Chap01 - Number Systems in Digital Logic.pptx

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Dr Yasir Awais Butt Digital Logic Design
DIGITAL SYSTEMS AND BINARY
NUMBERS
Chapter 1

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System
 The term system comes from the Latin word systēma, "whole concept made of
several parts or members, system", literary "composition“
 A system is a set of related components that work as a whole to achieve a goal.
 Mathematically, a system is a function that
 Takes input – Variable or signal 𝑥
 Applies transformation – Function 𝑓
 Gives output – Variable or signal 𝑦 = 𝑓(𝑥)
 For example a ball thrown in the air vs a rocket launched in the air are both
examples of systems

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Types of Variables

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Continuous vs Discrete Variable
 A continuous variable is a type
of quantitative variable
consisting of numerical values
that can be measured but not
counted, because there are
infinitely many values between
1 measurement and another.
 A discrete variable is a type of
quantitative variable consisting
of numerical values that can be
measured and counted, because
these values are separate or
distinct.
 In practice, all continuous
variables are discrete!

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Analog, Discrete, Mixed, Digital Signals
continuous time
continuous value
continuous time
discrete value
Digital Signal
Mixed
Mixed

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Digital Signal
continuous time
discrete value

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Analog, Discrete, Mixed, Digital Systems
 Given an input/ output pair of signals for a system, there are four types of
combinations:
 (1) Continuous 𝑥 and continuous 𝑓(𝑥)
 This is the most common analog system.
 (2) Continuous 𝑥 and discrete 𝑓(𝑥)
 Mixed System – (Digital System)
 Example –Baseband modulation in digital communication, ADCs
 (3) Discrete 𝑥 and continuous 𝑓(𝑥)
 Mixed system
 The output of the CCD sensor.
 (4) Discrete 𝑥 and discrete 𝑓(𝑥)
 This is the digital system
 Computers are the prime examples

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Digital Systems
 Digital systems are designed to store, process, and communicate information in
digital form
 A digital system refers to a system that processes, stores, and communicates
information using digital signals.

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Digital Systems
 Digital System play a prominent role in this digital age
 Communication, medical treatment, internet, DVD, CD, etc…
 Digital Computer follow a sequence of instructions, called programs, that
operate on given data
 User can specify and change program or data according to needs
 Digital Systems have the ability to Manipulate discrete elements of information.
 Any set that is restricted to a finite number of elements contains discrete
information
 10 Decimal digits
 26 Alphabet letters
 52 Playing cards
 64 squares of a chessboard
Lec 1

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Merits of Digital Systems
 Occupy minimum space
 Most digital devices are programmable
 Accuracy
 Cost reduction
 Efficient Processing & Data Storage
 Efficient & Reliable Transmission
 Detection and Correction of Errors
 Precise & Accurate Reproduction
 Easy Design and Implementation

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Digital Systems
 A Digital System is interconnection of digital modules
 To understand Digital module, we need to know about digital circuits and their
logical functions
 Hardware Description Language (HDL) is a programming language that is
suitable for describing digital circuit in a textual form
 Simulate a digital system to verify operation before it is built

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Logic Gates
 Logic gates are the building blocks of the digital circuits
 AND, OR and NOT Gates
 NAND, NOR, XOR and XNOR Gates
 Integrated Circuits (ICs)
Inverter
AND
OR

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Circuit Diagram
Lec 1

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1.2 BINARY NUMBER SYSTEM

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Definition
 A Number system is a system of mathematical notation for representing
numbers of a given set, using digits or other symbols.
 Example
 Roman number system – Symbolic System
 decimal system – numbers 0-9
 The first recorded zero appeared in Mesopotamia around 3 B.C.
 The Mayans invented it independently circa 4 A.D.
 It was later devised in India in the mid-fifth century, spread to Cambodia near the end
of the seventh century, and into China and the Islamic countries at the end of the
eighth.
 Indian Claim - The first modern equivalent of the numeral zero comes from a Hindu
astronomer and mathematician Brahmagupta in 628
 Numbers of distinct symbols is known as radix or base denoted as 𝑟
 In decimal system base is 10

