CT455: Computer Organization Boolean Algebra

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CT455 Computer OrganizationBoolean Algebra
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Lecture 3 Boolean Algebra

• Digital circuits
• Boolean Algebra
• Two-Valued Boolean Algebra
• Boolean Algebra Postulates
• Precedence of Operators
• Truth Table Proofs
• Duality

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Lecture 3 Boolean Algebra

• Basic Theorems of Boolean Algebra
• Boolean Functions
• Complement of Functions
• Standard Forms
• Minterm Maxterm
• Canonical Forms
• Conversion of Canonical Forms
• Binary Functions

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Digital Circuits

• Digital circuit can be represented by a black-box
  with inputs on one side, and outputs on the other.

The input/output signals are discrete/digital in
nature, typically with two distinct voltages (a
high voltage and a low voltage).
In contrast, analog circuits use continuous
signals.
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Digital Circuits

• Advantages of Digital Circuits over Analog
  Circuits
• more reliable (simpler circuits, less
  noise-prone)
• specified accuracy (determinable)
• but slower response time (sampling rate)
• Important advantages for two-valued Digital
  Circuit
• Mathematical Model Boolean Algebra
• Can help design, analyse, simplify Digital
  Circuits.

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Boolean Algebra
What is an Algebra? (e.g. algebra of
integers) set of elements (e.g. 0,1,2,..) set
of operations (e.g. , -, ,..) postulates/axioms
(e.g. 0xx,..)

• Boolean Algebra named after George Boole who used
  it to study human logical reasoning calculus of
  proposition.
• Events true or false
• Connectives a OR b a AND b, NOT a
• Example Either it has rained OR someone
  splashed water, must be tall AND good vision.

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Boolean Algebra
Later, Shannon introduced switching algebra
(two-valued Boolean algebra) to represent
bi-stable switching circuit.
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Two-valued Boolean Algebra

• Set of Elements 0,1
• Set of Operations ., ,

x.y
Signals High 5V 1 Low 0V 0
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Boolean Algebra Postulates
A Boolean algebra consists of a set of elements
B, with two binary operations and . and a
unary operation ', such that the following
axioms hold

• Commutative laws
• A B B A
• A . B B . A

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Boolean Algebra Postulates

• Associative laws
• (A B) C A (B C) A B C
• (A . B) . C A .( B . C) A . B . C
• Distributive laws
• A . (B C) (A . B) (A . C)
• A (B . C) (A B) . (A C)

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Boolean Algebra Postulates

• Complement For every x in B, there exists an
  element x' in B such that
• x x' 1
• x . x' 0

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Precedence of Operators

• To lessen the brackets used in writing Boolean
  expressions, operator precedence can be used.
• Precedence (highest to lowest) ' .
• Examples
• a . b c (a . b) c
• b' c (b') c
• a b' . c a ((b') . c)

