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Basics of Digital ElectronicsBasics of Digital Electronics
Prof.Anjali JagtapProf.Anjali Jagtap
anjalij@isquareit.edu.inanjalij@isquareit.edu.in
Prof.Anjali JagtapProf.Anjali Jagtap
anjalij@isquareit.edu.inanjalij@isquareit.edu.in
Department of Electronics andTelecommunicationDepartment of Electronics andTelecommunication
International Institute of InformationTechnology,International Institute of InformationTechnology,
PunePune –– 411057411057
Digital SignalsDigital Signals
An electrical signal with two discrete levels (high and low)
Two discrete levels are represented by binary digits 0 and 1
referred as Binary number system.
Gorge Boole introduced binary number system with algebra
developed “Boolean Algebra”
Represented in two different ways
• Positive logic system
• Negative logic system
Digital system typesDigital system types
Combinational logic system/circuits
• An output at any instant depends only on inputs applied at
that instant.
• Example – Adder, subtractor, Comparator etc
• Basic building block – logic gates
Sequential logic system/circuits
• An output at any instant depends only on inputs applied at
that instant as well as on past inputs/outputs.
• Example – counters, sequence generator/ detector etc
• Requires memory
• Basic building block – Flips and logic gates
Logic GatesLogic Gates
Basic logic gates
• AND gate
• Logical Multiplication
• Two input gate shown
Basic logic gates
•OR gate
• Logical Addition
• Two input gate shown
•NOT/Inversion gate
• Logical inversion
• Single input single output gate
••Universal logic gatesUniversal logic gates
• NAND gate
• Two input gate shown
• NOR gate
• Two input gate shown
••Special gatesSpecial gates
• Ex-OR gate
• Two input gate shown
• Ex-NOR gate
• Two input gate shown
Boolean AlgebraBoolean Algebra
Mathematician George Boole developed
rules for manipulation of binary variables.
Rules :
• A+0=A
• A+1=1• A+1=1
• A+A=A
• A+A’=1
• A.0=0
• A.1=A
• A.A=A
• A.A’=0
• A.(B+C)=AB+AC
Boolean AlgebraBoolean Algebra
• A+BC=(A+B).(A+C)
• A+A.B=A
• A.(A+B)=A
• A+A’.B=A+B
• A.(A’+B)=A.B
• A.B+A’.B’=A
• (A+B).(A+B’)=A
• A.B+A.C’=(A+C).(A’+B)
• (A+B).(A’+C)=AC+A’B
• AB+A’C+BC=AB+A’C
• (A+b).(A’+C).(B+C)=(A+B).(A’+C)
De Morgan’s TheoremDe Morgan’s Theorem
• (A.B)’=A’+B’
• (A+B)’=A’.B’
Number SystemNumber System
Number
System
Base or radix Symbols used
(di or d-f)
Weight assigned to
position
Example
i -f
Binary 2 0,1 2i 2-f 10101.10
Octal 8 0,1,2,3,4,5,6,7 8i 8-f 3547.25
Decimal 10 0,1,2,3,4,5,6,7
,8,9
10i 10-f 974.27
Hexadecimal 16 0,1,2,3,4,5,6,7
,8,9,A,B,C,D,
E,F
16i 16-f FA9.46
Quantities/CountingQuantities/Counting
Decimal Binary Octal Hexadecimal
0 0000 00 0
1 0001 01 1
2 0010 02 2
3 0011 03 3
4 0100 04 4
5 0101 05 5
6 0110 06 6
7 0111 07 7
8 1000 10 8
9 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
16 1111 17 F
Number System conversionNumber System conversion
•Binary to decimal
• Multiply each bit by 2n, n is the “weight” of the bit
• The weight is the position of the bit, starting from 0
on the right
• Add the results• Add the results
Example
(110101)2 = ( )10
= 1x25+1x24+0x23+1x22+0x21+1x20
= 32+16+0+4+0+1
= (53)10
Number System conversionNumber System conversion
•Binary to octal
• Group bits in threes, starting on right
• Convert to octal number
ExampleExample
(110101)2 = ( )8
= 110 101
= 6 5
= (65)8
Number System conversionNumber System conversion
•Binary to hexadecimal
• Group bits in fours, starting on right
• Convert to hexadecimal number
ExampleExample
(110101)2 = ( )16
= 11 0101
= 0011 0101
= (35)16
Number System conversionNumber System conversion
•Decimal to binary
• Divide by two, keep track of the remainder
• First remainder is bit 0 (LSB, least-significant bit)
• Second remainder is bit 1Group bits in fours, starting
on right
Example
(53) = ( )(53)10 = ( )2
= (110101)2
Number System conversionNumber System conversion
•Decimal to octal
• Divide by eight, keep track of the remainder
• First remainder is bit 0 (LSB, least-significant
bit)
• Second remainder is bit 1Group bits in fours,
starting on rightstarting on right
Example
(53)10 = ( )8
= (65)8
Number System conversionNumber System conversion
•Decimal to hexadecimal
• Divide by 16, keep track of the remainder
• First remainder is bit 0 (LSB, least-significant bit)
• Second remainder is bit 1Group bits in fours, starting on
right
Example
(53) = ( )(53)10 = ( )16
=
= (35)16 `
Number System conversionNumber System conversion
•Octal to binary
• Convert each octal digit to a 3-bit equivalent
binary representation
ExampleExample
(65)8 = ( )2
= 110 101
= (110101)2
Number System conversionNumber System conversion
•Octal to decimal
• Multiply each bit by 8n, n is the “weight” of the bit
• The weight is the position of the bit, starting from 0
on the right
• Add the results• Add the results
Example
(65)8 = ( )10
= 6x81+5x80
= 48+5
= (53)10
Number System conversionNumber System conversion
•Octal to hexadecimal
• Use binary as an intermediary.
