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
Signed numbers Arithmetic operations with signed numbers Binary-Coded-Decimal (BCD) ASCII codes Gray codes Digital codes & parity
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.
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.
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
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 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
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
ABCDEF ABCDEF BinaryOctalHexDec NUMBER SYSTEMSNUMBER SYSTEMS
Binary: Most significant digit Least significant digit Hexadecimal: 1D63A7A Most significant digit Least significant digit
Base 10 - (0,1,2,3,4,5,6,7,8,9) Example : X X X = = Weights for whole numbers are positive power of ten that increase from right to left, beginning with 10 0
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 : X X 2 1 1X 2 2 Weights in a binary number are based on power of two, that increase from right to right to left,beginning with = =5 10
Base 8 system – (0,1,2,3,4,5,6,7) multiplication and division algorithms for conversion to and from base 10 Example : convert to decimal: X X 8 1 7X 8 2 Weights in a binary number are based on power of eight that increase from right to right to left,beginning with = = Readily converts to binary Groups of three (binary) digits can be used to represent each octal number Example : convert to binary :
Base 16 system Uses digits 0 ~ 9 & letters A,B,C,D,E,F Groups of four bits represent each base 16 digit HEXADECIMAL DECIMAL BINARY A B C D E F151111
Base 16 system – multiplication and division algorithms for conversion to and from base 10 Example : A9F 16 convert to decimal: A 9 F 15 X X X 16 2 Weights in a hexadecimal number are based on power of sixteen that increase from right to right to left,beginning with = = Readily converts to binary Groups of four (binary) digits can be used to represent each hexadecimal number Example : A9F 8 convert to binary : A 9 F
Any Radix (base) to Decimal Conversion
Binary to Decimal Conversion
Convert ( ) 2 to its decimal equivalent: Binary Positional Values x x x xx x x x Products
Convert to its decimal equivalent: xx x Positional Values Products Octal Digits
Convert 3B4F 16 to its decimal equivalent: Hex Digits 3 B 4 F xx x , Positional Values Products x
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: to Binary
2 5 = = = = = MSBLSB = Remainder
Carry x 2 = x 2 = x 2 = x 2 = The Answer: MSBLSB
Convert to its octal equivalent: 427 / 8 = 53 R3Divide by 8; R is LSD 53 / 8 = 6 R5Divide Q by 8; R is next digit 6 / 8 = 0 R6Repeat until Q =
Convert to its hexadecimal equivalent: 830 / 16 = 51 R14 51 / 16 = 3 R3 3 / 16 = 0 R3 33E 16 = E in Hex
Binary to Octal Conversion (vice versa) 1. Grouping the binary position in groups of three starting at the least significant position.
Each octal number converts to 3 binary digits To convert to binary, just substitute code:
Example: Convert the following binary numbers to their octal equivalent (vice versa). a) b) c) Answer: a) b) c)
Binary to Hexadecimal Conversion (vice versa) 1. Grouping the binary position in 4-bit groups, starting from the least significant position.
The easiest method for converting binary to hexadecimal is to use a substitution code Each hex number converts to 4 binary digits
Example: Convert the following binary numbers to their hexadecimal equivalent (vice versa). a) b) 1F.C 16 Answer: a) b)
Convert to hex using the 4-bit substitution code : A E 6 A 56AE6A 16
Substitution code can also be used to convert binary to octal by using 3-bit groupings:
0 + 0 = 0Sum of 0 with a carry of = 1Sum of 1 with a carry of = 1 Sum of 1 with a carry of = 10Sum of 0 with a carry of 1 Example: ???
Changing all the 1s to 0s and all the 0s to 1s Example: Binary number ’s complement
2’s complement Step 1: Find 1’s complement of the number Binary # ’s complement Step 2: Add 1 to the 1’s complement
Sign bit 0 = positive 1 = negative 31 bits for magnitude This is your basic Integer format …
Left most is the sign bit 0 is for positive, and 1 is for negative Sign-magnitude = +25 sign bit magnitude bits 1’s complement The negative number is the 1’s complement of the corresponding positive number Example: +25 is is
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 in i. sign magnitude ii. 1’s complement iii. 2’s complement
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 BCD Convert 10 to binary and BCD.
ASCII (American Standard Code for Information Interchange) Code 1. 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 ?????