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Precedence of Operators

• Use brackets to overwrite precedence.
• Examples
• a . (b c)
• (a b)' . c

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Rules for Boolean Algebra
1. A 0 A
9. (A) A
2. A 1 1
10. A AB A
3. A . 0 0
11. A AB A B
4. A . 1 A
12. (A B)(A C) ABC
5. A A A
6. A A 1
7. A . A A
8. A . A 0
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10. A AB A A AB A(1 B) DISTRIBUTIVE
LAW A . 1 RULE 2 A RULE 4
A B AB AAB
0 0 1 1 0 1 0 1 0 0 0 1 0 0 1 1
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Theorem 10 (Absorption) (a) a ab a (b) a(a
b) a Examples (X Y) (X Y)Z X
Y T4(a) AB'(AB' B'C) AB' T4(b)
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11. A AB A B A AB (A AB)
AB (AA AB) AB AA AB AA
AB (A A)(A B) 1. (A B) ( A
B)
A B AB AAB AB
0 0 1 1 0 1 0 1 0 1 0 0 0 1 1 1 0 1 1 1
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Theorem 11 (a) a a'b a b (b) a(a' b)
ab Examples B AB'C'D B
AC'D T11(a) (X Y)((X Y)' Z) (X
Y)Z T11(b)
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11. A AB A B
??????????????????????
A AB A B
A AB A B
A ABC A BC
(AB) ABCD (AB) CD
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12. (A B)(B C) AA AC AB BC A
AC AB BC A(1C) AB BC A.1
AB BC A(1B) BC A.1 BC A .
BC
A B C AB AC (AB)(AC) BC ABC
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 1
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Theorem (a) ab ab' a (b) (a b)(a b')
a Examples ABC AB'C AC T6(a) (W' X'
Y' Z')(W' X' Y' Z)(W' X' Y Z')(W'
X' Y Z) (W' X' Y')(W' X' Y
Z')(W' X' Y Z) (W' X' Y')(W' X'
Y) (W' X')
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Theorem 7 (a) ab ab'c ab ac (b) (a
b)(a b' c) (a b)(a c) Examples wy'
wx'y wxyz wxz' wy' wx'y wxy
wxz' T7(a) wy' wy wxz' T7(a)
w wxz' T7(a) w T7(a) (x'y'
z)(w x'y' z') (x'y' z)(w
x'y') T7(b)
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Basic Theorems of Boolean Algebra
6. DeMorgan. (a) (x y)' x'.y' (b)
(x.y)' x' y' 7. Consensus. (a) x.y
x'.z y.z x.y x'.z (b)
(xy).(x'z).(yz) (xy).(x'z)
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Laws of Boolean Algebra
DeMorgan's Law
(X Y)' X' ? Y'
NOR is equivalent to AND with inputs complemented
(X ? Y)' X' Y'
NAND is equivalent to OR with inputs complemented
DeMorgan's Law can be used to convert AND/OR
expressions to OR/AND expressions or NAND/NAND
expression
Example
Z A' B' C A' B C A B' C A B C
AND - OR (ABC) ? (ABC) ? (ABC) ?
(ABC) NAND - NAND Z' (A B C') ?
(A B' C') ? (A' B C') ? (A' B' C)
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• Theorem 8 (DeMorgan's Theorem)
• (a) (a b)' a'b' (b) (ab)' a' b'
• Generalized DeMorgan's Theorem
• (a) (a b z)' a'b' z' (b) (ab z)'
  a' b' z'
• Examples
• (a bc)' (a (bc))'
• a'(bc)' T8(a)
• a'(b' c') T8(b)
• a'b' a'c' P5(b)
• Note (a bc)' ¹ a'b' c'

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• More Examples for DeMorgan's Theorem
• (a(b z(x a')))' a' (b z(x
  a'))' T8(b)
• a' b' (z(x a'))' T8(a)
• a' b' (z' (x a')') T8(b)
• a' b' (z' x'(a')') T8(a)
• a' b' (z' x'a) T3
• a' b' (z' x') T5(a)
• (a(b c) a'b)' (ab ac a'b)' P5(b)
• (b ac)' T6(a)
• b'(ac)' T8(a)
• b'(a' c') T8(b)

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Truth Table

• Provides a listing of every possible combination
  of inputs and its corresponding outputs.
• Example (2 inputs, 2 outputs)

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Switching Functions

• Switching algebra Boolean algebra with the set
  of elements K 0, 1
• If there are n variables, we can define
  switching functions.
• Sixteen functions of two variables (Table 2.3)
• A switching function can be represented by a
  table as above, or by a switching expression as
  follows
• f0(A,B) 0, f6(A,B) AB' A'B, f11(A,B) AB
  A'B A'B' A' B, ...
• Value of a function can be obtained by plugging
  in the values of all variables
• The value of f6 when A 1 and B 0 is
  0 1 1.

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Truth Tables (1)

• Shows the value of a function for all possible
  input combinations.
• Truth tables for OR, AND, and NOT (Table 2.4)

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Truth Tables (2)

• Truth tables for f(A,B,C) AB A'C AC' (Table
  2.5)

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Truth Table

• Example (3 inputs, 2 outputs)

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Proof using Truth Table

• Can use truth table to prove by perfect
  induction.
• Prove that x . (y z) (x . y) (x . z)
• (i) Construct truth table for LHS RHS of above
  equality.
• (ii) Check that LHS RHS
• Postulate is SATISFIED because output column 2
  5 (for LHS RHS expressions) are equal for all
  cases.

