Digital system and binary numbers (1'st update)

Sistem DigitalNugroho Adi P

nugnux@gmail.comAravir@me.com

Monday, August 25, 14

One characteristic of digital systems is their ability to represent and manipulate discrete elements of information.

Discrete InformationHuruf

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Jumlah Mahasiswa

Discrete Information26 Huruf

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Elemen informasi diskrit diwakili di sistem digital oleh sinyal

tegangan

Sinyal diskrit hanya mempunyai dua nilai

(Biner)

Bit bernilai 0 dan 1

Discrete elements of information are represented with groups of bits called binary codes

contoh kode biner

Kadang

(1000)2 ≠ (1000)10

tidak mungkin

(1008)2

tidak mungkin keliru

a digital system is a system that manipulates discrete elements of information represented internally in binary form.

Diskrit secara alami

Diambil dari proses kontinu (kuantisasi)

Informasi Diskrit

Diskrit alami

Kuantisasi

Kenapa Digital?

Dapat diprogram

Satu alat dapat digunakan untuk bermacam keperluan

Bilangan Biner

seribu tiga ratus sembilan puluh lima

1000 + 300 + 90 + 5

1 x 1000 + 3 x 100 + 9 x 10 + 5

1000 + 300 + 90 + 5

1 x 1000 + 3 x 100 + 9 x 10 + 5

1 x 103 + 3 x 102 + 9 x 101 + 5 x 100

1000 + 300 + 90 + 5

1 x 1000 + 3 x 100 + 9 x 10 + 5

1 x 103 + 3 x 102 + 9 x 101 + 5 x 100

a3 x 103 + a2 x 102 + a1 x 101 + a0 x 100

basis 10

basis 10 secara umum

...+103a3 + 102a2 + 101a1 + 100a0 + 10-1a-1 + 10-2a-2 + 10-3a-3 + ...

secara umum hanya ditulis koefisennya saja

...+23a3 + 22a2 + 21a1 + 20a0 + 2-1a-1 + 2-2a-2 + 2-3a-3 + ...

1 x 23 + 1 x 22 + 0 x 21 + 1 x 20

8 + 4 + 0 + 1 (desimal)

1 x 23 + 1 x 22 + 0 x 21 + 1 x 20

8 + 4 + 0 + 1 (desimal)

13 (desimal)

1 x 21 + 0 x 20 + 1 x 2-1

2 + 0 + 0.5 (desimal)

1 x 21 + 0 x 20 + 1 x 2-1

2 + 0 + 0.5 (desimal)

2.5 (desimal)

Basis-r

...+r3a3 + r2a2 + r1a1 + r0a0 + r-1a-1 + r-2a-2 + r-3a-3 + ...

Basis-r

...+r3a3 + r2a2 + r1a1 + r0a0 + r-1a-1 + r-2a-2 + r-3a-3 + ...

nilai ai antara 0 dan r-1

1 x 81 + 0 x 80 + 1 x 8-1

8 + 0 + 0.125 (desimal)

1 x 81 + 0 x 80 + 1 x 8-1

8 + 0 + 0.125 (desimal)

8.125 (desimal)

hexadesimal

1 x 161 + 0 x 160 + 1 x 16-1

hexadesimal

1 x 161 + 0 x 160 + 1 x 16-1

16 + 0 + 0.125 (desimal)

hexadesimal

1 x 161 + 0 x 160 + 1 x 16-1

16 + 0 + 0.125 (desimal)

16.0625 (desimal)

210 = 1024 = 1K

220 = M

210 = 1024 = 1K

220 = M

230 = G

210 = 1024 = 1K

220 = M

230 = G

240 = T

210 = 1024 = 1K

220 = M

230 = G

240 = T

Aritmatika bilangan berbasis r mengikuti aturan yang sama dengan bilangan desimal

