Digital electronics number system gates and boolean theorem

Digital Electronics
Number System, Logic Gates and Boolean Theorems
Dr. Nilesh Bhaskarrao Bahadure
nbahadure@gmail.com
https://www.sites.google.com/site/nileshbbahadure/home
June 30, 2021
Dr. Nilesh Bhaskarrao Bahadure (Ph.D.) Digital Electronics June 30, 2021 1 / 74

Overview
1 Binary logic functions
2 Number System Conversion
Conversion for fractional part
3 Other base number system to Decimal
Binary to Decimal Conversion
Convert Binary fraction to Decimal
4 Laws of Boolean Algebra
5 De Morgan’s Theorem
6 Logic Gates
7 Examples
8 Thank You

Syllabus

Syllabus
Binary logic functions, Boolean laws, truth tables, associative and
distributive properties, DeMorgans theorems, realization of switching
functions using logic gates

binary logic functions

binary logic functions
Binary logic consists of binary variables and logical operations. The
variables are designated by the alphabets such as A, B, C, x, y, z, etc.,
with each variable having only two distinct values: 1 and 0. ... Logic gates
are the basic elements that make up a digital system.

Number System
Figure : Number System

Decimal Number System

The number system that we use in our day-to-day life is the decimal
number system. Decimal number system has base 10 as it uses 10 digits
from 0 to 9. In decimal number system, the successive positions to the left
of the decimal point represents units, tens, hundreds, thousands and so on.

The number system that we use in our day-to-day life is the decimal
number system. Decimal number system has base 10 as it uses 10 digits
from 0 to 9. In decimal number system, the successive positions to the left
of the decimal point represents units, tens, hundreds, thousands and so on.
Each position represents a specific power of the base (10). For example,
the decimal number 1234 consists of the digit 4 in the units position, 3 in
the tens position, 2 in the hundreds position, and 1 in the thousands
position, and its value can be written as
(1 ∗ 1000) + (2 ∗ 100) + (3 ∗ 10) + (4 ∗ 1)
(1*103) + (2 ∗ 102) + (3 ∗ 101) + (4 ∗ 100)
1000 + 200 + 30 + 1
1234

Binary Number System
Characteristics
1 Uses two digits, 0 and 1.
2 Also called base 2 number system
3 Each position in a binary number represents a 0 power of the base
(2). Example: 20
4 Last position in a binary number represents an x power of the base
(2). Example: 2x where x represents the last position - 1.
Example
Binary Number: 10101
Calculating Decimal Equivalent –

Binary Number System...
Step Binary Number Decimal Number
Step 1 10101 ((1x24) + (0x23) + (1x22) + (0x21) + (1x20))10
Step 2 10101 (16 + 0 + 4 + 0 + 1)10
Step 3 10101 2110

Octal Number System
Characteristics
1 Uses eight digits, 0,1,2,3,4,5,6,7.
2 Also called base 8 number system
3 Each position in an octal number represents a 0 power of the base
(8). Example: 80
4 Last position in an octal number represents an x power of the base
(8). Example: 8x where x represents the last position - 1.
Example
Octal Number – 12570

Octal Number System...
Step Octal Number Decimal Number
Step 1 12570 ((1x84) + (2x83) + (5x82) + (7x81) + (0x80))10
Step 2 12570 (4096 + 1024 + 320 + 56 + 0)10
Step 3 12570 549610

Octal Number System...
Octal Number Binary Number
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111

Hexadecimal Number System
Characteristics
1 Uses 10 digits and 6 letters, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F.
2 Letters represents numbers starting from 10. A = 10, B = 11, C =
12, D = 13, E = 14, F = 15.
3 Also called base 16 number system.
4 Each position in a hexadecimal number represents a 0 power of the
base (16). Example 160.
5 Last position in a hexadecimal number represents an x power of the
base (16). Example 16x where x represents the last position - 1.
Example
Hexadecimal Number: 19FDE

Hexadecimal Number System...
Step Hexadecimal
Number
Decimal Number
Step 1 19FDE ((1x164
) + (9x163
) + (Fx162
) + (Dx161
) + (Ex160
))10
Step 2 19FDE ((1x164
) + (9x163
) + (15x162
) + (13x161
) + (14x160
))10
Step 3 19FDE (65536 + 36864 + 3840 + 208 + 14)10
Step 4 19FDE 10646210

Hexadecimal Number System...
Binary Number Hex. Number
0000 0
0001 1
0010 2
0011 3
0100 4
0101 5
0110 6
0111 7
1000 8
1001 9
1010 A
1011 B
1100 C
1101 D
1110 E
1111 F

Number System Conversion
There are many methods or techniques which can be used to convert
numbers from one base to another.

Decimal to Other Base System
1 Step 1 - Divide the decimal number to be converted by the value of
the new base.
2 Step 2 - Get the remainder from Step 1 as the rightmost digit (least
significant digit) of new base number.
3 Step 3 - Divide the quotient of the previous divide by the new base.
4 Step 4 - Record the remainder from Step 3 as the next digit (to the
left) of the new base number.

the new base.
Repeat Steps 3 and 4, getting remainders from right to left, until the
quotient becomes zero in Step 3.

the new base.
Repeat Steps 3 and 4, getting remainders from right to left, until the
quotient becomes zero in Step 3.
The last remainder thus obtained will be the Most Significant Digit (MSD)
of the new base number.