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Place Value System
 Also called Positional Numeral System
 Place value is the basis of our entire number system.
 The position of a digit in a number determines its value.
 The number 42,316 is different from 61,432 because the digits are in different
positions

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Place Value System Characteristics
 A general number may be written as 𝑎𝑛 … 𝑎3𝑎2𝑎1𝑎0. 𝑎−1𝑎−2𝑎−3 … 𝑎−𝑚 is
interpreted as
 𝑎𝑛𝑟𝑛
+ 𝑎𝑛−1𝑟𝑛−1
+ ⋯ + 𝑎3𝑟3
+ 𝑎2𝑟2
+ 𝑎1𝑟 +𝑎0 𝑟0
+ 𝑎−1𝑟−1
+ 𝑎−2𝑟−2
𝑎−3𝑟−3
+ ⋯ +
𝑎−𝑚𝑟−𝑚
 𝑎𝑗 are a series of coefficients weighted by exponents of the radix 𝑟
 Significance of Radix 𝑟
 There are (0,1,2 … , 𝑟 − 1) unique symbols in the number system
 means that the coefficients 𝑎𝑗 are weighted by powers of 𝑟
 that the 𝑎𝑗 coefficients are any of the digits (0,1,2 … , 𝑟 − 1)
 We will analyze all number systems on this basic criterion

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Decimal Number System
 Radix 𝑟 =?
 Unique symbols?
 Are all numbers writtend as: 𝑎𝑛 … 𝑎3𝑎2𝑎1𝑎0. 𝑎−1𝑎−2𝑎−3 … 𝑎−𝑚?
 Does it mean
𝑎𝑛 … 𝑎3𝑎2𝑎1𝑎0. 𝑎−1𝑎−2𝑎−3 … 𝑎−𝑚 = 𝑎𝑛𝑟𝑛 + 𝑎𝑛−1𝑟𝑛−1 + ⋯ + 𝑎3𝑟3 + 𝑎2𝑟2 +
𝑎1𝑟 +𝑎0 𝑟0
+ 𝑎−1𝑟−1
+ 𝑎−2𝑟−2
𝑎−3𝑟−3
+ ⋯ + 𝑎−𝑚𝑟−𝑚
 Does it means that the coefficients 𝑎𝑗 are weighted by powers of 𝑟
 Are the 𝑎𝑗 coefficients are any of the digits (0,1,2 … , 𝑟 − 1)

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Decimal Number System - Example
 7392 10 = 7 × 103 + 3x102 + 9 × 101 + 2 × 100
 Position of each digit in a decimal number indicates the magnitude of the quantity
represented and assigned a weight.
 The weights of the whole numbers are positive powers of 10 that increase form left
to right starting with 100 = 1
 For fractional numbers the weights are negative powers of 10 that decrease form
left to right

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Examples
 90136 10
 600942 10

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Fractions in Decimal System
 26.75 10 =2 × 101 + 6 × 100 + 7 × 10−1 + 5 × 10−2
 =2x10 + 6x1+ 7/10 + 5/100
 =20+6+0.7+0.05
 =26+0.75
 =26.75

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Digital Systems and Binary Numbers
 Digitization got its start as far back as 1679 when Gottfried Wilhelm Leibniz
developed the first-ever binary system
 He then authored a remarkable and well-crafted book known as the
Explanation of Binary Systems in 1703 to help other people understand the
binary system
 In 1847, Boolean algebra was introduced by George Boole.
 This unique algebra was used in the mathematical analysis of and has played a
massive role in discovering mathematical logic used in digitization today.
 Claude Shannon, an MIT tech student, compiled his master’s thesis that
exhaustively centered on developing and using digital circuits.
 Shannon studied Boolean algebra law, which he incorporated to show how digital
circuits can be made and used in the telecommunication industry.

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Digital Systems and Binary Numbers
 The advantage of the binary system is that it can be used to represent numbers
in systems which are capable of being in two mutually exclusive states.
 The Frenchman Raymond Louis André Valtat from Paris patented in 1932 in
Germany a calculator that worked on the binary system.
 In his paper he advocated the usage of the binary system in a calculating
apparatus in comparison to the decimal system, for instance that the computation
of a square root is especially simple in this system.
 In 1936 Alan Turing designed an electromechanical multiplier.
 In 1938, the American George Stibitz built an binary adder using
electromechanical relays.
 In 1946 A. W. Burks, H. H. Goldstine and J. von Neumann published a
memorandum in which they advocated abandoning the decimal representation
in favor of the binary system.

