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**CHAPTER 1 : INTRODUCTION**

EKT 121 / 4 ELEKTRONIK DIGIT 1 CHAPTER 1 : INTRODUCTION Lecturer : FAIRUL AFZAL Office : KKF 8 ( )

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**1.0 Number & codes Digital vs. 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

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**Number & codes Signed numbers**

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

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Digital vs. Analogue Digital signals are represented by only two possible - 1 (binary 1) 0r 0 (binary 0) Sometimes call these values “high” and “low” or “ true” and “false”” Example : light switch , it can be in just two position – “ on ” or “ off ” More complicated signals can be constructed from 1s and 0s by stringing them end-to end. Example : 3 binary digits, have 8 possible combinations : 000,001,010,011,100,101,110 and 111. The diagram : example a typical digital signal, represented as a series of voltage levels that change as time goes on.

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Digital vs. Analogue Analogue electronics can be any value within limits. Example : Voltage change simultaneously from one value to the next, like gradually turning a light dimmer switch up or down. The diagram below shows an analoq signal that changes with time.

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**Digital vs. Analogue Why digital ? Problem with all signals – noise**

Noise isn't just something that you can hear - the fuzz that appears on old video recordings also qualifies as noise. In general, noise is any unwanted change to a signal that tends to corrupt it. Digital and analogue signals with added noise: Analog : never get back a perfect copy of the original signal Digital : easily be recognized even among all that noise : either 0 or 1

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**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

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**Digital and analog systems**

Digital system: combination of devices designed to manipulate logical information or physical quantities that are represented in digital forms Example : digital computers and calculators, digital audio/video equipments, telephone system. Analog system: contains devices manipulate physical quantities that are represented in analog form Example : audio amplifiers, magnetic tape recording and playback equipment, and simple light dimmer switch

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Analog Quantities Continuous values

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Digital Waveform

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**Introduction to Numbering Systems**

We are all familiar with decimal number systems - use everyday :calculator, calendar, phone or any common devices use this numbering system : Decimal = Base 10 Some other number systems: Binary = Base 2 Octal = Base 8 Hexadecimal = Base 16

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**Numbering Systems Decimal Binary Octal Hexadecimal 0 ~ 9 0 ~ 1 0 ~ 7**

0 ~ F

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A B C D E F Binary Octal Hex Dec N U M B E R S Y T

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**Most significant digit Least significant digit**

Significant Digits Binary: Most significant digit Least significant digit Hexadecimal: 1D63A7A

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**Decimal numbering system (Base 10)**

Example : 39710 Weights for whole numbers are positive power of ten that increase from right to left , beginning with 100 3 X 102 + 9 X 101 + 7 X 100 = = 39710

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**Binary Number System 1X 22 + 0 X 21 + 1 X 20 = 4 + 0 + 1 = 510**

Base 2 system – (0 , 1) 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 Example : Weights in a binary number are based on power of two, that increase from right to right to left,beginning with 20 1X 22 + 0 X 21 + 1 X 20 = = 510

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**Octal Number System + + = = Base 8 system – (0,1,2,3,4,5,6,7) 7X 82**

multiplication and division algorithms for conversion to and from base 10 Example : convert to decimal: Weights in a binary number are based on power of eight that increase from right to right to left,beginning with 80 + 7X 82 5 X 81 + 6 X 80 = 49410 = Readily converts to binary Groups of three (binary) digits can be used to represent each octal number Example : convert to binary :

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**Hexadecimal Number System**

BINARY 0000 1 0001 2 0010 3 0011 4 0100 5 0101 6 0110 7 0111 8 1000 9 1001 A 10 1010 B 11 1011 C 12 1100 D 13 1101 E 14 1110 F 15 1111 Base 16 system Uses digits 0 ~ 9 & letters A,B,C,D,E,F Groups of four bits represent each base 16 digit

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**Hexadecimal Number System**