Example

Example
Convert decimal number 29 to binary
Step Operation Result Remainder
Step 1 29 / 2 14 1
Step 2 14 / 2 7 0
Step 3 7 / 2 3 1
Step 4 3 / 2 1 1
Step 5 1 / 2 0 1

Example
Step 1 29 / 2 14 1
Step 2 14 / 2 7 0
Step 3 7 / 2 3 1
Step 4 3 / 2 1 1
Step 5 1 / 2 0 1
As mentioned in Steps 2 and 4, the remainders have to be arranged in the
reverse order so that the first remainder becomes the Least Significant
Digit (LSD) and the last remainder becomes the Most Significant Digit
(MSD).

Example
Step 1 29 / 2 14 1
Step 2 14 / 2 7 0
Step 3 7 / 2 3 1
Step 4 3 / 2 1 1
Step 5 1 / 2 0 1
As mentioned in Steps 2 and 4, the remainders have to be arranged in the
reverse order so that the first remainder becomes the Least Significant
Digit (LSD) and the last remainder becomes the Most Significant Digit
(MSD).
Decimal Number (29)10 = Binary Number (11101)2

Example

Example
Convert decimal number 210 to Octal
Step 1 210 / 8 26 2
Step 2 26 / 8 3 2
Step 3 3 / 8 0 3

Example
Step 1 210 / 8 26 2
Step 2 26 / 8 3 2
Step 3 3 / 8 0 3
Now, write remainder from bottom to up (in reverse order), this will be
322 which is equivalent octal number of decimal integer 210.

Example
Step 1 210 / 8 26 2
Step 2 26 / 8 3 2
Step 3 3 / 8 0 3
322 which is equivalent octal number of decimal integer 210.
Decimal Number (210)10 = Octal Number (322)8

Example

Example
Convert decimal number 105 to Hexadecimal
Step 1 105 / 16 6 9
Step 2 6 / 16 0 6

Example
Step 1 105 / 16 6 9
Step 2 6 / 16 0 6
Now, write remainder from bottom to up (in reverse order), this will be 69
which is equivalent hexadecimal number of decimal integer 105.

Example
Step 1 105 / 16 6 9
Step 2 6 / 16 0 6
Now, write remainder from bottom to up (in reverse order), this will be 69
which is equivalent hexadecimal number of decimal integer 105.
Decimal Number (105)10 = Hexadecimal Number (69)16

Example

Example
Convert decimal number 540 into hexadecimal number.
Step 1 540 / 16 33 12 = C
Step 2 33 / 16 2 1
Step 3 2 / 16 0 2

Example
Step 1 540 / 16 33 12 = C
Step 2 33 / 16 2 1
Step 3 2 / 16 0 2
21C which is equivalent hexadecimal number of decimal integer 540.

Example
Step 1 540 / 16 33 12 = C
Step 2 33 / 16 2 1
Step 3 2 / 16 0 2
21C which is equivalent hexadecimal number of decimal integer 540.
Decimal Number (540)10 = Hexadecimal Number (21C)16

Convert decimal fraction to other base number
1 Multiply the fractional decimal number by base value number.
2 Integral part of resultant decimal number will be first digit of fraction
base number.
3 Repeat step 1 using only fractional part of decimal number and then
step 2.

Convert decimal fraction to binary number
1 Multiply the fractional decimal number by 2.
2 Integral part of resultant decimal number will be first digit of fraction
binary number.
3 Repeat step 1 using only fractional part of decimal number and then
step 2.

Example

Example
Convert decimal number 4.47 into binary number.
Part I: Conversion of Integer Part
Step 1 4 / 2 2 0
Step 2 2 / 2 1 0
Step 3 1 / 2 0 1
Part II: Conversion of Fractional Part
Step Operation Result Integer value
Step 1 0.47 * 2 0.94 0
Step 2 0.94 * 2 1.88 1
Step 3 0.88 * 2 1.76 1
Step 4 0.76 * 2 1.52 1

Example
Step 1 4 / 2 2 0
Step 2 2 / 2 1 0
Step 3 1 / 2 0 1
Step 1 0.47 * 2 0.94 0
Step 2 0.94 * 2 1.88 1
Step 3 0.88 * 2 1.76 1
Step 4 0.76 * 2 1.52 1
Now, write remainder from bottom to up (in reverse order) this will be
100, collect integer parts from top to bottom in fractional part this will be
0111 by combining integer and fractional part the results will be.

Example
Step 1 4 / 2 2 0
Step 2 2 / 2 1 0
Step 3 1 / 2 0 1
Step 1 0.47 * 2 0.94 0
Step 2 0.94 * 2 1.88 1
Step 3 0.88 * 2 1.76 1
Step 4 0.76 * 2 1.52 1
Now, write remainder from bottom to up (in reverse order) this will be
100, collect integer parts from top to bottom in fractional part this will be
0111 by combining integer and fractional part the results will be.
Decimal Number (4.47)10 = Binary Number (100.0111)2

Example

Example
Convert decimal fractional number 0.8125 into binary number.
Step 1 0.81252 x 2 1.625 1
Step 2 0.625 x 2 1.25 1
Step 3 0.25 x 2 0.50 0
Step 4 0.50 x 2 1.0 1

Example
Step 1 0.81252 x 2 1.625 1
Step 2 0.625 x 2 1.25 1
Step 3 0.25 x 2 0.50 0
Step 4 0.50 x 2 1.0 1
Now, collect integer parts from top to bottom this will be 0.1101 so the
results will be.

