cs1354_may2007
TRANSCRIPT
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B.E / B.Tech Degree Examination May/June 2007
C 3160
CS1354 Computer Graphics and Multimedia
Class : III CSE Staff Name : Ms.V.Pushparani
Part A
1. Explain the Disadvantages of DDA line drawing algorithm Round-off error in successive additions of the floating-point increment can
cause the calculated pixel positions to drift away from the true line path for
long line segments.
Rounding operations and floating-point arithmetic in procedure are still time-consuming.
2.
Define Vanishing point with an exampleThe point at which a set of projected parallel lines appears to converge
3. Define MorphingTransformation of object shapes from one form to another is called morphing.4. Define View port with an example
An area on a display device to which a window is mapped is called a view port.
5. Define MultimediaMultimedia is defined as a Computer based Interactive Communication process that
incorporates text, numeric data, record based data, graphic art, video and audio elements,
animation etc. It is used for describing sophisticated systems that support moving images and
audio. Eg.Personal Computer.
6. List the Component of multimedia Facsimile Document Images
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Photographic Images Geographic Information System Maps(GIS) Voice Commands and Voice Synthesis Audio Messages Video Messages Full motion stored and Live Video Holographic Images Fractals
7. List any four Hardware requirements of multimediaScanners, video cameras, sound equipment, software device driver, VCRs etc.
8. What is MIDI?This is the interface standard for file transfer between a computer and a musical
instrument such as a digital piano.
9. List 3 Multimedia authoring toolsDedicated Authoring system, Timeline-Based Authoring, Structured Multimedia
Authoring, Programmable Authoring Systems, Multisource Multi-User Authoringsystems, Telephone Authoring Systems.
10.What is meant by Media communicationMany multimedia application are developed in workgroups, comprising instructional
designers, writer, graphic artists, programmers and musicians located in the same
office space or building. The clients computer may be thousands of miles distant,
requiring methods of good communication. Communication can be made by cables,
AIR, satellite among the work group members.
Part B
11.a. Explain the basic concepts of Midpoint ellipse algorithm. Derive the decisionparameter for the algorithm and write down the algorithm steps. (16)
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Condition for dy/dx = -1 : at (x0,y0) where x0*(ry)2
= y0*(rx)2
At this point we switch from incrementing x to decrementing y!
Algorithm:
while ( x*(ry)2
< y*(rx)2
) Dx = 1;
Dy = -1;
FE(x,y) = (ry)2*x
2+ (rx)
2*y
2(rx)
2*(ry)
2
inside the ellipse FE(x,y) < 0
on the boundary FE(x,y) = 0
outside the ellipse FE(x,y) > 0
We apply the midpoint algorithm in two regions:
Region 1: Dx = 1
Region 2: Dy = -1
Start with Region 1, then continue in Region 2.
For Region 1, draw the very first pixel at (0,ry)
Suppose (xk,yk) has just been drawn
Decision parameter for (xk+1,yk+1):
p1k = FE(xk+1,yk0.5)= (ry)2*(xk+1)
2+ (rx)
2*(yk-0.5)
2-(rx)
2*(ry)
2
if p1k < 0 then the midpoint for the next x is inside the ellipse implying that yk is
closerto the boundary than yk-1 -> choose yk+1 = yk
if p1k > 0 then the midpoint for the next x is outside the ellipse implying that yk -1
is
closer to the boundary than yk -> choose yk+1 = yk -1
In region 2, Dy = -1 and we check for the midpoint in the x-direction:
p2k = FE(xk+0.5,yk1)= (ry)
2*(xk+0.5)
2+ (rx)
2*(yk-1)