11 + 0 = 11

1+1 = 10

11 + 0 = 11

1+1 = 10

10+11= 101

11 + 0 = 11

1+1 = 10

10+11= 101

11+11 = 110

1 + 1 = 2

7 + 1 = 10

1 + 1 = 2

7 + 1 = 10

17 +11 = 30

1 + 1 = 2

7 + 1 = 10

17 +11 = 30

17 + 1 = 20

hexadesimal

9 + 1 = A

hexadesimal

9 + 1 = A

9 + 2 = B

hexadesimal

9 + 1 = A

9 + 2 = B

F + 2 = 11

hexadesimal

9 + 1 = A

9 + 2 = B

F + 2 = 11

FF + 1 = 100

1 - 1 = 0

10 - 01 = 01

1 - 1 = 0

10 - 01 = 01

1011 - 1001 =0010

10 - 1 = 7

17 - 10 = 7

10 - 1 = 7

17 - 10 = 7

21 - 17 = 2

Hexadesimal

10 - 1 = F

Hexadesimal

10 - 1 = F

1A - 11 = 9

Hexadesimal

10 - 1 = F

1A - 11 = 9

A0 - 1 = 9F

Konversi

(1)2 = (1)8 = (1)16 = (1)5 = 1 (desimal)

Konversi

(1)2 = (1)8 = (1)16 = (1)5 = 1 (desimal)

(1 0000)2 = (20)8 = (10)16 = (31)5 = 16 (desimal)

Konversi

(1)2 = (1)8 = (1)16 = (1)5 = 1 (desimal)

(1 0000)2 = (20)8 = (10)16 = (31)5 = 16 (desimal)

(0.000110...)2 = 0.05 (desimal)

Konversi

(1)2 = (1)8 = (1)16 = (1)5 = 1 (desimal)

(1 0000)2 = (20)8 = (10)16 = (31)5 = 16 (desimal)

(0.000110...)2 = 0.05 (desimal)

(0.001)2 = 0.125 (desimal)

Desimal ke binary

Desimal ke binary102

Desimal ke binary102102 bagi dua

Desimal ke binary102102 bagi dua51 sisa 0

Desimal ke binary102102 bagi dua51 sisa 025 sisa 1

Desimal ke binary102102 bagi dua51 sisa 025 sisa 112 sisa 1

Desimal ke binary102102 bagi dua51 sisa 025 sisa 112 sisa 16 sisa 0

Desimal ke binary102102 bagi dua51 sisa 025 sisa 112 sisa 16 sisa 03 sisa 0

1 sisa 1

1 sisa 10 sisa 1

tulis dari bawah

1 sisa 10 sisa 1

tulis dari bawah1100110

1 sisa 10 sisa 1

tulis dari bawah1100110jadi

1 sisa 10 sisa 1

tulis dari bawah1100110jadi (102)10 = (1100110)2

1 sisa 10 sisa 1

tulis dari bawah1100110jadi (102)10 = (1100110)2

Desimal ke octal

Desimal ke octal102

Desimal ke octal102102 bagi delapan

Desimal ke octal102102 bagi delapan12 sisa 6

Desimal ke octal102102 bagi delapan12 sisa 61 sisa 4

Desimal ke octal102102 bagi delapan12 sisa 61 sisa 40 sisa 1

(102)10 = (146)2

Desimal ke heksadesimal

Desimal ke heksadesimal102

Desimal ke heksadesimal102102 bagi enambelas

Desimal ke heksadesimal102102 bagi enambelas6 sisa 6

Desimal ke heksadesimal102102 bagi enambelas6 sisa 60 sisa 6

Desimal ke heksadesimal102102 bagi enambelas6 sisa 60 sisa 6jadi

Desimal ke heksadesimal102102 bagi enambelas6 sisa 60 sisa 6jadi (102)10 = (6)16

Pecahan

Pecahan(0.2)10

0.2 x 2 = 0.4 + 0

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 1

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 1

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 0

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 1

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 1

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 0

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 0.....

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 0.....(0.22)10 = (0.0011001100...)2

Pecahan(0.2)10

0.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 00.4 x 2 = 0.8 + 0 0.8 x 2 = 0.6 + 10.6 x 2 = 0.2 + 10.2 x 2 = 0.4 + 0.....(0.22)10 = (0.0011001100...)2

Pecahan

Pecahan(0.25)10

0.25 x 2 = 0.5 + 0

Pecahan(0.25)10

0.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1

Pecahan(0.25)10

0.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1 .....

