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This first edition of Digital Electronics: Part 2 Boolean operations is to expose students with a simple method remembered for a basic logic gate and to simplified the expression in Boolean operation. It is part of the Diploma in Electrical and Electronic Engineering and Diploma in Electronic Engineering (Communication) course, which aim to emphasize students on the digital system fundamentals and applications.

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Published by samsawibujang, 2023-10-31 23:33:15

Digital Electronics Part 2 Boolean Operations

This first edition of Digital Electronics: Part 2 Boolean operations is to expose students with a simple method remembered for a basic logic gate and to simplified the expression in Boolean operation. It is part of the Diploma in Electrical and Electronic Engineering and Diploma in Electronic Engineering (Communication) course, which aim to emphasize students on the digital system fundamentals and applications.

Keywords: Digital Electronics,Boolean Operations

DIGITAL ELEC TRO NIC S PART 2 SamsawiBujang Ts.Dr.MuhammadSufyanSafwanMohamadBasir “Makeiteasy,don’ttakeiteasy” BOOLEAN OPERATIONS


DIGITAL ELEC TRO NIC S PART 2 SamsawiBujang Ts.Dr.MuhammadSufyanSafwanMohamadBasir “Makeiteasy,don’ttakeiteasy” BOOLEAN OPERATIONS


P R E F A C E | i Preface This first edition of Digital Electronics: Part 2 Boolean operations is to expose students with a simple method remembered for a basic logic gate and to simplified the expression in Boolean operation. It is part of the Diploma in Electrical and Electronic Engineering and Diploma in Electronic Engineering (Communication) course, which aim to emphasize students on the digital system fundamentals and applications. The approach used in Digital Electronics: Part 2 Boolean Operations allows students to master the all-important fundamental concepts, example, and exercise on how to design the conceptual logic circuit. We hope that from this e-book, students will have a strong grounding in the fundamentals of digital technology. The knowledge obtained will help students in solving the critical task that they will be facing. Make it easy, but don’t take it easy.


C O N T E N T S | i Contents PAGE PART 2 BOOLEAN OPERATIONS GENERAL OUTCOMES 1 1.0 BASIC LOGIC GATES 2 1.1 Problem Solving 1 8 1.2 Problem Solving 2 9 1.3 Problem Solving 3 10 1.4 Problem Solving 4 11 1.5 Problem Solving 5 12 1.6 Problem Solving 6 13 2.0 UNIVERSALITY OF NAND GATE 14 2.1 Problem Solving 1 15 2.2 Problem Solving 2 16 2.3 Problem Solving 3 17 3.0 STANDARD FORM OF BOOLEAN OPERATIONS 18 3.1 Problem Solving 1 20 3.2 Problem Solving 2 21 3.3 Problem Solving 3 22 3.4 Problem Solving 4 23 2.5 Problem Solving 5 24


C O N T E N T S | ii PAGE 4.0 SIMPLIFICATION OF BOOLEAN EXPRESSION 25 Algebraic Simplification (Using Boolean Theorems) 26 4.1 Problem Solving 1 27 4.2 Problem Solving 2 28 4.3 Problem Solving 3 29 4.4 Problem Solving 4 30 4.5 Problem Solving 5 32 5.0 SIMPLIFICATION OF BOOLEAN EXPRESSION Karnaugh Map 31 5.1 Problem Solving 1 33 5.2 Problem Solving 2 34 5.3 Problem Solving 3 35 5.4 Problem Solving 4 36 5.5 Problem Solving 5 37 5.6 Problem Solving 6 38 SELF-ASSESMENT 39 REFERENCES


B O O L E A N O P E R A T I O N S | 1 GENERAL outcomes 1) Remember symbols, operations and functions of logic gates 2) Apply basic logic gates 3) Apply Boolean algebra in logic circuits analysis and design 4) Construct logic expressions from the truth 2 table PART BOOLEAN OPERATIONS


B O O L E A N O P E R A T I O N S | 2 BASIC LOGIC GATES 01 OR Gate 02 AND Gate Symbol Operation Truth Table Timing Diagram X = A + B X = A . B A X A B B X


B O O L E A N O P E R A T I O N S | 3 BASIC LOGIC GATES 03 NOT Gate Symbol Operation Truth Table Timing Diagram X = A A X


