MC9211 Computer Organization

Unit 1 : Digital Fundamentals

Lesson2 : Boolean Algebra and Simplification

(KSB) (MCA) (2009- 12/ODD) (2009 - 10/1 A&B)

KSB-1601-07 2

Coverage – Lesson2

Introduces the basic postulates of Boolean Algebra and shows the correlation between Boolean expressions and their corresponding logic diagrams.All possible logic operations for two variables are investigated and from that, the most useful logic gates used in the design of digital systems are determined.

KSB-1601-07 3

Lesson2 – Digital Logic Circuits

1. Basic definitions2. Basic theorems and properties of Boolean

algebra3. Boolean Functions4. Canonical and standard forms5. Other logic operations6. Digital Logic Gates

KSB-1601-07 4

1.Basic Definitions Boolean Algebra deals with binary variables and

logic operationsBinary variables are represented by A, B, x, y etcLogic Operations are AND , OR , NOT etc

A Boolean function can be expressed algebraically with binary variables, the logic operation symbols, parentheses and equal sign ( ex: F = x + y’z )

For a given value of variables the function can be either 0 or 1

The relationship between a function and its binary variables can be represented in a truth table

KSB-1601-07 5

Boolean Algebra (contd..)From truth table a Boolean function can be obtained (and vice versa)A Boolean function can be transformed from an algebraic expression in to a logic diagram composed of AND, OR, NOT etc gates

KSB-1601-07 6

Boolean Algebra (contd..)x y z F0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 1

TruthTable

BooleanFunction

LogicDiagram

F = x + y’z

xy

z

F

KSB-1601-07 7

Boolean Algebra (contd..)

The purpose of Boolean Algebra is to facilitate the analysis and design of digital circuits

It provides a convenient tool to 1.Express in algebraic form a truth table

relationship between binary variables2.Express in algebraic form the input-output

relationship of logic diagrams3.Find simpler circuits for the same function

KSB-1601-07 8

2.Basic Theorems of Boolean Algebra1.Different letters in Boolean expressions are

called variablesExample: A.B’ + A’.C +A.(D+E) has 5 variables

2.Each occurrence of a variable or its complement is called a literal

Example: in the above expression there are 7 literals

3.Two expressions are equivalent if one expression equals 1 only when the other equals 1, and one equals 0 only when the other equals 0

KSB-1601-07 9

Basic Definitions (contd..)4.Two expressions are complements of each

other if one expression equals 1 only when the other equals 0, and vice versa

The complement of a Boolean expression is obtained bychanging all .’s to +’schanging all +’s to .’schanging all 1’s to 0’schanging all 0’s to 1’s

and complementing each literalExample: 1.A + B’C + 0 ?

KSB-1601-07 10

Basic Definitions (contd..)5.The dual of a Boolean expression is obtained

bychanging all .’s to +’schanging all +’s to .’schanging all 1’s to 0’schanging all 0’s to 1’s

but not complementing any literalExample: 1.A + B’C + 0 ?

KSB-1601-07 11

Postulates 1. X = 1 or else X = 02. AND 3. OR represented by . represented by +

X Y X+Y 1 + 1 = 11 + 0 = 10 + 1 = 10 + 0 = 0

X Y X.Y 1 . 1 = 11 . 0 = 00 . 1 = 00 . 0 = 0

KSB-1601-07 12

Basic Theorems1a. 0 . X = 0 1b. 1 + X = 1 2a. 1 . X = X 2b. 0 + X = X 3a. X X = X 3b.X + X = X4a. X X’ = 0 4b. X + X’ = 15a. X Y = Y X 5b. X + Y = Y + X6a.XYZ=X(YZ)=(XY)Z 6b.X+Y+Z= X+(Y+Z)= (X+Y)+Z DeMorgan’s Theorem7a. (X Y . . . Z)’ = X’ + Y’ + ……+Z’7b. (X + Y+ ……+Z) = X’ Y’ …..Z’

KSB-1601-07 13

Basic Theorems (contd..)8a.XY + XZ = X(Y+Z) 8b. (X+Y)(X+Z) = X + YZ9a. XY + XY’ = X 9b. (X+Y)(X+Y’) = X10a. X + XY = X 10b. X(X+Y) = X12a. X + X’Y = X + Y 12b. X (X’+Y) = XY13a. XY + X’Z + YZ = XY + X’Z13b. (X+Y)(X’+Z)(Y+Z) = (X+Y)(X’+Z)14a. XY + X’Z = (X+Z)(X’+Y)14b. (X+Y)(X’+Z) = XZ + X’Y15.a. X(Y+Z) = XY + XZ Distributive Law15b. X + YZ = (X+Y)(X+Z)

KSB-1601-07 14

Exercise 11.Determine by means of a truth table the

validity of DeMorgan’s theorem for three variables: (ABC)’ = A’ + B’ + C’

2.Simplify the following expressions using Boolean algebra a) A + AB b) AB + AB’c) A’BC + AC d) A’B + ABC’ + ABC

3.Simplify the following expressions using Boolean algebra a) AB + A(CD + CD’) b) (BC’ + A’D)(AB’ + CD’)

KSB-1601-07 15

Exercise 1 (contd..)

