Chapter 2 Combinational Logic Circuits · 2019-10-22 · Chapter 2 - Part 2 3 Overview Part 1...
Transcript of Chapter 2 Combinational Logic Circuits · 2019-10-22 · Chapter 2 - Part 2 3 Overview Part 1...
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Charles Kime & Thomas Kaminski
© 2008 Pearson Education, Inc.
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Chapter 2 – Combinational
Logic CircuitsPart 2 – Circuit Optimization
Logic and Computer Design Fundamentals
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Updated by Dr. Waleed Dweik
Chapter 2 - Part 2 2
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Chapter 2 - Part 2 3
Overview
▪ Part 1 – Gate Circuits and Boolean Equations
• Binary Logic and Gates
• Boolean Algebra
• Standard Forms
▪ Part 2 – Circuit Optimization
• Two-Level Optimization
• Map Manipulation
▪ Part 3 – Additional Gates and Circuits
• Other Gate Types
• Exclusive-OR Operator and Gates
• High-Impedance Outputs
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Chapter 2 - Part 2 4
Circuit Optimization
▪ Goal: To obtain the simplestimplementation for a given function
▪ Optimization is a more formal approach tosimplification that is performed using aspecific procedure or algorithm
▪ Optimization requires a cost criterion tomeasure the simplicity of a circuit
▪ Distinct cost criteria we will use:• Literal cost (L)
• Gate input cost (G)
• Gate input cost with NOTs (GN)
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Literal Cost
▪ Literal: a variable or its complement
▪ Literal cost (L): the number of literal
appearances in a Boolean expression
corresponding to the logic circuit diagram
▪ Examples:
• 𝐹 = 𝐵𝐷 + 𝐴𝐵′𝐶 + 𝐴𝐶′𝐷′▪ 𝐿 = 8 (Minimum cost → Best solution)
• 𝐹 = 𝐵𝐷 + 𝐴𝐵′𝐶 + 𝐴𝐵′𝐷′ + 𝐴𝐵𝐶′
▪ 𝐿 = 11
• 𝐹 = (𝐴 + 𝐵)(𝐴 + 𝐷)(𝐵 + 𝐶 + 𝐷′)(𝐵′ + 𝐶′ + 𝐷)▪ 𝐿 = 10
Chapter 2 - Part 2 5
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Gate Input Cost
▪ Gate input cost (G): the number of inputs to the gates in the
implementation corresponding exactly to the given equation or
equations. (G: inverters not counted, GN: inverters counted)
▪ For SOP and POS equations, it can be found from the equation(s) by
finding the sum of:
• All literal appearances
• The number of terms excluding single literal terms,(G) and
• optionally, the number of distinct complemented single literals (GN).
▪ Examples:
• 𝐹 = 𝐵𝐷 + 𝐴𝐵′𝐶 + 𝐴𝐶′𝐷′
▪ 𝐺 = 11 , 𝐺𝑁 = 14 (Minimum cost → Best solution)
• 𝐹 = 𝐵𝐷 + 𝐴𝐵′𝐶 + 𝐴𝐵′𝐷′ + 𝐴𝐵𝐶′
▪ 𝐺 = 15 , 𝐺𝑁 = 18
• 𝐹 = (𝐴 + 𝐵)(𝐴 + 𝐷)(𝐵 + 𝐶 + 𝐷′)(𝐵′ + 𝐶′ + 𝐷)
▪ 𝐺 = 14 , 𝐺𝑁 = 17
Chapter 2 - Part 2 6
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Chapter 2 - Part 2 7
▪ Example 1:
▪ F = A + B C +
Cost Criteria (continued)
A
BC
F
B CL = 5
▪ L (literal count) counts the AND inputs and the single
literal OR input.
G = L + 2 = 7
▪ G (gate input count) adds the remaining OR gate inputs
GN = G + 2 = 9
▪ GN(gate input count with NOTs) adds the inverter inputs
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Cost Criteria (continued)
▪ Example 2:
▪ 𝑭 = 𝑨,𝑩, 𝑪, 𝑫 = 𝑨𝑩𝑪 + 𝑫′ . 𝑪′
• 𝑳 = 𝟓
• 𝑮 = 𝟓 + 𝟐 = 𝟕
• 𝑮𝑵 = 𝟕 + 𝟐 = 𝟗
Chapter 2 - Part 2 8
D
AB
C
F
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Chapter 2 - Part 2 9
▪ Example 3:
▪ F = A B C +
▪ L = 6, G = 8, GN = 11
▪ F = (A + )( + C)( + B)
▪ L = 6 , G = 9, GN = 12
▪ Same function and same
literal cost
▪ But first circuit has better
gate input count and better
gate input count with NOTs
▪ Select it!
