ekt 121 / 4 elektronik digit 1 chapter 1 : introduction
TRANSCRIPT
EKT 121 / 4ELEKTRONIK DIGIT 1
CHAPTER 1 : INTRODUCTION
1.0 Number & Codes
Digital and analog quantities Decimal numbering system (Base 10) Binary numbering system (Base 2) Hexadecimal numbering system (Base 16) Octal numbering system (Base 8) Number conversion Binary arithmetic 1’s and 2’s complements of binary numbers
Signed numbers Arithmetic operations with signed numbers Binary-Coded-Decimal (BCD) ASCII codes Gray codes Digital codes & parity
Digital and analog quantities
Two ways of representing the numerical values of quantities : i) Analog (continuous) ii) Digital (discrete)
Analog : a quantity represented by voltage, current or meter movement that is proportional to the value that quantity
Digital : the quantities are represented not by proportional quantities but by symbols called digits
Digital and analog systems
Digital system: combination of devices designed to manipulate logical
information or physical quantities that are represented in digital forms
include digital computers and calculators, digital audio/video equipments, telephone system.
Analog system: contains devices manipulate physical quantities that are
represented in analog form audio amplifiers, magnetic tape recording and playback
equipment, and simple light dimmer switch
Analog Quantities
• Continuous values
Digital Waveform
Introduction to Numbering Systems
We are all familiar with the decimal number system (Base 10). Some other number systems that we will work with are:
Binary Base 2 Octal Base 8 Hexadecimal Base 16
Number Systems
0 ~ 9
0 ~ 1
0 ~ 7
0 ~ F
Decimal
Binary
Octal
Hexadecimal
Characteristics of Numbering Systems
1) The digits are consecutive.2) The number of digits is equal to the size of the
base.3) Zero is always the first digit.4) When 1 is added to the largest digit, a sum of zero
and a carry of one results.5) Numeric values determined by the implicit
positional values of the digits.
00000000000000010000001000000011000001000000010100000110000001110000100000001001000010100000101100001100000011010000111000001111
000001002003004005006007010011012013014015016017
0123456789ABCDEF
0123456789
101112131415
BinaryOctalHexDecNUMBER SYSTEMS
Significant Digits
Binary: 11101101
Most significant digit Least significant digit
Hexadecimal: 1D63A7A
Most significant digit Least significant digit
Binary Number System
Also called the “Base 2 system” The binary number system is used to model the
series of electrical signals computers use to represent information
0 represents the no voltage or an off state 1 represents the presence of voltage or an
on state
Binary Numbering ScaleBase 2 Number Base 10 Equivalent Power Positional Value
000 0 20 1
001 1 21 2
010 2 22 4
011 3 23 8
100 4 24 16
101 5 25 32
110 6 26 64
111 7 27 128
Octal Number System
Also known as the Base 8 System Uses digits 0 - 7 Readily converts to binary Groups of three (binary) digits can be used to
represent each octal digit Also uses multiplication and division
algorithms for conversion to and from base 10
Hexadecimal Number System
Base 16 system Uses digits 0-9 &
letters A,B,C,D,E,F Groups of four bits
represent eachbase 16 digit
Number Conversion
Any Radix (base) to Decimal Conversion
Number Conversion Binary to Decimal Conversion
Binary to Decimal Conversion
Convert (10101101)2 to its decimal equivalent:
Binary 1 0 1 0 1 1 0 1
Positional Values
xxxxxxxx2021222324252627
128 + 32 + 8 + 4 + 1Products
17310
Octal to Decimal Conversion
Convert 6538 to its decimal equivalent:
6 5 3xxx
82 81 80
384 + 40 + 3
42710
Positional Values
Products
Octal Digits
Hexadecimal to Decimal Conversion
Convert 3B4F16 to its decimal equivalent:
Hex Digits 3 B 4 Fxxx
163 162 161 160
12288 +2816 + 64 +15
15,18310
Positional Values
Products
x
Number Conversion Decimal to Any Radix (Base)
Conversion
1. INTEGER DIGIT: Repeated division by the radix & record the remainder
2. FRACTIONAL DECIMAL:Multiply the number by the radix until the answer is in integer
Example:25.3125 to Binary
Decimal to Binary Conversion
2 5 = 12 + 1 2
1 2 = 6 + 0 2
6 = 3 + 0 2
3 = 1 + 1 2
1 = 0 + 1 2 MSB LSB 2510 = 1 1 0 0 1 2
Remainder
Decimal to Binary Conversion
Carry . 0 1 0 10.3125 x 2 = 0.625 0 0.625 x 2 = 1.25 1
0.25 x 2 = 0.50 0
0.5 x 2 = 1.00 1
The Answer: 1 1 0 0 1.0 1 0 1
MSB LSB
Decimal to Octal Conversion
Convert 42710 to its octal equivalent:
427 / 8 = 53 R3 Divide by 8; R is LSD
53 / 8 = 6 R5 Divide Q by 8; R is next digit
6 / 8 = 0 R6 Repeat until Q = 0
6538
Decimal to Hexadecimal Conversion
Convert 83010 to its hexadecimal equivalent:
830 / 16 = 51 R14
51 / 16 = 3 R3
3 / 16 = 0 R3
33E16
= E in Hex
Number Conversion
Binary to Octal Conversion (vice versa)
