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Digital Video Solutions to Midterm Exam2 2012 Edited by Yang-Ting Chou

Digital Video Solutions to Midterm Exam2 2012 Edited by Yang-Ting Chou Confirmed by Prof. Jar-Ferr Yang LAB: 92923 R, TEL: ext. 621 E-mail: yangting115@gmail.com Page of MPL: http://mediawww.ee.ncku.edu.tw. I. (a). (b).

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Digital Video Solutions to Midterm Exam2 2012 Edited by Yang-Ting Chou

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  1. Digital Video Solutions to Midterm Exam2 2012 Edited by Yang-Ting Chou Confirmed by Prof. Jar-Ferr Yang LAB: 92923 R, TEL: ext. 621 E-mail: yangting115@gmail.com Page of MPL: http://mediawww.ee.ncku.edu.tw

  2. I (a)

  3. (b) • Luminance is a photometric measure of the luminous intensity per unit area of light travelling in a given direction. • It describes the amount of light that passes through or is emitted from a particular area, and falls within a given solid angle. (c) (d)

  4. (e) (f)

  5. II 2.1 (a) binary bits : {a2,a1,a0 } 3 3 6 (b) 0.25 0 1.0 0 (3) 0.0625 1 (3) 0.25 0 0.109375 1 (6) 0.25 0.0625 0.109375<W<0.25, 1/8, 001

  6. (c) 0.25 0 1.0 0 (3) 0.0625 1 (3) 0.25 0 0.109375 0 (6) 0.25 0.0625 0.0625<W<0.109375, 3/32, 00011

  7. 2.2 (a) p(A) = 0.8, p(B) = 0.1, and p(C) = 0.1 0 p(AA) = 0.64 • Level-2 VLC: • AA 0 (1) • AB 11 (00) • BA 101 (010) • AC 1001 (0110) • CA 10000 (01111) • BB 100011 (011100) • BC 1000101 (0111010) • CB 10001001 (01110110) • CC 10001000 (01110111) p(AB) = 0.08 1 1 p(BA) = 0.08 1 1 p(AC) = 0.08 0 0 p(CA) = 0.08 0 1 p(BB) = 0.01 0 1 p(BC) = 0.01 1 1 0 p(CB) = 0.01 p(CC) = 0.01 0 0

  8. (b) 0 1 1 0 10 11 100 101 1000 1001 10000 10001 100000 • Optimal symmetrical RVLC: • AA 0 • AB 11 • BA 101 • AC 1001 • CA 10001 • BB 100001 • BC 1000001 • CB 10000001 • CC 100000001 100001 1000000 1000001 10000000 10000001 10000001 100000000

  9. 100000001 (c) 10000001 000000001 1000001 00000001 100001 0000001 10001 000001 00001 1001 0001 101 11 001 Prefix conflict 01 0 • Optimal asymmetrical RVLC: • AA 0 • AB 11 • BA 101 • AC 1001 • CA 10001 • BB 100001 • BC 1000001 • CB 10000001 • CC 100000001 1 0 1

  10. 2.3 (a) Entropy: (b) 1 Huffman code: A 01 (10) B 1 (0) C 00 (11) B 0.7 1 A 0.2 C 0.1 0 0 (c) {ABABBC} 01 1 01 1 1 00

  11. (d) Occurrence symbols: {ABABBC} 0.2 A 0.0 1.0 0.04 0.18 B 0.0 0.2 0.068 A 0.04 0.18 0.0456 0.0652 0.04 B 0.068 0.04952 0.06324 0.0456 B 0.0652 0.061868 0.04952 C 0.06324 0.061868<W<0.06324 0001

  12. 2.4 題目為Haar Filter，所以直接使用以上三個公式完成填表，X0與X1不必填

  13. 2.5 (c) (a) They are not unitary transforms. 以quantization or scaling 來做補償 AA-1 = AAT ≠ I (b)

