Chap4_boolean Algebra and Logic Gates

CHAPTER 4

BOOLEAN ALGEBRA

AND

LOGIC GATES

Computer Systems Organization FlowComputer Systems Organization Flow

IPOS (Input, Process, Output, Storage)

Process and Storage (CPU, Registers, Memory)

CPU (Binary Numbers)

Binary Numbers (Encoded as Signals in Digital Form known as Data Encoding)

Data Encoding (Made and interpret by Transistors that made up the CPU and act like switches)

Switches (Act in accordance to Binary Values)

22 / 28 / 28

Binary Logic and GatesBinary Logic and Gates

Binary variables take on one of two values.

Logical operators operate on binary values and binary variables.

Basic logical operators are the logic functions AND, OR and NOT.

Logic gates implement logic functions.

Boolean Algebra: a useful mathematical system for specifying and transforming logic gates.

We study Boolean algebra as a foundation for designing and analyzing COMPUTER SYSTEMS!

LEVELS OF MODERN COMPUTING SYSTEMSLEVELS OF MODERN COMPUTING SYSTEMS

44 / 28 / 28

Binary VariablesBinary Variables

Recall that the two binary values have different names:

● True/False

● On/Off

● Yes/No

● 1/0

We use 1 and 0 to denote the two values.

Variable identifier examples:

● A, B, y, z, or X1 for now

● RESET, START_IT, or ADD1 later

Logical OperationsLogical Operations

The three basic logical operations are:

● AND

● OR

● NOT

AND is denoted by a dot (·).

OR is denoted by a plus (+).

NOT is denoted by an overbar ( ¯ ), a single quote mark (') after, or (~) before the variable.

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 77 / 28 / 28

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

88 / 28 / 28

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 = 199 / 28 / 28

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 1010 / 28 / 28

Operator PrecedenceOperator Precedence

Parentheses

( . . . ) • ( . . .)

NOT

x’ + y

AND

x + x • y

OR

1111 / 28 / 28

])([ xwzyx

])([

)(

)(

)(

)(

xwzyx

xwzy

xwz

xw

xw

DeMorgan’s TheoremDeMorgan’s Theorem

1212 / 28 / 28

)]([ 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

1313 / 28 / 28

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

Boolean Algebra and Logic Boolean Algebra and Logic Gates Gates 1414

Using Switches

● Inputs:

♦ logic 1 is switch closed

♦ logic 0 is switch open

● Outputs:

♦ logic 1 is light on

♦ logic 0 is light off.

● NOT input:

♦ logic 1 is switch open

♦ logic 0 is switch closed

Logic Function ImplementationLogic Function Implementation

Switches in series => AND

Switches in parallel => OR

CNormally-closed switch => NOT


Example: Logic Using Switches

Light is on (L = 1) for

L(A, B, C, D) =

and off (L = 0), otherwise.

Useful model for relay and CMOS gate circuits, the foundation of current digital logic circuits

Logic Function Implementation – cont’dLogic Function Implementation – cont’d

B

A

D

C

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


Logic GatesLogic Gates

In the earliest computers, switches were opened

and closed by magnetic fields produced by

energizing coils in relays. The switches in turn

opened and closed the current paths.

Later, vacuum tubes that open and close current

paths electronically replaced relays.

Today, transistors are used as electronic

switches that open and close current paths.


Logic Gate Symbols and BehaviorLogic Gate Symbols and Behavior Logic gates have special symbols:

And waveform behavior in time as follows:X 0 0 1 1

Y 0 1 0 1

X · Y(AND) 0 0 0 1

X+ Y(OR) 0 1 1 1

(NOT) X 1 1 0 0

OR gate

X

YZ= X+ Y

X

YZ= X · Y

AND gate

X Z= X

NOT gate orinverter

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)1818 / 28 / 28

Distributive law (+ over •)

Complement of a FunctionComplement of a Function

DeMorgan’s Theorm

Duality & Literal Complement

1919 / 28 / 28

CBAF

CBAF

CBAF

CBAF CBAF

CBAF

Canonical FormsCanonical Forms

Minterm

● Product (AND function)

● Contains all variables

● Evaluates to ‘1’ for aspecific combination

Example

A = 0 A B C

B = 0 (0) • (0) • (0)

C = 0

2020 / 28 / 28

1 • 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


Maxterm

● Sum (OR function)

● Contains all variables

● Evaluates to ‘0’ for aspecific combination

Example

A = 1 A B C

B = 1 (1) + (1) + (1)

C = 1

2121 / 28 / 28

0 + 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


Truth Table to Boolean Function

2222 / 28 / 28

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 1

1 1 0 0

1 1 1 1

CBAF CBA CBA ABC


Sum of Minterms

Product of Maxterms

2323 / 28 / 28

A B C F

0 0 0 0 0

1 0 0 1 1

2 0 1 0 0

3 0 1 1 0

4 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

1

0

0

1

0

Standard FormsStandard Forms

Sum of Products (SOP)

2424 / 28 / 28

ABCCBACBACBAF

AC

BBAC

)(

CB

AACB

)(

BA

BA

CCBA

)1(

)(

)()()( BBACCCBAAACBF

ACBACBF

Standard FormsStandard Forms

Product of Sums (POS)

2525 / 28 / 28

)( AACB

)( BBCA

)( CCBA

)()()( AACBCCBABBCAF

CBBACAF

CABBCACBACBAF

))()(( CBBACAF

TwoTwo -- Level ImplementationsLevel Implementations

Sum of Products (SOP)

Product of Sums (POS)

2626 / 28 / 28

ACBACBF

))()(( CBBACAF

B’C

FB’A

AC

AC

FB’A

B’C

Logic OperatorsLogic Operators

AND

NAND (Not AND)

2727 / 28 / 28

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


OR

NOR (Not OR)

2828 / 28 / 28

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


XOR (Exclusive-OR)

XNOR (Exclusive-NOR) (Equivalence)

2929 / 28 / 28

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


NOT (Inverter)

Buffer

3030 / 28 / 28

x NOT

0 1

1 0

x Buffer

0 0

1 1

x x

x x

Multiple Input GatesMultiple Input Gates

3131 / 28 / 28

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

3232 / 28 / 28 Change the “Shape” and “bubble” all lines

HomeworkHomework

Mano

● Chapter 2

♦ 2-4

♦ 2-5

♦ 2-6

♦ 2-8

♦ 2-9

♦ 2-10

♦ 2-12

♦ 2-15

♦ 2-18

♦ 2-19 3333 / 28 / 28

HomeworkHomework

Mano

3434 / 28 / 28

2-4 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

HomeworkHomework

3535 / 28 / 28

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.

HomeworkHomework

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

3636 / 28 / 28

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.

HomeworkHomework

3737 / 28 / 28

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).

HomeworkHomework

3838 / 28 / 28

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’)

Chap4_boolean Algebra and Logic Gates

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