EKT 121 / 4ELEKTRONIK DIGIT 1

CHAPTER 1 : INTRODUCTION

1.0 Number & Codes

Digital and analog quantities Decimal numbering system (Base 10) Binary numbering system (Base 2) Hexadecimal numbering system (Base 16) Octal numbering system (Base 8) Number conversion Binary arithmetic 1’s and 2’s complements of binary numbers

Signed numbers Arithmetic operations with signed numbers Binary-Coded-Decimal (BCD) ASCII codes Gray codes Digital codes & parity

Digital and analog quantities

Two ways of representing the numerical values of quantities : i) Analog (continuous) ii) Digital (discrete)

Analog : a quantity represented by voltage, current or meter movement that is proportional to the value that quantity

Digital : the quantities are represented not by proportional quantities but by symbols called digits

Digital and analog systems

Digital system: combination of devices designed to manipulate logical

information or physical quantities that are represented in digital forms

include digital computers and calculators, digital audio/video equipments, telephone system.

Analog system: contains devices manipulate physical quantities that are

represented in analog form audio amplifiers, magnetic tape recording and playback

equipment, and simple light dimmer switch

Analog Quantities

• Continuous values

Digital Waveform

Introduction to Numbering Systems

We are all familiar with the decimal number system (Base 10). Some other number systems that we will work with are:

Binary Base 2 Octal Base 8 Hexadecimal Base 16

Number Systems

0 ~ 9

0 ~ 1

0 ~ 7

0 ~ F

Decimal

Binary

Octal

Hexadecimal

Characteristics of Numbering Systems

1) The digits are consecutive.2) The number of digits is equal to the size of the

base.3) Zero is always the first digit.4) When 1 is added to the largest digit, a sum of zero

and a carry of one results.5) Numeric values determined by the implicit

positional values of the digits.

00000000000000010000001000000011000001000000010100000110000001110000100000001001000010100000101100001100000011010000111000001111

000001002003004005006007010011012013014015016017

0123456789ABCDEF

0123456789

101112131415

BinaryOctalHexDecNUMBER SYSTEMS

Significant Digits

Binary: 11101101

Most significant digit Least significant digit

Hexadecimal: 1D63A7A

Most significant digit Least significant digit

Binary Number System

Also called the “Base 2 system” The binary number system is used to model the

series of electrical signals computers use to represent information

0 represents the no voltage or an off state 1 represents the presence of voltage or an

on state

Binary Numbering ScaleBase 2 Number Base 10 Equivalent Power Positional Value

000 0 20 1

001 1 21 2

010 2 22 4

011 3 23 8

100 4 24 16

101 5 25 32

110 6 26 64

111 7 27 128

Octal Number System

Also known as the Base 8 System Uses digits 0 - 7 Readily converts to binary Groups of three (binary) digits can be used to

represent each octal digit Also uses multiplication and division

algorithms for conversion to and from base 10

Hexadecimal Number System

Base 16 system Uses digits 0-9 &

letters A,B,C,D,E,F Groups of four bits

represent eachbase 16 digit

Number Conversion

Any Radix (base) to Decimal Conversion

Number Conversion Binary to Decimal Conversion

Binary to Decimal Conversion

Convert (10101101)2 to its decimal equivalent:

Binary 1 0 1 0 1 1 0 1

Positional Values

xxxxxxxx2021222324252627

128 + 32 + 8 + 4 + 1Products

17310

Octal to Decimal Conversion

Convert 6538 to its decimal equivalent:

6 5 3xxx

82 81 80

384 + 40 + 3

42710

Positional Values

Products

Octal Digits

Hexadecimal to Decimal Conversion

Convert 3B4F16 to its decimal equivalent:

Hex Digits 3 B 4 Fxxx

163 162 161 160

12288 +2816 + 64 +15

15,18310

Positional Values

Products

x

Number Conversion Decimal to Any Radix (Base)