2-(rx)
2*(ry)
2
if p2k > 0 then the midpoint for the next y is outside the ellipse implying that xk is
closer to
the boundary than xk+1 -> choose xk+1 = xk
if p2k < 0 then the midpoint for the next y is inside the ellipse implying that xk +1 is
closer to the boundary than xk -> choose xk+1 = xk +1
Again calculate p2k+1 from p
2k: p
2k+1 = FE(xk+1+0.5,yk+1-1)
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= p2k2(rx)
2*(yk-1) + (rx)
2+(ry)
2*[(xk+1+0.5)
2 (xk+0.5)
2]
Initial value: last accepted point (x0,y0) in Region 1
11.b. i. Explain Two dimensional scaling and translation with an example. (8)Translation
Assume objects are constructed from points (x,y)
An point is translated in the 2D plane if the coordinates change according tox = x + tx , y = y + ty
T = (tx, ty) is the translation vector
Translations are rigid body motions Linear equations stay linear, hence straight linesremain straight: a*x + b*y +c = 0 -> a*x + b*y + c = 0
Reposition objects according to T and redraw the object (points, equations etc. havechanged)
Scaling
Scaling changes the size of an object Achieved by applying scaling factors sx and sy Scaling factors are applied to the X and Y
co-ordinates of points defining an objects
If the point (xf,yf) is to be the fixed point, the transformation is:x' = xf+ (x - xf) Sx
y' = yf+ (y - yf) SyThis can be rearranged to give:
x' = x Sx + (1 - Sx) xf
y' = y Sy + (1 - Sy) yfwhich is a combination of a scaling about the origin and a translation.
11.b.ii. Obtain a Transformation matrix for rotating an object about a specified
pivot point (8)
2D Rotation
A rotation R repositions points in the plane by moving them along a circular arc
centered at the rotation point (xr,yr) by a given positive rotation angle q:
Using polar coodinates we easily obtain
x = R*cos(f + q)
= R*cos(f)*cos(q) R*sin(f)*sin(q)
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y = R*sin(f + q)
= R*cos(f)*sin(q) + R*sin(f)*cos(q)
But x = R*cos(f) and y = R*sin(f), hence
x = x*cos(q) y*sin(q)
y = x*sin(q) + y*cos(q)
For a rotation with rotation point (xr,yr)
x = xr + (x-xr)*cos(q) (y-yr)*sin(q)y = yr + (x-xr)*sin(q) + (y-yr)*cos(q)
12.a. Explain 3D transformation with an example (16) Translation Scale Rotation Shear TP = (x + tx, y + ty, z + tz) SP = (sxx, syy, szz)
1000
100010
001
z
y
x
tt
t
1
zy
x
1000
000
000
000
z
y
x
s
s
s
1
z
y
x
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Positive Rotations are defined as follows:
Axis of rotation is Direction of positive rotation is
x y to z
y z to x
z x to yZ
Y
X Rotation about x-axis Rx()P
Rotation about y-axis Ry()P
Rotation about z-axis Rz()P
xy Shear: SHxyP
12.b.i. Design a Storyboard layout and accompanying key frame for an animation ofa single polyhedron. (9)A conventional animation is created in a fairly fixed sequence. The story for the
animation is written, then a story board is laid out. A story board is an animation in
outline form a high level sequence of sketches showing the structure and ideas of the
animation. The sound track is recorded, a detailed layout is produced and the sound track
is read that is the instants at which significant sounds occurs are recorded in order. The
detailed layout and the sound track are then correlated. Certain key frames of the
animation are drawn these are the frames in which the entities being animated are at
10000cossin0
0sincos0
0001
1
z
y
x
1000
0cos0sin
0010
0sin0cos
1
z
y
x
1000
010000cossin
00sincos
1
z
y
x
1000
0100
010
001
y
x
sh
sh
1
z
y
x
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extreme or characteristic positions, from which their intermediate positions can be
inferred.