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 ????? DecimalBinaryGray Code
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 b) Truth Table (Jadual Kebenaran) c) Timing Diagram (Rajah Pemasaan)
NAND Gate a) Gate Symbol, Boolean Equation a) Gate Symbol, Boolean Equation & Truth Table b) Timing Diagram
NOR Gate a) Gate Symbol, Boolean Equation a) Gate Symbol, Boolean Equation & Truth Table b) Timing Diagram
Exclusive-OR Gate a) Gate Symbol, Boolean Equation a) Gate Symbol, Boolean Equation & Truth Table b) Timing Diagram
Examples : Logic Gates IC NOT gate AND gate Note : x is referring to family/technology (eg : AS/ALS/HCT/AC etc.)
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
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 B X A B X
Commutative law of addition 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) = AB + AC (A+B)(C+D) = AC + AD + BC + BD
In math if you add 0 you have changed nothing In Boolean Algebra ORing with 0 changes nothing
ORing with 1 must give a 1 since if any input is 1 an OR gate will give a 1
In math if 0 is multiplied with anything you get 0. If you AND anything with 0 you get 0
ANDing anything with 1 will yield the anything
ORing with itself will give the same result
Either A or A must be 1 so A + A =1
ANDing with itself will give the same result
In digital Logic 1 =0 and 0 =1, so AA=0 since one of the inputs must be 0.
If you not something twice you are back to the beginning
Proof: A + AB = A(1 +B)DISTRIBUTIVE LAW = A∙1 RULE 2: (1+B)=1 = ARULE 4: A∙1 = A
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
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
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
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
16) (X+Y) = X. Y 17) (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 :
(a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgan’s Theorem (a) (b) Input Output X Y X+Y XY (c)
Implications of De Morgan’s Theorem (a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgan’s Theorem (a) (b) Input Output X Y XY X+Y (c)
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
ABC + ABC (A + B +C)D = (A+B+C).(A+B+C) = (A.B.C)+D
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.
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.
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
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.
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.
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
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
Example: Convert the following SOP expression to an equivalent POS expression: Example: Develop a truth table for the expression:
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)
AAAA 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 AAAA B 1 This show the expression true when A = 0 and B = Variables Karnaugh Map
If X=AB + AB then put an X in both of these cells AAAA 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 AAAA B Variables Karnaugh Map
Gray Code 00A B 01A B 11 A B 10A B 0 1 C C
Gray Code 00A B 01A B 11 A B 10A B 0 1 C C Each 3 variable term is one cell on a 4 X 2 Karnaugh map Variables Karnaugh Map (cont’d)
Gray Code 00A B 01A B 11 A B 10A B 0 1 C C One simplification could be X = A B + A B Variables Karnaugh Map (cont’d)
Gray Code 00A B 01A B 11 A B 10A B 0 1 C C Another simplification could be X = B C + B C A Karnaugh Map does wrap around Variables Karnaugh Map (cont’d)
Gray Code 00A B 01A B 11 A B 10A B 0 1 C C The Best simplification would be X = B Variables Karnaugh Map (cont’d)
One cell requires 3 Variables Two adjacent cells require 2 variables Four adjacent cells require 1 variable Eight adjacent cells is a 1
Gray Code 00A B 01A B 11A B 10A B C D C D
Gray Code 00A B 01A B 11A B 10A B C D C D X = ABD + ABC + CD Now try it with Boolean reductions
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
Gray Code 00A B 01A B 11A B 10A B C D C D Z = C + A B + B D
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 00A B 01A B 11A B 10A B C D C D Y = Simplify using Karnaugh map (cont’d)
Gray Code AB C
A B C D
Mapping a Standard SOP expression Example: Answer: Mapping a Standard POS expression Example: Using K-Map, convert the following standard POS expression into a minimum SOP expression Answer: Y = AB + AC or standard SOP :
Input Output Example : 3 variables with output “don’t care (X)”
4 variables with output “don’t care (X)”
“Don’t Care” Conditions Example: Determine the minimal SOP using K-Map: Answer: Answer: K-Map with “Don’t Care” Conditions (cont’d)
AB CD X 1 0 X X Minimum SOP expression is CD AD BC
Minimize this expression with a Karnaugh map ABCD + ACD + BCD + ABCD
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’)
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.
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’
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