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Basic Theorems of Boolean Algebra

• Apart from the axioms/postulates, there are other
  useful theorems.
• 1. Idempotency.
• (a) x x x (b) x . x x
• Proof of (a)
• x x (x x).1 (identity)
• (x x).(x x') (complementarity)
• x x.x' (distributivity)
• x 0 (complementarity)
• x (identity)

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Basic Theorems of Boolean Algebra

• 2. Null elements for and . operators.
• (a) x 1 1 (b) x . 0 0
• 3. Involution. (x')' x
• 4. Absorption.
• (a) x x.y x (b) x.(x y) x
• 5. Absorption (variant).
• (a) x x'.y xy (b) x.(x' y) x.y

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Basic Theorems of Boolean Algebra

• Theorem 4a (absorption) can be proved by
• x x.y x.1 x.y
  (identity)
• x.(1 y)
  (distributivity)
• x.(y 1)
  (commutativity)
• x.1
  (Theorem 2a)
• x
  (identity)
• By duality, theorem 4b
• x.(xy) x
• Try prove this by algebraic manipulation.

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Boolean Functions

• Boolean function is an expression formed with
  binary variables, the two binary operators, OR
  and AND, and the unary operator, NOT, parenthesis
  and the equal sign.
• Its result is also a binary value.
• We usually use . for AND, for OR, and ' or
  for NOT. Sometimes, we may omit the . if there
  is no ambiguity.

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Boolean Functions

• Examples
• F1 x.y.z'
• F2 x y'.z
• F3(x'.y'.z)(x'.y.z)(x.y')
• F4x.y'x'.z

From the truth table, F3F4. Can you also prove
by algebraic manipulation that F3F4?
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Simplifying logic expressions
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• EXAMPLE
• A(AB) AA AB
• A AB
• A(1 B)
• A
• A(A B) AA AB
• AB
• ( AB AB) (A B)(A B)
• AA AB AB BB
• AB AB

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X ABC ABC AB ( C C) AB.1 AB
A X(XYZ XYZ) XXYZ XXYZ 0.YZ
0.YZ 0
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EXAMPLE AB AB ABAAABBB
A(BA) B(AB) (AB)(AB)
(AB)(AB)
A B AB AB (AB) AB
0 0 0 1 1 0 1 1 0 0 1 0 0 1 0 0 1 1 1 0 0 1 1 1
ABAB (AB)(AB)
0 1 1 0 0 1 1 0
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EXAMPLE Y AB AB AB
Y B(A A) AB
Y B.1 AB
Y B AB
Y B A
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W AB(AC B) ABAC ABB 0.BC
AB 0 AB AB
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Z ( X.Y) ( X . ( Y X )
Z ( X.Y) ( X.Y) (X.X)
Z ( X.Y) ( X.Y) (0)
Z ( X.Y) ( X.Y)
Z Y. ( X X)
Z Y. 1
Z Y
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Z ( X.Y) ( X . Y ) Y )
Z ( X . Y ).(X . Y ) ( X . Y ) Y )
Z ( X . X ).(Y . Y ) X . (Y Y )
Z ( X . X ).( 0 ) X . (Y Y )
Z 0 X . Y
Z X . Y
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Z X X.Y Y.X
Z X X. (Y Y)
Z X X. (1)
Z 1
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Z X( X Y )(Y X)
Z (X.X XY )( Y X )
Z ( 0 XY)( Y X
Z XY( Y X)
Z XYY XYX
Z 0 0
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Z w( y w )( x y w ) (wy ww)( x y
w ) (wy w)( x y w ) (w wy)( x
y w ) w( x y w ) wx wy ww wx
( wy w ) wx ( w wy) wx w w
wx w
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Simplify
AB BC(B C) AB BBC BCC AB BC BC AB
BC B( A C)
AB(A BC C) ABA ABBC ABC AB ABBC
ABC AB ABC AB(1 C) AB
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EXAMPLE
SIMPLIFY THE EXPRESSION AB A(B C) B(B C)
USING BOOLEAN ALGEBRA

