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Malaysian Institute of Aviation Technology 
DIGITAL TECH (MECH) 
AKD 21102 
CHAPTER 1 
NUMBERING SYSTEM 
Revision 01 Issue02 Module 5.2 1
Malaysian Institute of Aviation Technology 
INTRODUCTION 
• In electronics, the data has to be transform from 
analogue to digital 
• The conversion of data is in ON state (binary 1) 
and OFF state (binary 0) 
• However, binary conversion has limitation when 
dealing with large value. 
• Therefore, the conversion can be in the form of 
decimal, octal, hexadecimal 
Revision 01 Issue02 Module 5.2 2
Malaysian Institute of Aviation Technology 
CHAPTER CONTENT 
1. Identifies binary, octal and hexadecimal system 
2. Convert and perform calculation on decimal 
and binary, octal and hexadecimal systems and 
vice versa. 
3. Identify BCD system. 
Revision 01 Issue02 Module 5.2 3
Malaysian Institute of Aviation Technology 
DECIMAL SYSTEM 
• The conventional system 
• Comprise of 10 digits from 0 to 9 
• Base 10 system 
• Positional value system numbering system 
• Example : 15110,254110 
Revision 01 Issue02 Module 5.2 4
Malaysian Institute of Aviation Technology 
DECIMAL SYSTEM 
• Example: 25410 consists of 2 HUNDREDS, 5 TENS and 4 
ONE units. 
• Written as : (2 X 100) + (5 X 10) + (4 X 1) 
= (2 X 102) + (5 X 101) + (4 X 100) 
• Digit 2 carries the MOST weight and is known as MOST 
SIGNIFICANT DIGIT (MSD) 
• Digit 4 carries the LEAST weight and is known as LEAST 
SIGNIFICANT DIGIT (LSD) 
Revision 01 Issue02 Module 5.2 5
Malaysian Institute of Aviation Technology 
• DISADVANTAGE for usage in digital 
computer example transistor due to : 
-Having 10 Discrete Value Level which is 
extremely difficult to operate due to: 
a. Any VARIATION of POWER SUPPLY would cause error 
b. Component TOLERANCE MUST be ZERO 
c. Component VALUE will change with AGE 
Revision 01 Issue02 Module 5.2 
6 
DECIMAL SYSTEM
Malaysian Institute of Aviation Technology 
• Comprise of 2 digits (0 & 1) known as BITS 
• Base 2 system. Example : 10112 
• POSITIONAL value system 
10112=(1X23)+(0X22)+(1X21)+(1X20) 
• BINARY to DECIMAL Conversion 
10112=(1X23)+(0X22)+(1X21)+(1X20) 
= 8+0+2+1 
= 11 
sum of each bits multiplied by its particular positional value 
Revision 01 Issue02 Module 5.2 
7 
BINARY SYSTEM
Malaysian Institute of Aviation Technology 
• DECIMAL to BINARY Conversion 
I) SUCCESSIVE POWER OF 2. 
Example: 27 =16+8+2+1 
=24+23+21+20 
=(1x24)+(1x23)+(0X22)+(1X21)+(1X20) 
=110112 
ii) SUCCESSIVE divide by 2 and record any remainder 
of division. 
• Suitable for SMALL number 
Revision 01 Issue02 Module 5.2 
8 
BINARY SYSTEM
Malaysian Institute of Aviation Technology 
• Any number converted into BINARY form, 
the binary numbers is known as a WORD. 
• Each word is formed of a numbers of BITS 
(BINARY DIGITS) and this represents the 
WORD LENGTH 
• Example : 34710 = 1010110112. 
• So 1010110112 is a WORD. Word length is 9 
because there is 9 bits 
Revision 01 Issue02 Module 5.2 
9 
BINARY SYSTEM
Malaysian Institute of Aviation Technology 
• Convert from binary to decimal 
• Convert from decimal to binary 
Revision 01 Issue02 Module 5.2 
10 
BINARY SYSTEM
Malaysian Institute of Aviation Technology 
TUTORIAL 
1. Convert the following binary numbers to decimal: 
(a) 101012 
(b) 1100112 
(c) 10010012 
(d) 101010112 
2. Convert the following decimal numbers to binary: 
(a) 25 
(b) 43 
(c) 65 
(d) 100. 
