Lecture Four

By

Dr. Noor Abdul Khaleq Z.

12/11/2023
 Content

❑ Review of AND, OR and NOT gates.

❑NAND, NOR, XOR, XNOR.

❑ Boolean Operation and Expressions.

❑Laws and Rules of Boolean Algebra.

❑De Morgan's Theorems.

 NAND Gate
❑ The NAND gate is a popular logic element because it can be used as a
universal gate such that NAND gates can be used in combination to
perform the AND, OR, and inverter operations.
❑ The NAND gate is the same as the AND gate except the output is
inverted.

❑ The Boolean expression for the output of a 2-input NAND gate is

Y=A.B
Where a bar over a variable or variables indicates an inversion.
 NAND Gate
❑ A NAND gate produces a LOW output only when all the inputs are HIGH. When
any of the inputs is LOW, the output will be HIGH.
❑ The table below is for the specific case of a 2-input NAND gate’

❑ Negative-OR Equivalent Operation of a NAND Gate

❑ For a 2-input NAND gate performing a negative-OR operation, output X is HIGH
when either input A or input B is LOW, or when both A and B are LOW.
The NOR Gate

❑ The NOR gate, like the NAND gate, is a useful logic element because it can
also be used as a universal gate.
❑ The NOR is the same as the OR except the output is inverted.

❑ The Boolean expression for the output of a 2-input NAND gate is

Y=A+B
 NOR Gate
❑ Operation of a NOR Gate

❑ Negative-AND Equivalent Operation of the NOR Gate

❑ For a 2-input NOR gate performing a negative-AND operation, output X is
HIGH only when both inputs A and B are LOW.
The Exclusive-OR Gate
❑ The output of an exclusive-OR gate is HIGH only when the two inputs are at
opposite logic levels.
 The Exclusive-NOR Gate

❑ For an exclusive-NOR gate, output X is LOW when input A is LOW and input
B is HIGH, or when A is HIGH and B is LOW; X is HIGH when A and B are
both HIGH or both LOW.
Timing Diagrams
❑ A timing diagram is basically a graph that accurately displays the relationship
of two or more waveforms with respect to each other on a time basis.
❑ Example:
A waveform is applied to an inverter in the figure below.

Determine the output waveform corresponding to the input and show the timing
diagram.
❑ Solution:
The output waveform is exactly opposite to the input (inverted).
Boolean Algebra and Logic Simplification
❑ Boolean algebra is the mathematics of digital logic.
❑ A variable is a symbol (usually an italic uppercase letter or word) used to
represent an action, a condition, or data. In Boolean algebra, any single
variable can have only a 1 or a 0 value.
❑ The complement is the inverse of a variable, and it is indicated by a bar
over the variable (overbar). For example, the complement of the variable
A is A. If A = 1, then A = 0. If A = 0, then A = 1.
❑ The complement of the variable A is read as “not A” or “A bar.”
Sometimes a prime symbol rather than an overbar is used to denote the
complement of a variable; for example, A’ indicates a complement of A.
❑ A literal is a variable or the complement of a variable.
Boolean Addition
❑ The Boolean addition is equivalent to the OR operation.
❑ In Boolean algebra, a sum term is a sum of literals.
❑ Some examples of sum terms are A + B, A + B, A + B + C, and A + B + C
+ D.
❑ A sum term is equal to 1 when one or more of the literals in the term are 1.
A sum term is equal to 0 only if each of the literals is 0.
❑ Example:
Determine the values of A, B, C, and D that make the sum term A + B + C +
D equal to 0.
❑ Solution:
❑ For the sum term to be 0, each of the literals in the term must be 0.
Therefore, A = 0, B = 1 so that B = 0, C = 0, and D = 1 so that D = 0.
❑ A+B+C+D=0+1+0+1=0+0+0+0=0
Boolean Multiplication

❑ Boolean multiplication is equivalent to the AND operation.

