logic circuits and switching theory

7/27/2019 Logic Circuits and Switching Theory

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Louie Angelo M. Jalandoni


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The mathematical representation of numerals

according to its bases.


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Binary base 2 ( 0 and 1 )

Octal base 8 ( 0 7 )

Decimal base 10 ( 0 9 ) Hexadecimal base 16 ( 0 9 ) ( A F )


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Binary Decimal Octal Hexadecimal

0000 0 0 0

0001 1 1 1

0010 2 2 2

0011 3 3 3

0100 4 4 4

0101 5 5 5

0110 6 6 6

0111 7 7 7

1000 8 10 8


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Binary Decimal Octal Hexadecimal

1001 9 11 9

1010 10 12 A

1011 11 13 B

1100 12 14 C

1101 13 15 D

1110 14 16 E

1111 15 17 F


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1. Divide the decimal number to its base

2. Write the remainder on the right side

3. Repeat Step 1 and 2 until the quotientbecame zero

4. Read all the remainders from bottom to top.


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Base Decimal Remainder

2 50 0

2 25 1

2 12 0

2 6 0

2 3 1

2 1 1

0 0

When the remainder will be read from bottom to top it

will be 01100102


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Convert the following decimal values in binary,

octal and hexadecimal

1. 425

2. 330

3. 927


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0110010

0 x 20

1 x 21

1 x 24

1 x 25

Add up the product of the 1s multiplied to base raise to thepositional value then the value will be

0110010 = 2 + 16 + 32 = 50


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0110010

Remember that the left most bit is the most

significant bit and the right most is the least

significant bit

Group the number by three starting from the LSB

until you reach the MSB

0 / 110 / 010 = 0 6 2 8


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0110010

Do the same process on Binary to Octal but groupthe term by four.

011 / 0010 Convert the number using the binary system to

decimal again.

011 = 3 0010 = 2

Then the answer is 3216


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Convert each character into binary composing 3

bits binary number

6 2

110 010

Then combine the converted number

1100102


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The same on Binary but the base will be 8

instead of

62

2 x 81 = 2

6 x 82

= 48 Add the values and the answer will be 5010


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Convert each character into binary composing 4

bits binary number

3 2

0011 0010

Then combine the converted number

1100102


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The same on Binary but the base will be 16

instead of

32

2 x 161 = 2

3 x 162

= 48 Add the values and the answer will be 5010


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CONVERSION of FRACTIONAL PART (From Decimal to any Bases)

STEPS:1. Multiply the given Decimal number by the base-r of the nummber into which the given decimal no. is to be

converted to.

2. Repeat the proceeding process until the fractions becomes Zero or until the number of digits have sufficientaccuracy

3. Generate the Final answer from the integral part from TOP to BOTTOM.

Example: 0.125 >>>Binary=(001)2

INTEGER FRACTIONAL PART0.125x2 0 0.25

0.25x2 0 0.5

0.5x2 1 0

INTEGER FRACTIONAL PART

0.125 >>>OCTAL = (0.1)8 1 0

INTEGER FRACTIONAL PART

0.125 >>>HEXADECIMAL =(0.2)16 2 0


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Used in Digital Computers for simplifying

subtraction operation and for logical

operation.

Radix Complementrs complement =10

Diminished Radix Complement r -1s

complement = 9


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To get the 1s complement, just invert the

values of the term

To get the 2s complement, just add 1 to the

1s complement.


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STEPS in Binary Subtraction Using 1s Complement

1. Copy the minuend

2. Get the 1s complement of the subtrahend then add it to

the minuend

3. If there is an end carry, Add 1 to the sum otherwise, get

the 1s complement of the sum then prefix a NEGATIVE

sign (-).


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STEPS in Binary Subtraction Using 2s

Complement

The twos complement of a Binary number

is obtained by getting its Ones complement

then adding 1 to Binary.


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Binary Logic Deals with variable that take on two discrete values and with

operations that assume logical meaning.

Used to describe in mathematical way the manipulation &

processing Binary information.

LOGIC 1 LOGIC 0

HIGH LOW

+5V 0V

OPEN CLOSE

ON OFF

TRUE FALSE


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Variables are represented by a single letter havingonly two values: 1 or 0.

There are three basic logic operation:

And Operation represented by a dot or anabsence of operation. Z = X.Y or Z = XY

Or Operation represented by plus sign.

