Chapter 2:Chapter 2:

Basic DefinitionsBasic 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

Boolean Algebra PostulatesBoolean Algebra Postulates

Commutative Law

x • y = y • x x + y = y + x

Identity Element

x • 1 = x x + 0 = x

Complement

x • x’ = 0 x + x’ = 1

Boolean Algebra TheoremsBoolean Algebra Theorems

Duality

● The dual of a Boolean algebraic expression is obtained by interchanging the AND and the OR operators and replacing the 1’s by 0’s and the 0’s by 1’s.

● 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

Applied to a valid equation produces a valid equation

Boolean Algebra TheoremsBoolean Algebra Theorems

Theorem 3: Involution● ( x’ )’ = x ( x ) = x

Theorem 4: Associative & Distributive● ( 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

Operator PrecedenceOperator Precedence

Parentheses

( . . . ) • ( . . .)

NOT

x’ + y

AND

x + x • y

OR

])([ xwzyx

])([

)(

)(

)(

)(

xwzyx

xwzy

xwz

xw

xw

DeMorgan’s TheoremDeMorgan’s Theorem

)]([ edcba

)]([ edcba

))(( edcba

))(( edcba

))(( edcba

)( edcba

Boolean FunctionsBoolean Functions

Boolean Expression

Example: F = x + y’ z

Truth Table

All possible combinationsof input variables

Logic Circuit

x y z F

0 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

xyz

F

Algebraic ManipulationAlgebraic 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 ( 3 Literals)

= x + ( x’ y )

= ( x + x’ ) ( x + y )

= ( 1 ) ( x + y ) = x + y ( 2 Literals)

Distributive law (+ over •)

Complement of a FunctionComplement of a Function

DeMorgan’s Theorm

Duality & Literal Complement

CBAF

CBAF

CBAF

CBAF CBAF

CBAF

Standard FormsStandard Forms

• A Boolean function can be written algebraically in a variety of ways

• Standard form: is a standard algebraic expression of the function:• Help simplification procedures and frequently results in

more desirable logic circuits (e.g. less number of gates)

• Standard form: contains product terms and sum terms• Product term: X’Y’Z (logical product with AND)

• Sum term: X + Y + Z’ (logical sum with OR)

Minterms and MaxtermsMinterms and Maxterms

• A minterm: a product term in which all variables (or literals) of the function appear exactly once (complemented or not complemented)

• A maxterm: a sum term in which all the variables (or literals) of the function appear exactly once (complemented or not complemented)

• A function of n variables – have 2n possible minterms and 2n possible maxterms

• Example: for the function F(X,Y,Z),

• the term X’Y is not a minterm, but XYZ’ is a minterm

• The term X’+Z is not a maxterm, but X+Y’+Z’ is maxterm

MintermsMinterms

Minterms for two variables F(X,Y)

X Y Product Terms

Symbol m0 m1 m2 m3

0 0 X’Y’ m0 1 0 0 0

0 1 X’Y m1 0 1 0 0

1 0 XY’ m2 0 0 1 0

1 1 XY m3 0 0 0 1

Variable complemented if 0

Variable uncomplemented if 1

mi indicated the ith minterm

For each binary combination of X and Y there is a minterm

The index of the minterm is specified by the binary combination

mi is equal to 1 for ONLY THAT combination

MaxtermsMaxterms

Maxterms for two variables F(X,Y)

X Y Sum

TermsSymbol M0 M1 M2 M3

0 0 X+Y M0 0 1 1 1

0 1 X+Y’ M1 1 0 1 1

1 0 X’+Y M2 1 1 0 1

1 1 X’+Y’ M3 1 1 1 0

Variable complemented if 1

Variable not complemented if 0

Mi indicated the ith maxterm

For each binary combination of X and Y there is a maxterm

The index of the maxterm is specified by the binary combination

Mi is equal to 0 for ONLY THAT combination

ExampleExample

• A Boolean function can be expressed algebraically from a give truth table by forming the logical sum of ALL the minterms that produce 1 in the function

