m9_tif305_sns_ubb

TIF305 Organisasi dan TIF305 Organisasi dan Arsitektur KomputerArsitektur Komputer

Ir. Sihar, M.T.T. Informatika / FTIBandung - 2016

Fractional-Number

Daftar PustakaDaftar Pustaka1) Irvine, Kip R. ”Assembly Language for Intel-Based

Computers Fifth Edition”. Florida International University, School of Computing and Information Sciences, Pearson Education, Inc., 2007.

2) Simamora, S.N.M.P. “Diktat Tuntunan Praktis Bahasa Rakitan menggunakan Turbo Assembler 2.0”. Program studi T. Komputer. Politeknik TELKOM. Bandung. 2008.

3) Simamora, S.N.M.P. ”Diktat Kuliah SK-303 Organisasi dan Arsitektur Komputer”. Dept. Teknik Elektro, Fak.Teknik, ITHB, Bandung, 2002.

4) Simamora, S.N.M.P. Modul Belajar Praktis Algoritma dan Pemrograman. Penerbit Deepublish, Yogyakarta. 2016. ISBN: 978-602-401-318-9.

5) Simamora, S.N.M.P. "Diktat Organisasi dan Arsitektur Komputer". Program studi T. Informatika. Fak. Teknologi Informasi. Univ. BALE. Bandung. 2012.

Tran

sfor

mas

i BI

N-to

-DEC

Kasus: (0.011)2 = DEC(...);

Solusi:

(0.011)2 = 0.(2)-1 + 1.(2)-2 + 1.(2)-3;

= 81

410

= 81

82

= ;83

10

Kasus: (0.0101)2 = DEC(...);Solusi:

(0.0111)2 = 0.(2)-1 + 1.(2)-2 + 0.(2)-3 + 1.(2)-4;

= 1610

410

= 161

164

= ;165

10

Algoritma Matematika Informasi dituliskan sbb:


Kasus: (0.101)2 = DEC(...);

Solusi:

(0.101)2 = 1.(2)-1 + 0.(2)-2 + 1.(2)-3;

= 810

21

= 81

84

= ;85

10

Kasus: (0.0111)2 = DEC(...);

Solusi:

(0.0111)2 = 0.(2)-1 + 1.(2)-2 + 1.(2)-3 + 1.(2)-4;

= 161

81

410

= 161

162

164

= ;167

10



Kasus: (010.011)2 = DEC(...);

Solusi:

(010.0)2 = 0.(2)2 + 1.(2)1 + 0.(2)0;

= 0 + 2 + 0

= DEC(2) = (2)10;

Karena (010.011)2 = (010.0)2 + (0.011)2;


Bag-I Bag-IIPenyelesaian Bag-I:

Penyelesaian Bag-II:

(010.0)2 = DEC(...);

(0.011)2 = DEC(...);(0.011)2 = 0.(2)-1 + 1.(2)-2 + 1.(2)-3;

= 81

410

= 81

82

= ;83

10

maka, di-concatenate-kan menjadi: Bag-I + Bag-II = ;832

832

10

Tran

sfor

mas

i DE

C-to

-BIN

Kasus:

Solusi:1

;...85

210


;411

452

85

;2102

41

;2212

21

0

1

0.101

;101.085

210

Maka:

stop!

Kasus:

Solusi:1

;...1611

210


;831

8112

1611

;4302

83

;2302

43

0

0 0.10011

;2112

23

1

stop!

;2212

21

1

;10011.01611

210

Maka:

Kasus:

Solusi:

1

;...8714 2

10


;431

472

87

;211

232

43

;2212

21

1

1

0.111

;111.11108714 2

10

Maka:

stop!

Bagian-I:

Bagian-II:

(14)10 = (...)2;14 2 = 7 sisa 0

7 2 = 3 sisa 13 2 = 1 sisa 11 2 = 0 sisa 1

1110

stop!

Tran

sfor

mas

i OC

T-to

-DEC

Kasus: (0.101)8 = DEC(...);Solusi:

(0.101)8 = 1.(8)-1 + 0.(8)-2 + 1.(8)-3;

= 51210

81

= 5121

51264

= ;51265

10

Kasus: (0.021)8 = DEC(...);

Solusi:

(0.0111)8 = 0.(8)-1 + 1.(8)-2 + 1.(8)-3 + 1.(8)-4;

= 40961

5121

6410

= 40961

40968

409664

= ;409673

10



Kasus: (123.210)8 = DEC(...);Solusi:

(0.210)8 = 2.(8)-1 + 1.(8)-2 + 0.(8)-3;

= 0641

82

= 641

6416

= ;6417

10

Karena (123.110)8 = (123.0)8 + (0.210)8;

Bag-I Bag-II


(123.0)8 = 1.(8)2 + 2.(8)1 + 3.(8)0;

= 64 + 16 + 3

= DEC(83) = (83)10;

Penyelesaian Bag-I:(123.0)8 = DEC(...);



641783

10

Tran

sfor

mas

i DE

C-to

-OCT

Kasus:

Solusi:1

;...51275

810


;64111

64758

51275

;831

8118

6411

;8838

83

1

3

0.113

;113.051275

810

Maka:

stop!

Kasus:

Solusi:

1

;...5128784 8

10


;64231

64878

51287

;872

8238

6423

;8878

87

2

7

0.127

;127.1245128784 8

10

Maka:

stop!

Bagian-I:

Bagian-II:

(84)10 = (...)8;84 8 = 10 sisa 4

10 8 = 1 sisa 21 8 = 0 sisa 1

124

stop!maka, di-concatenate-kan menjadi:Bag-I + Bag-II = 124.0 + 0.127

Tran

sfor

mas

i HE

X-to

-DEC

Kasus: (0.A1)16 = DEC(...);Solusi: (0.A1)16 = A.(16)-1 + 1.(16)-2;

= 2561

1610

= 2561

256160

= ;256161

10

Kasus: (0.59)16 = DEC(...);Solusi: (0.59)16 = 5.(16)-1 + 9.(16)-2 ;

= 2569

165

= 2569

25680

= ;25689

10

Kasus: (12C.610)16 = DEC(...);Solusi:

(0.610)16 = 6.(16)-1 + 1.(16)-2 + 0.(16)-3;

= 02561

166

= 2561

25696

= ;25697

10

Karena (12C.610)16 = (12C.0)16 + (0.610)16;

Bag-I Bag-II


(12C.0)16 = 1.(16)2 + 2.(16)1 + C.(16)0;

= 256 + 32 + 12

= DEC(300) = (300)10;

Penyelesaian Bag-I:(12C.0)16 = DEC(...);



25697300

10

Tran

sfor

mas

i DE

C-to

-HEX

Kasus:

Solusi:2

;...4096715

1610


;2562032

25671516

4096715

;161112

1620316

256203

;16161116

1611

12

11

0.2CB

;2.04096715

1610

CB

Maka:

stop!

C

B

Kasus:

Solusi:

0

;...4096177154 16

10


;2561770

25617716

4096177

;16111

1617716

256177

;1616116

161

11 B

1

0.0B1

;10.94096177154 16

10

BA

Maka:

stop!

Bagian-I:

Bagian-II:

(154)10 = (...)16;154 16 = 9 sisa 10 A

9 16 = 0 sisa 9 9A

stop!

maka, di-concatenate-kan menjadi:Bag-I + Bag-II = 9A.0 + 0.0B1