Zonal coding
keep
Information theory says coefficients
with maximum variance carry the
most information.
15 out of 64 transform coefficients
, with largest variance, are kept
discard
©2001 Bijan Mobasseri
76
Quantization
• Retained coefficients are quantized and
coded
• Coefficients are either normalized by their
standard deviation and uniformly quantized
or put through an optimum Lloyd-Max
quantizer
©2001 Bijan Mobasseri
77
Bit allocation
• How many bits should be assigned to each
coefficient?
• Information theory tells us, that a Gaussian
random variable, subject to distortion D,
cannot be represented by fewer than
 2 
1
log2   bits
2
 D 
• Therefore, large variance coefficients need
more bits
©2001 Bijan Mobasseri
78
Bit allocation table
8 7 6 4 3 2 1 0
7 6 5 4 3 2 1 0
6 5 4 3 2 1 1 0
4
3
2
1
0
Number of bits per coefficient
©2001 Bijan Mobasseri
79
Problem with zonal coding
• If we rope off a fixed area we run the risk of
cutting off occasional transform coefficients
that are large enough to be included
excluded
©2001 Bijan Mobasseri
80
Threshold coding
• One solution is to threshold coefficients and
possibly use different thresholds for
different subimages
• This way we will end up keeping those
coefficients that exceed threshold, wherever
they might be
©2001 Bijan Mobasseri
81
Picking the N-largest
transform coefficients
• In a 2D array, how do we pick N-largest
coefficients.
• Answer: zig-zag scan. Coefficients are
rearranged in a 1D- sequence
0 1 5 6 14
2 4 7 1316
3 8 121725
9 11182431
©2001 Bijan Mobasseri
82
Threshold details
• How do we threshold the transform
coefficients?
• There are 3 ways
– Single global threshold for all subimages
– Different thresholds for different subimages
– Different threshold for different locations in the
subimage
©2001 Bijan Mobasseri
83
Issue of code rate
• If we always keep the same number of
coefficients and bit allocation table, we will
be operating under a constant code rate that
is also known in advance.
• If we keep different number of coefficients
for different subimages, code rate will be
variable
©2001 Bijan Mobasseri
84
Combining thresholding and
quantization
• Let T(u,v) be the transform coefficient
matrix. Let Z(u,v) be the normalization
array. Then define
16 11 10 16 24 40 51 61
T(u, v) 
Tˆ u,v  round 