2c) Solution:A’BC + AC = A’BC + AC(B + B’)

= A’BC + ABC + AB’C= A’BC + ABC + ABC + AB’C= BC(A’ + A) + AC(B + B’)= BC + AC = C(A + B)

KSB-1601-07 16

Exercise 1 (contd..)

4.Using DeMorgan’s theorem , show that:a) (A + B )’(A’ + B’)’ = 0b) A + A’B + A’B’ = 1

5.Given the Boolean function F = x’y + xyz’ :a) Derive an algebraic expression for the complement F’b) Show that F.F’ = 0c) Show that F + F’ = 1

KSB-1601-07 17

Exercise 1 (contd..)6.Given the Boolean function

F = xy’z + x’y’z + xyza) List the truth table of the functionb) Draw the logic diagram using the original Boolean expressionc) Simplify the algebraic expression using Boolean algebrad) List the truth table of the function from the simplified expression and show that it is the same as the truth table 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)

KSB-1601-07 18

3. Boolean FunctionsA Binary variable can take value of 1 or 0A Boolean Function is an expression formed with

- Binary variables- Two binary operators OR and AND- A unary operator NOT- Parentheses and - An equal (=) sign

For a given value of the variables the function can be either 1 or 0

Example: F1 = xyz’

KSB-1601-07 19

Boolean Functions (contd..)Any Boolean function can be represented by a

TRUTH TABLEThe number of rows in the table is 2n where n is

the number of binary variables in the functionThe 1 and 0 combinations for each row is

obtained from binary numbers by counting from 0 to 2n -1

For each row of the table there is a value for the function equal to either 1 or 0

KSB-1601-07 20

Boolean Functions (contd..)Truth tables for Boolean functions

F1 = XYZ’F2 = X+Y’ZF3 = X’Y’ZF4 = XY’+X’Z

Is given on the next slide

KSB-1601-07 21

Boolean Functions (contd..)X Y Z F1 F2 F3 F4

0 0 0 0 0 0 0

0 0 1 0 1 1 1

0 1 0 0 0 0 0

0 1 1 0 0 0 1

1 0 0 0 1 0 1

1 0 1 0 1 0 1

1 1 0 1 1 0 0

1 1 1 0 1 0 0

KSB-1601-07 22

Boolean Functions (contd..)A Boolean Function may be transformed from

algebraic expression in to a logic diagram composed of AND, OR, and NOT gates

Exercise: Convert the four functions given previously in to logic gates

Algebraic manipulation: The minimization of number of literals and the number of terms results in a circuit with minimum equipment

KSB-1601-07 23

4. Canonical and Standard FormsThere are two canonical or standard forms by

which we can express any combinational logic network:

- The SUM of PRODUCTs form (SOP)- The PRODUCT of SUMs form (POS)

Before studying SOP and POS we have to study MINTERM and MAXTERM

KSB-1601-07 24

MINTERM A MINTERM of n variables is a product of the

variables in which each variable appears exactly once in true or complemented form

Each MINTERM is obtained from an AND term of the n variables , with each variable being

- primed if the corresponding bit of the binary number is 0 and

- unprimed if it is a 1The symbol for each MINTERM is of the form mj,

where j denotes the decimal equivalent of the binary number of the minterm designated

KSB-1601-07 25

MINTERM (contd..)A MINTERM of n variables is a product of the

variables in which each variable appears exactly once in true or complemented form

Each MINTERM is obtained from an AND term of the n variables , with each variable being

- primed if the corresponding bit of the binary number is 0 and

- unprimed if it is a 1The symbol for each MINTERM is of the form mj,

where j denotes the decimal equivalent of the binary number of the minterm designated

KSB-1601-07 26

MINTERM (contd..)X Y Z TERM Designation

0 0 0 X’Y’Z’ m0

0 0 1 X’Y’Z m1

0 1 0 X’YZ’ m2

0 1 1 X’YZ m3

1 0 0 XY’Z’ m4

1 0 1 XY’Z m5

1 1 0 XYZ’ m6

1 1 1 XYZ m7

KSB-1601-07 27

MINTERM (CONTD..)A Boolean function may be expressed

algebraically from a given truth table by - forming a minterm for each combination of the variables that produces a 1 in the function - and then taking OR of all those terms

