Ch. 4 Boolean Algebra and
Logic Simplification
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Boolean Operations and Expressions
Laws and Rules of Boolean Algebra
Boolean Analysis of Logic Circuits
Simplification Using Boolean Algebra
Standard Forms of Boolean Expressions
Truth Table and Karnaugh Map
Programmable Logic: PALs and GALs
Boolean Expressions with VHDL
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Introduction
2
Boolean Algebra
•
George Boole(English mathematician), 1854
“An Investigation of the Laws of Thought, on Which Are Founded
the Mathematical Theories of Logic and Probabilities”
•
•
Boolean Algebra
{(1,0), (NOT, AND, OR}
Mathematical tool to expression and analyze digital
(logic) circuits
Claude Shannon, the first to apply Boole’s work, 1938
–
•
“A Symbolic Analysis of Relay and Switching Circuits” at MIT
This chapter covers Boolean algebra, Boolean expression
and its evaluation and simplification, and VHDL program
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Basic Functions
Boolean functions : NOT, AND, OR,
exclusive OR(XOR) : odd function
exclusive NOR(XNOR) : even function(equivalence)
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Basic Functions (계속)
• AND
Z=X  Y or Z=XY
Z=1 if and only if X=1 and Y=1, otherwise Z=0
• OR
Z=X + Y
Z=1 if X=1 or if Y=1, or both X=1and Y=1. Z=0 if
and only if X=0 and Y=0
• NOT
Z=X or
Z=1 if X=0, Z=0 if X=1
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Basic Functions (계속)
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Boolean Operations and Expressions
•
Boolean Addition
–
Logical OR operation
Ex 4-1) Determine the values of A, B, C, and D that make the
sum term A+B’+C+D’
Sol) all literals must be ‘0’ for the sum term to be ‘0’
A+B’+C+D’=0+1’+0+1’=0 A=0, B=1, C=0, and D=1
•
Boolean Multiplication
–
Logical AND operation
Ex 4-2) Determine the values of A, B, C, and D for AB’CD’=1
Sol) all literals must be ‘1’ for the product term to be ‘1’
AB’CD’=10’10’=1 A=1, B=0, C=1, and D=0
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Basic Identities of Boolean Algebra
The relationship between
a single variable X, its
complement X, and the
binary constants 0 and 1
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Laws of Boolean Algebra
• Commutative
Law
the order of literals does not matter
–A + B = B + A
–A
B=BA
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Laws of Boolean Algebra (계속)
• Associative
Law
the grouping of literals does not matter
– A + (B + C) = (A + B) + C (=A+B+C)
– A(BC)
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= (AB)C (=ABC)
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Laws of Boolean Algebra (계속)
• Distributive
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Law : A(B + C) = AB + AC
A
B
C
X
Y
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X=Y
Laws of Boolean Algebra (계속)
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(A+B)(C+D) = AC + AD + BC + BD
A
B
C
D
X
Y
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X=Y
Rules of Boolean Algebra
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A+0=A
In math if you add 0 you have changed nothing
in Boolean Algebra ORing with 0 changes
nothing

A
X
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X=A+0
=A
Rules of Boolean Algebra (계속)
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A+1=1
ORing with 1 must give a 1 since if any
input is 1 an OR gate will give a 1
A
X
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X=A+1
=1
Rules of Boolean Algebra (계속)
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A•0=0
In math if 0 is multiplied with anything
you get 0. If you AND anything with 0
you get 0
A
X
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X=A0
=0
Rules of Boolean Algebra (계속)
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A•1 =A
ANDing anything with 1 will yield the anything
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A
A
X
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X=A1
=A
Rules of Boolean Algebra (계속)
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A+A = A
ORing with itself will give the same
result
A
A
X
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A=A+A
=A
Rules of Boolean Algebra (계속)
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A+A’=1
Either A or A’ must be 1 so A + A’ =1
A
A’
X
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X=+A’
=1
Rules of Boolean Algebra (계속)
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A•A = A
ANDing with itself will give the same
result
A
A
X
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A=AA
=A
Rules of Boolean Algebra (계속)
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A•A’ =0
In digital Logic 1’ =0 and 0’ =1, so AA’=0
since one of the inputs must be 0.