Cost Criteria (continued)
B C
A
A
B
C
F
C B
F
A
B
C
A
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Chapter 2 - Part 2 10
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
▪ Introduce a graphical technique using Karnaugh maps
(K-maps, for short)
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Chapter 2 - Part 2 11
Karnaugh Maps (K-map)
▪ A K-map is a collection of squares
• Graphical representation of the truth table
• Each square represents a minterm, or a maxterm, or a rowin the truth table
• For n-variable, there are 2n squares
• The collection of squares is a graphical representation of aBoolean function
• Adjacent squares differ in the value of one variable
• Alternative algebraic expressions for the same function arederived by recognizing patterns of squares
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Chapter 2 - Part 2 12
Some Uses of K-Maps
▪ Finding optimum or near optimum
• SOP and POS standard forms, and
• two-level AND/OR and OR/AND circuit implementations
for functions with small numbers of variables
▪ Visualizing concepts related tomanipulating Boolean expressions, and
▪ Demonstrating concepts used by computer-aided design programs to simplify large circuits
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Chapter 2 - Part 2 13
Two Variable Maps
▪ A 2-variable Karnaugh Map:
• Note that minterm m0 and
minterm m1 are “adjacent”
and differ in the value of the
variable y
• Similarly, minterm m0 and
minterm m2 differ in the x variable
• Also, m1 and m3 differ in the x variable as well
• Finally, m2 and m3 differ in the value of the variable y
𝐲 = 𝟏𝐲 = 𝟎
𝑚1 = ҧ𝑥𝑦𝑚0 = ҧ𝑥 ത𝑦𝐱 = 𝟎
𝑚3 = 𝑥𝑦𝑚2 = 𝑥ത𝑦𝐱 = 𝟏
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K-Map and Truth Tables
▪ The K-Map is just a different form of the truth table
▪ Example: Two variable function
• We choose a,b,c and d from the set {0,1} to implement
a particular function, 𝐹(𝑥, 𝑦)
Chapter 2 - Part 2 14
Input Values
(𝒙, 𝒚)𝐅(𝐱, 𝐲)
0 0 a
0 1 b
1 0 c
1 1 d
𝐲 = 𝟏𝐲 = 𝟎
𝒃𝒂𝐱 = 𝟎
𝒅𝒄𝐱 = 𝟏
Truth Table K-Map
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Chapter 2 - Part 2 15
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 = 𝑥ത𝑦 + 𝑥𝑦 = 𝑥
𝐲 = 𝟏𝐲 = 𝟎𝑭 𝒙, 𝒚 = 𝒙
𝟎𝟎𝐱 = 𝟎
𝟏𝟏𝐱 = 𝟏
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Chapter 2 - Part 2 16
K-Map Function Representation
▪ Example: 𝐺 𝑥, 𝑦 = 𝑥 + 𝑦
