Topic 2

TOPIC 2 : BOOLEAN
OPERATIONS

Example of timing diagram

Logic Gates Symbols & Boolean
Equation

Truth Table
•
A truth table is a mathematical table used in
logic – specifically in connection with Boolean
algebra, Boolean function and propositional
calculus.
•
Practically, a truth table is composed of one
column for each input variable (for example A
and B) and A one B A.B final column for all of the
possible
0results 0 0of the logical operation that
0 1 0
Truth Table
the table is meant to represent (for example A
1 0 0
AND B)
1 1 1

Example 1:

Example 2:
A
D = A + B.C
B.C
Example 3:
A + B
D = (A + B).(A + C)
A + C

Example 4:
Example 5:

NAND gates equivalent in combinational logic gates
Universal
combinational
gates
A
B
C
F
NAND gates
equivalent
combinational
gates
A
B
F
C
F = (A+B).(A+B).C

Exercise:
Boolean Expression =

Construct a Truth Table from Boolean Expression
•
Example: Output S(T+U) + TU
•
Solution:
S T U T+U S(T+U) TU S(T+U) + TU
0 0 0 0 0 0 0
0 0 1 1 0 0 0
0 1 0 1 0 0 0
0 1 1 1 0 1 1
1 0 0 0 0 0 0
1 0 1 1 1 0 1
1 1 0 1 1 0 1
1 1 1 1 1 1 1

Construct a Combinational Logic Circuit from
Boolean Expression:
•
Example: Y = AC + B + BC
•
Solution:

Construct a Boolean Expression from
Combinational Logic Circuit:

Construct A Boolean expression from truth
table
A B C Y
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 0
1 1 1 1
Answer:
Boolean expression: (SUM OF
PRODUCT – SOP)
+ +

Boolean Law & DeMorgan’s Theorems
Commutative
Law
Associative
Law
Distributive
Law
Consensus
Theorem
DeMorgan’s
22
X
X
9)
1
X
X
8)
X
X
X
7)
1
1
X
6)
X
0
X
5)
0
X
X
4)
X
X
X
3)
X
1
X
2)
0
0
X
1)
Y
X
Y
X
14B)
Y
X
X Y
14A)
Y
X
XY
X
13D)
Y
X
XY
X
13C)
Y
X
XY
X
13B)
Y
X
XY
X
13A)
YZ
YW
XZ
XW
Z
W
Y
X
12B)
XZ
XY
Z
Y
X
12A)
Z
Y
X
Z
Y
X
11B)
Z
XY
YZ
X
11A)
X
Y
Y
X
10B)
X
Y
Y
X
10A)

Simplify The Boolean Expression:
Example of simplification

Boolean Law Example

Boolean Law
example

DeMorgan Shortcut
BREAK THE LINE, CHANGE THE SIGN
Break the LINE over the two variables,
and change the SIGN directly under the line.
A
B
A
B
For Theorem #14A, break the line, and
change the AND function to an OR function.
Be sure to keep the lines over the variables.
A
B
A
B
For Theorem #14B, break the line, and
change the OR function to an AND function.
Be sure to keep the lines over the variables.
26

DeMorgan’s: Example #1
Example
Simplify the following Boolean expression and note the
Boolean or DeMorgan’s theorem used at each step. Put
the answer in SOP form.
Solution
F
1
F
1
F
1
F
1
F
1
(X Y) (Y
(X Y)
(X Y)
(X Y)
XY
YZ
(Y
F
1
Z)
Z)
(Y Z)
(Y Z)
(X Y) (Y
Z)
; Theorem #14A
; Theorem #9 & #14B
; Theorem #9
; Rewritten without AND symbols
and parentheses
27

DeMorgan’s: Example #2
So, where would such an odd Boolean expression come from? Take a
look at the VERY poorly designed logic circuit shown below. If you were
to analyze this circuit to determine the output function F2, you would
obtain the results shown.
X
Y
XY
X
XY
( X
Z)(XY)
F 2
(X Z)(XY)
Z
X Z
Example
Simplify the output function F2. Be sure to note the Boolean or
28

DeMorgan’s: Example #2
Solution
F
2
(X
Z)(XY)
F
2
(X
Z)
(XY)
; Theorem #14A
F
2
(X
Z)
(XY)
; Theorem #9
F
2
(X Z)
(XY)
; Theorem #14B
F
F
2
2
(X Z) (XY)
X Z X Y
; Theorem #9
; Rewritten without AND symbols
29

SUM OF PRODUCT (SOP)

Logic Expression from Truth table
•
SUM Of PRODUCT

PRODUCT OF SUM (POS)
POS
POS

CONVERSION OF SOP POS TERM
Take as an example the truth table of a three-variable function as shown below.
Three variables, each of which can take the values 0 or 1, yields eight possible
combinations of values for which the function may be true. These eight
combinations are listed in ascending binary order and the equivalent decimal value is
also shown in the table.
The function has a value 1 for the
combinations shown, therefore:

