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Momentary adjusting methods for simulating the color temperature, hues and brightness of daylight illumination with RGB LEDs for indoor lighting

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Abstract

Methods for simulating the color temperature, hue and brightness of daylight illumination for indoor lighting simply by adjusting the intensity of red, green, and blue light emitting diodes are proposed. We obtain uniform color mixing with a light box by adjusting the ratios between the intensities of red, green and blue LEDs. The intensity can be found by measuring the CIE chromaticity coordinates (x, y) and the luminance Y of the daylight with a chroma meter. After the chromaticity coordinates (x, y) and the luminance Y are found, the tristimulus values can be calculated and then transferred to red, green, and blue primaries by linear transformation. With the correct ratio of red, green, and blue intensities, the color temperature, hues and brightness of daylight can be rebuilt by red, green, and blue light emitting diodes.

©2011 Optical Society of America

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Corrections

Chih-Hsuan Tsuei and Wen-Shing Sun, "Momentary adjusting methods for simulating the color temperature, hues and brightness of daylight illumination with RGB LEDs for indoor lighting: errata," Opt. Express 19, 18318-18318 (2011)
https://opg.optica.org/oe/abstract.cfm?uri=oe-19-19-18318

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Figures (5)

Fig. 1
Fig. 1 Layout of the light box above the indoor space.
Fig. 2
Fig. 2 CIE1931 chromaticity diagram.
Fig. 3
Fig. 3 Simulated daylight illumination for indoor lighting in early morning.
Fig. 4
Fig. 4 Simulated daylight illumination for indoor lighting at 10 a.m.
Fig. 5
Fig. 5 Simulated daylight illumination for indoor lighting at noon.

Equations (9)

Equations on this page are rendered with MathJax. Learn more.

X = ( x y ) Y
Z = ( z y ) Y ,
[ X Y Z ] = [ 2.7689 1.7517 1.1302 1.0000 4.5907 0.0601 0.00000 0.0565 5.5943 ] [ R G B ] ,
= [ 0.041844 0.15866 0.08283 0.09117 0.25242 0.01570 0.00092 0.00255 0.17858 ] [ X Y Z ] .
x 3 = Y 1 Y 1 + Y 2 x 1 + Y 2 Y 1 + Y 2 x 2
y 3 = Y 1 Y 1 + Y 2 y 1 + Y 2 Y 1 + Y 2 y 2 ,
x 2 = ( Y 1 + Y 2 ) x 3 Y 1 x 1 Y 2 ,
y 2 = ( Y 1 + Y 2 ) y 3 Y 1 y 1 Y 2 ,
Y 2 = Y 3 Y 1 .
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