Engineering 2nd Digital design week 5.ppt

Digital Logic Design
KARNAUGH MAPS &
SIMPLIFICATION
USING KARNAUGH MAPS
By: Engr.Dr. M. Afzal

2
Introduction
• Gate-level minimization refers to the design
task of finding an optimal gate-level
implementation of Boolean functions
describing a digital circuit.

Chapter 2 - Part 2 4
Boolean Function Optimization
• Minimizing the gate input (or literal) cost of a (a set
of) Boolean equation(s) reduces circuit cost.
• We choose gate input cost.
• Boolean Algebra and graphical techniques are tools
to minimize cost criteria values.
• Some important questions:
– When do we stop trying to reduce the cost?
– Do we know when we have a minimum cost?
• Treat optimum or near-optimum cost functions
for two-level (SOP and POS) circuits first.
• Introduce a graphical technique using Karnaugh
maps (K-maps, for short)

April 10, 2025 5
The Map Method
• The complexity of the digital logic gates
– The complexity of the algebraic expression
• Logic minimization
– Algebraic approaches: lack specific rules
– The Karnaugh map
• A simple straight forward procedure
• A pictorial form of a truth table
• Applicable if the # of variables < 7
• A diagram made up of squares
– Each square represents one minterm

April 10, 2025 6
Review of Boolean Function
• Boolean function
– Sum of minterms
– Sum of products (or product of sum) in the
simplest form
– A minimum number of terms
– A minimum number of literals
– The simplified expression may not be unique

KARNAUGH MAPS
• A Karnaugh map provides a systematic method for
simplifying Boolean expressions and, if properly
used, will produce the simplest S O P or P O S
expression possible, known as the minimum
expression
• A Karnaugh map is similar to a truth table because
it presents all of the possible values of input
variables and the resulting output for each value.

Introduction to
KARNAUGH MAPS
(2,3&4 variable MAPS)

KARNAUGH MAPS
• Karnaugh maps an array of cells in which each cell
represents a binary value of the input variables.
• The cells are managed in a way so that
simplification of a given expression is simply a
matter of properly grouping the cells.
• Karnaugh map scan be used for expressions with
two, three, four. and five variables, but we will
discuss only 2,3-variable and 4-variable situations to
illustrate the principles.

Karnaugh Maps (K-map)
• A K-map is a collection of squares
– Each square represents a minterm
– The collection of squares is a graphical representation of
a Boolean function
– Adjacent squares differ in the value of one variable
– Alternative algebraic expressions for the same function
are derived by recognizing patterns of squares
• The K-map can be viewed as
– A reorganized version of the truth table
– A topologically-warped Venn diagram as used to
visualize sets in algebra of sets

Two variable KARNAUGH MAPS
• A two variable has four minterms, hence it
has 4 squares one for each term, as shown
below:

Two variable KARNAUGH MAPS: EXAMPLE
•Example: map the following functions into two variable
k-map,
1. F(x,y)=x.y
2. F(x,y)=x + y= x' y + xy' + xy = m1 + m2 + m3
Solution:

K-Map Function Representation
• Example: F(x,y) = x
• For function F(x,y), the two adjacent cells containing
1’s can be combined using the Minimization
Theorem:
F = x y = 0 y = 1
x = 0 0 0
x = 1 1 1
x
y
x
y
x
)
y
,
x
(
F =
+
=

April 10, 2025 17
Two-Variable Map
• A two-variable map
– Four minterms
– x' = row 0; x = row 1
– y' = column 0; y =
column 1
– A truth table in
square diagram
– Fig. 3.2(a): xy = m3
– Fig. 3.2(b): x+y =
x'y+xy' +xy =
m1+m2+m3
Figure 3.2 Representation of functions in the map
Figure 3.1 Two-variable Map

THREE VARIABLE KARNAUGH MAPS
• LET A, B, C ARE THREE BOOLEAN
VARIABLES

April 10, 2025 19
A Three-variable Map
• A three-variable map
– Eight minterms
– The Gray code sequence
– Any two adjacent squares in the map differ by only
one variable
• Primed in one square and unprimed in the other
• e.g., m5 and m7 can be simplified
• m5+ m7 = xy'z + xyz = xz (y'+y) = xz
Figure 3.3 Three-variable Map