B O O L E A N O P E R A T I O N S | 4 BASIC LOGIC GATES 04 NOR Gate 05 NAND Gate Symbol Operation Truth Table Timing Diagram X = A + B X = A . B A A B B X X


B O O L E A N O P E R A T I O N S | 5 BASIC LOGIC GATES 06 Exclusive OR Gate 07 Exclusive NOR Gate Symbol Operation Truth Table Timing Diagram X = A + B X = A + B A A B B X X


B O O L E A N O P E R A T I O N S | 6 COMBINATIONAL LOGIC GATE 01 Example 1 02 Example 2 Method 1 Logic circuit → Expression


B O O L E A N O P E R A T I O N S | 7 COMBINATIONAL LOGIC GATE 01 Example 1 02 Example 2 Method 2 Expression → Logic circuit


B O O L E A N O P E R A T I O N S | 8 Write the output X for combination logic circuit below Solution: 1 STEP 1 STEP 2 Find the output of each logic gate Obtain the final output base on input of final logic gate SOLVE ME…. “Improve yourself” A B C X = ( + ) .


B O O L E A N O P E R A T I O N S | 9 Write the output Y for combination logic circuit below Solution: 2 STEP 1 STEP 2 Find the output of each logic gate Obtain the final output base on input of final logic gate SOLVE ME…. “Improve yourself” A B C Y = ( + ). A


B O O L E A N O P E R A T I O N S | 10 v Write the output Z for combination logic circuit below Solution: 3 STEP 1 STEP 2 Find the output of each logic gate Obtain the final output base on input of final logic gate SOLVE ME…. “Improve yourself” A B C Z = + ( + )


B O O L E A N O P E R A T I O N S | 11 Draw logic circuit for following expression using combinational logic gates Solution: 4 SOLVE ME…. “Improve yourself” Draw the logic gate start from final output Obtain the input base on expression


B O O L E A N O P E R A T I O N S | 12 Draw logic circuit for following expression using combinational logic gates Solution: 5 SOLVE ME…. “Improve yourself” A B C X D E F Draw the logic gate start from final output Obtain the input base on expression


B O O L E A N O P E R A T I O N S | 13 Draw logic circuit for following expression using combinational logic gates Solution: 6 Draw the logic gate start from final output Obtain the final output base on input of final logic gate SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 14 Universality Of NAND Gate NOT GATE → AND GATE → OR GATE →


B O O L E A N O P E R A T I O N S | 15 Construct function of NOR gate using NAND gate Solution: 1 SOLVE ME…. “Improve yourself” Redraw the logic circuit using NAND gate only


B O O L E A N O P E R A T I O N S | 16 Construct combination logic circuit below using NAND gate only Solution: 2 SOLVE ME…. “Improve yourself” A B C X A B C X Replace the origin each logic gate using NAND gate accordingly Redraw the logic circuit using NAND gate only A B C X


B O O L E A N O P E R A T I O N S | 17 Draw logic circuit for following expression using NAND gate only Solution: 3 SOLVE ME…. “Improve yourself” Replace the origin each logic gate using NAND gate accordingly Redraw the logic circuit using NAND gate only = ( + ). A B C X A B C X Draw the logic circuit using basic logic gate A B C X


B O O L E A N O P E R A T I O N S | 18 STANDARD FORMS OF BOOLEAN EXPRESSION Standard Form SOP (Sum of Product) POS (Product of Sum)


B O O L E A N O P E R A T I O N S | 19 Let’s produce a Sum of Product (SOP) and Product of Sum (POS), which has 3 inputs and one output INPUT OUTPUT A B C Z SOP POS 0 0 0 0 + + 0 0 1 0 + + 0 1 0 1 0 1 1 1 1 0 0 0 + + 1 0 1 0 + + 1 1 0 1 1 1 1 1 Boolean Expression SOP = + + + POS = ( + + )( + + )( + + )( + + )


B O O L E A N O P E R A T I O N S | 20 INPUT OUTPUT A B C X 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 INPUT OUTPUT A B C X SOP 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 = + + + + Write the form of Sum of Product (SOP) logic expression from truth table below Solution: 1 SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 21 INPUT OUTPUT A B C X 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 INPUT OUTPUT A B C X SOP 0 0 0 1 0 0 1 0 + + 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 0 + + 1 1 0 1 1 1 1 0 + + = ( + + )( + + )( + + ) Write the form of Product of Sum (POS) logic expression from truth table below Solution: 2 SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 22 INPUT OUTPUT A B C X 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 = + + + INPUT OUTPUT A B C X SOP 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 Referring to the truth table below, draw the combinational logic circuit for Product of Sum (POS) equation Solution: 3 SOLVE ME…. “Improve yourself” A B C X A B C A B C A B C


