Digital logic gates are basic building blocks of digital circuits that make logical decisions based on input combinations. There are three basic logic gates: OR, AND, and NOT. Other common gates such as NAND, NOR, XOR, and XNOR are derived from these. Boolean algebra uses variables that can be 1 or 0, and logical operators like AND, OR, and NOT to represent logic functions. Logic functions can be expressed in canonical forms such as sum of minterms or product of maxterms. Standard forms like SOP and POS are also used. Conversions between these forms allow simplifying logic functions.

1. 1. DIGITAL LOGIC GATES
A logic gate is the basic building block of digital circuits. A digital logic gate is an
electronic device that makes logical decision based on the different combination of inputs.
There are three basic logic gates, namely the OR gate,the AND gate and the NOT gate.
Other logic gates that are derived from these basic gates are the NAND gate, the NOR gate,
the EXCLUSIVE-OR (XOR) gate and the EXCLUSIVE-NOR (XNOR) gate.
AND gate:
The AND gate gives a low output (0) if any one of its input is low. A dot (.) is used to
show the AND operation.
OR gate:
The OR gate gives a high output (1) if any one of its input is high. A plus (+) is used to
show the OR operation.
NOT gate:
The NOT gate produces an inverted (complement) version of the input at its output. It
is also known as an inverter.

2. 1
NAND gate:
The NAND gate gives a high output (1) if any one of its inputs is low (0). The
NAND gate operation is inversion of AND gate operation.
NOR gate:
The NOR gate gives a low output (0) if any one of its input is high. The NOR gate
operation is inversion of OR gate operation.
XOR:
Exclusive-OR gate gives a high output (1), if the inputs are at different logic levels i.e
either (0 and 1) or (1 and 0) and its output is low (0) if the inputs are at the same logic levels.
An encircled plus sign (⊕) is used to show the XOR operation.
A⊕B =A’B + AB’

3. 2
XNOR:
The Exclusive-NOR gate gives a high (1) output only if both of its inputs are same and
its output is low (0) if the inputs are at different logic levels, either (0 and 1) or (0 and 1). An
encircled dot sign (⊙) is used to show the XNOR operation.
A ⊙ B =AB + A’B’
2. BOOLEAN ALGEBRA
Boolean Algebra is an algebra that deals with binary variables and logic operations.
2.1. Rules in Boolean Algebra:
Following are the important rules used in Boolean algebra.
▪ Variable used can have only two values. Binary 1 for HIGH and Binary 0 for LOW.
▪ Logical AND operation of the variables is represented by a dot (.) sign between the
variables eg: (A.B.C) or (ABC).
▪ Logical OR operation of the variables is represented by a plus (+) sign between the
variables eg: (A+B+C).
▪ Logical NOT operation of a variable (eg) ‘x’ is represented by a bar ( 𝑥 ) or prime
(x').
2.2. Variables, Literals and constant in Boolean Functions:
Boolean function or Boolean Expression:
a. Boolean function consists of binary variables and logical operations (AND, OR,
NOT) and constants (0 and 1).
b. It has one of the two possible outputs either ‘0’ or ‘1’.
Eg: F(A,B,C) = AB+BC'+CA'
● Binary Variables or Variables:
The variables are designated by alphabets such as A, B, C, X, Y, Z and so on.
● Literals:
Each occurrence of a variable or its complement in the Boolean expression is called
as a literal.
Eg: F( X, Y ,Z) = XY+YZ'+X'Z
There are 6 literals in the above example.
● Constant :
Constant is a value that does not change.
Y = A + 1 or Y = A + 0 (here 0 and 1 are constants)

4. 3
2.3. Duality principle:
Dual of the Boolean expression is obtained by interchanging all AND’s and OR’s and
all 0’s and 1’s. The Boolean expression remains valid.
Eg: A+A' = 1
After applying duality principle,
A.A' = 0
2.4. Theorems and postulates of Boolean algebra
The theorems and postulates of Boolean algebra can be used to simplify the
Boolean expression by reducing the number of literals.
Some of the important Boolean postulates (Basic Assumptions) are
1. Identity: x + 0 = x
x . 1= x
2. Distributive: x . (y+ z) = (x . y) + (x . z)
x + (y . z) = (x + y) . (x + z)
3. Commutative: (x . y) = (y . x)
x + y = y + x
4. Complement: x + x' = 1
x . x' = 0
Some of the important Boolean theorems are
Theorem 1: Idempotency: x + x = x
x . x = x
Theorem 2: Annulment: x + 1 = 1
x . 0 = 0
Theorem 3: Involution: (x')' = x
Theorem 4: Associative: (x . y) . z = x . (y . z)
(x + y) + z = x + (y + z)
Theorem 5: Absorption x + x . y = x
x . (x + y) = x
Theorem 6: De-Morgan theorem:
(𝐴 + 𝐵) = 𝐴 . 𝐵
(𝐴 . 𝐵) = 𝐴 + 𝐵

