1. Chapter 2.0:
Data Representation on CPU
FP203 : Computer Organization

2. Topic Cover
2.1 Number System (decimal, binary,
octal, and hexadecimal)
2.2 Arithmetic Operation in number
system.
2.3 Convert Decimal, Binary, Octal
and Hexadecimal Numbers to
different bases.
2.4 Coding system:
Sign and magnitude,
1‟s Complement and 2‟s Complement
Binary Coded Decimal (BCD system)
ASCII and EBCDIC

3. INTRODUCTION
Examples
Real World Computer
Data Input device Data
Dear Mom: Keyboard 10110010…
Digital 10110010…
camera

4. 2.1 Number System
• Many number system are in use in digital
technology.
• Most common are:
– Decimal, N10
– Binary, N2
– Octal, N8
– Hexadecimal, N16

5. 2.2 ARITHMETIC OPERATION
• Arithmetic operation in number system
consist of:
– Addition Only cover this 2 topics
– Subtraction
– Multiplication
– Division

6. Decimal
number system
• Decimal system is composed of 10
numerals or symbol.
• Symbol: 0,1,2,3,4,5,6,7,8,9
10 Symbol
• Example: 23410
Multiplier:
103 102 101 100 . 10-1
= 1000 = 100 = 10 =1 . = 0.1

7. Example:
2746.210
This number is came from this calculation:
2 7 4 2 . 2
103 102 101 100 . 10-1
= 1000 = 100 = 10 =1 . = 0.1
2746.210 = (2x1000) + (7x100) + (4x10) + (2x1) + (2x0.1)
= 2000 + 700 + 40 + 2 + 0.2
= 2746.2
Decimal number = Natural Number

8. Arithmetic Operation
Decimal
+ -

9. Decimal Addition
Example:
a. 89310 + 32110 = b. 75710 + 24510 =
89310 75710
+ 24510
+32110
100210
121410
Try this : 73310 + 79910 = ?

10. Decimal Subtraction
Example:
a. 5410 - 1710 = b. 15710 - 8910 =
5410 15710
- 1710 - 8910
3710 6810
Try this : 533310 - 3710 = ?

11. Octal
number system
• Octal system is composed of 8 numerals
or symbol.
• Symbol: 0,1,2,3,4,5,6,7
8 Symbol
• Example: 658
Multiplier:
83 82 81 80 . 8-1
= 512 = 64 =8 =1 . = 0.125

12. Example:
107.158
This number can be convert to decimal value using this calculation:
1 0 7 . 1 5
82 81 80 . 8-1 8-2
= 64 =8 =1 . = 0.1250 = 0.0156
107.158 = (1x64)+(0x8)+(7x1)+(1x0.1250)+(5x0.0156)
= 64 + 0 + 7 + 0.1250 + 0.078
= 71.20310

13. Arithmetic Operation
Octal
+ -

14. Octal Addition
Sekiranya setiap hasil perjumlahan yang melebihi atau sama
dengan 8 mestilah ditolak dengan 8.
Example:
a. 1238 + 3218 =
b. 4578 + 2458 =
1238 4578
+3218 + 2458
4448 7248
Try this : 7338 + 748 = ?

15. Octal Subtraction
Sekiranya terdapat peminjam, nombor peminjam mestilah
dijumlahkan dengan 8.
Example:
a. 5248 - 1678 =
b. 1678 - 248 =
5248 1678
- 1678 - 248
3358 1438
Try this : 15238 - 3648 = ?

16. Binary
number system
• Binary system is composed of 2 numerals
or symbol.
• Symbol: 0,1
2 Symbol
• Example: 1012
Multiplier:
25 24 23 22 21 20
= 32 = 16 =8 =4 =2 =1

17. Example:
10.1012
This number can be convert to decimal value using this calculation:
1 0 . 1 0 1
21 20 . 2-1 2-2 2-3
=2 =1 . = 0.5000 = 0.2500 = 0.1250
10.1012 = (1x2)+(0x1)+(1x0.5)+(0x0.25)+(1x0.125)
= 2 + 0 + 0.5 + 0 + 0.125
= 2.62510

18. Arithmetic Operation
Binary
+ -

19. Binary Addition
The four basic rules for adding binary digits are as follows:
0+0=0
0+1=1
1+0=1
1 + 1 = 0 carry 1
Example:
110112 + 100012 =
110112
+ 100012
1011002
Try this : 101112 + 1112 = ?

