001 digital 315 slid
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DIGITAL ELECTRONICS
BASICS
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Introduction
to
Digital Electronics
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Increasing levels of development and complexity:
Transistors built from semiconductors
Logic gates built from transistors
Logic functions built from gates
Flip-flops built from logic
Counters and sequencers from flip-flops
Microprocessors from sequencers
Computers from microprocessors
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Most natural quantities that we see are analog and vary continuously.
Analog systems can generally handle higher power than digital systems.
Analog Quantities
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A digital quantity has a set of discrete values.
Digital Quantities
Digital systems can process, store, and transmit data more efficiently but can only assign discrete values to each point.
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Digital waveforms change between the LOW and HIGH levels.
A positive going pulse is one that goes from a normally LOW logic level to a HIGH level and then back again.
Digital waveforms are made up of a series of pulses.
Digital Waveforms
Falling orleading edge
(b) Negative–going pulse
HIGH
Rising ortrailing edge
LOW
(a) Positive–going pulse
HIGH
Rising orleading edge
Falling ortrailing edge
LOWt0 t1 t0 t1
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Analog ExampleA public address system, used to amplify sound so that it can be heard by large audience, is one example of an application of analog electronics.
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Digital Example• A computer system is one example of an application of
digital electronics.
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A Mixed System• The compact disk (CD) player is an example of a system in
which both digital and analog circuits are used.
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Analog Vs. Digital
• Analog– Continuous– Can take on any values in a given range– Very susceptible to noise
• Digital– Discrete– Can only take on certain values in a given range– Can be less susceptible to noise
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Advantages Over Analog
• Programmability• Predictable accuracy• Maintainability• Processed more efficiently and reliably• Compact storage• Does not affected by noise as well as analog
values
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Example
Controlling a storage tank system for a pancake syrup manufacturing
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Example
A key-coded deadbolt
K_L[11..0]
K_L9
K_L5
K_L1
K_L10
K_L6
K_L2 K_L3
K_L8
K_L4
K_L0
K_L11
K_L7
DEADLOCK CONTROL
ON-OFF-CONTROL
GS
Z
D1
Q1BC547
R112k
R2
1k
RELAY4V
B1120V
VCC
D41N4148
VCC
GND
ACTUATOR
120V
RA
100MEG
C[3..0]
KEY_ENCODER
ENCODER
COD[3..0]
K_L[11..0]
0 1 2 34 5 6 78 9 * #
GS
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What digital electronics do you use?• Computer• CD & DVD players• IPod• Cell phone• HDTV• Digital cameras
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What are digital electronics?
• Sound is an analog signal.• On a CD, digital sound is
encoded as 44.1 kHz, 16 bit audio. – The original wave is 'sliced' 44,100
times a second - and an average amplitude level is applied to each sample.
– 16 bit means that a total of 65,536 different values can be assigned, or quantized to each sample.
• DVD-Audio can be 96 or 192 kHz and up to 24 bits resolution
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ADC
Analog to digital converder-Sampling time/sampling frequency fs-Number of bits
Sample and hold
AnalogTo
Digital
10011001
1.4V
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Sampling
Time
Level
Level
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Data can be transmitted by either serial transfer or parallel transfer.
Serial and Parallel Data
Computer Modem
1 0 1 1 0 0 1 0
t0 t1 t2 t3 t4 t5 t6 t7
Computer Printer
0
t0 t1
1
0
0
1
1
0
1
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Serial communication between computers.
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Parallel communication between a computer and a printer.
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• A digital electronics device that combine hardware and software to accept the input of data, process and store the data, and produce some useful output.
A Computer is…
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Digital Electronics
• Digital electronics devices store and process bits electronically.– A bit represents data using 1’s and 0’s– Eight bits is a byte – the standard grouping in
digital electronics– Digitization is the process of transforming
information into 1’s and 0’s
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Microprocessor
• The computer you are using to read this page uses a microprocessor to do its work.
• The microprocessor is the heart of any normal computer .
• The microprocessor you are using might be a Pentium, a K6, a PowerPC, a Sparc or any of the many other brands and types of microprocessors.
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Microprocessor (Processor)• Designed to process instructions • Largest chip on motherboard • Intel: world’s largest chipmaker (Pentiums)• AMD: Cheaper chips (Athlons)
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Motherboard
• Main circuit board
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Processor Components
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Processor Performance
• Speed: processor clock set clock speed (MHz or GHz )
• Word Size: number of bits the processor can manipulate at one time (32-bit or 64-bit)
• Cache: high speed memory (kilobytes)
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Memory Types
• Random Access Memory (RAM)• Virtual Memory • Read-Only Memory (ROM)• CMOS
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Memory Cells
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Physical File Storage • Storage medium formatted into
tracks /sectors electronically
• File system keeps track of names and file locations.
• Clusters: a group of sectors that speeds up storage and retrieval
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Digital Data Representation • The form in which information
is conceived, manipulated and recorded on a digital device.
• Uses discrete digits/electronic signals
Byte = 8 bits = 1 character
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Numeric Data
• Consists of numbers representing quantities used in arithmetic operations.– Binary system, “Base 2”- 1,0 (bits - binary digits)- On/Off, Yes/No
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Digital Technology Metrics
Kilo, Mega, Giga, what comes next?
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Binary Digits• The two digits in the binary system, 1 and 0, are called bits,
which is a contraction of the words binary digit.
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How to represent 0 and 1?
• In digital circuits, two different voltage levels are used to represent the two bits.
