5.6 Problems

5.6 Problems

1. 

The luminance quantisation Table 5.1 is used in the baseline JPEG for a quality factor of 50 per cent. Find the quantisation table for the following quality factors:

  1. 25%

  2. 99%

  3. 100 %

 a. multiply all the luminance matrix elements by α = 50/25 = 2 b. , and small elements of the matrix will be 1 and larger ones become 2. c. α = 0, hence all the matrix elements will be 1.

2. 

In problem 1 find the corresponding tables for the chrominance.

the same as problem 1.

3. 

The DCT transform coefficients (luminance) of an 8 x 8 pixel block prior to quantisation are given by:

1000

-2

35

18

15

-8

62

5

-4

15

-21

-4

51

2

-11

1

9

-8

13

-11

43

-20

7

-3

-17

16

-11

3

-2

5

-13

6

-6

12

42

-15

31

-2

7

-3

12

-7

2

-11

15

-5

3

18

-19

52

6

13

4

-10

8

10

35

-11

-7

3

5

9

7

382

Find the quantisation indices for the baseline JPEG with the quality factors of:

  1. 50%

  2. 25%

 a. 62 0 3 1 0 0 1 0 0 1 -1 0 1 0 0 0 0 0 0 0 1 0 0 0 -1 0 0 0 0 0 0 0 0 0 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 3 b. 31 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1

4. 

In problem 3, if the quantised index of the DC coefficient in the previous block was 50, find the pairs of symbol-1 and symbol-2 for the given quality factors.

for 50% quality: diff = 62 - 50 = 12, symbol_1 = 4; symbol_2 = 12 scanned pairs (3,1)(0,3)(0,1)(0,-1)(1,-1)(6,1)(6,1)(1,1)(1,1)(35,3) and the resultant events: (3,1)(0,2)(0,1)(0,1)(1,1)(6,1)(6,1)(1,1)(1,1)(15,0)(15,0)(3,3) for 25% quality: diff= 31-50 = -19, symbol_1 = 5, symbol_2 = -19-1 =-20 scanned pairs: (4, 1)(57, 1) events: (4,1)(15,0)(15,0)(15,0)(9,1)

5. 

Derive the Huffman code for the 25 per cent quality factor of problem 4 and hence calculate the number of bits required to code this block.

for dc: diff= -19 → cat = 5; diff-1 = -20 vlc for cat = 5 is 110 and -20 in binary is 111 01100, hence the vlc for the dc coefficient is 11001100 for ac, using the ac vlc tables: for each (15,0) the vlc is 11111111001 and for (9, 1) the vlc is 111111001 total number of bits: 8 + 3 x 11 + 9 = 50 bits.

6. 

A part of the stripe of the wavelet coefficients of a band is given as:

20

30

-16

65

31

11

50

24

Assume the highest bit plane is 6. Using EBCOT identify which coefficient is coded at bit plane 6 and which one at bit plane 5. In each case identify the type of fractional bit plane used.

at bit-plane 6 coefficient 65 at clean-up pass. at bit-plane 5 coefficient 65 at all passes and coefficient 50 at clean-up pass.

Answers

1. 

  1. multiply all the luminance matrix elements by α = 50/25 = 2

  2. , and small elements of the matrix will be 1 and larger ones become 2.

  3. α = 0, hence all the matrix elements will be 1.

2. 

the same as problem 1.

3. 

  1. 62

    0

    3

    1

    0

    0

    1

    0

    0

    1

    -1

    0

    1

    0

    0

    0

    0

    0

    0

    0

    1

    0

    0

    0

    -1

    0

    0

    0

    0

    0

    0

    0

    0

    0

    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

    3

  2. 31

    0

    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

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    0

    1

4. 

for 50% quality: DIFF = 62 - 50 = 12, symbol_1 = 4; symbol_2 = 12 scanned pairs (3,1)(0,3)(0,1)(0,-1)(1,-1)(6,1)(6,1)(1,1)(1,1)(35,3) and the resultant events:

(3,1)(0,2)(0,1)(0,1)(1,1)(6,1)(6,1)(1,1)(1,1)(15,0)(15,0)(3,3)

for 25% quality: DIFF= 31-50 = -19, symbol_1 = 5, symbol_2 = -19-1 =-20

scanned pairs: (4, 1)(57, 1)

events: (4,1)(15,0)(15,0)(15,0)(9,1)

5. 

for DC: DIFF= -19 CAT = 5; DIFF-1 = -20

VLC for CAT = 5 is 110 and -20 in binary is 11101100, hence the VLC for the DC coefficient is 11001100

for AC, using the AC VLC tables:

for each (15,0) the VLC is 11111111001

and for (9, 1) the VLC is 111111001

total number of bits: 8 + 3 x 11 + 9 = 50 bits.

6. 

At bit-plane 6 coefficient 65 at clean-up pass. At bit-plane 5 coefficient 65 at all passes and coefficient 50 at clean-up pass.



Standard Codecs(c) Image Compression to Advanced Video Coding
Standard Codecs: Image Compression to Advanced Video Coding (IET Telecommunications Series)
ISBN: 0852967101
EAN: 2147483647
Year: 2005
Pages: 148
Authors: M. Ghanbari

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