1. Grouping the binary position in groups of three starting at the least significant position.
Octal to Binary Conversion
Each octal number converts to 3 binary digits
To convert 6538 to binary, just substitute code:
6 5 3
110 101 011
Number Conversion
Example: Convert the following binary numbers to
their octal equivalent (vice versa).
a) 1001.11112 b) 47.38
c) 1010011.110112
Answer:
a) 11.748
b) 100111.0112
c) 123.668
Number Conversion
Binary to Hexadecimal Conversion (vice versa)
1. Grouping the binary position in 4-bit groups, starting from the least significant position.
Binary to Hexadecimal Conversion
The easiest method for converting binary to hexadecimal is to use a substitution code
Each hex number converts to 4 binary digits
Number Conversion
Example: Convert the following binary numbers
to their hexadecimal equivalent (vice versa).a) 10000.12
b) 1F.C16
Answer:
a) 10.816
b) 00011111.11002
Convert 0101011010101110011010102 to hex using
the 4-bit substitution code :
0101 0110 1010 1110 0110 1010
Substitution Code
5 6 A E 6 A
56AE6A16
Substitution code can also be used to convert binary to octal by using 3-bit groupings:
010 101 101 010 111 001 101 010
Substitution Code
2 5 5 2 7 1 5 2
255271528
Binary Addition0 + 0 = 0 Sum of 0 with a carry of 0
0 + 1 = 1 Sum of 1 with a carry of 0
1 + 0 = 1 Sum of 1 with a carry of 0
1 + 1 = 10 Sum of 0 with a carry of 1
Example:
11001 111
+ 1101 + 11
100110 ???
Simple Arithmetic Addition Example:
100011002
+ 1011102
101110102
Substraction Example:
10001002
- 1011102
101102
Example:
5816
+ 2416
7C16
Binary Subtraction 0 - 0 = 0
1 - 1 = 0
1 - 0 = 1
10 -1 = 1 0 -1 with a borrow of 1
Example:
1011 101
- 111 - 11
100 ???
Binary Multiplication0 X 0 = 0
0 X 1 = 0 Example:
1 X 0 = 0100110
1 X 1 = 1 X 101
100110
000000
+ 100110
10111110
Binary Division
Use the same procedure as decimal division
1’s complements of binary numbers Changing all the 1s to 0s and all
the 0s to 1s
Example: 1 1 0 1 0 0 1 0 1 Binary
number
0 0 1 0 1 1 0 1 0 1’s complement
2’s complements of binary numbers
2’s complement Step 1: Find 1’s complement of the number
Binary # 110001101’s complement 00111001
Step 2: Add 1 to the 1’s complement00111001
+ 0000000100111010
Signed Magnitude Numbers
Sign bit
0 = positive
1 = negative
31 bits for magnitude
This is your basic
Integer format
110010.. …00101110010101
Sign numbers Left most is the sign bit
0 is for positive, and 1 is for negative Sign-magnitude
0 0 0 1 1 0 0 1 = +25sign bit magnitude bits
1’s complement The negative number is the 1’s
complement of the corresponding positive number
Example: +25 is 00011001 -25 is 11100110
Sign numbers 2’s complement
The positive number – same as sign magnitude and 1’s complement
The negative number is the 2’s complement of the corresponding positive number.
Example Express +19 and -19 ini. sign magnitudeii. 1’s complementiii. 2’s complement
Digital Codes BCD (Binary Coded Decimal) Code
1. Represent each of the 10 decimal digits (0~9) as a 4-bit binary code.
Example: Convert 15 to BCD.