1. AB AB AC BB BC
2. AB AB AC B BC
3. AB AC B BC
4. AB AC B(1 C)
5. AB AC B.1
6. AB AC B
7. B(A 1) AC
8. B.1 AC
9. B AC

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SIMPLIFY THE EXPRESSION AB(C BD) ABC
USING BOOLEAN ALGEBRA

1. AB(C BD) ABC
2. (ABC ABBD AB)C
3. (ABC A.0.D AB)C
4. (ABC 0 AB)C
5. (ABC AB)C
6. ABCC ABC
7. ABC ABC
8. BC(A A)
9. BC.1
10. BC

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SIMPLIFY THE EXPRESSION ABC ABC ABC
ABC ABC USING BOOLEAN ALGEBRA
ABC ABC ABC ABC ABC BC(A A)
ABC ABC ABC BC.1 AB(C C)
ABC BC AB.1 ABC BC B(A AC) BC
B(A C) BC AB BC
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Gate Logic Laws of Boolean Algebra
Apply the laws and theorems to simplify Boolean
equations
Example full adder's carry out function
identity
Cout A' B Cin A B' Cin A B Cin' A
B Cin
A' B Cin A B' Cin A B Cin' A B Cin
A B Cin A' B Cin A B Cin A B' Cin
A B Cin' A B Cin (A' A) B Cin A
B' Cin A B Cin' A B Cin (1) B Cin
A B' Cin A B Cin' A B Cin B Cin A
B' Cin A B Cin' A B Cin A B Cin B
Cin A B' Cin A B Cin A B Cin' A B
Cin B Cin A (B' B) Cin A B Cin'
A B Cin B Cin A (1) Cin A B Cin' A
B Cin B Cin A Cin A B (Cin' Cin)
B Cin A Cin A B (1) B Cin A Cin
A B
associative
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Z (x y) w (xy) ( x y ) w ( x
y ) ( x y ) w ( x y ) . 1 (
x y ) ( w 1) ( x y ) (1) x y
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M (A BC) (A) (B)(C) ABC
S (AB(C D)) (AB) (C D)
(A) (B) (C) (D) A B CD
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Complement of Functions

• Given a function, F, the complement of this
  function, F', is obtained by interchanging 1 with
  0 in the functions output values.

Example F1 xyz' Complement
F1' (x.y.z')' x' y'
(z')' DeMorgan x' y' z
Involution
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Complement of Functions

• More general DeMorgans theorems useful for
  obtaining complement functions
• (A B C ... Z)' A' . B' . C' .
  Z'
• (A . B . C ... . Z)' A' B' C'
  Z'

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Algebraic Forms of Switching Functions (1)

• Literal A variable, complemented or
  uncomplemented.
• Product term A literal or literals ANDed
  together.
• Sum term A literal or literals ORed together.
• SOP (Sum of Products)
• ORing product terms
• f(A, B, C) ABC A'C B'C
• POS (Product of Sums)
• ANDing sum terms
• f (A, B, C) (A' B' C')(A C')(B C')

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Algebraic Forms of Switching Functions (2)

• A minterm is a product term in which all the
  variables appear exactly once either complemented
  or uncomplemented.
• Canonical Sum of Products (canonical SOP)
• Represented as a sum of minterms only.
• Example f1(A,B,C) A'BC' ABC' A'BC
  ABC (2.1)
• Minterms of three variables

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Algebraic Forms of Switching Functions (3)

• Compact form of canonical SOP form
• f1(A,B,C) m2 m3 m6 m7 (2.2)
• A further simplified form
• f1(A,B,C) S m (2,3,6,7) (minterm list
  form) (2.3)
• The order of variables in the functional notation
  is important.
• Deriving truth table of f1(A,B,C) from minterm
  list

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Algebraic Forms of Switching Functions (4)

• Example Given f(A,B,Q,Z) A'B'Q'Z' A'B'Q'Z
  A'BQZ' A'BQZ, express f(A,B,Q,Z) and f
  '(A,B,Q,Z) in minterm list form.
• f(A,B,Q,Z) A'B'Q'Z' A'B'Q'Z A'BQZ' A'BQZ
• m0 m1 m6 m7
• S m(0, 1, 6, 7)
• f '(A,B,Q,Z) m2 m3 m4 m5 m8 m9 m10
  m11 m12
• m13 m14 m15
• S m(2, 3, 4, 5, 8, 9, 10, 11, 12, 13, 14,
  15)
• (2.6)
• AB (AB)' 1 and AB A' B' 1, but AB
  A'B' ¹ 1.