Z u,v
12 12 14 19 26 58 60 55
14 13 16 24 40 57 69 56
Z(u,v)=
14 17 22 29 51 87 80 62
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
49 64 78 87 103 121 120 101
©2001 Bijan Mobasseri
72 92 95 98 112 100 103 99
85
Quantization curve
Tˆ u,v
2
1
-2c
-c
c
2c
For Z=c, Tˆ u,v
assumes integer value k
if and only if
c
c
kc   T(u, v)  kc 
2
2
T(u,v)
If Z(u,v)>2T(u,v),Tˆ u,v=0
©2001 Bijan Mobasseri
86
Bit rate question
• As k increases, Tˆ u,v is represented by a
variable length code
• The number of bits used to represent T is
controlled by c
• Thus the elements of Z can be scaled to
achieve a variety of compression levels
©2001 Bijan Mobasseri
87
VECTOR
QUANTIZATION
©2001 Bijan Mobasseri
88
DEFINING THE “VECTOR” IN
VQ
• An n=lxm block of pixels is converted into
an n dimensional vector X
X
©2001 Bijan Mobasseri
89
THEORETICAL BASIS
• There is an information-theoretic basis for
vector quantization
• An extended source can be source encoded
with arbitrarily to the source entropy for
large vector length
1
H X   R  H X  
n
©2001 Bijan Mobasseri
90
Achieving the limit
• Let each pixel be represented by k bits.
Then there are a total of 2nk possible code
vectors
• To transmit/store all such vectors requires
log2(2nk)/n pixels which puts us right back
at k bits per pixel
• Solution: Codebook
©2001 Bijan Mobasseri
91
CODEBOOK CONCEPT
• Codebook approach is a many-to-one
transformation of image vectors.
©2001 Bijan Mobasseri
92
MAPPING RULE
• X is compared to a codebook given by
Xˆ i,i  1,...,N c
• Nc is a lot fewer than all possible image
vectors
• The best match codevector is chosen using a
minimum distortion rule; i.e.
dX, Xˆ k  dX, Xˆ j j 1,...N c
©2001 Bijan Mobasseri
93
DISTANCE MEASURE
• Distance measure is the Euclidean distance
between two vectors given by
1 n
2
ˆ
dX, X    x i  xˆ i 
n i1
©2001 Bijan Mobasseri
94
IMPACT ON BIT RATE
• Only the entry into the codebook, via index
K, is transmitted at the cost of log2 N c bits
per vector
• The resulting bit rate of a VQ is
R  log2 N c  n bits pixel
• With no VQ we would need k bits/pixel
©2001 Bijan Mobasseri
95
EXAMPLE: BINARY IMAGE
• Image vector X formed by grouping a 3x3
window
101011100
• There are a total of 29 possible image
vectors
[1 1 0 1 0 1 1 0 1],
[ 0 0 01 1 0 0 1 0 ],
[1 0 1 0 1 0 1 1 1 ]
?
101011100
Which entry is closest?
[0 0 0 0 0 0 0 0 0 ]
Example codebook
©2001 Bijan Mobasseri
96
COMPRESSION
• Straight encoding needs 1 bit/pixel
• Here we will transmit only an index into the
codebook; one of 4
• Effective bits/pixel then is
log 2 Nc  n bits pixel 
log 2 (4) / 9  0.222bits / pixel
©2001 Bijan Mobasseri
97
A NUMERICAL EXAMPLE
• Want to encode an 8-bit image using VQ at
a desired rate of 1 bit/pixel (8:1
compression)
• If the vectors are formed as 2x2 blocks(n=4)
of image pixels, and the codebook contains
16 codevectors, Nc=16.
• We will then have R=log216/4=1 bit/pixel
©2001 Bijan Mobasseri
98
INCREASING VECTOR SIZE
• Only 16 codevectors are available to
represent all the 2564 possible image
vectors
• If block size is increased to 4x4(n=16), the
codebook size has to increase to 216
codevectors to keep the bit rate at 1.0
bit/pixel
• The number of possible image vectors for a
Bijan16
Mobasseri
99
4x4 block is now©2001256
CONCLUSION
• It appears that ratio of codebook size to
image vectors decreases, pointing to
possibly large distortion
• It can be shown, however, that mean square
distortion actually goes down as vector
dimension increases.
VQ at an effective rate of 1bit/pixel
outperforms a straight 1bit/pixel
encoding
©2001 Bijan Mobasseri
100
A PRACTICAL EXAMPLE
• The most commonly used test image is that
of “Lena”.
©2001 Bijan Mobasseri
101
VQ PARAMETERS
•
•
•
•
Original image:8 bits/pixel
Block size: 4x4(16 pixels)
Possible number of vectors:25616=2128
Codebook size: 215
©2001 Bijan Mobasseri
102
COMPRESSION
• Mean of each block is quantized to 8 bits
for a rate of 8/16=0.5 bits/pixel
• Total of 215codevectors are represented by
15 bits.
• This amounts to 15 bits/16pixels=15/16
bits/pixel.
• Total bit rate: 0.5+15/16=1.438 bits/pixel
©2001 Bijan Mobasseri
103
Predictive Compression
