       α 
                                       1 
                                     
                                      
                                   
                                   2 
                                 
                                 + 
                                 
                                   
                                      
                                     
                                       α 
                                       2 
                                     
                                      
                                   
                                   2 
                                 
                                 + 
                                 
                                   
                                      
                                     
                                       α 
                                       3 
                                     
                                      
                                   
                                   2 
                                 
                               
                               ) 
                             
                           
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   4 
                   ) 
                 
               
             
           
         
       
     
         [0060]    Wherein |α 1 | 2 +|α 2 | 2 +|α 3 | 2 =K, by knowing 
         [0000]    
       
         
           
             
               
                 
                    
                   
                     α 
                     1 
                   
                    
                 
                 2 
               
               = 
               
                 
                   
                     
                       
                          
                         
                           h 
                           1 
                         
                          
                       
                       2 
                     
                     
                       
                          
                         
                           h 
                           2 
                         
                          
                       
                       2 
                     
                   
                    
                   
                     
                        
                       
                         α 
                         3 
                       
                        
                     
                     2 
                   
                    
                   
                       
                   
                    
                   and 
                    
                   
                       
                   
                    
                   
                     
                        
                       
                         α 
                         2 
                       
                        
                     
                     2 
                   
                 
                 = 
                 
                   
                     x 
                     2 
                   
                    
                   
                     
                        
                       
                         α 
                         3 
                       
                        
                     
                     2 
                   
                 
               
             
             , 
           
         
       
     
         [0000]    |α 3 | 2  represents as 
         [0000]    
       
         
           
             
               
                 
                   
                      
                     
                       h 
                       2 
                     
                      
                   
                   2 
                 
                  
                 K 
               
               
                 
                   
                      
                     
                       h 
                       1 
                     
                      
                   
                   2 
                 
                 + 
                 
                   
                     ( 
                     
                       1 
                       + 
                       
                         x 
                         2 
                       
                     
                     ) 
                   
                    
                   
                     
                        
                       
                         h 
                         2 
                       
                        
                     
                     2 
                   
                 
               
             
             . 
           
         
       
     
         [0061]    According to the above terms, the equation (4) can be obtained as the following equation: 
         [0000]    
       
         
           
             
               
                 
                   
                     γ 
                     
                       s 
                       1 
                     
                   
                   = 
                     
                    
                   
                     
                       
                         
                           
                             ( 
                             
                               
                                 
                                    
                                   
                                     h 
                                     1 
                                   
                                    
                                 
                                 2 
                               
                               + 
                               
                                 
                                    
                                   
                                     h 
                                     2 
                                   
                                    
                                 
                                 2 
                               
                             
                             ) 
                           
                           2 
                         
                         
                           
                              
                             
                               h 
                               2 
                             
                              
                           
                           2 
                         
                       
                       * 
                       
                         
                           
                             
                                
                               
                                 h 
                                 2 
                               
                                
                             
                             2 
                           
                            
                           K 
                         
                         
                           
                             
                                
                               
                                 h 
                                 1 
                               
                                
                             
                             2 
                           
                           + 
                           
                             
                               ( 
                               
                                 1 
                                 + 
                                 
                                   x 
                                   2 
                                 
                               
                               ) 
                             
                              
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                         
                       
                        
                       SNR 
                     
                     
                       
                         
                           ( 
                           
                             
                               
                                 x 
                                 2 
                               
                                
                               
                                 
                                    
                                   
                                     h 
                                     1 
                                   
                                    
                                 
                                 2 
                               
                             
                             - 
                             
                               2 
                                
                               x 
                                
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                                
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                           ) 
                         
                          
                         
                           
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                              
                             K 
                           
                           
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                 ( 
                                 
                                   1 
                                   + 
                                   
                                     x 
                                     2 
                                   
                                 
                                 ) 
                               
                                
                               
                                 
                                    
                                   
                                     h 
                                     2 
                                   
                                    
                                 
                                 2 
                               
                             
                           
                         
                          
                         SNR 
                       
                       + 
                       K 
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           ( 
                           
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                           ) 
                         
                         2 
                       
                        
                       SNR 
                       * 
                       K 
                     
                     
                       
                         
                           ( 
                           
                             
                               
                                 x 
                                 2 
                               
                                
                               
                                 
                                    