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Binary Number
 Digital Systems manipulate discrete quantities of information in binary form
 Strings of binary digits (“bits”)
 Two possible values 0 and 1 which represent Two States
 On/Off
 Black/White
 Hot/Cold
 Stationary/Moving
 Combination of 0v & 5v
+5
V
–5
1 0 1
Time

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Binary Number System
 Each digit represents a power of 2
 Coefficient have two possible values 0 and 1
 𝑛 bits can store numbers from 0 to 2𝑛−1
 If 𝑛=5
 =25−1
 =32 − 1
 =31
 𝑛 bits can store 2𝑛
distinct combinations( permutations to be exact) of 1’s and 0’s
 Each coefficient 𝑎𝑗 is multiplied by 2𝑗
 So 101 binary is
 1 x 22 + 0 x 21 + 1 x 20
 Or
 1 x 4 + 0 x 2 + 1 x 1 = 5

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Binary Numbers
 Each digit represents a power of 2
 Coefficient have two possible values 0 and 1
 𝑛 bits can store numbers from 0 to 2𝑛−1
 If 𝑛=5
 =25−1
 =32 − 1
 =31
 𝑛 bits can store 2𝑛
distinct combinations( permutations to be exact) of 1’s and 0’s
 Each coefficient aj is multiplied by 2𝑗
 So 101 2 binary is
 1 × 22
+ 0 × 21
+ 1 × 20
 Or
 1 × 4 + 0 × 2 + 1 × 1 = 5 10

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1.3 NUMBER BASE CONVERSIONS

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Number Base Conversions
 Representations of a number in a different radix are said to be equivalent if they
have the same decimal representation.
 For example, 0011 8 and 1001 2 are equivalent—both have decimal value 9.
 The conversion of a number in base 𝑟 to decimal is done by expanding the
number in a power series and adding all the terms as shown previously.
 We now present a general procedure for the reverse operation of converting a
decimal number to a number in base 𝑟.
 If the number includes a radix point, it is necessary to separate the number into
an integer part and a fraction part, since each part must be converted differently

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Binary  Decimal Example
1 0 0 1 1 1 0 0
7 6 5 4 3 2 1 0
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 1
128 + 0 + 0 + 16 + 8 + 4 + 0 + 0 = 156
What is 10011100 in decimal?

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Binary  Decimal Example
1 0 0 1 1 1 0 0
Lec 1
4 3 2 1 0 -1 -2 -3
24 23 22 21 20 2-1 2-2 2-3
16 8 4 2 1 1/2 1/4 1/8
16 + 0 + 0 + 2 + 1 + 0.5 + 0 + 0 = 19.5
What is 10011.100 in decimal?

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Decimal to Binary
 Long Division Method
 19 10 = ? 2 45 10 = ? 2
2 19
2 9-1
2 4-1
2 2-0
1-0
(10011)2

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Decimal(Fractional)  Binary (Fraction)
 Convert decimal 0.6875 to binary
 IntegerFraction Coefficient
 0.6875 x 2= 1.3750 0.3750 𝑎−1 = 1
 0.3750 x 2= 0.7500 0.7500 𝑎−2 = 0
 0.7500 x 2= 1.5000 0.5000 𝑎−3 = 1
 0.5000 x 2= 1.0000 0.0000 𝑎−4 = 1
 0.6875 10 = 0.1011 2

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RULES
 Begin with the decimal fraction and multiply by 2. The whole number part of
the result is the first binary digit to the right of the decimal point.
 .625 x 2 = 1.25, the first binary digit to the right of the point is 1.
 Next discard the whole number part of the previous result (the 1 in this case)
and multiply by 2 once again. The whole number part of this new result is
the second binary digit to the right of the point.
 .25 x 2 = 0.50, the 2nd binary digit to the right of the point is 0.
 Continue this process until we get a zero as our decimal part or until we
recognize an infinite repeating pattern.
 0.50 x 2=1.00, 3rd digit to the right of decimal point is 1. Now the fractional part
becomes zero so algorithm will stop here.
 (0.625)10= (0.101)2

