ECEG-3201 Digital Logic Design
Addis Ababa Institute of Technology
(AAIT) Department of Electrical and
Computer Engineering
Learning Outcomes

At the end of the lecture, students should able to:




Perform proper groupings of 1‟s or 0‟s in K-map
Obtain the minimized SOP or POS expression from a
K-map
Convert SOP expression to POS using a K-map and
vice versa
Apply “don‟t cares” on K-maps to minimize Boolean
expressions.
AAIT, Department of
Electrical and Computer
Engineering
2
Nebyu Yonas Sutri
Definitions


Minterms: is defined as a product term that is 1 in
exactly one row of the truth table or cell of a K-map.
Maxterms: is defined as a sum term that is 0 in
exactly one row of the truth table or cell of a K-map.
AAIT, Department of
Electrical and Computer
Engineering
3
Nebyu Yonas Sutri
Definitions


Canonic Sum: is sum of minterms corresponding
to truth-table rows (input combinations) for which
the function produces a 1 output.
Canonic Product: is product of the maxterms
corresponding to input combinations for which the
function produces a 0 output.
Canonic sum
Canonic product
AAIT, Department of
Electrical and Computer
Engineering
4
Nebyu Yonas Sutri
Introduction to K-Map





An alternate approach to represent a
simplified Boolean function.
Similar to a truth table.
Can be used to minimize Boolean functions
graphically.
Each cell‟s position is fixed by a binary value
of the input variables.
Number of cells = 2N where N is the number
of input variables.
AAIT, Department of
Electrical and Computer
Engineering
5
Nebyu Yonas Sutri
2-Variable K-Map

For a 2 inputs system, 22 = 4 cells.
2 inputs truth table
A
0
0
1
1
B Q
0
1
0
1
AAIT, Department of
Electrical and Computer
Engineering
2-variable K-Map
A
B
0
1
0
1
00
01
10
B
0
1
0
A’B’
A’B
1
AB’
AB
A
11
6
Nebyu Yonas Sutri
3-Variable K-Map

For a 3 inputs system, 23 = 8 cells.
3 inputs truth table
A
B
C
0
0
0
0
0
1
0
1
0
Q
3-variable K-Map
A
BC
00
01
11
10
Grey code form
0
OR
1
AB
C
0
1
BC
A
0
1
1
1
0
0
1
0
1
01
1
1
0
11
11
1
1
1
10
10
AAIT, Department of
Electrical and Computer
Engineering
00
7
0
1
00
OR
01
Nebyu Yonas Sutri
3-Variable K-Map
C
0
1
00
A’B’C’
A’B’C
01
A’BC’
A’BC
11
ABC’
ABC
10
AB’C’
AB’C
AB
AAIT, Department of
Electrical and Computer
Engineering
8
Nebyu Yonas Sutri
4-Variable K-Map
For a 4 inputs system, 24 = 16 cells.
4 inputs truth table

A
B
C
D
0
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
0
0
0
1
0
0
1
1
0
1
0
1
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
AAIT, Department of
Electrical and Computer
Engineering
Q
4-variable K-Map
Grey code form
AB
CD
00
01
11
10
00
01
11
10
Grey code form
9
Nebyu Yonas Sutri
4-Variable K-Map
CD
00
AB
00
01
11
10
A’B’C’D’ A’B’C’D A’B’CD A’B’CD’
01
A’BC’D’ A’BC’D
A’BCD
A’BCD’
11
ABC’D’
ABC’D
ABCD
ABCD’
10
AB’C’D’ AB’C’D
AB’CD
AB’CD’
AAIT, Department of
Electrical and Computer
Engineering
10
Nebyu Yonas Sutri
Cell Adjacency



The cells are arranged in a way that there is only one
single variable change between adjacent cells.
Cells that are different by only 1 variable  “adjacent”.
Cell adjacency rules:




Each cell adjacent to cells immediately next to it, on any
four sides.
A cell is NOT adjacent to the cells that diagonally touch any
of its corners.
Cells in top row adjacent to corresponding cells in bottom
row.
Cells in outer left column adjacent to corresponding cells in
outer right column.
AAIT, Department of
Electrical and Computer
Engineering
11
Nebyu Yonas Sutri
Cell Adjacency

For example, let‟s look at the cell adjacency
of a 3-variable K-map:
C
0
1
AB
00
01
11
10
AAIT, Department of
Electrical and Computer
Engineering
12
Nebyu Yonas Sutri
Filling in the K-Map




From any Boolean expression or a problem
statement, write the truth table first.
Determine how many inputs and outputs, and
therefore how many K-maps cells are
required.
Draw the K-map and label it accordingly.
Do the mapping on the K-map according to
the inputs and outputs condition in the truth
table.
AAIT, Department of
Electrical and Computer
Engineering
13
Nebyu Yonas Sutri
Grouping in K-Map





Group all 1s in the map into group of 2n
where n = 1,2,4,8 ...
The grouping of the 1s must be in adjacent
cell and can overlap.
Every square containing 1 must be
considered at least once.
A square containing 1 can be included in as
many groups as desired.
A group must be as large as possible.
AAIT, Department of
Electrical and Computer
Engineering
14
Nebyu Yonas Sutri
Grouping in K-Map

AAIT, Department of
Electrical and Computer
Engineering

15
Nebyu Yonas Sutri
Karnaugh Map SOP Minimization

SOP minimization using Karnaugh Map:

Step 1: Map SOP expression onto the K-map.



For a non-standard SOP expression, must convert into
standard form first.
Step 2: Group the 1‟s.
Step 3: Determine minimum SOP expression
AAIT, Department of
Electrical and Computer
Engineering
16
Nebyu Yonas Sutri
Mapping a Standard SOP Expression



A 1 is placed in the cell corresponding to the
binary value of each product term.
E.g. product term A‟BC, a 1 is placed in the cell
corresponding to 011.
The steps are:


Step 1: Determine binary value of each product term.
Step 2: Place a 1 in the cell corresponding to the
binary value of the product term.
AAIT, Department of
Electrical and Computer
Engineering
17
Nebyu Yonas Sutri
Mapping a Standard SOP Expression

E.g. Map the standard SOP expression
A‟BC+AB‟C+AB‟C‟ on a K-map
011
101
100
C
0
1
AB
00
1
01
11
10
AAIT, Department of
Electrical and Computer
Engineering
1
1
18
Nebyu Yonas Sutri
Mapping a Standard SOP Expression

E.g. Map the standard SOP expression
A‟BCD+ABCD‟+ABC‟D‟+ABCD on a K-map
0111
1110
1100
1111
CD
AB
00
01
11
10
00
1
01
11
1
1
1
10
AAIT, Department of
Electrical and Computer
Engineering
19
Nebyu Yonas Sutri
Mapping a Non Standard SOP
Expression


Convert to standard SOP form and map.
e.g. Map the expression AD+A‟BCD‟ on a K-map
AD  AD( B  B)(C  C )
 ABCD  ABC D  ABCD  AB C D
 ABCD  ABC D  ABCD  AB C D
 ABC D
AAIT, Department of
Electrical and Computer
Engineering
1111
1101
1011
1001
0110
20
CD 00
01
11
10
AB
00
01
11
10
1
1
1
1
1
Nebyu Yonas Sutri
Grouping the 1’s
The rules to group the 1‟s are:

1.
2.
3.
A group must contain either 1,2,4,8, or 16 cells.
In the case of a 3-variable map, 23= 8 cells is the
max. group.
Each cell in a group must be adjacent to one or
more cells in that same group, but all cells in the
group do not have to be adjacent to each other.
Always include the largest possible number of 1s
in a group in accordance to rule 1.
AAIT, Department of
Electrical and Computer
Engineering
21
Nebyu Yonas Sutri
Grouping of two 1’s