                                   
                                     h 
                                     1 
                                   
                                    
                                 
                                 2 
                               
                             
                             - 
                             
                               2 
                                
                               x 
                                
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                                
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                           ) 
                         
                          
                         
                           
                              
                             
                               h 
                               2 
                             
                              
                           
                           2 
                         
                          
                         K 
                         * 
                         SNR 
                       
                       + 
                       
                         K 
                          
                         
                           ( 
                           
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                 ( 
                                 
                                   1 
                                   + 
                                   
                                     x 
                                     2 
                                   
                                 
                                 ) 
                               
                                
                               
                                 
                                    
                                   
                                     h 
                                     2 
                                   
                                    
                                 
                                 2 
                               
                             
                           
                           ) 
                         
                       
                     
                   
                 
               
             
             
               
                 
                   = 
                     
                    
                   
                     
                       
                         
                           ( 
                           
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                           ) 
                         
                         2 
                       
                       * 
                       SNR 
                     
                     
                       
                         
                           
                             
                               
                                 ( 
                                 
                                   
                                     SNR 
                                      
                                     
                                       
                                          
                                         
                                           h 
                                           1 
                                         
                                          
                                       
                                       2 
                                     
                                      
                                     
                                       
                                          
                                         
                                           h 
                                           2 
                                         
                                          
                                       
                                       2 
                                     
                                   
                                   + 
                                   
                                     
                                        
                                       
                                         h 
                                         2 
                                       
                                        
                                     
                                     2 
                                   
                                 
                                 ) 
                               
                                
                               
                                 
                                   { 
                                   
                                     x 
                                     - 
                                     
                                       
                                         
                                            
                                           
                                             h 
                                             1 
                                           
                                            
                                         
                                          
                                         
                                           
                                              
                                             
                                               h 
                                               2 
                                             
                                              
                                           
                                           3 
                                         
                                          
                                         SNR 
                                       
                                       
                                         
                                           
                                             
                                                
                                               
                                                 h 
                                                 1 
                                               
                                                
                                             
                                             2 
                                           
                                            
                                           
                                             
                                                
                                               
                                                 h 
                                                 2 
                                               
                                                
                                             
                                             2 
                                           
                                            
                                           SNR 
                                         
                                         + 
                                         
                                           
                                              
                                             
                                               h 
                                               2 
                                             
                                              
                                           
                                           2 
                                         
                                       
                                     
                                   
                                   } 
                                 
                                 2 
                               
                             
                             - 
                           
                         
                       
                       
                         
                           
                             
                               
                                 
                                   
                                      
                                     
                                       h 
                                       1 
                                     
                                      
                                   
                                   2 
                                 
                                  
                                 
                                   
                                      
                                     
                                       h 
                                       2 
                                     
                                      
                                   
                                   6 
                                 
                                  
                                 
                                   SNR 
                                   2 
                                 
                               
                               
                                 
                                   
                                     
                                        
                                       
                                         h 
                                         1 
                                       
                                        
                                     
                                     2 
                                   
                                    
                                   
                                     
                                        
                                       
                                         h 
                                         2 
                                       
                                        
                                     
                                     2 
                                   
                                    
                                   SNR 
                                 
                                 + 
                                 
                                   
                                      
                                     
                                       h 
                                       2 
                                     
                                      
                                   
                                   2 
                                 
                               
                             
                             + 
                             
                               
                                 
                                   
                                     
                                        
                                       
                                         h 
                                         1 
                                       
                                        
                                     
                                     2 
                                   
                                    
                                   
                                     
                                        
                                       
                                         h 
                                         2 
                                       
                                        
                                     
                                     4 
                                   
                                 
                                 
                                   
                                      
                                     
                                       h 
                                       1 
                                     
                                      
                                   
                                   2 
                                 
                               
                                
                               SNR 
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                         
                       
                     
                   
                 
               
             
           
         
       
     
         [0062]    The left most term in the dominator of the above equation should be zero for obtaining a maximum SINR. 
         [0063]    In other words, 
         [0000]    
       
         
           
             
               x 
               = 
               
                 
                   
                      
                     
                       h 
                       1 
                     
                      
                   
                    
                   
                     
                        
                       
                         h 
                         2 
                       
                        
                     
                     3 
                   
                    
                   SNR 
                 
                 
                   