▪ For 𝐺(𝑥, 𝑦) , two pairs of adjacent cells
containing 1’s can be combined using the
Minimization Theorem:
𝐺 𝑥, 𝑦 = 𝑥ത𝑦 + 𝑥𝑦 + ҧ𝑥𝑦 + 𝑥𝑦
𝐺 𝑥, 𝑦 = 𝑥 + 𝑦
𝐲 = 𝟏𝐲 = 𝟎𝑮 𝒙, 𝒚 = 𝒙 + 𝒚
𝟏𝟎𝐱 = 𝟎
𝟏𝟏𝐱 = 𝟏
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Chapter 2 - Part 2 17
Three Variable Maps
▪ A three-variable K-map:
▪ Where each minterm corresponds to the product terms:
▪ Note that if the binary value for an index differs in onebit position, the minterms are adjacent on the K-Map
𝐲𝐳 = 𝟏𝟎𝐲𝐳 = 𝟏𝟏𝐲𝐳 = 𝟎𝟏𝐲𝐳 = 𝟎𝟎
𝑚2𝑚3𝑚1𝑚0𝐱 = 𝟎
𝑚6𝑚7𝑚5𝑚4𝐱 = 𝟏
𝐲𝐳 = 𝟏𝟎𝐲𝐳 = 𝟏𝟏𝐲𝐳 = 𝟎𝟏𝐲𝐳 = 𝟎𝟎
ҧ𝑥𝑦 ҧ𝑧ҧ𝑥𝑦𝑧ҧ𝑥 ത𝑦𝑧ҧ𝑥 ത𝑦 ҧ𝑧𝐱 = 𝟎
𝑥𝑦 ҧ𝑧𝑥𝑦𝑧𝑥 ത𝑦𝑧𝑥 ത𝑦 ҧ𝑧𝐱 = 𝟏
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Alternative Map Labeling
▪ Map use largely involves:
• Entering values into the map, and
• Reading off product terms from the map
▪ Alternate labelings are useful:
Chapter 2 - Part 2 18
ഥ𝒀 𝒀
ഥ𝑿 0 1 3 2
𝑿 4 5 7 6
ഥ𝒁 𝒁 ഥ𝒁
YZ
X00 01 11 10
00 1 3 2
14 5 7 6
Y
Z
X
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Example Functions
▪ By convention, we represent the minterms of 𝐹 by a "1" in
the map and leave the minterms of ത𝐹 blank
▪ Example:
• 𝐹 𝑥, 𝑦, 𝑧 = σ𝑚(2,3,4,5)
▪ Example:
• 𝐺 𝑎, 𝑏, 𝑐 = σ𝑚(3,4,6,7)
▪ Learn the locations of the 8 indices based on the
variable order shown (X, most significant and Z,
least significant) on the map boundaries
Chapter 2 - Part 2 19
𝒀
0 1 3
12
1
𝑿 4
15
17 6
𝒁
𝒃
0 1 3
12
𝒂 4
15 7
16
1
𝒄
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Steps for using K-Maps to Simplify Boolean
Functions
▪ Enter the function on the K-Map
• Function can be given in truth table, shorthand notation, SOP,…etc
• Example:
▪ 𝐹 𝑥, 𝑦 = ҧ𝑥 + 𝑥𝑦
▪ 𝐹 𝑥, 𝑦 = σ𝑚(0,1,3)
▪ Combining squares for simplification
• Rectangles that include power of 2 squares {1, 2, 4, 8, …}
• Goal: Fewest rectangles that cover all 1’s → as large as possible
▪ Determine if any rectangle is not needed
▪ Read-off the SOP terms
Chapter 2 - Part 2 20
𝐱 𝐲 𝐅(𝐱, 𝐲)
0 0 1
0 1 1
1 0 0
1 1 1
𝒚
0
11
1
𝒙 2 3
1
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Chapter 2 - Part 2 21
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 2-variable K-Map:
• One square represents a minterm with two variables
• Two adjacent squares represent a product term with one variable
• Four “adjacent” terms is the function of all ones (no variables) = 1.
▪ 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.
