about lossy compression in image JPEG..

1. Lossy Compression
By:
Mahmoud Hikmet Bzhar Omer
Supervisor:
Dr.Roojwan Sdiq

2. Overview
• What's Comparession ?
• What’s lossless and lossy compression ?
• What’s JPEG?
• The Major Steps in JPEG Coding involve:
 Transform RGB to YIQ or YUV and subsample color.
 DCT(Discrete Cosine Transformation).
 Quantization.
 Zig-zag ordering
 DPCM on DC component
 Run-length encoding.
 Entropy coding.

3. What is Comparession ?
• Compression is the reduction in size of data
in order to save space or transmission time.
Learn how files are compressed .

4. What’s lossless and lossy
compression ?
• Lossless: The compression of a file, all original
data can be recovered when the file is
uncompressed.
• Lossy : -The compressed data is not the
same as the original data, but a close
approximation of it.

5. What is JPEG?
• "Joint Photographic Expert Group" -- an
international standard in 1992.
• Works with colour and greyscale images, Many
applications e.g., satellite, medical, ..

6. JPEG compression involves the following:

7. DCT : Discrete Cosine Transform
• DCT converts the information contained in a block(8x8)
of pixels from spatial domain to the frequency domain.
1-D DCT: 1-D Inverese DCT:





1N
0n
2N
1)(2n
f(n)cos
2
a(u))F( 
 




1N
0
2N
1)(2n
)cosF(
2
a(u))(f’


n
 0p1a(p)
2
1a(0)



8. Examplesubimage2*2
154 123
192 186
D
Subtract 128 from each value to convert to signed
26 -5
64 58
D
First Row= 2
1
secondRow= 2N
1)(2n
cos

N/2

9. 1- = 0.7071
2-
2
1
N*2
1)(2n
cos

N/2
D(1,0)= 2*2
*1)0*(2
cos
1
2/2
D(1,0)= 4
cos

D(1,0)=0.7071
D(1,1)= 2*2
*1)1*(2
cos
1
2/2
D(1,1)= 4
cos
3
D(1,1)=-0.7071
0.7071 0.7071
0.7071 -0.7071
T

10. 0.7071 0.7071
0.7071 -0.7071
0.7071 0.7071
0.7071 -0.7071
26 -5
64 58* *
T D -T
63.639 33.234
-26.87 -40.305
0.7071 0.7071
0.7071 -0.7071*
68.5 21.5
-47.5 9.5DCT=

11. Quantization:
• The quantization step is the main source for loss in JPEG
compression
• Encoder: Each value in the current block is divided by 16
and rounded down to create the quantised block.
• Round(DCT/Q)
68.5 21.5
-47.5 9.5
16 11
12 12
Q
4 2
-4 1
4 2
-4 1
QDCT

12. Quantization:
• The quantization step is the main source for loss in JPEG
compression.
• Decoder: Each value in the quantised block is multiplied
by quntize block.
4 2
-4 1
QDCT
16 11
12 12*
64 22
-48 12
Q-1

13. • DCT-1=round(T*Q-1*T)+128
DCT-inverse
0.7071 0.7071
0.7071 -0.7071
0.7071 0.7071
0.7071 -0.7071
26 -5
64 58
T Q-1 -T
* *
11.314 24.041
79.195 7.071 *
0.7071 0.7071
0.7071 -0.7071
=
=
= 25 -9
61 51
+
128 128
128 128
153 119
189 179
=

14. 154 123
192 186
D
- 153 119
189 179
DCT-1
1 4
3 7
=

15. Zig-Zag Scan
• Why? to group low frequency coefficients in top of vector and high
frequency coefficients at the bottom
−26, −3, 0, −3, −2, −6, 2, −4, 1, −4, 1, 1, 5, 1, 2, −1, 1, −1, 2, 0, 0, 0,
0, 0, −1, −1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, EOB

16. • The 1x64 vectors have a lot of zeros in them, more so towards the end
of the vector.
• Higher up entries in the vector capture higher frequency (DCT) components
which tend to be capture less of the content.
• Could have been as a result of using a quantization table
• Encode a series of 0s as a (skip,value) pair, where skip is the number of
zeros and value is the next non-zero component.
• Send (0,0) as end-of-block sentinel value.
. . .
1x64
0 0 0 0 0 1 1 0 0 0 0 0
5,1
0 0
7,2
0 . . .2
RLEon AC Components
Run-length encode:
−26, −3, 0, −3, −2, −6, 2, −4, 1, −4, {2 x 1}, 5, 1, 2, −1, 1, −1, 2, {5 x
0} , −1, −1,

17. • Is based on the frequency of occurance of data item(pixel in image).
• The principle is to use a lower number of bits to encode the data occurs more
frequently.
H(x)= 𝑖=1
𝑛
𝐿𝑃𝑥𝑖
𝑃𝑥𝑖=log 2 xi
𝐿 number of bit for each character.
𝑃𝑥𝑖 entropy for each character.
Huffman Coding

18. Example
Symbol Xi sorting Xi Symol Code number ofbit
______ __ ______ _____ _____ ___________
A 0.3 0.3 A 00 2
B 0.2 0.23 C 01 2
C 0.23 0.2 B 11 2
D 0.07 0.15 E 010 3
E 0.15 0.07 D 0110 4
F 0.05 0.05 F 1110 4
H(x)= 𝑖=1
𝑛
𝐿𝑃𝑥𝑖
=2log 2 0.3 + 2log 2 0.23 + 2log 2 0.2 + 3log 2 0.15 +
4log 2 0.07 + 4log 2 0.05= -0.21

19. Thank you