                     
                       
                          
                         
                           h 
                           1 
                         
                          
                       
                       2 
                     
                      
                     
                       
                          
                         
                           h 
                           2 
                         
                          
                       
                       2 
                     
                      
                     SNR 
                   
                   + 
                   
                     
                        
                       
                         h 
                         2 
                       
                        
                     
                     2 
                   
                 
               
             
             , 
             
                 
             
              
             for 
              
             
                 
             
             , 
             
                 
             
              
             
               SNR 
               &gt;&gt; 
               
                 
                    
                   
                     h 
                     2 
                   
                    
                 
                 2 
               
             
             , 
             
                 
             
              
             
               x 
               ≈ 
               
                 
                    
                   
                     h 
                     2 
                   
                    
                 
                 
                    
                   
                     h 
                     1 
                   
                    
                 
               
             
             , 
           
         
       
     
         [0000]    and γ s     1   ≈(|h 1 | 2 +h 1 | 2 )SNR, which has the same gain as CC. Consequently, the coefficient α 1 , α 2 , α 3 , which maximize the SINR of S 1  (i.e. the first symbol S 201 ) should have the following relationship: 
         [0000]    
       
         
           
             
               
                 α 
                 1 
               
               = 
               
                 
                   
                     h 
                     1 
                     * 
                   
                   
                     h 
                     2 
                     * 
                   
                 
                  
                 
                   α 
                   3 
                 
               
             
             , 
             
                 
             
              
             
               
                 x 
                 * 
                 
                   α 
                   3 
                 
               
               ≈ 
               
                 
                   
                      
                     
                       h 
                       2 
                     
                      
                   
                   
                      
                     
                       h 
                       1 
                     
                      
                   
                 
                  
                 
                   α 
                   3 
                 
                 * 
                 
                    
                   
                     j 
                      
                     
                       ( 
                       
                         
                           ∠ 
                            
                           
                               
                           
                            
                           
                             h 
                             2 
                           
                         
                         - 
                         
                           ∠ 
                            
                           
                               
                           
                            
                           
                             h 
                             1 
                           
                         
                       
                       ) 
                     
                   
                 
               
             
           
         
       
     
