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

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Page 1: lossy compression JPEG

Lossy Compression

By:Mahmoud Hikmet Bzhar Omer

Supervisor:Dr.Roojwan Sdiq

Page 2: lossy compression JPEG

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.

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What is Comparession ?

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

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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.

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What is JPEG?

• "Joint Photographic Expert Group" -- an international standard in 1992.• Works with colour and greyscale images, Many

applications e.g., satellite, medical, ..

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JPEG compression involves the following:

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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

0n2N

1)(2nf(n)cos2

a(u))F(

1N

02N

1)(2n)cosF(2

a(u))(f’

n

0p 1a(p)21a(0)

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Example subimage 2*2

154 123

192 186D

Subtract 128 from each value to convert to signed

26 -5

64 58D

First Row=21

second Row= 2N1)(2ncos

N/2

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1- = 0.7071

2-

21

N*21)(2ncos

N/2

D(1,0)=2*2

*1)0*(2cos 12/2

D(1,0)=4

cos

D(1,0)=0.7071

D(1,1)=2*2

*1)1*(2cos 12/2

D(1,1)=4

cos 3

D(1,1)=-0.70710.7071 0.7071

0.7071 -0.7071T

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0.7071 0.7071

0.7071 -0.7071

0.7071 0.7071

0.7071 -0.707126 -564 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=

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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

Q4 2

-4 1

4 2

-4 1QDCT

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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 1QDCT16 11

12 12*

64 22

-48 12Q-1

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• DCT-1=round(T*Q-1*T)+128

DCT-inverse

0.7071 0.7071

0.7071 -0.7071

0.7071 0.7071

0.7071 -0.707126 -564 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=

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154 123

192 186

D

-153 119

189 179

DCT-1

1 4

3 7=

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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

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• 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

RLE on 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 x0} , −1, −1,

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• 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)==log 2 xi

number of bit for each character. entropy for each character.

Huffman Coding

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Example

Symbol Xi sorting Xi Symol Code number ofbit______ __ ______ _____ _____ ___________A 0.3 0.3 A 00 2B 0.2 0.23 C 01 2C 0.23 0.2 B 11 2D 0.07 0.15 E 010 3E 0.15 0.07 D 0110 4F 0.05 0.05 F 1110 4

H(x)= =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

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Thank you