20. Binary Subtraction
The four basic rules for subtracting binary digits are as follows:-
0-0=0
0 - 1 = 1 borrow 1
1-0=1
1-1=0
Example:
10012 -102 =
10012
- 102
1112
Try this : 1010112 – 11112 =?

21. Binary Subtraction
Have previously looked at the subtraction operation. A
quick review.
Just like subtraction in any other base
10110
-10010
00100
• And when a borrow is needed. Note that the borrow
gives us 2 in the current bit position.
.

22. Example

23. In General
• When there is no borrow into the msb position, then the
subtrahend in not larger than the minuend and the result is
positive and correct.
• If a borrow into the msb does occur, then the subtrahend is
larger than the minuend.

24. Consider
• Now do the operation 4 – 6
• Correct difference is -2 or -0010
• Different because 2n was brought in and made the operation M-
N+2n

25. Hexadecimal
number system
• Hexadecimal system is composed of 16
numerals or symbol. 10 11 12 13 14 15
• Symbol: 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
16 Symbol
• Example: 7A16
Multiplier:
163 162 161 160 . 16-1
= 4096 = 256 = 16 =1 . = 0.0626

26. Example:
B6F.7C16
This number can be convert to decimal value using this calculation:
B 6 F . 7 C
162 161 160 . 16-1 16-2
= 256 = 16 =1 . 0.0625 = 0.0039
B6F.7C16 = (11x256) + (6x16) + (15x1) + (7x0.0625) + (12x0.0039)
= 2816 + 96 + 15 + 0.4375 + 0.0468
= 2927.484310

27. Arithmetic Operation
Hexadecimal
+ -

28. Hexadecimal Addition
Sekiranya setiap hasil perjumlahan yang melebihi atau sama
dengan 16 mestilah ditolak dengan 16.
Example:
a. 3316 + 4716 = b. 20D316 + 12BC16 =
3316 20D316
+ 4716 + 12BC16
338F16
7A16
Try this : DF16 + AB16 = ?

29. Hexadecimal Subtraction
Nilai yang kecil daripada 16 boleh dipinjam dari sebelah dengan
nilai 16.
Example:
a. 4416 - 1716 = b. 20D316 - 12BC16 =
20D316
4416
- 12BC16
- 1716 0E1716
2D16
Try this : DF16 - AB16 = ?

30. 2.3: Convert Decimal, Binary, Octal
and Hexadecimal Numbers to
different bases

31. Convert Binary to Decimal (N2 – N10)
Example:
1111012
This number can be convert to decimal value using this calculation:
1 1 1 1 0 1
25 24 23 22 21 20
= 32 = 16 =8 =4 =2 =1
1111012 = (1x32)+(1x16)+(1x8)+(1x4)+(0x2)+(1x1)
= 32 + 16 + 8 + 4 + 0 + 1
= 6110
Try this: Convert 1100.10112 to decimal?
Convert 100.10112 to decimal?

32. Convert Binary to Octal (N2 - N8)
Convert Binary to Octal adalah dengan membahagikan nombor
Binary tersebut kepada 3 bit bermula dari sebelah kanan (LSB)
LSB
1111012
1 1 1 1 0 1
22 21 20 22 21 20
=4 =2 =1 =4 =2 =1
1111012 = [(1x4)+(1x2)+(1x1)] [(1x4)+(0x2)+(1x1)]
= [4 + 2 + 1][ 4 + 0 + 1]
= 758
Try this: Convert 110010112 to Octal?

33. Convert Binary to Hexadecimal
(N2 – N16)
Convert Binary to Hexadecimal adalah dengan membahagikan
nombor binary kepada 4 bit bermula dari LSB. Sekiranya bit tersebut
tidak mencukupi, maka digit „0‟ perlu ditambah pada MSB
LSB
01012
0 1 0 1
23 22 21 20
=8 =4 =2 =1
01012 = (0x8)+(1x4)+(0x2)+(1x1)
=0+4+0+1
= 516
Try this: Convert 101111012 to Hexadecimal?

34. Convert Decimal to Binary (N10 – N2)
Example: Convert 1810 to binary
2
2
18
9
181
0
2 4 1
2 2 0
2 1 0
0 1
1810 = 100102
Try this: Convert 32.20210 to binary?
Convert 8910 to binary?