         [0064]    |α 1 | 2 +|α 2 | 2 +|α 3 | 2 =K, wherein K is an arbitrary positive value. 
         [0065]    Those skilled in the art can understand the corresponding approach of decoding S 2  (i.e. the second symbol S 202 ) by the explanation of the above description, and thus more detailed description is unnecessary. 
         [0066]    In brief, the MGC will set α 1 +α 2 +α 3 =K, wherein the K is a fixed value, and eliminate other symbols, e.g. set |α 1 ∥h 1 −|α 3 ∥h 3 |=0 to eliminate S 2  (i.e. the second symbol S 202 ), so the processing module  232  can estimate S 1  (i.e. the first symbol S 201 ) according to the two set conditions to max the gain of S 1  (i.e. the first symbol S 201 ), i.e. |α 1 ∥h 1 +|α 3 ∥h 2 |. 
         [0067]    Approach 5: Cross Feedback 
         [0068]    In this approach, S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) may be first decoded by some algorithms (e.g. one of the above approaches 1˜4), then the decoding result of S 1  and S 2  will be feedback to other symbols. According to the feedback of other symbols, a symbol could have another decoding result by using decision feedback (a.k.a. interference cancellation). Thus each symbol has a pair of decoding results. The processing module  232  may adopt a decision criteria to make a final decision for the values of S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ). 
         [0069]    For example, if the processing module  232  estimates S 1  (i.e. the first symbol S 201 ) to be X by MGC approach, then the processing module  232  can obtain S 2  (i.e. the second symbol S 202 ) to be Y according to the equation (1) and X (i.e. the estimated S 1 ). Next, the processing module  232  estimates S 2  (i.e. the second symbol S 202 ) to be X′ by MGC approach, then the processing module  232  can obtain S 1  (i.e. the first symbol S 201 ) to be Y′ according to the equation (1) and Y′ (i.e. the estimated S 2 ). The processing module  232  will obtain a pair estimated result of the S 1 , i.e. (X, X′), and a pair estimated result of S 2 , i.e. (Y, Y′). 
         [0070]    The processing module  232  will make a final decision for the values of S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) according to the two pair (X, X′) and (Y, Y′) based on the decision criteria. The decision criteria can be determined according to wireless channel status, used modulation, empirical adjustment, etc. 
         [0071]    Preferably, if one of the two pairs is matched, but the other pair is not matched, the processing module  232  will adopt the pair which the result is matched. Oppositely, when the pair is not matched, the processing module  232  will adopt the result of decision feedback. For example, if X=X′ and Y≠Y′, the processing module  232  will adopt S 1  (i.e. the first symbol S 201 ) to be X and adopt S 2  (i.e. the second symbol S 202 ) to be Y′. 
         [0072]    If the two pairs are not matched, the processing module  232  will adopt the pair with a larger gain, and for the other pair, adopt the result of decision feedback. For example, if X≠X′, Y≠Y′ and a gain of (X, X′) is larger than a gain of (Y, Y′), the processing module  232  adopt S 1  (i.e. the first symbol S 201 ) to be X and adopt S 2  (i.e. the second symbol S 202 ) to be Y′. 
         [0073]    If both of the two pairs are matched, the processing module  232  will adopt the estimated results. For example, if X=X′ and Y=Y′, the processing module  232  adopt S 1  (i.e. the first symbol S 201 ) to be X and adopt S 2  (i.e. the second symbol S 202 ) to be Y. 
         [0074]    It should be noted that above decision criteria is only for illustration, not to limit the present invention, people skilled in this art can design other decision criteria depend on the practical requirements. 
         [0075]    The estimated approaches of the first embodiment are described as above. The processing module  232  will further determines whether the estimated S 1  (i.e. the first symbol S 201 ) and S 2  the second symbol is correct or not according to the CRC C 20 . If the estimated symbols are correct, the MS  23  will feedback a acknowledgement to the BS  21 , otherwise, the MS  23  will feedback a negative acknowledge to the BS  21  and continues the above steps. 
         [0076]    The following descriptions are to illustration other embodiments that can also be applied the above estimated approach. 
         [0077]    In the first embodiments, the processing module  212  generates the third symbol S 221  by performing a subtraction, for other embodiments, the processing module  212  can shift a first predetermined phase for S 1  (i.e. the first symbol S 201 ), shift a second predetermined phase for S 2  (i.e. the second symbol S 202 ), and then generates the third symbol S 221  by proceeding the linear combination of the shifted S 1  (i.e. the shifted first symbol) and the shifted S 2  (i.e. the shifted second symbol). 
         [0078]    For example, the processing module  212  shifts S 1  (i.e. the first symbol S 201 ) and S 2  (i.e. the second symbol S 202 ) a phase θ in opposite directions, wherein the shifted phase θ is a predetermined value and known by both the BS  21  and the MS  23 . Then, the processing module  212  generates the third symbol S 203  by linear combination function of the shifted S 1  (i.e. the shifted first symbol) and shifted S 2  (i.e. the shifted second symbol), e.g. e −jθ S 1 −e jθ S 2 . It should be noted that the first embodiment can be regarded a special case of the above method with the shifted phase θ equal to zero. 
         [0079]    Accordingly, the equation (2) may be modified as the following equation: 
         [0000]    
       
         
           
             
               
                 
                   
                     X 
                     ′ 
                   
                   = 
                   
                     
                       
                         
                           [ 
                           
                             
                               
                                 
                                   h 
                                   1 
                                 
                               
                               
                                 
                                   h 
                                   1 
                                 
                               
                             
                             
                               
                                 
                                   
                                     h 
                                     2 
                                   
                                    
                                   
                                      
                                     
                                       - 
                                       jθ 
                                     
                                   
                                 
                               
                               
                                 
                                   
                                     - 
                                     
                                        
                                       jθ 
                                     
                                   
                                    
                                   
                                     h 
                                     2 
                                   
                                 
                               
                             
                           
                           ] 
                         
                          
                         
                           [ 
                           
                             
                               
                                 
                                   S 
                                   1 
                                 
                               
                             
                             
                               
                                 
                                   S 
                                   2 
                                 
                               
                             
                           
                           ] 
                         
                       
                       + 
                       N 
                     
                      
                     
                       
 
                     
                      
                     
                         
                     
                     = 
                     
                       
                         
                           
                             [ 
                             
                               
                                 