Example
Step 1 0.81252 x 2 1.625 1
Step 2 0.625 x 2 1.25 1
Step 3 0.25 x 2 0.50 0
Step 4 0.50 x 2 1.0 1
results will be.
Decimal Number (0.81252)10 = Binary Number (0.1101)2

Example

Example
Convert decimal fractional number 0.015625 into octal form.
Step 1 0.015625 x 8 0.125 0
Step 2 0.125 x 8 1.000 1

Example
Step 1 0.015625 x 8 0.125 0
Step 2 0.125 x 8 1.000 1
When we multiply 0.125 by 8 and take the integer part
0.125 x 8 = 1.000
Integer part = 1
Fractional part = 0
Now the fractional part is 0 so, we stop here.

Example
Step 1 0.015625 x 8 0.125 0
Step 2 0.125 x 8 1.000 1
0.125 x 8 = 1.000
Integer part = 1
Fractional part = 0
results will be.

Example
Step 1 0.015625 x 8 0.125 0
Step 2 0.125 x 8 1.000 1
0.125 x 8 = 1.000
Integer part = 1
Fractional part = 0
results will be.
Decimal Number (0.015625)10 = Octal Number (0.01)8

Example

Example
Convert decimal number 7.16 into octal form
Step 1 7 / 8 0 7
Step 1 0.16 x 8 1.28 1
Step 2 0.28 x 8 2.24 2
Step 3 0.24 x 8 1.92 1
Step 4 0.92 * 2 7.36 7
Step 4 0.36 * 2 2.88 2

Example
Step 1 7 / 8 0 7
Step 1 0.16 x 8 1.28 1
Step 2 0.28 x 8 2.24 2
Step 3 0.24 x 8 1.92 1
Step 4 0.92 * 2 7.36 7
Step 4 0.36 * 2 2.88 2
in this case, we have 5 digits as answer and the fractional part is still not 0
so, we stop here.

Example
Step 1 7 / 8 0 7
Step 1 0.16 x 8 1.28 1
Step 2 0.28 x 8 2.24 2
Step 3 0.24 x 8 1.92 1
Step 4 0.92 * 2 7.36 7
Step 4 0.36 * 2 2.88 2
in this case, we have 5 digits as answer and the fractional part is still not 0
so, we stop here.
Decimal Number (7.16)10 = Octal Number (7.12172)8

Example

Example
Convert decimal fractional number 0.06640625 into hexadecimal number.
Step 1 0.06640625 x 16 1.0625 1
Step 2 0.0625 x 16 1.0 1

Example
Step 1 0.06640625 x 16 1.0625 1
Step 2 0.0625 x 16 1.0 1
0.0625 x 16 = 1.0000
Integer part = 1
Fractional part = 0

Example
Step 1 0.06640625 x 16 1.0625 1
Step 2 0.0625 x 16 1.0 1
0.0625 x 16 = 1.0000
Integer part = 1
Fractional part = 0
Decimal Number (0.0625)10 = Hexadecimal Number (0.11)16

There are two methods to apply a binary to decimal conversion. The first
one uses positional representation of the binary. The second method is
called double dabble and is used for converting longer binary strings faster.
It doesn’t use the positions.
Method 1: Using Positions
1 Step 1: Write down the binary number.
2 Step 2: Starting with the least significant digit (LSB - the rightmost
one), multiply the digit by the value of the position. Continue doing
this until you reach the most significant digit (MSB - the leftmost
one).
3 Step 3: Add the results and you will get the decimal equivalent of the
given binary number.

Example

Example
Convert binary number 1010 into Decimal number.
Solution:
1 Step 1: Write down (1010)2 and determine the positions, namely the
powers of 2 that the digit belongs to.
2 Step 2: Represent the number in terms of its positions.
(1 ∗ 23) + (0 ∗ 22) + (1 ∗ 21) + (0 ∗ 20)
3 Step 3: (1 * 8) + (0 * 4) + (1 * 2) + (0 * 1) = 8 + 0 + 2 + 0 = 10
4 Therefore, (1010)2 = (10)10

Example
Solution:
1 Step 1: Write down (1010)2 and determine the positions, namely the
powers of 2 that the digit belongs to.
2 Step 2: Represent the number in terms of its positions.
(1 ∗ 23) + (0 ∗ 22) + (1 ∗ 21) + (0 ∗ 20)
3 Step 3: (1 * 8) + (0 * 4) + (1 * 2) + (0 * 1) = 8 + 0 + 2 + 0 = 10
4 Therefore, (1010)2 = (10)10
Binary Number (1010)2 = Decimal Number (10)10

Method 2: Double Dabble
Also called doubling, this method is actually an algorithm that can be
applied to convert from any given base to decimal. Double dabble helps
converting longer binary strings in your head and the only thing to
remember is ’double the total and add the next digit’.
1 Step 1: Write down the binary number. Starting from the left, you
will be doubling the previous total and adding the current digit. In the
first step the previous total is always 0 because you are just starting.
Therefore, double the total (0 * 2 = 0) and add the leftmost digit.
2 Step 2: Double the total and add the next leftmost digit.
3 Step 3: Double the total and add the next leftmost digit. Repeat this
until you run out of digits.
4 Step 4: The result you get after adding the last digit to the previous
doubled total is the decimal equivalent.