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Example: Combining Squares
▪ Example: 𝐹 𝐴, 𝐵 = σ𝑚(0,1,2)
𝐹 𝐴, 𝐵 = ҧ𝐴 ത𝐵 + ҧ𝐴𝐵 + 𝐴 ത𝐵
▪ Using Distributive law
• 𝐹 𝐴, 𝐵 = ҧ𝐴 + 𝐴 ത𝐵
▪ Using simplification theorem
• 𝐹 𝐴, 𝐵 = ҧ𝐴 + ത𝐵
▪ Thus, every two adjacent terms that form a 2×1
rectangle correspond to a product term with
one variable
Chapter 2 - Part 2 22
𝑩
0
11
1
𝑨 2
13
0
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Example: Combining Squares
▪ Example: 𝐹 𝑥, 𝑦, 𝑧 = σ𝑚(2,3,6,7)
▪ 𝐹 𝑥, 𝑦, 𝑧 = ҧ𝑥𝑦 ҧ𝑧 + ҧ𝑥𝑦𝑧 + 𝑥𝑦 ҧ𝑧 + 𝑥𝑦𝑧
▪ Using Distributive law
• 𝐹 𝑥, 𝑦, 𝑧 = ҧ𝑥𝑦 + 𝑥𝑦
▪ Using Distributive law again
• 𝐹 𝑥, 𝑦, 𝑧 = 𝑦
▪ Thus, the four adjacent terms that form a 2×2
square correspond to the term "y"
Chapter 2 - Part 2 23
𝒚
0 1 3
12
1
𝒙 4 5 7
16
1
𝒛
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Chapter 2 - Part 2 24
Three-Variable Maps
▪ Reduced literal product terms for SOP standard
forms correspond to rectangles on K-maps
containing cell counts that are powers of 2
▪ Rectangles of 2 cells represent 2 adjacent minterms
▪ Rectangles of 4 cells represent 4 minterms that form
a “pairwise adjacent” ring
▪ Rectangles can contain non-adjacent cells as
illustrated by the “pairwise adjacent” ring above
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▪ Example shapes of 2-cell rectangles:
Chapter 2 - Part 2 25
Three-Variable Maps
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Chapter 2 - Part 2 26
Three-Variable Maps
▪ Example shapes of 2-cell rectangles:
▪ Read-off the product terms for the rectangles
shown:
• 𝑅𝑒𝑐𝑡 0,1 = ത𝑋 ത𝑌
• 𝑅𝑒𝑐𝑡 0,2 = ത𝑋 ҧ𝑍
• 𝑅𝑒𝑐𝑡 3,7 = 𝑌𝑍
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▪ Example shapes of 4-cell Rectangles:
Chapter 2 - Part 2 27
Three-Variable Maps
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Chapter 2 - Part 2 28
Three-Variable Maps
▪ Example shapes of 4-cell Rectangles:
▪ Read off the product terms for the rectangles
shown:
• 𝑅𝑒𝑐𝑡 1,3,5,7 = 𝑍
• 𝑅𝑒𝑐𝑡 0,2,4,6 = ҧ𝑍
• 𝑅𝑒𝑐𝑡 4,5,6,7 = 𝑋
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Three Variable Maps
▪ K-maps can be used to simplify Boolean functions
by systematic methods. Terms are selected to
cover the “1s”in the map.
▪ Example: Simplify 𝐹 𝑥, 𝑦, 𝑧 = σ𝑚( 1,2,3,5,7)
𝐹 𝑥, 𝑦, 𝑧 = 𝑧 + ҧ𝑥𝑦
Chapter 2 - Part 2 29
𝒚
0 1
13
12
1
𝒙 4 5
17
16
𝒛
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Chapter 2 - Part 2 30
Three-Variable Map Simplification
▪ Use a K-map to find an optimum SOP equation
for 𝐹 𝑋, 𝑌, 𝑍 = σ𝑚(0,1,2,4,6,7)
𝒀0
11
13 2
1
𝑿 4
15 7
16
1
𝒁
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Chapter 2 - Part 2 31
Three-Variable Map Simplification
▪ Use a K-map to find an optimum SOP equation
for 𝐹 𝑋, 𝑌, 𝑍 = σ𝑚(0,1,2,4,6,7)
𝐹 𝑋, 𝑌, 𝑍 = ҧ𝑍 + ത𝑋 ത𝑌 + 𝑋𝑌
𝒀0
11
13 2
1
𝑿 4
15 7
16
1
𝒁
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Four Variable Maps
▪ Map and location of minterms
𝐹(𝑊, 𝑋, 𝑌, 𝑍):
Chapter 2 - Part 2 32
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Four Variable Terms
▪ Four variable maps can have rectangles
corresponding to:
• A single 1: 4 variables (i.e. Minterm)
• Two 1’s: 3 variables
• Four 1’s: 2 variables
• Eight 1’s: 1 variable
• Sixteen 1’s: zero variables (function of all ones)
Chapter 2 - Part 2 33
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Chapter 2 - Part 2 34
Four-Variable Maps