                                   
                                     
                                        
                                       jθ 
                                     
                                      
                                     
                                       h 
                                       1 
                                     
                                   
                                 
                                 
                                   
                                     h 
                                     1 
                                   
                                 
                               
                               
                                 
                                   
                                     h 
                                     2 
                                   
                                 
                                 
                                   
                                     
                                       - 
                                       
                                          
                                         jθ 
                                       
                                     
                                      
                                     
                                       h 
                                       2 
                                     
                                   
                                 
                               
                             
                             ] 
                           
                            
                           
                             [ 
                             
                               
                                 
                                   
                                     
                                        
                                       
                                         - 
                                         jθ 
                                       
                                     
                                      
                                     
                                       S 
                                       1 
                                     
                                   
                                 
                               
                               
                                 
                                   
                                     S 
                                     2 
                                   
                                 
                               
                             
                             ] 
                           
                         
                         + 
                         N 
                       
                        
                       
                         
 
                       
                        
                       
                           
                       
                       = 
                       
                         
                           HS 
                           + 
                           N 
                         
                          
                         
                           
 
                         
                          
                         
                             
                         
                         = 
                         
                           QRS 
                           + 
                           N 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   5 
                   ) 
                 
               
             
           
         
       
     
         [0080]    The matrix Q is a unitary matrix as shown in the following equation: 
         [0000]    
       
         
           
             Q 
             = 
             
               [ 
               
                 
                   
                     
                       
                         
                           h 
                           1 
                         
                         
                           
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                         
                       
                        
                       
                           
                       
                        
                       … 
                     
                   
                 
                 
                   
                     
                       
                         
                           
                             h 
                             2 
                           
                            
                           
                              
                             
                               - 
                               jθ 
                             
                           
                         
                         
                           
                             
                               
                                  
                                 
                                   h 
                                   1 
                                 
                                  
                               
                               2 
                             
                             + 
                             
                               
                                  
                                 
                                   h 
                                   2 
                                 
                                  
                               
                               2 
                             
                           
                         
                       
                        
                       
                           
                       
                        
                       … 
                     
                   
                 
               
               ] 
             
           
         
       
     
         [0081]    The matrix R is an upper triangular matrix as shown in the following equation: 
         [0000]    
       
         
           
             R 
             = 
             
               [ 
               
                 
                   
                     
                       
                         
                            
                           
                             h 
                             1 
                           
                            
                         
                         2 
                       
                       + 
                       
                         
                            
                           
                             h 
                             2 
                           
                            
                         
                         2 
                       
                     
                   
                   
                     … 
                   
                 
                 
                   
                     0 
                   
                   
                     g 
                   
                 
               
               ] 
             
           
         
       
     
         [0082]    People skilled in this art may rapidly estimate S 1  (i.e. the first symbol S 201 ) and S 2  (the second symbol S 202 ) according to equation (5) and above said approaches, the details will not be further mentioned here. 
         [0083]    It should be mentioned that the gain of the HARQ retransmission scheme with shifting phase of symbols before performing the linear combination is the same as that of the HARQ retransmission scheme without shifting phase of symbols. Beyond that, the HARQ retransmission scheme with shifting phase of symbols can also mitigate the self-cancellation problem. 
         [0084]    Table 1 is another example for the pre-shifting phase of symbols if there is more than one retransmission. 
         [0000]    
       
         
               
             
               
               
               
               
               
             
               
               
               
               
               
             
           
               
                 TABLE 1 
               
             
             
               
                   
               
               
                 the retransmission pattern 
               
             
          
           
               
                   
                 Symbol 1 
                 Symbol 2 
                 Symbol 3 
                 Symbol 4 
               
               
                   
                   
               
             
          
           
               
                 Original sub-burst 
                 S 1   
                 S 2   
                 S 3   
                 S 4   
               
               
                 Odd Retransmitted 
                 e −jθ  S 1  − e jθ  S 2   
                 e −jθ  S 3  − e jθ  S 4   
                 N/A 
                 N/A 
               
               
                 sub-burst 
               
               
                 Even Retransmitted 
                 e jθ  S 1  + e −jθ  S 2   
                 e jθ  S 3  + e −jθ  S 4   
                 N/A 
                 N/A 
               
               
                 sub-burst 
               
               
                   
               
             
          
         
