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Optical Amplifiers                                                                 289




           Using Parseval’s theorem, the mean noise power is
                                                      T∕2
                                                  1        2
                                       N    = lim        < I (t) > dt
                                        s−sp  T→∞ T ∫      F
                                                     −T∕2
                                                      ∞
                                                  1            2
                                                          ̃
                                            = lim       < |I (f)| > df.                    (6.216)
                                             T→∞ T ∫       F
                                                     −∞
           From Eq. (6.215), we have
                                2
                            ̃
                                             2
                                                                               2
                                      2 ̃
                                                   2
                                                                     2
                                                            2
                         < |I (f)| > = R |H (f)| [| | < |̃n (f)| > +| | < |̃n (−f)| >
                                                                          F
                                                        F
                                                                  out
                                         e
                                                out
                            F
                                                                   ∗
                                         ∗
                                           2
                                                                        ∗
                                     +( ) <̃n (f)̃n (−f) > + 2  <̃n (−f)̃n (f) >].     (6.217)
                                         out   F    F        out   F    F
           The contribution of the third and fourth terms on the right-hand side of Eq. (6.217) is zero. This can be shown
           by writing
                                           n (f)= |n (f)| exp[i (f)],                    (6.218)
                                                   F
                                            F
                                                            F
                                             out  = | | exp[i ],                     (6.219)
                                                   out
                                                           out
                       ∗  2
                     ( ) <̃n (f)̃n (−f) > = < |̃n (f)||̃n (−f)| exp(i[ (f)+  (−f)− 2 ] >)
                       out    F   F            F    F           F     F        out
                         2    ∗     ∗
                         <̃n (−f)̃n (f) > = < |̃n (f)||̃n (−f)| exp(i[− (f)−  (−f)+ 2 ]) >,
                         out  F     F          F    F            F     F        out
                                                                                           (6.220)
                      ∗  2                 2    ∗    ∗            2
                    ( ) <̃n (f)̃n (−f) > +
                      out    F   F        out  <̃n (−f)̃n (f) > = 2| | |̃n (f)||̃n (−f)|
                                                                out
                                                                     F
                                                                          F
                                                     F
                                                F
                                                             < cos [ (f)+  (−f)− 2 ] >
                                                                   F      F        out
                                                           = 0,                            (6.221)
           since  (f) is a random variable with uniform distribution in the interval [0, 2]. By definition, the power
                F
           spectral densities are
                                                     2
                                              < |̃n(f)| >
                                      = lim           ,                                  (6.222)
                                     ASE
                                          T→∞     T
                                                      2
                                              < |̃n (f)| >            2
                                                  F
                                                                ̃
                                       = lim           =   |H (f)| ,                   (6.223)
                                      n F                   ASE  opt
                                          T→∞      T
                                                 ̃
                                                     2
                                              < |I (f)| >
                                                 F
                                      = lim           .                                  (6.224)
                                      I F
                                          T→∞     T
           Using Eqs. (6.217), (6.223), and (6.224), we find
                                                 2 ̃    2    2
                                           = 2R |H (f)| | | 
                                           I F      e      out  n F
                                                 2              2
                                                          ̃
                                             = 2R P    |H (f)| ,                         (6.225)
                                                   out ASE  eff
           where
                                                           ̃
                                             ̃
                                                     ̃
                                             H (f)= H (f)H (f)                             (6.226)
                                              eff     opt   e
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