▪ Example shapes of 4-cell rectangles:
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Chapter 2 - Part 2 35
Four-Variable Maps
▪ Example shapes of 4-cell rectangles:
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Chapter 2 - Part 2 36
Four-Variable Maps
▪ Example shapes of 8-cell rectangles:
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Chapter 2 - Part 2 37
Four-Variable Maps
▪ Example shapes of 8-cell rectangles:
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Four-Variable Map Simplification
▪ 𝐹 𝑊,𝑋, 𝑌, 𝑍 = σ𝑚( 0,2,4,5,6,7,8,10,13,15)
Chapter 2 - Part 2 38
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Four-Variable Map Simplification
▪ 𝐹 𝑊,𝑋, 𝑌, 𝑍 = σ𝑚( 0,2,4,5,6,7,8,10,13,15)
𝐹 𝑊,𝑋, 𝑌, 𝑍 = 𝑋𝑍 + ത𝑋 ҧ𝑍 + ഥ𝑊𝑋Chapter 2 - Part 2 39
𝒀0
11 3 2
14
15
17
16
1𝑿
𝑾
12 13
115
114
8
19 11 10
1
𝒁
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Chapter 2 - Part 2 40
Four-Variable Map Simplification
▪ 𝐹 𝑊,𝑋, 𝑌, 𝑍 = σ𝑚( 3,4,5,7,9,13,14,15)
𝐹 𝑊,𝑋, 𝑌, 𝑍 = ഥ𝑊𝑌𝑍 + ഥ𝑊𝑋ത𝑌 +𝑊𝑋𝑌 +𝑊ത𝑌𝑍
𝒀0 1 3 2
4 5 7 6
𝑿
𝑾
12 13 15 14
8 9 11 10
𝒁
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Chapter 2 - Part 2 41
Four-Variable Map Simplification
▪ 𝐹 𝑊,𝑋, 𝑌, 𝑍 = σ𝑚( 3,4,5,7,9,13,14,15)
𝐹 𝑊,𝑋, 𝑌, 𝑍 = ഥ𝑊𝑌𝑍 + ഥ𝑊𝑋ത𝑌 +𝑊𝑋𝑌 +𝑊ത𝑌𝑍
𝒀0 1 3
12
4
15
17
16
𝑿
𝑾
12 13
115
114
18 9
111 10
𝒁
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Systematic Simplification
▪ Prime Implicant: is a product term obtained by
combining the maximum possible number of adjacent
squares in the map into a rectangle with the number of
squares a power of 2
▪ A prime implicant is called an Essential Prime Implicant
if it is the only prime implicant that covers (includes) one
or more minterms
▪ Prime Implicants and Essential Prime Implicants can be
determined by inspection of a K-Map
▪ A set of prime implicants "covers all minterms" if, for each
minterm of the function, at least one prime implicant in the
set of prime implicants includes the minterm
Chapter 2 - Part 2 42
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Chapter 2 - Part 2 43
DB
CB
1 1
1 1
1 1
B
D
A
1 1
1 1
1
Example of Prime Implicants
▪ Find ALL Prime Implicants
ESSENTIAL Prime Implicants
C
BD
CD
BD
Minterms covered by single prime implicant
DB
1 1
1 1
1 1
B
C
D
A
1 1
1 1
1
AD
BA
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Prime Implicant Practice
▪ Find all prime implicants for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(0,2,3,8,9,10,11,12,13,14,15)
▪ Prime Implicants:
Chapter 2 - Part 2 44
𝑪0
11 3
12
1
4 5 7 6
𝑩
𝑨
12
113
115
114
1
8
19
111
110
1
𝑫
![Page 45: Chapter 2 Combinational Logic Circuits · 2019-10-22 · Chapter 2 - Part 2 3 Overview Part 1 –Gate Circuits and Boolean Equations • Binary Logic and Gates • Boolean Algebra](https://reader030.fdocuments.net/reader030/viewer/2022040211/5e72cfdc9532ee379d4b9cac/html5/thumbnails/45.jpg)
Prime Implicant Practice
▪ Find all prime implicants for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(0,2,3,8,9,10,11,12,13,14,15)
▪ Prime Implicants:
• 𝐴
• ത𝐵𝐶
• ത𝐵ഥ𝐷
Chapter 2 - Part 2 45
𝑪0
11 3
12
1
4 5 7 6
𝑩
𝑨
12
113
115
114
1
8
19
111
110
1
𝑫
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Chapter 2 - Part 2 46
Another Example
▪ Find all prime implicants for:
𝐺 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(0,2,3,4,7,12,13,14,15)
▪ Hint: There are seven prime implicants!