Conversion

1. INTEGER DIGIT: Repeated division by the radix & record the remainder

2. FRACTIONAL DECIMAL:Multiply the number by the radix until the answer is in integer

Example:25.3125 to Binary

Decimal to Binary Conversion

2 5 = 12 + 1 2

1 2 = 6 + 0 2

6 = 3 + 0 2

3 = 1 + 1 2

1 = 0 + 1 2 MSB LSB 2510 = 1 1 0 0 1 2

Remainder

Decimal to Binary Conversion

Carry . 0 1 0 10.3125 x 2 = 0.625 0 0.625 x 2 = 1.25 1

0.25 x 2 = 0.50 0

0.5 x 2 = 1.00 1

The Answer: 1 1 0 0 1.0 1 0 1

MSB LSB

Decimal to Octal Conversion

Convert 42710 to its octal equivalent:

427 / 8 = 53 R3 Divide by 8; R is LSD

53 / 8 = 6 R5 Divide Q by 8; R is next digit

6 / 8 = 0 R6 Repeat until Q = 0

6538

Decimal to Hexadecimal Conversion

Convert 83010 to its hexadecimal equivalent:

830 / 16 = 51 R14

51 / 16 = 3 R3

3 / 16 = 0 R3

33E16

= E in Hex

Number Conversion

Binary to Octal Conversion (vice versa)

1. Grouping the binary position in groups of three starting at the least significant position.

Octal to Binary Conversion

Each octal number converts to 3 binary digits

To convert 6538 to binary, just substitute code:

6 5 3

110 101 011

Number Conversion

Example: Convert the following binary numbers to

their octal equivalent (vice versa).

a) 1001.11112 b) 47.38

c) 1010011.110112

Answer:

a) 11.748

b) 100111.0112

c) 123.668

Number Conversion

Binary to Hexadecimal Conversion (vice versa)

1. Grouping the binary position in 4-bit groups, starting from the least significant position.

Binary to Hexadecimal Conversion

The easiest method for converting binary to hexadecimal is to use a substitution code

Each hex number converts to 4 binary digits

Number Conversion

Example: Convert the following binary numbers

to their hexadecimal equivalent (vice versa).a) 10000.12

b) 1F.C16

Answer:

a) 10.816

b) 00011111.11002

Convert 0101011010101110011010102 to hex using

the 4-bit substitution code :

0101 0110 1010 1110 0110 1010

Substitution Code

5 6 A E 6 A

56AE6A16

Substitution code can also be used to convert binary to octal by using 3-bit groupings:

010 101 101 010 111 001 101 010

Substitution Code

2 5 5 2 7 1 5 2

255271528

Binary Addition0 + 0 = 0 Sum of 0 with a carry of 0

0 + 1 = 1 Sum of 1 with a carry of 0

1 + 0 = 1 Sum of 1 with a carry of 0

1 + 1 = 10 Sum of 0 with a carry of 1

Example:

11001 111

+ 1101 + 11

100110 ???

Simple Arithmetic Addition Example:

100011002

+ 1011102

101110102

Substraction Example:

10001002

- 1011102

101102

Example:

5816

+ 2416

7C16

Binary Subtraction 0 - 0 = 0

1 - 1 = 0

1 - 0 = 1

10 -1 = 1 0 -1 with a borrow of 1

Example:

1011 101

- 111 - 11

100 ???

Binary Multiplication0 X 0 = 0

0 X 1 = 0 Example:

1 X 0 = 0100110

1 X 1 = 1 X 101

100110

000000

+ 100110

10111110

Binary Division

Use the same procedure as decimal division

1’s complements of binary numbers Changing all the 1s to 0s and all

the 0s to 1s

Example: 1 1 0 1 0 0 1 0 1 Binary

number

0 0 1 0 1 1 0 1 0 1’s complement

2’s complements of binary numbers

2’s complement Step 1: Find 1’s complement of the number

Binary # 110001101’s complement 00111001

Step 2: Add 1 to the 1’s complement00111001

+ 0000000100111010

Signed Magnitude Numbers

Sign bit

0 = positive

1 = negative

31 bits for magnitude

This is your basic

Integer format

110010.. …00101110010101

Sign numbers Left most is the sign bit

0 is for positive, and 1 is for negative Sign-magnitude

0 0 0 1 1 0 0 1 = +25sign bit magnitude bits

1’s complement The negative number is the 1’s

complement of the corresponding positive number

Example: +25 is 00011001 -25 is 11100110

Sign numbers 2’s complement

The positive number – same as sign magnitude and 1’s complement

The negative number is the 2’s complement of the corresponding positive number.