35. Convert Decimal to Octal (N10 – N8)
Example: Convert 300.3410 to Octal
i. 300 Divide by 8
ii. 0.34 Multiply by 8
8 300
8 37 4
8 4 5
8 0 4 454
0.3410 = 0.34 x 8 = 2.72 ( 2+0.72 )
0.72 x 8 = 5.76 ( 5+0.76 )
0.76 x 8 = 6.08 ( 6+0.08 )
0.08 x 8 = 0.64 ( 0+0.64 )
0.25605
0.64 x 8 = 5.12 ( 5+0.12 )
300.3410 = 454.256058
Try this: Convert 32.20210 to Octal?

36. Convert Decimal to Hexadecimal
(N10 – N16)
Example: Convert 2010 to Hexadecimal
16 20 Balance
16 1 4
0 1
2010= 1416
Try this: Convert 343410 to hexadecimal?

37. Convert Octal to Binary (N8 – N2)
Convert Octal to Binary adalah dengan menukar setiap digit oktal
kepada nilai 3 bit binary nya
MSB
4 58 LSB
1 0 0 1 0 1
22 21 20 22 21 20
=4 =2 =1 =4 =2 =1
1001018 = [(1x4)+(0x2)+(0x1)] [(1x4)+(0x2)+(1x1)]
= [4 + 0 + 0][ 4 + 0 + 1]
= 100 1012
Try this: Convert 110010112 to Octal?

38. Convert Hexadecimal to Binary
(N16 – N2)
Convert Octal to Binary adalah dengan menukar setiap digit
hexadecimal kepada nilai 4 bit binary nya
MSB
3 A16 LSB
0 0 1 1 1 0 1 0
23 22 21 20 23 22 21 20
=8 =4 =2 =1 =8 =4 =2 =1
3A16 = [(0x8)+(0x4)+(1x2)+(1x1)][(1x8)+(0x4)+(1x2)+(0x1)]
= [0 + 0 + 2 + 1][ 8 + 0 + 2 + 0]
= 0011 10102
Try this: Convert EFA16 to Binary?

39. 2.4: Coding System
 Sign and magnitude,
 1‟s Complement & 2‟s Complement

40. 8-Bit Binary Number System
Apply what you have learned to the +127 01111111
pos(+)
binary number systems. How do you +126 01111110
represent negative numbers in this 8-bit
+125 01111101
binary system?
Cut the number system in half. +1 00000001
0 00000000
Use 00000001 – 01111111 to indicate
-1 11111111
positive numbers.
-2 11111110
Use 10000000 – 11111111 to indicate
negative numbers. -127 10000001
neg(-)
-128 10000000
Notice that 00000000 is not positive or
negative.

41. Representing Negative Numbers
• As there is no third symbol available to
store a negative symbol explicitly we must
use a bit to show if a number is negative
or not.
– We name this bit the „Sign Bit‟
– We use the leftmost bit.
– If the „Sign Bit‟ is 1 then the number is
negative, if it is 0 then it is positive.

42. Sign Bit
• What did do you notice about the +127 01111111
pos(+)
most significant bit of the binary +126 01111110
numbers? +125 01111101
• The MSB is (0) for all positive
numbers. +1 00000001
• The MSB is (1) for all negative 0 00000000
numbers. -1 11111111
-2 11111110
• The MSB is called the sign bit.
• In a signed number system, this -127 10000001
allows you to instantly determine -128 10000000
neg(-)
whether a number is positive or
negative.

43. 1‟s Complement
• This is just inverting each bit.
flip the
0000010 number.
1 1 11 1 0 1
 1‟s compliment of 00000010
is 1111101

44. 2‟S Complement Process
The steps in the 2’s Complement process
First, complement all of the digits in a number.
– A digit‟s complement is the number you add to the digit to
make it equal to the largest digit in the base (i.e., 1 for
binary). In binary language, the complement of 0 is 1, and
the complement of 1 is 0.
Second, add 1.
– Without this step, our number system would have two
zeroes (+0 & -0), which no number system has.