       
     
         [0085]    In still another embodiments, the at least one MS may comprise a plurality of MSs (e.g. MS 1  and MS 2 ). If the BS is set to be a broadcast mode, MS 1  and MS 2  are able to receive the at least one first burst meant to transmit to each other. In other words, if the BS transmits a first burst to one of the MSs, the other MS can also receive it. Similarly, if the BS received at least one negative acknowledgement from the MSs, the BS will also generate at least one second burst according to the at least one first burst. 
         [0086]    For example, S 1   1  and S 2   1  represent two symbols of the first burst meant to send to the MS 1 , and S 1   2  and S 2   2  represent two symbols of the first burst meant to send to the MS 2 . If the MS 1  is able to decode S 1   2  and S 2   2  successfully but fails to decode S 1   1  and S 2   1 , and the MS 2  is able to decode S 1   1  and S 2   1  successfully but fails to decode S 1   2  and S 2   2 , the BS will receive two negative acknowledgements form MS 1  and MS 2 . 
         [0087]    After receiving the negative acknowledgements from MS 1  and MS 2 , similarly, the BS may generate at least one second burst according to S 1   2 , S 2   2 , S 1   1  and S 2   1 , e.g. the second burst having symbols of S 1   1 +S 1   2  and S 2   1 +S 2   2 , or the second burst having symbols of e −jθ S 1   1 −e jθ S 1   2  and e −jθ S 2   1 −e jθ S 2   2 . The BS will transmit the second burst to the MS 1  and the MS 2 . 
         [0088]    Next, the MS 1  and MS 2  may estimate their symbols (i.e. the MS 1  estimates S 1   1  and S 2   1  and the MS 2  estimates S 1   2  and S 2   2 ) according to the second burst and the above approaches individually. Then, for the MS 1 , since it has decoded S 1   2  and S 2   2  successfully, thus it can simply cancel the interference term of the second burst to obtain S 1   1  and S 2   1 , for the MS 2 , since it has decoded S 1   1  and S 2   1  successfully, thus it can simply cancel the interference term of the second burst to obtain S 1   2  and S 2   2 . Those skilled in the art can understand the corresponding approach of estimation of the symbols S 1   2 , S 2   2 , S 1   1  and S 2   1 . by the explanation of the above description, and thus no necessary detail is given. 
         [0089]    Table 2 is another example of the retransmission pattern for the MS 1  and the MS 2  if there is more than one retransmission. 
         [0000]    
       
         
               
             
               
               
               
               
               
             
               
               
               
               
               
             
           
               
                 TABLE 2 
               
             
             
               
                   
               
               
                 the retransmission pattern 
               
             
          
           
               
                   
                 Symbol 1 
                 Symbol 2 
                 Symbol 3 
                 Symbol 4 
               
               
                   
                   
               
             
          
           
               
                 Original sub-burst 
                 S 1   1  (S 1   2 ) 
                 S 1   2  (S 2   2 ) 
                 S 3   1  (S 3   2 ) 
                 S 4   1  (S 4   2 ) 
               
               
                 Odd Retransmitted 
                 e −jθ  S 1   1  − e jθ  S 1   2   
                 e jθ  S 2   1  − e jθ  S 2   2   
                 e −jθ  S 3   1  − e jθ  S 3   2   
                 e −jθ  S 4   1  − e jθ  S 4   2   
               
               
                 sub-burst 
               
               
                 Even Retransmitted 
                 e jθ S 1   1  + e −jθ  S 1   2   
                 e jθ  S 1   1  + e −jθ  S 1   2   
                 e jθ  S 1   1  + e −jθ  S 1   2   
                 e jθ  S 1   1  + e −jθ  S 1   2   
               
               
                 sub-burst 
               
               
                   
               
             
          
         
       
     