Example Express +19 and -19 ini. sign magnitudeii. 1’s complementiii. 2’s complement

Digital Codes BCD (Binary Coded Decimal) Code

1. Represent each of the 10 decimal digits (0~9) as a 4-bit binary code.

Example: Convert 15 to BCD.

1 5

0001 0101BCD

Convert 10 to binary and BCD.

Digital Codes ASCII (American Standard Code for

Information Interchange) Code

1. Used to translate from the keyboard characters to computer language

Digital Codes The Gray Code

Only 1 bit changes Can’t be used in

arithmetic circuits Binary to Gray Code

and vice versa.

Decimal Binary Gray Code

0 0000 0000

1 0001 0001

2 0010 0011

3 0011 0010

4 0100 0110

5 0101 0111

6 0110 0101

3.0 LOGIC GATES

• Inverter (Gate Not)• AND Gate• OR Gate• NAND Gate• NOR Gate• Exclusive-OR and Exclusive-NOR• Fixed-function logic: IC Gates

Introduction

Three basic logic gates AND Gate – expressed by “ . “ OR Gates – expressed by “ + “ sign (not an ordinary

addition) NOT Gate – expressed by “ ‘ “ or “¯”

NOT Gate (Inverter)

a) Gate Symbol & Boolean a) Gate Symbol & Boolean

EquationEquation

b) Truth Table (Jadual b) Truth Table (Jadual Kebenaran) Kebenaran) c) Timing Diagram (Rajah c) Timing Diagram (Rajah

Pemasaan)Pemasaan)

OR Gate

a) Gate Symbol & Boolean a) Gate Symbol & Boolean EquationEquation

b) Truth Table (Jadual b) Truth Table (Jadual Kebenaran) Kebenaran)

c) Timing Diagram (Rajah c) Timing Diagram (Rajah Pemasaan)Pemasaan)

AND Gate

a) Gate Symbol & Boolean a) Gate Symbol & Boolean EquationEquation

b) Truth Table (Jadual b) Truth Table (Jadual Kebenaran) Kebenaran)

c) Timing Diagram (Rajah c) Timing Diagram (Rajah Pemasaan)Pemasaan)

NAND Gate

a) Gate Symbol, Boolean a) Gate Symbol, Boolean Equation Equation

& Truth Table& Truth Table

b) Timing Diagram b) Timing Diagram

NOR Gate

a) Gate Symbol, Boolean a) Gate Symbol, Boolean Equation Equation

& Truth Table& Truth Table

b) Timing Diagram b) Timing Diagram

Exclusive-OR Gate

BABABA

a) Gate Symbol, Boolean a) Gate Symbol, Boolean Equation Equation

& Truth Table & Truth Table

b) Timing Diagram b) Timing Diagram

Examples : Logic Gates IC

NOT gateNOT gate AND gateAND gate

Note : x is referring to family/technology (eg : AS/ALS/HCT/AC etc.)

4.0 BOOLEAN ALGEBRA

Boolean Operations & expression Laws & rules of Boolean algebra DeMorgan’s Theorems Boolean analysis of logic circuits Simplification using Boolean Algebra Standard forms of Boolean Expressions Boolean Expressions & truth tables The Karnaugh Map

Karnaugh Map SOP minimization Karnaugh Map POS minimization 5 Variable K-Map Programmable Logic

•Boolean Operations & expression

Expression : Variable – a symbol used to represent logical

quantities (1 or 0)

ex : A, B,..used as variable Complement – inverse of variable and is indicated by

bar over variable

ex : Ā

Operation : Boolean Addition – equivalent to the OR operation

X = A + B

- Boolean Multiplication – equivalent to the AND operation X = A∙B

A

BX

A

BX

Laws & rules of Boolean algebra

Commutative law of addition

Commutative law of addition,

A+B = B+A

the order of ORing does not matter.

Commutative law of Multiplication

Commutative law of Multiplication

AB = BA

the order of ANDing does not matter.