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(834.153)10 -------> Binary
2 834
2 417-0
2 208-1
2 104-0
2 52-0
2 26-0
2 13-0
2 6-1
2 3-0
2 1-1
0.153 x 2=0.306 0
0.306 x 2=0.612 0
0.612 x 2=1.224 1
0.224 x 2=0.448 0
0.448 x 2=0.896 0
0.896 x 2=1.792 1
0.792 x 2=1.584 1
0.584 x 2=1.168 1
0.168 x 2=0.336 0
0.336 x 2=0.672 0
(1101000010.0010011100)2

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1.4 OCTAL/ HEXADECIMAL SYSTEMS

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Decimal to Octal
8 175
8 21-7
2-5
Example :
(175)10
Ans :
(257)8
8 10745
8 1343-
1
8 167-7
8 20-7
2-4
Example: (10745)10
Ans : (24771)8

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Decimal to Octal (Fraction)

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Decimal to Octal Conversion
Example: (0.3125)10

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Binary - Octal
 Binary decimal Octal
 Octal decimal Binary

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Binary − Octal Conversion
 8 = 23
 Each group of 3 bits represents an octal digit
Octal Binary
0 0 0 0
1 0 0 1
2 0 1 0
3 0 1 1
4 1 0 0
5 1 0 1
6 1 1 0
7 1 1 1
Example:
( 1 0 1 1 0 . 0 1 )2
( 2 6 . 2 )8
Assume Zeros
Works both ways (Binary to Octal & Octal to Binary)

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Decimal- Hexadecimal
 Example (10068)10
 Example (9054)10
16 10068
16 629-4
16 39-5
2-7
16 1529
16 95-9
16 5-F
16 9054
16 565-E
16 35-5
16 2-3
Ans: (2754)16
Ans: (5F9)16
Ans: (235E)16
Example (1529)10

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Decimal-Hexadecimal Conversion
 Example 2𝐶5𝐸 16
 = 2 × 163 + 𝐶 × 162 + 5 × 161 + 𝐸 × 160
 = 2 × 4096 + 12 × 256 + 5 × 16 + 14 × 1
 = 8192 + 3072 + 80 + 14
 = 11358 10

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Binary − Hexadecimal Conversion
 16 = 24
 Each group of 4 bits represents a hexadecimal digit
Hex Binary
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
4 0 1 0 0
5 0 1 0 1
6 0 1 1 0
7 0 1 1 1
8 1 0 0 0
9 1 0 0 1
A 1 0 1 0
B 1 0 1 1
C 1 1 0 0
D 1 1 0 1
E 1 1 1 0
F 1 1 1 1
Example:
( 1 0 1 1 0 . 0 1 )2
( 1 6 . 4 )16
Assume Zeros
Works both ways (Binary to Hex & Hex to Binary)

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Octal to Hexadecimal via Binary
Conversion from Octal to Hexadecimal and vice versa
2138
010 00 1 011
8 B16
Convert 2138 into an hexadecimal number:

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Octal − Hexadecimal Conversion
 Convert to Binary as an intermediate step
Example:
( 0 1 0 1 1 0 . 0 1 0 )2
( 1 6 . 4 )16
Assume Zeros
Works both ways (Octal to Hex & Hex to
Octal)
( 2 6 . 2 )8
Assume Zeros

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Decimal, Binary, Octal and Hexadecimal
Decimal Binary Octal Hex
00 0000 00 0
01 0001 01 1
02 0010 02 2
03 0011 03 3
04 0100 04 4
05 0101 05 5
06 0110 06 6
07 0111 07 7
08 1000 10 8
09 1001 11 9
10 1010 12 A
11 1011 13 B
12 1100 14 C
13 1101 15 D
14 1110 16 E
15 1111 17 F

