Abstract:
A new class of biphase complementary code sets is proposed. Individual codes in each set have peak sidelobes of unity magnitude. The individual codes could have gaps of zeros but they are interlaced together without gaps in the final scheme. A number of such codes, as introduced here, use Barker codes as building blocks with additional elements from {+1,−1,0}. The main drawback of regular complementary codes longer than 4 is that they have sidelobe magnitude greater than unity. In the presence of frequency selective fading, inexact sidelobe cancellation results in non-zero sidelobes at the output. These sidelobes are minimized by using sets with individual codes that have peak sidelobes of unity magnitude. The constituent codes are transmitted at different frequencies. They are transmitted in parallel or using an appropriate combination of frequency division multiplexing (FDM) and time division multiplexing (TDM) such that the final transmission scheme is free of gaps.

Description:
FIELD OF THE INVENTION 
       [0001]    The invention relates to radar and sonar systems, wireless communication systems and other systems using complementary codes. 
       BACKGROUND OF THE INVENTION 
       [0002]    Complementary codes or sequences were introduced by Golay for use in radar astronomy. A complementary set is a set of finite sequences whose autocorrelation functions, when added together, cancel out the sidelobes of each other. Any complementary set can be used to generate longer ones. The root sequences which are defined directly and not generated from shorter ones, are also referred to as kernels. Golay came up with kernels of binary complementary pairs of length  2 ,  10  and  26 . This is found in: M. J. E. Golay,  Complementary series , IRE Transactions on Information Theory, Vol IT-7, April 1961, pp. 82-87. 
         [0003]    Sivaswamy proposed polyphase complementary sequences and gave kernels of complementary triplets of length  2  and  3 . This work is found in: R. Sivaswamy,  Multiphase complementary codes , IEEE Transactions on Information Theory, Vol IT-24, No. 5, March 1973, pp 214-218. 
         [0004]    Additional polyphase codes, and their related kernels, based on the discrete Fourier transform (DFT) are introduced in: R. L. Frank,  Polyphase complementary codes , IEEE Transactions on Information Theory, Vol IT-26, no. 6, October 1980, pp. 641-647. 
         [0005]    Biphase and polyphase complementary codes are attractive in radar applications because they can achieve zero sidelobes. However, in the presence of frequency selective fading, unequal fading of the individual codes prevents exact cancellation of their sidelobes, thus degrading their performance. Hence, a logical prerequisite for a good complementary set is that the sidelobes of the individual codes be as small as possible. This is satisfied by the new class of Interlaced Complementary Codes (ICC) introduced next. 
         [0006]    Let us consider the class of codes that are constructed from the alphabet          ={+1, −1, 0} and have peak sidelobes of unity magnitude. We denote this class by           The constituent codes of the ICC are members of           Let {t n } be a code of length N in           Its aperiodic autocorrelation function (ACF) is given by: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       A 
                       t 
                     
                      
                     
                       ( 
                       τ 
                       ) 
                     
                   
                   = 
                   
                     
                       ∑ 
                       
                         n 
                         = 
                         0 
                       
                       
                         N 
                         - 
                         τ 
                         - 
                         1 
                       
                     
                      
                     
                       
                         t 
                         n 
                       
                        
                       
                         t 
                         
                           n 
                           + 
                           τ 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   1 
                   ) 
                 
               
             
           
         
       
     
         [0000]    where 0≦τ≦N−1.
 
The peak sidelobe constraint on the members of           specifies that:
 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       max 
                       
                         τ 
                         ≠ 
                         0 
                       
                     
                      
                     
                        
                       
                         
                           A 
                           t 
                         
                          
                         
                           ( 
                           τ 
                           ) 
                         
                       
                        
                     
                   
                   = 
                   1 
                 
               
               
                 
                   ( 
                   2 
                   ) 
                 
               
             
           
         
       
     