Base 16 system – multiplication and division algorithms for conversion to and from base 10 Example : A9F16 convert to decimal: A F Weights in a hexadecimal number are based on power of sixteen that increase from right to right to left,beginning with 160 10X 162 + 9 X 161 + 15 X 160 = 271910 = Readily converts to binary Groups of four (binary) digits can be used to represent each hexadecimal number Example : A9F8 convert to binary : A F16

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Number Conversion Any Radix (base) to Decimal Conversion

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**Number Conversion (BASE 2 –BASE 10)**

Binary to Decimal Conversion

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**Binary to Decimal Conversion**

Convert ( )2 to its decimal equivalent: Binary Positional Values x x x x x x x x 27 26 25 24 23 22 21 20 Products 17310

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**Octal to Decimal Conversion**

Convert 6538 to its decimal equivalent: Octal Digits x x x Positional Values Products 42710

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**Hexadecimal to Decimal Conversion**

Convert 3B4F16 to its decimal equivalent: Hex Digits 3 B F x x x x Positional Values Products 15,18310

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**Number Conversion INTEGER DIGIT:**

Decimal to Any Radix (Base) Conversion INTEGER DIGIT: Repeated division by the radix & record the remainder FRACTIONAL DECIMAL: Multiply the number by the radix until the answer is in integer Example: to Binary

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**Decimal to Binary Conversion**

Remainder 2 5 = 2 1 2 = 6 = 3 = 1 = MSB LSB 2510 =

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**Decimal to Binary Conversion**

MSB LSB Carry x 2 = x 2 = x 2 = 0.5 x 2 = The Answer:

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**Decimal to Octal Conversion**

Convert 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

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**Decimal to Hexadecimal Conversion**

Convert to its hexadecimal equivalent: 830 / 16 = 51 R14 51 / 16 = 3 R3 3 / 16 = 0 R3 = E in Hex 33E16

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Number Conversion Binary to Octal Conversion (vice versa) Grouping the binary position in groups of three starting at the least significant position.

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**Octal to Binary Conversion**

Each octal number converts to 3 binary digits To convert 6538 to binary, just substitute code:

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Number Conversion Example: Convert the following binary numbers to their octal equivalent (vice versa). b) 47.38 Answer: 11.748

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Number Conversion Binary to Hexadecimal Conversion (vice versa) Grouping the binary position in 4-bit groups, starting from the least significant position.

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**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

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**Number Conversion Example:**

Convert the following binary numbers to their hexadecimal equivalent (vice versa). 1F.C16 Answer: 10.816

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**Substitution Code 56AE6A16 0101 0110 1010 1110 0110 1010 5 6 A E 6 A**

Convert to hex using the 4-bit substitution code : A E A 56AE6A16

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

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**Binary Addition 0 + 0 = 0 Sum of 0 with a carry of 0**

Example: ???

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**Simple Arithmetic Addition Example: Example: 100011002 5816 + 1011102**

Substraction 101102 Example: 5816 7C16

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Binary Subtraction 0 - 0 = 0 1 - 1 = 0 1 - 0 = 1 10 -1 = with a borrow of 1 Example: ???

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**Binary Multiplication**

0 X 0 = 0 0 X 1 = 0 Example: 1 X 0 = 1 X 1 = X 100110 000000

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Binary Division Use the same procedure as decimal division

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**1’s complements of binary numbers**

Changing all the 1s to 0s and all the 0s to 1s Example: Binary number ’s complement

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**2’s complements of binary numbers**

Step 1: Find 1’s complement of the number Binary # 1’s complement Step 2: Add 1 to the 1’s complement

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**Signed Magnitude Numbers**

… Sign bit 31 bits for magnitude 0 = positive 1 = negative This is your basic Integer format

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**Sign numbers Left most is the sign bit Sign-magnitude 1’s complement**

0 is for positive, and 1 is for negative Sign-magnitude = +25 sign bit magnitude bits 1’s complement The negative number is the 1’s complement of the corresponding positive number Example: +25 is is

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**Sign numbers 2’s complement Example Express +19 and -19 in**

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 in i. sign magnitude ii. 1’s complement iii. 2’s complement

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Digital Codes BCD (Binary Coded Decimal) Code Represent each of the 10 decimal digits (0~9) as a 4-bit binary code. Example: Convert 15 to BCD. BCD Convert 10 to binary and BCD.

