lossy compression JPEG
Download as PPTX, PDF4 likes2,804 views
The document discusses JPEG image compression, which involves lossy compression. It describes the major steps in JPEG coding as: transforming RGB to YIQ/YUV color space and subsampling color; applying discrete cosine transformation (DCT); quantization; zig-zag ordering; DPCM on DC component; run-length encoding; and entropy coding like Huffman coding. Quantization is the main source of loss in JPEG compression, where each DCT value is divided by a quantization value and rounded down.
lossy compression 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