         [0007]    Barker codes are the only biphase codes that satisfy (2). Let the class of Barker codes be           Obviously,          ⊂          The constituent codes in the proposed ICC may contain Barker codes or Barker codes modified with additional elements from          . In this work, we present several schemes involving two or more codes from           that form complementary sets. 
         [0008]    All constituent codes are transmitted at different frequencies and those with gaps of zeros are properly interlaced to result in a final scheme free of gaps. The final scheme could be purely time division multiplexed (TDM) or allow some amount of parallel transmission depending on the structure of the constituent codes. If the purely TDM scheme can be broken down into equal length blocks in a way such that each constituent code is contained only in one particular block, then these equal length blocks can be transmitted simultaneously. This gives rise to a hybrid transmission scheme using both TDM and simultaneous transmission. The advantages of TDM are low peak to average power ratio resulting in a low probability of interception. In addition, only one transmitter with frequency hopping capabilities is required. However, using TDM alone results in longer codes that are more susceptible to multipath and eclipsing problems. The parallel transmission scheme results in shorter codes and hence is better suited to counter eclipsing and multipath problems. These come at the cost of a higher peak to average power ratio, higher detectability and increased number of transmitters or a wider band transmitter with higher power output. A hybrid scheme achieves a trade-off between these two transmission modalities. However, special attention has to be paid to carefully select the modulation schemes to optimize the parameters mentioned above. Orthogonal frequency division multiplexing (OFDM) techniques could be used with judiciously selected frequencies to enhance the performance of the overall system, in particular to reduce peak to average power ratio. An example of such a scheme is discussed in: N. Levanon,  Multifrequency complementary phase - coded radar signal , IEE Proceedings on Radar, Sonar and Navigation, Vol 147, No. 6, pp. 276-284. 
         [0009]    We introduce several examples of the proposed Interlaced Complementary Codes (ICC). As mentioned before, the constituent codes of the ICC&#39;s are constructed from           and hence may contain zeros. However, the final transmission scheme is free of zeros and contains only elements from {−1, +1}. In these examples, we use Barker codes as building blocks with additional elements from           to obtain the constituent codes belonging to           However, using Barker codes is not a necessary condition for constructing codes belonging to           and consequently, the class of ICC. Let the z-transform of a Barker code of length N be denoted by B N (z). If (k−1) zeros are placed between every two elements of the code, then the modified code is represented by B N (z k ). The ACF of B N (z) is given by: 
         [0000]        R   N ( z )= B   N ( z ) B   N ( z   −1 )  (3) 
         [0000]    from which we get: 
         [0000]        R   N ( z   k )= B   N ( z   k ) B   N ( z   −k )  (4) 
         [0010]    The following steps are involved in the construction of valid interlaced codes free of any gaps. The challenge is to have the individual codes with unity peak sidelobes fit together such that the final scheme is composed of {+1, −1} without any gaps of zeros.
       Two or more constituent codes should be chosen from           such that their sidelobes cancel out when the ACF&#39;s are added together.   Constituent codes with gaps are interlaced with each other such that there are no gaps in the final transmission scheme. The constituent codes may have to be delayed with respect to each other so that they fit together without any gaps. Correspondingly, appropriate delays would also be needed at the receiver end for each constituent code, so that the ACF&#39;s (or the output of the matched filters) are aligned properly before addition.   Each constituent code is transmitted at a different frequency. The number of frequencies used is therefore equal to the number of constituent codes.       
 
     
    
     
       BRIEF DESCRIPTION OF THE DRAWINGS 
         [0014]      FIG. 1  is a diagram illustrating the autocorrelation of length  13  Barker code. 
           [0015]      FIG. 2  is a diagram illustrating the autocorrelation of length  11  Barker code. 
           [0016]      FIG. 3  is a diagram illustrating the autocorrelation of the modified length  2  Barker code. 
           [0017]      FIG. 4  is a diagram illustrating the autocorrelation of the modified length  7  Barker code. 
           [0018]      FIG. 5  is a diagram illustrating the autocorrelation of the modified length  5  Barker code. 
       
    
    