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1.5 BINARY ARITHMATIC

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Arithmetic Addition
Rules for binary
0+0=0
0+1=1
1+1=0 and carry 1 to the next more significant bit
00011010 +
00001100 =
00100110
augen
d
adden
d
1 1 carries
0 0 0 1 1 0 1 0 = 26(base 10)
+ 0 0 0 0 1 1 0 0 = 12(base 10)
0 0 1 0 0 1 1 0 = 38(base 10)
00010011 + 00111110
= 01010001
1 1 1 1 1
carries
0 0 0 1 0
0 1 1
= 19(base 10)
+ 0 0 1 1 1 1 1
0
= 62(base 10)
0 1 0 1 0 0 0
1
= 81(base 10)

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Rules for Subtraction
00100101 -
00010001 =
00010100
minuend
subtrahend
0 borrows
0 0 1 10 0 1 0 1 = 37(base 10)
- 0 0 0 1 0 0 0 1 = 17(base 10)
0 0 0 1 0 1 0 0 = 20(base 10)
00110011 - 00010110 =
00011101
0 10 borrows
0 0 1 1 1 0 10 1
1
= 51(base 10)
-
0 0 0 1 0 1 1 0
= 22(base 10)
0 0 0 1 1 1 0 1 = 29(base 10)
0 - 0 = 0
0 - 1 = 1, and borrow 1 from the next more significant bit
1 - 0 = 1
1 - 1 = 0

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Multiplication
 Four Basic rules for multiplication:
 0 x 0 = 0
 0 x 1 = 0
 1 x 0 = 0
 1 x 1 = 1
 Example
11 x 11 111 x 101
11 (Multiplicand) 111
x 11(Multiplier) x 101
11 111
11 000
1001 111
100011

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Division
 110 ÷11 110÷10


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Hexadecimal Addition
Carry 1
2 A C 6 6+5=11d  Bh
+ 9 2 B 5 C+B=23d  17h
B D 7 B A+2+1=13d  Dh
2+9=11d  Bh

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Hexadecimal Subtraction
Borrow 1 1 1
9 2 B 5 21-6=(15)10  (F)16
- 2 A C 6 26-C=(14)10 (E)16
6 7 E F 17-A=(7)10 (7)16
8-2=(6)10 (6)16

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Binary Number Representation
 Our computer can understand only (0, 1)
language.
 The positive numbers are easy to
represent in machine language and
perform arithmetic
 Negative numbers
 need a sign representation for storage
 Subtraction circuitry is more complex
than addition
 The solution is signed number
representation and complements

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Signed Binary Number Representation
 Sign-Magnitude form
 A binary number has a bit for a sign symbol.
 If this bit is set to 1, the number will be negative else the number will be positive if it is set
to 0. Apart from this sign-bit, the n-1 bits represent the magnitude of the number.
 1's Complement – Primarily used for subtraction
 Invert each bit of a number
 The negative numbers can also be represented in the form of 1's complement.
 In this form, the binary number also has an extra bit for sign representation as a sign-
magnitude form.
 2's Complement – Primarily used for subtraction
 Invert each bit of a number and add 1 to its least significant bit
 The negative numbers can also be represented in the form of 2's complement.
 In this form, the binary number also has an extra bit for sign representation as a sign-
magnitude form.

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Complements
 Complementation in mathematics and computing is a technique for encoding symmetric
ranges of positive and negative integers in such a way that the same algorithm (or
mechanism) can be used for addition over the entire range
 For a given number of digits, half of the possible number representations encode
positive numbers and the other half represent their additive reciprocals.
 A pair of mutually additive reciprocals is called a complement.
 Therefore, subtraction of any number is implemented by adding its complement.
 A change of sign of any number is encoded by generating its complement.
 Among other things, it is used by most digital computers to perform subtraction,
represent negative numbers in base-2 or binary arithmetic, and test computations for
underflow and overflow.

