English Department Syllabus College English I

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• cours - matière potentielle : for course information
• expression écrite - matière potentielle : assignments
• cours - matière potentielle : work
• expression écrite - matière potentielle : assignment
• expression écrite - matière potentielle : faculty blackboard community
• leçon - matière potentielle : from the engl 1201 textbook
• exposé
• expression écrite
• leçon - matière potentielle : as the basis
• cours - matière : literature
• cours - matière potentielle : first year writing
• expression écrite - matière potentielle : class
• revision
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• expression écrite - matière potentielle : freshman
• cours - matière potentielle : syllabus

English Department Syllabus College English I 2006-2007 The department syllabus is designed to help instructors put together their own syllabi for ENGL 1201, College English I. It includes specific guidelines for teaching the course, entrance requirements, a model course, and suggested assignments. There will be an interactive on-line version of this syllabus available later in the summer. We suggest that faculty save this syllabus to their computers and use it as a template for creating their own syllabi.

• readings from the engl 1201 textbook
• writing
• assignments
• texts
• engl
• 3 engl
• journals
• class
• students
• work
• course

Sujets

• Assignment
• Exposé
• Literature

Informations

• Publié par: jazat
• Nombre de lectures: 34
• Langue: English
• Poids de l'ouvrage: 2 Mo

Extrait

¨ UNIVERSITAT DES SAARLANDES Prof.Dr.Philipp Slusallek Dr.Vincent Pegoraro Computer Graphics Group Piotr Danilewski ([e-mail])

12. December2011

Introduction to Computer Graphics Assignment 6

Solutions 6.1 Gammacorrection (25 Points) The worst case is obviously if settingγ= 2.highest accuracy is required near zero input, there we5. The get:

2.5 f(0.00000000b) = 0= 0 2.5 f(0.00000001b) = (1/5128) =.39E−6 −6 To represent the step from 0 to 5.39E-6 we need at least 18 bits for the fractional part (log 2 (5.39∙10 )= −17.5).

6.2 Transformsin color space (50 points)

a)

We can’t change direction of vectors red, green and blue since they depend on the monitor’s hardware characteristics.So the only way remaining to change the position of the white point is by changing the length of the vectors. Simplyspeaking: Wedimm the maximum brightness of one or more primary colors.

b) Overview: letC rgb = (r, g, b) be the color on the first monitor, then we want to know which color 0 Cwe need to use on the second monitor so that both monitors show the same color.With rgb formulars: 0 0 C cie =M∙C rgb andC cie =M∙C rgb 0 00 0−1 C⇔C M ⇒M∙C rgb =M∙ rgb rgb =∙M∙C rgb

Given:x r , y r , x g , y g , x b , y b , x w , y w and due to the normalization of the white point:Y w = 1 (see slide of computer graphics lecture)

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Color Transformations (Cont.) the constants C • Computing X – Per definition the white point is given as (R, G, B) = (1, 1, 1) X • ( w , Y , Z )= M*(1,1,1) w w X xC xC xC w rr gg bb 1 = Y yC yC yC1 w rr gg bb [ ] [ ][ ] 1 Z1−x−yC1−x−yC1−x−yC w rr rg gg bb b

(X ,Y ,Z )can be computed from (x , y ) – w w w x x • Usingthe normalization constant Y= 1 w

Computer Graphics WS2009/2010 – Color

Y w Y w Y w Then:X w =x w ∙ ≈0.9505 andZ w =z w ∙= (1−x w −y w )∙ ≈1.0891 y w y w y w and for the second monitor we get:X w2 ≈1.2121 andZ w2 = 0.8182 We need to computeC r , C g , C b : In the equation above, setz r = (1−x r −y r ), z g andz b analog for convenience issues.

        X w x r C r x g C g x b C b 1X w x r x g x b C r 10 0         Y w = y r C r y g C g y b C b 1 ⇔ Y w = y r y g y b 0C g 0 1 ⇔ Z w z r C r z g C g z b C b 1Z w z r z g z b 0 0C b 1      −1    X w x r x g x b C r x r x g x b X w C r        ⇔ Y w = y r y g y b C g ⇔ y r y g y b Y w = C g Z w z r z g z b C b z r z g z b Z w C b

Solving these equations yields: for the first white point:⇒C r = 0.6444, C g = 1.1919, C b = 1.2033 for the second white point:⇒C r = 1.2778, C g = 0.8763, C b = 0.8763

0 0−1 Now one can directly computeM, MandM:    0.4124 0.3576 0.1805 0.8178 0.2629 0.1314 0    M = 0.2127 0.7151 0.0722 M = 0.4217 0.5258 0.0526 0.0193 0.1192 0.9506 0.0383 0.0876 0.6922   1.63435−0.775302−0.251333   0−1 M= −1.31842,2.55168 0.0563736 0.07642−0.280025 1.45144

0−1 Another way to arrive at the final solution (M M) is to observe that since the primary colors are 0 the sameMandMcan be expressed as:

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     0 x r x g x b C r 0 0x r x g x b C0 0 r 0 0      M= y r y g y b 0C g 0 , M= y r y g y b 0C0 g 0 z r z g z b 0 0C b z r z g z b 0 0C b Thus :        C r −1 −1 0 0 0 0 0 x x xC0 0 r C r 0 x r x g x rb rg b C C g 0−1 0        0 0 0 M M = 0C0 y r y g y b y r y g y b 0C g 0 =  C  g g 0 C b 0 0 C z r z g z b z r z g z b 0 0C b b 0 0 0 C b Finally :   0.5043 00 0−1   M M = 01.36015 0 0 01.37315

6.3 ColorSpace (25 Points) For the CIE-XYZ color space one simply computes (X, Y, Z) =M∙(1,0,1) = (0.5929,0.2849,0.9699). By normalizing we get the coordinates in the CIE-xy color space:X/(X+Y+Z) = 0.32,Y /(X+Y+Z) = 0.15. Gammacorrection on (1,0,1) yelds (1,0,1) regardless of the gamma, which is why it is not necessary.

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