Revision 01 Issue02 Module 5.2 11
Malaysian Institute of Aviation Technology 
• Base 8 systems 
• Composed of 8 digits from 0 to 7 
• OCTAL to DECIMAL conversion 
Example : 
2758=(2x82)+(7x81)+(5x80) 
=128+56+5 = 18910 
• DECIMAL to OCTAL conversion 
 Divide by 8 and Record any REMAINDER 
of division 
Revision 01 Issue02 Module 5.2 
12 
OCTAL SYSTEM
Malaysian Institute of Aviation Technology 
• OCTAL to BINARY conversion 
- Convert each OCTAL number into 3 bits BINARY equivalent. 
• Example : 
a) 6358 TO BINARY. b) 
6 3 5 
110 011 101 
Thus, 6358= 1100111012 
Revision 01 Issue02 Module 5.2 
13 
OCTAL SYSTEM
Malaysian Institute of Aviation Technology 
• BINARY to OCTAL conversion 
- Divide BINARY number into groups of 3 BITS starting 
from LSB. 
Example : 
1001110112 TO OCTAL. 
100 111 011 
4 7 3 
Thus, 1001110112=4738 
• If the FINAL group of MSB does NOT have 3 BITS, ADD 
enough ZERO to make up 3 BITS. 
Revision 01 Issue02 Module 5.2 
14 
OCTAL SYSTEM
Malaysian Institute of Aviation Technology 
TUTORIAL 
1. Convert the following binary numbers to octal: 
(a) 101012 
(b) 1100112 
(c) 10010012 
(d) 101010112 
2. Convert the following decimal numbers to octal: 
(a) 25 
(b) 43 
(c) 65 
(d) 100. 
Revision 01 Issue02 Module 5.2 15
Malaysian Institute of Aviation Technology 
• BASE 16. 
• Composed of 16 digit Symbols 
0 1 2 3 4 5 6 7 8 9 A B C D E F 
• Example: 85D1B16 
• HEX to DECIMAL 
conversion 
Example :B2F16=(11x162)+(2x161)+(15x160) 
=2816+32+15 
=286310 
Revision 01 Issue02 Module 5.2 
16 
HEXADECIMAL SYSTEM
Malaysian Institute of Aviation Technology 
HEXADECIMAL SYSTEM 
• DECIMAL to HEX Conversion 
Divide with 16 and take the REMAINDER of division 
• HEX to BINARY Conversion 
-Convert each HEX digit into 
4 bits BINARY equivalent. 
i.e. B2F16 TO BINARY 
B 2 F 
1011 0010 1111 
THUS, B2F16= 1011001011112 
Revision 01 Issue02 Module 5.2 
17
Malaysian Institute of Aviation Technology 
• BINARY to HEX Conversion 
- Divide BINARY number into groups of 4 bits STARTING at LSB. 
i.e. 1101101010012 TO HEX 
1101 1010 1001 
13 10 9 
D A 9 
Thus, 1101101010012=DA916 
Revision 01 Issue02 Module 5.2 
18 
HEXADECIMAL SYSTEM
Malaysian Institute of Aviation Technology 
• HEX to OCTAL Conversion and vice versa 
i) Convert HEX to BINARY 
ii)Convert BINARY to OCTAL 
 To Convert OCTAL to HEX, just REVERSE the process 
Example : 3D16 convert to OCTAL 
i)Convert HEX to BINARY, 3 D 
0011 1101 3D16=1111012 
ii)Convert BINARY to OCTAL, 111 101 
7 5 1111012=758 
Thus, 3D16=758 
Revision 01 Issue02 Module 5.2 
19 
HEXADECIMAL SYSTEM
Malaysian Institute of Aviation Technology 
TUTORIAL 
1. Find the decimal equivalent of the octal number 41. 
2. Find the octal equivalent of the decimal number 139. 
3. Find the binary equivalent of the octal number 537. 
4. Find the octal equivalent of the binary number 
111001100. 
5. Convert the hexadecimal number 3F to: 
(a) decimal and (b) binary. 
6. Convert the binary number 101111001 to 
(a) octal and (b) hexadecimal. 
7. Which of the following numbers is the largest? 
(a) C516 (b) 110000012 (c) 3038. 
