Digital electronics lesson 2 part 2

Digital Electronics
Lesson 2: Logic Gates and
Boolean Algebra (contd.)
Sukriti Dhang
04/06/2021

Content 1. Standard SOP and POS terms
2. Minterms and Maxterms
3. Karnaugh Map

Standard SOP and POS terms
Logical functions are expressed in terms of logical variables. Values that are taken by
logical functions and logical variables are in the binary form i.e. ‘0’ or ‘1’.
An logic functions can be expressed as :
(a) Sum of Product (SOP) - The logical sum of two or more logical product terms. It is
an OR operation of AND operated variable.
(b) Product of Sum (POS) - The logical product of two or more logical sum terms. It is
an AND operation of OR operated variable.

Product Term and Sum Term
Product Term -
● The AND function of two or more logical variables is referred to as
product.
● The logical product of two or more variables on which a function depends
is considered to be a product term.
● The variables in the product term can be either in complemented form or
uncomplemented form.
● Eg. of product term- ABC
̅

Sum Term -
● The OR function of two or more logical variables is referred to as sum.
● The logical sum of two or more variables on which a function depends is
considered to be a sum term.
● The variables in the sum term can be either in complemented form or
uncomplemented form.
● Eg. of sum term- A+B+C
̅

Minterm and Maxterm
Minterm -
● The product term containing all variables of the function in either in
complemented form or uncomplemented form is called minterm.
● A n-variable function has 2n
possible combinations.
● A 2-variable function has four possible combinations, A
̅ B
̅ , A
̅ B, AB
̅ , AB. These
product terms are called minterms or standard products.
● In the minterm, variables appears either in uncomplemented form, if it
contains the value ‘1’, or in complemented form , if it contains the value ‘0’.
● A 3-variable function has eight possible combinations. The minterms of
3-variable function can be represented by m0
, m1
, m2
, m3
, m4
, m5
, m6
, m7
● The suﬃx indicates the decimal code corresponding the minterm
condictins.

The minterm Table
A B C Minterm
0 0 0 A
̅ B
̅ C
̅
0 0 1 A
̅ B
̅ C
0 1 0 A
̅ BC
̅
0 1 1 A
̅ BC
1 0 0 AB
̅ C
̅
1 0 1 AB
̅ C
1 1 0 ABC
̅
1 1 1 ABC

Property of Minterm -
● It possesses the value 1 for only one combination of K input variables.
● For K variable function of the 2k
minterms, only one minterm will have
the value ‘1’, while the remaining 2k
-1 minterms will possess the value ‘0’.
● As shown in the truth table,
For input combination 010, i.e., for A=0, B=1, C=0, only the minterm A
̅ BC
̅ will
have value ‘1’, while the remaining seven minterms will have the value ‘0’.

Maxterm -
● The sum term containing all variables of the function in either in
complemented form or uncomplemented form is called maxterm.
● A n-variable function has 2n
possible combinations.
● A 2-variable function has four possible combinations, A+B, A+B
̅ , A
̅ +B, A
̅ +B
̅ .
These sum terms are called maxterms or standard sums.
● In the maxterm, variables appears either in uncomplemented form, if it
contains the value ’0’, or in complemented form , if it contains the value
‘1’.
● A 3-variable function has eight possible combinations. The minterms of
3-variable function can be represented by M0
, M1
, M2
, M3
, M4
, M5
, M6
, M7
● The suﬃx indicates the decimal code corresponding the minterm
condictins.

The Maxterm Table
A B C Minterm
0 0 0 A+B+C
0 0 1 A+B+C
̅
0 1 0 A+B
̅ +C
0 1 1 A+B
̅ +C
̅
1 0 0 A
̅ +B+C
1 0 1 A
̅ +B+C
̅
1 1 0 A
̅ +B
̅ +C
1 1 1 A
̅ +B
̅ +C
̅

Property of Maxterm -
● It possesses the value 0 for only one combination of K input variables.
● For K variable function of the 2k
maxterms, only one maxterm will have the value ‘0’,
while the remaining 2k
-1 maxterms will possess the value ‘1’.
● As shown in the truth table,
For input combination 101, i.e., for A=1, B=0, C=1, only the maxterm A
̅ + B
̅ +C
̅ will have value
‘0’, while the remaining seven maxterms will have the value ‘1’.
● Each maxterm is the complement of the corresponding minterm. For eg. for 2 variable
function, if the maxterm is (A+B), then its corresponding minterm is its complement i.e.
A
̅ B
̅ .
Similarly for 3 variable function , if the maxterm is (A+B+C), then its corresponding
minterm is its complement i.e. A
̅ B
̅ C
̅ .
Note: If there are ‘n’ Boolean variables, then there will be 2n
min terms and 2n
Max terms.

