CHAPTER 1 : INTRODUCTION EKT 121 / 4 ELEKTRONIK DIGIT 1 CHAPTER 1 : INTRODUCTION Lecturer : FAIRUL AFZAL Office : KKF 8 (04-9854150) Email :fairul@unimap.edu.my

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

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

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.

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.

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

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 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

Analog Quantities Continuous values

Digital Waveform

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

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

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

Decimal numbering system (Base 10) Example : 39710 3 9 7 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 = 300 + 90 + 7 = 39710

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 : 1012 1 0 1 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 = 4 + 0 + 1 = 510

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 : 7568 convert to decimal: 7 5 6 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 = 448 + 40 + 6 49410 = Readily converts to binary Groups of three (binary) digits can be used to represent each octal number Example : 7568 convert to binary : 7 5 68 1111011102

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

Hexadecimal Number System Base 16 system – multiplication and division algorithms for conversion to and from base 10 Example : A9F16 convert to decimal: A 9 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 = 2560 + 144 + 15 271910 = Readily converts to binary Groups of four (binary) digits can be used to represent each hexadecimal number Example : A9F8 convert to binary : A 9 F16 1010100111112

Number Conversion Any Radix (base) to Decimal Conversion

Number Conversion (BASE 2 –BASE 10) 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

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

Sign numbers Left most is the sign bit Sign-magnitude 1’s complement 0 is for positive, and 1 is for negative Sign-magnitude 0 0 0 1 1 0 0 1 = +25 sign 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 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

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. 1 5 0001 0101BCD Convert 10 to binary and BCD.

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 ?????

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 ?????

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 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

AND gate NOT gate Examples : Logic Gates IC 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 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) Negative- AND (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) Negative-OR (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

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.

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 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

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 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

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 C 1 1 Each 3 variable term is one cell on a 4 X 2 Karnaugh map 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 C One 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 C Another 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 C The Best simplification would be X = B 1 1 1 1

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 2 4 5 7 6 12 13 15 14 8 9 11 10

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

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

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

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)

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 :

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

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’

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

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

END OF Chapter 1