April 10, 2025 20
A Three-variable Map
– m0 and m2 (m4 and m6) are adjacent
– m0+ m2 = x'y'z' + x'yz' = x'z' (y'+y) = x'z'
– m4+ m6 = xy'z' + xyz' = xz' (y'+y) = xz'

April 10, 2025 21
Example 3.1
• Example 3.1: simplify the Boolean function
F(x, y, z) = S(2, 3, 4, 5)
– F(x, y, z) = S(2, 3, 4, 5) = x'y + xy'
Figure 3.4 Map for Example 3.1, F(x, y, z) = Σ(2, 3, 4, 5) = x'y + xy'

22
Example 3.2
• Example 3.2: simplify F(x, y, z) = S(3, 4, 6, 7)
– F(x, y, z) = S(3, 4, 6, 7) = yz+ xz'
Figure 3.5 Map for Example 3-2; F(x, y, z) = Σ(3, 4, 6, 7) = yz + xz'

Combining Squares
• By combining squares, we reduce number of literals
in a product term, reducing the literal cost, thereby
reducing the other two cost criteria
• On a 3-variable K-Map:
– One square represents a minterm with three variables
– Two adjacent squares represent a product term with
two variables
– Four “adjacent” terms represent a product term with
one variable
– Eight “adjacent” terms is the function of all ones (no
variables) = 1.

April 10, 2025 24
Four adjacent Squares
• Consider four adjacent squares
– 2, 4, and 8 squares
– m0+m2+m4+m6 = x'y'z'+x'yz'+xy'z'+xyz' = x'z'(y'+y)
+xz'(y'+y) = x'z' + xz‘ = z'
– m1+m3+m5+m7 = x'y'z+x'yz+xy'z+xyz =x'z(y'+y) +
xz(y'+y) =x'z + xz = z
Figure 3.3 Three-variable Map

April 10, 2025 25
Example 3.3
 Example 3.3: simplify F(x, y, z) = S(0, 2, 4, 5, 6)
– F(x, y, z) = S(0, 2, 4, 5, 6) = z'+ xy'
Figure 3.6 Map for Example 3-3, F(x, y, z) = Σ(0, 2, 4, 5, 6) = z' +xy'

April 10, 2025 26
Example 3.4
• Example 3.4: let F = A'C + A'B + AB'C + BC
a) Express it in sum of minterms.
b) Find the minimal sum of products expression.
Ans: F(A, B, C) = S(1, 2, 3, 5, 7) = C + A'B
Figure 3.7 Map for Example 3.4, A'C + A'B + AB'C + BC = C + A'B

Three Variable Maps
z)
y,
F(x, =
y
1
1
x
z
1 1
1
z
z
y
x
+
y
x
 K-Maps can be used to simplify Boolean functions by
systematic methods. Terms are selected to cover the
“1s”in the map.
 Example: Simplify )
(1,2,3,5,7
z)
y,
F(x, m



Three-Variable Map Simplification
• Use a K-map to find an optimum SOP
equation for ,7)
(0,1,2,4,6
Z)
Y,
F(X, m



THREE VARIABLE KARNAUGH MAPS: EXAMPLE
• F(A,B,C)=∑(5,7)=m5+m7= (A.B’.C) + (A.B.C )
• F(A,B,C)=∑(5,7)=m5+m7= A.C by k-map

FOUR VARIABLE KARNAUGH MAPS: EXAMPLE
• F(A,B,C,D)=A'B'CD+A'BC'D'+ABC'D+ABCD
+ABC'D’+A'B'C'D+AB' CD'

April 10, 2025 32
3.3 Four-Variable Map
• The map
– 16 minterms
– Combinations of 2, 4, 8, and 16 adjacent squares
Figure 3.8 Four-variable Map

April 10, 2025 33
Example 3.5
• Example 3.5: simplify F(w, x, y, z) = S(0, 1, 2, 4,
5, 6, 8, 9, 12, 13, 14)
F = y'+w'z'+xz'
Figure 3.9 Map for Example 3-5; F(w, x, y, z) = Σ(0, 1, 2, 4, 5, 6, 8, 9, 12, 13, 14) = y' + w' z' +xz'

