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English Department Syllabus College English I

3 pages
• 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 : process
• 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
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¨ UNIVERSITAT DES SAARLANDES Prof.Dr.Philipp Slusallek Dr.Vincent Pegoraro Computer Graphics Group Piotr Danilewski (danilewski@cs.uni-saarland.de)
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) =.39E6 6 To represent the step from 0 to 5.39E-6 we need at least 18 bits for the fractional part (log2(5.3910 )= 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: letCrgb= (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 Ccie=MCrgbandCcie=MC rgb 0 00 0−1 CC M MCrgb=Mrgb rgb=MCrgb
Given:xr, yr, xg, yg, xb, yb, xw, ywand due to the normalization of the white point:Yw= 1 (see slide of computer graphics lecture)
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Color Transformations (Cont.) the constants C ComputingX – Perdefinition 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 bb1 = Y yC yC yC1 w rr gg bb [ ] [ ][ ] 1 Z1xyC1xyC1xyC 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
YwYwYw Then:Xw=xw∙ ≈0.9505 andZw=zw= (1xwyw)∙ ≈1.0891 ywywyw and for the second monitor we get:Xw21.2121 andZw2= 0.8182 We need to computeCr, Cg, Cb: In the equation above, setzr= (1xryr), zgandzbanalog for convenience issues.
       XwxrCrxgCgxbCb1XwxrxgxbCr10 0        Yw=yrCrygCgybCb1Yw=yrygyb0Cg0 1ZwzrCrzgCgzbCb1Zwzrzgzb0 0Cb1     1   XwxrxgxbCrxrxgxbXwCr        Yw=yrygybCgyrygybYw=Cg ZwzrzgzbCbzrzgzbZwCb
Solving these equations yields: for the ﬁrst white point:Cr= 0.6444, Cg= 1.1919, Cb= 1.2033 for the second white point:Cr= 1.2778, Cg= 0.8763, Cb= 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 0.0722M= 0.4217 0.5258 0.0526 0.0193 0.1192 0.9506 0.0383 0.0876 0.6922   1.634350.7753020.251333   0−1 M=1.31842,2.55168 0.0563736 0.076420.280025 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 xrxgxbCr0 0xrxgxbC0 0 r 0 0      M=yrygyb0Cg0, M=yrygyb0C0 g 0 zrzgzb0 0Cbzrzgzb0 0C b Thus :       Cr 11 000 0 0x x xC0 0r Cr0xrxgxrb rg b C Cg 0−10      000 M M= 0C0yrygybyrygyb0Cg0 =Cg g 0 Cb 0 0C zrzgzbzrzgzb0 0Cb b0 00 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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