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1. Bilangan P ecahan

1. Bilangan P ecahan. Dengan : (A) r = Bilangan A 1 = Digit ke 1 di belakang koma A 2 = Digit ke 2 di belakang koma A 3 = Digit ke 3 di belakang koma A n = Digit ke n di belakang koma r = radik. Bilangan pecahan Biner. Contoh 1.1 :

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1. Bilangan P ecahan

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  1. 1. Bilangan Pecahan

  2. Dengan : (A)r = Bilangan A1 = Digit ke 1 dibelakangkoma A2 = Digit ke 2 dibelakangkoma A3 = Digit ke 3 dibelakangkoma An = Digit ke n dibelakangkoma r = radik

  3. BilanganpecahanBiner Contoh 1.1: (0,10011) 2 = (………)10 Solusi:

  4. Contoh 1.2: (0,59375)10 = (……….)2 Solusi: 0,59375 × 2 = 1,1875 0,1875 × 2 = 0,375 0,375 × 2 = 0,75 0,75 × 2 = 1,5 0,5 × 2 = 1,0 Jadi, (0,5937)10 = (0,10011)2

  5. Contoh 1.3:Konversidaribinerke Octal danke Hexadecimal • 10110001101011, 1111002 = 26153, 748 2 6 1 5 3 7 4 • 10111001101011,111100102 = 2C6B,F216 2 C 6 B F 2

  6. Contoh 1.4:Konversidari Octal dan Hexadecimal kebiner • 673,1248 = 1101110110010101002 6 7 3 1 2 4 • 306,D16 = 00110000011011012 3 0 6 D

  7. 2. COMPLEMENT ARITHMETIC

  8. a. Binary 1’s complement for substraction To take the 1’s complement of binary number, Sweply change each bit. The 1’s complement of 1 is 0 and vice versa. The 1’s complement of 1001010 is 0110101. To substract 1’s complement : 1. Take the 1’s complement of the substrahend ( bottom number ) 2. Add the 1’s complement to the minu end ( top number ) 3. Overflow indicated that the answers is positive. Add the overflow to the least significant bit. This operation is called end – around carry ( EAC ).

  9. Lanjutan … 4.If there is no overflow then the answers is negatif. Take the 1’s complement of the original addition to obtain the true magnitude of the answer.

  10. Contoh. 2.1 • Substract 110012 – 100012 Jawab : 11001 11001 -10001 + 01110 1 00111 00111 + 1 1000 Jawabannya adalah : +1000 • Periksa : 2510 – 1710 = 810 EAC + - + Overflow

  11. Contoh. 2-1( Lanjutan ) 2. Substract 100002 – 111012 Jawab : 10000 10000 11101 00010 10010 - 01101 Jawabannyaadalah : - 1101 • Periksa : 1610– 2910 = -1310 1’s Complement + - No overflow

  12. Binary 2’s complement for subtraction the 2’s complement is 1’s complement and then add 1. The 2’s complement of 10110 is 01001+1= 01010 To subtract using 2’s complement idem 1’s complement Contoh. 1. 10112 – 1002 = Jawab. 1011 1011 - 0100 + 1100 overflow 10111 + 111 Jadi 10112 – 1002 = + 1112

  13. Lanjutan ….. 2. 100102 – 110002 = ……….. 2 Jawab. 10010 10010 - 11000 + 01000 11010 101 + 1 110 Jadi 100102 – 11002 = - 1102 2’s comp No overflow

  14. Ignore overflow Sign + b. Operasi adder/subtracter bilangan signed 2’sc Jawaban adder/subtracter diindikasikan oleh bit sign, jika jawaban positif maka bit lainnya merupakan true magnitude dan jika negatif maka bit lainnya merupakan bentuk 2’sc. Contoh ! 1. add untuk bilangan 8 bit 2’sc 01011001 + 10101101 Jawab. 01011001 (+89) + 10101101 (-83) 1 00000110 (+ 6) Jadi true mag = +6

  15. 2. Add 11011001 + 10101101 Jawab. 1011001 (- 39) + 10101101 (- 83) 1 10000110 (-122) jadi true mag 10000110 1111010(-122) 3. Subtract bilangan 8 bit signed 2’sc 01011011 11100101 (+91) (-27) Ignore overflow Sign - 2’sc

  16. 2’sc 2’sc Sign bit + No overflow Sign bit - No overflow 2’sc Jawab. 01011011 01011011 - 11100101 + 00011011 01110110 jadi true mag 01110110 (+118) 4. Subtract 10001010 11111100 Jawab. 10001010 10001010 - 11111100 + 00000100 10001110 jadi true mag 10001110 01110010(-114)

  17. 2. Rubah 10010011 kedalam bilangan decimal menggunakan sistem signed 2’sc. Jawab. 1 0010011 Sign bit 64 32 16 8 4 2 1 = 64+32+8+4+1 1 1 0 1 1 0 1 = 99 true magnitude Jadi true magnitude = -99

  18. 3. Tunjukkan -7810 sebagai bilangan 8 bit signed 2’sc. Jawab. 7810 = 0 1 0 0 1 1 1 0 128 64 32 16 8 4 2 1 true magnitude 01001110 2’sc 10110010 jadi -7810 = 10110010 (signed 2’sc).

  19. 3. BINARY CODE

  20. Pada Binary Code Decimal ( BCD ) setiap digit decimal direpresentasikandenganempat bit biner. Contoh 2-2 Konversibilangan decimal ke BCD • 390610 = ….. BCD Jawab : 3 9 0 6 0011 1001 0000 0110 396010 = 0011100100000110 BCD

  21. Lanjutan ….. 2. 543710 = ….. BCD Jawab : 5 4 3 7 0101 0100 0011 0111 543710 = 0101010000110111 BCD

  22. BCD (Binary Code to Desimal)

  23. odd parity Even parity 4. OTHER DECIMAL CODES 1. BCD, 2421, EXCESS–3(XS-3), 84-2-1 2. Gray Codes 3. ASCII character code 5. ERROR DETECTING CODE Untukmendeteksi error padakomunikasidanprosessing data indikasideteksi error untuksetiapkarakterinformasi / ASCII ditambah 1 bit parity (even, add) Contoh. ASCII A = 1000001 01000001 11000001 T = 1010100 11010100 01010100

  24. Kode Gray

  25. Kode ASCII

  26. Sign bit  6. BINARY STORAGE AND REGISTER Bilangan signed 2’s complement indikasibilangan decimal diletakkanpada Most Significant Bit atau MSB dan bit sisanyasebagai true magnitude. Untuk sign bit 0 true magnitude positif 1 true magnitude negatif Contoh ! 1. Rubah 00101101 kedalambilangan decimal menggunakansistem signed 2’s C. 0 0 1 01101 0432168421 32 + 8 + 4 +1 = 45 Jadi true magnitude adalah +45

  27. Soallatihan ! 1. Tunjukkanbilanganbiner 8 bit signed 2’sc untuk : a. -75 c. -150 b. +47 d. +93 2. Add bilangan 8 bit signed 1’sc dan 2’sc a. 00011110 + 00111001 b. 00110011 + 11001000 3. Subtract bilangan 8 bit signed 1’sc dan 2’sc a. 00111001 – 11001110 b. 10101010 - 10011011

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