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Digital Codes ASCII (American Standard Code for Information Interchange) Code Used to translate from the keyboard characters to computer language Can convert from ASCII code to binary / hexadecimal/ or any numbering systems and vice versa How to convert ?????

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**Digital Codes Decimal Binary Gray Code 0000 1 0001 2 0010 0011 3 4**

0000 1 0001 2 0010 0011 3 4 0100 0110 5 0101 0111 6 The Gray Code Only 1 bit changes Can’t be used in arithmetic circuits Can convert from Binary to Gray Code and vice versa. How to convert ?????

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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

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**Introduction Three basic logic gates AND Gate – expressed by “ . “**

OR Gates – expressed by “ + “ sign (not an ordinary addition) NOT Gate – expressed by “ ‘ “ or “¯”

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**NOT Gate (Inverter) a) Gate Symbol & Boolean Equation**

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

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**OR Gate a) Gate Symbol & Boolean Equation**

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

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**AND Gate a) Gate Symbol & Boolean Equation**

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

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**a) Gate Symbol, Boolean Equation**

NAND Gate a) Gate Symbol, Boolean Equation & Truth Table b) Timing Diagram

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**a) Gate Symbol, Boolean Equation**

NOR Gate a) Gate Symbol, Boolean Equation & Truth Table b) Timing Diagram

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**a) Gate Symbol, Boolean Equation**

Exclusive-OR Gate a) Gate Symbol, Boolean Equation & Truth Table b) Timing Diagram

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**AND gate NOT gate Examples : Logic Gates IC**

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

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**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

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**Karnaugh Map SOP minimization**

Karnaugh Map POS minimization 5 Variable K-Map Programmable Logic

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**Boolean Operations & 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

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**Operation : Boolean Addition – equivalent to the OR operation**

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

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**Laws & rules of Boolean algebra**

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**Commutative law of addition**

A+B = B+A the order of ORing does not matter.

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**Commutative law of Multiplication**

AB = BA the order of ANDing does not matter.

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**Associative law of addition**

A + (B + C) = (A + B) + C The grouping of ORed variables does not matter

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**Associative law of multiplication**

A(BC) = (AB)C The grouping of ANDed variables does not matter

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

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

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Boolean Rules 1) A + 0 = A In math if you add 0 you have changed nothing In Boolean Algebra ORing with 0 changes nothing

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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

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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

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Boolean Rules 4) A • 1 = A ANDing anything with 1 will yield the anything

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Boolean Rules 5) A + A = A ORing with itself will give the same result

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Boolean Rules 6) A + A = 1 Either A or A must be 1 so A + A =1

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Boolean Rules 7) A • A = A ANDing with itself will give the same result

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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.

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Boolean Rules 9) A = A If you not something twice you are back to the beginning

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Boolean Rules 10) A + AB = A Proof: A + AB = A(1 +B) DISTRIBUTIVE LAW = A∙ RULE 2: (1+B)=1 = A RULE 4: A∙1 = A

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**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

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**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

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De Morgan’s Theorem

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**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

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**Theorems of Boolean Algebra**

9) 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

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De Morgan’s Theorems 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 : 16) (X+Y) = X . Y 17) (X.Y) = X + Y

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**Implications of De Morgan’s Theorem**

Input Output X Y X+Y XY (b) (c) (a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgan’s Theorem

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**Implications of De Morgan’s Theorem**

Input Output X Y XY X+Y (b) (c) (a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgan’s Theorem

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**De Morgan’s Theorem Conversion**

Step 1: Change all ORs to ANDs and all ANDs to Ors Step 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

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**De Morgan’s Theorem Conversion**

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

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**Examples: Analyze the circuit below**

2. Simplify the Boolean expression found in 1

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**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.