There are four ways to group two 1‟s:
CD 00
AB
11
CD 00
10
AB
00
01
1
1
1
01
1
11
10
10
00
01
11
1
CD 00
10
AB
01
11
11
11
10
10
1
1
10
1
AAIT, Department of
Electrical and Computer
Engineering
01
11
00
01
10
01
00
11
CD 00
AB
01
22
Nebyu Yonas Sutri
Grouping of four 1’s

There are six ways to group four 1‟s:
CD 00
AB
11
CD 00
10
AB
00
01
00
1
01
1
11
11
1
10
10
1
01
1
CD 00
AB
01
1
01
1
11
1
CD 00
10
AB
00
00
01
1
1
01
11
11
1
1
11
10
10
AAIT, Department of
Electrical and Computer
Engineering
23
11
01
11
11
1
1
1
1
10
10
Nebyu Yonas Sutri
Grouping of four 1’s
CD 00
AB
01
11
CD 00
10
AB
00
00
01
1
1
01
11
1
1
11
10
10
AAIT, Department of
Electrical and Computer
Engineering
24
01
11
10
1
1
1
1
Nebyu Yonas Sutri
Grouping of eight 1’s

There are four ways to group eight 1‟s:
CD 00
AB
01
11
CD 00
10
AB
00
11
00
1
1
01
1
1
1
1
01
1
1
11
1
1
1
1
11
1
1
10
1
1
01
11
10
CD 00
AB
01
00
11
10
1
1
1
CD 00
AB
10
00
1
1
01
01
1
1
11
11
1
1
10
1
1
10
1
01
10
1
1
1
AAIT, Department of
Electrical and Computer
Engineering
1
25
Nebyu Yonas Sutri
Determining the Minimum SOP
Expression from the Map


Each group of cells  one product term
composed of all variables that occur in
only one form (either complemented or
uncomplemented) within the group.
Variables that occur both complemented and
uncomplemented within the group are
eliminated.
AAIT, Department of
Electrical and Computer
Engineering
26
Nebyu Yonas Sutri
Examples – (given truth table)

Example: Given the truth table below. Find
the simplified expression using K-map.
Step 1: Fill in the K-map with the respective output values.
A
B
Q
0
0
1
0
1
1
1
0
0
1
1
0
AAIT, Department of
Electrical and Computer
Engineering
A
B
0
1
27
0
1
1
1
0
0
Nebyu Yonas Sutri
Examples – (given truth table)
Step 3: Derive the Boolean expression
Step 2: Group the 1s together.
A
B
0
1
B
0
1
0
Q A
1
1
0
AAIT, Department of
Electrical and Computer
Engineering
Know why is this so?
Since 1s is located at the
area outside A.
A
28
Nebyu Yonas Sutri
Examples – (given truth table)

Example: Given the truth table below. Find
the simplified expression using K-map.
A
B
C
Q
0
0
0
1
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
1
1
0
1
0
1
1
0
1
1
1
1
1
AAIT, Department of
Electrical and Computer
Engineering
Step 1: Fill in the K-map with the
respective output values.
A
29
BC
00
01
11
0
1
0
0
0
1
1
0
1
1
10
Nebyu Yonas Sutri
Examples – (given truth table)
Step 2: Group the 1s together.
Step 3: Derive the Boolean expression
C
A
A
BC
00
01
11
0
1
0
0
0
1
1
0
1
1
10
Q  AB  BC
B
AAIT, Department of
Electrical and Computer
Engineering
30
Nebyu Yonas Sutri
Examples – (given truth table)
A
B
C
D
Q
0
0
0
0
0
0
0
0
1
1
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
0
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
1
AAIT, Department of
Electrical and Computer
Engineering