Example

Example
Solution:
1 Step 1: Your previous total 0. Your leftmost digit is 1. Double the
total and add the leftmost digit (0 ∗ 2) + 1 = 1
2 Step 2: Double the previous total and add the next leftmost digit.
(1 ∗ 2) + 0 = 2
(2 ∗ 2) + 1 = 5
(5 ∗ 2) + 0 = 10

Example
Solution:
1 Step 1: Your previous total 0. Your leftmost digit is 1. Double the
total and add the leftmost digit (0 ∗ 2) + 1 = 1
(1 ∗ 2) + 0 = 2
(2 ∗ 2) + 1 = 5
(5 ∗ 2) + 0 = 10

Example

Example
Convert binary number (1110010)2 into Decimal number.
Solution: Method 1
(1 ∗ 26
) + (1 ∗ 25
) + (1 ∗ 24
) + (0 ∗ 23
) + (0 ∗ 22
) + (1 ∗ 21
) + (0 ∗ 20
)
= (1 ∗ 64) + (1 ∗ 32) + (1 ∗ 16) + (0 ∗ 8) + (0 ∗ 4) + (1 ∗ 2) + (0 ∗ 1)
= 64 + 32 + 16 + 0 + 0 + 2 + 0 = 114

Example
Solution: Method 1
(1 ∗ 26
) + (1 ∗ 25
) + (1 ∗ 24
) + (0 ∗ 23
) + (0 ∗ 22
) + (1 ∗ 21
) + (0 ∗ 20
)
= (1 ∗ 64) + (1 ∗ 32) + (1 ∗ 16) + (0 ∗ 8) + (0 ∗ 4) + (1 ∗ 2) + (0 ∗ 1)
= 64 + 32 + 16 + 0 + 0 + 2 + 0 = 114

Example

Example
Solution: Method 2
0 (previous sum at starting point)
0 ∗ 2 + 1 = 1
1 ∗ 2 + 1 = 3
3 ∗ 2 + 1 = 7
7 ∗ 2 + 0 = 14
14 ∗ 2 + 0 = 28
28 ∗ 2 + 1 = 57
57 ∗ 2 + 0 = 114

Example
Solution: Method 2
0 ∗ 2 + 1 = 1
1 ∗ 2 + 1 = 3
3 ∗ 2 + 1 = 7
7 ∗ 2 + 0 = 14
14 ∗ 2 + 0 = 28
28 ∗ 2 + 1 = 57
57 ∗ 2 + 0 = 114

Example

Example
Solution: Method 1
(1 ∗ 24
) + (1 ∗ 23
) + (0 ∗ 22
) + (1 ∗ 21
) + (1 ∗ 20
)
= (1 ∗ 16) + (1 ∗ 8) + (0 ∗ 4) + (1 ∗ 2) + (1 ∗ 1)
= 16 + 8 + 0 + 2 + 1 = 27

Example

Example
Solution: Method 2
0 ∗ 2 + 1 = 1
1 ∗ 2 + 1 = 3
3 ∗ 2 + 0 = 6
6 ∗ 2 + 1 = 13
13 ∗ 2+ = 27

Example

Example
Convert binary number (110.101)2 into Decimal number.
Step 1: Conversion of 110 to decimal
(1 ∗ 22
) + (1 ∗ 21
) + (0 ∗ 20
)
= (1 ∗ 4) + (1 ∗ 2) + (0 ∗ 1)
= (4 + 2 + 0) = 6
Step 2: Conversion of .101 to decimal
0.101 = (1 ∗ 2−1
) + (0 ∗ 2−2
) + (1 ∗ 2−3
)
0.101 = (1/2) + (0/4) + (1/8)
0.101 = (0.5 + 0 + 0.125) = 0.625
Step 3: Add result of step 1 and 2.

Example
(1 ∗ 22
) + (1 ∗ 21
) + (0 ∗ 20
)
= (1 ∗ 4) + (1 ∗ 2) + (0 ∗ 1)
= (4 + 2 + 0) = 6
0.101 = (1 ∗ 2−1
) + (0 ∗ 2−2
) + (1 ∗ 2−3
)
0.101 = (1/2) + (0/4) + (1/8)
0.101 = (0.5 + 0 + 0.125) = 0.625
Binary Number (110.101)2 = Decimal Number (6.625)10

Example

Example
(1 ∗ 22
) + (0 ∗ 21
) + (1 ∗ 20
)
= (1 ∗ 4) + (0 ∗ 2) + (1 ∗ 1)
= (4 + 0 + 1) = 5
0.1101 = (1 ∗ 2−1
) + (1 ∗ 2−2
) + (0 ∗ 2−3
) + (1 ∗ 2−4
)
0.1101 = (1/2 + 1/4 + 0/8 + 1/16)
0.1101 = (0.5 + 0.25 + 0 + 0.0625) = 0.8125

Example
(1 ∗ 22
) + (0 ∗ 21
) + (1 ∗ 20
)
= (1 ∗ 4) + (0 ∗ 2) + (1 ∗ 1)
= (4 + 0 + 1) = 5
0.1101 = (1 ∗ 2−1
) + (1 ∗ 2−2
) + (0 ∗ 2−3
) + (1 ∗ 2−4
)
0.1101 = (1/2 + 1/4 + 0/8 + 1/16)
0.1101 = (0.5 + 0.25 + 0 + 0.0625) = 0.8125
Binary Number (101.1101)2 = Decimal Number (5.8125)10

Laws of Boolean Algebra

A set of rules or Laws of Boolean Algebra expressions have been invented
to help reduce the number of logic gates needed to perform a particular
logic operation resulting in a list of functions or theorems known
commonly as the Laws of Boolean Algebra.

Boolean Algebra is the mathematics we use to analyse digital gates and
circuits. We can use these Laws of Boolean to both reduce and simplify a
complex Boolean expression in an attempt to reduce the number of logic
gates required. Boolean Algebra is therefore a system of mathematics
based on logic that has its own set of rules or laws which are used to
define and reduce Boolean expressions.