Representation of minterms and maxterms for 2 variable
For 2-variable function, four possible combinations.
A B Minterms Maxterms
0 0 m0
=A
̅ B
̅ M1
=A+B
0 1 m1
=A
̅ B M2
=A+B
̅
1 0 m2
=AB
̅ M3
=A
̅ +B
1 1 m3
=AB M4
=A
̅ +B
̅
If the binary variable is ‘0’, then it is represented as complement of variable in
minterm and as the variable itself in Max term. Similarly, if the binary variable is
‘1’, then it is represented as complement of variable in Max term and as the
variable itself in minterm.

Canonical Sum of Product Expression
Deﬁnation: It is deﬁned as the logical sum of all the minterms derived from the truth table, for
which the value of the function is ‘1’.
The canonical sum of product expression can be given by listing the decimal codes in
correspondence with the minterms containing the function value of ‘1’.
For eg. :If the canonical SOP form of 3-variable logic function Y has three minters A
̅ B
̅ C
̅ , A B
̅ C
̅ ,
ABC
̅ . This can be expressed as the sum of decimal codes corresponding to these minterms as
explained below:
Y = ∑m
(0,5,6)
= m0
+ m5
+ m6
= A
̅ B
̅ C
̅ + A B
̅ C
̅ + ABC
̅

Canonical Product of Sum Expression
Deﬁnation: It is deﬁned as the logical product of all the maxterms derived from the truth
table, for which the value of the function is ‘0’.
The canonical product of sum expression can be given by listing the decimal codes in
correspondence with the maxterms containing the function value of ‘0’.
For eg. :If the canonical SOP form of 3-variable logic function Y has three minters (A+B+C),
(A+B
̅ +C), (A
̅ +B+C), (A
̅ +B
̅ +C
̅ ). This can be expressed as the product of decimal codes
corresponding to these maxterms as explained below:
Y = 𝚷(0, 2, 4, 7)
= M0
. M2
.M4
. M7
= (A+B+C) . (A+B
̅ +C) . (A
̅ +B+C) . (A
̅ +B
̅ +C
̅ )

Deriving SOP Expression from Truth Table
The SOP expression for a boolean function can be derived from its truth table by
summing (OR function) the product terms that correspond to the combinations
containing a function value (output Y) ‘1’.
For 3 input function Y, here the Y value is ‘1’ for the input combination 010, 011, 101 and
111. Their corresponding product term are A
̅ BC
̅ , A
̅ BC, AB
̅ C, ABC respectively.
The SOP expression for the output Y is obtained by summing the four product terms.
The Product terms are A
̅ BC
̅ , A
̅ BC, AB
̅ C, ABC. So by summing these product terms we will
get the SOP.
Y= A
̅ BC
̅ + A
̅ BC + AB
̅ C + ABC
Note: The product terms are obtained by AND operated. All the product terms are then
OR operated together in order to produce the ﬁnal SOP expression of the output

Inputs Output
Y
Product terms
(Minterm)
Sum terms
(Maxterm)
A B C
0 0 0 0 A+B+C
0 0 1 0 A+B+C
̅
0 1 0 1 A
̅ BC
̅
0 1 1 1 A
̅ BC
1 0 0 0 A
̅ +B+C
1 0 1 1 AB
̅ C
1 1 0 0 A
̅ +B
̅ +C
1 1 1 1 ABC
Truth Table of Product term and Sum term

Deriving POS Expression from Truth Table
The POS expression for a boolean function can be derived from its truth table by AND
operation of the sum terms that correspond to the combinations containing a function
value (output Y) ‘0’.
For 3 input function Y, here the Y value is ‘0’ for the input combination 000, 001, 100 and
110. Their corresponding product term are A+B+C, A+B+C
̅ , A
̅ +B+C, A
̅ +B
̅ +C respectively.
The POS expression for the output Y is obtained by AND operation of the four sum
terms.
The Product terms are A+B+C, A+B+C
̅ , A
̅ +B+C, A
̅ +B
̅ +C . So by AND operation to these
sum terms we will get the POS.
Y= (A+B+C) (A+B+C
̅ ) (A
̅ +B+C) (A
̅ +B
̅ +C)
Note: The sum terms are obtained by OR operated. All the sum terms are then AND
operated together in order to produce the ﬁnal POS expression of the output.

Karnaugh Map (K-Map)
The simplification of the functions using boolean laws and theorems becomes
complex with the increase in the number of variables and terms. The
Karnaugh map technique provides a systematic method for simplifying
expressions. K-map can take two forms Sum of Product (SOP) and Product of
Sum (POS). K-map is a modified form of truth table. K-map is represented
with grids filled with ‘0’ and ‘1’.
Note: In n-variable K-map, there are 2n
cells. Each cells corresponds to one
combination of n-variables. Therefore, for each row of the truth, i.e. for each
minterm and for each maxterm, there is one specific cell in the K-map.