Example : Given the truth table below.
Find the simplified expression using
K-map.
Step 1: Fill in the K-map with the
respective output values.
AB
CD
00
01
11
10
31
00
01
0
1
0
0
0
1
0
0
0
1
1
0
0
0
0
0
11
10
Nebyu Yonas Sutri
Examples – (given truth table)
Step 3: Derive the Boolean expression
Step 2: Group the 1s together.
AB
CD
00
01
A
11
10
C
00
01
0
1
0
0
0
1
0
0
0
1
1
0
0
0
0
0
11
10
Q  ACD  ABD
B
D
AAIT, Department of
Electrical and Computer
Engineering
32
Nebyu Yonas Sutri
Examples – (given expression)

Example: Given the Boolean expression below. Find the
simplified expression using K-map.
Q  A B  C  D  A B  C  D  A B  C  D  A B  C  D  A B  C  D  A B  C  D
AB
CD
00
01
11
10
00
01
11
10
0
1
1
0
0
1
1
0
0
0
0
0
1
1
0
0
Step 1: Fill in the K-map with 1 according to the expression.
AAIT, Department of
Electrical and Computer
Engineering
33
Nebyu Yonas Sutri
Examples – (given expression)
Step 3: Derive the Boolean expression
Step 2: Group the 1s together.
C
AB
CD
00
01
A
11
10
00
01
11
10
0
1
1
0
0
1
1
0
0
0
0
0
1
1
0
0
Q  AD  ABC
B
D
AAIT, Department of
Electrical and Computer
Engineering
34
Nebyu Yonas Sutri
Examples – (given expression)

Example: Given the Boolean expression below. Find the
simplified expression using K-map.
Q  A  AB  ABC
Step 1:
Convert the non standard terms
to standard forms!
Q  A( B  B )(C  C )  AB(C  C )  ABC
Pay Attention to the White
board note!!!!
AAIT, Department of
Electrical and Computer
Engineering
35
Q  A  AB  ABC
000
001
010
011
101
100
110
Inputs condition that
satisfy the output
(Q = 1)
Nebyu Yonas Sutri
Examples – (given expression)
Step 2: Fill in the K-map with 1
where applicable
Step 3: Derive the Boolean expression
B
A
A
BC
00
01
11
0
1
1
1
1
1
1
1
0
1
10
Q  A B  C
C
AAIT, Department of
Electrical and Computer
Engineering
36
Nebyu Yonas Sutri
Examples – (given circuit)

Example : Given the circuit
below. Find the simplified
circuit using K-map.
A
Q
B
C
Step 1:
Find the expression at the output of the circuit.
Q  A  A B  A B  C
AAIT, Department of
Electrical and Computer
Engineering
37
Nebyu Yonas Sutri
Examples – (given circuit)

From the expression, derive the truth table.
Q  A  A B  A B  C
100
110
111
AAIT, Department of
Electrical and Computer
Engineering
001
011
101
38
A
B
C
Q
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
Nebyu Yonas Sutri
Examples – (given circuit)
Step 2: Fill in the K-map with the respective output values.
A
B
C
Q
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
1
1
1
1
1
BC
B
00
01
0
0
1
1
0
A 1
1
1
1
1
A
11
10
C
Step 3: Derive the Boolean expression
AAIT, Department of
Electrical and Computer
Engineering
Q  A C
39
Nebyu Yonas Sutri
Examples – (given circuit)

Therefore, simplified circuit is:
A
Q
C

Hmmmm?... So where does B go?
AAIT, Department of
Electrical and Computer
Engineering
40
Nebyu Yonas Sutri
5 Variable K-Map


Boolean functions with 5 variables can be
simplified using a 32-cell K-Map.
Two 4-variable maps are used to construct a 5
variable map.
DE
BC
00
00
01
11
DE
10
BC
00
01
01
11
11
10
10
A=0
AAIT, Department of
Electrical and Computer
Engineering
00
01
11
10
A=1
41
Nebyu Yonas Sutri
5 Variable K-Map