Associative law of addition

Associative law of addition

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

The grouping of ORed variables does not matter

Associative law of multiplication

Associative law of multiplication

A(BC) = (AB)C

The grouping of ANDed variables does not matter

Distributive Law

A(B + C) = AB + AC

(A+B)(C+D) = AC + AD + BC + BD

Boolean Rules

1) A + 0 = A

In math if you add 0 you have changed nothing In Boolean Algebra ORing with 0 changes nothing

Boolean Rules

2) A + 1 = 1

ORing with 1 must give a 1 since if any input

is 1 an OR gate will give a 1

Boolean Rules

3) A • 0 = 0

In math if 0 is multiplied with anything you

get 0. If you AND anything with 0 you get 0

Boolean Rules

4) A • 1 = A

ANDing anything with 1 will yield the anything

Boolean Rules

5) A + A = A

ORing with itself will give the same result

Boolean Rules

6) A + A = 1

Either A or A must be 1 so A + A =1

Boolean Rules

7) A • A = A

ANDing with itself will give the same result

Boolean Rules

8) A • A = 0

In digital Logic 1 =0 and 0 =1, so AA=0 since one of the inputs must be 0.

Boolean Rules

9) A = A

If you not something twice you are back to the beginning

Boolean Rules

10) A + AB = A

Proof:

A + AB = A(1 +B)DISTRIBUTIVE LAW

= A∙1 RULE 2: (1+B)=1

= A RULE 4: A∙1 = A

Boolean Rules 11) A + AB = A + B

If A is 1 the output is 1 , If A is 0 the output is B

Proof:

A + AB = (A + AB) + AB RULE 10

= (AA +AB) + AB RULE 7

= AA + AB + AA +AB RULE 8

= (A + A)(A + B) FACTORING

= 1∙(A + B) RULE 6

= A + B RULE 4

Boolean Rules

12) (A + B)(A + C) = A + BC

PROOF

(A + B)(A +C) = AA + AC +AB +BC DISTRIBUTIVE LAW

= A + AC + AB + BC RULE 7

= A(1 + C) +AB + BC FACTORING

= A.1 + AB + BC RULE 2

= A(1 + B) + BC FACTORING

= A.1 + BC RULE 2

= A + BC RULE 4

De Morgan’s Theorem,De Morgan’s Theorem,

Theorems of Boolean Algebra

1) A + 0 = A

2) A + 1 = 1

3) A • 0 = 0

4) A • 1 = A

5) A + A = A

6) A + A = 1

7) A • A = A

8) A • A = 0

Theorems of Boolean Algebra9) A = A

10) A + AB = A

11) A + AB = A + B

12) (A + B)(A + C) = A + BC

13) Commutative : A + B = B + A

AB = BA

14) Associative : A+(B+C) =(A+B) + C

A(BC) = (AB)C

15) Distributive : A(B+C) = AB +AC

(A+B)(C+D)=AC + AD + BC + BD

De Morgan’s Theorems

16) (X+Y) = X . Y17) (X.Y) = X + Y

• Two most important theorems of Boolean Algebra were contributed by De Morgan.

• Extremely useful in simplifying expression in which product or sum of variables is inverted.

• The TWO theorems are :

Implications of De Morgan’s Theorem

(a) Equivalent circuit implied by theorem (16) (b) Alternative symbol for the NOR function (c) Truth table that illustrates DeMorgan’s Theorem

(a)

(b)

Input Output

X Y X+Y XY

0 0 1 1

0 1 0 0

1 0 0 0

1 1 0 0

(c)

Implications of De Morgan’s Theorem

(a) Equivalent circuit implied by theorem (17) (b) Alternative symbol for the NAND function (c) Truth table that illustrates DeMorgan’s Theorem

(a)

(b)

Input Output

X Y XY X+Y

0 0 1 1

0 1 1 1

1 0 1 1

1 1 0 0

(c)

De Morgan’s Theorem ConversionStep 1: Change all ORs to ANDs and all ANDs to OrsStep 2: Complement each individual variable (short overbar)Step 3: Complement the entire function (long overbars)Step 4: Eliminate all groups of double overbars

Example : A . B A .B. C= A + B = A + B + C= A + B = A + B + C= A + B = A + B + C= A + B

ABC + ABC (A + B +C)D

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

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

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

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

De Morgan’s Theorem Conversion

Examples: Analyze the circuit below

Y

1. Y=???2. Simplify the Boolean expression found in 1

Follow the steps list below (constructing truth table) List all the input variable combinations of 1 and 0

in binary sequentially Place the output logic for each combination of

input Base on the result found write out the boolean

expression.

Exercises: Simplify the following Boolean expressions

1. (AB(C + BD) + AB)C

2. ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit.

