Lesson 2 : Logic Gates and Boolean Algebra Part 1 Content: 1 .Boolean Theorem 2. Logic gates and Universal gates Part 2 Content : 1. Standard SOP and POS forms 2. Minterms and Maxterms 3. Karnaugh Map P.S. Part 2 content will be uploaded later

1. Digital Electronics
Digital
Electronics
Lesson 2 : Logic Gates and
Boolean Algebra
Sukriti Dhang
19/06/2020

2. Part 1

3. Content
1. Boolean Theorem
2. Logic gates and Universal
gates
3. Standard SOP and POS
forms
4. Minterms and Maxterms
5. Karnaugh Map

4. Boolean Logic Operations
A boolean function is an algebraic expression formed using binary
variables and basic logical operation symbols.
Basic logical
operations
AND function
(logical multiplication)
OR function
(logical addition)
NOT function
(logical complementation)

5. Logical AND function
Let us consider two variables A and B, where A and B are the inputs.
Y be the output, Y=A ᐧ B
A B Y=AB
0 0 0
0 1 0
1 0 0
1 1 1

6. Logical OR function
Let us consider two variables A and B, where A and B are the inputs.
Y be the output, Y=A + B
A B Y=A+B
0 0 0
0 1 1
1 0 1
1 1 1

7. Logical NOT function
Let us consider one variables A, where A is the input.
Y be the output, Y=A’
A Y=A’
0 1
1 0

8. Properties of Boolean Algebra
It is mathematical system consisting of a set of two or more distinct
elements.
➢ Commutative property
➢ Associative property
➢ Distributive property
➢ Absorption law
➢ Consensus law
➢ Idempotency property

9. Boolean laws
A+0=A A 1=Aᐧ -
A+1=A A 0=0ᐧ -
A+A=A A A=Aᐧ Idempotency
A+A’=1 A Aᐧ ’=0 Full set or null set
A’’=A Double complement

10. Commutative Property
Boolean addition is commutative
A+B = B+A
According to this property, the order of OR operation on variable makes
no diference. For eg. 0+1=1+0 (i.e. 0+1=1 and 1+0=1)
Boolean multiplication is commutative
AᐧB = BᐧA
According to this property, the order of AND operation on variable
makes no diference. For eg. 0ᐧ1=1ᐧ0 (i.e. 0ᐧ1=0 and 1ᐧ0=0)

11. Associative Property
Boolean addition is associative
A+(B+C) = (A+B)+C
According to this property, the OR operation of several variable makes no diference,
regardless of the grouping of the variables. For eg. 0+(1+1)=(0+1)+1
(i.e.0+(1+1)=1 and (0+1)+1 =1)
Boolean multiplication is associative
Aᐧ(BᐧC) = (AᐧB)ᐧC
According to this property, the order of AND operation of several variable makes no
diference, regardless of the grouping of the variables. For eg. 0ᐧ(1ᐧ1)=(0ᐧ1)ᐧ1 (i.e.
0ᐧ(1ᐧ1)=0 and (0ᐧ1)ᐧ1=0)

12. Distributive Property
Boolean addition is distributive
A+BᐧC = (A+B)ᐧ(A+C)
According to this property, the AND operation of several variables and then OR
operation of the result with single variable is equivalent to OR operation of
single variable with each of the several variables and then the AND operation
of the sums.
Boolean multiplication is distributive
Aᐧ(B+C) = (AᐧB)+(AᐧC)
According to this property, the OR operation of several variables and then AND
operation of the result with single variable is equivalent to AND operation of
single variable with each of the several variables and then the OR operation of

13. Absorption Law
Boolean addition is absorption
A+AᐧB = A
Proof: A+AᐧB=Aᐧ1+AᐧB
=Aᐧ(1+B) [since 1+B=1]
=Aᐧ1=A
Boolean multiplication is absorption
Aᐧ(A+B) = A
Proof: Aᐧ(A+B)=AᐧA+AᐧB
=A+(AᐧB)
=Aᐧ(1+B) [since 1+B=1]
=Aᐧ1=A