▪ Prime Implicants: 𝑪0
11 3
12
1
4
15 7
16
𝑩
𝑨
12
113
115
114
1
8 9 11 10
𝑫
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Chapter 2 - Part 2 47
Another Example
▪ Find all prime implicants for:
𝐺 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(0,2,3,4,7,12,13,14,15)
▪ Hint: There are seven prime implicants!
▪ Prime Implicants:
• 𝐴𝐵
• 𝐵𝐶𝐷
• 𝐵 ҧ𝐶ഥ𝐷
• ҧ𝐴𝐶𝐷
• ҧ𝐴 ҧ𝐶ഥ𝐷
• ҧ𝐴 ത𝐵𝐶
• ҧ𝐴 ത𝐵ഥ𝐷
𝑪0
11 3
12
1
4
15 7
16
𝑩
𝑨
12
113
115
114
1
8 9 11 10
𝑫
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Chapter 2 - Part 2 48
Optimization Algorithm
1. Find all prime implicants
2. Include all essential prime implicants in the
solution
3. Select a minimum cost set of non-essential
prime implicants to cover all minterms not yet
covered
• Selection Rule: Minimize the overlap among prime
implicants as much as possible. In particular, in the
final solution, make sure that each prime implicant
selected includes at least one minterm not included in
any other prime implicant selected
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Selection Rule Example
▪ Simplify F(A, B, C, D) given on the K-map
Chapter 2 - Part 2 49
𝑪0
11 3
12
4
15
17
16
1𝑩
𝑨
12 13
115 14
8
19
111 10
𝑫
Prime Implicants
𝑪0
11 3
12
4
15
17
16
1𝑩
𝑨
12 13
115 14
8
19
111 10
𝑫
Essential and Selected Non-essential Prime Implicants
Selected Non-essential Prime
Implicants
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Selection Rule Example
▪ Simplify F(A, B, C, D) given on the K-map
Chapter 2 - Part 2 50
𝑪0
11 3
12
4
15
17
16
1𝑩
𝑨
12 13
115 14
8
19
111 10
𝑫
Prime Implicants
𝑪0
11 3
12
4
15
17
16
1𝑩
𝑨
12 13
115 14
8
19
111 10
𝑫
Essential and Selected Non-essential Prime Implicants
Essential Prime
Implicants
Selected Non-essential Prime
Implicants
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Chapter 2 - Part 2 51
Product of Sums Example
▪ Find the optimum POS solution for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(1,3,9,11,12,13,14,15)
▪ Solution:
• Find optimized SOP for ത𝐹 by combining 0’s in K-Map of 𝐹
• Complement ത𝐹 to obtain optimized POS for 𝐹 𝑪0
01
13
12
0
4
05
07
06
0𝑩
𝑨
12
113
115
114
1
8
09
111
110
0
𝑫
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Chapter 2 - Part 2 52
Product of Sums Example
▪ Find the optimum POS solution for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(1,3,9,11,12,13,14,15)
▪ Solution:
• Find optimized SOP for ത𝐹 by combining 0’s in K-Map of 𝐹
• Complement ത𝐹 to obtain optimized POS for 𝐹
▪ ത𝐹 𝐴, 𝐵, 𝐶, 𝐷 = ҧ𝐴𝐵 + ത𝐵ഥ𝐷
▪ Using Demorgan’s Law:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 = (𝐴 + ത𝐵)(𝐵 + 𝐷)
𝑪0
01
13
12
0
4
05
07
06
0𝑩
𝑨
12
113
115
114
1
8
09
111
110
0
𝑫
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Example
▪ Find the optimum POS and SOP solution for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =ෑ
𝑀
( 0, 2, 4, 5, 6, 7)
▪ POS solution (Red):
• Find optimized SOP for ത𝐹 by combining 0’s in K-Map of 𝐹
• Complement ത𝐹 to obtain optimized POS for 𝐹
▪ SOP solution (Blue):
Chapter 2 - Part 2 53
𝑪0
01
13
12
0
4
05
07
06
0𝑩
𝑨
12
113
115
114
1
8
19
111
110
1
𝑫
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Example
