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Princess Sumaya Univ.Computer Engineering Dept.
Chapter 2:
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Basic Definitions
Binary Operators
AND
z= x y= x y z=1 if x=1 AND y=1
OR
z= x+ y z=1 if x=1 OR y=1
NOT
z= x=x z=1 if x=0
Boolean Algebra
Binary Variables: only 0 and 1 values
Algebraic Manipulation
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Boolean Algebra Postulates
Commutative Law
x y= y x x+ y= y+ x
Identity Element
x 1 = x x+ 0 = x
Complement
xx= 0 x+x= 1
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Boolean Algebra Theorems
Duality
The dualof a Boolean algebraic expression is obtained
by interchanging the ANDand the ORoperators and
replacing the 1s by 0s and the 0s by 1s.
x ( y+ z) = ( x y) + ( x z)
x+ ( y z) = ( x+ y) ( x+ z)
Theorem 1
x x= x x+ x= x
Theorem 2
x 0 = 0 x+ 1 = 1
App l ied to a validequat ion prod uces
a validequat ion
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Boolean Algebra Theorems
Theorem 3: Involution
(x)= x ( x) = x
Theorem 4: Associative & Distr ibutive
( x y) z= x ( y z) ( x+ y) + z= x+ ( y+ z)
x ( y+ z) = ( x y) + ( x z)
x+ ( y z) = ( x+ y) ( x+ z)
Theorem 5: DeMorgan
( x y)=x +y ( x+ y)=x y ( x y)= x + y ( x+ y)= x y
Theorem 6: Absorption
x ( x+ y) = x x+ ( x y) = x
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Operator Precedence
Parentheses
(. . . ) (. . .)
NOT
x+ y
AND
x+ xy
OR
])([ xwzyx
])([
)(
)()(
)(
xwzyx
xwzy
xwz
xw
xw
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DeMorgans Theorem
)]([ edcba
)]([ edcba
))(( edcba
))(( edcba
))(( edcba
)( edcba
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Boolean Functions
Boolean Expression
Example: F= x+yz
Truth Table
All possible combinationsof input variables
Logic Circuit
x y z F0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 1
1 1 0 1
1 1 1 1
x
yz
F
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Algebraic Manipulation
Literal:
A single variable within a term that may be complemented
or not.
Use Boolean Algebra to simplify Boolean functions
to produce simpler circuits
Example: Simplify to a minimum number of literals
F= x+x y ( 3Literals)
= x+ (x y )
= ( x+x) ( x+ y)
= ( 1 ) ( x+ y) = x+ y ( 2Literals)
Distr ibut iv e law(+ over )
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Complement of a Function
DeMorgans Theorm
Duality & Literal Complement
CBAF
CBAF
CBAF
CBAF
CBAF
CBAF
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Canonical Forms
Minterm
Product (ANDfunction)
Contains all variables
Evaluates to 1 for a
specific combination
Example
A= 0 A B C
B= 0 (0) (0) (0)
C= 01 1 1= 1
A B C Minterm
0 0 0 0 m0
1 0 0 1 m1
2 0 1 0 m2
3 0 1 1 m3
4 1 0 0 m4
5 1 0 1 m5
6 1 1 0 m6
7 1 1 1 m7
CBA
CBA
CBA
CBA
CBA
CBA
CBA
CBA
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Canonical Forms
Maxterm
Sum (ORfunction)
Contains all variables
Evaluates to 0 for a
specific combination
Example
A= 1 A B C
B= 1 (1) + (1) + (1)
C= 10 + 0 + 0= 0
A B C Maxterm
0 0 0 0 M0
1 0 0 1 M1
2 0 1 0 M2
3 0 1 1 M3
4 1 0 0 M4
5 1 0 1 M5
6 1 1 0 M6
7 1 1 1 M7
CBA
CBA
CBA
CBA
CBA
CBA
CBA
CBA
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Canonical Forms
Truth Table toBoolean Function
A B C F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 1
1 0 1 11 1 0 0
1 1 1 1
CBAF CBA CBA ABC
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Canonical Forms
Sum of Minterms
Product of Maxterms
A B C F
0 0 0 0 0
1 0 0 1 1
2 0 1 0 0
3 0 1 1 04 1 0 0 1
5 1 0 1 1
6 1 1 0 0
7 1 1 1 1
ABCCBACBACBAF
7541 mmmmF
)7,5,4,1(F
CABBCACBACBAF
CABBCACBACBAF
CABBCACBACBAF
))()()(( CBACBACBACBAF
6320 MMMMF
(0,2,3,6)F
F
1
0
1
10
0
1
0
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Standard Forms