Boolean Algebra is the mathematics we use to analyse digital gates and
circuits. We can use these Laws of Boolean to both reduce and simplify a
complex Boolean expression in an attempt to reduce the number of logic
gates required. Boolean Algebra is therefore a system of mathematics
based on logic that has its own set of rules or laws which are used to
define and reduce Boolean expressions.
The variables used in Boolean Algebra only have one of two possible
values, a logic ”0” and a logic ”1” but an expression can have an infinite
number of variables all labelled individually to represent inputs to the
expression, For example, variables A, B, C etc, giving us a logical
expression of A + B = C, but each variable can ONLY be a 0 or a 1.

Laws of Boolean Algebra...
Annulment Law
A term AND’ed with a ”0” equals 0 or OR’ed with a ”1” will equal 1

Annulment Law
A . 0 = 0 A variable AND’ed with 0 is always equal to 0

Annulment Law
A + 1 = 1 A variable OR’ed with 1 is always equal to 1
Identity Law
A term OR’ed with a ”0” or AND’ed with a ”1” will always equal that
term

Annulment Law
Identity Law
term
A + 0 = A A variable OR’ed with 0 is always equal to the variable

Annulment Law
Identity Law
term
A + 0 = A A variable OR’ed with 0 is always equal to the variable
A . 1 = A A variable AND’ed with 1 is always equal to the variable

Idempotent Law
An input that is AND’ed or OR’ed with itself is equal to that input

Idempotent Law
A + A = A A variable OR’ed with itself is always equal to the variable

Idempotent Law
A . A = A A variable AND’ed with itself is always equal to the variable
Complement Law
A term AND’ed with its complement equals ”0” and a term OR’ed with
its complement equals ”1”

Idempotent Law
Complement Law
A.Ā = 0 A variable AND’ed with its complement is always equal to 0

Idempotent Law
Complement Law
A.Ā = 0 A variable AND’ed with its complement is always equal to 0
A + Ā = 1 A variable OR’ed with its complement is always equal to 1

Commutative Law
The order of application of two separate terms is not important

Commutative Law
A.B = B.A The order in which two variables are AND’ed makes no difference

Commutative Law
A+B = B+A The order in which two variables are OR’ed makes no difference
Double Negation Law
A term that is inverted twice is equal to the original term

Commutative Law
A+B = B+A The order in which two variables are OR’ed makes no difference
Double Negation Law
A term that is inverted twice is equal to the original term
¯
Ā = A A double complement of a variable is always equal to the variable

Associative Law
This law allows the removal of brackets from an expression and regrouping
of the variables

Associative Law
of the variables
A + (B + C) = (A + B) + C = A + B + C (OR Associate Law)

Associative Law
of the variables
A(B.C) = (A.B)C = A . B . C (AND Associate Law)
Absorptive Law
This law enables a reduction in a complicated expression to a simpler one
by absorbing like terms.

Associative Law
of the variables
Absorptive Law
A + (A.B) = A (OR Absorption Law)

Associative Law
of the variables
Absorptive Law
A(A + B) = A (AND Absorption Law)
Distributive Law
This law permits the multiplying or factoring out of an expression.

Associative Law
of the variables
Absorptive Law
Distributive Law
A(B + C) = A.B + A.C (OR Distributive Law)

Associative Law
of the variables
Absorptive Law
Distributive Law
A(B + C) = A.B + A.C (OR Distributive Law)
A + (B.C) = (A + B).(A + C) (AND Distributive Law)

De Morgan’s Theorem

De Morgan has suggested two theorems which are extremely useful in
Boolean Algebra.

Boolean Algebra.
De Morgan’s first theorem states that two (or more) variables NOR’ed
together is the same as the two variables inverted (Complement) and
AND’ed, while the second theorem states that two (or more) variables
NAND’ed together is the same as the two terms inverted (Complement)
and OR’ed.
Theorem 1

Boolean Algebra.
and OR’ed.
Theorem 1
AB = Ā + B̄

Boolean Algebra.
and OR’ed.
Theorem 1
AB = Ā + B̄
NAND gate = Bubbled OR

Boolean Algebra.
and OR’ed.
Theorem 1
AB = Ā + B̄
1 The left hand side (LHS) of this theorem represents a NAND gate
with inputs A and B, whereas the right hand side (RHS) of the
theorem represents an OR gate with inverted inputs.

Boolean Algebra.
and OR’ed.
Theorem 1
AB = Ā + B̄
1 The left hand side (LHS) of this theorem represents a NAND gate
with inputs A and B, whereas the right hand side (RHS) of the
theorem represents an OR gate with inverted inputs.
2 This OR gate is called as Bubbled OR.

De Morgan’s Theorem...

Table 1 showing verification of the De Morgan’s first theorem
A B AB Ā B̄ Ā + B̄

0 0 1 1 1 1

0 0 1 1 1 1
0 1 1 1 0 1

0 0 1 1 1 1
0 1 1 1 0 1
1 0 1 0 1 1

0 0 1 1 1 1
0 1 1 1 0 1
1 0 1 0 1 1
1 1 0 0 0 0
Table : Truth Table for De Morgan’s Theorem 1

Theorem 2

Theorem 2
A + B = Ā.B̄

Theorem 2
A + B = Ā.B̄
NOR gate = Bubbled AND

Theorem 2
A + B = Ā.B̄
1 The LHS of this theorem represents a NOR gate with inputs A and B,
whereas the RHS represents an AND gate with inverted inputs.