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Algebraic Forms of Switching Functions (5)

• A maxterm is a sum term in which all the
  variables appear exactly once either complemented
  or uncomplemented.
• Canonical Product of Sums (canonical POS)
• Represented as a product of maxterms only.
• Example f2(A,B,C) (ABC)(ABC')(A'BC)(A'B
  C') (2.7)
• Maxterms of three variables

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Algebraic Forms of Switching Functions (6)

• f2(A,B,C) M0M1M4M5 (2.8)
• PM(0,1,4,5) (maxterm list
  form) (2.9)
• The truth table for f2(A,B,C)

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Algebraic Forms of Switching Functions (7)

• Truth tables of f1(A,B,C) of Eq. (2.3) and
  f2(A,B,C) of Eq. (2.7) are identical.
• Hence, f1(A,B,C) S m (2,3,6,7)
• f2(A,B,C)
• PM(0,1,4,5)
  (2.10)
• Example Given f(A,B,C) ( ABC')(AB'C')(A'B
  C')(A'B'C'), construct the truth table and
  express in both maxterm and minterm form.
• f(A,B,C) M1M3M5M7 PM(1,3,5,7) S m (0,2,4,6)
  

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Algebraic Forms of Switching Functions (8)

• Relationship between minterm mi and maxterm Mi
• For f(A,B,C), (m1)' (A'B'C)' A B C' M1
• In general, (mi)' Mi (2.11)
• (Mi)' ((mi)')'
  mi (2.12)

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Algebraic Forms of Switching Functions (9)

• Example Relationship between the maxterms for a
  function and its complement.
• For f(A,B,C) ( ABC')(AB'C')(A'BC')(A'B'C
  ')
• The truth table is

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Algebraic Forms of Switching Functions (10)

• From the truth table
• f '(A,B,C) PM(0,2,4,6) and f(A,B,C)
  PM(1,3,5,7)
• Since f(A,B,C) f '(A,B,C) 0,
• (M0M2M4M6)(M1M3M5M7) 0 or
• In general, (2.13)
• Another observation from the truth table
• f(A,B,C) S m (0,2,4,6) PM(1,3,5,7)
• f '(A,B,C) S m (1,3,5,7) PM(0,2,4,6)

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Derivation of Canonical Forms (1)

• Derive canonical POS or SOP using switching
  algebra.
• Theorem 10. Shannon's expansion theorem
• (a). f(x1, x2, , xn) x1 f(1, x2, , xn)
  (x1)' f(0, x2, , xn)
• (b). f(x1, x2, , xn) x1 f(0, x2, , xn)
  (x1)' f(1, x2, , xn)
• Example f(A,B,C) AB AC' A'C
• f(A,B,C) AB AC' A'C A f(1,B,C) A'
  f(0,B,C)
• A(1B 1C' 1'C) A'(0B 0C'
  0'C) A(B C') A'C
• f(A,B,C) A(B C') A'C BA(1C') A'C
  B'A(0 C') A'C
• BA A'C B'AC' A'C AB A'BC
  AB'C' A'B'C
• f(A,B,C) AB A'BC AB'C' A'B'C
• CAB A'B1 AB'1' A'B'1 C'AB
  A'B0 AB'0' A'B'0
• ABC A'BC A'B'C ABC' AB'C'

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Derivation of Canonical Forms (2)