• The higher/lower voltage level is referred to as a HIGH/LOW, or H/L.
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Logic Levels
• The voltages used to represent a 1 and a 0 are called logic levels.
• HIGH can be any voltage between a specified minimum value and a specified maximum value.
• LOW can be any voltage between a specified minimum and a specified maximum.
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Binary Representations
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Positive Logic System• A 1 is represented by HIGH and a 0 is represented by LOW.• Also called ACTIVE HIGH LOGIC
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Negative Logic System• A 0 is represented by HIGH and a 1 is represented by LOW.• Also called ACTIVE LOW LOGIC
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Codes• Groups of bits (combination of 1s and 0s), called codes, are
used to represent numbers, letters, symbols, instructions, and anything else required in a given application.
• The American Standard Code for Information Interchange (ASCII) – pronounced “askee” – is a universally accepted alphanumeric code used in most computers and other electronic equipment.
1
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ASCII
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Digital Waveforms
• Digital waveforms consists of a series of pulses, (voltage levels that are changing back and forth between the HIGH and LOW levels).
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Pulse Train
• Digital waveforms are sometimes called pulse trains.
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Ideal Pulses
• A single positive-going pulse is generated when the voltage goes from its normally LOW level to its HIGH level and then back to its LOW level.
• A single negative-going pulse is generated when the voltage goes from its normally HIGH level to its LOW level and then back to its HIGH level.
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Nonideal Pulse• The time required for the pulse to go from its LOW (HIGH) level
to its HIGH (LOW) level is called the rise (fall) time. • In practice, it is common to measure rise (fall) time from
10%(90%) of the pulse amplitude to 90%(10%) of the pulse amplitude.
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Nonideal Pulse• The pulse width is a measure of the duration of the
pulse and is often defined as the time interval between the 50% points on the rising and falling edges.
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Periodic Pulse• A periodic waveform is one that repeats itself at a fixed
interval, called a period (T ). • The frequency (f ) is the rate at which it repeats itself and is
measured in hertz (Hz). • The relationship between f and T is expressed as follows:
fT
Tf
1,
1
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Periodic Pulse• The duty cycle (D ) is defined as the ratio of the pulse width (tw )
to the period (T ) and can be expressed as a percentage.
%100
T
tD W
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Nonperiodic Pulse
• A nonperiodic waveform, of course, does not repeat itself at fixed intervals.
• They are composed of pulses of randomly differing pulses widths and/or randomly differing time intervals between the pulses.
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A Digital Waveform Carries Binary Information• Binary information that is handled by digital systems appears
as waveforms that represent sequences of bits. • When the waveform is HIGH, a binary 1 is present; when the
waveform is LOW, a binary 0 is present. • Each bits in a sequence occupies a defined time interval called
a bit time, or bit interval.
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The Clock• In digital systems, all digital waveforms are
synchronized with a basic timing waveform, called the clock.
• The clock is a periodic waveform. • The clock waveform itself does not carry information.
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Timing Diagrams• A timing diagram is a graph of digital waveforms
showing the actual time relationship of two or more waveforms and how each changes in relation to the others.
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Data Transfer• Data refers to groups of
bits that convey some type of information.
• Binary data, which are
represented by digital waveforms, must be transferred from one circuit to another within a digital system or from one system to another in order to accomplish a given purpose.
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Serial Data Transfer• When bits are transferred in serial form from one point to
another, they are sent one bit at a time along a single conductor.
• To transfer n bits in series, it takes n time intervals.
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Parallel Data Transfer• When bits are transferred in parallel form, all the bits in a
group are sent out on separate lines at the same time. • There is one line for each bit. • To transfer n bits in parallel, it takes one time interval.
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Number System
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It depends on the number system
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Number Systems
• To talk about binary data, we must first talk about number systems
• The decimal number system (base 10) you should be familiar with!
• Positional number system
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Positional Notation
•Value of number is determined by multiplying each digit by a weight and then summing.
•The weight of each digit is a POWER of the RADIX (also called BASE) and is determined by position.
22
11
00
11
22
33
210123 .
rararararara
aaaaaa
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Radix (Base) of a Number System
• Decimal Number System (Radix = 10)Eg:- 7392 = 7x103+3x102+9x101+2x100
• Binary Number System (Radix = 2)Eg:- 101.101 = 1x22+0x11+1x20+1x2-1+0x2-11x2-2
• Octal number system (radix = 8)
• Hexadecimal number system (radix = 16)
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Radix (Base) of a Number System• When counting upwards in base-10, we increase the
units digit until we get to 10 when we reset the units to zero and increase the tens digit.
• So, in base-n, we increase the units until we get to n when we reset the units to zero and increase the n-s digit.
• Consider hours-minutes-seconds as an example of a base-60 number system:– Eg. 12:58:43 + 00:03:20 = 13:02:03
NB. The base of a number is often indicated by a subscript. E.g. (123)10 indicates the base-10 number 123.
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Decimal Number Systems • Base 10
– Ten digits, 0-9– Columns represent (from right to left) units, tens,
hundreds etc.
123
1´102 + 2´101 + 3´100
or 1 hundred, 2 tens and 3 units
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Binary Number System
• Base 2– Two digits, 0 & 1– Columns represent (from right to left) units, twos,
fours, eights etc.
1111011
1´26 + 1´25 + 1´24 + 1´23 + 0´22 + 1´21 + 1´20
= 1´64 + 1´32 + 1´16 + 1´8 + 0´4 + 1´2 + 1´1
= 123
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Binary Numbers• Each binary digit (called bit) is either 1 or 0
• Bits have no inherent meaning, can represent
– Unsigned and signed integers
– Characters
– Floating-point numbers
– Images, sound, etc.
