ANALOGUE AND DIGITAL ELECTRONICS

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

1.1.9 HEXADECIMAL NUMBER SYSTEM.

The hexadecimal system uses base 16. Thus, it has 16 possible digit symbols. It uses the
digits 0-9 plus the letters A,B,C,D,E and F as the 16 digit symbols.

Table 7.7.0

Hexadecimal Decimal Binary
0 0 0000
1 1 0001
2 2 0010
3 3 0010
4 4 0100
5 5 0101
6 6 0110
7 7 0111
8 8 1000
9 9 1001
A 10 1010
B 11 1011
C 12 1100
D 13 1101
E 14 1110
F 15 1111

Table 7.7.0 shows the relationships among hexadecimal, decimal and binary.

101

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

1.1.10 BINARY TO HEXADECIMAL CONVERSION.

Example 1.10

011101001102 = 0011 1010 0110
3 A 6

= 3A616

1.1.11 HEXADECIMAL TO BINARY CONVERSION.

Example 1.11

9F216 =9 F2

1001 1111 0100
= 1001111101002

1.1.12 BINARY CODED DECIMAL (BCD)

When numbers, letters or words are represented by special group of symbols,this
is called encoding, and the group of symbol is called a code. When a decimal number is
represented by its equivalent binary number,we call it straight binary coding.

If each digit of decimal number is represented by its binary equivalent,this is
produce code called binary-code-decimal (BCD).

102

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Example 1.12 ,

874 to BCD-code

87 4

1000 0111 0100
94.3 to BCD-code

9 4.3

1001 0100 . 0011

Once again , each decimal digit is changed to its binary equivalent. Note that 4 bits
are always use for each digit. Clearly, only the 4-bit binary numbers from 0000 through
1001 are used.

Example 1.13 :

Convert the BCD number 0110100000111001 to its decimal equivalent.

0110 1000 0011 1001
6 83 9

103

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Convert the BCD number 011111000001 to its decimal equivalent.

0111 1100 0001
71

Forbidden code group indicates error in BCD number.

1.1.13 BINARY ADDITION

Adding binary numbers is a very simple task, and very similar to the longhand
addition of decimal numbers. As with decimal numbers, you start by adding the bits (digits)
one column, or place weight, at a time, from right to left. Unlike decimal addition, there is
little to memorize in the way of rules for the addition of binary bits:
0+0=0
1+0=1
0+1=1
1 + 1 = 10
1 + 1 + 1 = 11

Just as with decimal addition, when the sum in one column is a two-bit (two-digit)
number, the least significant figure is written as part of the total sum and the most
signifcant figure is "carried" to the next left column. Consider the following
examples :

104

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

. 11 1 <--- Carry bits -----> 11
. 1001101
. + 0010010 1001001 1000111
. ---------
. 1011111 + 0011001 + 0010110

--------- ---------

1100010 1011101

The addition problem on the left did not require any bits to be carried, since the
sum of bits in each column was either 1 or 0, not 10 or 11. In the other two problems, there
definitely were bits to be carried, but the process of addition is still quite simple.

As we'll see later, there are ways that electronic circuits can be built to perform this
very task of addition, by representing each bit of each binary number as a voltage signal
(either "high," for a 1; or "low" for a 0). This is the very foundation of all the arithmetic which
modern digital computers perform.

1.1.14 BINARY SUBTRACTION

We can subtract one binary number from another by using the standard techniques
adapted for decimal numbers (subtraction of each bit pair, right to left, "borrowing" as
needed from bits to the left). However, if we can leverage the already familiar (and easier)
technique of binary addition to subtract, that would be better. As we just learned, we can
represent negative binary numbers by using the "two's complement" method and a
negative place-weight bit. Here, we'll use those negative binary numbers to subtract
through addition. Here's a sample problem:

Subtraction: 710 – 510 Addition equivalent: 710 + (-510)

105

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

If all we need to do is represent seven and negative five in binary (two's
complemented) form, all we need is three bits plus the negative-weight bit:

positive seven = 01112
negative five = 10112

Now, let's add them together:
. 1111 <--- Carry bits
. 0111
. + 1011
. ------
. 10010
.
.

Discard extra bit
.
. Answer = 00102

Since we've already defined our number bit field as three bits plus the negative-
weight bit, the fifth bit in the answer (1) will be discarded to give us a result of 00102, or
positive two, which is the correct answer. Another way to understand why we discard that
extra bit is to remember that the leftmost bit of the lower number possesses a negative
weight, in this case equal to negative eight. When we add these two binary numbers
together, what we're actually doing with the MSBs is subtracting the lower number's MSB
from the upper number's MSB. In subtraction, one never "carries" a digit or bit on to the
next left place-weight.