Theorem 2
A + B = Ā.B̄
1 The LHS of this theorem represents a NOR gate with inputs A and B,
whereas the RHS represents an AND gate with inverted inputs.
2 This AND gate is called as Bubbled AND.

A B A + B Ā B̄ Ā.B̄

A B A + B Ā B̄ Ā.B̄
0 0 1 1 1 1

A B A + B Ā B̄ Ā.B̄
0 0 1 1 1 1
0 1 0 1 0 0

A B A + B Ā B̄ Ā.B̄
0 0 1 1 1 1
0 1 0 1 0 0
1 0 0 0 1 0

A B A + B Ā B̄ Ā.B̄
0 0 1 1 1 1
0 1 0 1 0 0
1 0 0 0 1 0
1 1 0 0 0 0
Table : Truth Table for De Morgan’s Theorem 2

Logic Gates

Logic Gates
Logic gates are the basic building blocks of any digital system. It is an
electronic circuit having one or more than one input and only one output.
The relationship between the input and the output is based on a certain
logic. Based on this, logic gates are named as AND gate, OR gate, NOT
gate etc.
AND Gate

Logic Gates
gate etc.
AND Gate
A circuit which performs an AND operation is shown in figure. It has n
input (n >= 2) and one output.

Logic Gates
gate etc.
AND Gate
Y = A AND B AND C....N

Logic Gates
gate etc.
AND Gate
Y = A.B.C....N

Logic Gates
gate etc.
AND Gate
Y = A.B.C....N
Y = ABC...N

AND Gate

AND Gate
AND gate logic diagram:

AND Gate
Truth Table:
Inputs Output
A B AB

AND Gate
Truth Table:
Inputs Output
A B AB
0 0 0

AND Gate
Truth Table:
Inputs Output
A B AB
0 0 0
0 1 0

AND Gate
Truth Table:
Inputs Output
A B AB
0 0 0
0 1 0
1 0 0

AND Gate
Truth Table:
Inputs Output
A B AB
0 0 0
0 1 0
1 0 0
1 1 1
Table : Truth Table of AND Gate

Logic Gates
OR Gate

Logic Gates
OR Gate
A circuit which performs an OR operation is shown in figure. It has n

Logic Gates
OR Gate
Y = A OR B OR C....N

Logic Gates
OR Gate
Y = A OR B OR C....N
Y = A + B + C + ...N

OR Gate

OR Gate
OR Gate logic diagram:

OR Gate
Truth Table:
Inputs Output
A B A + B

OR Gate
Truth Table:
Inputs Output
A B A + B
0 0 0

OR Gate
Truth Table:
Inputs Output
A B A + B
0 0 0
0 1 1

OR Gate
Truth Table:
Inputs Output
A B A + B
0 0 0
0 1 1
1 0 1

OR Gate
Truth Table:
Inputs Output
A B A + B
0 0 0
0 1 1
1 0 1
1 1 1
Table : Truth Table of OR Gate

Logic Gates
NOT Gate

Logic Gates
NOT Gate
NOT gate is also known as Inverter. It has one input A and one output Y.

Logic Gates
NOT Gate
Y = NOT A

Logic Gates
NOT Gate
Y = NOT A
Y = Ā

NOT Gate

NOT Gate
NOT Gate logic diagram:

NOT Gate
Truth Table:
Inputs Output
A Y

NOT Gate
Truth Table:
Inputs Output
A Y
0 1

NOT Gate
Truth Table:
Inputs Output
A Y
0 1
1 0
Table : Truth Table of NOT Gate

Logic Gates
NAND Gate

Logic Gates
NAND Gate
A NOT-AND operation is known as NAND operation. It has n input
(n >= 2) and one output.

Logic Gates
NAND Gate
Y = A NOT AND B NOT AND C....N

Logic Gates
NAND Gate
Y = A NAND B NAND C +...N

Logic Gates
NAND Gate
Y = A NAND B NAND C +...N
Y = A.B.C...N

NAND Gate

NAND Gate
NAND Gate logic diagram:

NAND Gate
Truth Table:
Inputs Output
A B AB

NAND Gate
Truth Table:
Inputs Output
A B AB
0 0 1

NAND Gate
Truth Table:
Inputs Output
A B AB
0 0 1
0 1 1

NAND Gate
Truth Table:
Inputs Output
A B AB
0 0 1
0 1 1
1 0 1

NAND Gate
Truth Table:
Inputs Output
A B AB
0 0 1
0 1 1
1 0 1
1 1 0
Table : Truth Table of NAND Gate

Logic Gates
NOR Gate

Logic Gates
NOR Gate
A NOT-OR operation is known as NOR operation. It has n input

Logic Gates
NOR Gate
Y = A NOT OR B NOT OR C....N

Logic Gates
NOR Gate
Y = A NOR B NOR C +...N

Logic Gates
NOR Gate
Y = A NOR B NOR C +...N
Y = A + B + C + ...N

NOR Gate

NOR Gate
NOR Gate logic diagram:

NOR Gate
Truth Table:
Inputs Output
A B A + B

NOR Gate
Truth Table:
Inputs Output
A B A + B
0 0 1

NOR Gate
Truth Table:
Inputs Output
A B A + B
0 0 1
0 1 0

NOR Gate
Truth Table:
Inputs Output
A B A + B
0 0 1
0 1 0
1 0 0

NOR Gate
Truth Table:
Inputs Output
A B A + B
0 0 1
0 1 0
1 0 0
1 1 0
Table : Truth Table of NOR Gate

Logic Gates
XOR Gate

Logic Gates
XOR Gate
XOR or Ex-OR gate is a special type of gate. It can be used in the half
adder, full adder and subtractor. The exclusive-OR gate is abbreviated as
EX-OR gate or sometime as X-OR gate. It has n input (n >= 2) and one
output.