• Bit Numbering
– Least significant bit (LSB) is rightmost (bit 0)
– Most significant bit (MSB) is leftmost (bit 7 in an 8-bit number)
1 0 0 1 1 1 0 1
27 26 25 24 23 22 21 20
01234567
MostSignificant Bit
LeastSignificant Bit
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Data Organization
• Bits – or one, true or false, on or off, male or female, and
right or wrong. • Nibbles
– Group of 4 bits
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Bytes
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Words
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Double Words
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Octal Number System
• Base 8– Eight digits, 0-7– Columns represent (from right to left) units, 8s,
64s, 512s etc.
173
1´82 + 7´81 + 3´80 = 123
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Hexadecimal Number System
• Base 16– Sixteen digits, 0-9 and A-F (ten to fifteen)– Columns represent (from right to left) units, 16s,
256s, 4096s etc.
7B
7´161 + 11´160 = 123
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Hexadecimal Integers• More convenient to use than binary numbers
Binary, Decimal, and Hexadecimal Equivalents
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Decimal to Binary Conversion
123 ¸ 2 = 61 remainder 1 61 ¸ 2 = 30 remainder 1 30 ¸ 2 = 15 remainder 0 15 ¸ 2 = 7 remainder 1 7 ¸ 2 = 3 remainder 1 3 ¸ 2 = 1 remainder 1 1 ¸ 2 = 0 remainder 1
Least significant bit (LSB) (rightmost)
Most significant bit (MSB) (leftmost)
Answer : (123)10 = (1111011)2
Example – Converting (123)10 into binary
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Decimal to Binary Conversion
The quotient is divided by 2 until the new quotient becomes 0
Integer Remainder
41
20 1
10 0
5 0
2 1
1 0
0 1 101001 answer
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Converting Decimal to Binary• To convert a fraction, keep multiplying the fractional part by
2 until it becomes 0. Collect the integer parts in forward order
• Example: 162.375:• So, (162.375)10 = (10100010.011)2
162 / 2 = 81 rem 0 81 / 2 = 40 rem 1 40 / 2 = 20 rem 0 20 / 2 = 10 rem 0 10 / 2 = 5 rem 0 5 / 2 = 2 rem 1 2 / 2 = 1 rem 0 1 / 2 = 0 rem 1
0.375 x 2 = 0.7500.750 x 2 = 1.5000.500 x 2 = 1.000
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Why does this work?• This works for converting from decimal to
any base• Why? Think about converting 162.375 from
decimal to decimal
162 / 10 = 16 rem 2 16 / 10 = 1 rem 6 1 / 10 = 0 rem 1
0.375 x 10 = 3.7500.750 x 10 = 7.5000.500 x 10 = 5.000
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Binary to Decimal Conversion• Each bit represents a power of 2
• Every binary number is a sum of powers of 2
• Decimal Value = (dn-1 2n-1) + ... + (d1 21) + (d0 20)
• Binary (10011101)2 = 27 + 24 + 23 + 22 + 1 = 157
1 0 0 1 1 1 0 1
27 26 25 24 23 22 21 20
01234567
Some common powers of 2
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Converting Binary to Decimal• For example, here is 1101.01 in binary:
1 1 0 1 . 0 1 Bits 23 22 21 20 2-1 2-2 Weights (in base 10)
(1 x 23) + (1 x 22) + (0 x 21) + (1 x 20) + (0 x 2-1) + (1 x 2-2) =
8 + 4 + 0 + 1 + 0 + 0.25 = 13.25
(1101.01)2 = (13.25)10
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Unsigned Binary Numbers
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Binary and Octal Conversions• Converting from octal to binary: Replace each octal digit with
its equivalent 3-bit binary sequence
= 6 7 3 . 1 2 = 110 111 011 . 001 010=
11170113110601021015001110040000
BinaryOctalBinaryOctal
8)12.673(
2)001010.110111011(
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Binary and Octal Conversions• Converting from binary to octal: Make groups of 3 bits,
starting from the binary point. Add 0s to the ends of the number if needed. Convert each bit group to its corresponding octal digit.
10110100.0010112 = 010 110 100 . 001 0112
= 2 6 4 . 1 38
11170113110601021015001110040000
BinaryOctalBinaryOctal
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Binary and Hex Conversions• Converting from hex to binary: Replace each hex digit with its
equivalent 4-bit binary sequence
261.3516 = 2 6 1 . 3 516
=0010 0110 0001 . 0011 01012
1111F1011B01117001131110E1010A01106001021101D1001901015000111100C100080100400000
BinaryHexBinaryHexBinaryHexBinaryHex
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Binary and Hex Conversions• Converting from binary to hex: Make groups of 4 bits, starting
from the binary point. Add 0s to the ends of the number if needed. Convert each bit group to its corresponding hex digit
10110100.0010112 = 1011 0100 . 0010 11002
= B 4 . 2 C16
1111F1011B01117001131110E1010A01106001021101D1001901015000111100C100080100400000
BinaryHexBinaryHexBinaryHexBinaryHex
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Decimal Binary Octal Hex 0 0000 0 0 1 0001 1 1 2 0010 2 2 3 0011 3 3 4 0100 4 4 5 0101 5 5 6 0110 6 6 7 0111 7 7 8 1000 10 8 9 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
Numbers with Different Bases
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Two Interpretations
101001112 16710 -8910
• Signed vs. unsigned is a matter of interpretation; thus a single bit pattern can represent two different values.
