Gernot Hoffmann CIE Color Space

11.2 Color Space Calculations / General
For the white point correction, the basis vectors R,G,B are scaled by u,v,w. This does not
change their coordinates in xyY .The mapping from XYZ to xyY is a central planar projection.
T
( 12) X = L( x, y, z) = ruR+ gvG+ bwB
For the white point we have r = g = b = 1.
( 13) W = Lx ( , y , z ) = Lux ( , y, z) + Lvx ( , y, z) + Lwx ( , y, z
( 14)
⎡x
⎢y
⎣
⎢z
w
w
w
⎤
⎥
⎦
⎥ =
( 15) w = 1−u−v
w w w T
⎡x
x x
⎢y
y y
⎢
⎣z
z z
r g b
r g b
r g b
⎤ ⎡ u ⎤
⎥ ⎢ v ⎥
⎥
w
⎦ ⎣
⎢
⎦
⎥
r r r T
This can be re-arranged, L cancels on both sides.:
⎡ u ⎤
= P ⎢ v ⎥
⎣
⎢w⎦
⎥
⎡x
w ⎤
( 16)
⎢y
w ⎥
⎣
⎢
⎦
⎥
xr yr xg yg xb
u
yb
v
1 u v
=
⎡
⎢
⎢
⎣
⎤ ⎡ ⎤
⎥ ⎢ ⎥
⎥
⎦ ⎣
⎢ − − ⎦
⎥
( 17) x = ( x − x ) u + ( x − x ) v + x
w r b g b b
y = ( y − y ) u+ ( y − y ) v+ y
w r b g b b
These linear equations are solved by Cramer’s rule.
( 18) D = ( xr −xb) ( yg −yb) −( yr −yb) ( xg −xb)
U = ( x −x )( y −y ) −( y − y )( x − x )
w b g b w b
g b
V = ( xr −xb) ( yw −yb) −( yr −yb) ( xw −xb)
( 19)
u = U/ D
v = V/ D
w = 1−u−v 13
g g g T
b b b) T
It is not necessary to invert the whole matrix numerically. We can simplify the calculation by
adding the first two rows to the third row and find so immediately Eq.(15), which is anyway
clear:
In the next step we assume that u,v,w are already calculated and we use the general color
transformation Eq.(12) and furtheron Eq.(8). We get the matrices C xr and C rx .
⎡X⎤
( 20)
⎢Y⎥
⎣
⎢Z⎦
⎥
ux / y
V uy / y
uzr / y
vx
vy
/ y
/ y
wx
wy
/ y
/ y
=
⎡
⎢
⎢
⎣ vz / y wz / y
( 21 X = VC
r
xr
−
xr
r w g w b w
r w g w b w
w g w b w
( 22) r = ( 1/
V)
C 1
X = (/ 1 ) C X
V rx
⎤ ⎡r
⎤
⎥ ⎢g⎥
⎥
b
⎦ ⎣
⎢
⎦
⎥