     DETAILED DESCRIPTION OF THE INVENTION 
       [0019]    ICC set I Involving B 13 (z), B 11 (z) and B 2 (z 12 ) 
         [0020]    The autocorrelation function R 13 (z) of the Barker code of length  13  is shown in  FIG. 1 . It can be seen that on each side of the mainlobe, there are 12 sidelobes of alternating +1&#39;s and 0&#39;s. 
         [0021]    The autocorrelation function R 11  (z) for the Barker code of length  11  is shown in  FIG. 2 . R 11 (z) has 10 sidelobes on each side of the mainlobe of alternating −1&#39;s and 0&#39;s. Furthermore, it can also be observed that when they are aligned, −1 sidelobes of R 11 (z) coincides exactly with +1 sidelobes of R 13 (z). Thus, when added together, the sidelobes cancel out with the exception of one sidelobe each at either end of R 13 (z). We also note that the Barker code of length  2 , {1,−1}, produces two sidelobes of height − 1 . The code of length  2 , with 11 zeros inserted between its two elements, is represented by B 2 (z 12 ). The corresponding autocorrelation function R 2 (z 12 ) can be used to cancel out the two remaining sidelobes of R 13 (z) that R 1  (z) did not cancel. R 2 (z 12 ) is shown in  FIG. 3 . 
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                 TABLE 1 
               
             
             
               
                   
               
               
                 Transmission scheme for ICC-I 
               
             
          
           
               
                   
                 t 1   
                 t 2   
                 t 3   
                 t 4   
                 t 5   
                 t 6   
                 t 7   
                 t 8   
                 t 9   
                 t 10   
                 t 11   
                 t 12   
                 t 13   
                 t 14   
                 t 15   
                 t 16   
                 t 17   
                 t 18   
                 t 19   
                 t 20   
                 t 21   
                 t 22   
                 t 23   
                 t 24   
                 t 25   
                 t 26   
               
               
                   
                   
               
             
          
           
               
                 f 1   
                 −1 
                 −1 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 −1 
                 −1 
                 1 
                 −1 
                 1 
                 −1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
               
               
                 f 2   
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
               
               
                 f 3   
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 1 
                 −1 
                 1 
                 1 
                 −1 
                 1 
                 0 
               
               
                   
               
             
          
         
       
     
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                 TABLE 2 
               
             
             
               
                   
               
               
                 Hybrid transmission scheme for ICC-I 
               
             
          
           
               
                   
                 t 1   
                 t 2   
                 t 3   
                 t 4   
                 t 5   
                 t 6   
                 t 7   
                 t 8   
                 t 9   
                 t 10   
                 t 11   
                 t 12   
                 t 13   
               
               
                   
                   
               
             
          
           
               
                 f 1   
                 −1 
                 −1 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 −1 
                 −1 
                 1 
                 −1 
                 1 
                 −1 
               
               
                 f 2   
                 1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
               
               
                 f 3   
                 0 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 1 
                 −1 
                 1 
                 1 
                 −1 
                 1 
                 0 
               
               
                   
               
             
          
         
       
     
         [0022]    It is easy to see that the transmission scheme should be such that the 11 gaps between the elements of B 2 (z 12 ) should be filled by code bits from some other code so that the final transmission scheme is free of gaps. We also observe that these gaps could be filled by placing B 11 (z) as it is in these gaps. The transmission scheme is therefore given by: 
         [0000]        T ( z )= B   13 ( z )+ z   −13   B   2 ( z   12 )+ z   −14   B   11 ( z )  (5) 
         [0000]    Thus, in this transmission scheme B 13 (z) is transmitted first. The first bit of B 2 (z) is transmitted next followed by the entire B 11 (z). Finally the remaining bit of B 2 (z 12 ) is transmitted resulting in transmission of all the codes in an interlaced fashion and free of any gaps. 
         [0023]    Table I depicts the transmission scheme graphically. The columns t 1 -t 26  depict the time slots whereas f 1 -f 3  depict the three different frequencies used to modulate B 13 (z), B 2 (z 12 ) and B 11 (z), respectively. Since no gaps (or 0&#39;s) are allowed in the transmission scheme and only one code bit can be transmitted in a given time slot, each column in the table should have only a single 1 or −1. Also, no columns can have all 0&#39;s. Similar tables can be constructed for all the transmission schemes described in this paper. 
         [0024]    There are some obvious alternative schemes to the one mentioned above. Each of the B 13 (z), B 11 (z) and B 2 (z 12 ) as well as the entire transmission scheme can be flipped to produce 16 obvious alternative schemes. It is also to be noted that B 13 (z) comprises the first 13 bits of the scheme. The combination of B 11  (z) and B 2 (z 12 ) comprise the remaining 13 bits. Hence, B 13 (z) and the combination of B 11 (z) and B 2 (z 12 ) can be transmitted simultaneously to reduce the overall length of the complementary code. This transmission scheme contains both simultaneous and TDM transmissions. This hybrid scheme is depicted in Table II. The parallel components for this case are given as: 
         [0000]        T   P1 ( z )= B   13 ( z )  (6) 
         [0000]        T   P2 ( z )= B   2 ( z   12 )+ z   −1   B   11 ( z )  (7) 
         [0025]    All ICC sets are constructed in a way that they can always be transmitted in a sequential bit by bit fashion. However, some amount of parallel transmission is possible in certain code sets depending on their structure. ICC set I is an example of such a code. 
         [0000]    ICC Set II Involving B 13 (z 2 ), B 11 (z 2 ) and B 2 (z 24 ) 
         [0026]    If we put one gap in between the elements of B 13 (z) and B 11 (z), we come up with two modified Barker codes B 13 (z 2 ) and B 11 (z 2 ), respectively. The corresponding autocorrelation functions are now R 13 (z 2 ) and R 11 (z 2 ), respectively. When lined up with one another, all the sidelobes will cancel each other except the ones at the extreme ends of R 13 (z 2 ). We note that in this case, we will require R 2 (z 24 ) to cancel these two remaining sidelobes. Hence, we need to transmit B 2 (z 24 ) in the transmission scheme. 
         [0027]    In the transmission scheme, 11 of the 12 gaps created in B 13 (z 2 ) are filled by the elements of B 11  (z 2 ). The one remaining gap just before the last element of B 13 (z 2 ) needs to be filled somehow using the code bits of B 2 (z 24 ). We observe that if we place the first bit of B 2 (z 24 ) just before the first element of B 13 (z 2 ), the last bit of B 2 (z 24 ) occupies the gap to be filled. Thus, the gap-free transmission scheme is given by: 
         [0000]        T ( z )= B   2 ( z   24 )+ z   −1   B   13 ( z   2 )+ z   −2   B   11 ( z   2 )  (8) 
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                 TABLE 3 
               