To visualize adjacency
imagine as A=0 map is
placed on top of A=1
map.
Example: Determine the simplified
expression of the example given
Solution
Q  DE  BCE  ABD  BC DE
AAIT, Department of
Electrical and Computer
Engineering
42
Nebyu Yonas Sutri
Exercise – 5 Variables

Minimize/Simplify the following Boolean expression
F  ABC DE  ABC DE  ABC DE  ABC DE  ABC DE
 ABC DE  ABCDE  ABC DE  ABC DE  ABC DE
 ABCDE  ABCDE
DE
BC
00
00
01
1
1
01
1
11
1
10
1
11
DE
10
BC
00
00
01
1
1
01
1
11
1
11
10
1
1
1
10
A=0
AAIT, Department of
Electrical and Computer
Engineering
A=1
43
Nebyu Yonas Sutri
Exercise – 5 Variables
DE
DE
00
01
00
1
1
01
1
11
1
10
1
BC
11
10
BC
00
00
01
1
1
01
1
11
1
11
10
1
1
1
10
A=0
A=1
Q  ADE  BC D  BCE  ACDE
AAIT, Department of
Electrical and Computer
Engineering
44
Nebyu Yonas Sutri
Don’t Care Conditions




Don‟t care conditions are those that have
input conditions for which there are no
specified output levels.
In other words, there are certain input
conditions that may occur that we don‟t care
whether the output is high or low.
In the K Map, we place an „X‟ wherever
output is a Don‟t Care condition.
The „X‟ is used as either a 1 or a 0 in the K
map simplification process to our advantage.
AAIT, Department of
Electrical and Computer
Engineering
45
Nebyu Yonas Sutri
Usage of “Don’t Cares”




Don‟t cares can be either 0 or 1.
, can be used to find simplified Boolean
expression in SOP or POS form.
When grouping 1‟s, treat X‟s as 1‟s.
When grouping 0‟s, treat X‟s as 0‟s.

Include “don’t cares” in grouping ONLY
if doing so simplifies the expression.
Otherwise, ignore them.
AAIT, Department of
Electrical and Computer
Engineering
46
Nebyu Yonas Sutri
Usage of “Don’t Cares”

Example
YZ 00
01
11
10
WX
00
1
0
1
1
01
0
X
1
1
11
1
0
0
0
10
1
1
0
0
Without don’t cares : W’Y+WY’Z’+WX’Y’+W’X’Z’
With don’t cares : W’Y+WY’Z’+WX’Y’+W’X’Z’+W’XZ
AAIT, Department of
Electrical and Computer
Engineering
47
Nebyu Yonas Sutri
Example
A
B
C
D
X
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
0
0
0
0
0
0
1
1
1
X
X
X
X
X
X
AAIT, Department of
Electrical and Computer
Engineering
CD 00
01
11
10
AB
00
0
0
0
0
01
0
0
1
0
A’BCD
11
X
X
X
X
BCD
10
1
1
X
X
A
AB’C’