• Allowing both interpretations is useful:
Some data (e.g., count, age) can never be negative, and having a greater range is useful.
unsigned signed
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Changing the Sign
+6 = 0110
-6 = 1110
Sign+Magnitude: 2’s Complement:
+6 = 0110
+4 = 1001 +1
-6 = 1010
Invert
Increment
Change 1 bit
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Why Not Sign+Magnitude?
• Complicates addition :– To add, first check the signs. If
they agree, then add the magnitudes and use the same sign; else subtract the smaller from the larger and use the sign of the larger.
– How do you determine which is smaller/larger?
• Complicates comparators:– Two zeroes!
+3 0011+2 0010+1 0001+0 0000-0 1000-1 1001-2 1010-3 1011
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Which is Greater: 1001 or 0011?
Answer: It depends!
It’s a matter of interpretation, and depends on how x and y were declared: signed? Or unsigned?
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Why Not Sign+Magnitude?
9
3
+
12
+
-1
+3
- 4
0011
HardwareAdder
1100
1001
Right! Wrong!
Manipulates bit patterns, not
numbers!
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Why 2’s Complement?
+3 0011+2 0010+1 00010 0000-1 1111-2 1110-3 1101-4 1100
1. Just as easy to determine sign as in sign+magnitude.
2. Almost as easy to change the sign of a number.
3. Addition can proceed w/out worrying about which operand is larger.
4. A single zero!
5. One hardware adder works for both signed and unsigned operands.
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Easier Hand Method
+6 = 0110
-6 = 1010
Step 1: Copy the bits from right to left, through and including the first 1.
Step 2: Copy the inverse of the remaining bits.
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One Hardware Adder Handles Both!(or subtractor)
9
3
+
12
+
-7
+3
- 4
0011
HardwareAdder
1100
1001
Manipulates bit patterns, not
numbers!
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Twos Complement
• Most common scheme of representing negative numbers in computers
• Affords natural arithmetic (no special rules!)• To represent a negative number in 2’s
complement notation…1. Decide upon the number of bits (n)2. Find the binary representation of the +ve value in n-
bits3. Flip all the bits (change 1’s to 0’s and vice versa)4. Add 1
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Twos Complement Example• Represent -5 in binary using 2’s complement
notation1. Decide on the number of bits
2. Find the binary representation of the +ve value in 6 bits
3. Flip all the bits
4. Add 1
6 bits(for example)
111010
111010+ 1 111011
-5
000101+5
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“Complementary” Notation
• Conversions between positive and negative numbers are easy
• For binary (base 2)…
+ve
-ve
2’s C
2’s C
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Example+5
2’s C
-5
2’s C
+5
0 0 0 1 0 1
1 1 1 0 1 0+ 1
1 1 1 0 1 1
0 0 0 1 0 0+ 1
0 0 0 1 0 1
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Properties of Two's Complement Numbers
• X plus the complement of X equals 0.
• There is one unique 0.
• Positive numbers have 0 as their leading bit (MSB); while negatives have 1 as their MSB.
• The range for an n-bit binary number in 2’s complement representation is:
from -2(n-1) to 2(n-1) - 1
• The complement of the complement of a number is the original number.
• Subtraction is done by addition to the complement of the number.
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The Two’s Complement Representation
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Range of Unsigned IntegersTotal no. of patterns of n bits = 2 2 2… 2
‘n’ 2’s
= 2n
If n-bits are used to represent an unsigned integer value:
Range: 0 to 2n-1 (2n different values)
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Range of Signed Integers
• Half of the 2n patterns will be used for positive values, and half for negative.
• Half is 2n-1.
• Positive Range: 0 to 2n-1-1 (2n-1 patterns)
• Negative Range: -2n-1 to -1 (2n-1 patterns)
• 8-Bits (n = 8): -27 (-128) to +27-1 (+127)
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Binary Coded Decimal (BCD)
0111 0011
0000 0111 0000 0011
7 3
7 3
1. Packed BCD (2 digits per byte):
2.Unpacked BCD (1 digit per byte):
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What Values Can Be Represented in N Bits?
• Unsigned: 0 to 2N - 1• 2s Complement: - 2 N-1 to 2 N-1 - 1• BCD 0 to 10 N/4 - 1
• For 32 bits:
Unsigned: 0 to4,294,967,295
2s Complement: - 2,147,483,648 to 2,147,483,647 BCD: 0 to 99,999,999
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• But, what about
– Very large numbers?
9,369,396,989,487,762,367,254,859,087,678
– . . . or very small number?
0.0000000000000000000000000318579157
What Values Can Be Represented in N Bits?
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Only Solution is ……..
