Page 107 - Computational Colour Science Using MATLAB
P. 107

94 CHROMATIC-ADAPTATION TRANSFORMS AND COLOUR APPEARANCE

               361 column matrix of normalized tristimulus values by the 363 matrix M BFD .
               The XYZ values of the sample are used with Equations (6.6) to compute the
               RGB values for the sample; X WT , Y WT , Z WT  and X WR , Y WR , Z WR  are used to
               compute R WT , G WT , B WT  and R WR , G WR , B WR , respectively.


               Step 2: Calculate the corresponding RGB values for the test sample (R , G , B )
                                                                              C
                                                                                 C
                                                                                     C
               and for the reference white (R WC , G WC , B WC ) using Equations (6.10).
               Step 3: Calculate the luminance level adaptation factor (F ), the chromatic
                                                                      L
               background induction factor (N ) and the brightness background induction
                                             CB
               factor (N ),
                       BB
                          4               4 2    1=3
                    F L ¼ K ðL A Þþ 0:1ð1   K Þ ð5L A Þ  ,                       ð6:16Þ
               where K ¼ 1/(5L +1), N   ¼ N   ¼ 0.725(1/n) 0.2  and n ¼ Y /Y .
                              A       CB    BB                       B  W
               Step 4: Calculate the corresponding tristimulus values for the test sample (R , G ,
                                                                                     0
                                                                                  0
               B ) and for the reference white (R , G , B ),
                                                  0
                                                      0
                0
                                              0
                                                  W
                                                      W
                                              W
                    2   3           2     3
                     R 0              R C Y
                     G 5 ¼ M H M      G C Y 5,
                    6  0 7        1 6     7
                    4            BFD 4                                           ð6:17Þ
                     B 0              B C Y
               where
                           2                               3
                               0:9870      0:1471    0:1600
                    M  1   6   0:4323      0:5184          7
                      BFD  ¼ 4                       0:0493 5
                              0:0085       0:0400    0:9685
               and
                             0:38971       0:68898      0:07868
                          2                                    3
                            0:22981        1:18340      0:04641 5.
                          6                                    7
                    M H ¼ 4
                             0:00000       0:00000      1:00000
               Step 5: Calculate the cone responses after adaptation for the test sample (R , G ,
                                                                                     0
                                                                                  0
                                                                                     a
                                                                                  a
               B ) and for the reference white (R aW , G  0 aW , B aW ),
                                              0
                                                       0
                0
                a
                                         0:73           0:73
                      0             0              0       þ 2Š,
                    R a ¼ 1 þ½40ðF L R =100Þ  Š=½ðF L R =100Þ
                                         0:73          0:73
                      0             0             0       þ 2Š,
                    G a ¼ 1 þ½40ðF L G =100Þ  Š=½F L G =100Þ                     ð6:18Þ
                                         0:73           0:73
                      0             0             0        þ 2Š,
                    B a ¼ 1 þ½40ðF L B =100Þ  Š=½ðF L B =100Þ
               and where, if R 5 0,
                             0
                             a
                                        0:73          0:73
                     0             0             0       þ 2Š.
                    R a ¼ 1  ½40ð R =100Þ  Š=½ð R =100Þ
   102   103   104   105   106   107   108   109   110   111   112