Page 326 - Satellite Communications, Fourth Edition
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306  Chapter Ten

                                Equation (10.18) is sometimes expressed in the alternative form

                                                                2E b
                                                        P e   Qa   b                     (10.20)
                                                              Å
                                                                 N 0
                              Here, the Q(⋅) function is simply an alternative way of expressing the
                              complementary error function, and in general
                                                     erfc(x)   2Q( 22x)                 (10.21)

                              These relationships are given for reference only and will not be used fur-
                              ther in this book.
                                An important parameter for carrier systems is the ratio of the aver-
                              age carrier power to the noise power density, usually denoted by [C/N ].
                                                                                            0
                              The [E /N 0 ] and [C/N 0 ] ratios can be related as follows. The average
                                    b
                              carrier power at the receiver is P W. The energy per symbol is there-
                                                            R
                              fore P /R sym  J, with R sym  in symbols per second. Since each symbol contains
                                   R
                              m bits, the energy per bit is P /mR sym  J. But mR sym    R , and therefore,
                                                                                b
                                                        R
                              the energy per bit, E , is
                                                b
                                                               P R
                                                          E                             (10.22)
                                                            b
                                                               R b
                                              represent the noise power density. Then E /N   P /R N .
                                As before, let N 0                               b  0   R  b  0
                              But P /N is the carrier-to-noise density ratio, usually denoted by C/N , and
                                   R
                                      0
                                                                                         0
                              therefore,
                                                               C/N
                                                         E b      0
                                                                                        (10.23)
                                                         N 0    R b
                              Rearranging this and putting it in decibel notation gives
                                                     c  C  d   c  E b  d   [R ]         (10.24)
                                                                      b
                                                      N 0    N 0
                              It should be noted that whereas [E /N ] has units of decibels, [C/N ]has
                                                                0
                                                                                         0
                                                             b
                              units of dBHz, as explained in App. G.
                                Example 10.2 The downlink transmission rate in a satellite circuit is 61 Mb/s, and
                                the required [E b /N 0 ] at the ground station receiver is 9.5 dB. Calculate the required
                                [C/N 0 ].
                                                                                  6
                                Solution The transmission rate in decibels is  [R b ]   10 log(61   10 )   77.85 dBb/s
                                Hence

                                                 s  C  t   77.85   9.5   87.35 dBHz
                                                  N 0
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