Using the past to predict the future
©2001 Bijan Mobasseri
104
What is the idea?
• The familiar PCM encoding does A/D
conversion but very inefficiently. Voice, for
example, is coded at 64 Kb/sec.
• The reason is samples are treated without
regard to past sample values
• Therefore, huge redundancy is built into
PCM
©2001 Bijan Mobasseri
105
Differential PCM(DPC M)
• Concept of differential encoding is of great
importance in communications
• The underlying idea is not to look at
samples individually but to look at past
values as well.
• Often, samples change very little thus a
substantial compression can be achieved
©2001 Bijan Mobasseri
106
Why differential?
• Let’s say we have a DC signal and blindly
go about PCM-encoding it. Is it smart?
• Clearly not. What we have failed to realize
is that samples don’t change. We can send
the first sample and tell the receiver that the
rest are the same
©2001 Bijan Mobasseri
107
Definition of differential
encoding
• We can therefore say that in differential
encoding, what is recorded and ultimately
transmitted is the change in sample
amplitudes not their absolute values
• We should send only what is NEW.
©2001 Bijan Mobasseri
108
Where is the saving?
• Consider the following two situations
2
2
1.6
1.6
2
2
1.6
2
0.4 -0.8 -0.4
0.8
0
-0.4
0.4
0.8
• The right samples are adjacent sample
differences with much smaller dynamic
range requiring fewer quantization levels
©2001 Bijan Mobasseri
109
Implementation of
DPCM:prediction
• At the heart of DPCM is the idea of
prediction
• Based on n-1 previous samples, encoder
generates an estimate of the nth sample.
Since the nth sample is known, prediction
error can be found. This error is then
transmitted
©2001 Bijan Mobasseri
110
Illustrating prediction
• Here is what is happening at the transmitter
To be trasmited
Prediction error
Past samples(already sent)
Prediction of the
Current sample
Only Prediction error is sent
©2001 Bijan Mobasseri
111
What does the receiver do?
• Receiver has the identical prediction
algorithm available to it. It has also
received all previous samples so it can make
a prediction of its own
• Transmitter helps out by supplying the
prediction error which is then used by the
receiver to update the predicted value
©2001 Bijan Mobasseri
112
Interesting speculation
• What if our power of prediction was
perfect? In other words, what if we could
predict the next sample with no error?.
What kind of communication system would
be looking at?
©2001 Bijan Mobasseri
113
Prediction error
• Let m(t) be the message and Ts sample
interval, then prediction error is given
e(nTs )  m(nTs )  mˆ nTs 
Prediction error
©2001 Bijan Mobasseri
114
Prediction filter
• Prediction is normally done using a
weighted sum of N previous samples
N
mˆ nTs    wi mn  i Ts 
i1
• The quality of prediction depends on the
good choice of weights wi
©2001 Bijan Mobasseri
115
Finding the optimum filter
• How do you find the “best” weights?
• Obviously, we need to minimize the
prediction error. This is done statistically
2
e
Min  nTs 
over w
• Choose a set of weights that gives the
lowest (on average) prediction error
©2001 Bijan Mobasseri
116
Prediction gain
• Prediction provides an SNR improvement
by a factor called prediction gain
 2M
message power
Gp  2 
 e prediction error power
©2001 Bijan Mobasseri
117
How much gain?
• On average, this gain is about 4-11 dB.
• Recall that 6 dB of SNR gain can be
exchanged for 1 bit per sample
• At 8000 samples/sec(for speech) we can
save 1 to 2 bits per sample thus saving 8-16
Kb/sec.
©2001 Bijan Mobasseri
118
DPCM encoder
Input sample +
+
Prediction
- error
quantizer
Prediction error
encoder
+
Prediction
N-tap
prediction
Updated prediction
• Prediction error is used to correct the
estimate in time for the next round of
prediction
©2001 Bijan Mobasseri
119
Closer look at prediction
engine
• Two questions need to be addressed: 1):
How many past samples to use and 2): what
are the best weights?
• Notations:
x(n)
sample to be estimated
x(n  1), x(n  2),
previous samples
xˆ n  estimated sample
d n
prediction error = x n   xˆ n
©2001 Bijan Mobasseri
120
Encoder diagram
x n 
d n