             
             
               
                   
               
               
                 Transmission scheme for ICC-II 
               
             
          
           
               
                   
                 t 1   
                 t 2   
                 t 3   
                 t 4   
                 t 5   
                 t 6   
                 t 7   
                 t 8   
                 t 9   
                 t 10   
                 t 11   
                 t 12   
                 t 13   
                 t 14   
                 t 15   
                 t 16   
                 t 17   
                 t 18   
                 t 19   
                 t 20   
                 t 21   
                 t 22   
                 t 23   
                 t 24   
                 t 25   
                 t 26   
               
               
                   
                   
               
             
          
           
               
                 f 1   
                 1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
                 0 
               
               
                 f 2   
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 −1 
               
               
                 f 3   
                 0 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 0 
                 0 
               
               
                   
               
             
          
         
       
     
         [0028]    Table III shows the graphical representation of the transmission scheme. Once again, t 1 -t 26  represent the time slots and f 1 -f 3  represent the frequencies used to modulate B 2 (z 24 ), B 13 (z 2 ) and B 11 (z 2 ), respectively. Once again, reversal of the transmission order of the individual codes or the entire scheme produce several equivalent ICC sets. Unlike the ICC set I, simultaneous transmission is not possible for this set since the purely TDM scheme shown in table III cannot be divided into equal length blocks containing complete constituent codes. 
         [0000]    ICC Set III Involving B 13 (z), B 11 (z), B 7 (z 2 ) and B 5 (z 2 ) 
         [0029]    This set is based on Barker codes of length  13 ,  11 ,  7  and  5 . We have shown the autocorrelation functions of the length  13  and length  11  codes in  FIGS. 2 and 3 , respectively. When these two autocorrelation functions R 13 (z) and R 11 (z) are added together, the sidelobes cancel each other out except the two positive sidelobes at the two extremes of R 13 (z). In this set, we seek to cancel out this sidelobe with the autocorrelation function of suitably modified length  7  Barker code. We insert a zero between each element of the length  7  Barker code. In other words we use B 7 (z 2 ). The corresponding autocorrelation function R 7 (z 2 ) is shown in  FIG. 4 . 
         [0030]    This autocorrelation function, R 7  (z 2 ), cancels the two residual sidelobes of R 13  (z), but introduces two new negative sidelobes. We observe that these sidelobes can easily be cancelled out by the autocorrelation function of the length  5  Barker code when one zero is inserted between each of its elements. Using similar notations as before, this modified length  5  Barker code is represented as B 5 (z 2 ) and its autocorrelation function R 5 (z 2 ) is shown in  FIG. 5 . 
         [0031]    When these autocorrelation functions, i.e. R 13 (z), R 1  (z), R 7 (z 2 ) and R 5 (z 2 ) are summed together, The sidelobes cancel each other out and we get a mainlobe of height  72 . Next, we propose a time division multiplexed and interlaced scheme for the transmission of this code set. 
         [0032]    For this code set, we transmit B 13 (z) and B 11 (z) back to back. Then we send B 7 (z 2 ) and B 5 (z 2 ) interlaced with each other. We observe that this results in a gap just before the last bit. To avoid the gap, we flip this entire transmission scheme and the first bit of the flipped scheme is made to occupy the aforementioned gap. The final transmission scheme is given by: 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                      
                     