Pecahan(0.25)10

0.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1 .....(0.25)10 = (0.01)2

Pecahan(0.25)10

0.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1 .....(0.25)10 = (0.01)2

Pecahan

Pecahan(0.5)10

0.5 x 2 = 0.0 + 1

Pecahan(0.5)10

0.5 x 2 = 0.0 + 1

Pecahan(0.5)10

0.5 x 2 = 0.0 + 1

(0.5)10 = (0.1)2

Pecahan(0.5)10

0.5 x 2 = 0.0 + 1

(0.5)10 = (0.1)2

Pecahan

Pecahan(0.125)10

0.125 x 2 = 0.25 + 0

Pecahan(0.125)10

0.125 x 2 = 0.25 + 00.25 x 2 = 0.5 + 0

Pecahan(0.125)10

0.125 x 2 = 0.25 + 00.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1

Pecahan(0.125)10

0.125 x 2 = 0.25 + 00.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1

Pecahan(0.125)10

0.125 x 2 = 0.25 + 00.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1

(0.5)10 = (0.001)2

Pecahan(0.125)10

0.125 x 2 = 0.25 + 00.25 x 2 = 0.5 + 00.5 x 2 = 0.0 + 1

(0.5)10 = (0.001)2

Oktal dan Heksadesimal

(111111111111)2

(111 111 111 111)2 = (7 7 7 7)8

(111111111111)2

(111 111 111 111)2 = (7 7 7 7)8

(1111 1111 1111)2 = (F F F )16

Bilangan biner sulit untuk digunakan

Mengunakan digit yang lebih banyak dari desimal

Bilangan desimal lebih mudah digunakan

namun komputer menggunakan biner

konversi desimal ke biner rumit

Octal dan Hexadecimal memperkecil digit

Complements

‘negatif ’-nya sebuah bilangan

menyederhanakan pengurangan

(dilakukan dengan penjumlahan)

hemat sirkuit

Komplemen bilangan berbasis-r

komplemen r (radix)

komplemen r-1 (diminished radix)

jika sebuah bilangan N dengan basis r memiliki n digit,

maka akan memiliki diminished radix komplemen (rn-1)-N

mis 7 desimal akan memiliki diminished radix komplemen (101-1)-7 = 2

07 desimal akan memiliki diminished radix komplemen (102-1)-7 = 92

diminised radix complement tuk binary

100 -> (23-1)-100

= (1000-1)-100

100 -> (23-1)-100

= (1000-1)-100

= (111)-100

100 -> (23-1)-100

= (1000-1)-100

= (111)-100

111 -> (23-1)-111

= (1000-1)-111

111 -> (23-1)-111

= (1000-1)-111

= (111)-111

111 -> (23-1)-111

= (1000-1)-111

= (111)-111

101 -> (23-1)-101

= (1000-1)-101

101 -> (23-1)-101

= (1000-1)-101

= (111)-101

101 -> (23-1)-101

= (1000-1)-101

= (111)-101

maka akan memiliki radix komplemen (rn)-N

mis 7 desimal akan memiliki radix komplemen (101)-7 = 3

07 desimal akan memiliki radix komplemen (102)-7 = 93

radix complement tuk binary

100 -> (23)-100

= (1000)-100

100 -> (23)-100

= (1000)-100

100 -> (23)-100

= (1000)-100

111 -> (23)-111

= (1000)-111

111 -> (23)-111

= (1000)-111

101 -> (23-1)-101

= (1000)-101

101 -> (23-1)-101

= (1000)-101

Jika memiliki radix point, hilangkan!

Kembalikan lagi setelah hitungan

Pengurangan dengan komplemen

(lebih mudah bagi hardware)

(tak ada pinjaman)

pengurangan bilangan M-N berdigit n berbasis r

tambahkan M dengan radix komplemen dari N

M + ( rn - N ) = M - N + rn

jika M≥N maka penjumlahan akan menghasilkan (end carry)

buang end carry

yang tersisa merupakan hasilnya

Jika M < N, maka penjumlahan tidak menghasilkan end carry

hasilnya merupakan komplemen (N-M)

agar ‘enak’ dilihat, ambil komplemen hasil penjumlahan

beri tanda negatif

Selesai

Dan dia hidup bahagia selama-lamanya...Monday, August 25, 14

Digital system and binary numbers (1'st update)

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