Logic Gates
XOR Gate
output.
Y = A XOR B XOR C....N

Logic Gates
XOR Gate
output.
Y = A ⊕ B ⊕ C ⊕ ...N

Logic Gates
XOR Gate
output.
Y = A ⊕ B ⊕ C ⊕ ...N
Y = AB̄ + ĀB

XOR Gate

XOR Gate
XOR Gate logic diagram:

XOR Gate
Truth Table:
Inputs Output
A B A ⊕ B

XOR Gate
Truth Table:
Inputs Output
A B A ⊕ B
0 0 0

XOR Gate
Truth Table:
Inputs Output
A B A ⊕ B
0 0 0
0 1 1

XOR Gate
Truth Table:
Inputs Output
A B A ⊕ B
0 0 0
0 1 1
1 0 1

XOR Gate
Truth Table:
Inputs Output
A B A ⊕ B
0 0 0
0 1 1
1 0 1
1 1 0
Table : Truth Table of XOR Gate

Logic Gates
XNOR Gate

Logic Gates
XNOR Gate
XNOR gate is a special type of gate. It can be used in the half adder, full
adder and subtractor. The exclusive-NOR gate is abbreviated as EX-NOR
gate or sometime as X-NOR gate. It has n input (n >= 2) and one output.

Logic Gates
XNOR Gate
Y = A XNOR B XNOR C....N

Logic Gates
XNOR Gate
Y = A B C ...N

Logic Gates
XNOR Gate
Y = A B C ...N
Y = AB + ĀB̄

XNOR Gate

XNOR Gate
XNOR Gate logic diagram:

XNOR Gate
Truth Table:
Inputs Output
A B A B

XNOR Gate
Truth Table:
Inputs Output
A B A B
0 0 1

XNOR Gate
Truth Table:
Inputs Output
A B A B
0 0 1
0 1 0

XNOR Gate
Truth Table:
Inputs Output
A B A B
0 0 1
0 1 0
1 0 0

XNOR Gate
Truth Table:
Inputs Output
A B A B
0 0 1
0 1 0
1 0 0
1 1 1
Table : Truth Table of XNOR Gate

Example - 1
Problem

Example - 1
Problem
Simplify: C + BC
Solution

Example - 1
Problem
Simplify: C + BC
Solution
1 Step 1: DeMorgan’s Theore C + B̄ + C̄

Example - 1
Problem
Simplify: C + BC
Solution
2 Step 2: Commutative, Associative Laws C + C̄ + B̄

Example - 1
Problem
Simplify: C + BC
Solution
3 Step 3: Complement Laws 1 + B̄

Example - 1
Problem
Simplify: C + BC
Solution
4 Step 4: Annulment/Identity Law 1

Example - 1
Problem
Simplify: C + BC
Solution
4 Step 4: Annulment/Identity Law 1
C + BC = 1

Example - 2
Problem

Example - 2
Problem
Simplify: AB(Ā + B)(B̄ + B)
Solution

Example - 2
Problem
Solution
1 Step 1: Identity Law (B + B̄ = 1), so the equation becomes
AB(Ā + B)

Example - 2
Problem
Solution
AB(Ā + B)
2 Step 2: DeMorgan’s Theorem (Ā + B̄)(Ā + B)

Example - 2
Problem
Solution
AB(Ā + B)
3 Step 3: Distributive law Ā.Ā + Ā.B + B̄.Ā + B̄.B

Example - 2
Problem
Solution
AB(Ā + B)
4 Step 4: Annulment/Identity Law Ā + Ā.B + B̄.Ā = Ā(1 + B + B̄)

Example - 2
Problem
Solution
AB(Ā + B)
5 Step 5: Complement/Identity Law Ā

Example - 2
Problem
Solution
AB(Ā + B)
5 Step 5: Complement/Identity Law Ā
AB(Ā + B)(B̄ + B) = Ā

Example - 3
Problem

Example - 3
Problem
Simplify: (A + C)(AD + AD̄) + AC + C
Solution

Example - 3
Problem
Solution
1 Step 1: Distributive Law (A + C)A(D + D̄) + C(A + 1), so the
equation becomes
(A + C)A + C

Example - 3
Problem
Solution
equation becomes
(A + C)A + C
2 Step 2: Distributive/Identity Law A.A + A.C + C = A + A.C + C

Example - 3
Problem
Solution
equation becomes
(A + C)A + C
3 Step 3: Distributive/Commutative law A(1 + C) + C

Example - 3
Problem
Solution
equation becomes
(A + C)A + C
4 Step 4: Annulment/Identity Law A + C

Example - 3
Problem
Solution
equation becomes
(A + C)A + C
4 Step 4: Annulment/Identity Law A + C
(A + C)(AD + AD̄) + AC + C = A + C

Example - 4
Problem

Example - 4
Problem
Simplify: Ā(A + B) + (B + AA)(A + B̄)
Solution

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
2 Step 2: Ā.A + Ā.B + B.A + B.B̄ + A.A + A.B̄

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
3 Step 3: Ā.B + B.A + A + A.B̄

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
3 Step 3: Ā.B + B.A + A + A.B̄
4 Step 4: Ā.B + A(B + 1 + B̄)

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
3 Step 3: Ā.B + B.A + A + A.B̄
4 Step 4: Ā.B + A(B + 1 + B̄)
5 Step 5: Ā.B + A