Example:X Y Z m F0 0 0 m0 1

0 0 1 m1 0

0 1 0 m2 1

0 1 1 m3 0

1 0 0 m4 0

1 0 1 m5 1

1 1 0 m6 0

1 1 1 m7 1

Consider the function defined by the truth table

F(X,Y,Z) 3 variables 8 mintermsF can be written asF = X’Y’Z’ + X’YZ’ + XY’Z + XYZ, or = m0 + m2 + m5 + m7

= m(0,2,5,7)

Minterms and Maxterms Minterms and Maxterms

• In general, a function of n variables has• 2n minterms: m0, m1, …, m2

n-1

• 2n maxterms: M0, M1, …, M2n

-1

• mi’ = Mi or Mi’ = mi

Example: for F(X,Y): m2 = XY’ m2’ = X’+Y = M2

Example (Cont.)Example (Cont.)

• A Boolean function can be expressed algebraically from a give truth table by forming the logical product of ALL the maxterms that produce 0 in the function

X Y Z M F F’0 0 0 M0 1 0

0 0 1 M1 0 1

0 1 0 M2 1 0

0 1 1 M3 0 11 0 0 M4 0 11 0 1 M5 1 0

1 1 0 M6 0 1

1 1 1 M7 1 0

• Example:Consider the function defined by the

truth tableF(X,Y,Z) in a manner similar to

the previous example, F’ can be written as

F’ = m1 + m3 + m4 + m6

= m(1,3,4,6)Now apply DeMorgan’s rule F = F’’ = [m1 + m3 + m4 + m6]’ = m1’.m3’.m4’.m6’ = M1.M3.M4.M6

= M(1,3,4,6)

Note the indices in this list are those that are

missing from the previous list in m(0,2,5,7)

Expressing Functions with Minterms Expressing Functions with Minterms and Maxtermsand Maxterms

F = X . ( Y’ + Z)

X Y Z F=X. (Y’+Z)

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 0

1 1 1 1

F = XY’Z’ + XY’Z + XYZ

= m4 + m5 + m7

= m(4,5,7)

F = M(0,1,2,3,6)

SummarySummary

• A Boolean function can be expressed algebraically as:• The logical sum of minterms• The logical product of maxterms

• Given the truth table, writing F as• mi – for all minterms that produce 1 in the table,

or Mi – for all maxterms that produce 0 in the table

• Another way to obtain the mi or Mi is to use ALGEBRA – see next example

ExampleExample

• Write E = Y’ + X’Z’ in the form of mi and Mi?

• Solution: Method1 First construct the

Truth Table as shown

Second:E = m(0,1,2,4,5), andE = M(3,6,7)

X Y Z m M E0 0 0 m0 M0 1

0 0 1 m1 M1 1

0 1 0 m2 M2 1

0 1 1 m3 M3 0

1 0 0 m4 M4 1

1 0 1 m5 M5 1

1 1 0 m6 M6 0

1 1 1 m7 M7 0

Example (Cont.)Example (Cont.)

Solution: Method2_aE = Y’ + X’Z’

= Y’(X+X’)(Z+Z’) + X’Z’(Y+Y’)= (XY’+X’Y’)(Z+Z’) + X’YZ’+X’Z’Y’= XY’Z+X’Y’Z+XY’Z’+X’Y’Z’+ X’YZ’+X’Z’Y’= m5 + m1 + m4 + m0 + m2 + m0

= m0 + m1 + m2 + m4 + m5

= m(0,1,2,4,5)

To find the form Mi, consider the remaining indices

E = M(3,6,7)

Solution: Method2_bE = Y’ + X’Z’E’ = Y(X+Z)

= YX + YZ= YX(Z+Z’) + YZ(X+X’)= XYZ+XYZ’+X’YZ

E = (X’+Y’+Z’)(X’+Y’+Z)(X+Y’+Z’)= M7 . M6 . M3 = M(3,6,7)

To find the form mi, consider the remaining indices

E = m(0,1,2,4,5)

SOP vs POSSOP vs POS

• Boolean function can be represented as Sum of Products (SOP) or as Product of Sums (POS) of their input variables• AB+CD = (A+C)(B+C)(A+D)(B+D)

• The sum of minterms is a special case of the SOP form, where all product terms are minterms