Revision 01 Issue02 Module 5.2 20
Malaysian Institute of Aviation Technology 
BCD SYSTEM 
• Binary Coded Decimal 
• Number represented in 4 bits binary code 
• Leaving a space between each group of 4 
digits 
• Example : 
a) 1110 to BCD is 0001 0001BCD 
b) 1000 0101BCD in BCD to Decimal is 8510 
Revision 01 Issue02 Module 5.2 
21
Malaysian Institute of Aviation Technology 
TUTORIAL 
a) Convert from decimal to BCD 
i. 94 
ii. 529 
iii. 2947 
b) Convert from BCD to decimal 
i. 0111 0000 1001BCD 
ii. 0011 0110 0100BCD 
Revision 01 Issue02 Module 5.2 
22
Malaysian Institute of Aviation Technology 
ONE’S COMPLIMENT 
• The one’s complement of a binary number is 
formed by inverting the value of each digit of the 
original binary number (i.e. replacing 1s with 0s 
and 0s with 1s) 
• Example: the one’s complement of the binary 
number 1010 is 0101. 
• Similarly, the one’s complement of 01110001 is 
10001110. 
– Note: if you add the one’s complement of a number to 
the original number the result will be all 1s, 
Revision 01 Issue02 Module 5.2 
23
Malaysian Institute of Aviation Technology 
ONE’S COMPLIMENT 
• The one’s complement of a binary number is formed by 
inverting the value of each digit of the original binary 
number (i.e. replacing 1s with 0s and 0s with 1s) 
• Example: the one’s complement of the binary number 
1010 is 0101. 
• Similarly, the one’s complement of 01110001 is 
10001110. 
– Note: if you add the one’s complement of a number to the 
original number the result will be all 1s, 
Revision 01 Issue02 Module 5.2 
24
Malaysian Institute of Aviation Technology 
TWO’S COMPLIMENT 
• Two’s complement notation is frequently used to represent negative 
numbers in computer mathematics (with only one possible code for 
zero—unlike one’s complement notation). 
• The two’s complement of a binary number is formed by inverting the 
digits of the original binary number and then adding 1 to the result. 
• Example: the two’s complement of the binary number 1001 is 0111. 
Similarly, the two’s complement of 01110001 is 10001111. 
• When two’s complement notation is used to represent negative 
numbers the most significant digit (MSD) is always a 1 
Revision 01 Issue02 Module 5.2 
25
Malaysian Institute of Aviation Technology 
TUTORIAL 
1. Find the one’s complement of the binary 
number 100010. 
2. Find the two’s complement of the binary 
number 101101. 
Revision 01 Issue02 Module 5.2 26

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Topic 1 Digital Technique Numbering system

  • 1. Malaysian Institute of Aviation Technology DIGITAL TECH (MECH) AKD 21102 CHAPTER 1 NUMBERING SYSTEM Revision 01 Issue02 Module 5.2 1
  • 2. Malaysian Institute of Aviation Technology INTRODUCTION • In electronics, the data has to be transform from analogue to digital • The conversion of data is in ON state (binary 1) and OFF state (binary 0) • However, binary conversion has limitation when dealing with large value. • Therefore, the conversion can be in the form of decimal, octal, hexadecimal Revision 01 Issue02 Module 5.2 2
  • 3. Malaysian Institute of Aviation Technology CHAPTER CONTENT 1. Identifies binary, octal and hexadecimal system 2. Convert and perform calculation on decimal and binary, octal and hexadecimal systems and vice versa. 3. Identify BCD system. Revision 01 Issue02 Module 5.2 3
  • 4. Malaysian Institute of Aviation Technology DECIMAL SYSTEM • The conventional system • Comprise of 10 digits from 0 to 9 • Base 10 system • Positional value system numbering system • Example : 15110,254110 Revision 01 Issue02 Module 5.2 4