KSB-1601-07 28

MINTERM ExampleX Y Z f1 f20 0 0 0 00 0 1 1 00 1 0 0 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

f1 = X’Y’Z + XY’Z’ + XYZ = m1 + m4 + m7

f1(X,Y,Z) = ∑ ( 1, 4 , 7)f2

f2 = X’YZ + XY’Z + XYZ’ + XYZ = m3 + m5 + m6 + m7

f2(X,Y,Z) = ∑ ( 3, 5 , 6, 7)

Note: Any Boolean function can be expressed as a sum of MITERM s

KSB-1601-07 29

MINTERM (contd..)Examples:1.Express Boolean Function F = A+B’C in a sum

of MINTERMs

KSB-1601-07 30

MAXTERM A MAXTERM of n variables is a sum of the

variables in which each variable appears exactly once in true or complemented form

Each MAXTERM is obtained from an OR term of the n variables , with each variable being- unprimed if the corresponding bit of the

binary number is 0 and- primed if it is a 1

The symbol for each MAXTERM is of the form Mj, where j denotes the decimal equivalent of the binary number of the maxterm designated

KSB-1601-07 31

MAXTERM (contd..)X Y Z TERM Designation

0 0 0 X+Y+Z M0

0 0 1 X+Y+Z’ M1

0 1 0 X+Y’+Z M2

0 1 1 X+Y’+Z’ M3

1 0 0 X’+Y+Z M4

1 0 1 X’+Y+Z’ M5

1 1 0 X’+Y’+Z M6

1 1 1 X’+Y’+Z’ M7

KSB-1601-07 32

MAXTERM ( contd..)A Boolean function may be expressed

algebraically from a given truth table by - forming a maxterm for each combination of the variables that produces a 0 in the function - and then taking AND of all those terms

KSB-1601-07 33

MAXTERM ExampleX Y Z f1 f20 0 0 0 00 0 1 1 00 1 0 0 00 1 1 0 11 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

f1 = (X+Y+Z)( X+Y’+Z)(X+Y’+Z’)(X’+Y+Z’)(X’+Y’+Z)= M0 + M2 +M3+M5+M6

f1(X,Y,Z) = π ( 0,2,3,5,6)

f2 = ?

Note: Any Boolean function can be expressed as a productof MAXTERM s

KSB-1601-07 34

MAXTERM (contd..)Examples:1.Express Boolean Function F = XY+X’Z in a

product of of MAXTERMs

KSB-1601-07 35

Conversion between canonical formsTo convert from canonical form to another

- interchange the symbols ∑ and π and- list those numbers missing from the original form

Example: F(X,Y,Z) = ∑ (1,3,6,7) = π ( 0,2,4,5)

Standard Form : The terms that form the function may contain one, two, or any number of literals in sum of product form or product of sum form Ex: F=Y’+XY+X’YZ’ (SoP) or

F =X(X’+Y)(X’Y’Z) (PoS)

KSB-1601-07 36

5 Other Logic Operations

X Y F0

F1

F2

F3

F4

F5

F6

F7

F8

F9

F10

F11

F12

F13

F14

F15

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

KSB-1601-07 37

KSB-1601-07 38

6. Digital Logic Gates

A gate is a circuit with one or more input signals but only one output signal

Gates are two-state digital circuits because the input and output signals are either low (0 Volts) or high (+5 Volts)

Gates are usually called logic circuits because they can be analyzed with Boolean algebra

KSB-1601-07 39

BASIC LOGIC GATEGate.

..

BinaryDigitalInputSignal

BinaryDigitalOutputSignal

Types of Basic Logic GatesCombinational Logic Gates

Logic Gates whose output logic value depends only on the input logic values

Sequential Logic GatesLogic Gates whose output logic value depends on the input values and the state (stored information) of the Gates

Functions of Gates can be described byTruth TableBoolean FunctionKarnaugh Map

There are two types of Combinational Logic Gates Basic gates – NOT , OR and ANDDerived gates – NOR, NAND, XOR, XNOR

KSB-1601-07 40

Basic Gates - AND GateDefinition: An AND gate has two or more input signals but only one output signal: all inputs must be high to get a high outputTruth table Three Input

A B C x

0 0 0 0

0 0 1 0

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 0

1 1 0 0

1 1 1 1

Two Input

A B x

0 0 0

0 1 0

1 0 0

1 1 1

Algebraic Function : x = A . B

A

BSymbol x

KSB-1601-07 41

Basic Gates - OR GateDefinition: An OR gate has two or more input signals but only one output signal: if any input signal is high , the output signal is highTruth table

