Spectral power distribution

(1)

Color and Radiometry

Digital Image Synthesis Yung-Yu Chuang

10/19/2006

with slides by Pat Hanrahan and Matt Pharr

(2)

Radiometry

• Radiometry: study of the propagation of

electromagnetic radiation in an environment

• Four key quantities: flux, intensity, irradiance and radiance

• These radiometric quantities are described by their spectral power distribution (SPD)

• Human visible light ranges from 370nm to 730nm

1 1 02

1 04

1 06

1 08

1 01 0

1 01 2

1 01 4

1 01 6

1 01 8

1 02 0

1 02 2

1 02 4

1 02 6

1 01 6

1 01 4

1 01 2

1 01 0 1 08

1 06

1 04

1 02

1 1 0-2

1 0-4

1 0-6

1 0-8 Cosm ic

Ra y s Ga m m a

Ra y s X -Ra y s

Ultra - V iolet Infra -

Red Ra d io

H ea t Pow er

7 0 0 6 0 0 5 0 0 4 0 0

IR R G B UV

W a velength (N M )

(3)

Spectral power distribution

fluorescent light (日光燈)

400nm (bluish)

650nm (red) 550nm

(green)

(4)

Spectral power distribution

lemmon skin

(5)

Color

• Need a compact, efficient and accurate way to represent functions like these

• Find proper basis functions to map the infinite- dimensional space of all possible SPD functions to a low-dimensional space of coefficients

• For example, B(λ)=1, a bad approximation

(6)

Spectrum

• In core/color.*

• Not a plug-in, to use inline for performance

• Spectrum stores a fixed number of samples at a fixed set of wavelengths. Better for smooth functions.

#define COLOR_SAMPLE 3 class COREDLL Spectrum { public:

<arithmetic operations>

private:

float c[COLOR_SAMPLES];

...

}

component-wise

+ - * / comparison…

Why is this possible? Human vision system We actually sample RGB

(7)

Human visual system

• Tristimulus theory: all visible SPDs can be accurately represented for human observers with three values, x_λ, y_λ and z_λ.

• The basis are the spectral matching curves, X(λ), Y(λ) and Z(λ).

∫

=

λ λ λ λ λ λ

λ λ

λ

λ λ

λ

λ λ

λ

d Z

S z

d Y

S y

d X

S x

) ( ) (

) ( )

(

(8)

XYZ basis

360 830

pbrt has discrete versions (sampled every 1nm) of these bases in core/color.cpp

(9)

XYZ color

• It’s, however, not good for spectral computation. A

product of two SPD’s XYZ values is likely different from the XYZ values of the SPD which is the product of the two original SPDs.

• Hence, we often have to convert our samples (RGB) into XYZ

void XYZ(float xyz[3]) const { xyz[0] = xyz[1] = xyz[2] = 0.;

for (int i = 0; i < COLOR_SAMPLES; ++i) { xyz[0] += XWeight[i] * c[i];

xyz[1] += YWeight[i] * c[i];

xyz[2] += ZWeight[i] * c[i];

} }

(10)

Conversion between XYZ and RGB

float Spectrum::XWeight[COLOR_SAMPLES] = { 0.412453f, 0.357580f, 0.180423f

};

float Spectrum::YWeight[COLOR_SAMPLES] = { 0.212671f, 0.715160f, 0.072169f

};

float Spectrum::ZWeight[COLOR_SAMPLES] = { 0.019334f, 0.119193f, 0.950227f

};

Spectrum FromXYZ(float x, float y, float z) { float c[3];

c[0] = 3.240479f * x + -1.537150f * y + - 0.498535f * z;

c[1] = -0.969256f * x + 1.875991f * y + 0.041556f * z;

c[2] = 0.055648f * x + -0.204043f * y + 1.057311f * z;

return Spectrum(c);

}

(11)

Basic radiometry

• pbrt is based on radiative transfer: study of the transfer of radiant energy based on radiometric principles and operates at the geometric optics level (light interacts with objects much larger than the light’s wavelength)

• It is based on the particle model. Hence,

diffraction and interference can’t be easily accounted for.

(12)

Basic assumptions about light behavior

• Linearity: the combined effect of two inputs is equal to the sum of effects

• Energy conservation: scattering event can’t produce more energy than they started with

• No polarization: that is, we only care the frequency of light but not other properties

• No fluorescence or phosphorescence:

behavior of light at a wavelength doesn’t affect the behavior of light at other wavelengths

• Steady state: light is assumed to have reached equilibrium, so its radiance distribution isn’t changing over time.