Let's try another example, this time with larger numbers. If we want to add -2510
to 1810, we must first decide how large our binary bit ¯eld must be. To represent the
largest (absolute value) number in our problem, which is twenty-¯ve, we need at least five
bits, plus a sixth bit for the negative-weight bit.

106

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Let's start by representing positive twenty-five, then finding the two's complement
and putting it all together into one numeration:

+2510 = 0110012 (showing all six bits)
One's complement of 110012 = 1001102
One's complement + 1 = two's complement = 1001112

-2510 = 1001112

Essentially, we're representing negative twenty-five by using the negative-weight
(sixth) bit with a value of negative thirty-two, plus positive seven (binary 1112).
Now, let's represent positive eighteen in binary form, showing all six bits:

. 1810 = 0100102
.
. Now, let's add them together and see what we get:
.
. 11 <--- Carry bits
. 100111
. + 010010
. --------
. 111001

Since there were no "extra" bits on the left, there are no bits to discard. The leftmost
bit on the answer is a 1, which means that the answer is negative, in two's complement
form, as it should be. Converting the answer to decimal form by summing all the bits times
their respective weight
values, we get:

(1 x -3210) + (1 x 1610) + (1 x 810) + (1 x 110) = -710

Indeed -710 is the proper sum of -2510 and 1810.

107

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

EXERCISE 7.0 :

1. Convert the fractional binary number 0.1011 to decimal.
2. Convert decimal number 45.25 to binary.
3. Convert each decimal number to binary.

a. 19
b. 57
c. 40
4. Convert each decimal number to binary.
a. 14
b. 21
c. 0.375
5. Convert 11010.101112 to octal.
6. Convert 4538 to binary.
7. Convert 111101001102 to hexadecimal.
8. Convert 7B316 to binary.
9. Convert 74.3 to BCD code.

108

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

REFERENCE

1. DIGITAL FUNDAMENTALS
- BY THOMAS L.FLOYD
- 8TH EDITION

2. DIGITAL SYSTEMS : PRINCIPLES AND APPLICATION
- BY RONALD J. TOCCI
- MONROE COMMUNITY COLLEGE

3. DIGITAL LOGIC CIRCUIT ANALYSIS AND DESIGN
- BY NELSON, NAGLE, CARROL, IRWIN
- PRENTICE-HALL INTERNATIONAL INC.

109

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

MODULE NO 7: GATES AND INVERTERS_

INTRODUCTION

It is a simplest of the logic circuit. It have more of logic circuit such
as the inverter, and gates, or gates, nand gates, nor gates and other.

LEARNING OBJECTIVES

The objectives of this module are to:
1. Define the type of logic gate.
2. Explain the function of logic gates.
3. Define the step of Boolean Algebra.

LEARNING OUTCOMES

After completing the module, students should be able to:
4. Define the type of logic gate.
5. Explain the function of logic gates.
6. Define the step of Boolean Algebra.

110

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.1LOGIC GATES

7.1.1 AN INVERTER

The INVERTER (NOT circuit) performs the operation called inversion
or complementation. The purpose of the inverter is to change one logic level
to the opposite level. In term of bits, it changes a 1 to 0 and a 0 to 1.

Standard Logic Symbol : Bubble
Output
Input

Truth Table : Output
1 (HIGH)
Input 0 (LOW)
0 (LOW)
1 (HIGH)

111

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Inverter Operations :

t1 t2 t1 t2
Input pulse Output

Pulse

t1 and t2 indicate the corresponding points on the input and output pulse
waveform.

Timing Diagram :

Input waveform
Output waveform

112

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

t1 t2

Logical Expression :

A X=A

7.1.2 AND GATE
The AND Gate is one of the basic gates from which all logic functions

are constructed. An AND gate can have 2 or more inputs and performs what
is known as logical multiplication.

LOGIC SYMBOL :

113

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

TRUTH TABLE : B OUTPUT
0 X
INPUTS 1 0
A 0 0
0 1 0
0 1
1
1

The total number of possible combinations of binary inputs to a gate is
determined by the following formula :

N = 2n

N = possible combination
n = number of binary input variables

For 2 input variables : N = 2n =4
For 3 input variables : N = 2n =8
For 4 input variables : N = 2n = 16

114

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Example :
a) Develop the truth table for a input AND gate .
b) Determine the total number of possible input combinations for a 5 input
AND gate.