A
A’
X
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X=AA’
=0
Rules of Boolean Algebra (계속)
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A = (A’)’
If you not something twice you are back to the
beginning
A
X
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X=(A’)’
=A
Rules of Boolean Algebra (계속)
 A + AB = A
A
B
X
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Rules of Boolean Algebra (계속)
A + A’B = A + B
If A is 1 the output is 1
is B
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If A is 0 the output
A
B
X
Y
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X=Y
Rules of Boolean Algebra (계속)
 (A + B)(A + C) = A + BC
A
B
C
X
Y
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DeMorgan’s Theorems
•
DeMorgan’s Theorem
F(A,A, , + , 1,0) = F(A, A, + , ,0,1)
(A • B)’ = A’ + B’ and (A + B)’ = A’ • B’
– DeMorgan’s theorem will help to simplify digital
circuits using NORs and NANDs his theorem
states
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Look at (A +B +C + D)’ = A’ • B’ • C’ • D’
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Ex 4-3) Apply DeMorgan’s theorems to (XYZ)’ and
(X+Y+Z)’
Sol) (XYZ)’=X’+Y’+Z’ and (X+Y+Z)’=X’Y’Z’
Ex 4-5) Apply DeMorgan’s theorems to
(a) ((A+B+C)D)’ (b) (ABC+DEF)’ (c) (AB’+C’D+EF)’
Sol) (a) ((A+B+C)D)’= (A+B+C)’+D’=A’B’C’+D’
(b) (ABC+DEF)’=(ABC)’(DEF)’=(A’+B’+C’)(D’+E’+F’)
(c) (AB’+C’D+EF)’=(AB’)’(C’D)’(EF)’=(A’+B)(C+D’)(E’+F’)
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Boolean Analysis of Logic Circuits
• Boolean
Expression for a Logic Circuit
Figure 4-16 A logic circuit showing the development of the
Boolean expression for the output.
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•
Constructing a Truth Table for a Logic
Circuit
Convert the expression into the min-terms
containing all the input literals
– Get the numbers from the min-terms
– Putting ‘1’s in the rows corresponding to the minterms and ‘0’s in the remains
–
Ex) A(B+CD)=AB(C+C’) (D+D’) +A(B+B’)CD
=ABC(D+D’) +ABC’(D+D’) +ABCD+AB’CD
=ABCD+ABCD’+ABC’D+ABC’D’ +ABCD+AB’CD
=ABCD+ABCD’+ABC’D+ABC’D’ +AB’CD
=m11+m12+m13+m14+m15=(11,12,13,14,15)
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Truth Table from Logic Circuit
A(B+CD)=m11+m12+m13+m14+m15
=(11,12,13,14,15)
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A
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
Input
B
C
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
0
0
0
0
0
1
0
1
1
0
1
0
1
1
1
1
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D
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
Output
A(B+CD)
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
1
Simplification Using Boolean Algebra
Ex 4-8) Using Boolean algebra, simplify this
expression
AB+A(B+C)+B(B+C)
Sol) AB+AB+AC+BB+BC =B(1+A+A+C)+AC=B+AC
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Ex 4-9) Simplify the following Boolean expression
(AB’(C+BD)+A’B’)C
Sol) (AB’C+AB’BD+A’B’)C=AB’CC+A’B’C=(A+A’)B’C=B’C
Ex 4-10) Simplify the following Boolean expression
A’BC+AB’C’+A’B’C’+AB’C+ABC
Sol) (A+A’)BC+(A+A’)B’C’+AB’C=BC+B’C’+AB’C
=BC+B’(C’+AC)=BC+B’(C’+A)=BC+B’C’+AB’
Ex 4-11) Simplify the following Boolean expression
(AB +AC)’+A’B’C
Sol) (AB)’(AC)’+A’B’C=(A’+B’)(A’+C’)+A’B’C=A’+A’B’
+A’C’+B’C+A’B’C =A’(1+B’+C’+B’C)+B’C=A’+B’C’
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Standard Forms of Boolean Expressions
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The Sum-of-Products(SOP) Form
Ex) AB+ABC, ABC+CDE+B’CD’
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The Product-of-Sums(POS) Form
Ex) (A+B)(A+B+C), (A+B+C)(C+D+E)(B’+C+D’)
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Principle of Duality : SOP  POS
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Domain of a Boolean Expression
The set of variables contained in the expression
Ex) A’B+AB’C : the domain is {A, B, C}
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Implementation of a SOP Expression
AND-OR logic