1111

1011

1101

1001

1110

1010

1100

0000

XCBA

Three Input

Two InputA B X0 0 00 1 11 0 11 1 1

Algebraic Function : x = A + BA

BSymbol x

Basic gates – NOT or InverterDefinition:

An Inverter (NOT) is a gate with only one input signal and one output signal the output state is always the opposite of the input state

Algebraic Function : x = A’

Symbol

A x

A x0 11 0

Truth Table

KSB-1601-07 42

KSB-1601-07 43

Basic gates – Buffer

Definition: A Buffer is gate with only one input signal and one output signalthe output state is always the same as the input state the circuit is used only for power amplification

Symbol

A x

Truth Table

A x

0 0

1 1Algebraic Function : x = A

KSB-1601-07 44

Exercise 11. The following register has an output of 100101. Show

how to complement each bit.

6-Bit Register

1

0

0

1

0

1

2.Show the truth table of a 4-input OR gate

KSB-1601-07 45

3.What are the values of Y3 Y2 Y1 Y0 when each of the switches is pressed ?

KSB-1601-07 46

4. What does the following circuit do?

Y5 Y4 Y3 Y2 Y1 Y0

5.Write down the truth table for the following circuit

KSB-1601-07 47

KSB-1601-07 48

6. BCD Decoder: For each ofthe combinations of ABCDwhich output line is high

KSB-1601-07 49

Derived Gates - NOR GateDefinition: A NOR gate has two or more input signals but only one output signal: all input signals must be low to get a high outputTruth table

0111

0011

0101

0001

0110

0010

0100

1000

XCBA

Three InputTwo Input

A B X0 0 10 1 01 0 01 1 0

Algebraic Function : x = (A + B)’=A’.B’

SymbolA

Bx

KSB-1601-07 50

NOR Gate Application

xA

Control

Control = 0

A x0 11 0

Control = 1

A x0 01 0

Observations:1.When Control is 0, the output is the complement of input2.When Control is 1, the output is always 0

KSB-1601-07 51

Derived Gates - NAND GateDefinition: A NAND gate has two or more input signals but only one output signal: all input signals must be high to get a low outputTruth table

Algebraic Function : x = (A . B)’=A’+B’

Symbol

Two Input

A B X

0 0 1

0 1 1

1 0 1

1 1 0

0111

1011

1101

1001

1110

1010

1100

1000

XCBA

Three Input

AB x

KSB-1601-07 52

NAND Gate Application

xA

ControlControl = 0

A x0 11 1

Control = 1

A x0 11 0

Observations:1.When Control is 0, the output is always 12.When Control is 1, the output is the complement of input

KSB-1601-07 53

Derived Gates – XOR (Exclusive OR) GateDefinition: An XOR gate has two or more input signals but only one output signal: it recognizes words that have odd number of 1’sTruth table

1111

0011

0101

1001

0110

1010

1100

0000

XCBA

Three Input

Two Input

A B X

0 0 0

0 1 1

1 0 1

1 1 0

Algebraic Function : x = A B = AB’+A’B+

AB

xSymbol

KSB-1601-07 54

XOR Gate Application

xA

INVERT

INVERT = 1

A x0 11 0

INVERT = 0

A x0 01 1

Observations:1.When INVERT is 0, the output is same as input2.When Control is 1, the output is the complement of input

KSB-1601-07 55

XOR Application (contd..)Odd Parity Generator

KSB-1601-07 56

XOR Application (contd..)INVERTER

Derived Gates - XNOR (Exclusive NOR) GateDefinition: An XNOR gate has two or more input signals but only one output signal: it recognizes words that have even number of 1’s (also number with all 0’s)Truth table

0111

1011

1101

0001

1110

0010

0100

1000

XCBA

Three InputTwo Input

A B X

0 0 1

0 1 0

1 0 0

1 1 1

SymbolKSB-1601-07 57

AB x

Algebraic Function X = (A ⊕ B)’

Digital Logic Gates - Summary

A X X = (A + B)’

B

Name Symbol Function Truth Table

AND A X = A • B

X orB X = AB

0 0 00 1 01 0 01 1 1

0 0 00 1 11 0 11 1 1

OR A X X = A + B

B

NOT A X X = A 0 11 0

Buffer A X X = AA X0 01 1

NAND A X X = (AB)’

B

0 0 10 1 11 0 11 1 0

NOR 0 0 10 1 01 0 01 1 0

XORExclusive OR

A X = A ⊕ BX or

B X = A’B + AB’0 0 00 1 11 0 11 1 0

A X = (A ⊕ B)’X or

B X = A’B’+ AB

0 0 10 1 01 0 01 1 1

XNORExclusive NORor Equivalence

A B X

A B X

A X

A B X

A B X

A B X

A B X

KSB-1601-07 58