CONVERSION OF SOP POS TERM

Examples of function in Boolean
•
Consider the function:
•
In binary form:
f(A, B, C, D) = ∑ (0101, 1011, 1100, 0000,
1010, 0111)
•
In decimal form:
f(A, B, C, D) = ∑ (5, 11, 12, 0, 10, 7)

KARNAUGH MAP (K-MAP)
•
A Karnaugh Map is a grid-like representation
of a truth table.
•
It is really just another way of presenting a
truth table, but the mode of presentation
gives more insight.
•
A Karnaugh map has zero and one entries at
different positions.
•
Each position in a grid corresponds to a truth
table entry.

Here's an example taken from the voting circuit presented in the lesson on
Minterms. The truth table is shown first. The Karnaugh Map for this truth
table is shown after the truth table.
Truth Table
K-Map
representing

K-Map Condition:
(i) 2 variables : Use 22 mapping condition (use either 1 only)
A B F
0 0 a
A\B 0 1
B\A 0 1
0 1 b
1 0 c
0 a b
1 c d
Or
0 a c
1 b d
1 1 d
Truth table
K-Map

K-Map Condition:
(ii) 3 variables : Use 23 mapping condition (use either 1 only)
A B C F
0 0 0 a
0 0 1 b
0 1 0 c
0 1 1 d
1 0 0 e
1 0 1 f
1 1 0 g
1 1 1 h
Truth table
(i)
AB\C 0 1
00 a b
01 c d
11 g h
10 e f
A\BC 00 01 11 10
0 a b d c
1 e f h g
(iii)
(ii)
K-Map
BC\A 0 1
00 a e
01 b f
11 d h
10 c g
C\AB 00 01 11 10
0 a c g e
1 b d h f
(iv)

A B C D F
0 0 0 0 a
0 0 0 1 b
0 0 1 0 c
0 0 1 1 d
0 1 0 0 e
0 1 0 1 f
0 1 1 0 g
0 1 1 1 h
1 0 0 0 i
1 0 0 1 j
1 0 1 0 k
1 0 1 1 l
1 1 0 0 m
1 1 0 1 n
1 1 1 0 o
1 1 1 1 p
Truth table
K-Map Condition:
(iii) 4 variables : Use 24 mapping condition
(use either 1 only)
(i)
CD
AB\
00 01 11 10
00 a b d c
01 e f h g
11 m n p o
10 i j l k
AB
CD\
K-Map
(ii)
00 01 11 10
00 a e m i
01 b f n j
11 d h p l
10 c g o k

K-Map Looping Rules
v
The Karnaugh map uses the following rules for the
simplification of expressions by grouping together
adjacent cells containing ones.
1. Groups may not include any cell containing a zero
2. Groups may be horizontal or vertical, but not diagonal.

3. Groups must contain 1, 2, 4, 8, or in general 2n cells.
That is if n = 1, a group will contain two 1's since 21 = 2.
If n = 2, a group will contain four 1's since 22 = 4.
4. Each group should be as large as possible.

5. Groups may overlap.
6. Each cell containing a one must be in at least one group.

7. Groups may wrap around the table. The leftmost cell in a row may be grouped with
the rightmost cell and the top cell in a column may be grouped with the bottom cell.
8. There should be as few groups as possible, as long as this does not contradict
any of the previous rules.

K-Map looping example:
Example 1 Example 2

K-Map solution from truth table
A B C F
0 0 0 1
0 0 1 0
A’B’C’
0 1 0 0
0 1 1 1
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 0
Truth table
A’BC
AB’C’
AB’C
ABC’
F = AB’ + AC’ + B’C’ + A’BC

K-Map solution from Boolean Expression
F = A’B’C’ + ABC’ + AB’C’
+ A’BC + AB’C
Allocate in digital logic:
000 + 110 + 100 +
011 + 101
F = AB’ + AC’ + B’C’ + A’BC

Simplify Boolean expression using K-Map
F = BD + AB’CD + ABC
Represent the hidden variable (don’t care
condition)
è
BD = ABCD, A’BCD, ABC’D, A’BC’D
è
ABC = ABCD, ABCD’
Thus, the expression will become as:
F = ABCD + A’BCD + ABC’D + A’BC’D +
AB’CD + ABCD + ABCD’
AB\CD 00 01 11 10
00 0 0 0 0
01 0 1 1 0
11 0 1 1 1
10 0 0 1 0
Output = BD + ACD + ABC

A B C D F
0 0 0 0 0
0 0 0 1 0
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 0
0 1 1 1 1
1 0 0 0 1
1 0 0 1 0
1 0 1 0 1
1 0 1 1 0
1 1 0 0 1
1 1 0 1 1
1 1 1 0 0
1 1 1 1 0

K-Map solution from FUNCTION
SOP

POS