April 10, 2025 34
Example 3.6
• Example 3-6: simplify F = ABC + BCD +
ABCD + ABC
Figure 3.9 Map for Example 3-6; ABC + BCD + ABCD + ABC= BD + BC
+ACD

Four Variable Terms
 Four variable maps can have rectangles
corresponding to:
• A single 1 = 4 variables, (i.e. Minterm)
• Two 1s = 3 variables,
• Four 1s = 2 variables
• Eight 1s = 1 variable,
• Sixteen 1s = zero variables (i.e.
Constant "1")

Four-Variable Maps
• Example Shapes of Rectangles:
8 9 10
11
12 13 14
15
0 1 3 2
5 6
4 7
X
Y
Z
W

Four-Variable Maps
• Example Shapes of Rectangles:
X
Y
Z
8 9 10
11
12 13 14
15
0 1 3 2
5 6
4 7
W

MAP FOLLOWING
• F(A, B, C) = Σ(0,2,6,7)
• F(A, B, C) = Σ(0,2,3,4,6)
• F(A, B, C, D) = Σ(1,3,9,13,11)

SIMPLIFICATION
USING
KARNAUGH MAPS

KARNAUGH MAPS
• Steps for Grouping and Simplification:
• A group must contain either 1, 2, 4, 8, or 16 cells, which
are all powers of two. In the case of a 3-variable map,
23= 8 cells is the maximum 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 1’s in a
group in accordance with rule 1.
• Each 1 on the map must be included in at least one
group. The Is already in a group can be included in
another group as long as the overlapping groups include
noncommon1’s.

EXAMPLE: SIMPLIFICATION
• F(x, y, z) = ∑(0, 2, 4, 5, 6)
• F = Z’+ X.Y’

KARNAUGH MAPS
• Determining the minimum term for each group
• FOR A 3-VARIABLE MAP.
• A 1-cell group yields a 3-variable product term
• A 4-cell group yields a 1-variable term
• An 8-cell group yields a value of 1 for the
expression

KARNAUGH MAPS
• Determining the minimum term for each group
• FOR A 4-VARIABLE MAP
• An 8-cell group yields a 1-variable term
• A 16-cell group yields a value of 1 for the
expression.

KARNAUGH MAPS EXAMPLE
• Simplify the following SOP expression,
F(x, y, z) = Σ(0,2,6,7)  1
F(x, y, z) = Σ(0,2,3,4,6)  2
F(x, y, z) = Σ(0,1,2,3,7)  3
F(x, y, z) = Σ(3,5,6,7)  4
F(x, y, z) = Σ(0,1,5,7)  5
F(x, y, z) = Σ(0,1,6,7)  6
F(x, y, z) = Σ(1,2,3,6,7)  7

Solution
 F(x, y, z) = Σ(0,2,6,7)  1
F = x y + x’ z'

 Solution
F(x, y, z) = Σ(0,2,3,4,6) 
2
F = z' + x'y

 Solution
F(x, y, z) = Σ(0,1,2,3,7)  3
F = x' + y z

Solution
 Solution
F(x, y, z) = Σ(3,5,6,7) 
4
F = xy + xz + yz

Solution of 5, 6, 7
5. F = x‘ y' + x z
7. F = y + x‘ z
6. F = x‘ y' + x y

53
Example
F (A,B,C,D)= A’B’C’ + B’CD’ + A’BCD’ + AB’C’
= B’.D’ B’.C’
+ A’.C.D’
+

ATTEMPT FOLLOWING
Simplify the following SOP Expression
• F (A,B,C,D) = Σ(1,2,3,5,7,9,10,11,13,15)
• F (A,B,C,D) = Σ(1,2,3,5,9,10,11,12,13)
• F (A,B,C,D) = Σ(0,2,3,5,7,8,10,11,14,15)
• F (A,B,C,D) = Σ(2,3,7,10,11,12,13,14,15)
• F (A,B,C,D) = Σ (1, 3, 5, 9, 12, 13, 14)
• F (A,B,C,D) = Σ (0, 2, 4,5,6,7, 8, 10, 13, 15)

Engineering 2nd Digital design week 5.ppt

Engineering 2nd Digital design week 5.ppt

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