• Alternative Use Theorem 6 to add missing
  literals.
• Example f(A,B,C) AB AC' A'C to canonical
  SOP form.
• AB ABC' ABC m6 m7
• AC' AB'C' ABC' m4 m6
• A'C A'B'C A'BC m1 m3
• Therefore,
• f(A,B,C) (m6 m7) (m4 m6) (m1 m3)
  ?m(1, 3, 4, 6, 7)
• Example f(A,B,C) A(A C') to canonical POS
  form.
• A (AB')(AB) (AB'C')(AB'C)(ABC')(ABC)
  
• M3M2M1M0
• (AC') (AB'C')(ABC') M3M1
• Therefore,
• f(A,B,C) (M3M2M1M0)(M3M1) ?M(0, 1, 2, 3)

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Incompletely Specified Functions

• A switching function may be incompletely
  specified.
• Some minterms are omitted, which are called
  don't-care minterms.
• Don't cares arise in two ways
• Certain input combinations never occur.
• Output is required to be 1 or 0 only for certain
  combinations.
• Don't care minterms di Don't
  care maxterms Di
• Example f(A,B,C) has minterms m0, m3, and m7 and
  don't-cares d4 and d5.
• Minterm list is f(A,B,C) ?m(0,3,7) d(4,5)
• Maxterm list is f(A,B,C) ?M(1,2,6)D(4,5)
• f '(A,B,C) ?m(1,2,6) d(4,5)
  ?M(0,3,7)D(4,5)
• f (A,B,C) A'B'C' A'BC ABC d(AB'C' AB'C)
• B'C' BC (use d4 and omit d5)

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Standard Forms

• Sum Term a single literal or a logical sum (OR)
  of several literals.
• Examples x, xyz', A'B, AB
• Sum-of-Products (SOP) Expression a product term
  or a logical sum (OR) of several product terms.
• Examples x, xy.z', x.y'x.y.z, A.BA'.B'
• Product-of-Sums (POS) Expression a sum term or a
  logical product (AND) of several sum terms.
• Exampes x, x.(yz'), (xy').(x'yz),
  (AB).(A'B')

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Standard Forms

• Every Boolean expression can either be expressed
  as sum-of-products or product-of-sums expression.
• Examples
• SOP x?.y x.y? x.y.z
• POS (x y?).(x? y).(x? z?)
• both x? y z or x.y.z?
• neither x.(w? y.z) or z? w.x?.y v.(x.z
  w?)

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Two Level Canonical Forms
Sum of Products
product term / minterm
Minterms
ANDed product of literals in which each variable
appears exactly once, in true or complemented
form (but not both!)
F in canonical form
F(A,B,C) ?m(3,4,5,6,7)
m3 m4 m5 m6 m7 A' B C A B' C'
A B' C A B C' A B C
Shorthand Notation for Minterms of 3 Variables
canonical form/minimal form
F A B' (C C') A' B C A B (C' C)
A B' A' B C A B A (B' B) A' B
C A A' B C A B C
B
F
C
A
F (A B C)' A' (B' C') A' B' A' C'
2-Level AND/OR Realization
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2 Level Canonical Forms
Product of Sums / Conjunctive Normal Form /
Maxterm Expansion
Maxterm
ORed sum of literals in which each variable
appears exactly once in either true or
complemented form, but not both!
Maxterm form
Find truth table rows where F is 0 0 in input
column implies true literal 1 in input column
implies complemented literal
Maxterm Shorthand Notation for a Function of
Three Variables
(?m(0,1,2)) (ABC ABC ABC)
F(A,B,C) ?M(0,1,2)
(A B C) (A B C') (A B' C)
F(A,B,C) ?M(3,4,5,6,7)
(A B' C') (A' B C) (A' B C') (A'
B' C) (A' B' C')
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Two Level Canonical Forms
F' A' B' C' A' B' C A' B C'
Apply DeMorgan's Law to obtain F
(F')' (A' B' C' A' B' C A' B C')'
F (A B C) (A B C') (A B' C)
F' (A B' C') (A' B C) (A' B C') (A'
B' C) (A' B' C')
Apply DeMorgan's Law to obtain F
(F')' (A B' C') (A' B C) (A' B C')
(A' B' C) (A' B' C')'
F A' B C A B' C' A B' C A B C'
A B C