                       ( 
                       z 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         B 
                         13 
                       
                        
                       
                         ( 
                         z 
                         ) 
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           13 
                         
                       
                        
                       
                         
                           B 
                           11 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           24 
                         
                       
                        
                       
                         
                           B 
                           7 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           25 
                         
                       
                        
                       
                         
                           B 
                           5 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           35 
                         
                       
                        
                       
                         
                           B 
                           7 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           38 
                         
                       
                        
                       
                         
                           B 
                           5 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           48 
                         
                       
                        
                       
                         
                           B 
                           11 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           59 
                         
                       
                        
                       
                         
                           B 
                           13 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   9 
                   ) 
                 
               
             
           
         
       
     
         [0033]    The long code that results can be shortened significantly by using the hybrid transmission schemes. We observe that the TDM transmission scheme given by (9) can be broken into 3 parallel components each of length  24 . The parallel components are given by: 
         [0000]        T   P1 ( z )= B   13 ( z )+ z   −13   B   11 ( z )  (10) 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       T 
                       
                         P 
                          
                         
                             
                         
                          
                         2 
                       
                     
                      
                     
                       ( 
                       z 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         z 
                         
                           - 
                           24 
                         
                       
                        
                       
                         
                           B 
                           7 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           1 
                         
                       
                        
                       
                         
                           B 
                           5 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           11 
                         
                       
                        
                       
                         
                           B 
                           7 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           14 
                         
                       
                        
                       
                         
                           B 
                           5 
                         
                          
                         
                           ( 
                           
                             z 
                             2 
                           
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   11 
                   ) 
                 
               
             
           
         
       
     
         [0000]        T   P3 ( z )= B   13 ( z )+ z   −13   B   11 ( z )  (12) 
         [0000]    The hybrid simultaneous and TDM transmission scheme is shown in table IV. 
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
               
             
           
               
                 TABLE 4 
               
             
             
               
                   
               
               
                 Hybrid transmission scheme for ICC-III 
               
             
          
           
               
                   
                 t 1   
                 t 2   
                 t 3   
                 t 4   
                 t 5   
                 t 6   
                 t 7   
                 t 8   
                 t 9   
                 t 10   
                 t 11   
                 t 12   
                 t 13   
                 t 14   
                 t 15   
                 t 16   
                 t 17   
                 t 18   
                 t 19   
                 t 20   
                 t 21   
                 t 22   
                 t 23   
                 t 24   
               
               
                   
                   
               
             
          
           
               
                 f 1   
                 −1 
                 −1 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 −1 
                 −1 
                 1 
                 −1 
                 1 
                 −1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
               
               
                 f 2   
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 1 
                 −1 
                 1 
                 1 
                 −1 
                 1 
               
               
                 f 3   
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
               
               
                 f 4   
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
               
               
                 f 5   
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 1 
                 0 
                 −1 
                 0 
                 1 
               
               
                 f 6   
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 −1 
                 0 
                 1 
                 0 
                 −1 
                 0 
               
               
                 f 7   
                 −1 
                 −1 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 −1 
                 −1 
                 1 
                 −1 
                 1 
                 −1 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
               
               
                 f 8   
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 0 
                 −1 
                 −1 
                 −1 
                 1 
                 1 
                 1 
                 −1 
                 1 
                 1 
                 −1 
                 1 
               
               
                   
               
             
          
         
       
     