1 5
0001 0101BCD
Convert 10 to binary and BCD.
Digital Codes ASCII (American Standard Code for
Information Interchange) Code
1. Used to translate from the keyboard characters to computer language
Digital Codes The Gray Code
Only 1 bit changes Can’t be used in
arithmetic circuits Binary to Gray Code
and vice versa.
Decimal Binary Gray Code
0 0000 0000
1 0001 0001
2 0010 0011
3 0011 0010
4 0100 0110
5 0101 0111
6 0110 0101
3.0 LOGIC GATES
• Inverter (Gate Not)• AND Gate• OR Gate• NAND Gate• NOR Gate• Exclusive-OR and Exclusive-NOR• Fixed-function logic: IC Gates
Introduction
Three basic logic gates AND Gate – expressed by “ . “ OR Gates – expressed by “ + “ sign (not an ordinary
addition) NOT Gate – expressed by “ ‘ “ or “¯”
NOT Gate (Inverter)
a) Gate Symbol & Boolean a) Gate Symbol & Boolean
EquationEquation
b) Truth Table (Jadual b) Truth Table (Jadual Kebenaran) Kebenaran) c) Timing Diagram (Rajah c) Timing Diagram (Rajah
Pemasaan)Pemasaan)
OR Gate
a) Gate Symbol & Boolean a) Gate Symbol & Boolean EquationEquation
b) Truth Table (Jadual b) Truth Table (Jadual Kebenaran) Kebenaran)
c) Timing Diagram (Rajah c) Timing Diagram (Rajah Pemasaan)Pemasaan)
AND Gate
a) Gate Symbol & Boolean a) Gate Symbol & Boolean EquationEquation
b) Truth Table (Jadual b) Truth Table (Jadual Kebenaran) Kebenaran)
c) Timing Diagram (Rajah c) Timing Diagram (Rajah Pemasaan)Pemasaan)
NAND Gate
a) Gate Symbol, Boolean a) Gate Symbol, Boolean Equation Equation
& Truth Table& Truth Table
b) Timing Diagram b) Timing Diagram
NOR Gate
a) Gate Symbol, Boolean a) Gate Symbol, Boolean Equation Equation
& Truth Table& Truth Table
b) Timing Diagram b) Timing Diagram
Exclusive-OR Gate
BABABA
a) Gate Symbol, Boolean a) Gate Symbol, Boolean Equation Equation
& Truth Table & Truth Table
b) Timing Diagram b) Timing Diagram
Examples : Logic Gates IC
NOT gateNOT gate AND gateAND gate
Note : x is referring to family/technology (eg : AS/ALS/HCT/AC etc.)
4.0 BOOLEAN ALGEBRA
Boolean Operations & expression Laws & rules of Boolean algebra DeMorgan’s Theorems Boolean analysis of logic circuits Simplification using Boolean Algebra Standard forms of Boolean Expressions Boolean Expressions & truth tables The Karnaugh Map
Karnaugh Map SOP minimization Karnaugh Map POS minimization 5 Variable K-Map Programmable Logic
•Boolean Operations & expression
Expression : Variable – a symbol used to represent logical
quantities (1 or 0)
ex : A, B,..used as variable Complement – inverse of variable and is indicated by
bar over variable
ex : Ā
Operation : Boolean Addition – equivalent to the OR operation
X = A + B
- Boolean Multiplication – equivalent to the AND operation X = A∙B
A
BX
A
BX
Laws & rules of Boolean algebra
Commutative law of addition
Commutative law of addition,
A+B = B+A
the order of ORing does not matter.
Commutative law of Multiplication
Commutative law of Multiplication
AB = BA
the order of ANDing does not matter.
Associative law of addition
Associative law of addition
A + (B + C) = (A + B) + C
The grouping of ORed variables does not matter
Associative law of multiplication
Associative law of multiplication
A(BC) = (AB)C
The grouping of ANDed variables does not matter
Distributive Law
A(B + C) = AB + AC
(A+B)(C+D) = AC + AD + BC + BD
Boolean Rules
1) A + 0 = A
In math if you add 0 you have changed nothing In Boolean Algebra ORing with 0 changes nothing
Boolean Rules
2) A + 1 = 1
ORing with 1 must give a 1 since if any input
is 1 an OR gate will give a 1
Boolean Rules
3) A • 0 = 0
In math if 0 is multiplied with anything you
get 0. If you AND anything with 0 you get 0
Boolean Rules
4) A • 1 = A
ANDing anything with 1 will yield the anything
Boolean Rules
5) A + A = A
ORing with itself will give the same result
Boolean Rules
6) A + A = 1
Either A or A must be 1 so A + A =1
Boolean Rules
7) A • A = A
ANDing with itself will give the same result
Boolean Rules
8) A • A = 0
In digital Logic 1 =0 and 0 =1, so AA=0 since one of the inputs must be 0.