Without “don‟t cares”:
AB’C’+A’BCD

With “don‟t cares”:
A+BCD
48
Nebyu Yonas Sutri
Don’t Care Conditions
Example : Given the truth table with don’t care condition, find the
simplified expression using K-map.
A
B
C
Q
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
X
1
0
0
X
1
0
1
1
1
1
0
1
1
1
1
1
AAIT, Department of
Electrical and Computer
Engineering
C
0
1
00
0
0
01
0
X
11
1
1
10
X
1
AB
0
1
00
0
0
01
0
0
11
1
1
10
1
1
AB
A
C
B
Q A
49
Nebyu Yonas Sutri
Don’t Care Conditions
Example : We are dealing with an elevator in a three floor building.
An indicator is placed at the elevator and is active (ON) when
the elevator stops and one of the floor is aligned with the
indicator. The indicator will be OFF when it is aligned at one of
the floor while it is still moving. Derive the digital controller
expression to control the indicator.
First: simplify the problems and set what are the inputs and output.
Inputs: F1, F2, F3 (floor) and M (moving)
Output: O (indicator)
O = 1 when M = 0, one of the F = 1
O = 0 when M = 1, one of the F = 1
Indicator can not be at two or three floors at the same time  don’t care condition!
AAIT, Department of
Electrical and Computer
Engineering
50
Nebyu Yonas Sutri
Don’t Care Conditions
Second: Derive the truth table
M
F1
F2
F3
O
0
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
X
0
1
0
0
1
0
1
0
1
X
0
1
1
0
X
0
1
1
1
X
1
0
0
0
0
1
0
0
1
0
1
0
1
0
0
1
0
1
1
X
1
1
0
0
0
1
1
0
1
X
1
1
1
0
X
1
1
1
1
x
AAIT, Department of
Electrical and Computer
Engineering
Third: Map into the K-map
F2 F3
00
M F1
00
01
M
11
10
F2
01
11
10
0
1
X
1
1
X
X
X
0
X
X
X
0
0
X
0
F1
F3
51
Nebyu Yonas Sutri
Don’t Care Conditions
Forth: Group the 1‟s
F2 F3
00
M F1
00
01
M
11
10
Fifth: Derive the Boolean expression
F2
01
11
O  M  F1  M  F 2  M  F 3
10
0
1
X
1
1
X
X
X
0
X
X
X
0
0
X
0
O  M ( F1  F 2  F 3 )
F1
F3
AAIT, Department of
Electrical and Computer
Engineering
52
Nebyu Yonas Sutri
Karnaugh Map POS Minimization

Almost similar to that for an SOP expression.

Step 1: Map the POS expression onto the K-map.



For a non-standard POS expression, must convert into
standard form first.
Step 2: Group the 0‟s.
Step 3: Determine the minimum POS expression
AAIT, Department of
Electrical and Computer
Engineering
53
Nebyu Yonas Sutri
Mapping a Standard POS Expression



A 0 is placed in the cell corresponding to the
binary value of each sum term in the expression.
E.g. sum term (A‟+B+C), a 0 is placed in the cell
corresponding to 100.
The steps are:


Step 1: Determine the binary value of each sum term.
Step 2: Place a 0 in the cell corresponding to the
binary value of the sum term.
AAIT, Department of
Electrical and Computer
Engineering
54
Nebyu Yonas Sutri
Mapping a Standard POS Expression

E.g. Map the standard POS expression
(A+B‟+C‟+D)(A+B+C+D‟)(A+B+C+D)
(A‟+B+C‟+D) on a K-map
0110
0001
0000
1010
CD 00
01
0
0
11
10
AB
00
0
01
11
10
AAIT, Department of
Electrical and Computer
Engineering
0
55
Nebyu Yonas Sutri
Mapping a Non Standard POS
Expression


Convert to standard POS form and map.
E.g. Map the expression
(B+C+D)(A+B+C‟+D)(A‟+B+C+D‟)(A+B‟+C+D)(A‟+B‟+C
+D) on a K-map
CD 00
( B  C  D)  ( B  C  D)  A A
 ( A  B  C  D)( A  B  C  D)
 ( A  B  C  D)( A  B  C  D)
( A  B  C  D)( A  B  C  D)
( A  B  C  D)( A  B  C  D)
AAIT, Department of
Electrical and Computer
Engineering
01
11
10
AB
0000
1000
0010
1001
0100
1100
56
00
0
01
0
11
0
10
0
0
0
Nebyu Yonas Sutri
Simplification of POS Expressions



Group the 0‟s in the K-map.
The rules to group the 0‟s is similar to the
rules of grouping 1‟s in SOP minimization.
The process of determining the minimum
POS expression is similar to SOP, except
now you find the SUM term from the
groups of 0’s.
AAIT, Department of
Electrical and Computer
Engineering
57
Nebyu Yonas Sutri
Example