5. 4
2.5. Boolean Functions
Boolean function can be represented in a truth table, which lists all binary combinations
of input variables (2n , where n is the number of variables) and their corresponding output
values. Consider the following Boolean function: F (A,B,C) = A+ B'C
Truth table: Implementation using logic circuit:
Inputs Output
A
B C F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
Other examples of implementation of Boolean function using logic circuit:
F1 = AB + CD + E F2 = W'Z + WZ'(X+Y')
2.6. Minimization of Boolean Functions
using Boolean theorems
Examples:
1. x(x' + y)
x(x' + y) = xx' + xy
= 0 + xy
= xy
2. (x + y)(x + y')
(x + y)(x + y') = x + xy + xy' + yy'
= x(1 + y + y') [i.e., 1+x=1]
= x
3. (x + y)(x' + z)(y + z)
= (x+y)(z+x'y) [i.e., (x+y).(y+z) = y+(x.z)]
= xz+xx'y+yz+yx'y [i.e., x.x'=0 & y.y=y]
= xz+0+yz+x'y
= xz+ x'y+yz

6. 5
Examples:
Minimization of Boolean function using demorgan’s theorem
1. (A + B + C) ' = A' .(B + C) ' by DeMorgan’s theorem
= A' . (B' . C') by DeMorgan’s theorem
= A'B'C'
2. (A. B .C) ' = A' + (B . C) ' by DeMorgan’s theorem
= A'+ (B'+C') by DeMorgan’s theorem
= A'+B' +C'
2.7. Complement ofa Boolean Function:
Examples:
1. Find complements using De-Morgans theorem: F1 = x'yz'+ x'y'z ; F2 = x(y'z'+ yz).
F1' = (x'yz' + x'y'z) ' = (x'yz')' . (x'y'z)'
= (x + y' + z) . (x + y + z')
F2' = [x(y'z' + yz)] ' = x' + (y'z' + yz) ' = x' + (y'z') ' (yz) '
= x' + (y + z)(y' + z')
= x' +y y'+ yz' + y'z +z z' [i.e., y . y'=0]
= x' + yz' + y'z
2. Find the complement of the functions F1 and F2 by applying duality principle and
complementing each literal.
F1 = x'yz' + x'y'z.
The duality of F1 = (x' + y + z')(x' + y' + z).
Complement each literal F1 = (x + y' + z)(x + y + z')
F1' = (x + y' + z)(x + y + z')
F2 = x(y'z' + yz)
The duality of F2 = x + (y'+ z')(y + z).
Complement each literal F2 = x'+ (y + z)(y' + z')
F2' = x'+ (y + z)(y' + z')
3. CANONICAL AND STANDARD FORMS
There are two ways to express a Boolean expression
1. Standard form
2. Canonical form
There are two types of standard forms. They are
● Sum of products (SOP)
● Products of sums (POS)

7. 6
3.1. Sum of products:
A Sum of products (SOP) expressions will appear as two or more AND terms ORed
together.
3.2. Product of sums:
Product of Sums (POS) expressions will appear as two or more OR terms ANDed
together.
3.3. CANONICAL FORM:
Boolean functions expressed as a Sum of Minterms (or) Product of Maxterms are
said to be in canonical form.
All the terms in the given expression should be present in either minterm or maxterm
with each variable in either primed (complemented) or unprimed (True) form.
Example: F(A,B,C) = ABC+AB'C+A'BC'
In the above example, A,B,C are variables. All variables are present in each term.
3.3.1. MINTERM (Standard Products):
Minterm are called standard products because they are the logical ‘AND’ of a set of
variables.
The pr
oduct t
er
m cont
ains all ‘n’ var
iables of t
he funct
ion in
eit
her pr
im
ed (com
plem
ent
) or unpr
im
ed (nor
m
al) for
m
. Each
m
int
er
m is obt
ained by an AND oper
at
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he var
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heir
primed or unprimed form. It is denoted by a symbol (∑)
If the bit is 0, the variable is primed. If the bit is 1, the variable is unprimed.