Boolean Rules
9) A = A
If you not something twice you are back to the beginning
Boolean Rules
10) A + AB = A
Proof:
A + AB = A(1 +B)DISTRIBUTIVE LAW
= A∙1 RULE 2: (1+B)=1
= A RULE 4: A∙1 = A
Boolean Rules 11) A + AB = A + B
If A is 1 the output is 1 , If A is 0 the output is B
Proof:
A + AB = (A + AB) + AB RULE 10
= (AA +AB) + AB RULE 7
= AA + AB + AA +AB RULE 8
= (A + A)(A + B) FACTORING
= 1∙(A + B) RULE 6
= A + B RULE 4
Boolean Rules
12) (A + B)(A + C) = A + BC
PROOF
(A + B)(A +C) = AA + AC +AB +BC DISTRIBUTIVE LAW
= A + AC + AB + BC RULE 7
= A(1 + C) +AB + BC FACTORING
= A.1 + AB + BC RULE 2
= A(1 + B) + BC FACTORING
= A.1 + BC RULE 2
= A + BC RULE 4
De Morgan’s Theorem,De Morgan’s Theorem,
Theorems of Boolean Algebra
1) A + 0 = A
2) A + 1 = 1
3) A • 0 = 0
4) A • 1 = A
5) A + A = A
6) A + A = 1
7) A • A = A
8) A • A = 0
Theorems of Boolean Algebra9) A = A
10) A + AB = A
11) A + AB = A + B
12) (A + B)(A + C) = A + BC
13) Commutative : A + B = B + A
AB = BA
14) Associative : A+(B+C) =(A+B) + C
A(BC) = (AB)C
15) Distributive : A(B+C) = AB +AC
(A+B)(C+D)=AC + AD + BC + BD
De Morgan’s Theorems
16) (X+Y) = X . Y17) (X.Y) = X + Y
• Two most important theorems of Boolean Algebra were contributed by De Morgan.
• Extremely useful in simplifying expression in which product or sum of variables is inverted.
• The TWO theorems are :
Implications of De Morgan’s Theorem
(a) Equivalent circuit implied by theorem (16) (b) Alternative symbol for the NOR function (c) Truth table that illustrates DeMorgan’s Theorem
(a)
(b)
Input Output
X Y X+Y XY
0 0 1 1
0 1 0 0
1 0 0 0
1 1 0 0
(c)
Implications of De Morgan’s Theorem
(a) Equivalent circuit implied by theorem (17) (b) Alternative symbol for the NAND function (c) Truth table that illustrates DeMorgan’s Theorem
(a)
(b)
Input Output
X Y XY X+Y
0 0 1 1
0 1 1 1
1 0 1 1
1 1 0 0
(c)
De Morgan’s Theorem ConversionStep 1: Change all ORs to ANDs and all ANDs to OrsStep 2: Complement each individual variable (short overbar)Step 3: Complement the entire function (long overbars)Step 4: Eliminate all groups of double overbars
Example : A . B A .B. C= A + B = A + B + C= A + B = A + B + C= A + B = A + B + C= A + B
ABC + ABC (A + B +C)D
= (A+B+C).(A+B+C) = (A.B.C)+D
= (A+B+C).(A+B+C) = (A.B.C)+D
= (A+B+C).(A+B+C) = (A.B.C)+D
= (A+B+C).(A+B+C) = (A.B.C)+D
De Morgan’s Theorem Conversion
Examples: Analyze the circuit below
Y
1. Y=???2. Simplify the Boolean expression found in 1
Follow the steps list below (constructing truth table) List all the input variable combinations of 1 and 0
in binary sequentially Place the output logic for each combination of
input Base on the result found write out the boolean
expression.