Standard Forms of Boolean Expressions Sum of Products (SOP) Products of Sum (POS)

Notes: SOP and POS expression cannot have more than one variable combined in a term with an inversion bar There’s no parentheses in the expression

Standard Forms of Boolean Expressions Converting SOP to Truth Table

Examine each of the products to determine where the product is equal to a 1. Set the remaining row outputs to 0.

Standard Forms of Boolean Expressions

Converting POS to Truth Table Opposite process from the SOP expressions. Each sum term results in a 0. Set the remaining row outputs to 1.

Standard Forms of Boolean Expressions

BCACABCBACBAf ),,(

632),,( mmmCBAf

The standard SOP Expression All variables appear in each product term. Each of the product term in the expression is called as minterm. Example:

In compact form, f(A,B,C) may be written as

)6,3,2(),,( mCBAf

Standard Forms of Boolean Expressions

)()()(),,( CBACBACBACBAf

The standard POS Expression All variables appear in each product term. Each of the product term in the expression is called as maxterm. Example:

541),,( MMMCBAf

In compact form, f(A,B,C) may be written as

)5,4,1(),,( MCBAf

Standard Forms of Boolean Expressions

)()()()(),,( CBACBACBACBACBAf

Example:

Convert the following SOP expression to an equivalent POS expression:

Example:

Develop a truth table for the expression:

CBACBACABABCCBAf ),,(

THE K-MAP

Karnaugh Map (K-Map)

Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit.

This will replace Boolean reduction when the circuit is large.

Write the Boolean equation in a SOP form first and then place each term on a map.

• The map is made up of a table of every possible SOP using the number of variables that are being used.

• If 2 variables are used then a 2X2 map is used

• If 3 variables are used then a 4X2 map is used

• If 4 variables are used then a 4X4 map is used

• If 5 Variables are used then a 8X4 map is used

Karnaugh Map (K-Map)

K-Map SOP Minimization

A

A

B B

Notice that the map is going false to true, left to right and top to bottom

The upper right hand cell is A B if X= A B then put an X in that cell

A

A

B B

1

This show the expression true when A = 0 and B = 0

0 10 1

2 32 3

2 Variables Karnaugh Map

If X=AB + AB then put an X in both of these cells

A

A

B B

1

1

From Boolean reduction we know that A B + A B = B

From the Karnaugh map we can circle adjacent cell and find that X = B

A

A

B B

1

1

2 Variables Karnaugh Map

3 Variables Karnaugh Map

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C

0 10 1

2 32 3

6 76 7

4 54 5

X = A B C + A B C + A B C + A B C

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C

Each 3 variable term is one cell on a 4 X 2 Karnaugh map

1 1

1 1

3 Variables Karnaugh Map (cont’d)

X = A B C + A B C + A B C + A B C

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C COne simplification could be

X = A B + A B

1 1

1 1

3 Variables Karnaugh Map (cont’d)

X = A B C + A B C + A B C + A B C

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C CAnother simplification could be

X = B C + B C

A Karnaugh Map does wrap around

1 1

1 1

3 Variables Karnaugh Map (cont’d)

X = A B C + A B C + A B C + A B C

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C CThe Best simplification would be

X = B

1 1

1 1

3 Variables Karnaugh Map (cont’d)

On a 3 Variables Karnaugh Map• One cell requires 3 Variables

• Two adjacent cells require 2 variables

• Four adjacent cells require 1 variable

• Eight adjacent cells is a 1

4 Variables Karnaugh Map

Gray Code

00 A B

01 A B

11 A B

10 A B

0 0 0 1 1 1 1 0

C D C D C D C D

0 1 3 20 1 3 2

4 5 7 64 5 7 6

12 13 15 1412 13 15 14

8 9 11 108 9 11 10

Gray Code

00 A B

01 A B

11 A B

10 A B

0 0 0 1 1 1 1 0

C D C D C D C D

1

1

1

1

1

1

X = ABD + ABC + CD

Now try it with Boolean reductions

Simplify : X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D

On a 4 Variables Karnaugh map

• One Cell requires 4 variables

• Two adjacent cells require 3 variables

• Four adjacent cells require 2 variables

• Eight adjacent cells require 1 variable

• Sixteen adjacent cells give a 1 or true

Simplify :