Logical Expression :

115

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.1.3 OR GATE

An OR gate can have 2 or more inputs and performs what is known as
logical addition

LOGIC SYMBOL :

TRUTH TABLE :

INPUTS B OUTPUT
A 0 X=A+B
0 1 0
0 0 1
1 1 1
1 1

116

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

LOGICAL EXPRESSION:

7.1.4 NAND GATE
NAND gate can be used to perform the AND , OR and inverter

operations or any combination of these operations. The term NAND is a
contraction of NOT – AND and implies an AND function with a complements
output.

117

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

LOGIC SYMBOL :

AND Gate + INVERTER

118

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

TRUTH TABLE :

INPUTS B OUTPUT

A 0 X = A.B
1
0 0 1
0 1 1
1 1
1 0

LOGICAL EXPRESSION:

119

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.1.5 NOR GATE.
The term of NOR gate is a contraction of NOT – OR and implies an

OR function with an inverted output.

LOGIC SYMBOL :

OR Gate + INVERTER

120

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

TRUTH TABLE :

INPUTS B OUTPUT

A 0 X = A+B
1
0 0 1
0 1 0
1 0
1 0

LOGICAL EXPRESSION:

121

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.1.6 THE EXCLUSIVE OR GATE
The output of an EX-OR gate is HIGH only when two inputs are at

apposite levels.

LOGIC SYMBOL :

X = 1 if A = 0 and B = 1 or if A = 1 and B = 0
X= 0 if A and B = 1 or if A and B = 0

122

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

TRUTH TABLE :

INPUTS B OUTPUT

A 0 X
1
0 0 0
0 1 1
1 1
1 0

7.1.7 THE EXCLUSIVE NOR GATE
When 2 inputs level are opposite the output of EX-NOR gate is LOW

123

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

LOGIC SYMBOL :

X = 1 if A and B = 1 or if A and B = 0
X= 0 if A =0 and B = 1 or if A=1 and B = 0

TRUTH TABLE :

INPUTS B OUTPUT

A 0 X
1
0 0 1
0 1 0
1 0
1 1

124

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.3 BOOLEAN ALGEBRA
LAWS AND RULES OF BOOLEAN ALGEBRA

1. Boolean Algebra
The basic of Boolean algebra: -
1. A + B = B + A
2. AB = BA
3. A + (B + C) = (A + B) + C

4. A (BC) = (AB) C

5. A (B + C) = AB + AC

125

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.3.1 Twelve Basic Rules of Boolean Algebra

There are 12 basic rules that are useful in manipulating and
simplifying Boolean expressions.

Rule 1. A+0=A

Rule 2. A+1=1

Rule 3. A.0=0

Rule 4. A.1=A

Rule 5. A+A=A

126

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Rule 6. A+A=1

Rule 7. A.A=A

Rule 8. A.A=0

Rule 9. A=A

Rule 10. A + AB = A

A + AB = A (1 +B) - factoring
=A.1 - rule 2: (1 + B) = 1
=A - rule4: A . 1 = A

127

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Rule 11. A + AB = A + B
A + AB = (A + AB) + AB
= (AA + AB) + AB - rule 10: A = A + B
= AA + AB + A A+ AB - rule 7: A = AA
- rule 8: adding
= (A + A)(A + B)
= 1 . (A + B) AA = 0
=A+B - factoring
- rule 6: A + A = 1
- rule 4: drop the 1

Rule 12. (A + B)(A + C) = A + BC

(A + B)(A + C) = AA + AC + AB + BC

= A + AC + AB + BC - rule 7: AA = A

= A (1 + C) + AB + BC - factoring

= A . 1 + AB + BC - rule 2:1 + C=1

= A(1 + B) + BC - factoring

= A . 1 + BC - rule 2:1 + B=1

= A + BC - rule 4: A .1 =A

128

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.3.2 BOOLEAN ALGEBRA APPLICATION

Boolean algebra is the mathematics of digital systems. A basic
knowledge of Boolean algebra is indispensable to the study and
analysis of logic circuits.

i. Variable, complement and literal
Variable, complement and literal are the terms used in Boolean
algebra.