The K-map for 2 variable shown below. A is the MSB and B is the LSB. The
decimal codes corresponding to the combination of variables are given inside
the cells.
0 1
2 3
A
B
0
1
0
1
A B Decimal
code
0 0 0
0 1 1
1 0 2
1 1 3
Note: The decimal value in the ﬁrst cell is 0 as for combination of 2 variables A B (0 0)
represents 0. Similarly, decimal value in the second cell is 1 as for combination of 2 variables
A B (0 1) represents 1 and so on.
00 01
10 11

The K-map for 3 variable shown below. A is the MSB and C is the LSB. The
the cells.
0 1 3 2
4 5 7 6
A B C Decimal
Code
0 0 0 0
0 0 1 1
0 1 0 2
0 1 1 3
1 0 0 4
1 0 1 5
1 1 0 6
1 1 1 7
A
BC
00 01 11 10
0
1
000
111
001
101
011
100 110
010
Note: Rule of adjacency is followed to build the K-map
grid. According to rule of adjacency, in adjacent cells only
one variable will change.
For eg. Let us consider the second cell in the ﬁrst row, the
decimal value is 1 and binary value is 001, so according to
rule of adjacency, the decimal value in third cell in the
ﬁrst row will be 3 (011) i.e. only one variable is changing
i.e. variable B.

The K-map for 4 variable shown below. A is the MSB and D is the LSB. The
the cells.
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
CD
AB 00
00
01 11 10
01
11
10
0000 0011
1100
0101
0001 0010
0100
1110
0110
0111
1000
1111
1010
1011
1001
1101
Note: Rule of adjacency is followed to build the K-map
grid. According to rule of adjacency, in adjacent cells only
one variable will change.
For eg. Let us consider the second cell in the first row,
the decimal value is 1 and binary value is 0001, so
according to rule of adjacency, the decimal value in third
cell in the first row will be 3 (0011) i.e. only one variable
is changing i.e. variable B.
Similarly, in the second row first cell, the decimal value is
4 (0100), the decimal value in the third row first cell will
be 12 (1100) i.e. only one variable is changing i.e.
variable A.
Dec A B C D
4 0 1 0 0
12 1 1 0 0

Steps to solve expression using K-map
1. Select K-map according to the number of variables.
2. Identify minterms or maxterms as given in problem.
3. For SOP put 1’s in blocks of K-map respective to the minterms (0’s elsewhere).
4. For POS put 0’s in blocks of K-map respective to the maxterms(1’s elsewhere).
5. Make rectangular groups containing total terms in power of two like 2,4,8 ..(except 1) and try to cover as
many elements as you can in one group.
6. From the groups made in step 5 find the product terms and sum them up for SOP form.

K-map for 3 variable using SOP form
0 1 3 2
4 5 7 6
A B C Y
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
A
BC
00 01 11 10
1
0 0
0
1
0
1
1
0
1
Group of two elements
No need to take this
group as those 1’s
has already been
covered
From 1st group of two 1’s (brown border), ﬁrst product term = A
̅ C
From 2nd group of two 1’s (orange border), second product term = AB
Summing of these product term is the ﬁnal expression
Y=A
̅ C + AB

Explanation:
From 1st group of two 1’s (brown border), ﬁrst product term = A
̅ C
We have
A B C
0 0 1
0 1 1
A
̅ C
From 2nd group of two 1’s (orange border), second product term = AB
A B C
1 1 0
1 1 1
A B
Summing of these product term is the ﬁnal expression
Y=A
̅ C + AB

K-map for 4 variable using SOP form
Y(A,B,C,D)= ∑(0, 2, 5, 7, 8, 10, 13, 15)
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
AB
CD
00 01 11 10
00
01
11
10
1
1
1 1
1
1 1
1
From 1st group of four 1’s (brown
border), ﬁrst product term = BD
From 2nd group of four 1’s (orange
border), second product term = B
̅ D
̅
Summing of these product term is
the ﬁnal expression
Y= BD + B
̅ D
̅

Explanation:
From 1st group of four 1’s (brown border),
ﬁrst product term =BD
We have
A B C D
0 1 0 1
0 1 1 1
1 1 0 1
1 1 1 1
B D
Summing of these product term is the ﬁnal
expression
Y=BD + B
̅ D
̅
From 2nd group of four 1’s (orange border),
second product term = B
̅ D
̅
We have
A B C D
0 0 0 0
0 0 1 0
1 0 0 0
1 0 1 0
B
̅ D
̅

Digital electronics lesson 2 part 2

More Related Content

What's hot (12)

Similar to Digital electronics lesson 2 part 2 (20)

Recently uploaded (20)

Digital electronics lesson 2 part 2