❑ In Boolean algebra, a product term is the product of literals.
❑ Some examples of product terms are AB, AB, ABC, and ABCD.
❑ A product term is equal to 1 only if each of the literals in the term is 1.
A product term is equal to 0 when one or more of the literals are 0.
❑ Example:
Determine the values of A, B, C, and D that make the product term ABCD
equal to 1.
❑ Solution:
For the product term to be 1, each of the literals in the term must be 1.
Therefore, A = 1, B = 0 so that B = 1, C = 1, and D = 0 so that D = 1.
ABCD = 1 . 0 . 1 . 0 = 1 . 1 . 1 . 1 =
Laws and Rules of Boolean Algebra

❑ The commutative law of addition for two variables is written as:

A+B=B+A
❑ The commutative law of multiplication for two variables is:
AB = BA
❑ The associative law of addition is written as follows for three variables:
A +(B + C) = (A + B)+ C

❑ The associative law of multiplication is written as follows for three variables:

A(BC)=(AB)C
Laws and Rules of Boolean Algebra

❑ Distributive Law
❑ The distributive law is written for three variables as follows:
A(B + C) = AB + AC
Rules of Boolean Algebra
❑ Rule 1: A + 0 = A. A variable ORed with 0 is always equal to the variable.
❑ Rule 2: A + 1 = 1. A variable ORed with 1 is always equal to 1.
❑ Rule 3: A.0 = 0. A variable ANDed with 0 is always equal to 0.
❑ Rule 4: A.1=A. A variable ANDed with 1 is always equal to the variable.
❑ Rule 5: A+A=A. A variable ORed with itself is always equal to the
variable.
❑ Rule 6: A+A=1. A variable ORed with its complement is always equal to
1.
❑ Rule 7: A.A =A. A variable ANDed with itself is always equal to the
variable.
❑ Rule 8: A.A=0. A variable ANDed with its complement is always equal to
0.
❑ Rule 9: A= A. The double complement of a variable is always equal to the
variable.
Rules of Boolean Algebra

❑ Rule 10: A + AB = A. This rule can be proved by applying the distributive

law, rule 2, and rule 4 as follows:
A + AB = A.1 + AB =A(1+B) Factoring (distributive law)
= A .1 Rule 2 (1 + B) = 1
=A Rule 4 (A . 1) = A
Rules of Boolean Algebra

❑ Rule 11: A + AB=A + B. This rule can be proved as follows:

Rules of Boolean Algebra

❑ Rule 12: (A + B)(A +C)= A + BC. This rule can be proved as follows:
DeMorgan’s Theorems

❑ DeMorgan’s first theorem is stated as follows:

❑ The complement of a product of variables is equal to the sum of the
complements of the variables.
❑ The formula for expressing this theorem for two variables is:

XY = X + Y

❑ DeMorgan’s second theorem is stated as follows:

❑ The complement of a sum of variables is equal to the product of the
complements of the variables.
❑ The formula for expressing this theorem for two variables is:

X + Y = XY
 Applying De Morgan’s Theorems
❑ The following procedure illustrates the application of De Morgan’s theorems and
Boolean algebra to the specific expression
 Applying De Morgan’s Theorems

❑ Example: Apply DeMorgan’s theorems to the following expression:

( A + B + C) D
❑ Solution:
 Applying De Morgan’s Theorems

❑ Example: Apply DeMorgan’s theorems to the following expression:

ABC + DEF
❑ Solution:
 Applying De Morgan’s Theorems

❑ Example: Apply DeMorgan’s theorems to the following expression:

AB + CD + EF
❑ Solution:
 Applying De Morgan’s Theorems

❑ Example: Apply DeMorgan’s theorems to each of the following expressions:

 Applying De Morgan’s Theorems

❑ Example: The Boolean expression for an exclusive-OR gate is AB + AB. With this
as a starting point, use DeMorgan’s theorems and any other rules or laws that are
applicable to develop an expression for the exclusive-NOR gate.

❑ Solution: