CHAPTER 1 : INTRODUCTION EKT 121 / 4 ELEKTRONIK 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 Decimal Binary Octal Hexadecimal 0 ~ 9 0 ~ 1 0 ~ 7 0 ~ F

Characteristics of Numbering Systems The digits are consecutive. The number of digits is equal to the size of the base. Zero is always the first digit. When 1 is added to the largest digit, a sum of zero and a carry of one results. Numeric values determined by the implicit positional values of the digits.

00000000 00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 00001011 00001100 00001101 00001110 00001111 000 001 002 003 004 005 006 007 010 011 012 013 014 015 016 017 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Binary Octal Hex Dec N U M B E R S Y T

Most significant digit Least significant digit Significant Digits Binary: 11101101 Most significant digit Least significant digit Hexadecimal: 1D63A7A

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 Scale Base 2 Number Base 10 Equivalent Power Positional Value 000 20 1 001 21 2 010 22 4 011 3 23 8 100 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 each base 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 x x x x x x x x 27 26 25 24 23 22 21 20 Products 128 + 32 + 8 + 4 + 1 17310

Octal to Decimal Conversion Convert 6538 to its decimal equivalent: Octal Digits 6 5 3 x x x Positional Values 82 81 80 Products 384 + 40 + 3 42710

Hexadecimal to Decimal Conversion Convert 3B4F16 to its decimal equivalent: Hex Digits 3 B 4 F x x x x Positional Values 163 162 161 160 Products 12288 +2816 + 64 +15 15,18310

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: 25.3125 to Binary

Decimal to Binary Conversion Remainder 2 5 = 12 + 1 2 1 2 = 6 + 0 6 = 3 + 0 3 = 1 + 1 1 = 0 + 1 2 MSB LSB 2510 = 1 1 0 0 1 2

Decimal to Binary Conversion MSB LSB Carry . 0 1 0 1 0.3125 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

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 = E in Hex 33E16

Number Conversion Binary to Octal Conversion (vice versa) 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). 1001.11112 b) 47.38 1010011.110112 Answer: 11.748 100111.0112 123.668

Number Conversion Binary to Hexadecimal Conversion (vice versa) 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). 10000.12 1F.C16 Answer: 10.816 00011111.11002

Substitution Code 56AE6A16 0101 0110 1010 1110 0110 1010 5 6 A E 6 A Convert 0101011010101110011010102 to hex using the 4-bit substitution code : 0101 0110 1010 1110 0110 1010 5 6 A E 6 A 56AE6A16

Substitution Code Substitution code can also be used to convert binary to octal by using 3-bit groupings: 010 101 101 010 111 001 101 010 2 5 5 2 7 1 5 2 255271528

Binary Addition 0 + 0 = 0 Sum of 0 with a carry of 0 Example: 11001 111 + 1101 + 11 100110 ???

Simple Arithmetic Addition Example: Example: 100011002 5816 + 1011102 + 1011102 101110102 Substraction 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 Multiplication 0 X 0 = 0 0 X 1 = 0 Example: 1 X 0 = 0 100110 1 X 1 = 1 X 101 100110 000000 + 100110 10111110

Binary Division Use the same procedure as decimal division

Numbering Systems Decimal numbering system (Base 10) Binary numbering system (Base 2) Hexadecimal numbering system (Base 16) Octal numbering system (Base 8) Binary-Coded-Decimal (BCD) ASCII codes Gray codes

BCD (Binary Coded Decimal) Code Represent each of the 10 decimal digits (0~9) as a 4-bit binary code. Example: Convert 1510 to BCD. 1 5 0001 0101BCD

Convert from decimal to BCD 4510 17010 246910

Convert from BCD to decimal 1001 0110 11 0110 1001 0100 0111 0000

ASCII (American Standard Code for Information Interchange) Code Used to translate from the keyboard characters to computer language. ASCII has 128 characters and symbols represented by a 8-bit binary code, with MSB always 0, i.e. 0016 to 7F16. Refer table on page 91.

ASCII Table

Exercise Determine the ASCII code sequence (in both binary and hex) when we type this on the keyboard: 71 INPUT Z

Gray Code Decimal Binary Gray Code 0000 1 0001 2 0010 0011 3 4 0100 0000 1 0001 2 0010 0011 3 4 0100 0110 5 0101 0111 6 Only 1 bit changes at a time. Can’t be used in arithmetic circuits. Binary to Gray Code and vice versa.

Gray Codes Not an arithmetic code Exhibits only a single change from one code word to the next in sequence Need to know : Conversion binary  gray code Conversion gray code  binary

Binary  Gray Codes The MSB in Gray code is the same as the corresponding MSB in the binary number. Going from left to right, add each adjacent pair of binary code bits to get the next Gray code bit. Discard carries.

Binary  Gray Codes 1 0 1 1 0 1 1 1 0 1 Gray Codes + + + + Convert binary number 10110 to gray code. 1 0 1 1 0 1 1 1 0 1 Gray Codes + + + +

Binary  Gray Codes 1 1 1 1 1 1 0 0 0 0 Gray Codes + + + + Convert binary number 11111 to gray code. 1 1 1 1 1 1 0 0 0 0 Gray Codes + + + +

Gray Codes  Binary The MSB in binary code is the same as the corresponding bit in the Gray code. Add each binary code bit generated to the Gray code bit in the next adjacent position. Discard carries.

Gray Codes  Binary 1 0 1 1 0 1 1 0 1 1 Binary + + + + Convert gray code 10110 to binary. 1 0 1 1 0 1 1 0 1 1 Binary + + + +

Gray Codes  Binary 1 1 0 1 1 1 0 0 1 0 Binary + + + + Convert gray code 11011 to binary. 1 1 0 1 1 1 0 0 1 0 Binary + + + +

Next … 1’s and 2’s complements of binary numbers Signed numbers Arithmetic operations with signed numbers Digital codes & parity

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 Step 1: Find 1’s complement of the number Binary # 11000110 1’s complement 00111001 Step 2: Add 1 to the 1’s complement 00111001 + 00000001 00111010

SIGNED NUMBERS

The Sign Bit A 0 sign bit indicates a positive number, and A 1 sign bit indicates a negative number. 01001 versus 11001

1. The Sign Magnitude Form 110010.. …00101110010101 Sign bit 31 bits for magnitude 0 = positive 1 = negative This is your basic Integer format

Signed numbers Left most is the sign bit Sign-magnitude Exercise 0 is for positive, and 1 is for negative Sign-magnitude 0 0 0 1 1 0 0 1 = +25 sign bit magnitude bits Exercise Write the representation for -25

2. The 1’s Complement Form Positive numbers are represented the same way as the positive sign-magnitude numbers. A negative number is the 1’s complement of the corresponding positive number. Example: +25 is 00011001 -25 is 11100110

3. The 2’s Complement Form 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 form ii. 1’s complement form iii. 2’s complement form

Solution Decimal No. Sign-Magnitude 1’s Complement 2’s Complement +19 0001 0011 -19 1001 0011 1110 1100 1110 1101

Error Detection & Correction Codes PARITY METHOD: Even Parity Odd Parity

Parity Bit Is an extra bit included with the data bits. To detect errors in data communication & processing.

Error-Detection Codes Redundancy (e.g. extra information), in the form of extra bits, can be incorporated into binary code words to detect and correct errors. A simple form of redundancy is parity, an extra bit appended onto the code word to make the number of 1’s odd or even. Parity can detect all single-bit errors and some multiple-bit errors. A code word has even parity if the number of 1’s in the code word is even. A code word has odd parity if the number of 1’s in the code word is odd.

4-Bit Parity Code Example Fill in the even and odd parity bits: The codeword "1111" has even parity and the codeword "1110" has odd parity. Both can be used to represent 3-bit data. Even Parity Odd Parity Message Parity Message Parity - - 000 000 - - 001 001 - - 010 010 - - 011 011 - - 100 100 - - 101 101 - - 110 110 - - Even Parity Bits: 0, 1, 1, 0, 1, 0, 0, 1 Odd Parity Bits: 1, 0, 0, 1, 0, 1, 1, 0 111 111 - -

BASIC LOGIC GATES

Exclusive-OR and Exclusive-NOR Fixed-function logic: IC Gates 3.0 LOGIC GATES Inverter (NOT Gate) 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 Equation b) Truth Table (Jadual Kebenaran) c) Timing Diagram (Rajah Pemasaan)

OR Gate a) Gate Symbol & Boolean Equation b) Truth Table (Jadual Kebenaran) c) Timing Diagram (Rajah Pemasaan)

AND Gate a) Gate Symbol & Boolean Equation c) Timing Diagram (Rajah Pemasaan) b) Truth Table (Jadual Kebenaran)

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

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

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

Exclusive-NOR Gate

Symbols For XOR and XNOR XOR symbol: XNOR symbol: Symbols exist only for two inputs

AND gate NOT gate Examples : Logic Gates IC Note : x is referring to family/technology (eg : AS/ALS/HCT/AC etc.)

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 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 X B A X B

Laws & rules of Boolean algebra

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

Commutative law of Multiplication AB = BA the order of ANDing does not matter.

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

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

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

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 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

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

Implications of De Morgan’s Theorem Input Output X Y X+Y XY 0 0 1 1 0 1 0 0 0 0 0 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (16) (b) Alternative symbol for the NOR function (c) Truth table that illustrates DeMorgan’s Theorem

Implications of De Morgan’s Theorem Input Output X Y XY X+Y 0 0 1 1 0 1 1 1 0 1 1 1 1 0 0 (b) (c) (a) Equivalent circuit implied by theorem (17) (b) Alternative symbol for the NAND function (c) Truth table that illustrates DeMorgan’s Theorem

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 .. Proving of De Morgan’s theorems

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

Examples: Analyze the circuit below 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 (AB(C + BD) + AB)C ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit.

Boolean Algebra 1. 3. 5. 7. 9. 2. 4. 6. 8. 10. 12. 14. 16. X + Y Y + X + 0 = + 1 X + X X = X 2. 4. 6. 8. X . 1 = . 0 X . X 10. 12. 14. 16. X + Y Y + X = (X + Y) Z + X + (Y Z) X(Y + XY XZ X . Y 11. 13. 15. 17. XY YX = (XY) Z X(Y Z) X + YZ (X + Y) (X + Z) X . Y X + Y Commutative Associative Distributive DeMorgan ’ s

More on … De Morgans’ theorems From Logic Circuit to Truth Table From Truth Table to Boolean Expression, and vice versa

De Morgan’s Theorems

Figure 4–15 Gate equivalencies and the corresponding truth tables that illustrate DeMorgan’s theorems. Notice the equality of the two output columns in each table. This shows that the equivalent gates perform the same logic function. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Apply De Morgan’s theorems:

From Logic Circuit to Boolean Algebra Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Effect of Simplification using Boolean Algebra Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

From Truth Table to Boolean Expression, and vice versa

Tutorial 2 Q2, Q3 & Q4 Extra: Given truth table Get the output equation, minimize using Boolean rules Sketch the circuit Extra: Apply K-map to output equation obtained from truth table Compare this equation with the one obtained from using Boolean rules

Tutorial 2 Q3 & Q4 From TT, get the output equation Minimize using Boolean rules Minimize using K-map Compare both results Sketch the logic circuit

Tutorial 2 Q5 & Q6 Given a Boolean expression Minimize using Boolean rules Minimize using K-map Compare both results Sketch the truth table Sketch the logic 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.

Determine the SOP and POS

Determine the SOP and POS

Minterm & Maxterm

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

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

Determine the minterms and maxterms representation

Determine the minterms and maxterms representation

Converting Product Terms to Standard SOPs Each product term that does not contain all variables in the domain has to be expanded. Use: Multiply each nonstandard product term by a term made up of the sum of a missing variable and its complement (use Boolean rule above). Repeat (1) until all resulting product terms contain all variables in the domain (either in complemented or noncomplemented forms). Tip: the no. of product terms is doubled for each missing variable

Exercise: Convert to standard SOP and Minterm Expression

Converting Sum Terms to Standard POS Each sum term that does not contain all variables in the domain has to be expanded. Use: Add to each nonstandard product term, a term made up of the product of the missing variable and its complement (use Boolean rule above). Apply Repeat (1) until all resulting sum terms contain all variables in the domain (either in complemented or noncomplemented forms).

Exercise: Convert to standard POS and Maxterm Expression

Converting standard SOP to standard POS Evaluate each product term in the SOP expression, to determine the binary numbers represented. Determine all of the binary numbers not included in the evaluation in step 1. Write the equivalent sum term for each binary number from step 2 and express in POS form. Similar procedure, in going from POS to SOP

Exercise: SOP to POS

Boolean Expression & Truth Table Converting standard SOP to TT Converting standard POS to TT Determining Standard SOP and POS Expressions from TT

Standard Forms of Boolean Expressions Example: Convert the following SOP expression to an equivalent POS expression: Example: Develop a truth table for the expression:

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.

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

K-Map SOP Minimization

2 Variables Karnaugh Map B B A Notice that the map is going false to true, left to right and top to bottom 0 1 2 3 B B The upper right hand cell is A B if X= A B then put an X in that cell A 1 a “1” This show the expression true when A = 0 and B = 0

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

3 Variables Karnaugh Map 0 1 C C Gray Code 00 A B 01 A B 11 A B 10 A B 0 1 2 3 6 7 4 5

Figure 4–21 A 3-variable Karnaugh map showing product terms. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Standard SOP on a 3 Variables K-map 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 1 1 Each 3 variable term is one cell on a 4 X 2 Karnaugh map 1 1

Simplification of standard SOP using K-map 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 One simplification could be X = A B + A B 1 1 1 1

Another method of looping for simplification 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 Another simplification could be X = B C + B C A Karnaugh Map does wrap around 1 1 1 1

A 2nd method of looping for simplification 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 The Best simplification would be X = B 1 1 1 1

Now… Compare the results from the 3 methods of looping..