Variable - Symbol (usually an italic uppercase letter to represent
a logical quantity)

- Single variable can have a 1 or 0 value.

ii. Complement - Is the inverse of a variable and is
indicated by a bar over the variable (over
bar)

- Example: A = A, if A = 1 then A = 0

iii. Literal - A literal is a variable or the complement of
variable.

- For example: B’ indicates the complement of B.

7.3.3 Boolean Addition

Boolean addition is equivalent to the OR operation and the basic
rules are illustrated with their relation to the OR gate as follows:

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 are1. A sum term is equal to 0 only if each of the literals is
0.

129

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Example 1:
Determine the values of A, B, C and D that make the sum term
A + B + C + D equal to 0.
Solution: A + B + C + D = 0 + 1 + 0 + 1 = 0 + 0 + 0 + 0 = 0
7.3.4 Boolean Multiplication
Boolean Multiplication is the equivalent to the AND operation and
the basic rules are illustrated with their relation to the AND gate
as follows:

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.

Example2:
Determine the values of A, B, C and D that make the product
term ABCD equal to 1.
Solution: ABCD = 1 . 0 . 1 . 0 = 1 . 1 . 1 . 1 = 1

130

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.3.5 Laws And Rules Of Boolean Algebra
1. Laws of Boolean Algebra
The basic of Boolean algebra: -
6. A + B = B + A

7. AB = BA

8. A + (B + C) = (A + B) + C

9. A (BC) = (AB) C

10. A (B + C) = AB + AC

131

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

7.3.6 Twelve Basic Rules of Boolean Algebra

There are 12 basic rules that are useful in manipulating and
simplifying Boolean expressions.

Rule 1. A+0=A

Rule 2. A+1=1

Rule 3. A.0=0

Rule 4. A.1=A

Rule 5. A+A=A

Rule 6. A+A=1

132

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Rule 7. A.A=A

Rule 8. A.A=0

Rule 9. A=A

Rule 10. A + AB = A

A + AB = A (1 +B) - factoring

=A.1 - rule 2: (1 + B) = 1

= A - rule4: A . 1 = A

Rule 11. A + AB = A + B

A + AB= A + (AB + AB) - rule 10: A = A + AB

= A + (AB + AB) - factoring

= A+ B(A + A) - rule 6: adding A+A = 1

= (A + B)(1) - rule 4: drop the 1

=A+B

Rule 12. (A + B)(A + C) = A + BC

133

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

(A + B)(A + C) = AA + AC + AB + BC

= A + AC + AB + BC - rule 7: AA = A

= A (1 + C) + AB + BC - factoring

= A . 1 + AB + BC - rule 2:1 + C=1

= A(1 + B) + BC - factoring

= A . 1 + BC - rule 2:1 + B=1

= A + BC - rule 4: A .1 =A

7.4 DE MORGAN’S THEOREMS
DeMorgan’s theorem is important because it show us how to find the
equivalent of NAND and NOR gates.

ii. State DeMorgan’s Theorem
Two DeMorgan’s theorem are: -
a. XY = X + Y
b. X + Y = X Y

iii. Relate DeMorgan’s Theorem to the equivalency of the: -
i. NAND and negative-OR gates

ii. NOR and negative-AND gates

134

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Gate equivalencies and the corresponding truth tables that
illustrate DeMorgan’s theorems. Notice the equality of the two
output columns in each table. This shows that the equivalent
gates perform the same logic function.
Example 1:
Apply DeMorgan’s theorems to the expression XYZ and X + Y
+Z
Solution:

Example 2:
Apply DeMorgan’s theorems to the expressions WXYZ and
W + X + Y + Z.
Solution:

135

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

iv. Applying DeMorgan’s Theorem
Example 1:
Applying DeMorgan’s theorem to the expression below:

Solution:

7.4 KARNAUGH MAP APPLICATION
The Karnaugh map provides a systematic method for simplifying
Boolean expressions and if properly used, will produce the simplest
SOP or POS expression possible, known as the minimum expression.
The Karnaugh map, on the other hand, basically provides a “cookbook”
method for simplification.

7.4.1 The Karnaugh Map (K-Map)
i. A Karnaugh map is similar to a truth table because it
presents all of the possible values of input variables and
resulting output for each value. Instead of being organized
into columns and rows like a truth table, the Karnaugh
map is an array of cells in which each cell represents a
binary value of the input variables.