B O O L E A N O P E R A T I O N S | 23 INPUT OUTPUT A B C D X SOP 0 0 0 0 1 0 0 0 1 1 0 0 1 0 0 + + + 0 0 1 1 0 + + + 0 1 0 0 1 0 1 0 1 1 0 1 1 0 0 + + + 0 1 1 1 0 + + + 1 0 0 0 0 + + + 1 0 0 1 0 + + + 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 = ( + + + )( + + + )( + + + )( + + + )( + + + ) ( + + + ) Develop a truth table for Boolean expression given. Write the form of Product of Sum (POS) = + + + + + + + + + Solution: 4 SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 24 = + + + + + + + + INPUT OUTPUT A B C D X SOP 0 0 0 0 0 + + + 0 0 0 1 0 + + + 0 0 1 0 1 0 0 1 1 1 0 1 0 0 0 + + + 0 1 0 1 0 + + + 0 1 1 0 0 + + + 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 0 + + + 1 0 1 1 0 + + + 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 = ( + + + )( + + + )( + + + )( + + + )( + + + ) ( + + + ) ( + + + ) Develop a truth table for Boolean expression given. Write the form of Product of Sum (POS) = + + + + + Solution: 5 SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 25 Simplification of Boolean Expressions Karnaugh 02 Maps Mapping 2 3 Algebraic Simplification (Using Boolean Theorems) 01 Boolean Laws 2 3 i ii iii iv v i ii iii


B O O L E A N O P E R A T I O N S | 26 Let’s simplify expression below using Boolean laws = + ( + ) + ( + ) = + + + + = + + + = + + + = + + ( 1 + ) = + + (1) = + + = + + = ( + 1) + = (1) + = + Algebraic Simplification (Using Boolean Theorems) 01 Boolean Laws 2 3 . = + = 1 + = 1 + 1 = 1


B O O L E A N O P E R A T I O N S | 27 = + + = + ( + ) = + ( + ) = + (1 ) = + Using Boolean algebra techniques, simplify expression below = + + Solution: 1 STEP 1 STEP 2 Find the input or term in expression that can be simplify using Boolean Theorem Obtain the final output base on input simplification SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 28 = + + + = + + = ( + ) + = (1) + = + Using Boolean algebra techniques, simplify expression below = + + + Solution: 2 STEP 1 STEP 2 Find the input or term in expression that can be simplify using Boolean Theorem Obtain the final output base on input simplification SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 29 Output F is summation of minterms m0, m1, m2, m5, m7 m0 = 000 = m1 = 001 = m2 = 010 = m5 = 101 = m7 = 111 = = + + + + = ( + ) + + ( + ) = ( 1 ) + + ( 1 ) = + + = ( + ) + = ( + ) + = + + Simplify the following 3 variable expression using Boolean algebra (, , ) = ∑ ( , , , , ) Solution: 3 SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 30 Output F is summation of Maxterms M1, M3, M5, M7 M1 = 001 = ( + + ) M3 = 011 = ( + + ) M5 = 101 = ( + + ) M7 = 111 = ( + + ) = ( + + )( + + )( + + )( + + ) TRY TO SOLVE THIS EXPRESSION USING POS FORM !!!! Simplify the following 3 variable expression using Boolean algebra (, , ) = ∏ ( , , , ) Solution: 4 SOLVE ME…. “Improve yourself”


B O O L E A N O P E R A T I O N S | 31 A Karnaugh map provides a pictorial method of grouping together expressions with common factors and therefore eliminating unwanted variables. The Karnaugh map can also be described as a special arrangement of a truth table. The diagram below illustrates the correspondence between the Karnaugh map and the truth table for the general case. Karnaugh 02 Maps Mapping 2 3