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Conversion of General Expression to SOP Form
A(B+CD)=AB +ACD
Ex 4-12) Convert each of the following expressions to
SOP form: (a) AB+B(CD+EF) (b) (A+B)(B+C+D)
Sol) (a) AB+B(CD+EF)=AB+BCD+BEF
(b) (A+B)(B+C+D)=AB+AC+AD+ BB+BC+BD
=B(1+A+C+D)+ AC+AD=B+AC+AD
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Standard SOP Form (Canonical SOP Form)
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For all the missing variables, apply (x+x’)=1 to the
AND terms of the expression
– List all the min-terms in forms of the complete set of
variables in ascending order
–
Ex 4-13) Convert the following expression into standard
SOP form: AB’C+A’B’+ABC’D
Sol) domain={A,B,C,D},
AB’C(D’+D)+A’B’(C’+C)(D’+D)+ABC’D
=AB’CD’+AB’CD+A’B’C’D’+A’B’C’D+A’B’CD’+A’B’CD+ABC’D
=1010+1011+0000+0001+0010+0011+1101
=0+1+2+3+10+11+13 = (0,1,2,3,10,11,13)
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Product-of-Sums Form
•
Implementation of a POS Expression
OR-AND logic
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Standard POS Form (Canonical POS Form)
For all the missing variables, apply (x’x)=0 to the
OR terms of the expression
– List all the max-terms in forms of the complete
set of variables in ascending order
–
Ex 4-15) Convert the following expression into
standard POS form: (A+B’+C)(B’+C+D’)(A+B’+C’+D)
Sol) domain={A,B,C,D},
(A+B’+C)(B’+C+D’)(A+B’+C’+D)
=(A+B’+C+D’D)(A’A+B’+C+D’)(A+B’+C’+D)
=(A+B’+C+D’)(A+B’+C+D)(A’+B’+C+D’)(A+B’+C+D’)(A
+B’+C’+D)=(0100) )(0101)(0110)(1101)= (4,5,6,13)
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Converting Standard SOP to Standard POS
Step 1. Evaluate each product term in the SOP
expression. Determine the binary numbers
that represent the product terms
Step 2. Determine all of the binary numbers not
included in the evaluation in Step 1
Step 3. Write in equivalent sum term for each binary
number Step 2 and expression in POS form
Ex 4-17) Convert the following SOP to POS
Sol) SOP= A’B’C’+A’BC’+A’BC+AB’C+ABC=0+2+3+5+7
=(0,2,3,5,7)
POS=(1)(4)(6) = (1, 4, 6)
(=(A+B+C’)(A’+B+C)(A’+B’+C))
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Boolean Expressions and Truth Tables
•
Converting SOP Expressions to Truth Table
Format
Ex 4-18) A’B’C+AB’C’+ABC =(1,4,7)
Inputs
A B C
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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Output
X
0
1
0
0
1
0
0
1
Product
Term
A’B’C
AB’C’
ABC
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Converting POS Expressions to Truth Table Format
Ex 4-19) (A+B+C)(A+B’+C)(A+B’+C’)(A’+B+C’)(A’+B’+C)
= (000)(010)(011)(101)(110) = (0,2,3,5,6)
Inputs
A B C
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
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Output
X
0
1
0
0
1
0
0
1
Sum Term
A+B+C
A+B’+C
A+B’+C’
A’+B+C’
A’+B’+C
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Ex 4-20) Determine standard SOP and POS from the
truth table
Sol) (a) Standard SOP
Inputs
Output
A B C
X
F=A’BC+AB’C’+ABC’+ABC
0 0 0
0
(b) Standard POS
0 0 1
0
F=(A+B+C)(A+B+C’)(A+B’+C)
0 1 0
0
0 1 1
1
(A’+B+C’)
1
1
1
1
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0
0
1
1
0
1
0
1
1
0
1
1
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Boolean
Expression
Truth
Table
Logic
Diagram
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Karnaugh Map
• Simplification
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methods
– Boolean
algebra(algebraic method)
– Karnaugh map(map method))
– Quine-McCluskey(tabular method)
XY+XY=X(Y+Y)=X
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Three- and Four-input Kanaugh maps
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Gray code
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Gray code sequence generation
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F(X,Y,Z)=m(0,1,2,6) =(XY+YZ)=X’Y’ + YZ’
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Example) F(X,Y,Z)=m(2,3,4,5) =XY+XY
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0
1
3