(13)

Fluorescent materials

(14)

Flux (Φ)

• Radiant flux, power

• Total amount of energy passing through a surface per unit of time (J/s,W)

(15)

Irradiance (E)

• Area density of flux (W/m²)

dA E dΦ

=

4 r2

E π

= Φ

E ΦA

= E ΦcosA θ

= Lambert’s law Inverse square law

(16)

Angles and Solid Angles

• Angle

• Solid angle

The solid angle subtended by a surface is defined as the surface area of a unit sphere covered by the surface's projection onto the sphere.

l θ = r

2

A Ω = R

⇒ circle has 2π radians

⇒ sphere has 4π steradians

(17)

Intensity (I)

• Flux density per solid angle

• Intensity describes the directional distribution of light

ω d I dΦ

=

( ) d I ω d

ω

≡ Φ

(18)

Radiance (L)

• Flux density per unit area per solid angle

• Most frequently used,

remains constant along ray.

• All other quantities can be derived from radiance

⊥

= Φ

dA d

L d

ω

(19)

Calculate irradiance from radiance

dA

θ

^d

ω

) ,

( x ω

L

x

Ω

∫

Φ =

= L x

ω θ

d

ω

dA x d

E( ) ( , )cos

Light meter

(20)

Irradiance Environment Maps

Radiance

Environment Map

Irradiance

Environment Map

R N

( , )

L θ ϕ E ( , ) θ ϕ

(21)

Differential solid angles

dφ

dθ θ

φ

r sin

r θ

Goal: find out the relationship between dω and dθ, dψ

(22)

Differential solid angles

dφ

dθ θ

φ

r sin

r θ

2

( )( sin )

sin

dA r d r d

r d d

θ θ φ

=

Goal: find out the relationship between dω and dθ, dψ

By definition, we know that

1 when =

= dA r

dω

2 sin

d dA d d

ω = r = θ θ φ φ θd d cos

−

=

(23)

Differential solid angles

dφ

dθ θ

φ

r sin

r θ

We can prove that ^Ω ⁼

∫

⁼

2

4

S

dω π

(24)

Differential solid angles

dφ

dθ θ

φ

r sin

r θ

We can prove that ^Ω ⁼

∫

⁼

2

4

S

dω π

π

θ π

θ θ φ

φ θ θ ω

π π

4

cos 2

sin

1 1

0 2

0

0 2

0

2

=

−

=

= Ω

∫

− d

d d

d d d

S

(25)

Isotropic point source

If the total flux of the light source is Φ,

what is the intensity?

(26)

Isotropic point source

2

4

S

I d I

ω π

Φ =

=

∫

I 4

π

= Φ

If the total flux of the light source is Φ,

what is the intensity?

(27)

Warn’s spotlight

θ

If the total flux is Φ, what is the intensity?

θ

ω

^S

I ( ) ∝ cos

(28)

Warn’s spotlight

θ

If the total flux is Φ, what is the intensity?

⎩ ⎨

⎧ ≥

= otherwise

I c

S

0 ) cos

(

²

θ

π

ω θ

1 S

c 2 0

1 1

S c y 2

cos cos

c 2 cos

cos

1 S

1

0 1

0 2

0

= +

=

= +

=

= Φ

+

∫

π π

θ θ

π φ

θ

π

θ

y y

d d

d

c

^S ^S

+ Φ

= 2 π

1 c S

(29)

Irradiance: isotropic point source

θ _h

Φ

What is the irradiance for this point?

(30)

Irradiance: isotropic point source

θ _h

Φ

r

θ cos r = h

2

cos 4

4 dA r

d dA

Id dA

E d θ

π ω

π

ω Φ

Φ =

= Φ =

=

π ω 4

= Φ

= Φ d I d

Lambert law

Inverse square law

(31)

Irradiance: isotropic point source

θ _h

Φ

r

θ cos r = h

2 3 2

cos 4

4 dA r h

d dA

Id dA

E d θ

π θ

π ω

π

ω Φ

Φ = Φ =

= Φ =

=

π ω 4

= Φ

d

I d

(32)

Photometry

( ) ( )

Y = ∫ V λ L λ λ d

Luminance