Design a Single-Output Combinational Switching Circuit Minterm ...

Chapter 4
Applications of Boolean Algebra
Minterm and Maxterm
Expansions
Xiaojun Qi
Conversion of English Sentences
to Boolean Equations
• For simple problems, it may be possible to
go directly from a word description of the
desired behavior of the circuit to an
algebraic expression for the output function.
– Break down each sentence into phrases and
associate a Boolean variable with each phrase.
– Define a two-valued variable to indicate the
truth or falsity of each phrase.
– Construct the Boolean expression by using the
defined Boolean variable, which is associated
with a phrase.
• Example 2: You will get a good mark if and only if you
understand the material in the class, do exercises and
read the problems of the exams carefully.
• Step 1: Break down the sentence into phrases and
associate a Boolean variable with each phrase.
– G: you will get good mark
– M: you understand the material in the class
– E: you do exercises
– R: you read questions carefully
• Step 2: Define a two-valued variable to indicate the truth
or falsity of each phrase.
– G=1 if you will get good mark is true; otherwise G=0
– M=1 if you understand the material in the class is true;
otherwise M=0
– E=1 if you do exercises is true; otherwise E=0
– R=1 if you read questions carefully is true; otherwise R=0.
• Step 3: Construct the Boolean expression.
G = MER
Design a Single-Output
Combinational Switching Circuit
• Three main steps in designing a singleoutput
combinational switching circuit are:
– Find a switching function which specifies the
desired behavior of the circuit.
– Find a simplified algebraic expression for the
function.
– Realize the simplified function using available
logic elements.
• Example 1: There will be a party if and only if you have
time and money, or you have a very good friend come to
visit you.
• Step 1: Break down the sentence into phrases and
associate a Boolean variable with each phrase.
– T: you have time
– M: you have money
– V: you have a very good friend come to visit you
– P: there will be a party
• Step 2: Define a two-valued variable to indicate the truth
or falsity of each phrase.
– T=1 if you have time is true; otherwise, T=0
– M=1 if you have money is true; otherwise, M=0
– V=1 if you have a very good friend come to visit you is true;
otherwise V=0
– P=1 if there will be a party is true; otherwise P=0
• Step 3: Construct the Boolean expression.
P = TM+V
Minterm Expansion
• Minterm: A minterm of n variables is a
product of n literals in which each
variable appears exactly once in either
true or complemented form, but not both
• Based on the truth table, the minterm
expansion can be obtained by:
– 1. Generate a product term corresponding to
each row in which the value of the function is 1.
– 2. In each product term, if the value is 1, then
use the variable, if is 0, then use its
complement.
• Minterm expansion also is referred to as
a canonical sum of products (SOP)
or standard sum of products (SOP).

Maxterm Expansion
• Maxterm: A maxterm of n variables is a
sum of n literals in which each variable
appears exactly once in either true or
complemented form, but not both
• Based on the truth table, the maxterm
expansion can be obtained by:
– 1. Generate a sum term corresponding to each
row in which the value of the function is 0.
– 2. In each sum term, if the value is 1, then use
its complement, if is 0, then use the variable
itself.
• Maxterm expansion also is referred to as
a canonical product of sums (POS) or
standard product of sums (POS).
Combinational Logic Design
Using a Truth Table
Example 1:
• The canonical SOP form F=A’B’+A’B
• The canonical POS form
F = (A’+B)(A’+B’)
A B F
Row 0 0 0 1
Row 1 0 1 1
Row 2 1 0 0
Row 3 1 1 0
• Simplify the Minterm Expansion:
F(A,B,C) = A’BC+AB’C’+AB’C+ABC’+ABC
=A’BC + AB’ + ABC’ + ABC
=A’BC + AB’ + AB
=A’BC + A
=A + BC
• The circuit based on this simplified
expression is:
Minterms and Maxterms for
Three Variables
Combinational Logic Design
Using a Truth Table
Example 2: Minterm Expansion:
F(A,B,C) =
A’BC+AB’C’+AB’C+ABC’+ABC
=m3 +m4 +m5 +m6 +m7 ∑ = ∑ m(3, 4, 5, 6, 7)
Maxterm Expansion:
F(A,B,C)
=(A+B+C)(A+B+C’)(A+B’+C)
=M0M1M2 = ∏ M(0, 1, 2)
• Simplify the Maxterm Expansion:
F(A,B,C) =(A+B+C)(A+B+C’)(A+B’+C)
=(A+B)(A+B’+C)
=A+B(B’+C)
=A+BC
• This result is the same as the one
obtained from the simplification of the
maxterm expansion.

Combinational Logic Design
Using a Truth Table
Example 1: Minterm Expansion:
F’(A,B,C) = A’B’C’+A’B’C+A’BC’
=m0 +m1 +m2 = ∑ m(0, 1, 2)
∑ Maxterm Expansion:
F’(A,B,C)
=(A+B’+C’)(A’+B+C)(A’+B+C’)
(A’+B’+C)(A’+B’+C’)
=M3M4M5M6M7 = ∏ M(3, 4, 5, 6, 7)
Conversion Between Minterm
and Maxterm -- In General
(4-9)
(4-10)
Minterm Expansion (p. 89)
Conversion of Minterm and
Maxterm of the Above Example
The Method to Get
Canonical Form
Algebraic Method:
Using distribution law and adding some
terms having value 0 or multiply some
terms having value 1.
Q =AB+AC=AB(C+C’)+A(B+B’)C
=ABC+ABC’+ABC+AB’C
=ABC+ABC’+AB’C
P = (A’+B)(A+B+C)
=(A’+B+CC’)(A+B+C)
=(A’+B+C)(A’+B+C’)(A+B+C)
f = a'(b'+d) + acd'
(4-11)
Maxterm Expansion (p. 90)

F
' '
' '
= ABC D + A BCD + ABC D
1101
= ∑ ( 7,
12 , 13 )
0101
0111
1001
1100
'
' '
'
'
P = ( A+
B + C + D )( A + B + C + D )( A+
B + C + D)
0010
= ∏(
2,
5,
9)
EX:
Given
Q(
A,
B,
C,
D)
= ∑(
0,
1,
7,
8,
10,
11,
12,
15)
Note: Q is four variable function and there will be 2 4
= 16 combinations whose decimal values range from
0 to 15. It has 8 minterms and should have 16-8=8 16 8=8
maxterms
Q(
A,
B,
C,
D)
= ∏(
2,
3,
4,
5,
6,
9,
13,
14)
Incompletely Specified Function
Truth Table with Don't Cares which are indicated
by X in the output
Case 2: We do not care
whether it is 0 or 1 for certain
input combination.
F = ∑ m(
0,
3,
7)
+ ∑ d(
1,
6)
F = ∏ M ( 2,
4,
5)
• ∏(
1,
6)
It is desirable to choose values
for the don’t care terms which
will help simplify the function!
X Y Cin Cout Sum
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
Truth Table for a Full Adder
Incompletely Specified Function
First Case: Assume that the output of N1 does not
generate all possible combinations of values for A,
B, and C. This will lead to the incompletely
specified function for N2 since certain combination
of circuit inputs did not occur.
Examples of Truth Table
Construction
•Example 1
•Example 2
•Example 3
•Example 4
On Pages 94-97.
Truth Table for Binary Full Subtracter