8. 7
Minterms for 3 variables is shown in the table:
For example:
Inputs Output
X Y Z F
0 0 0 0
0 0 1 0
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1
F(X,Y,Z) = XYZ+X'YZ+XY'Z' = ∑ (7,3,4)
= ∑ (3,4,7)

9. 8
3.3.2. MAXTERM (Standard Sums):
Maxterms are called standard sums because they are the logical ‘OR’ of a set of variables.
The sum term containing all n variables of the function in either primed (complemented)
or unprimed (Normal) form. Each maxterm is obtained by an OR operation of the variables in
their primed or unprimed form. It is denoted by a symbol (П).
If the bit is 1 the variable is primed. If the bit is 0 the variable is unprimed.
Maxterm for 3 variables as shown in below table:
For example:
Inputs Output
X Y Z F
0 0 0 1
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
F(X,Y,Z)= (X+Y'+Z).(X'+Y+Z) = П( 2, 4)

10. 9
Problems:
1. Express the function F(X,Y,Z)= X'Y'Z'+X'Y'Z+XYZ' in sum of minterms
= 000+001+110
=m0+m1+m6
Answer =∑ (0,1,6)
2. Express the function F(A,B,C,D) =(A+B+C+D').(A'+B+C+D').(A'+B'+C'+D) in
product of maxterm
= (0+0+0+1).(1+0+0+1).(1+1+1+0)
=M1.M9.M14
Answer=П(1,9,14)
4. CONVERSIONS – SOP, POS & CANONICAL FORM
a) CONVERSION BETWEEN CANONICAL FORMS:
Problems:
Convert the following sum of minterms to the product of maxterm.
F(X,Y,Z) =∑(1,3,7)
Answer = П(0,2,4,5,6)
Convert the following product of maxterm to the sum of minterms form.
F(X,Y,Z) =П(2,5,7)
Answer= ∑(0,1,3,4,6)
b) SOP TO CANONICAL FORM:
Procedure:
Step 1: Check whether the given expression contains all the variables.
Step 2: If there is a missing of variables in any minterm then “AND” the
corresponding minterm with expression (x+x'), where x be the missing
variable.
Step 3: Expand the terms by applying distributive law and reorder the literals
in the product terms.
Step 4: If two minterms are identical, eliminate one of the minterms.
Problems:
1. Express the Boolean function F (A,B,C) = A + B'C as a sum of minterms
Step 1: B, C variables is missed in 1st term and A variable is missed in 2nd term
Step 2: Missing variables are ‘AND’ operation with the corresponding minterms
= A(B+B')(C + C') + B'C(A+A')
Step 3:Expand the expression
= ABC + ABC' + AB'C + AB'C' + A'B'C+ AB'C
Step 4:Eliminate one of the repeated terms
= ABC + ABC' + AB'C + AB'C' + A'B'C+ AB'C
= ABC + ABC' + AB'C + AB'C' + A'B'C [i.e., x+x=x]
= m1 + m4 + m5 + m6 + m7
Answer: F(A, B, C) = ∑(1, 4, 5, 6, 7)

11. 10
2. Express the Boolean function F (X,Y,Z) = X'Y+YZ'+XZ' as a sum of
minterms
= X'Y(Z+Z')+YZ'(X+X')+XZ'(Y+Y') [by Step 2]
=X'YZ+X'YZ'+XYZ'+X'YZ'+XYZ'+XY'Z' [by Step 3]
= X'YZ+X'YZ'+XYZ' +XY'Z' [by Step 4]
=m3+m2+m6+m4
Answer: F(X,Y,Z)=∑(2,3,4,6)
c) POS TO CANONICAL FORM:
Procedure:
Step 1: Check whether the given expression contains all the variables.
Step 2: If there is missing of variables in any maxterm then “OR” the
corresponding maxterm with expression (x.x'), where x be the missing variable
Step 3: Expand the terms by applying distributive law and reorder the literals in
the sum terms.
Step 4: If two maxterms are identical, eliminate one of the maxterms.
Problems:
1. Express the Boolean function F (A,B,C) = (A+B') .( B+C').(A'+C) as a
product of maxterms
Step 1: C, variable is missed in 1st term and A, variable is missed in 2nd
term and B variable is missed in 3rd term.
Step 2: Missing variables are ‘OR’ operation with the corresponding
minterms
= (A+B' + (C.C')). (B+C' + (A.A')). ( A' +C +(B.B'))
Step 3: Expand the expression
= (A+B'+C). (A+B'+C'). (A+B+C').(A'+B+C').(A'+B+C).(A'+B'+C)
Answer: F(A, B, C) = П(1, 2, 3,4,5, 6)
2. Express the Boolean function F (X,Y,Z) = (X+Z') .( Y'+Z) as a product of
maxterms
=(X+Z'+(Y.Y')).(Y'+Z+(X.X')) [ by step 2]
=(X+Y+Z').(X+Y'+Z').(X+Y'+Z).(X'+Y'+Z) [ by step 3]
Answer: F(X,Y,Z) = П(1,2, 3, 6)
Limitations of Boolean algebra:
1. To reduce the Boolean expression it is necessary to have the knowledge of all the
Boolean laws, rules and theorems.
2. It lacks specific rules to predict each succeeding step in the minimization process.