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Digital Video Solutions to Midterm Exam 2013 Edited by Yang-Ting Chou

Digital Video Solutions to Midterm Exam 2013 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. Announcement. AVG: 106.172 STDEV: 21.8878. MAX: 143 MIN: 43. Announcement.

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Digital Video Solutions to Midterm Exam 2013 Edited by Yang-Ting Chou

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  1. Digital Video Solutions to Midterm Exam 2013 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. Announcement AVG: 106.172 STDEV: 21.8878 MAX: 143 MIN: 43

  3. Announcement • Course Website: http://mediawww.ee.ncku.edu.tw/dvideo.htm • Past Examinations • TA E-mail: yangting115@gmail.com • TA Time: Monday 9-10 am. MediaCore Lab, National Cheng Kung University, Tainan, Taiwan

  4. I (a)

  5. (b)

  6. (c)

  7. (d)

  8. (e)

  9. (f)

  10. 2.1 (a) binary bits : {a2,a1,a0 } 6 4 0 (b) 0.125 0 1.0 1 (6) 0.34375 1.0 0 (4) 0.125 0 (0) 0.34375 0.125 0.321875 0.125<W<0.321875, 1/4, 01

  11. (c) 0.125 0 1.0 (6) 0 0.03125 (4) 0.125 1 0 (0) 1 0.125 0.03125 0.115625 0.115625<W<0.125 15/128 0001111

  12. 2.2 (a) p(A) = 0.75, p(B) = 0.15, p(C) = 0.05, p(D) = 0.03, p(E) = 0.02 p(F)=1- p(A)- p(B)- p(C)- p(D)- p(E)=0 Huffman code: A 0 (1) B 10 (01) C 110 (001) D 1110 (0001) E 11110 (00001) F 11111 (00000) 0 A 0.75 0 B0.15 0 C0.05 1 D0.03 0 1 0 E0.02 1 F0 1 1 0 A 0.75 Huffman code: A 0 (1) B 10 (01) C 110 (001) D 1110 (0001) E 1111 (0000) 0 B0.15 0 C0.05 1 0 D0.03 1 1 E0.02 1

  13. (b) 0 1 1 0 10 11 100 101 1000 1001 10000 10001 100000 100001 Optimal symmetrical RVLC A 0 B 11 C 101 D 1001 E 10001 F 100001 A 0 B 11 C 101 D 1001 E 10001

  14. (c) 100001 10001 000001 00001 1001 0001 101 11 001 Prefix conflict 01 0 1 0 1 Optimal asymmetrical RVLC A 0 B 11 C 101 D 1001 E 10001 F 100001 A 0 B 11 C 101 D 1001 E 10001

  15. 2.3 (a) orthogonal transform: TT*T=I T2 is orthogonal transform (b) unitary transforms: T*T=I T2 is unitary transform (c) 以quantization or scaling 來做補償

  16. 2.4 Maximum: 49 Minimum: 13

  17. Maximum: 41 Minimum: 17

  18. Maximum: 38 Minimum: 11

  19. 2.5

  20. 2.6

  21. 2.7

  22. 2.8 (a) (b) (c) They are not unitary transforms. 以quantization or scaling 來做補償 AA-1 = AAT ≠ I

  23. EZW 2.9 42 17 4 5 -19 13 -4 -5 sp:11 sn:10 iz: 01 zr: 00 6 -7 3 -4 (a) 7 8 0 0 n=5  101 (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 zr zr zr zr zr zr zr zr reconstructed value for sp: 1.5T1 = 24 (101) 11 00 00 00 0 11 10 00 00 00 00 00 00 (c) 40 24 0 0 48 0 0 0 -24 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

  24. 2.10 10 000 000 10 11 0 0 1 10 000 0 0 10 48 0 0 0 0 0 0 0 T0=25=32 1.5T0=48 0 0 0 0 0 0 0 0 10 000 000 10 11 0 0 1 10 000 0 0 10 40 24 0 0 -24 0 0 0 T1=16 1.5T1=24 24 0 0 0 0 0 0 0 10 000 000 10 11 0 0 1 10 000 0 0 10 40 24 0 0 T2=8 1.5T2=12 -24 12 0 0 24 0 0 0 0 0 0 0

  25. 2.11

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