• FLOATING POINT REPRESENTATION
• Floating point representation allows much larger range at the expense of accuracy
• Floating point representation is otherwise called as SCIENTIFIC REPRESENTATION
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Floating Point Representation
1.02 x 10 -1.673 x 1023 -24
Radix (Base)
Mantissa(Significand): Exponent: It contain both Sign and magnitude
Decimal
It contain both Sign and magnitude
Decimal point
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Binary
1.xxxxx X 2 yyyyyyy
Number of ‘x’s determines accuracy
Number of ‘y’s determines range
Radix (Base)Binary point
Mantissa(Significand): May be Unsigned or Signed
Exponent: It contain both Sign and magnitude
Floating Point Representation
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IEEE floating point format
• IEEE defines two formats with different precisions:
1. IEEE Single Precision format
2. IEEE Double Precision format
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Single & Double Precision
8 bits 23 bits
11 bits 52 bits
Sign(1 bit) Exponent Significan
d
32 bits
64 bits
SinglePrecision
DoublePrecision
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IEEE floating point format
23.8510 = 10111.1101102 =1.0111110110 x 24
e = 127+4 = 13110 = 100000112 = 8316 or 83h
0 100 0001 1 011 1110 1100 1100 1100 1100
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Binary Arithm
etic
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Binary Arithmetic Operations
1. Binary Addition
2. Binary Subtraction1. 1’s Complement Subtraction2. 2’s Complement Subtraction
3. Binary Multiplication
4. Binary Division
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Decimal Addition Explanation
1 1 1
3 7 5 8
+ 4 6 5 7
8 4 1 5
What just happened?
1 1 1 (carry)
3 7 5 8
+ 4 6 5 7 8 14 11 15 (sum)
- 10 10 10 (subtract the base)
8 4 1 5
So when the sum of a column is equal to or greater than the base, we subtract the base from the sum, record the difference, and carry one to the next column to the left.
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Binary Addition RulesRules:
0 + 0 = 0 0 + 1 = 1 (just like in decimal) 1 + 0 = 1
1 + 1 = 210
= 102 = 0 with 1 to
carry
1 + 1 + 1 = 310
= 112 = 1 with 1 to carry
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Binary Addition Example 1
1 1 0 1 1 1 + 0 1 1 1 0 0
1
1
111
0 1 0 0 11
Example 1: Add binary 110111 to 11100
Col 1) Add 1 + 0 = 1 Write 1
Col 2) Add 1 + 0 = 1 Write 1
Col 3) Add 1 + 1 = 2 (10 in binary) Write 0, carry 1
Col 4) Add 1+ 0 + 1 = 2 Write 0, carry 1
Col 6) Add 1 + 1 + 0 = 2 Write 0, carry 1
Col 5) Add 1 + 1 + 1 = 3 (11 in binary) Write 1, carry 1
Col 7) Bring down the carried 1 Write 1
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Binary Addition Explanation
1 1 0 1 1 1 + 0 1 1 1 0 0 - .
1
1
111
0 1 0 0 11
In the first two columns, there were no carries.
In column 3, we add 1 + 1 = 2 Since 2 is equal to the base, subtract
the base from the sum and carry 1.
In column 4, we also subtract the base from the sum and carry 1.
In column 6, we also subtract the base from the sum and carry 1.
In column 5, we also subtract the base from the sum and carry 1.
In column 7, we just bring down the carried 1
22 2 23
222
What is actually happened when we carried in binary?
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Binary Addition Verification
Verification
1101112 5510
+0111002 + 2810
8310
1 0 1 0 0 1 12
= 64+0+16+0+0+2+1
= 8310
1 1 0 1 1 1 + 0 1 1 1 0 0
1 0 1 0 0 11
You can always check your answer by converting the figures to decimal, doing the addition, and comparing the answers.
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Binary Addition Example 2
Verification
1110102 5810
+ 0011112 +1510
7310
64 32 16 8 4 2 1
1 0 0 1 0 0 1
= 64 + 8 +1
= 7310
1 1 1 0 1 0 + 0 0 1 1 1 1
1
1
11
0 0 1 0 10
Example 2:Add 1111 to 111010.
11
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BCD Addition
965 - 1001 0110 0101 +
672 - 0110 0111 0010
1111 1101 0111 +
0110 0110
0001 0110 0011 0111 (1637)10
Greater than 9.So add 6 with the nibble
Add 965 and 672
Greater than 9.So add 6 with the nibble
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Decimal Subtraction Example
8 0 2 5- 4 6 5 7
Subtract 4657 from 8025:
7 9 1111
8633
1) Try to subtract 5 – 7 can’t.Must borrow 10 from next column.
4) Subtract 7 – 4 = 3
3) Subtract 9 – 6 = 3
2) Try to subtract 1 – 5 can’t. Must borrow 10 from next column.
But next column is 0, so must go to column after next to borrow.
Add the borrowed 10 to the original 0. Now you can borrow 10 from this column.
Add the borrowed 10 to the original 5.Then subtract 15 – 7 = 8.
Add the borrowed 10 to the original 1..Then subract 11 – 5 = 6
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Decimal Subtraction Explanation
So when you cannot subtract, you borrow from the column to
the left.
The amount borrowed is 1 base unit, which in decimal is 10
The 10 is added to the original column value, so you will be
able to subtract.
8633
8 0 2 5- 4 6 5 7
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Binary Subtraction Explanation
In binary, the base unit is 2
So when you cannot subtract, you borrow from the column to the left.
The amount borrowed is 2.
The 2 is added to the original column value, so you will be able to subtract.
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Binary Subtraction Example 1
1 1 0 0 1 1- 1 1 1 0 0
Example 1: Subtractbinary 11100 from 110011
20 0 2
12
1101
Col 1) Subtract 1 – 0 = 1
Col 5) Try to subtract 0 – 1 can’t. Must borrow from next column.
Col 4) Subtract 1 – 1 = 0
Col 3) Try to subtract 0 – 1 can’t. Must borrow 2 from next column.
But next column is 0, so must go to column after next to borrow.
Add the borrowed 2 to the 0 on the right. Now you can borrow from this column
(leaving 1 remaining).