Use a K-map to minimize
(A+B+C)(A+B+C‟)(A+B‟+C)(A+B‟+C‟)(A‟+B‟+C)
Step 1: Fill in the K-map with the
respective output values.
C
000
001
010
011
110
0
1
00
0
0
01
0
0
11
0
AB
10
AAIT, Department of
Electrical and Computer
Engineering
58
Nebyu Yonas Sutri
Example
Step 2: Group the 0’s together.
Step 3: Derive the Boolean expression
C
0
1
00
0
0
01
0
0
11
0
AB
F  A( B  C )
10
AAIT, Department of
Electrical and Computer
Engineering
59
Nebyu Yonas Sutri
Exercise

Use a K-map to minimize
(B+C+D)(A+B+C’+D)(A’+B+C+D’)(A+B’+C+D)
(A’+B’+C+D)
Step 1: Fill in the K-map with the
respective output values.
CD 00
( B  C  D)  ( B  C  D)  A A
 ( A  B  C  D)( A  B  C  D)
 ( A  B  C  D)( A  B  C  D)
( A  B  C  D)( A  B  C  D)
( A  B  C  D)( A  B  C  D)
AAIT, Department of
Electrical and Computer
Engineering
01
11
10
AB
0000
1000
0010
1001
0100
1100
60
00
0
01
0
11
0
10
0
0
0
Nebyu Yonas Sutri
Exercise
Step 2: Group the 0’s together.
Step 3: Derive the Boolean expression
CD 00
01
11
10
AB
00
0
01
0
11
0
10
0
0
F  (C  D )( A  B  D )( A  B  C )
0
AAIT, Department of
Electrical and Computer
Engineering
61
Nebyu Yonas Sutri
Converting Between POS and SOP
using K-map





Recall that it‟s possible to convert from POS to SOP and
vice-versa.
The Karnaugh map can be used for this purpose.
For a POS expression, all cells that do not contain 0‟s
contain 1‟s,  SOP equivalent can be derived.
The reverse can be said about an SOP expression.
Importance? To determine which form can be
implemented using less number of gates.
AAIT, Department of
Electrical and Computer
Engineering
62
Nebyu Yonas Sutri
Example

Use a Karnaugh map to convert the following
POS expression to minimum POS , standard
SOP and 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)
CD
AB
00
Step 1: Fill in the K-map with the
respective output values.
01
0
11
0
10
AAIT, Department of
Electrical and Computer
Engineering
00
63
01
11
10
0
0
0
0
Nebyu Yonas Sutri
Example
CD
00
AB
00
01
0
11
0
10
01
11
10
0
0
0
Minimum POS
F  ( B  C  D )( B  C  D )( A  B  C )
0
CD
00
01
11
10
00
1
0
0
0
01
0
1
1
1
11
0
1
1
1
10
1
0
1
1
AB
Standard SOP
F= A‟B‟C‟D‟ + A‟BC‟D + A‟BCD+
A‟BCD‟+ABC‟D+ABCD+ABCD‟+
AB‟C‟D‟+AB‟CD+AB‟CD‟
AAIT, Department of
Electrical and Computer
Engineering
64
Nebyu Yonas Sutri
Example
CD
00
01
11
10
00
1
0
0
0
01
0
1
1
1
11
0
1
1
1
10
1
0
1
1
AB
AAIT, Department of
Electrical and Computer
Engineering
Minimum SOP
F  BD  BC  AC  BC D
65
Nebyu Yonas Sutri
Homework

Use a Karnaugh map to convert the following
SOP expression to POS form:
A‟B‟C‟+AB‟C‟+A‟BC‟+ABC‟+A‟BC
C
0
1
AB
00
01
11
10
AAIT, Department of
Electrical and Computer
Engineering
66
Nebyu Yonas Sutri
What to do this week?

Reading assignment.

Digital Fundamentals, Thomas L. Floyd, Chapter 4,
Pages 210 – 228.
AAIT, Department of
Electrical and Computer
Engineering
67
Nebyu Yonas Sutri