Key frame animation Route sheet Exposure sheet
12.b.ii. How to specify Object motion in an animation system? (7) Direct motion specification
Specifying the geometric transformation parameters (ex. the rotation anglesand translation vectors)
Goal-directed systems Abstractly describing the actions in terms of the final results (the final state of
the motions)
Kinematics and dynamics Kinematics: motion parameters (position, velocity, and acceleration) without
reference to causes or goals of the motion
Inverse kinematics: specify the initial and final positions of object at specifiedtimes system computes motion parameters
Dynamics: specifyingforces that produce the velocities and accelerations13.a.i List the Multimedia application> Explain them briefly. (12)
Document Imagingo Document Image Hardware requirements
Image Processing and Image Recognitiono Image Enhancemento Image Calibrationo Real time alignmento Gray-Scale normalizationo RGB hue intensity adjustmento
Color Separation
o Frame averaging Image Animation Image Annotation Optical Character Recognition Handwriting Recognition Non-Textual Image Recognition Full-Motion Digital Video Applications
o Electronic Messaging
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A universal Multimedia Application Full-Motion Video Messages
o Viewer Interactive Live Video Audio and Video Indexing
13.a.ii. Briefly discuss the History and future of multimedia (4) Integrate different media, especially the audio and video in computers Interactive multimedia will be delivered to many homes throughout the world. Entertainment companies that own content early converted to multimedia projects
are teaming up with cable TV companies such as QVC or via com.
Film studios such as Disney and warner brothers are creating new divisions toproduce interactive multimedia.
13.b.i Explain the Characteristics of MDBMs (10)
RDBMS Extensions for Multimedia Object-Oriented Databases for Multimedia Extensibilityo Encapsulation
o Associationo Classification
Encapsulation inheritance
13.b.ii. Write short note on Multimedia system architecture (6)
High Resolution Graphics Displayo VGA mixingo VGA mixing with scalingo Dual-buffered VGA/Mixing/Scaling
14.a. List the type of Fixed and removable storage devices available for multimedia
and discuss the strengths and weakness of each one. (16)o Magnetic media technologieso Physical constructions of CDROMso Magnetic Disk organizationo CDROM standards
CD-DA(DD-Digital audio) Red book CD-ROM Mode 1 Yellow book CD-ROM Mode 2 Yellow book CD-I Green book CD-ROM XA CD-MO orange book Part I CD-R Orange Book Part II
o Mini disko WORM optical drives
Recording of information Reading information from disk Worm drive applications
o Rewritable optical disk technologies
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Magnetic optical Phase change
14.b. Explain the Data compression used in multimedia. (16)
CCITT Group 3 1-D Compressiono Huffman encoding
The principle of Huffman coding is to assign shorter code for symbol that has higherprobability of occurring in the data stream. The length of the Huffman code is optimal.
For example, a data stream has only five symbols ABCDE with the following
probabilities: A Huffman code tree is created using the following procedures: _ two
characters with the lowest probabilities are combined to form a binary tree. The two
entries in the probability table is replaced by a new entry whose value is the sum of the
probabilities of the two characters. Repeat the two steps above until there is only one
entry and a binary tree containing all characters is formed. Assign 0 to the left branches
and 1 to the right branches of the binary tree. The Huffman code of each character can be
read from the tree starting from the root.
A 011
B 1
C 000D 010
E 001
CCITT Group 3 2-D Compressiono Necessity of k factoro Data formatting for CCITT group 3 2D
CCITT Group 4 2-D Compression15.a.i Distinguish between Multimedia and hypermedia system (4)Multimedia- The combination of text graphic and audio elements into a single collection
of presentation becomes interactive multimedia when you give the user some control over
what information is viewed and when it is viewed.Hypermedia Interactive multimedia becomes hypermedia when its designer provides a
structure of linked elements through which a user can navigate and interact.
15.a.ii. List the main attributes, benefits and drawbacks of 3 Types of authoring
system (12)
Dedicated authoring system Structured media authoring system
o Construction of structure of a presentationo Assignment of detailed timing constraint
Programmable authoring system15.b.i. Write short notes on Mobile messaging (8)
Represents the major new dimensions into users interaction with the managingsystem.
With the power to handle email and manage calendars, reminders and schedulesthese devices will require increasingly complex multimedia solutions.
UMA covers voice mail, email, EDI, telex and enclosures which can be images,voice, video or other binaries.
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15.b.ii. Write short notes on Document management (8)
Primary document storage Linked object storage Linked object management Image and still video store Audio and full motion video store