45. 2‟s Complement Examples
Example #1
5 = 00000101
Complement Digits

11111010
+1
Add 1
-5 = 11111011
Example #2
-13 = 11110011
Complement Digits

00001100
+1
13 = 00001101

46. Using The 2‟s Compliment Process
Use the 2‟s complement process to add together
the following numbers.
POS 9 NEG (-9)
+ POS  + 5 + POS  + 5
POS 14 NEG -4
POS 9 NEG (-9)
+ NEG  + (-5) + NEG  + (-5)
POS 4 NEG -4

47. POS + POS → POS Answer
If no 2‟s complement is needed, use regular binary
addition.
9  00001001
+ 5  + 00000101
14  00001110

48. POS + NEG → POS Answer
Take the 2‟s complement of the negative number and
use regular binary addition.
9  00001001
+ (-5) + 11111011
4  1]00000100
8th Bit = 0: Answer is Positive
Disregard 9th Bit
00000101
 2’s
11111010 Complement
Process
+1
11111011

49. POS + NEG → NEG Answer
Take the 2‟s complement of the negative number and
use regular binary addition.
(-9) 11110111
+ 5  + 00000101
-4  11111100
8th Bit = 1: Answer is Negative
11111100 00001001
To Check:   2’s
Perform 2’s
Complement
00000011 11110110 Complement
Process
On Answer +1 +1
00000100 11110111

50. NEG + NEG → NEG Answer
Take the 2‟s complement of both negative numbers and
use regular binary addition.
2’s Complement
(-9)  11110111 Numbers, See
Conversion Process
+ (-5)  + 11111011 In Previous Slides
-14  1]11110010
8th Bit = 1: Answer is Negative
Disregard 9th Bit
11110010
To Check: 
Perform 2’s
Complement
00001101
On Answer +1
00001110

51. 2.4: Coding System
 Binary Coded Decimal (BCD System)
 ASCII and EBCDIC

52. Binary-Coded Decimal (BCD)
Four bits per digit Digit Bit pattern
Note: the following bit 0 0000
patterns are not used: 1 0001
1010 2 0010
1011
1100 3 0011
1101 4 0100
1110
5 0101
1111
6 0110
7 0111
8 1000
9 1001

53. Example
• 709310 = ? (in BCD)
7 0 9 3
0111 0000 1001 0011

54. ASCII
• ASCII = American National Standard
Code for Information Interchange
• 7-bit code
• 8th bit is unused (or used for a parity bit)
• 27 = 128 codes
• Two general types of codes:
– 95 are “Graphic” codes (displayable on a
console)
– 33 are “Control” codes (control features of the
console or communications channel)

55. ASCII Chart
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

56. 000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 Most significant bit
$ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011Least significant ESC
VT bit + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

57. e.g., ‘a’ = 1100001
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

58. 95 Graphic codes
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

59. 33 Control codes
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

60. Alphabetic codes
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

61. Numeric codes
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

62. Punctuation, etc.
000 001 010 011 100 101 110 111
0000 NULL DLE 0 @ P ` p
0001 SOH DC1 ! 1 A Q a q
0010 STX DC2 " 2 B R b r
0011 ETX DC3 # 3 C S c s
0100 EDT DC4 $ 4 D T d t
0101 ENQ NAK % 5 E U e u
0110 ACK SYN & 6 F V f v
0111 BEL ETB ' 7 G W g w
1000 BS CAN ( 8 H X h x
1001 HT EM ) 9 I Y i y
1010 LF SUB * : J Z j z
1011 VT ESC + ; K [ k {
1100 FF FS , < L l |
1101 CR GS - = M ] m }
1110 SO RS . > N ^ n ~
1111 SI US / ? O _ o DEL

63. The Problem
• Representing text strings, such as
“Hello, world”, in a computer

64. “Hello, world” Example
Binary Hexadecimal Decimal
H = 01001000 = 48 = 72
e = 01100101 = 65 = 101
l = 01101100 = 6C = 108
l = 01101100 = 6C = 108
o = 01101111 = 6F = 111
, = 00101100 = 2C = 44
= 00100000 = 20 = 32
w = 01110111 = 77 = 119
o = 01100111 = 67 = 103
r = 01110010 = 72 = 114
l = 01101100 = 6C = 108
d = 01100100 = 64 = 100

65. EBCDIC
Extended BCD Interchange Code
(pronounced ebb’-se-dick)
• 8-bit code
• Developed by IBM
• Rarely used today
• IBM mainframes only

66. EBCDIC “Extended Binary Coded
Decimal Interchange Code” code table

67. EBCDIC “Extended Binary Coded
Decimal Interchange Code” code table
Example: MSB LSB
1111 1111 1110 1001 1111 0111 1101 0111EBCDIC CODE
Z 6 P
Message below are represented in EBCDIC code. What is the message?
Please convert by using EBCDIC Code table given:
i) 1111 1100 1011 0101 1101 1001 EBCDIC CODE