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**Exercises: Simplify the following Boolean expressions**

(AB(C + BD) + AB)C ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit.

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**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

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**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.

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**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.

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**Standard Forms of Boolean Expressions**

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

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**Standard Forms of Boolean Expressions**

The standard POS Expression All variables appear in each product term. Each of the product term in the expression is called as maxterm. Example: In compact form, f(A,B,C) may be written as

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**Standard Forms of Boolean Expressions**

Example: Convert the following SOP expression to an equivalent POS expression: Example: Develop a truth table for the expression:

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THE K-MAP

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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.

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Karnaugh Map (K-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

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**K-Map SOP Minimization**

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**2 Variables Karnaugh Map**

B B A Notice that the map is going false to true, left to right and top to bottom B B The upper right hand cell is A B if X= A B then put an X in that cell A 1 This show the expression true when A = 0 and B = 0

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**2 Variables Karnaugh Map**

B B If X=AB + AB then put an X in both of these cells A 1 1 From Boolean reduction we know that A B + A B = B B B From the Karnaugh map we can circle adjacent cell and find that X = B A 1 1

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**3 Variables Karnaugh Map**

C C Gray Code 00 A B 01 A B 11 A B 10 A B

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**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 C C 1 1 Each 3 variable term is one cell on a 4 X 2 Karnaugh map 1 1

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**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 C C One simplification could be X = A B + A B 1 1 1 1

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**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 C C Another simplification could be X = B C + B C A Karnaugh Map does wrap around 1 1 1 1

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**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 C C The Best simplification would be X = B 1 1 1 1

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**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

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**4 Variables Karnaugh Map**

Gray Code 00 A B 01 A B 11 A B 10 A B C D C D C D C D

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**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**

Gray Code 00 A B 01 A B 11 A B 10 A B C D C D C D C D Now try it with Boolean reductions 1 1 1 1 1 1 X = ABD + ABC + CD

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**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

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**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 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

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**Simplify using Karnaugh map**

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

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**Simplify using Karnaugh map (cont’d)**

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 SOP expression

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**Simplify using Karnaugh map (cont’d)**

Gray Code 00 A B 01 A B 11 A B 10 A B C D C D C D C D Y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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**K-Map POS Minimization**

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**3 Variables Karnaugh Map**

Gray Code 0 0 0 1 1 1 1 0 C AB

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**3 Variables Karnaugh Map (cont’d)**

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**4 Variables Karnaugh Map**

0 0 0 1 1 1 1 0 C D A B

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**4 Variables Karnaugh Map (cont’d)**

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**4 Variables Karnaugh Map (cont’d)**

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**Karnaugh Map - Example Mapping a Standard SOP expression Example:**

Answer: Mapping a Standard POS expression Using K-Map, convert the following standard POS expression into a minimum SOP expression Y = AB + AC or standard SOP :

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**K-Map with “Don’t Care” Conditions**

Example : Input Output 3 variables with output “don’t care (X)”

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

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

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

Example: Determine the minimal SOP using K-Map: Answer:

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**Minimum SOP expression is**

Solution : CD AB X X X X X 00 01 11 10 AD BC CD Minimum SOP expression is

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**Extra Exercise Minimize this expression with a Karnaugh map**

ABCD + ACD + BCD + ABCD

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5 variable K-map 5 variables -> 32 minterms, hence 32 squares required

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

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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.

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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’

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6 variable K-map 6 variables -> 64 minterms, hence 64 squares required

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Tutorial 1.5 ICS217-Digital Electronics - Part 1.5 Combinational Logic 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

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END OF Chapter 1

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