Exercises: Simplify the following Boolean expressions
1. (AB(C + BD) + AB)C
2. ABC + ABC + ABC + ABC + ABC Write the Boolean expression of the following circuit.
Standard Forms of Boolean Expressions Sum of Products (SOP) Products of Sum (POS)
Notes: SOP and POS expression cannot have more than one variable combined in a term with an inversion bar There’s no parentheses in the expression
Standard Forms of Boolean Expressions Converting SOP to Truth Table
Examine each of the products to determine where the product is equal to a 1. Set the remaining row outputs to 0.
Standard Forms of Boolean Expressions
Converting POS to Truth Table Opposite process from the SOP expressions. Each sum term results in a 0. Set the remaining row outputs to 1.
Standard Forms of Boolean Expressions
BCACABCBACBAf ),,(
632),,( mmmCBAf
The standard SOP Expression All variables appear in each product term. Each of the product term in the expression is called as minterm. Example:
In compact form, f(A,B,C) may be written as
)6,3,2(),,( mCBAf
Standard Forms of Boolean Expressions
)()()(),,( CBACBACBACBAf
The standard POS Expression All variables appear in each product term. Each of the product term in the expression is called as maxterm. Example:
541),,( MMMCBAf
In compact form, f(A,B,C) may be written as
)5,4,1(),,( MCBAf
Standard Forms of Boolean Expressions
)()()()(),,( CBACBACBACBACBAf
Example:
Convert the following SOP expression to an equivalent POS expression:
Example:
Develop a truth table for the expression:
CBACBACABABCCBAf ),,(
THE K-MAP
Karnaugh Map (K-Map)
Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit.
This will replace Boolean reduction when the circuit is large.
Write the Boolean equation in a SOP form first and then place each term on a map.
• The map is made up of a table of every possible SOP using the number of variables that are being used.
• If 2 variables are used then a 2X2 map is used
• If 3 variables are used then a 4X2 map is used
• If 4 variables are used then a 4X4 map is used
• If 5 Variables are used then a 8X4 map is used
Karnaugh Map (K-Map)
K-Map SOP Minimization
A
A
B B
Notice that the map is going false to true, left to right and top to bottom
The upper right hand cell is A B if X= A B then put an X in that cell
A
A
B B
1
This show the expression true when A = 0 and B = 0
0 10 1
2 32 3
2 Variables Karnaugh Map
If X=AB + AB then put an X in both of these cells
A
A
B B
1
1
From Boolean reduction we know that A B + A B = B
From the Karnaugh map we can circle adjacent cell and find that X = B
A
A
B B
1
1
2 Variables Karnaugh Map
3 Variables Karnaugh Map
Gray Code
00 A B
01 A B
11 A B
10 A B
0 1
C C
0 10 1
2 32 3
6 76 7
4 54 5
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
0 1
C C
Each 3 variable term is one cell on a 4 X 2 Karnaugh map
1 1
1 1
3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
0 1
C COne simplification could be
X = A B + A B
1 1
1 1
3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
0 1
C CAnother simplification could be
X = B C + B C
A Karnaugh Map does wrap around
1 1
1 1
3 Variables Karnaugh Map (cont’d)
X = A B C + A B C + A B C + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
0 1
C CThe Best simplification would be
X = B
1 1
1 1
3 Variables Karnaugh Map (cont’d)
On a 3 Variables Karnaugh Map• One cell requires 3 Variables
• Two adjacent cells require 2 variables
• Four adjacent cells require 1 variable
• Eight adjacent cells is a 1
4 Variables Karnaugh Map
Gray Code
00 A B
01 A B
11 A B
10 A B
0 0 0 1 1 1 1 0
C D C D C D C D
0 1 3 20 1 3 2
4 5 7 64 5 7 6
12 13 15 1412 13 15 14
8 9 11 108 9 11 10
Gray Code
00 A B
01 A B
11 A B
10 A B
0 0 0 1 1 1 1 0
C D C D C D C D
1
1
1
1
1
1
X = ABD + ABC + CD