  14. EZW 2.6 42 17 4 5 -19 13 -4 -5 (a) 6 -7 3 -4 7 8 0 0 (b) first pass: sp zr zr zr 8 bits reconstructed value for sp: 1.5T0 = 48 T1=16 refine: T1/2 = 8, 1 bit “0” for refinement second pass (T1=16): 0 sp sn zr 7 bits reconstructed value for sp: 1.5T1 = 24 T2=8 refine: T1/2 = 4, 1 bit “0” for refinement second pass (T1=16): 1 0 0 sp zr iz iz iz sp zr 17 bits reconstructed value for sp: 1.5T1 = 24 Generated bitstream: 11 00 00 00 0 11 10 00 1 0 0 11 00 01 0

  15. (c) (1) Generated bitstream: 11 00 00 00 0 11 10 00 1 0 0 11 00 01 0 48 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (2) Generated bitstream: 11 00 00 00 0 11 10 00 1 0 0 11 00 01 0 40 24 0 0 -24 0 0 0 0 0 0 0 0 0 0 0 (3) Generated bitstream: 11 00 00 00 0 11 10 00 1 0 0 11 00 01 0 44 20 0 0 -20 12 0 0 0 0 0 0 0 0 0 0

  16. 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 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 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2.7 Significance Propagation Pass (Pass 1) (a) ZC: Zero Coding SC: Sign Coding zc zc zc zc zc zc sc zc zc zc zc zc zc zc zc zc zc zc zc sc zc zc zc zc sc zc zc : Coefficient which is already significant : Significance Propagation Pass (Pass 1)

  17. 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 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 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (a) Magnitude Refinement Pass (Pass 2) zc zc zc zc zc zc sc zc zc zc zc zc zc zc zc zc zc zc zc sc zc zc zc zc sc zc zc : Magnitude Refinement Pass (Pass 2) : Significance Propagation Pass (Pass 1)

  18. 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 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 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (a) Clean-up Pass (Pass 3) zc zc zc zc zc zc sc zc zc zc zc zc zc zc zc zc sc zc zc zc zc zc zc zc zc zc sc sc zc zc zc zc sc zc zc zc zc zc zc zc : Pass 1 : Pass 3 (ZC & SC) : Pass 2 : Pass 3(RLC)

  19. 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 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 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 (b) zc zc zc zc zc zc sc (a) (b) zc zc zc zc zc (c) zc zc zc zc sc zc zc zc zc zc zc zc zc zc sc sc (d) zc zc zc zc sc zc zc zc zc zc zc zc a: MR kh[j] = 1, kv[j] = 1, kd[j] = 0, ksig[j]=7 kmag[j] = 15 or 16 b: SC h[j] = 1, v[j] = 0, ksign[j] = 12 c: ZC, LL band kh[j] = 0, kv[j] = 0, kd[j] = 1, ksig[j]=1 d: SC h[j] = 1, v[j] = 0, ksign[j] = 12

  20. (b) Assignment of context labels for significant coding “x” means “don’t care.”

  21. (b) Assignment of context labels and flipping factor for sign coding Current sample ch[j] , cv[j]: neighborhood sign status -1: one or both negative. 0: both insignificant or both significant but opposite sign. 1: one or both positive.

  22. (b) Assignment of context labels and flipping factor for magnitude refinement coding s [j]: remains zero until after the first magnitude refinement bit has been coded. For subsequent refinement bits, s [j] = 1. ksig[j]: context label for significant coding of sample j

  23. 2.8 (a) Cross Search

  24. -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 (-1, 2): 5+3+8 = 16 points

  25. -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 (2, -7): 5+3+3+8 = 19 points

  26. (b) Novel TSS

  27. -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 (-1, 2): 9+5+8 = 22 points

  28. -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 (2, -7): 41points

  29. (c) Hexagon Search

  30. -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 (-1, 2): 7+3+4 = 14 points

  31. -2 -1 0 1 2 3 4 5 6 7 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 (2, -7): 7+3+3+3+3+4 = 23 points

  32. 2.9

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