Z = B C D + B C D + C D + B C D + A B C

Gray Code

00 A B

01 A B

11 A B

10 A B

00 01 11 10

C D C D C D C D

1 1 1 1

1 1

1 1

1 1

1

1

Z = C + A B + B D

Simplify using Karnaugh map

First, we need to change the circuit to an SOP expression

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

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

Y = A B + B C + A B C + A B D

Y = A B + B C + A B C A B D

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

Y = A B + B C + A + A B + A D + B + B D + AC + C D

Simplify using Karnaugh map (cont’d)

SOP expression

Gray Code

00 A B

01 A B

11 A B

10 A B

00 01 11 10

C D C D C D C D

1 1

1 1

1 1 1 1

Y = 1 Y = 1

1 1 1 1

1 1

1 1

Simplify using Karnaugh map (cont’d)

K-Map POS Minimization

3 Variables Karnaugh Map

Gray Code

0 0

0 1

1 1

1 0

0 1

ABC

0 10 1

2 32 3

6 76 7

4 54 5

3 Variables Karnaugh Map (cont’d)

4 Variables Karnaugh Map

0 0

0 1

1 1

1 0

0 0 0 1 1 1 1 0 A B

C D

0 1 3 20 1 3 2

4 5 7 64 5 7 6

12 13 15 1412 13 15 14

8 9 11 108 9 11 10

4 Variables Karnaugh Map (cont’d)

4 Variables Karnaugh Map (cont’d)

Mapping a Standard SOP expression Example:

Answer:Answer:

Mapping a Standard POS expression Example: Using K-Map, convert the following standard POS expression into a minimum SOP expression

Answer:Answer:

Y = AB + AC or standard SOP :

Karnaugh Map - Example

DCBADCBABCDACDBADCBADCBAY

CDADBY

ABCCBACBAY

)( CBAY

K-Map with “Don’t Care” Conditions

Input Output

Example :

3 variables with output “don’t care (X)”

K-Map with “Don’t Care” Conditions (cont’d)

4 variables with output “don’t care (X)”

“Don’t Care” Conditions Example: Determine the minimal SOP using K-Map:

Answer:Answer:

DACBCDDCBAF ),,,(

14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A,

K-Map with “Don’t Care” Conditions (cont’d)

Solution : 14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A,

ABCD

00

01

11

10

00 01 11 10

0 1 1 0

1 X 1 0

X X X X

0 0 1 0

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

DACBCDDCBAF ),,,(

Minimum SOP expression is CD

ADBC

Extra Exercise

Minimize this expression with a Karnaugh map

ABCD + ACD + BCD + ABCD

5 variable K-map

5 variables -> 32 minterms, hence 32 squares required

K-map Product of Sums simplification

Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in

(a) S-of-p (b) P-of-s

Using the minterms (1’s)F(ABCD)= B’D’+B’C’+A’C’D

Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De Morgen’s theorem to get F.

F’(ABCD)= BD’+CD+AB

F(ABCD)= (B’+D)(C’+D’)(A’+B’)

5 variable K-map

Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11.

The centre line must be considered as the centre of a book, each half of the K-map being a page

The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.

5 variable K-map

Example: Simplify the Boolean function

F(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31)

Soln: F(ABCDE) = BE+AD’E+A’B’E’

6 variable K-map

6 variables -> 64 minterms, hence 64 squares required

ICS217-Digital Electronics - Part 1.5 Combinational Logic

Tutorial 1.5

1. Simplify the Boolean function F(ABCDE) = (0,1,4,5,16,17,21,25,29)

Soln: F(ABCDE) = A’B’D’+AD’E+B’C’D’

2. Simplify the following Boolean expressions using K-maps.

(a) BDE+B’C’D+CDE+A’B’CE+A’B’C+B’C’D’E’

Soln: DE+A’B’C’+B’C’E’

(b) A’B’CE’+A’B’C’D’+B’D’E’+B’CD’+CDE’+BDE’

Soln: BDE’+B’CD’+B’D’E’+A’B’D’+CDE’

(c) F(ABCDEF) = (6,9,13,18,19,27,29,41,45,57,61)

Soln: F(ABCDEF) = A’B’C’DEF’+A’BC’DE+CE’F+A’BD’EF

END OF Chapter 1