d˜ n 
Quantizer
xˆ n 
N-tap
predictor
©2001 Bijan Mobasseri
x˜ n 

121
DPCM Decoder
dˆn
x˜ n 

xˆ n 
N-tap predictor
Using the same predictor, encoder produces an estimate of x(n).
Then uses received prediction error to update its estimate
©2001 Bijan Mobasseri
122
One-tap prediction
• This is the simplest predictor. We use the
current sample to predict the next one
x n n  1  axn  1n  1
notation:
x n m  estimate of the nth sample given
samples collected up through time m
• What is a?
©2001 Bijan Mobasseri
123
Minimizing prediction error
• Pick “a” so that prediction error is
minimized
dn  xn  xˆn  x n  ax n  1n  1
• Must minimize the power of prediction
error by choosing the right multiplier
2
Ed 2 n Ex n   axn  1n  1 
 Rx 0   2a Rx 1  a 2 Rx 0 
signalpower
onesample
correlation
©2001 Bijan Mobasseri
124
Finding a
• Take the derivative of error and set it to
zero. The optimum a is
Rx 1 onesmapelcorrelatin
a

Rx 0 
signalpower
• The resulting minimum error
  R 1 2 
Rx 01   x  

 Rx 0 

©2001 Bijan Mobasseri
125
Example
• Clearly we need source statistics to perform
DPCM. There is 80% sample correlation
0.8
Rx
a=0.8,…x(n|n-1)=0.8x(n-1|n-1)
1
n
Rx 1 onesmapelcorrelatin
a

Rx 0 
signalpower
©2001 Bijan Mobasseri
126
N-tap prediction
• N consecutive samples are weighted and
summed. Samples are T seconds apart
delay
x(n)
T
T
a1
T
a2
aN
N
xˆ n    a j x n  j 
j 1
 aT X

©2001 Bijan Mobasseri
127
Finding a
• Take N derivatives of error and set it to
zero. The result is the following matrix
equation
Rx 1
Rx 2
 Rx 1  Rx 0
R 2    R 1
Rx 0 
Rx 1
 x   x
Rx 3  Rx 2
Rx 1
Rx 0 

 

 

Rx  N 
 
Rx  N  Rx  N  2  Rx  N  3
Rx 1  N  a1 
Rx 2  N a2 
 
Rx 3  N a3 
 
 
Rx 0  
a N 


 Rx 1, N   R x a  a  R 1
x R x 1, N 
©2001 Bijan Mobasseri
128
DPCM for images
• For images we have to do 2D prediction.
We also need to know the 2D correlation
function. A bit more involved!
?
• Want to find the center pixel from its
neighboring values
©2001 Bijan Mobasseri
129
Special case
• For a 2D Markov source the autocorrelation
function at shifts (i,j) is defined
Ri, j   E f x, y f x  i, y  j 
  2 ix  yj
(i,j)
(0,0)
©2001 Bijan Mobasseri
130
Prediction for Markov source
• It can be shown that the optimal 2D linear
predictor is
fˆ x, y  1 f x, y  1  2 f x  1, y  1 
3 f x  1, y   4 f x  1, y  1
• Where
horizontal
correlation
1  x , 2   x  y ,3  y , 4  0
©2001 Bijan Mobasseri
131
JPEG
©2001 Bijan Mobasseri
132
JPEG standard
• The concept of JPEG had been known for at
least 30 years before it became a widely
adopted standard in the early 90’s
• JPEG is a block transform compression that
uses DCT
• OF all practical transforms, DCT has the
most energy compaction property
©2001 Bijan Mobasseri
133
Elements of JPEG
• JPEG closely follows DCT with a few
caveats.
–
–
–
–
–
–
8x8 DCT block transform
Perceptual masking
Zigzag ordering
Transform coefficient quantization
Run-length encoding compression
Variable-length encoding
©2001 Bijan Mobasseri
134
8x8 block DCT
• Image is subdivided into non overlapping
8x8 blocks
• Pixels are level shifted by subtracting 2n-1
where 2n is the maximum number of gray
levels
©2001 Bijan Mobasseri
135
Quantization
• The standard quantization table is applied to
each 8x8 DCT block
DC block
T(u, v) 
Tˆ u,v  round 