  • 5. Malaysian Institute of Aviation Technology DECIMAL SYSTEM • Example: 25410 consists of 2 HUNDREDS, 5 TENS and 4 ONE units. • Written as : (2 X 100) + (5 X 10) + (4 X 1) = (2 X 102) + (5 X 101) + (4 X 100) • Digit 2 carries the MOST weight and is known as MOST SIGNIFICANT DIGIT (MSD) • Digit 4 carries the LEAST weight and is known as LEAST SIGNIFICANT DIGIT (LSD) Revision 01 Issue02 Module 5.2 5
  • 6. Malaysian Institute of Aviation Technology • DISADVANTAGE for usage in digital computer example transistor due to : -Having 10 Discrete Value Level which is extremely difficult to operate due to: a. Any VARIATION of POWER SUPPLY would cause error b. Component TOLERANCE MUST be ZERO c. Component VALUE will change with AGE Revision 01 Issue02 Module 5.2 6 DECIMAL SYSTEM
  • 7. Malaysian Institute of Aviation Technology • Comprise of 2 digits (0 & 1) known as BITS • Base 2 system. Example : 10112 • POSITIONAL value system 10112=(1X23)+(0X22)+(1X21)+(1X20) • BINARY to DECIMAL Conversion 10112=(1X23)+(0X22)+(1X21)+(1X20) = 8+0+2+1 = 11 sum of each bits multiplied by its particular positional value Revision 01 Issue02 Module 5.2 7 BINARY SYSTEM
  • 8. Malaysian Institute of Aviation Technology • DECIMAL to BINARY Conversion I) SUCCESSIVE POWER OF 2. Example: 27 =16+8+2+1 =24+23+21+20 =(1x24)+(1x23)+(0X22)+(1X21)+(1X20) =110112 ii) SUCCESSIVE divide by 2 and record any remainder of division. • Suitable for SMALL number Revision 01 Issue02 Module 5.2 8 BINARY SYSTEM
  • 9. Malaysian Institute of Aviation Technology • Any number converted into BINARY form, the binary numbers is known as a WORD. • Each word is formed of a numbers of BITS (BINARY DIGITS) and this represents the WORD LENGTH • Example : 34710 = 1010110112. • So 1010110112 is a WORD. Word length is 9 because there is 9 bits Revision 01 Issue02 Module 5.2 9 BINARY SYSTEM
  • 10. Malaysian Institute of Aviation Technology • Convert from binary to decimal • Convert from decimal to binary Revision 01 Issue02 Module 5.2 10 BINARY SYSTEM
  • 11. Malaysian Institute of Aviation Technology TUTORIAL 1. Convert the following binary numbers to decimal: (a) 101012 (b) 1100112 (c) 10010012 (d) 101010112 2. Convert the following decimal numbers to binary: (a) 25 (b) 43 (c) 65 (d) 100. Revision 01 Issue02 Module 5.2 11
  • 12. Malaysian Institute of Aviation Technology • Base 8 systems • Composed of 8 digits from 0 to 7 • OCTAL to DECIMAL conversion Example : 2758=(2x82)+(7x81)+(5x80) =128+56+5 = 18910 • DECIMAL to OCTAL conversion  Divide by 8 and Record any REMAINDER of division Revision 01 Issue02 Module 5.2 12 OCTAL SYSTEM
  • 13. Malaysian Institute of Aviation Technology • OCTAL to BINARY conversion - Convert each OCTAL number into 3 bits BINARY equivalent. • Example : a) 6358 TO BINARY. b) 6 3 5 110 011 101 Thus, 6358= 1100111012 Revision 01 Issue02 Module 5.2 13 OCTAL SYSTEM
  • 14. Malaysian Institute of Aviation Technology • BINARY to OCTAL conversion - Divide BINARY number into groups of 3 BITS starting from LSB. Example : 1001110112 TO OCTAL. 100 111 011 4 7 3 Thus, 1001110112=4738 • If the FINAL group of MSB does NOT have 3 BITS, ADD enough ZERO to make up 3 BITS. Revision 01 Issue02 Module 5.2 14 OCTAL SYSTEM
  • 15. Malaysian Institute of Aviation Technology TUTORIAL 1. Convert the following binary numbers to octal: (a) 101012 (b) 1100112 (c) 10010012 (d) 101010112 2. Convert the following decimal numbers to octal: (a) 25 (b) 43 (c) 65 (d) 100. Revision 01 Issue02 Module 5.2 15
  • 16. Malaysian Institute of Aviation Technology • BASE 16. • Composed of 16 digit Symbols 0 1 2 3 4 5 6 7 8 9 A B C D E F • Example: 85D1B16 • HEX to DECIMAL conversion Example :B2F16=(11x162)+(2x161)+(15x160) =2816+32+15 =286310 Revision 01 Issue02 Module 5.2 16 HEXADECIMAL SYSTEM
