Assignment 1
Chapter 1
1.5 For the following colors in the digital RGB coordinate, determine their valuse
in the YCbCr coordinate.
(a)(255, 255, 255) (b)(0, 255, 0) (c)(255, 255, 0) (d)(0, 255, 255)
Y
Cb
Cr
0.257
= -0.148
0.439
0.504
-0.291
-0.368
(a)
0.098
0.439
-0.071
R
G
B
(b)
Y
(c)
235.045
169.51
Cb =
128
=
165.74
Cr
128
16.055
+
16
128
128
Y
Cb
Cr
=
(d)
144.52
Y
53.795
Cb
34.16
Cr
=
210.055
Y
16.055
Cb
146.105
Cr
Please calculate the needed bit-rate (bits/s) of a non-coded video stream
with following formats:
1.
Color Coordinate: YCbCr
2.
Frame size: 352 x 288
3.
Frame rate: 30 fps
4.
Sampling format: (a) 4:4:4 (b) 4:2:2 (c) 4:2:0
(a) 352*288*(1+1+1)*30*8=727990720 bits/sec.
(b) 352*288*(1+1/2+1/2)*30*8=48660480 bits/sec.
(c) 352*288*(1+1/4+1/4)*30*8=36495360 bits/sec.
Chapter 2
2.5 Consider a horizontal bar pattern on a TV screen with 100 cycle/picture-height.
If the picture-height is 1 meter, and the viewer sits at 3 meters from the screen,
what is the equivalent angular frequency in cpd? What if the viewer sits at 1 meter
or 5 meters distance? In either case, would the viewer be able to perceive the vertical
variation properly?
When viewer sits at 3 meter from the screen, fθ=5.236 (cycle/degree)
When viewer sits at 1 meter from the screen, fθ=1.7453 (cycle/degree)
When viewer sits at 5 meter from the screen, fθ=8.7266 (cycle/degree)
The maximum sensitivity occurs at about 3~5 cpd, so d = 3 m will be best
2.6 Consider an object that has a flat, homogeneously textured surface with maximum
spatial frequency of (fx,fy)=(3,4) cycle/meter, and that is moving at a constant speed
of (vx,vy)=(1,1) meter/second. What is the temporal frequency of the object surface
at any point? What are the results for the following speeds (in
meter/second):(4,-3),(4,0),(0,1)?
When (fx, fy) = (3, 4), (vx, vy) = (1, 1)
ft,max = -(vx fx+vy fy) = -(1*3+1*4) = -7 cycle/second
When (fx, fy) = (3, 4), (vx, vy) = (4, -3)
ft,max = -(vx fx+vy fy) = -(4*3+(-3)*4) = 0 cycle/second
When (fx, fy) = (3, 4), (vx, vy) = (4, 0)
ft,max = -(vx fx+vy fy) = -(4*3+0*4) = -12 cycle/second
When (fx, fy) = (3, 4), (vx, vy) = (0, 1)
ft,max = -(vx fx+vy fy) = -(0*3+1*4) = -4 cycle/second
Assignment 2
Chapter 5
5.2 Show that, under orthographic projection, the projected 2-D motion of a planar patch
undergoing translation, rotation, and scaling (because of camera zoom) can be described by
an affine function.
x=X , y=Y
(5.1.3)
 X '  r 1 r 2 r 3   X 
Tx 
Y '  = 
Y  + Ty 

  r 4 r 5 r 6   
 
 Z '  r 7 r 8 r 9   Z 
Tz 
(5.5.12)
 orthographic projection  將(5.1.3)代入(5.5.12)
 planar patch  將 aX+bY+cZ=1 代入(5.5.12)
則 x’=r1X＋r2Y＋r3Z＋Tx
= r1x＋r2y＋r3 (
=( r1－
1  ax  by
)＋Tx
c
r 3a
r 3b
r3
)x＋( r2－
)y＋( Tx＋ )
c
c
c
dx(x, y)=x’－x=( r1－
=( Tx＋
r 3a
r 3b
r3
)x＋( r2－
)y＋( Tx＋ )－x
c
c
c
r3
r 3a
r 3b
)＋( r1－
－1)x＋( r2－
)y
c
c
c
r3
r 3a
=a0, r1－
－1=a1,
c
c
則 dx(x, y)= a0＋a1x＋a2y
令 Tx＋
r2－
而 y’= r4X＋r5Y＋r6Z＋Ty
= r4x＋r5y＋r6 (
1  ax  by
)＋Ty
c
r 3b
=a2
c
=( r4－
r 6a
r 6b
r6
)x＋( r5－
)y＋( Ty＋ )
c
c
c
dy(x, y)=y’－y=( r4－
=( Ty＋
r 6a
r 6b
r6
)x＋( r5－
)y＋( Ty＋ )－y
c
c
c
r6
r 6a
r 6b
)＋( r4－
)x＋( r5－
－1)y
c
c
c
r6
r 6a
=b0, r4－
= b1,
c
c
則 dy(x, y)= b0＋b1x＋b2y
令 Ty＋
dx( x, y )  a0  a1x  a2y 
推得 
=

dy ( x, y ) b0  b1x  b2y 
r5－
r 6b
－1= b2
c
(Affine Motion)
Chapter 6
6.1 Describe the advantages and disadvantages of different motion representation methods
(pixel-based, block-based, mesh-based, region-based, and global).
See sec 6.2.1
6.9 In Section 6.9.2, we derived the number of operations required by the HBMA, if the
search range at every level is R/2L-1. What will be the number if one uses a search range of ±1
pel in every level, except at the first level, where the search range is set to R/2L-1? Is this
parameter set-up appropriate?
L
( M / 2 L 1 ) 2 (2 R / 2 L 1  1) 2   (M / 2 L l ) 2 (2 R / 2 L 1  1) 2
l 2
 (M / 2
L 1 2
) (2 R / 2
L 1
 1)  12M 2 (1  41 L )
2
For higher levels, 1 is too small for sequences with high motions. Please discuss
how to solve this problem.