Simplification Results of Looping on a 3 Variables K-map Looping of: one cell produces a 3 variables term 2 adjacent cells produces a 2 variables term 4 adjacent cells produces a 1 variable term 8 adjacent cells results in 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 2 4 5 7 6 12 13 15 14 8 9 11 10

Figure 4–22 A 4-variable Karnaugh map. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–23 Adjacent cells on a Karnaugh map are those that differ by only one variable. Arrows point between adjacent cells. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

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 0 0 0 1 1 1 1 0 C D C D C D C D Now try it with Boolean reductions 1 1 1 1 1 1 X = ABD + ABC + CD

Simplification Results of Looping on a 4 Variables K-map Looping of: One Cell produces a 4 variables term 2 adjacent cells produces a 3 variables term 4 adjacent cells produces a 2 variables term 8 adjacent cells produces 1 variable term 16 adjacent cells results in a “1”

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

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

Simplify using Karnaugh map (cont’d) 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 Y = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

K-Map POS Minimization Looping of the zeroes …

3 Variables Karnaugh Map Gray Code 0 0 0 1 1 1 1 0 C 0 1 AB 0 1 2 3 6 7 4 5

3 Variables Karnaugh Map (cont’d)

4 Variables Karnaugh Map 0 0 0 1 1 1 1 0 C D 0 0 0 1 1 1 1 0 A B 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

4 Variables Karnaugh Map (cont’d)

4 Variables Karnaugh Map (cont’d)

Exercises K-Map

Figure 4–29 Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–30 Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

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 :

Mapping with … “Don’t Cares”

K-Map with “Don’t Care” Conditions Example : Input Output 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)”

K-Map with “Don’t Care” Conditions (cont’d) Example: Determine the minimal SOP using K-Map: Answer:

Minimum SOP expression is Solution : CD 00 01 11 10 AB 0 1 1 0 1 X 1 0 X X X X 0 0 1 0 00 01 11 10 0 1 3 2 5 7 6 13 15 14 8 9 11 10 AD BC CD Minimum SOP expression is

Extra Exercise Minimize this expression with a Karnaugh map ABCD + ACD + BCD + ABCD

The 7-segment LED display PROJECT The 7-segment LED display

Figure 4–47 Seven-segment display format showing arrangement of segments. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–48 Display of decimal digits with a 7-segment device. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–49 Arrangements of 7-segment LED displays. Active-Low Active-high Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–50 Block diagram of 7-segment logic and display. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

To light up the “a” segment Figure 4–51 Karnaugh map minimization of the segment-a logic expression. To light up the “a” segment Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–52 The minimum logic implementation for segment a of the 7-segment display. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

The 5-variables K-Map

Figure 4–42 A 5-variable Karnaugh map. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–43 Illustration of groupings of 1s in adjacent cells of a 5-variable map. Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

Figure 4–44 Thomas L. Floyd Digital Fundamentals, 9e Copyright ©2006 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.

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 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

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’

How many k-map is needed? If you have 5 variables, you’ll need 2 k-map… Let’s say the variables are A, B, C, D and E. 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 E = 0

How many k-map is needed? 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10

Try this out… Simplify the Boolean function F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29) Soln: F(A,B,C,D,E) = A’B’D’+AD’E+B’C’D’

1st step – convert the minterm into Boolean equation F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29) How to convert … ?? 0 = 00000  1 = 00001 

1st step – convert the minterm into Boolean equation F(A,B,C,D,E) = (0,1,4,5,16,17,21,25,29)

2nd – prepare 2 k-map E = 0 E = 1 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D E = 0 E = 1

3rd- plug in the Boolean term/minterm into k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1

4th- look for similar grouping that can be done in both k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That will give us

4th- look for similar grouping that can be done in both k-map 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That will give us

The combination will look like this… 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That will give us

5th- look for the remaining 1’s that has not been included yet 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1

6th- group the remaining 1’s 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 That new grouping will give us Note that this time, E need to be included in the term, since the grouping is in the E = 1 k-map only.

6th- group the remaining 1’s 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 A B C D 1 1 1 1 1 1 1 1 1 E = 0 E = 1 Full expression will be,

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 ICS217-Digital Electronics - Part 1.5 Combinational Logic

END OF Chapter 1