  • 17. Malaysian Institute of Aviation Technology HEXADECIMAL SYSTEM • DECIMAL to HEX Conversion Divide with 16 and take the REMAINDER of division • HEX to BINARY Conversion -Convert each HEX digit into 4 bits BINARY equivalent. i.e. B2F16 TO BINARY B 2 F 1011 0010 1111 THUS, B2F16= 1011001011112 Revision 01 Issue02 Module 5.2 17
  • 18. Malaysian Institute of Aviation Technology • BINARY to HEX Conversion - Divide BINARY number into groups of 4 bits STARTING at LSB. i.e. 1101101010012 TO HEX 1101 1010 1001 13 10 9 D A 9 Thus, 1101101010012=DA916 Revision 01 Issue02 Module 5.2 18 HEXADECIMAL SYSTEM
  • 19. Malaysian Institute of Aviation Technology • HEX to OCTAL Conversion and vice versa i) Convert HEX to BINARY ii)Convert BINARY to OCTAL  To Convert OCTAL to HEX, just REVERSE the process Example : 3D16 convert to OCTAL i)Convert HEX to BINARY, 3 D 0011 1101 3D16=1111012 ii)Convert BINARY to OCTAL, 111 101 7 5 1111012=758 Thus, 3D16=758 Revision 01 Issue02 Module 5.2 19 HEXADECIMAL SYSTEM
  • 20. Malaysian Institute of Aviation Technology TUTORIAL 1. Find the decimal equivalent of the octal number 41. 2. Find the octal equivalent of the decimal number 139. 3. Find the binary equivalent of the octal number 537. 4. Find the octal equivalent of the binary number 111001100. 5. Convert the hexadecimal number 3F to: (a) decimal and (b) binary. 6. Convert the binary number 101111001 to (a) octal and (b) hexadecimal. 7. Which of the following numbers is the largest? (a) C516 (b) 110000012 (c) 3038. Revision 01 Issue02 Module 5.2 20
  • 21. Malaysian Institute of Aviation Technology BCD SYSTEM • Binary Coded Decimal • Number represented in 4 bits binary code • Leaving a space between each group of 4 digits • Example : a) 1110 to BCD is 0001 0001BCD b) 1000 0101BCD in BCD to Decimal is 8510 Revision 01 Issue02 Module 5.2 21
  • 22. Malaysian Institute of Aviation Technology TUTORIAL a) Convert from decimal to BCD i. 94 ii. 529 iii. 2947 b) Convert from BCD to decimal i. 0111 0000 1001BCD ii. 0011 0110 0100BCD Revision 01 Issue02 Module 5.2 22
  • 23. Malaysian Institute of Aviation Technology ONE’S COMPLIMENT • The one’s complement of a binary number is formed by inverting the value of each digit of the original binary number (i.e. replacing 1s with 0s and 0s with 1s) • Example: the one’s complement of the binary number 1010 is 0101. • Similarly, the one’s complement of 01110001 is 10001110. – Note: if you add the one’s complement of a number to the original number the result will be all 1s, Revision 01 Issue02 Module 5.2 23
  • 24. Malaysian Institute of Aviation Technology ONE’S COMPLIMENT • The one’s complement of a binary number is formed by inverting the value of each digit of the original binary number (i.e. replacing 1s with 0s and 0s with 1s) • Example: the one’s complement of the binary number 1010 is 0101. • Similarly, the one’s complement of 01110001 is 10001110. – Note: if you add the one’s complement of a number to the original number the result will be all 1s, Revision 01 Issue02 Module 5.2 24
  • 25. Malaysian Institute of Aviation Technology TWO’S COMPLIMENT • Two’s complement notation is frequently used to represent negative numbers in computer mathematics (with only one possible code for zero—unlike one’s complement notation). • The two’s complement of a binary number is formed by inverting the digits of the original binary number and then adding 1 to the result. • Example: the two’s complement of the binary number 1001 is 0111. Similarly, the two’s complement of 01110001 is 10001111. • When two’s complement notation is used to represent negative numbers the most significant digit (MSD) is always a 1 Revision 01 Issue02 Module 5.2 25
  • 26. Malaysian Institute of Aviation Technology TUTORIAL 1. Find the one’s complement of the binary number 100010. 2. Find the two’s complement of the binary number 101101. Revision 01 Issue02 Module 5.2 26