Class Notes
(809,204)

Canada
(493,579)

McMaster University
(38,382)

Computer Engineering
(55)

COMPENG 2DI4
(46)

Xudong Zhu
(4)

Lecture

# application of Boolean algebra

Unlock Document

McMaster University

Computer Engineering

COMPENG 2DI4

Xudong Zhu

Fall

Description

APPLICATIONS OF BOOLEAN ALGEBRA
MINTERM AND MAXTERM EXPANSIONS – Module 4
MULTIPLEXERS AND DECODERS – Module 9
A. Combinational Logic Design Using a Truth Table
B. Minterm and Maxterm Expansions
C. General Minterm and Maxterm Expansions
D. Incompletely Specified Functions
Click the mouse to move to the next page.
Use the ESC key to exit this chapter. Conversion of English Sentences
to Boolean Equations
The three main steps in designing a single-output combinational
switching circuit are
1. Find a switching function that specifies the desired behavior of
the circuit.
2. Find a simplified algebraic expression for the function.
3. Realize the simplified function using available logic elements. Combinational Logic Design using a
Truth Table
Suppose we want the output of a circuit to be
f = 1 if N ≥ 021 and f = 0 if N <2011 . Then
the truth table is: Next, we will derive an algebraic expression for f from the
truth table by using the combinations of values of A, B, and C
for which f = 1. For example, the term A′BC is 1 only if A = 0,
B = 1, and C = 1. Finding all terms such that f = 1 and ORing
them together yields:
f = A′BC + AB′C′ + AB′C + ABC′ + ABC (4-1) The equation can be simplified by first combining terms
and then eliminating A′:
f = A′BC + AB′ + AB = A′BC + A = A + BC (4-2)
This equation leads directly to the following circuit: Instead of writing f in terms of the 1’s of the function, we
may also write f in terms of the 0’s of the function. Observe
that the term A + B + C is 0 only if A = B = C = 0. ANDing
all of these ‘0’ terms together yields:
f = (A + B + C)(A + B + C′)(A + B′ + C) (4-3) By combining terms and using the second distributive law, we
can simplify the equation:
f = (A + B + C)(A + B + C′)(A + B′ + C) (4-3)
f = (A + B)(A + B′ + C) = A + B(B′ + C) = A + BC (4-4) Minterm and Maxterm
Expansions
f = A′BC + AB′C′ + AB′C + ABC′ + ABC(4-1)
Each of the terms in Equation (4-1) is referred to as a
minterm. In general, 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.
(A literal is a variable or its complement)
Section 4.3 (p. 93) Table 4-1 Minterms and
Maxterms for Three
Variables Minterm expansion for a function is unique. Equation (4-1)
can be rewritten in terms of m-notation as:
f = A′BC + AB′C′ + AB′C + ABC′ + ABC(4-1)
f (A, B, C) = m 3 m + 4 + m5+ m 6 7 (4-5)
This can be further abbreviated by listing only the decimal
subscripts in the form:
f (A, B, C) = Ʃ m(3, 4, 5, 6, 7) (4-5) Minterm Expansion
Find the minterm expansion of f(a,b,c,d) = a'(b' + d) + acd'.
Section 4.3 (p. 95) Design a comparator
• Design a circuit returns 1 if number (a1
a0)2 >= (b1 b0)2.
• We assume that the numbers are > 0
• Build the truth table of the comparator
• Write the minterm expansion formula
• Write the maxterm expansion formula Table 4-2. General Truth Table
for Three Variables
Table 4-2 represents a
truth table for a general
function of three
variables. Each aiis a
constant with a value of
0 or 1. General Minterm and
Maxterm Expansions
of three variables as follows:sion for a general function
The maxterm expansion for a general function of three
variables is:
Section 4.4 (p. 97) Conversion of Forms QUIZ
• Let’s assume that f( A, B, C) = ∑ m(0, 1, 2,
3); g( A, B, C) = ∏M(0, 1, 6, 7). What is
the minimal form of h = f g
A)A’B
B)A’ + B’
C)AB + A’B’
D)B’C’
E)None of the above QUIZ
• Let’s assume that f( A, B, C) = ∑ m(0, 3, 4,

More
Less
Related notes for COMPENG 2DI4