Col 2) Subtract 1 – 0 = 1
Add the borrowed 2 to the original 0.Then subtract 2 – 1 = 1
1 Add the borrowed 2 to the remaining 0.Then subtract 2 – 1 = 1
Col 6) Remaining leading 0 can be ignored.
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Binary Subtraction Verification
Verification
1100112 5110
- 111002 - 2810
2310
64 32 16 8 4 2 1
1 0 1 1 1
= 16 + 4 + 2 + 1
= 2310
1 1 0 0 1 1- 1 1 1 0 0
1101 1
Subtract binary 11100 from 110011:
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Binary Subtraction Example 2
1 0 1 0 0 1- 1 0 1 0 0
Example 2: Subtractbinary 10100 from 101001
20 0 2
1101 0
Verification
1010012 4110
- 101002 - 2010
2110
64 32 16 8 4 2 1
1 0 1 0 1
= 16 + 4 + 1
= 2110
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1’s complement Subtraction• Steps:
a. First find out the 1’s Complement of the subtrahend.
b. Then do unsigned addition on the numbers.
c. If there is a carry, then take the carry out and add it to the sum for getting the final result.
d. If there is no carry, the result will be negative and we should take the 1’s Complement of the sum to get final result
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1’s complement Subtraction• Example: 7 – 4 = ?
0 1 1 1 (+7)+ 1 0 1 1 + (- 4)1 0 0 1 0
0 0 1 0+ 1
0 0 1 1 (+3)
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1’s complement Subtraction• Example: 4 – 7 = ?
0 1 0 0 (+4)+ 1 0 0 0 + (- 7) 1 1 0 0
No carry, so take the 1’s complement of the result and will be negative.
1’s complement of 1 1 0 0 is 0 0 1 1 = (-3)
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• To Subtract two binary numbers, take the 2’s Complement of the Subtrahend and add it with the Minuend.
• If there is a final carry after the leftmost column addition, discard it and the final result is the obtained one.
• If there is no carry, the result is negative. So the final result will be the 2’s Complement of the obtained and put a “ – “ sign
2’s Complement Subtraction
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• Example without CARRY:-• Q: 20 - 25
Write -25 in two's complement format.1 1 1 0 0 1 1 0 one's complement1 1 1 0 0 1 1 1 two's complement
1 1 1 0 0 1 1 1 (-25)0 0 0 1 0 1 0 0 ( 20)1 1 1 1 1 0 1 1
•No carry , so take 2’s Complement of the result.•Final Result = 0 0 0 0 0 1 0 1 = (- 5)
2’s Complement Subtraction
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• Example with CARRY:-• Q: 10 – 5• 2’s Complement of 5 = 1 1 1 1 1 0 1 1
0 0 0 0 1 0 1 0 +1 1 1 1 1 0 1 1
1 0 0 0 0 0 1 0 1
• Now discard the carry and the obtained is the result• Final result = 0 0 0 0 0 1 0 1 = (5)
2’s Complement Subtraction
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• The difference, (a – b) is computed as:
(a – b) = a + [2’s complement(b)]
In General ……………….
2’s Complement Subtraction
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• Since the negative of any number is its two's complement, the sum of a number and its two's complement is always 0
• Add +12 and -12
+12 = 000011002
-12 = 111101002
0 000000002
2’s Complement Subtraction
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Binary Multiplication
• Multiplication can’t be that hard!
– It’s just repeated addition– If we have adders, we can do multiplication also
• Remember that the AND operation is equivalent to multiplication on two bits:
a b ab
0 0 0
0 1 0
1 0 0
1 1 1
a b ab
0 0 0
0 1 0
1 0 0
1 1 1
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Binary multiplication example
• Since we always multiply by either 0 or 1, the partial products are always either 0000 or the multiplicand (1101 in this example)
• There are four partial products which are added to form the result
1 1 0 1 Multiplicandx 0 1 1 0 Multiplier
0 0 0 0 Partial products1 1 0 1
1 1 0 1+ 0 0 0 0
1 0 0 1 1 1 0 Product
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Unsigned Multiplication
1 1 0 1 x 1 0 1 1 1 1 0 1 1 1 0 1 0 0 0 01 1 0 1
1 0 0 0 1 1 1 1
Add
Shift, then add
Shift
Shift, then add
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Signed Multiplication
1 1 1 1 (-1)10
x 0 0 0 1 (+1)10
1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 (+15)10X
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Signed Multiplication
1 1 1 1 1 1 1 1 (-1)10
x 0 0 0 1 (+1)10
1 1 1 1 1 1 1 10 0 0 0 0 0 00 0 0 0 0 00 0 0 0 0 1 1 1 1 1 1 1 1 (-1)10
Sign extended
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0 0 0 0 1 1 1 1
Division of Unsigned Binary Integers
1 0 1 1
0 0 0 0 1 1 0 1
1 0 0 1 0 0 1 1
1 0 1 1
0 0 1 1 1 0 1 0 1 1
1 0 1 1
1 0 0
Quotient
Dividend
Remainder
PartialRemainder 1
Divisor
PartialRemainder 2
Q: Divide 1 0 0 1 0 0 1 1 by 1 0 1 1
Quotient = 1 1 0 1Remainder = 1 0 0
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Logic Functions
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Introductory Paragraph• In its basic form, logic is the realm of human
reasoning that tells you a certain proposition (declarative statement) is true if certain conditions are true.
• Propositions can be classified as true or false.
• Many situations and processes that you encounter in your daily life can be expressed in the form of propositional, or logic, functions.