Z u,v
16 11 10 16 24 40 51 61
12 12 14 19 26 58 60 55
14 13 16 24 40 57 69 56
Z(u,v)=
14 17 22 29 51 87 80 62
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
49 64 78 87 103 121 120 101
©2001 Bijan Mobasseri
72 92 95 98 112 100 103 99
136
Re-ordering
• DCT coefficients are reordered according to
the following pattern
0 1 5 6 14
2 4 7 1316
3 8 121725
9 11182431
©2001 Bijan Mobasseri
137
Coding
• Zigzag scanning following quantization
insures long runs of AC coefficients
(indicating lack of high frequency content)
• DC coefficient is differentially encoded
• Non-zero AC coefficients are variable
length encoded (Huffman)
©2001 Bijan Mobasseri
138
JPEG example
• Consider the following hypothetical 8x8
block
52 55 61
66
70
61
64 73
63 59 66
90
109
85
69 72
1
2
62 59 68 113 144 104 66 73
63 58 71 122 154 106 70 69
67 61 68 104 126
88
68 70
79 65 60
70
77
68
58 75
85 71 64
59
55
61
65 83
87 79 69
68
65
76
78 94
3
4
5
6
7
©2001 Bijan Mobasseri
8
1
2
3
4
5
6
7
8
139
Level shifting
• The coding process begins by level shifting
the pixels by -128 gray levels
-76 -73 -67 -62 -58 -67 -64 -55
-65 -69 -62 -38 -19 -43 -59 -56
-66 -69 -60 -15 16 -24 -62 -55
-65 -70 -57 -6 26 -22 -58 -59
-61 -67 -60 -24 -2 -40 -60 -58
-49 -63 -68 -58 -51 -60 -70 -53
-43 -57 -64 -69 -73 -67 -63 -45
-41 -49 -59 -60 -63 -52 -50 -34
©2001 Bijan Mobasseri
140
Do the DCT
• Here are the transform coefficients
-415
7
-46
-50
11
-10
-4
-1
-29
-21
8
13
-8
1
-1
-1
-62
-62
77
35
-13
3
2
-1
25 55 -20 -1 3
9 11 -7 -6 6
-25 -30 10 7 -5
-15 -9 6 0 3
-2 -1 1 -4 1
-3 -1 0 2 -1
-1 2 -3 1 -2
-2 -1 -1 0 -1
©2001 Bijan Mobasseri
141
Quantization, scaling and
truncation
• JPEG standard normalization array is given by the following(left)
truncated coefficients are shown on the right
16 11 10 16 24 40 51 61
12 12 14 19 26 58 60 55
14 13 16 24 40 57 69 56
14 17 22 29 51 87 80 62
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
49 64 78 87 103 121 120 101
72 92 95 98 112 100 103 99
-26 -3 -6 2 2
1 -2 -4 0 0
-3 1 5 -1 -1
-4 1 2 -1 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
©2001 Bijan Mobasseri
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
142
Zigzag ordering
• We then make a 1D sequence according to
the zigzag pattern
-26 -3 1 -3 -2 -6 2 -4 1 -4 1 1 5 0 2 0 0 -1 2 0 0 0 0 0 -1 -1 EOB
• Anything after EOB
isMobasseri
taken as zero
©2001 Bijan
143
Coding the DC coefficient
• The DC coefficient is differentially encoded
relative the DC average of the previous
block
• Let the block to the immediate left have a
DC coefficient of -17
• Differential is [-27-(-16)]=-9
• What do we do with this now?
©2001 Bijan Mobasseri
144
JPEG coefficient coding table
JPEG Coefficients Coding Categories
RANGE
0
-1,1
-3, -2, 2, 3
-7,…, -4, 4, …, 7
-15,…, -8, 8,…, 15
-31,…,-16, 16,…, 31
-63,…, -32, 32,…, 63
DC difference
category
A/C category
0
1
2
3
4
5
6
N/A
1
2
3
4
5
6
©2001 Bijan Mobasseri
145
Huffman coding of categories
– DC block differential was -9. This lies in
difference category 4. But where do we go from
here?
– Difference categories are variable length
Huffman coded as shown in the next table
©2001 Bijan Mobasseri
146
Huffman coding of categories
Category
Base code
Length
Category
Base Code
Length
0
1
2
3
010
011
100
00
3
4
5
5
6
7
8
9
10
12
14
16
4
101
7
A
5
110
8
B
1110
11110
111110
111111
0
111111
10
111111
110
18
20
Length is the final length in bits of each category
©2001 Bijan Mobasseri
147
Huffman code of DC block
differential
• We had -9. This value belongs to category 4
giving rise to 101
• The remaining 4 bits come from the least
significant bits(LSB) of the difference value
• For category k, an additional k bits are
computed as k LSBs of of the positive
difference (+9), 0110 In this case
• Full DPCM coded DC block is 1010110
©2001 Bijan Mobasseri
148
AC coefficients code table
Each coeff.depends on the number of zero-valued coeff.
preceding the non-zero ones
-26 -3 1 -3 -2 -6 2 -4 1 -4 1 1 5 0 2 0
0 -1 2 0 0 0 0 0 -1 -1 EOB
JPEG Default AC Code
Run/Category
0/0
0/1
0/2
0/3
0/4
0/5
Base Code
1010 (=EOB)
00
01
100
1011
11010
.
.
Coded as 0100
.
.
.
.
First 2bits (01):category 2 0/A
1111111110000011
1/1
1100
with no zero-valued coeff 1/2
111001
1/3
1111001
preceding it
Last two are found the same 1/A
1111111110001000
©2001 Bijan Mobasseri
way as in DC
Length
4
3
4
6
8
10
.
.
.
26
5
8
10
26
149
Final JPEG bitstream
• For the 8x8 block the following represents
its JPEG bitstream
[1010110 0100 001 0100 0101 100001 0110 100011 001 100011
001 001 100101 11100110 110110 0110 11110100 000 1010]
• Total number of bits: 92
• Original block: 512
• Compression ratio: 5.6:1
©2001 Bijan Mobasseri
150
Decoding JPEG
• Via table lookup operation, the quantized
coefficient matrix is quickly recreated
• Denormalization is done by simply
multiplying the normalized matrix above by
the normalization matrix
T(u, v) 
Tˆ u,v  round 