       ICC Set IV Involving Modified B 13 (z), B 11 (z) and B 2 (z) 
       [0034]    In this set, we start with a Barker code of length  13 . In the time domain the Barker code B N (Z) is denoted as b N (n). It should be noted that the time reversed or flipped version of this code is denoted by b N (N−n) while the corresponding z-transform is given by z −N B N (z −1 ). Starting with b 13 (n) and appending a 1 and thirteen 0&#39;s at the beginning, the first constituent code is constructed as: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         1 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             13 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             13 
                           
                            
                           
                             ( 
                             n 
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   13 
                   ) 
                 
               
             
           
         
       
     
         [0035]    Next, the Barker code of length  13  is flipped and thirteen 0&#39;s and a −1 are appended at the end. Thus, the second constituent code looks like: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         2 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               b 
                               13 
                             
                              
                             
                               ( 
                               
                                 13 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               13 
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   14 
                   ) 
                 
               
             
           
         
       
     
         [0036]    It is easy to see that c 1 (n) and c 2 (n) will interlace with each other with the b 13 (13−n) of c 2 (n) occupying the positions of the thirteen 0&#39;s in c 1 (n). To counter the sidelobes due to these two codes, we employ the Barker codes of lengths  11  and  2  using each of them twice. The length  2  and length  3  codes are modified to produce the following constituent codes: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         3 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             14 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             11 
                           
                            
                           
                             ( 
                             n 
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   15 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         4 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               b 
                               11 
                             
                              
                             
                               ( 
                               
                                 11 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               14 
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   16 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         5 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       
                         
                           c 
                           6 
                         
                          
                         
                           ( 
                           n 
                           ) 
                         
                       
                       = 
                       
                         [ 
                         
                           
                             1 
                              
                             
                                 
                             
                              
                             
                               
                                 0 
                                  
                                 
                                     
                                 
                                  
                                 0 
                                  
                                 
                                     
                                 
                                  
                                 ⋯ 
                                  
                                 
                                     
                                 
                                  
                                 0 
                               
                               
                                  
                                 11 
                               
                             
                           
                            
                           
                               
                           
                           - 
                           1 
                         
                         ] 
                       
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   17 
                   ) 
                 
               
             
           
         
       
     
         [0000]    which corresponds to: 
         [0000]        C   5 ( z )= C   6 ( z )= B   2 ( z   12 )  (18) 
         [0037]    Also, c i (n)ε         ∀i. The autocorrelation functions produced by c 3 (n), c 4 (n), c 5 (n) and c 6 (n) cancel out the sidelobes due to c 1 (n) and c 2 (n). All these codes can be interlaced with each other perfectly and the transmission scheme in the z-domain notations is given as: 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                      
                     
                       ( 
                       z 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         C 
                         1 
                       
                        
                       
                         ( 
                         z 
                         ) 
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           1 
                         
                       
                        
                       
                         
                           C 
                           2 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           28 
                         
                       
                        
                       
                         
                           C 
                           3 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           30 
                         
                       
                        
                       
                         
                           C 
                           4 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           29 
                         
                       
                        
                       
                         
                           C 
                           5 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           42 
                         
                       
                        
                       
                         
                           C 
                           6 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   19 
                   ) 
                 
               
             
           
         
       
     
         [0038]    This ICC set uses six different frequencies to produce a mainlobe of height  56 . A parallel scheme is also possible for this scheme apart from the sequential scheme described by equation (19). The combination of c 1 (n) and c 2 (n) is completely contained in the first 28 bits of the sequential transmission scheme. On the other hand the combination of c 3 (n), c 4 (n), c 5 (n) and c 6 (n) is completely contained in the remaining 28 bits of sequential transmission scheme. Hence, the two combination schemes can be transmitted in parallel. 
       ICC Set V Obtained by Modifying Set III 
       [0039]    Some ICC sets could be modified to obtain higher mainlobes using the same number of frequencies. Consider the Barker code of length n. If we add to it a 1 or −1 before or after n or more zeros, the resultant autocorrelation function will still have unity peak sidelobe. Hence, if a code contains two copies of any code block of length N from          , the following trick can be applied to enhance the mainlobe further.
       Add a 1 before at least (N−1) zeros preceding one of the code blocks thus yielding:       
 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         1 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             
                               ≥ 
                               