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Complements
 There are two types of complements for each base-r system Where r
represents the base and n represents the number of digits
1. the radix complement ( r’s complement) and
rn-N
For Binary numbers : 2’s complement
For decimal numbers: 10’s complement
2. diminished radix complement (r-1 complement).
( rn-1)-N
For Binary numbers : 1’s complement
For decimal numbers: 9’s complement

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r and (r-1)’s Complement with r = 2
 1's complement of the binary number by simply inverting the given number.
 Let us take N = 010010 and r = 2 and n = 6. So directly 1’s complement of N is
 111111-010010 = 101101
 2’s complement is 1’s complement +1 i.e. change 1 to 0 and 0 to 1 and then add 1
to the number.
 2’s complement of 010101 is
 101010+1 = 101011

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r and (r-1)’s Complement with r = 10
 If we are given a number N in base-r having n digits the (r-1)’s complement or
Diminished Radix complement is defined as
 (𝑟𝑛 − 1) − 𝑁
 Let us take N =1988. Here, r = 10 and n = 4, so 9’s complement of 1988 is
 9999 - 1998 = 8001
 Let us take N = 01234. Here, r = 10 and n = 5, so 9’s complement of 01234 is
 99999 - 01234 = 98765
 If we are given a number N in base-r having n digits the (r)’s Complement or
Radix Complement is defined as:
 𝑟𝑛
− 𝑁
 This is also the same as adding 1 to (r-1)’s complement to get r’s complement.
 Let N = 12345 and n = 5 and r = 10. So 10’s complement of N is
 100000 - 12345 = 98765

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Examples
 Diminished Radix Complement
 The 9’s complement of 546700 is 999999 – 546700 = 453299
 The 9’s complement of 012398 is 999999 – 012398 = 987601
 The 1’s complement of 1011000 is 0100111
 Radix Complement

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Complements
■ Subtraction with r Complements
The subtraction of two n-digit unsigned numbers M – N in base r can be done as follows:

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Complements
Example
Using 10's complement, subtract 72532 – 3250.
Example
Using 10's complement, subtract 3250 – 72532
There is no end carry.
Therefore, the answer is – (10's complement of 30718) =  69282.

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Subtraction Using 2’s Complement
 Subtract (2)10 from (7)10 using 2’s complement
 Steps
1. Convert the minuend and subtrahend into binary
1. 2 10 = 0010
2. 7 10 = 0111
2. Now the statements become
1. 0111 – 0010
3. Taking the 2’scomplement of subtrahend
1110
0 1 1 1
+1 1 1 0
0 1 0 1

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Complements
Example
Given the two binary numbers X = 1010100 and Y = 1000011,
perform the subtraction (a) X – Y and (b) Y  X by using 2's
complement.
There is no end
carry. Therefore,
the answer is Y – X
=  (2's
complement of
1101111) = 
0010001.

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Complements
 Subtraction of unsigned numbers can also be done by
means of the (r  1)'s complement. Remember that the (r  1) 's
complement is one less then the r's complement.
Example
There is no end carry,
Therefore, the answer is Y – X
=  (1's complement of 1101110)
=  0010001.

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1.6 SIGNED BINARY NUMBERS

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Signed Binary Numbers
 A signed binary number consists of both magnitude and sign
 The sign indicates whether a number is positive or negative and the magnitude
is the value of the number
 There are three ways to represent the signed numbers
 Sign magnitude
 1’s Complement
 2’s Complement
 The left-most bit in a signed number is the sign bit
 0 for +ve number
 1 for –ve number

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Sign Magnitude System
 Left-most bit represents the sign
 And the remaining bits are the magnitude bits
 For example +25 decimal number is expressed as an 8 bit binary number as
» 0 0 0 1 1 0 0 1
 Decimal -25
» 10011001
 In the sign magnitude system , a negative number has the same magnitude bits
as the corresponding positive number except the sign bit
Magnitude
Sign
Bit

72
1’s Complement
 Negative numbers are 1’s complement of the corresponding positive numbers
 +25=(00011001)2
 1’s complement= 11100110
 Decimal value of the signed number
-27 +26 +25 +24+23+22+21+20 27 +26 +25 +24+23+22+21+20
=-128+64+32+0+0+4+2+0 =0+0+0+16+8+0+0+1
=-26 =+25
Adding 1 to the result
=-25

73
2’s Complement
 In the 2’s complement system a negative number is the 2’s complement of the
corresponding positive number
 +25 = (00011001)
 Its 2’s complement =11100111
Decimal value of the signed number
-27 +26 +25 +24+23+22+21+20 =27 +26 +25 +24+23+22+21+20
=-128+64+32+0+0+4+2+1 =0+0+0+16+8+0+0+1
=-25 =+25

