Subsurface Scattering - The Magic of Computer Graphics - page 107

P i

T 1 T 1 R 2 T 1 R 2 R 1

T 1 R 2 P o P o T 1 R 2 R 1 T 2

Figure 4.20 A multilayered model. Light scattering between layers is a significant complication.

tance for transmittance functions T 1 ( r ) and T 2 ( r ) is

y 2 T 2

2 dx dy .
(4.17)

∞ ∞

T 1 x 2

T 1 ∗ T 2 ( x , y )= + ( x − x )

2

+( y − y )

− ∞ − ∞

Note that T 1 ∗ , as is the integral on the

right side of Equation (4.17). This dependence on position is omitted hereafter.
The convolution T 1 ∗

T 2 ( x , y ) is a function of the position ( x , y )

T 2 is not quite the same as the transmittance function
of two adjacent layers, because it does not account for scattering between the
layers. For example, some of the light transmitted through layer 1 into layer 2 is
scattered back into layer 1, and some of that light is scattered back into layer 2,
where some of it gets transmitted ( Figure 4.20 ) . Light can scatter between layers
like this arbitrarily many times. The actual transmittance function T 12 for two
layers is therefore the infinite series

T 12 = T 1 ∗ T 2 + T 1 ∗ R 2 ∗ R 1 ∗ T 2 + T 1 ∗ R 2 ∗ R 1 ∗ R 2 ∗ R 1 ∗ T 2 + ···. (4.18)

Considering that each convolution function is a double integral, Equation (4.18)
involves a lot of computation even when the series is truncated to just a few terms.
Fortunately, there is a shortcut. The convolution theorem states that the convo-
lution of two functions can be computed from the product of their Fourier trans-
forms. The Fourier transform of T 12 is therefore

T 12 = T 1 T 2 + T 1 R 2 R 1 T 2 + T 1 R 2 R 1 R 2 R 1 T 2 + ···

where denote the Fourier transforms of T and R , respectively, and the

product indicates pointwise multiplication of the transformed functions. Rear-

T and R