Page 139 - Phase Space Optics Fundamentals and Applications
P. 139

120   Chapter Four


                   Cylindrical
                     lens
                           1D input  FrFT of order p
                           y 0       y
                                                        Varifocal
                                                     y   lens (L)  Output
                                                                    plane
                                 x 0     y(R )                 y
                                           0
                                         x
                                                           y(R )
                                                             0
                                                           x          y(R )
                                                                        0
                                                                      x
                                R p
                                       z
                                             a p
                                        l
                                                          a'
               FIGURE 4.4 RWD setup (multichannel continuous FrFT transformer).




               for each fractional order p is
                                    −a            a
                            M L ( p) =  =   s    tan( p /2)         (4.45)
                                             2 −1
                                     a p                − l
                                          1+s (z )  −1  tan( p /2)
                                             2
               However, for the p-order slice of the RWT of the input function to be
               achieved, the lens L should counterbalance the magnification of the
               FRT located at R p to restore its proper magnification at the output
               plane. Therefore, by using Eq. (4.43), the magnification provided by
                L should be
                                              2
                                 −1       1 + s (z ) −1  tan( p /2)
                         M L ( p) =  =−                             (4.46)
                                 M p    1 + tan( p /2)  tan( p /4)
               Comparing Eqs. (4.45) and (4.46), we note that the functional depen-
               dence of both equations on p is different, and, consequently, we are
               unable to obtain an exact solution for all fractional orders. However,
               an approximate solution can be obtained by choosing the parameters

               of the system, namely, s, z, l,  , and a , in such a way that they mini-
               mize the difference between these functions in the interval p ∈ [0, 1].
               One way to find the optimum values for these parameters is by a least-
                                            6
               square method. This optimization leads to the following constraint
               conditions.
                                          	             2
                                    1                −ls
                              a = l   +    ,    z =                 (4.47)

                                    2   4            l + s 2
   134   135   136   137   138   139   140   141   142   143   144