• Since such functions are true/false or yes/no statements, digital circuits with their two-state characteristics are applicable.
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Logic Functions• Several propositions, when combined, form propositional,
or logic functions. • For example, the propositional statement “The light is
on” will be true if the “The bulb is not burned out” is true and if “The switch is on” is true.
• The first statement is then the basic proposition, and the other two statements are the conditions on which the proposition depends.
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Basis for digital computers.
• The true-false nature of logic makes it compatible with binary logic used in digital computers.
• Electronic circuits can produce logic operations.
• Circuits are called gates.– NOT– AND– OR
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Basic Logic Operations• There are three basic logic operations: NOT, AND, and OR.
• Each of the three basic logic operations produces a unique response (output) to a given set of conditions (inputs).
• The standard distinctive shape symbols for the three basic logic operations are shown below.
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Logic Gates• A circuit that performs a specified basic logic operation is
called a logic gate. • Logic gates form the building blocks for digital systems.• The true/false statements mentioned earlier are represented
by a HIGH (true) and a LOW (false).• AND & OR gates can have any number of inputs.
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The AND operator (both, all)
• rivers AND salinity
• dairy products AND export AND Europe
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The OR operator (either, any)
• fruit OR vegetables
• fruit OR vegetables OR cereal
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The NOT operator
• fruit NOT apples
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Let’s use logic to examine class.
• Please stand up if you are:– girl– AND black hair– AND left handed
• Please stand up if you are:– girl– OR black hair– OR left handed
• Please stand up if you are a girl NOT left handed• How has the group changed depending on the logical
operator used.
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Nesting• When more than one element is in parentheses, the
sequence is left to right. This is called "nesting."
– (foxes OR rabbits) AND pest control
– foxes OR rabbits AND pest control
– (animal pests OR pest animals) NOT rabbits
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Order of precedence of logic operators
• The order of operations is: AND, NOT, OR
• Parentheses are used to override priority.
• Expressions in parentheses are processed first.
• Parentheses are used to organize the sequence and
groups of concepts.
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Write out logic statements using logic operators for these.
• You have a buzzer in your car that sounds when your keys are in the ignition and the door is open.
• You have a fire alarm installed in your house. This alarm will sound if it senses heat or smoke.
• There is an election coming up. People are allowed to vote if they are a citizen and they are 18.
• To complete an assignment the students must do a presentation or write an essay.
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Truth Tables
• Truth tables provide a way to describe the relationship between inputs and outputs of a logic circuit.
• Typically, “A” is considered to be the least significant variable in a truth table.
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Timing Diagrams• Timing diagrams also show relationships between
input and output conditions in a logic circuit.
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Logic Gates
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NOT gate• The simplest possible gate is called an "inverter" or a NOT gate.
• One bit as input produces its opposite as output.
• The symbol for a NOT gate is shown below.
• The truth table for the NOT gate shows input and output.
A Q
0 1
1 0
A X
0 1
1 0
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Inside NOT GateTransistor as a Switch
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Timing analysis of an inverter gate
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NOT gate application
Binary number
1’s complement
1 0 0 0 1 1 0 1
0 1 1 1 0 0 1 0
A group of inverters can be used to form the 1’s complement of a binary number
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AND gate• The AND gate has the following symbol and truth table.
• Two or more input bits produce one output bit.
• Both inputs must be true (1) for the output to be true.
• Otherwise the output is false (0).
A B X0 0 00 1 01 0 01 1 1
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AND gate
Multiple Input AND Gate
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Inside the AND gate
A B X0 0 00 1 01 0 01 1 1
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Timing analysis of an AND gate
A B X0 0 00 1 01 0 01 1 1
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Timing analysis of 3 input AND gate
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Timing analysis of 3 input AND gate
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AND Gate ApplicationThe AND operation is used in computer programming as a selective mask.
If you want to retain certain bits of a binary number but reset the other bits to 0, you could set a mask with 1’s in the position of the retained bits.
If the binary number 10100011 is ANDed with the mask code - 00001111, what is the result? 0000001
1
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Application of the AND Gate for calculating frequency
AND Gate Application
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OR gate
• The OR gate has the following symbol and truth table.
• Two or more input bits produce one output bit.
• Either inputs must be true (1) for the output to be true.
A B X0 0 00 1 11 0 11 1 1
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OR gate
Multiple Input OR Gate
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Inside the OR gate
A B X0 0 00 1 11 0 11 1 1
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Timing analysis of OR gate
A B X0 0 00 1 11 0 11 1 1
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Timing analysis of OR gate
A B X0 0 00 1 11 0 11 1 1
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OR Gate ApplicationOR operation can be used in computer programming to SET / RESET certain bits of a binary number.
Example: ASCII letters have a 1 in the bit 5 position for all lower case letters and a 0 in this position for all upper case letters.(Bit positions are numbered from right to left starting with 0)
What will be the result if you OR an ASCII letter with the 8-bit mask 00100000?
The resulting letter will be lower case.
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Basic AND & OR gate operation
OR
AND
OR
AND
= 0 = 10 + 0 = 0 0 + 1 = 1
= 0 = 11 • 0 = 0 1 • 1 = 1
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Timing analysis of an AND gateLogic analyzer display.
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Using an AND gate to enable/disable a clock oscillator.
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Using an OR gate to enable/disable a clock oscillator.
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An Example: A Burglar Alarm
• This circuit shows how the elements discussed so far could be used to build a burglar alarm.