Sum of Products (SOP)
ABCCBACBACBAF
AC
BBAC
)(
CB
AACB
)(
BA
BA
CCBA
)1(
)(
)()()( BBACCCBAAACBF
ACBACBF
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Standard Forms
Product of Sums (POS)
)( AACB
)( BBCA
)( CCBA
)()()( AACBCCBABBCAF
CBBACAF
CABBCACBACBAF
))()(( CBBACAF
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Two-Level Implementations
Sum of Products (SOP)
Product of Sums (POS)
ACBACBF
))()(( CBBACAF
B
C
FBA
AC
AC
FBA
B
C
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Logic Operators
AND
NAND (NotAND) x y NAND0 0 1
0 1 11 0 1
1 1 0
x y AND
0 0 0
0 1 0
1 0 0
1 1 1
x
yx
y
xy
x y
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Logic Operators
OR
NOR (NotOR)
x y OR
0 0 0
0 1 1
1 0 1
1 1 1
x y NOR
0 0 1
0 1 01 0 0
1 1 0
xy
x+ y
x
y x+ y
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Logic Operators
XOR (Exclusive-OR)
XNOR (Exclusive-NOR)
(Equivalence)
x y XOR
0 0 0
0 1 1
1 0 1
1 1 0
x y XNOR
0 0 1
0 1 0
1 0 0
1 1 1x
y
x y
x yxy + xy
x
yx y
xy + xy
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Logic Operators
NOT (Inverter)
Buffer
x NOT
0 1
1 0
x Buffer
0 0
1 1
x x
x x
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Multiple Input Gates
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DeMorgans Theorem on Gates
AND Gate
F= x y F= (x y) F= x+ y
OR Gate
F= x+ y F= (x+ y) F= x y
Change the Shape and bubble all lines
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Homework
Mano
Chapter 2
2-4
2-5
2-6
2-8
2-9
2-10
2-12
2-15
2-18
2-19
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Homework
Mano
2-4 Reduce the following Boolean expressions to the indicated
number of literals:
(a) AC + ABC+ AC to three literals
(b) (xy + z) + z+ xy+ wz to three literals
(c) AB(D + CD) + B(A+ ACD) to one literal
(d) (A + C) (A + C) (A+ B+ CD) to four literals
2-5 Find the complement of F= x+ yz; then show that
FF = 0 and F+ F = 1
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Homework
2-6 Find the complement of the following expressions:
(a) xy + xy (b) (AB + C)D + E
(c) (x+ y + z) (x + z) (x+ y)
2-8 List the truth table of the function:
F= xy+ xy + yz
2-9 Logical operations can be performed on strings of bits by
considering each pair of corresponding bits separately
(this is called bitwise operation). Given two 8-bit stringsA= 10101101 and B= 10001110, evaluate the 8-bit result
after the following logical operations: (a) AND, (b) OR, (c)
XOR, (d) NOT A, (e) NOT B.
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Homework
2-10 Draw the logic diagrams for the following Boolean
expressions:
(a) Y= AB + B(A+ C) (b) Y= BC+ AC
(c) Y= A+ CD (d) Y= (A+ B) (C + D)
2-12 Simplify the Boolean function T1and T2to a minimum
number of literals.A B C T1 T2
0 0 0 1 0
0 0 1 1 0
0 1 0 1 00 1 1 0 1
1 0 0 0 1
1 0 1 0 1
1 1 0 0 1
1 1 1 0 1
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Homework
2-15 Given the Boolean function
F= xyz+ xyz+ wxy+ wxy+ wxy
(a) Obtain the truth table of the function.
(b) Draw the logic diagram using the original Boolean
expression.
(c) Simplify the function to a minimum number of literalsusing Boolean algebra.
(d) Obtain the truth table of the function from the
simplified expression and show that it is the same as
the one in part (a)(e) Draw the logic diagram from the simplified expression
and compare the total number of gates with the
diagram of part (b).
P S U 4241 D l L D C E D
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Homework
2-18 Convert the following to the other canonical form:
(a) F(x, y, z) = (1, 3, 7)
(b) F(A, B, C, D) = (0, 1, 2, 3, 4, 6, 12)
2-19 Convert the following expressions into sum of products
and product of sums:
(a) (AB+ C) (B+ CD)
(b) x + x(x+ y) (y+ z)