Z = X+Y Not Operation represented by a prime or bar

Z = X


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LOGIC GATES

blocks of hardware that produce a logic output signal if

the input requirement has been satisfied.

LOGIC CIRCUIT The interconnection of gates to achieve a prescribed

outcome

TRUTH TABLE Tabulations of all possible combinations of input & its

corresponding output


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The answer is false when one is false

Truth Table

Equation: Z = XY

X Y Z (OUTPUT)

0 0 0

0 1 0

1 0 0

1 1 1


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The answer is true when one is true

Truth table

Equation: Z = X+Y

X Y Z (OUTPUT)

0 0 0

0 1 1

1 0 1

1 1 1


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Inverse value

Truth Table

Equation: Z = X

X Z (OUTPUT)

0 1

1 0


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Nand Gate

Nor Gate


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The answer is true when one is false

Truth Table

Equation: Z = (XY)

X Y Z (OUTPUT)0 0 1

0 1 1

1 0 1

1 1 0


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The answer is false when one is true

Truth table

Equation: Z = (X+Y)

X Y Z (OUTPUT)

0 0 1

0 1 0

1 0 0

1 1 0


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EX OR

EX NOR


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The answer is true when the inputs are different

Truth table

Equation: Z = XY + XY

X Y Z (OUTPUT)0 0 0

0 1 1

1 0 1

1 1 0


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The answer is true when the inputs are same

Truth table

Equation: Z = XY + XY

X Y Z (OUTPUT)0 0 1

0 1 0

1 0 0

1 1 1


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Summary truth tables

The summary truth tables below show the output states

input for all types of 2-input and 3-gates:

A B AND NAND OR NOR EX-OR EX-NOR

0 0 0 1 0 1 0 1

0 1 0 1 1 0 1 0

1 0 0 1 1 0 1 0

1 1 1 0 1 0 0 1

A B C AND NAND OR NOR

0 0 0 0 1 0 1

0 0 1 0 1 1 0

0 1 0 0 1 1 0

0 1 1 0 1 1 0

1 0 0 0 1 1 0

1 0 1 0 1 1 0

1 1 0 0 1 1 0

Summary for all 2-input gates

Inputs Output of each gate

Summary for all 3-input gates

Note : that EX-OR and EX-NOR

gates can only have 2 inputs.

Inputs Output of each gate


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Type Distinctive shape Rectangular shape

Booleanalgebra

between A

& B

Truth table

AND

INPUT OUTPUT

A B A AND B

0 0 0

0 1 0

1 0 0

1 1 1

OR A + B

INPUT OUTPUT

A B A OR B

0 0 0

0 1 1

1 0 1

1 1 1

NOT

INPUT OUTPUT

A NOT A

0 1

1 0

In electronics a NOT gate is more commonly called an inverter. The circle on the symbol iscalled a bubble , and is generally used in circuit diagrams to indicate an inverted input oroutput.

NAND

INPUT OUTPUT

A B A NAND B

0 0 1

0 1 1

1 0 1

1 1 0

NOR

INPUT OUTPUT

A B A NOR B

0 0 1

0 1 0

1 0 0

1 1 0


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XOR

INPUTOUTPUT

A B A XOR B

0 0 0

0 1 1

1 0 1

1 1 0

XNOR

INPUT OUTPUT

A B A XNOR B

0 0 1

0 1 0

1 0 0

1 1 1

Two more gates are the exclusive-OR or XOR function and its inverse, exclusive-NOR or XNOR. The two input Exclusive-OR is true only when the two input valuesare different, false if they are equal, regardless of the value. If there are more thantwo inputs, the gate generates a true at its output if the number of trues at its input isodd.. In practice, these gates are built from combinations of simpler logic gates.


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BASIC

GATES

NAND Gate

ImplementationNOR Gate Implementation

NOT Gate

OR Gate

AND Gate


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A Mathematical notation used to represent the function

of the Digital circuit.

A notation that allows variables & constants to have only

2 possible values 0 & 1.

The Term Boolean Algebra honors a fascinating

English mathematician; George Boole


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- An expression formed with Binary variables

the two operators OR & AND & a

UNARY operator not parenthesis & equal

sign for the given variables the functioncan either be One or Zero.