74

75
Range of Signed Integer Numbers
 For 2’s complement signed numbers , the range of values for n-bit numbers is
 −(2𝑛−1
) to (2𝑛−1
− 1)
 Let n=4
 −(24−1) to (24−1 − 1)
 -8 to 7
 For 1’s complement signed numbers , the range of values for n-bit numbers is
 − 2𝑛−1 − 1 to (2𝑛−1 − 1)
 Let n=4
 −(24−1 − 1) to (24−1 − 1)
 -7 to +7

76
Signed Binary Numbers
 The addition of two numbers in the signed-magnitude system follows the rules
of ordinary arithmetic.
 If the signs are the same, we add the two magnitudes and give the sum the
common sign.
 If the signs are different, we subtract the smaller magnitude from the larger and
give the difference the sign of the larger magnitude.
 The addition of two signed binary numbers with negative numbers represented
in signed-2's-complement form is obtained
from the addition of the two numbers,
including their sign bits.
 A carry out of the sign-bit position is
discarded.

77
Arithmetic Operations with Signed Numbers
 Negative Number must be in 2’s complement
 Any carry out sign bit must be discarded
 Negative result is already in 2’s complement
 To get the correct answer, there must be sufficient bits to accommodate the
answer

78
Overflow
 Place of the sign bit is fixed from the beginning of the problem.
 Number of bits for magnitude is also fixed
 If the magnitude exceeds the allotted bits, overflow occurs
 Overflow occurs when:
 Two negative numbers are added and an answer comes positive or
 Two positive numbers are added and an answer comes as negative.

79
Overflow Example
 Let total number of bits =6 (including the sign bit)
 Range (−26−1 𝑡𝑜 26−1)
 -32 to +31
 Perform (-17)10 + (-19)10
 -17=101111
 -19=101101
 As per rule discard carry
 011100----- is it’s a positive number ?? (+28)10
1 0 1 1 1 1
1 0 1 1 0 1
0 1 1 1 0 0
1

80
Overflow Example
 Repeat the example using 7 bits
 -17=1101111
 -19=1101101
 Discard carry
 1011100 -ve number
 (-36)10
1 1 0 1 1 1 1
1 1 0 1 1 0 1
1 0 1 1 1 0 0
1

81
Overflow Detection
 MSB
 Cin≠Cout

82
BCD
 Binary Coded Decimal
 Decimal digits stored in binary
 Four bits/digit (Use 10 instead of 16)
 Like hex, except stops at 9
 Example
 931 is coded as 1001 0011 0001
 People understand decimal system better
 Written differently but decimal value is same
 Decimal 15 in BCD 0001 0101 and in Binary it was 1111

83
BCD Addition
 Since each digit is max 9 Sum will always be less than 19= 9+9+1(previous
carry)
 Two BCD digits are added as binary numbers
 When binary sum is more than binary 1001 2, result is invalid
 Addition of 6 = 0110 2 make a correct BCD and produces a carry
 Binary Sum carry and Decimal Carry differ by 16-10=6

84
BCD Addition
 Multi-digit BCD numbers can be added together
23 0010 0011
45 0100 0101
68 0110 1000
23 0010 0011
48 0100 1000
71 0110 1011
 1011 is illegal BCD number

85
BCD Addition
 Add a 0110 (6) to an invalid BCD number
 Carry added to the most significant BCD digit
23 0010 0011
48 0100 1000
71 0110 1011
0110
0111 0001
 BCD adders add BCD values directly, digit by digit, without converting the
numbers to binary. However, it is necessary to add 6 to the result if it is greater
than 9.
 BCD adders require significantly more hardware and no longer have a speed
advantage of conventional binary adders

86
Gray Code
 Gray Code is employed in many electronic devices that use rotary switches for
position encoders.
 The Gray Code also called “reflected binary code” or (RBC) was named after
Bell Labs researcher Frank Gray
 Mechanical switches that output binary results suffer from errors caused by
multiply contacts not being in synchronicity so that the switches don’t make
and brake at precisely the same time
 It is possible to get wrong readings from them when switching from one
number to the next.
 Take the number 3 to 4 transition in binary switching as a prime example, in
binary the switch contacts would go from “011” binary to “100” binary that is a
change of all three switch contacts with different make and break times,
 A data system could read any one of 8 different combinations of numbers at the
time of switch transition.