B O O L E A N O P E R A T I O N S | 32 01 02 03 04 05


B O O L E A N O P E R A T I O N S | 33 = + + + INPUT OUTPUT A B C W 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 0 0 1 0 0 0 0 0 1 1 1 1 1 0 0 10 1 1 INPUT OUTPUT A B C W SOP 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 0 1 1 1 0 Table below shows the output of security alarm system. From the table, produce Sum of Product (SOP) expression. Simplify the expression using Karnaugh Map method. Solution: 1 SOLVE ME…. “Improve yourself” C A B = +


B O O L E A N O P E R A T I O N S | 34 = + + + + 0 1 0 0 1 0 0 1 1 1 1 1 1 1 10 0 0 INPUT OUTPUT A B C X SOP 0 0 0 1 0 0 1 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 1 0 1 1 0 1 1 1 1 1 A combinational circuit has 3 inputs A, B, C and output X. X is HIGH for following input combinations (use term True = 1 and False = 0). A is False, B is True, C is False A is False, B is True, C is True A is True, B is True, C is False A, B, C are False A, B, C are True Produce the truth table for above input combinations and use K-Map method to simplify expression in Sum of Product (SOP) form. Solution: 2 SOLVE ME…. “Improve yourself” C A B = +


B O O L E A N O P E R A T I O N S | 35 = + 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 0 0 0 1 1 1 0 0 0 1 10 1 0 0 1 Simplify the following expression using Karnaugh Map. = + + + + + Solution: 3 STEP 1 STEP 2 Map the expression into KMap & loop the ‘1’ according to KMap rules Obtain the expression group by group SOLVE ME…. “Improve yourself” CD A B X


B O O L E A N O P E R A T I O N S | 36 = + 0 0 0 1 1 1 1 0 0 0 1 0 0 1 0 1 1 0 0 1 1 1 1 1 0 1 10 1 0 0 1 Simplify the following expression using Karnaugh Map. = + + + + + + + Solution: 4 STEP 1 STEP 2 Map the expression into KMap & loop the ‘1’ according to KMap rules Obtain the expression group by group SOLVE ME…. “Improve yourself” CD A B Y


B O O L E A N O P E R A T I O N S | 37 = + 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 1 1 0 0 10 1 1 0 0 Simplify the following expression using Karnaugh Map. = + + + + + Solution: 5 STEP 1 STEP 2 Map the expression into KMap & loop the ‘1’ according to KMap rules Obtain the expression group by group SOLVE ME…. “Improve yourself” CD A B Z


B O O L E A N O P E R A T I O N S | 38 = + 0 0 0 1 1 1 1 0 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 1 1 10 0 0 1 1 Simplify the following expression using Karnaugh Map. From simplification of expression, draw the logic circuit = + + + + + + + + Solution: 6 STEP 1 STEP 2 Map the expression into KMap & loop the ‘1’ according to KMap rules Obtain the expression group by group SOLVE ME…. “Improve yourself” CD A B Q STEP 2 Draw the logic circuit A B Q C D


B O O L E A N O P E R A T I O N S | 39 SELF-ASSESMENT 1. State the type of logic gate and logic expression for figure below. i. ii. iii. iv. v. vi. iv. Answ : refer to lecturer 2. Write the output combinational circuit as figure below. i) ii) Answ: A B Y C ) = ( + ) A B Z C ) = +


B O O L E A N O P E R A T I O N S | 40 3. Draw logic circuit for following expression using combinational logic gates. i) = ( + ) ii) = + ( + + ) iii) = ( + ) + ( + ) Answ: refer to lecturer 4. Simplify the expression below using Karnaugh Map method. i) = + + + + + ii) = + + + + + + + iii) = + + + + + Answ: 5. Simplify the expression below using Karnaugh Map and from simplification, draw the logic circuit. = + + + + + + + Answ: ) = + ) = ) = + = + C D W B


R e f e r e n c e s | R e f e r e n c e s Ronald J.Tocci and Neal S.Widmer (1998). Digital systems-Principles and Applications (7th Ed.). Prentice Hall, Inc. (ISBN: 0-13-649492-7). Thomas L. Floyd (1997), Digital Fundamentals (6th Ed.). Prentice Hall, Inc. (ISBN: 0- 13-573478-9) Salina Muhamad and Muhammad Nazir Mohammad Khalid (2017). Fundamental of Digital Electronics. Oxford Fajar Sdd. Bhd. (ISBN: 978 983 47 2139 8)


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