Z u,v
Not the same
T˜ u,v  Tˆ u,vZ u,v
©2001 Bijan Mobasseri
151
Recovered coefficients
-415
7
-46
-50
11
-10
-4
-1
-29
-21
8
13
-8
1
-1
-1
-62
-62
77
35
-13
3
2
-1
25 55 -20 -1 3
9 11 -7 -6 6
-25 -30 10 7 -5
-15 -9 6 0 3
-2 -1 1 -4 1
-3 -1 0 2 -1
-1 2 -3 1 -2
-2 -1 -1 0 -1
-416
12
-42
-56
18
0
0
0
original
-33 -60
-24 -56
13 80
17 44
0
0
0
0
0
0
0
0
32 48 0 0 0
0 0 0 0 0
-24 -40 0 0 0
-29 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
recovered
©2001 Bijan Mobasseri
152
Recovering block
52 55 61
66
70
61
64 73
58 64 67
64
59
62
70 78
63 59 66
90
109
85
69 72
56 55 67
89
98
88
74 69
62 59 68 113 144 104 66 73
60 50 70 119 141 116 80
64
63 58 71 122 154 106 70 69
69
67 61 68 104 126
88
68 70
74 53 64 105 115
84
65 72
79 65 60
70
77
68
58 75
76 57 56
74
75
57
57 74
85 71 64
59
55
61
65 83
83 69 59
60
61
61
67 78
87 79 69
68
65
76
78 94
93 81 67
62
69
80
84 84
51 71 128 149 115 77 68
Original
JPEG
©2001 Bijan Mobasseri
153
Color JPEG
• A color image can be represented in a
variety of spaces including RGB and YUV
– RGB image consists of 3 primary color
channels- red, green, blue
– YU V (also known as YIQ or YCbCr) color
model consists of an intensity component Y
(grayscale and two chrominance channels
©2001 Bijan Mobasseri
154
RGB planes
©2001 Bijan Mobasseri
155
Another look
R
G
B
Recording the color image in gray using red, green and blue filters
©2001 Bijan Mobasseri
156
YUV model
©2001 Bijan Mobasseri
157
Color quantization table
• Color bands are much more aggressively
compressed than luminance. Here is the
quantization table
16 11 10 16 24 40 51 61
17 18 24 47 99 99 99 99
12 12 14 19 26 58 60 55
18
14 13 16 24 40 57 69 56
24 26 56 99 99 99 99 99
14 17 22 29 51 87 80 62
47 66 99 99 99 99 99 99
18 22 37 56 68 109 103 77
24 35 55 64 81 104 113 92
99 99 99 99 99 99 99 99
99 99 99 99 99 99 99 99
49 64 78 87 103 121 120 101
99 99 99 99 99 99 99 99
72 92 95 98 112 100 103 99
99 99 99 99 99 99 99 99
Luminance table©2001 Bijan Mobasseri
21 26 66 99 99 99 99
Color table 158
Examples of JPEG-original
84,197 bytes
©2001 Bijan Mobasseri
159
JPEG-75%
36,401 bytes
©2001 Bijan Mobasseri
160
JPEG - 50%
22,112 bytes
©2001 Bijan Mobasseri
161
JPEG - 25%
13,907 bytes
©2001 Bijan Mobasseri
162
JPEG -15%
9,973 bytes
©2001 Bijan Mobasseri
163
JPEG -5%
5,662 bytes
©2001 Bijan Mobasseri
164
Foundation for
JPEG2000
Compression at multiples of scale
©2001 Bijan Mobasseri
165