Example
(65)8 = ( )16(65)8 = ( )16
= 110 101
= 11 0101
= 0011 0101
= (35)16
Number System conversionNumber System conversion
•Hexadecimal to binary
• Convert each hexadecimal digit to a 4-bit
equivalent binary representation.
ExampleExample
(6A)16 = ( )2
= 0110 1010
= (1101010)2
Number System conversionNumber System conversion
•Hexadecimal to octal
• Use binary as an intermediary.
Example
(6A)16 = ( )8(6A)16 = ( )8
= 0110 1010
= 1 101 010
= 001 101 010
= (152)8
Number System conversionNumber System conversion
•Hexadecimal to decimal
• Multiply each bit by 8n, n is the “weight” of the bit
• The weight is the position of the bit, starting from 0
on the right
• Add the results
ExampleExample
(6A)16 = ( )10
= 6x161+Ax160
= 6x161+10x160
= 96 + 10
= (106)10
Complement representationComplement representation
• One’s complement format
• In binary number system, each bit is complimented.
Example
(0100101)2 = (1011010) one’s complement form
• Two’s complement format• Two’s complement format
• In binary number system, each bit is complimented and
binary 1 is added.
• Used to represent negative number.
Example
(0100101)2 = (90)10
= (1011010)2 one’s complement form
= (1011010 + 1)2
= (1011011)2 = (-90)10 2’s complement

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Basics of Digital Electronics

  • 1. Basics of Digital ElectronicsBasics of Digital Electronics Prof.Anjali JagtapProf.Anjali Jagtap anjalij@isquareit.edu.inanjalij@isquareit.edu.in Prof.Anjali JagtapProf.Anjali Jagtap anjalij@isquareit.edu.inanjalij@isquareit.edu.in Department of Electronics andTelecommunicationDepartment of Electronics andTelecommunication International Institute of InformationTechnology,International Institute of InformationTechnology, PunePune –– 411057411057
  • 2. Digital SignalsDigital Signals An electrical signal with two discrete levels (high and low) Two discrete levels are represented by binary digits 0 and 1 referred as Binary number system. Gorge Boole introduced binary number system with algebra developed “Boolean Algebra” Represented in two different ways • Positive logic system • Negative logic system
  • 3. Digital system typesDigital system types Combinational logic system/circuits • An output at any instant depends only on inputs applied at that instant. • Example – Adder, subtractor, Comparator etc • Basic building block – logic gates Sequential logic system/circuits • An output at any instant depends only on inputs applied at that instant as well as on past inputs/outputs. • Example – counters, sequence generator/ detector etc • Requires memory • Basic building block – Flips and logic gates
  • 4. Logic GatesLogic Gates Basic logic gates • AND gate • Logical Multiplication • Two input gate shown
  • 5. Basic logic gates •OR gate • Logical Addition • Two input gate shown •NOT/Inversion gate • Logical inversion • Single input single output gate
  • 6. ••Universal logic gatesUniversal logic gates • NAND gate • Two input gate shown • NOR gate • Two input gate shown
  • 7. ••Special gatesSpecial gates • Ex-OR gate • Two input gate shown • Ex-NOR gate • Two input gate shown
  • 8. Boolean AlgebraBoolean Algebra Mathematician George Boole developed rules for manipulation of binary variables. Rules : • A+0=A • A+1=1• A+1=1 • A+A=A • A+A’=1 • A.0=0 • A.1=A • A.A=A • A.A’=0 • A.(B+C)=AB+AC
  • 9. Boolean AlgebraBoolean Algebra • A+BC=(A+B).(A+C) • A+A.B=A • A.(A+B)=A • A+A’.B=A+B • A.(A’+B)=A.B • A.B+A’.B’=A • (A+B).(A+B’)=A • A.B+A.C’=(A+C).(A’+B) • (A+B).(A’+C)=AC+A’B • AB+A’C+BC=AB+A’C • (A+b).(A’+C).(B+C)=(A+B).(A’+C)
  • 10. De Morgan’s TheoremDe Morgan’s Theorem • (A.B)’=A’+B’ • (A+B)’=A’.B’
  • 11. Number SystemNumber System Number System Base or radix Symbols used (di or d-f) Weight assigned to position Example i -f Binary 2 0,1 2i 2-f 10101.10 Octal 8 0,1,2,3,4,5,6,7 8i 8-f 3547.25 Decimal 10 0,1,2,3,4,5,6,7 ,8,9 10i 10-f 974.27 Hexadecimal 16 0,1,2,3,4,5,6,7 ,8,9,A,B,C,D, E,F 16i 16-f FA9.46