Now try it with Boolean reductions
Simplify : X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C D
On a 4 Variables Karnaugh map
• One Cell requires 4 variables
• Two adjacent cells require 3 variables
• Four adjacent cells require 2 variables
• Eight adjacent cells require 1 variable
• Sixteen adjacent cells give a 1 or true
Simplify :
Z = B C D + B C D + C D + B C D + A B C
Gray Code
00 A B
01 A B
11 A B
10 A B
00 01 11 10
C D C D C D C D
1 1 1 1
1 1
1 1
1 1
1
1
Z = C + A B + B D
Simplify using Karnaugh map
First, we need to change the circuit to an SOP expression
Y= A + B + B C + ( A + B ) ( C + D)
Y = A B + B C + A B ( C + D )
Y = A B + B C + A B C + A B D
Y = A B + B C + A B C A B D
Y = A B + B C + (A + B + C ) ( A + B + D)
Y = A B + B C + A + A B + A D + B + B D + AC + C D
Simplify using Karnaugh map (cont’d)
SOP expression
Gray Code
00 A B
01 A B
11 A B
10 A B
00 01 11 10
C D C D C D C D
1 1
1 1
1 1 1 1
Y = 1 Y = 1
1 1 1 1
1 1
1 1
Simplify using Karnaugh map (cont’d)
K-Map POS Minimization
3 Variables Karnaugh Map
Gray Code
0 0
0 1
1 1
1 0
0 1
ABC
0 10 1
2 32 3
6 76 7
4 54 5
3 Variables Karnaugh Map (cont’d)
4 Variables Karnaugh Map
0 0
0 1
1 1
1 0
0 0 0 1 1 1 1 0 A B
C D
0 1 3 20 1 3 2
4 5 7 64 5 7 6
12 13 15 1412 13 15 14
8 9 11 108 9 11 10
4 Variables Karnaugh Map (cont’d)
4 Variables Karnaugh Map (cont’d)
Mapping a Standard SOP expression Example:
Answer:Answer:
Mapping a Standard POS expression Example: Using K-Map, convert the following standard POS expression into a minimum SOP expression
Answer:Answer:
Y = AB + AC or standard SOP :
Karnaugh Map - Example
DCBADCBABCDACDBADCBADCBAY
CDADBY
ABCCBACBAY
)( CBAY
K-Map with “Don’t Care” Conditions
Input Output
Example :
3 variables with output “don’t care (X)”
K-Map with “Don’t Care” Conditions (cont’d)
4 variables with output “don’t care (X)”
“Don’t Care” Conditions Example: Determine the minimal SOP using K-Map:
Answer:Answer:
DACBCDDCBAF ),,,(
14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A,
K-Map with “Don’t Care” Conditions (cont’d)
Solution : 14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A,
ABCD
00
01
11
10
00 01 11 10
0 1 1 0
1 X 1 0
X X X X
0 0 1 0
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
DACBCDDCBAF ),,,(
Minimum SOP expression is CD
ADBC
Extra Exercise
Minimize this expression with a Karnaugh map
ABCD + ACD + BCD + ABCD
5 variable K-map
5 variables -> 32 minterms, hence 32 squares required
K-map Product of Sums simplification
Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in
(a) S-of-p (b) P-of-s
Using the minterms (1’s)F(ABCD)= B’D’+B’C’+A’C’D
Using the maxterms (0’s) and complimenting F Grouping as if they were minterms, then using De Morgen’s theorem to get F.
F’(ABCD)= BD’+CD+AB
F(ABCD)= (B’+D)(C’+D’)(A’+B’)
5 variable K-map
Adjacent squares. E.g. square 15 is adjacent to 7,14,13,31 and its mirror square 11.
The centre line must be considered as the centre of a book, each half of the K-map being a page
The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.
5 variable K-map
Example: Simplify the Boolean function
F(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31)
Soln: F(ABCDE) = BE+AD’E+A’B’E’
6 variable K-map
6 variables -> 64 minterms, hence 64 squares required
ICS217-Digital Electronics - Part 1.5 Combinational Logic
Tutorial 1.5
1. Simplify the Boolean function F(ABCDE) = (0,1,4,5,16,17,21,25,29)
Soln: F(ABCDE) = A’B’D’+AD’E+B’C’D’
2. Simplify the following Boolean expressions using K-maps.
(a) BDE+B’C’D+CDE+A’B’CE+A’B’C+B’C’D’E’
Soln: DE+A’B’C’+B’C’E’
(b) A’B’CE’+A’B’C’D’+B’D’E’+B’CD’+CDE’+BDE’
Soln: BDE’+B’CD’+B’D’E’+A’B’D’+CDE’
(c) F(ABCDEF) = (6,9,13,18,19,27,29,41,45,57,61)
Soln: F(ABCDEF) = A’B’C’DEF’+A’BC’DE+CE’F+A’BD’EF
END OF Chapter 1