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Transport aspects                                              2


                                          Table 9-21 – Derived framing parameters

                  Parameter                                        Definition

             fDMT                 Symbol rate of transmission expressed in Hz as specified in clause 10.4.4 (same for
                                  upstream and downstream).
              f D DS              The downstream data symbol rate:
                                                                                1  
                                                                      M ds  1    
                                                        f D DS   f DMT       M SF  
                                                                     
                                                                                    
                                                                          M  F     
                                                                                   
                                  where:
                                  1 = overhead due to one RMC symbol per TDD frame
                                  1/MSF = overhead due to one sync symbol per superframe
                                  MF = number of symbol periods per TDD frame
              f  US               The upstream data symbol rate:
               D
                                                                                1  
                                                                      M us  1    
                                                        f D US   f DMT      M SF  
                                                                                    
                                                                          M  F     
                                                                                   

              f RMC               The RMC symbol rate:
                                                                           1   
                                                             f RMC   f DMT      
                                                                                
                                                                           M  F  

             Beoc                 The maximum number of eoc bytes per direction per logical frame period

                                                            6×                      +125000×       
                                             = min {         −       ,                (           1−                      )} (Note 3)

             DPR                  DTU payload rate:
                                  DPR   DPR     DPR
                                              D       DR
             DPRD                 DTU payload rate part corresponding to data symbols:
                                                           K    
                                  DPR  8(    B )   f     FEC     1  DTUframing OH     (Note 1)
                                                         
                                                                 
                                                      D
                                                D
                                       D
                                                           N FEC 
             DPReoc               The maximum DTU payload rate corresponding to eoc:
                                  DPReoc = ( 8 x Beoc ) / (MF / fDMT )
             DPRDR                DTU payload rate part corresponding to the data portion of the RMC symbol:
                                                                      K   
                                                                    
                                                                  
                                           DPR  DR   8(  B )   f RMC   FEC     1  DTUframing OH 
                                                                           
                                                        DR
                                                                      N FEC 

             DTUframingOH         The relative overhead due to DTU framing:
                                                                               7
                                                         DTUframing  OH 
                                                                            Q   K  FEC
             NDR                  The net data rate (for each direction):
                                                       NDR   DPR   1000   kbit/s  (Note 2)






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