2
4
5
7
6
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Example) F(X,Y,Z)=m(0,2,4,6) = XZ+XZ 
=Z(X+X)=Z
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Four-Variable Map
16 minterms : m0 ~ m15
Rectangle group
–
2-squares(minterms) : 3-literals product
term
–
4-squares : 2-literals product term
–
8-squares : 1-literals product term
–
16-squares : logic 1
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F(W, X,Y,Z)=m(0,2,7,8,9,10,11) = WX’ + X’Z’ + W’XYZ
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Karnaugh Map SOP Minimization
•
Mapping a Standard SOP Expression
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Ex 4-21)
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Ex 4-22)
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Mapping a Nonstandard SOP Expression
Numerical Expression of a Nonstandard Product
Term
Ex 4-23) A’+AB’+ABC’
A’
AB’ ABC’
000 100 110
001 101
010
011
–
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Ex 4-24) B’C’+AB’+ABC’+AB’CD’+A’B’C’D+AB’CD
B’C’
AB’
ABC’ AB’CD’ A’B’C’D AB’CD
0000 1000 1100 1010
0001
1011
0001 1001
1101
1000 1010
1001 1011
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Karnaugh Map Simplification of SOP Expressions
•
•
•
Group 2n adjacent cells including the largest possible
number of 1s in a rectangle or square shape, 1<=n
Get the groups containing all 1s on the map for the
expression
Determine the minimum SOP expression form map
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Ex 4-26) F=B+A’C+AC’D
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Ex 4-27) (a) AB+BC+A’B’C’ (b) B’+AC+A’C’
(c) A’C’+A’B+AB’D (d) D’+BC’+AB’C
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Ex 4-28) Minimize the following expression
AB’C+A’BC+A’B’C+A’B’C’+AB’C’
Sol) B’+A’C
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Ex 4-29) Minimize the following expression
B’C’D’+A’BC’D’+ABC’D’+A’B’CD+AB’CD+A’B’CD’+A’BCD’
+ABCD’+AB’CD’
Sol) D’+B’C
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Mapping Directly from a Truth Table
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Don’t Care Conditions
• it really does not matter since they
will never occur(its output is either
‘0’ or ‘1’)
• The don’t care terms can be used
to advantage on the Karnaugh map
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Karnaugh Map POS Minimization
• Use
the Duality Principle
F(A,A, , + , 1,0)  F*(A,A, + , ,0,1)
SOP  POS
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Ex 4-30) (A’+B’+C+D)(A’+B+C’+D’)(A+B+C’+D)
(A’+B’+C’+D’)(A+B+C’+D’)
Sol)
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Ex 4-31) (A+B+C)(A+B+C’)(A+B’+C)(A+B’+C’)(A’+B’+C)
Sol) (0+0+0)(0+0+1)(0+1+0)(0+1+1)(1+1+0)=A(B’+C)
AC+AB’=A(B’+C)
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Ex 4-32)
(B+C+D)(A+B+C’+D)(A’+B+C+D’)(A+B’+C+D)(A’+B’+C+D)
Sol) (B+C+D)=(A’A+B+C+D)=(A’+B+C+D)(A+B+C+D)
(1+0+0+0)(0+0+0+0)(0+0+1+0)(1+0+0+1)(0+1+0+0)(1+1+0+0)
F=(C+D)(A’+B+C)(A+B+D)
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Converting Between POS and SOP Using the K-map71
Ex 4-33) (A’+B’+C+D)(A+B’+C+D)(A+B+C+D’)(A+B+C’+D’)
(A’+B+C+D’)(A+B+C’+D)
Sol)
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Five/Six –Variable K-Maps
• Five
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Variable K-Map : {A,B,C,D,E}
BC 00
DE
0
00 16
11
3
17
19
4
5
7
01
20
21
23
12
13
15
11
28
29
31
8
9
11
10
24
25
27
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01
1
10
18
22
30
26
2
6
14
10
A=0
A=1
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• Six
Variable K-Map : {A,B,C,D,E,F}
AB
00
10
01
11
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01
11
10
CD 00
EF
0
1
3
2
00 32 16 33 17 35 19 34 18
48
49
51
50
4
5
7
6
01 36 20 37 21 39 23 38 22
52
53
55
54
12
13
15
14
11 44 28 45 29 47 31 46 30
60
61
62
63
10 40 8 24 41 9 25 43 11 27 421026
56
57
59
58
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Ex 4-34)
Sol) A’D’E’+B’C’D’+BCD+ACDE
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