1 / 29

Bilangan dan Kode

Bilangan dan Kode. By: Moch. Rif’an. 0. 2. 7. 8. 9. 4. 1. 3. 5. 6. 10. Berapa Anjing ?. 1.1 Bilangan desimal. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 …. 1.1 Bilangan desimal. 1973

duer
Download Presentation

Bilangan dan Kode

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Bilangan dan Kode By: Moch. Rif’an

  2. 0 2 7 8 9 4 1 3 5 6 10 Berapa Anjing ?

  3. 1.1 Bilangan desimal 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 …

  4. 1.1 Bilangan desimal 1973 = 1000 + 900 + 70 + 3 = 1×1000 + 9×100 + 7×10 + 3 = 1×103 + 9×102 + 7×101 + 3×100 (1973)r = 1×r3 + 9×r2 + 7×r1 + 3×r0

  5. 1.2 Bilangan heksadesimal 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F 10, 11, 12, …

  6. 1.2 Bilangan heksadesimal (N)r = N0 × r0 + N1 × r1 +N2 × r2 + … Dengan : N = bilangan r = radik/basis N0 = bilangan digit ke 0 N1 = bilangan digit ke 1 N2 = bilangan digit ke 2

  7. 1.2 Bilangan heksadesimal

  8. 1.2 Bilangan heksadesimal (ED1)16 = E × 162 + D × 161 + 1 × 160 = 14 × 162 + 13 × 161 + 1 × 160 = 14 ×256 + 13 ×16 + 1×1 = 3584 + 208 + 1 = 3793 Sehingga (ED1)16 = (3793)10

  9. 1.3 Bilangan oktal

  10. 1.3 Bilangan oktal (1371)8 = 1 × 83 + 3 × 82 + 7 × 81 + 1 × 80 = 512 + 192 + 56 + 1 = 761 Sehingga (1371)8 = (761)10

  11. 1.4 Bilangan biner 0, 1, 10, 11, 110, 111, …

  12. 1.4 Bilangan biner (11101)2 = 1 × 24 + 1 × 23 + 1 × 22 + 0 × 21 + 1 × 20 = 1×16 + 8 + 4 + 0 + 1 = 29 Sehingga (11101)2 = (29)10

  13. 1.4 Bilangan biner

  14. Konversi dari bilangan desimal ke radik lain By: Moch. Rif’an

  15. Konversi dari desimal ke heksadesimal Contoh: (3793)10 = (……..)16 Penyelesaian: 3793 ÷ 16 = 237 sisa 1 237 ÷ 16 = 14 sisa 13 14 ÷ 16 = 0 sisa 14 (3793)10 = (ED1)16

  16. Konversi dari desimal ke oktal Contoh 1.6: (761)10 = (……..)8 Penyelesaian: 761 ÷ 8 = 95 sisa 1 95 ÷ 8 = 11 sisa 7 11 ÷ 8 = 1 sisa 3 1 ÷ 8 = 0 sisa 1 (761)10 = (1371)8

  17. Konversi dari desimal ke biner Contoh 1.7: (978)10 = (……..)2 Penyelesaian: 978 ÷ 2 = 479 sisa 0 479 ÷ 2 = 239 sisa 1 239 ÷ 2 = 119 sisa 1 119 ÷ 2 = 59 sisa 1 59 ÷ 2 = 29 sisa 1

  18. Penyelesaian: 29 ÷ 2 = 19 sisa 1 19 ÷ 2 = 9 sisa 1 9 ÷ 2 = 4 sisa 1 4 ÷ 2 = 2 sisa 0 2 ÷ 2 = 1 sisa 0 1 ÷ 2 = 0 sisa 1 (978)8 = (10011111110)2

  19. Konversi heksadesimal ke biner Contoh 1.8: (1DA)16 = (……)2 Penyelesaian: 1 D A . 0001 1101 1010 Sehingga hasil konversi: (1DA)16 = (111011010)2

  20. Konversi dari biner ke heksadesimal Contoh 1.9: (101011010)2 = (………)16 Penyelesaian: 101011010 000101011010 1 5 A Sehingga: (101011010)2 = (15A)16

  21. Konversi dari bilangan oktal ke bilangan Biner Contoh 1.10: (751)8 = (……..)2 Penyelesaian: 7 5 1 . 111 101 001 Sehingga: (751)8 = (111101001)2

  22. Konversi dari biner ke oktal Contoh: (101011010)2 = (………)8 Penyelesaian: 101011010 101011010 5 3 2 Sehingga: (101011010)2 = (532)8

  23. Konversi dari bilangan heksadesimal ke bilangan oktal Contoh 1.12: (A51)16 = (……)8 Penyelesaian: Cara 1. A51 = A×162 + 5×161 + 1×160 = 10×256 + 5×16 + 1×1 = 2641 (A51)16 = (2641)10

  24. Dari desimal baru dikonversi ke oktal 2641 ÷ 8 = 330 sisa 1 330 ÷ 8 = 41 sisa 2 41 ÷ 8 = 5 sisa 1 5 ÷ 8 = 0 sisa 5 (a51)16 = (5121)8

  25. Cara 2. Heksadesimal A 5 1 . Biner 1 0 10 0 10 1 00 0 1 Oktal 5 1 2 1 Sehingga: (a51)16 = (5121)8

  26. Konversi dari oktal ke heksadesimal Contoh 1.13: (1746)8 = (……)16 Penyelesaian: Cara 1. (1746)8 = 1×83 + 7×82 + 4×81 + 6×80 = 1×512 + 7×64 +4×8 +6×1 = (998)10 998 ÷ 16 = 62 sisa 6 62 ÷ 16 = 3 sisa 14 3 ÷ 16 = 0 sisa 3 Jadi, (1746)8 = (3e6)16

  27. Cara 2. Oktal 1 7 4 6 . Biner 0 0 1 11 1 1 00 1 1 0 Heksadesimal 3 e 6 Sehingga: (1746)8 = (3e6)16

  28. Kita Lanjutkan Pertemuan Berikutnya

More Related