▪ Find the optimum POS and SOP solution for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =ෑ
𝑀
( 0, 2, 4, 5, 6, 7)
▪ POS solution (Red):
• Find optimized SOP for ത𝐹 by combining 0’s in K-Map of 𝐹
• Complement ത𝐹 to obtain optimized POS for 𝐹
ത𝐹 𝐴, 𝐵, 𝐶, 𝐷 = ҧ𝐴𝐵 + ҧ𝐴ഥ𝐷𝐹 𝐴, 𝐵, 𝐶, 𝐷 = (𝐴 + ത𝐵)(𝐴 + 𝐷)
▪ SOP solution (Blue):
• Combining 1’s in K-Map of 𝐹
𝐹 𝐴, 𝐵, 𝐶, 𝐷 = 𝐴 + ത𝐵𝐷
Chapter 2 - Part 2 54
𝑪0
01
13
12
0
4
05
07
06
0𝑩
𝑨
12
113
115
114
1
8
19
111
110
1
𝑫
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Chapter 2 - Part 2 55
▪ Incompletely specified functions: Sometimes a function table or
map contains entries for which it is known:
• the input values for the minterm will never occur, or
• The output value for the minterm is not used
▪ In these cases, the output value is defined as a “don't care”
▪ By placing “don't cares” ( an “x” entry) in the function table or
map, the cost of the logic circuit may be lowered
▪ Example: A logic function having the binary codes for the
BCD digits as its inputs. Only the codes for 0 through 9 are
used. The six codes, 1010 through 1111 never occur, so the
output values for these codes are “x” to represent “don’t cares”
▪ “Don’t care” minterms cannot be replaced with 1’s or 0’s
because that would require the function to be always 1 or 0
for the associated input combination
Don't Cares in K-Maps
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Example: BCD “5 or More”
▪ The map below gives a function 𝐹(𝑤, 𝑥, 𝑦, 𝑧) which is defined as "5 or more" over
BCD inputs. With the don't cares used for the 6 non-BCD combinations:
▪ If don’t cares are treated as 1’s (Red):
▪ 𝐹1 𝑤, 𝑥, 𝑦, 𝑧 = 𝑤 + 𝑥𝑦 + 𝑥𝑧
• 𝐺 = 7
▪ If don’t cares are treated as 0’s (Blue):
▪ 𝐹2 𝑤, 𝑥, 𝑦, 𝑧 = ഥ𝑤𝑥𝑧 + ഥ𝑤𝑥𝑦 + 𝑤 ҧ𝑥 ത𝑦
• 𝐺 = 12
▪ For this particular function, cost G for the POS solution for 𝑭 𝒘, 𝒙, 𝒚, 𝒛 is
not changed by using the don't cares
• Choose the one less inverters (i.e. less GN)
Chapter 2 - Part 2 56
𝒚0 1 3 2
4 5
17
16
1𝒙
𝒘
12
X13
X15
X14
X
8
19
111
X10
X
𝒛
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Chapter 2 - Part 2 57
Selection Rule Example with Don't Cares
▪ Simplify F(A, B, C, D) given on the K-map. Selected
Minterms covered by essential prime implicants
1
1
x
x
x x
x
1
B
D
A
C
1
1 1
1
x
x
x x
x
1
B
D
A
C
1
1
Essential
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Chapter 2 - Part 2 58
Product of Sums with Don’t Care
Example
▪ Find the optimum POS solution for:
𝐹 𝐴, 𝐵, 𝐶, 𝐷 =
𝑚
(3,9,11,12,13,14,15) +
𝑑
(1,4,6)
ത𝐹 𝐴, 𝐵, 𝐶, 𝐷 = ҧ𝐴𝐵 + ത𝐵ഥ𝐷
𝐹 𝐴, 𝐵, 𝐶, 𝐷 = (𝐴 + ത𝐵)(𝐵 + 𝐷)
𝑪
0
0
1
X
3
1
2
0
4
X
5
0
7
0
6
X𝑩
𝑨
12
1
13
1
15
1
14
1
8
0
9
1
11
1
10
0
𝑫
𝑪
0
0
1
X
3
1
2
0
4
X
5
0
7
0
6
X𝑩
𝑨
12
1
13
1
15
1
14
1
8
0
9
1
11
1
10
0
𝑫
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Chapter 2 - Part 2 59
Five Variable or More K-Maps
▪ For five variable problems, we use two adjacent K-maps.
It becomes harder to visualize adjacent minterms for
selecting PIs. You can extend the problem to six variables
by using four K-Maps.
X
Y
Z
W
V = 0
X
Z
W
V = 1
Y
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Chapter 2 - Part 2 60
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