136

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

ii. The cells are arranged in a way so that simplification of a
given expression is simply a matter of properly grouping
the cells.

iii. Karnaugh map can be used for expressions with two,
three, four and five variables, but we will discuss only 3-
variable and 4-variable situations to illustrate the
principles.

iv. The number of cells in a Karnaugh map is equal to the
total number of possible input variable combinations as is
the number of rows in a truth table. For three variables,
the number of cells is 23 = 8. For four variables, the
number of cells is 24 = 16.

7.4.1.1 The 3-Variables Karnaugh Map

The 3-variable Karnaugh map is an array of eight cells
as shown in Figure 6(a). In this case A, B and C are
used for the variables although others letters could be
used. For example the cell in the upper left corner has a
binary value of 000 and the cell in the lower right corner
has a binary 101. Figure 6 (b) shows the standard
product terms that are represented by each cell in the
Karnaugh map.

(a)
(b)

Figure 6 : A 3-variable Karnaugh map

7.4.1.2 The 4-Variables Karnaugh Map

The 4-variable Karnaugh map is an array of sixteen cells
as shown in Figure 7 (a). For example, the cell in the
upper right corner has a binary value of 0010 and the
cell in the lower right corner has a binary value of 1010.

137

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Figure 7(b) shows the standard product terms that is
represented by each cell in the 4-variable Karnaugh
map.

(a) (b)
Figure 7: A 4-variable Karnaugh map
7.4.1.3 Looping
The expression for output Y can be simplified by
properly combining those squares in the Karnaugh map
which contain is. The process for combining these is
called looping.
i. Pairs
Looping a pair of adjacent 1s in a Karnaugh map
eliminates the variables that appear in
complemented and un-complemented form. Here
we are giving the different examples for
understanding it clearly in Figure 8.

138

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Figure 8 : Looping a pair of adjacent

ii. Quads

139

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

A Karnaugh map may contain a group of four is that
are adjacent to each other. This group is called a
quad. Looping a quad of adjacent is eliminates the
two variables that appear in both complemented and
un complemented form.
Figure 9 shows several examples of quads. In Figure
9(a) the four is vertically adjacent, and in Figure 9(b)
they are horizontally adjacent. When a quad is
looped, the resultant term will contain only the
variables that do not change form for all the squares
in the quad. For example in Figure 9(a) the four
squares that contain a ABC, ABC, ABC and ABC.
Examination of these terms reveals that only the
variable C remains unchanged. This can be proved
as follows:

140

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

)

Figure 9 : Looping a quads of adjacent
iii. Octests

A group of eight 1s that are adjacent to one other is called
octet. Several examples for octet are shown in Figure 10.

141

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Figure 10 : Looping a octets of adjacent
7.4.1.4 Karnaugh Map POS and SOP Minimization

Example :
Using a Karnaugh map, convert the following standard
POS expression into a minimum POS expression, a
standard SOP expression, and a minimum SOP
expression:
(A+B+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D)(A+B+C+D)

142

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

solution:
The O’s for the standard POS expression are mapped
and grouped to obtain the minimum POS expression in
Figure 11 (a). In Figure 11(b), is added to the cells that do
not contain O’s. From each cell containing a 1, a standard
product term is obtained as indicated. These product
terms form the standard SOP expression. In Figure 10(c),
the is grouped and a minimum SOP expression is
obtained.

143

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Figure 11 : Karnaugh Map POS and SOP Minimization
7.4.2 Combinational Logic Circuit After Simplify Using Karnaugh Map

Example1:

144

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

Logic circuit

Example 2 :

Logic circuit

145

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

QUESTIONS

1. When is the output of an OR gate HIGH?
________________________________

2. When is the output of an OR gate LOW?
________________________________

3. Describe the truth table for 3 input OR gate.
_____________________________________________________________
________
_____________________________________________________________
________

4. If the two input waveforms A and B in figure 3 are applied to the OR gate,whwt
is the resulting output waveform?
FIGURE 3

146

CERTIFICATE IN ELECTRICAL POWER
CEV 20042 DIGITAL AND ANALOGE

REFERENCE

4. DIGITAL FUNDAMENTALS
- BY THOMAS L.FLOYD
- 8TH EDITION

5. DIGITAL SYSTEMS : PRINCIPLES AND APPLICATION
- BY RONALD J. TOCCI
- MONROE COMMUNITY COLLEGE

6. DIGITAL LOGIC CIRCUIT ANALYSIS AND DESIGN
- BY NELSON, NAGLE, CARROL, IRWIN
- PRENTICE-HALL INTERNATIONAL INC.

147