                                 ( 
                                 
                                   N 
                                   - 
                                   1 
                                 
                                 ) 
                               
                             
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             N 
                           
                            
                           
                             ( 
                             n 
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   20 
                   ) 
                 
               
             
           
         
       
       
         
           
             Add a −1 before at least N−1 zeros preceding the other code and flip the final product. This gives: 
           
         
       
     
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         2 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               b 
                               N 
                             
                              
                             
                               ( 
                               
                                 N 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               
                                 ≥ 
                                 
                                   ( 
                                   
                                     N 
                                     - 
                                     1 
                                   
                                   ) 
                                 
                               
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   21 
                   ) 
                 
               
             
           
         
       
       
         
           
             The insertion of 1&#39;s and −1&#39;s should be done only if it is possible to accommodate them in a scheme without gaps. 
           
         
       
     
         [0043]    Equations (13)-(16) makes use of this concept. In this example we extend the concept to ICC set III to show how the mainlobe can be enhanced further without using more frequencies. The constituent codes for these sets are formed as follows: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         1 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             64 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             13 
                           
                            
                           
                             ( 
                             n 
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   22 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         2 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               b 
                               13 
                             
                              
                             
                               ( 
                               
                                 13 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               64 
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   23 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         3 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             13 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             11 
                           
                            
                           
                             ( 
                             n 
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   24 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         4 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               b 
                               11 
                             
                              
                             
                               ( 
                               
                                 11 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               13 
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   25 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         5 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             14 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             5 
                           
                            
                           
                             ( 
                             
                               n 
                               ↑ 
                               2 
                             
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   26 
                   ) 
                 
               
             
           
         
       
     
         [0000]    where b N (n↑m) is a modified version of b N (n) having (m−1) zeros between every two successive bits. 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       
                         c 
                         6 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               b 
                               5 
                             
                              
                             
                               ( 
                               
                                 5 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               14 
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   27 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         7 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         1 
                          
                         
                             
                         
                          
                         
                           
                             0 
                              
                             
                                 
                             
                              
                             0 
                              
                             
                                 
                             
                              
                             ⋯ 
                              
                             
                                 
                             
                              
                             0 
                           
                           
                              
                             12 
                           
                         
                          
                         
                             
                         
                          
                         
                           
                             b 
                             7 
                           
                            
                           
                             ( 
                             
                               n 
                               ↑ 
                               2 
                             
                             ) 
                           
                         
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   28 
                   ) 
                 
               
             
             
               
                 
                   
                     
                       
                         c 
                         8 
                       
                        
                       
                         ( 
                         n 
                         ) 
                       
                     
                     = 
                     
                       [ 
                       
                         
                           
                             
                               
                                 b 
                                 ^ 
                               
                               7 
                             
                              
                             
                               ( 
                               
                                 13 
                                 - 
                                 n 
                               
                               ) 
                             
                           
                            
                           
                               
                           
                            
                           
                             
                               0 
                                
                               
                                   
                               
                                
                               0 
                                
                               
                                   
                               
                                
                               ⋯ 
                                
                               
                                   
                               
                                
                               0 
                             
                             
                                
                               12 
                             
                           
                         
                          
                         
                             
                         
                         - 
                         1 
                       
                       ] 
                     
                   
                    
                   
                       
                   
                 
               
               
                 
                   ( 
                   29 
                   ) 
                 
               
             
           
         
       
     
         [0044]    It can be shown that these codes could be interlaced with each other yielding a gap-free transmission scheme which in the z-domain notations, is given by: 
         [0000]    
       
         
           
             
               
                 
                   
                     T 
                      
                     
                       ( 
                       z 
                       ) 
                     
                   
                   = 
                   
                     
                       
                         C 
                         1 
                       
                        
                       
                         ( 
                         z 
                         ) 
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           1 
                         
                       
                        
                       
                         
                           C 
                           3 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           2 
                         
                       
                        
                       
                         
                           C 
                           2 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           26 
                         
                       
                        
                       
                         
                           C 
                           7 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           27 
                         
                       
                        
                       
                         
                           C 
                           5 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         Z 
                         
                           - 
                           28 
                         
                       
                        
                       
                         
                           C 
                           8 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           29 
                         
                       
                        
                       
                         
                           C 
                           6 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                     + 
                     