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EQUATION No. BOOLEAN EQUATION DESCRIPTION

1X + Y = Y + X

Commutative Property

XY = YX

2X + (Y + Z) = (X + Y) + Z

Associative PropertyX(YZ) = (XY)Z

3X X = X

Idempotent PropertyX + X = X

4 X 1 = X Identity PropertyX + 1 = 1

5X 0 = 0

Null PropertyX + 0 = X

6X (Y + Z) = XY + XZ

Distributive Property

(XY) + (XZ) = X + YZ

7X X = 0

Negation PropertyX + X = 1

8 (X) = X Double Negation Property

9

X + XY = X

Absorption PropertyX (X + Y) = XX + (XY) = X + Y


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Y= C+CA + CB +C

Using commutative law

Y=CA + CB +C+C

Using Idempotent law where

Y = CA + CB + 1

OR Condition of 1Y = 1


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Is equivalent to

xy = x + y

(x+y ) = xy


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Canonical forms boolean functions

expressed in sum of minterms or product of

maxterms.


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a single variable or product of variable which

may or may not be complemented. Denoted

by a lower case m, the equation is anded

F= AB +AC+ ABC + AB


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A single variable or sum of variables, denoted

by an Uppercase M and the equation is ored.

F = (A+C+D)(A+D)(D+A)


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X Y Z MINTERMS MAXTERMS

DESIGNATION DESIGNATION

0 0 0 X'Y'Z' m0 X+Y+Z M0

0 0 1 X'Y'Z m1 X+Y'+Z' M1

0 1 0 X'YZ' m2 X+Y+Z' M2

0 1 1 X'YZ m3 X+Y'+Z M3

1 0 0 XYZ m4 X'+Y+Z M4

1 0 1 X'Y'Z' m5 X+Y+Z' M5

1 1 0 XY'Z' m6 X'+Y'+Z M61 1 1 XYZ m7 X'+Y'+Z' M7

0's to express 1's to express

MINTERMS & MAXTERMS WITH 3 BINARY VARIABLES


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MINTERM:

F = ABC + ABC + A B C

F(A,B,C) =m3 + m4 + m1

MAXTERM: F = (A+B+C) +( A+B+C )+ (A+ B +C)

F(A,B,C) =M4 . M7 . M0


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Standard Form of Equation:

The term that form the function may be one or any number or literal.

TWO TYPES:

Sum of Product (SOP)

A Boolean expression containing AND terms called PRODUCT of TERMS (one

or more literals)

Ex: F=Y + XY+XYZ

PRODUCT OF SUM (POS)

A Boolean expression containing OR terms called sum term.

Each term may have any number of literals

Ex: F=(X+Y)(Y+Z)(X+YZ)


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Map a diagram made up of squares represents one minterm

KARNAUGH MAP

A chart or grid containing boxes called cells; each which represents one minterm.

TYPES of MAP

1. Two Variable Map

Consists of two variables

2. Three Variable Map

A three variable map plotted in a map

3. FOUR Variable Map


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Steps: Construct the K-Map & place 1s in the squares corresponding to 1s in truth table ;

place zeros in the other squares.

Examine the map for adjacent 1s & loop those 1s which are not adjacent to any other

ones.

Looping continue as there are pairs octet or quad that contains 1. You can still loop the

one that is already looped if there are still other 1s left.

Form the OR sum of all terms generated by each loop.


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2 Variable Karnaugh Map

2 INPUT OR GATE


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3-variable Karnaugh maps 3-variable examples

Here is the truth table for a 3-input system 1. Simplify the following expression using a Karnaugh map:

inpu t C in pu t B inpu t A o utpu t

0 0 0 0

0 0 1 0 You may be able to tell what is going to happen by completing the

0 1 0 0

0 1 1 1 The Boolean statement is:

1 0 0 0

1 0 1 1

1 1 0 11 1 1 1 The truth table is:

This is converted into a Karnaugh map, as follows: in pu t C in pu t B in pu t A o utpu t

0 0 0 0

0 0 1 10 1 0 1

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 0

The Karnaugh map is:

This is the exclusive OR function. The value of C is irrelevant.


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B B

A m0 m1

A m2 m3


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CD CD CD CDAB m0 m1 m3 m2

AB m4 m5 m7 m6

AB m12 m13 m15 m14

AB m8 m9 m11 m10


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logic circuits and switching theory

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