  • 12. Quantities/CountingQuantities/Counting Decimal Binary Octal Hexadecimal 0 0000 00 0 1 0001 01 1 2 0010 02 2 3 0011 03 3 4 0100 04 4 5 0101 05 5 6 0110 06 6 7 0111 07 7 8 1000 10 8 9 1001 11 9 10 1010 12 A 11 1011 13 B 12 1100 14 C 13 1101 15 D 14 1110 16 E 16 1111 17 F
  • 13. Number System conversionNumber System conversion •Binary to decimal • Multiply each bit by 2n, n is the “weight” of the bit • The weight is the position of the bit, starting from 0 on the right • Add the results• Add the results Example (110101)2 = ( )10 = 1x25+1x24+0x23+1x22+0x21+1x20 = 32+16+0+4+0+1 = (53)10
  • 14. Number System conversionNumber System conversion •Binary to octal • Group bits in threes, starting on right • Convert to octal number ExampleExample (110101)2 = ( )8 = 110 101 = 6 5 = (65)8
  • 15. Number System conversionNumber System conversion •Binary to hexadecimal • Group bits in fours, starting on right • Convert to hexadecimal number ExampleExample (110101)2 = ( )16 = 11 0101 = 0011 0101 = (35)16
  • 16. Number System conversionNumber System conversion •Decimal to binary • Divide by two, keep track of the remainder • First remainder is bit 0 (LSB, least-significant bit) • Second remainder is bit 1Group bits in fours, starting on right Example (53) = ( )(53)10 = ( )2 = (110101)2
  • 17. Number System conversionNumber System conversion •Decimal to octal • Divide by eight, keep track of the remainder • First remainder is bit 0 (LSB, least-significant bit) • Second remainder is bit 1Group bits in fours, starting on rightstarting on right Example (53)10 = ( )8 = (65)8
  • 18. Number System conversionNumber System conversion •Decimal to hexadecimal • Divide by 16, keep track of the remainder • First remainder is bit 0 (LSB, least-significant bit) • Second remainder is bit 1Group bits in fours, starting on right Example (53) = ( )(53)10 = ( )16 = = (35)16 `
  • 19. Number System conversionNumber System conversion •Octal to binary • Convert each octal digit to a 3-bit equivalent binary representation ExampleExample (65)8 = ( )2 = 110 101 = (110101)2
  • 20. Number System conversionNumber System conversion •Octal to decimal • Multiply each bit by 8n, n is the “weight” of the bit • The weight is the position of the bit, starting from 0 on the right • Add the results• Add the results Example (65)8 = ( )10 = 6x81+5x80 = 48+5 = (53)10
  • 21. Number System conversionNumber System conversion •Octal to hexadecimal • Use binary as an intermediary. Example (65)8 = ( )16(65)8 = ( )16 = 110 101 = 11 0101 = 0011 0101 = (35)16
  • 22. Number System conversionNumber System conversion •Hexadecimal to binary • Convert each hexadecimal digit to a 4-bit equivalent binary representation. ExampleExample (6A)16 = ( )2 = 0110 1010 = (1101010)2
  • 23. Number System conversionNumber System conversion •Hexadecimal to octal • Use binary as an intermediary. Example (6A)16 = ( )8(6A)16 = ( )8 = 0110 1010 = 1 101 010 = 001 101 010 = (152)8
  • 24. Number System conversionNumber System conversion •Hexadecimal to decimal • Multiply each bit by 8n, n is the “weight” of the bit • The weight is the position of the bit, starting from 0 on the right • Add the results ExampleExample (6A)16 = ( )10 = 6x161+Ax160 = 6x161+10x160 = 96 + 10 = (106)10
  • 25. Complement representationComplement representation • One’s complement format • In binary number system, each bit is complimented. Example (0100101)2 = (1011010) one’s complement form • Two’s complement format• Two’s complement format • In binary number system, each bit is complimented and binary 1 is added. • Used to represent negative number. Example (0100101)2 = (90)10 = (1011010)2 one’s complement form = (1011010 + 1)2 = (1011011)2 = (-90)10 2’s complement