87
Gray Code
 Gray Code
 The advantage is that only bit in the code group
changes in going from one number to the next.
 Error detection.
 Representation of analog data.
 Low power design.
000 001
010
100
110 111
101
011
1-1 and onto!!

88
Gray Code
Decimal Gray Binary
0 0000 0000
1 0001 0001
2 0011 0010
3 0010 0011
4 0110 0100
5 0111 0101
6 0101 0110
7 0100 0111

90
Excess-3 – Self Complementing Code
 Binary codes for decimal digits require 4 bits per digit
 Many codes use 4 bits in 10 distinct possible combinations (out of 16)
 Dec Binary BCD Excess-3
 0 0 0000 0011
 1 1 0001 0100
 2 10 0010 0101
 3 11 0011 0110
 4 100 0100 0111
 5 101 0101 1000
 6 110 0110 1001
 7 111 0111 1010
 8 1000 1000 1011
 9 1001 1001 1100
 10 1010 - -
 11 1011 - -

91
ALPHANUMERIC CODES - ASCII Character Codes
 American Standard Code for Information Interchange
 This code is a popular code used to represent information sent as character-
based data. It uses 7-bits to represent:
 94 Graphic printing characters.
 34 Non-printing characters
 Some non-printing characters are used for text format (e.g. BS = Backspace, CR
= carriage return)

92
ASCII Code: B7B6B5 B4B3B2B1
H=(1001000)

93
Even Parity
 Sometimes high-order bit of ASCII coded to enable detection of errors
 Even parity – set bit to make number of 1’s even
 Examples
A (01000001) with even parity is 01000001
C (01000011) with even parity is 11000011

94
Odd Parity
 Similar except make the number of 1’s odd
 Examples
A (01000001) with odd parity is 11000001
C (01000011) with odd parity is 01000011

95
Error Detection
 Note that parity detects only simple errors
 One, three, etc. bits
 More complex methods exist
 Some that enable recovery of original info
 Cost is more redundant bits

96
Odd Parity Error Detection
 Original data 10011010
 With Odd Parity 110011010
 1-bit error 110111010
 Number of 1s even indicates 1-bit error
 2-bit error 110110010
 Number of 1s odd no error indicated
 3-bit error 100110010
 Number of 1s even indicates error

97
1.8 BINARY STORAGE AND REGISTERS

98
Binary Storage and Registers
 Physical existence in information storage medium for storing individual bits
 A binary cell is a device that posses stable stages and is capable of storing one
bit of information
 A Register is a group of binary cells.
 Can store any discrete quantity of information that contains n bits.
 1100001111001001 is a 16 bit register
 2n possible states to store 0 to 2n -1 number
 Contents can be interpreted differently

99
Register Transfers
 Basic Operation in digital systems
 When Key is pressed 8 bit alphanumeric character code in to Input Register
 Contents of Input Register are transferred to eight least significant cells of a
Processor Register
 After every transfer input register is cleared for new keystroke
 Each eight bit character transfer to the processor register is preceded by shift of
previous character to next eight cells on its left
 When Processor Register is full, its contents are transferred to the Memory
Register

100
Manipulation of binary variable
 Adding two 10 bit binary numbers
 Memory Unit
 Processor Unit

101
1.9 Binary Logic
 Definition of Binary Logic
 Binary logic consists of binary variables and a set of logical operations.
 The variables are designated by letters of the alphabet, such as A, B, C, x, y, z, etc, with each variable having
two and only two distinct possible values: 1 and 0,
 Three basic logical operations: AND, OR, and NOT.

102
1.9 BINARY LOGIC

103
Binary Logic
x
y z
 Truth Tables, Boolean Expressions, and Logic Gates
x y z
0 0 0
0 1 0
1 0 0
1 1 1
x y z
0 0 0
0 1 1
1 0 1
1 1 1
x z
0 1
1 0
AND OR NOT
x
y z
z = x • y = x y z = x + y z = x = x’
x z

104
‫ه‬‫للا‬ ‫جزاك‬
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