                       
                         z 
                         
                           - 
                           54 
                         
                       
                        
                       
                         
                           C 
                           4 
                         
                          
                         
                           ( 
                           z 
                           ) 
                         
                       
                     
                   
                 
               
               
                 
                   ( 
                   30 
                   ) 
                 
               
             
           
         
       
     
         [0045]    Using this ICC set, we can achieve a mainlobe height of 80 using the same 8 frequencies used in ICC set III. 
       Frequency Diversity Efficiency 
       [0046]    We define the frequency diversity efficiency (FDE) as the ratio of the height of the mainlobe H m  and the number of frequencies N f . The FDE is given by: 
         [0000]    
       
         
           
             
               
                 
                   η 
                   = 
                   
                     
                       H 
                       m 
                     
                     
                       N 
                       f 
                     
                   
                 
               
               
                 
                   ( 
                   31 
                   ) 
                 
               
             
           
         
       
     
         [0000]    
       
         
               
             
               
               
               
               
               
               
               
             
               
               
               
               
               
               
               
             
           
               
                 TABLE 5 
               
             
             
               
                   
               
               
                 FDE for the proposed schemes 
               
             
          
           
               
                   
                 Scheme 
                 I 
                 II 
                 III 
                 IV 
                 V 
               
               
                   
                   
               
             
          
           
               
                   
                 H m   
                 26 
                 26 
                 72 
                 56 
                 80 
               
               
                   
                 N f   
                 3 
                 3 
                 8 
                 6 
                 8 
               
               
                   
                 η 
                 8.67 
                 8.67 
                 9 
                 9.33 
                 10 
               
               
                   
                   
               
             
          
         
       
     
         [0047]    The FDE for the different schemes proposed in this paper are summarized in table 5. Iterative application of the idea mentioned for ICC set V would result in higher η provided it could be done without gaps. Let S N ={C N } be the set of all ICC codes of length N. Let η N  be the maximum η achieved by any code in S N . It is our conjecture that: 
         [0000]    
       
         
           
             
               
                 
                   
                     
                       lim 
                       
                         N 
                         → 
                         ∞ 
                       
                     
                      
                     
                       
                         η 
                         ^ 
                       
                       N 
                     
                   
                   = 
                   ∞ 
                 
               
               
                 
                   ( 
                   32 
                   ) 
                 
               
             
           
         
       
     
         [0048]    It is an important topic of further research to prove or disprove this conjecture. Also, given a certain number of frequencies, it is important to find the maximum achievable η or equivalently, the maximum N possible with the given number of frequencies. These open questions are of both theoretical and practical significance. 
         [0000]    Comparison with Orthogonal Matrix Codes 
         [0049]    Similar to complementary codes, orthogonal matrices could be used to generate codes with zero sidelobes. This could be done either with biphase elements or polyphase elements as in the DFT matrix. The case for biphase elements has been discussed in the publication: F. F. Kretschemer, Jr. and K. Gerlach, “Low sidelobe radar waveforms derived from orthogonal matrices, “ IEEE Trans. Aerospace and Electronics Systems ,” vol. 27, No. 1, pp. 92-101, January, 1991. The polyphase case of the DFT matrix has been described in:R. L. Frank, “Polyphase complementary codes,”  IEEE Trans. Inform. Theory ,” vol. IT-26, no. 6, pp. 641-647, October 1980. 
         [0050]    In such cases, the matrix elements of a N×N matrix are transmitted row-wise using N frequencies resulting in a mainlobe of N 2 . The conjugate transpose or the Hermitian of the transmitted matrix forms its matched filter in such cases. This approach is more susceptible to frequency selective fading effects since each inner-product contribution to the output is affected by all frequencies. It is also more vulnerable to jamming at any of the frequencies. The codes proposed in this paper results in more graceful degradation in the presence of fading and jamming effects since each element code has unity peak sidelobe and depends only on one frequency. If only one frequency survives, the ICC will still have a code with unity peak sidelobe, while orthogonal matrix codes will could result in much higher sidelobes. 
         [0051]    Also, since coherence is easier to maintain at each frequency but more difficult to maintain simultaneously at all frequencies, the ICC codes are more resistant to partial loss in coherence. The 5 examples of ICC sets in this paper have larger η&#39;s than the nearest matrix codes greater or equal in length.