• The product of maxterms is a special case of the POS form, where all sum terms are maxterms

• SOP and POS are referred to as the standard forms for representing Boolean functions

SOP vs POSSOP vs POS

SOP POS

F = AB + CD

= (AB+C)(AB+D)

= (A+C)(B+C)(AB+D)

= (A+C)(B+C)(A+D)(B+D)

Hint 1: Used X+YZ=(X+Y)(X+Z)

Hint 2: Factor

POS SOP

F = (A’+B)(A’+C)(C+D)

= (A’+BC)(C+D)

= A’C+A’D+BCC+BCD

= A’C+A’D+BC+BCD

= A’C+A’D+BCHint 1: Used (X+Y)(X+Z)=X+YZ

Hint 2: Multiply

Implementation of SOPImplementation of SOP

Any SOP expression can be implemented using 2-levels of gates

The 1st level consists of AND gates, and the 2nd level consists of a single OR gate

Also called 2-level Circuit

Implementation of POSImplementation of POS

Any POS expression can be implemented using 2-levels of gates

The 1st level consists of OR gates, and the 2nd level consists of a single AND gate

Also called 2-level Circuit

Implementation of SOPImplementation of SOP

• Consider F = AB + C(D+E)• This expression is NOT in the sum-of-

products form• Use the identities/algebraic manipulation to

convert to a standard form (sum of products), as in F = AB + CD + CE

• Logic Diagrams:

F

E

C

D

C

B

A

FC

B

A

E

D

3-level circuit 2-level circuit

Logic OperatorsLogic Operators

AND

NAND (Not AND) x y NAND0 0 10 1 11 0 11 1 0

x y AND0 0 00 1 01 0 01 1 1

xy

x • y

xy

x • y

Logic OperatorsLogic Operators

OR

NOR (Not OR)

x y OR0 0 00 1 11 0 11 1 1

x y NOR0 0 10 1 01 0 01 1 0

xy

x + y

xy

x + y

Logic OperatorsLogic Operators

XOR (Exclusive-OR)

XNOR (Exclusive-NOR) (Equivalence)

x y XOR0 0 00 1 11 0 11 1 0

x y XNOR0 0 10 1 01 0 01 1 1

xy

x Å yx � y

x y + x y

xy

x Å yx y + x y

Logic OperatorsLogic Operators

NOT (Inverter)

Buffer

x NOT

0 1

1 0

x Buffer

0 0

1 1

x x

x x

DeMorgan’s Theorem on GatesDeMorgan’s 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

AssignmentsAssignments

Mano

● Chapter 2

♦ 2-4

♦ 2-5

♦ 2-6

♦ 2-8

♦ 2-9

♦ 2-10

♦ 2-12

♦ 2-15

♦ 2-18

♦ 2-19

AssignmentsAssignments

Mano2-4 Reduce the following Boolean expressions to the indicated

number of literals:

(a) A’C’ + ABC + AC’ to three literals

(b) (x’y’ + z)’ + z + xy + wz to three literals

(c) A’B (D’ + C’D) + B (A + A’CD) to one literal

(d) (A’ + C) (A’ + C’) (A + B + C’D) to four literals

2-5 Find the complement of F = x + yz; then show thatFF’ = 0 and F + F’ = 1

AssignmentsAssignments

2-6 Find the complement of the following expressions:

(a) xy’ + x’y (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’ + y’z

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.

AssignmentsAssignments

2-10 Draw the logic diagrams for the following Boolean expressions:

(a) Y = A’B’ + B (A + C) (b) Y = BC + AC’

(c) Y = A + CD (d) Y = (A + B) (C’ + D)

2-12 Simplify the Boolean function T1 and T2 to a minimum number of literals.

A B C T1 T2

0 0 0 1 00 0 1 1 00 1 0 1 00 1 1 0 11 0 0 0 11 0 1 0 11 1 0 0 11 1 1 0 1

AssignmentsAssignments

2-15 Given the Boolean functionF = xy’z + x’y’z + w’xy + wx’y + 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 literals using 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).

AssignmentsAssignments

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 + C’D)

(b) x’ + x (x + y’) (y + z’)