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
3 Step 3: Ā.B + B.A + A + A.B̄
4 Step 4: Ā.B + A(B + 1 + B̄)
5 Step 5: Ā.B + A
6 Step 6: (A + Ā)(A + B)

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
3 Step 3: Ā.B + B.A + A + A.B̄
4 Step 4: Ā.B + A(B + 1 + B̄)
5 Step 5: Ā.B + A
6 Step 6: (A + Ā)(A + B)
7 Step 7: (A + B)

Example - 4
Problem
Solution
1 Step 1: Ā(A + B) + (B + A)(A + B̄)
3 Step 3: Ā.B + B.A + A + A.B̄
4 Step 4: Ā.B + A(B + 1 + B̄)
5 Step 5: Ā.B + A
6 Step 6: (A + Ā)(A + B)
7 Step 7: (A + B)
Ā(A + B) + (B + AA)(A + B̄) = (A + B)

Example - 5
Problem

Example - 5
Problem
Minimize the following Boolean expression using Boolean identities –
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
1 Step 1: F(A, B, C) = A0B + B(C0 + C) + AB0C0

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
2 Step 2: F(A, B, C) = A0B + B + AB0C0

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
2 Step 2: F(A, B, C) = A0B + B + AB0C0
3 Step 3: F(A, B, C) = B(A0 + 1) + AB0C0

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
2 Step 2: F(A, B, C) = A0B + B + AB0C0
3 Step 3: F(A, B, C) = B(A0 + 1) + AB0C0
4 Step 4: F(A, B, C) = B + AB0C0

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
2 Step 2: F(A, B, C) = A0B + B + AB0C0
3 Step 3: F(A, B, C) = B(A0 + 1) + AB0C0
4 Step 4: F(A, B, C) = B + AB0C0
5 Step 5: F(A, B, C) = (B + B̄)(B + AC̄)

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
2 Step 2: F(A, B, C) = A0B + B + AB0C0
3 Step 3: F(A, B, C) = B(A0 + 1) + AB0C0
4 Step 4: F(A, B, C) = B + AB0C0
5 Step 5: F(A, B, C) = (B + B̄)(B + AC̄)
6 Step 6: F(A, B, C) = (B + AC0)

Example - 5
Problem
F(A, B, C) = A0B + BC0 + BC + AB0C0
Solution
2 Step 2: F(A, B, C) = A0B + B + AB0C0
3 Step 3: F(A, B, C) = B(A0 + 1) + AB0C0
4 Step 4: F(A, B, C) = B + AB0C0
5 Step 5: F(A, B, C) = (B + B̄)(B + AC̄)
6 Step 6: F(A, B, C) = (B + AC0)
F(A, B, C) = A0B + BC0 + BC + AB0C0 = (B + AC0)

Example - 6
Problem

Example - 6
Problem
F(A, B, C) = (A + B)(A + C)
Solution

Example - 6
Problem
F(A, B, C) = (A + B)(A + C)
Solution
1 Step 1: F(A, B, C) = A.A + A.C + B.A + B.C

Example - 6
Problem
F(A, B, C) = (A + B)(A + C)
Solution
2 Step 2: F(A, B, C) = A + A.C + B.A + B.C

Example - 6
Problem
F(A, B, C) = (A + B)(A + C)
Solution
3 Step 3: F(A, B, C) = A(1 + C + A) + B.C

Example - 6
Problem
F(A, B, C) = (A + B)(A + C)
Solution
3 Step 3: F(A, B, C) = A(1 + C + A) + B.C
4 Step 4: F(A, B, C) = A + B.C

Example - 6
Problem
F(A, B, C) = (A + B)(A + C)
Solution
3 Step 3: F(A, B, C) = A(1 + C + A) + B.C
4 Step 4: F(A, B, C) = A + B.C
F(A, B, C) = (A + B)(A + C) = A + B.C

Example - 7
Problem

Example - 7
Problem
Using Boolean algebra techniques, simplify this expression:
AB + A(B + C) + B(B + C)
Solution

Example - 7
Problem
AB + A(B + C) + B(B + C)
Solution
1 Step 1:Apply the distributive law to the second and third terms in the expression, as
follows:
AB + AB + AC + BB + BC

Example - 7
Problem
AB + A(B + C) + B(B + C)
Solution
follows:
2 Step 2: (BB=B, AB+AB=AB)
AB + AC + B + BC

Example - 7
Problem
AB + A(B + C) + B(B + C)
Solution
follows:
AB + AC + B + BC
3 Step 3: B(A + 1 + C) + AC

Example - 7
Problem
AB + A(B + C) + B(B + C)
Solution
follows:
AB + AC + B + BC
3 Step 3: B(A + 1 + C) + AC
4 Step 4: B + AC

Example - 7
Problem
AB + A(B + C) + B(B + C)
Solution
follows:
AB + AC + B + BC
3 Step 3: B(A + 1 + C) + AC
4 Step 4: B + AC
AB + A(B + C) + B(B + C) = B + AC

Example - 7

Example - 8
Problem

Example - 8
Problem
Simplify the Boolean expressions:

Example - 8
Problem
1 AB̄ + A(B + C) + B(B + C)

Example - 8
Problem
1 AB̄ + A(B + C) + B(B + C)
2 [AB̄(C + BD) + ĀB̄]C

Example - 8
Problem
1 AB̄ + A(B + C) + B(B + C)
2 [AB̄(C + BD) + ĀB̄]C
3 ĀBC + AB̄C̄ + ĀB̄C̄ + AB̄C + ABC

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