14. A+A’ᐧB=A+B
Proof : A+A’ᐧB=(A+A’)ᐧ(A+B)
=1ᐧ(A+B) [since A+A’=1 ]
=(A+B)
Aᐧ(A’+B)=AᐧB
Proof : Aᐧ(A’+B)=(AᐧA’)+(AᐧB)
=0+(AᐧB) [since AᐧA’=0 ]
=(AᐧB)

15. Consensus laws
AᐧB+A’ᐧC+BᐧC= AᐧB + A’ᐧC
Proof: AᐧB+A’ᐧC+BᐧC=AᐧB+A’ᐧC+BᐧCᐧ1
=AᐧB+A’ᐧC+BᐧCᐧ(A+A’) [since A+A’=1]
=AᐧB+A’ᐧC+AᐧBᐧC+A’ᐧBᐧC
=AᐧBᐧ(1+C) +A’ᐧCᐧ(1+B) [since 1+C=1, 1+B=1 ]
=AᐧB+A’ᐧC
(A+B)ᐧ(A’+C)ᐧ(B+C)=( A+B) ᐧ (A’+C)
Proof:(A+B)ᐧ(A’+C)ᐧ(B+C) =(A+B)ᐧ(A’+C)ᐧ(B+C+0)
=(A+B)ᐧ(A’+C)ᐧ(B+C+AᐧA’)
=(A+B)ᐧ(A’+C)ᐧ(B+C+A)ᐧ(B+C+A’) [since A+BC=(A+B)(A+C)]
=(A+B)ᐧ(A+B+C)ᐧ(A’+C)ᐧ(A’+B+C)
=(A+B)ᐧ(A’+C)

16. Principle of Duality
DeMorgan’s Theorem
(AᐧB)’=A’+B’
The theorem states that complement of product is equal to the sum of the
complements
(A+B)’=A’ᐧB’
The theorem states that complement of sum is equal to the product of the
complements

17. Proof:
1 2 3 4 5 6 7 8 9 10
A B A’ B’ A+B A Bᐧ (A+B)’ A’ Bᐧ ’ (A B)ᐧ ’ A’+B’
0 0 1 1 0 0 1 1 1 1
0 1 1 0 1 0 0 0 1 1
1 0 0 1 1 0 0 0 1 1
1 1 0 0 1 1 0 0 0 0
From the table, it is clear that columns number 7 and 8 are equal, thus satisfies
(A+B)’=A’ Bᐧ ’
And columns number 9 and 10 are equal, thus satisfies (A B)ᐧ ’=A’+B’

18. Examples
Y=A’B’C’+A’BC’+AB’C’+ABC’
Proof: A’B’C’+A’BC’+AB’C’+ABC’
=A’C’(B+B’)+AC’(B+B’)
=A’C’+AC’ [since B+B’=1]
=(A’+A)C’
=C’
Y=(A’+B)(A+B)
Proof: (A’+B)(A+B)
=B+AA’
=B+0 [since AA’=0]
=B

19. Examples contd.
Y=A+A’B’+A’B’C+A’B’C’D
Proof: A+A’B’+A’B’C+A’B’C’D
= A+A’B’C+B+A’B’C’D [since A+A’X=A+X, here X=B’C]
= A+B’C+B+A’B’C’D
= A+A’B’C’D+B+B’C [since A+A’X=A+X, here X=B’C’D; B+B’C=B+C]
= A+B’C’D +B+C
= A+C+B+B’C’D [since B+B’X=B+X, here X=C’D]
= A+C+B+C’D
=A+B+C+C’D [since C+C’D=C+D]
= A+B+C+D

20. Practise
Problem
1. Draw truth table for
a) Y=AC+AB
b) Y=(B’+A)C’
1. Simplify
a) Y=AB’+(A’+B)C
b) Y=A’B+BD+ACD’
c) Y=A+AB+A’BC
d) Y=AB+A(B+C)+B(B+C)
1. Apply Demorgan’ theroem
a) (A(B+C))’
b) ((AB)’+(CD)’)’

21. Part 2
contd.
Lesson 2 : Logic Gates
and Boolean Algebra
1. Standard SOP and POS
forms
2. Minterms and Maxterms
3. Karnaugh Map

22. ThankYou