         [0090]    It should be noted that the first embodiment of the present invention can be adopted in both uplink and downlink. However, to be simplified, the first embodiment only illustrates the downlink case, since those skilled in this art can understand the corresponding operation of the present invention for the uplink case after the above description. 
         [0091]    The second embodiment in accordance with the present invention shown in  FIG. 3A  is a transmission method of a wireless network system for HARQ, wherein the wireless network comprises at least one MS, e.g. the MS  23  described in the first embodiment. More specifically, the transmission method of the second embodiment can be implemented by a computer program product. The computer program product can be stored in a tangible machine-readable medium, such as a floppy disk, a hard disk, an optical disc, a flash disk, a tape, a database accessible from a network or any other storage media with the same functionality that can be easily thought by those skilled in the art. 
         [0092]    Initially, in step  301 , at least one first burst having a first symbol and a second symbol is transmitted to the at least one MS. Then, it is determined whether at least one negative acknowledgement is received from the at least one MS via step  302 . If there is no negative acknowledgement received in step  302 , the transmission method will go back to step  301  to transmit another first burst. 
         [0093]    If there is a negative acknowledgement received in step  302 , the transmission method proceeds to step  303 , a third symbol is generated by proceeding a liner combination according to the first symbol and the second symbol of the at least one first burst. 
         [0094]    The third symbol can be generated by performing a subtraction between the first symbol and the second symbol of the at least one first burst. Or, the transmission method shifts a first predetermined phase for the first symbol, shifts a second predetermined phase for the second symbol, and then generates the third symbol by proceeding the liner combination of the shifted first symbol and the shifted second symbol. The details about how to generate the third symbol are already described in the first embodiment, so the details will not be mentioned here. 
         [0095]    Then, at least one second burst having third symbol is transmitted to the at least one MS in step  304 . In step  305 , it is determined that whether at least one negative acknowledgement is received. If there is no negative acknowledgement received in step  305 , the transmission proceeds to step  301  to transmit another first burst. If there is at least one negative acknowledgement received in step  305 , the transmission method proceeds to step  303  to generate another third symbol according to the above descriptions. 
         [0096]    The third embodiment in accordance with the present invention shown in  FIG. 3B  is a receiving method of a wireless network system for HARQ, wherein the wireless network comprises a BS, e.g. the BS  21  described in the first embodiments. More specifically, the receiving method of the third embodiment can be implemented by a computer program product. The computer program product can be stored in a tangible machine-readable medium, such as a floppy disk, a hard disk, an optical disc, a flash disk, a tape, a database accessible from a network or any other storage media with the same functionality that can be easily thought by those skilled in the art. 
         [0097]    Initially, in step  306 , at least one first burst having a first symbol and a second symbol is received from the BS. It is determined whether the at least one first burst is incorrect in step  307 , wherein the at least one first burst further comprises a CRC and the at least one NAK is performed according to the CRC. The details of how to determine the at least one first burst is incorrect is described in the first embodiment and will not be mentioned here. 
         [0098]    If the at least one first burst is determined to be not incorrect, the receiving method will go back to step  306  to receive another first burst. If the at least one first burst is determined to be incorrect in step  307 , at least one NAK is transmitted to the BS after determining that the at least one first burst is incorrect in step  308 . Then, in step  309 , at least one second burst having a third symbol is received from the BS, wherein the third symbol is generated by proceeding a liner combination according to the first symbol and the second symbol of the at least one first burst. 
         [0099]    The first symbol and the second symbol are estimated according to the at least one first burst and the at least one second burst via step  310 . For example, the first symbol and the second symbol are estimated by performing via a QR decomposition. Or, the first symbol is estimated by maxing a gain of the first symbol and eliminating the second symbol of the at least one first burst, and a second symbol is estimated according to the first symbol. The details about how to estimate first symbol and the second symbol are described in the first embodiment and will not be mentioned here. Finally, it is determined whether the estimated first symbol and the second symbol are correct according to the CRC. If the estimated first symbol and the second symbol is correct in step  311 , the receiving method proceeds to step  306 , otherwise, the receiving method proceeds to step  308 . 
         [0100]    Accordingly, the present invention can provide a retransmission scheme of saving bandwidth resource by transmitting half symbol number than that of the prior art. The spectrum efficient and system capacity therefore will be approved with low gain loss by appropriate decoding methods. Also, the pre-code algorithm does not need high complexity and the close gain can be maintained compared with the gain of CC. 
         [0101]    The above disclosure is related to the detailed technical contents and inventive features thereof. People skilled in this field may proceed with a variety of modifications and replacements based on the disclosures and suggestions of the invention as described without departing from the characteristics thereof. Nevertheless, although such modifications and replacements are not fully disclosed in the above descriptions, they have substantially been covered within the scope of the theory and spirit of the present invention.