• If any of the switches A or B or C is
ON and the Alarm SET switch, D is
ON then the Siren E is ON.
• This is written as :- DCBAE .
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Combine gates.• Gates can be combined.• The output of one gate can become the input of another. • Try to determine the logic table for this circuit.
p q Y = NOT((p AND q) OR q)
0 0 10 1 01 0 11 1 0Y
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Construct the logic table for these circuits.
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What happens when you add a NOT to an AND gate?
“Not AND” = NAND
A B X
0 0 1
0 1 1
1 0 1
1 1 0
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Symbols for 3 and 8-input NAND gates.
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Timing analysis of a NAND gate.
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Timing analysis of a NAND gate.
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What happens when you add a NOT to an OR gate?
“Not OR” = NOR
A B X
0 0 1
0 1 0
1 0 0
1 1 0
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NOR gate timing analysis.
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NOR gate timing analysis.
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NOR Gate Application
When is the LED is ON for the circuit shown?
The LED will be on when any of the four inputs are HIGH.
A
CB
D
X
330 W
+5.0 V
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“Exclusive” gates
• There are 2 Exclusive Gates in Digital Electronics.
• Exclusively OR Gate and Exclusive NOR Gate
Exclusive OR Gate – XOR Gate
&
Exclusive NOR Gate – XNOR Gate
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XOR Gate• Exclusive OR gate are true if either input is true but not both.
• The Symbol and Truth Table is shown below.
• Output will be high for different inputs
A B X
0 0 0
0 1 1
1 0 1
1 1 0
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XNOR Gate• The Symbol and Truth Table is shown below.
• Output will be high for same inputs
A B X
0 0 1
0 1 0
1 0 0
1 1 1
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Logic Gates Review
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NOT Operator
Y = A
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AND Operator
Y = A B
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OR Operator
Y = A + B
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NAND Operator
Y = A B
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NOR Operator
Y = A + B
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Exclusive OR (XOR) Operator
Y = A + B
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Exclusive NOR (XNOR) Operator
Y = A + B
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Universal Gates
• NAND Gate and NOR Gate are known as Universal Gates.
• All other basic Gates can be constructed using NAND or NOR Gates.
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NOT gate from a NAND
A Q
0 1
1 0
Universal Gates
A Q A Q
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AND gate from a NAND
Universal Gates
A B Q
0 0 0
0 1 0
1 0 0
1 1 1
AQ
B
A
BQ
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OR gate from a NAND
Universal Gates
We will study after
proving DE- MORGAN’S
LAW
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NOT gate from a NOR
A Q
0 1
1 0
Universal Gates
A Q
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AND gate from a NOR
Universal Gates
We will study after
proving DE- MORGAN’S
LAW
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OR gate from a NOR
Universal Gates
A B X0 0 00 1 11 0 11 1 1
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Logic Gates summary
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Electronic 2 input Logic Gate ICs
1B
Vcc 4B 4A 4Y 3B 3A 3Y
1A 1Y 2B2A 2Y
14 13 12 11 10 9 8
7654321GND
7400: Y = ABQuadruple two-input NAND gates
1A
Vcc 4Y 4B 4A 3Y 3B 3A
1Y 1B 2A2Y 2B
14 13 12 11 10 9 8
7654321GND
7402: Y = A + BQuadruple two-input NOR gates
1B
Vcc 4B 4A 4Y 3B 3A 3Y
1A 1Y 2B2A 2Y
14 13 12 11 10 9 8
7654321GND1Y
Vcc 6A 6Y 5A 5Y 4A 4Y
1A 2A 3A2Y 3Y
14 13 12 11 10 9 8
7654321GND
7404: Y = AHex inverters
7408: Y = ABQuadruple two-input AND gates
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Electronic 2 input Logic Gate ICs
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1B
Vcc 1C 1Y 3C 3B 3A 3Y
1A 2A 2C2B 2Y
14 13 12 11 10 9 8
7654321
GND
7410: Y = ABCTriple three-input NAND gates
1B
Vcc 2D 2C NC 2B 2A 2Y
1A NC 1D1C 1Y
14 13 12 11 10 9 8
7654321
GND
7420: Y = ABCDDual four-input NAND gates
Electronic 3 input Logic Gate ICs
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Electronic 8 input Logic Gate IC
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Selected Key Terms
Inverter
Truth table
Timing diagram
Boolean algebra
AND gate
A logic circuit that inverts or complements its inputs.
A table showing the inputs and corresponding output(s) of a logic circuit.
A diagram of waveforms showing the proper time relationship of all of the waveforms.
The mathematics of logic circuits.
A logic gate that produces a HIGH output only when all of its inputs are HIGH.
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OR gate
NAND gate
NOR gate
Exclusive-OR gate
Exclusive-NOR gate
A logic gate that produces a HIGH output when one or more inputs are HIGH.
A logic gate that produces a LOW output only when all of its inputs are HIGH.
A logic gate that produces a LOW output when one or more inputs are HIGH.
A logic gate that produces a HIGH output only when its two inputs are at opposite levels.
A logic gate that produces a LOW output only when its two inputs are at opposite levels.
Selected Key Terms
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Class Question:Basic Components
AND NAND OR NOR XOR XNOR NOT
Name and
Symbol
Truth table
Notation A·B=out
AB
out
AB out00 001 010 011 1
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1. The truth table for a 2-input AND gate is
0 00 11 01 1
InputsA B X
Output
0 00 11 01 1
10 00
InputsA B X
Output
0 00 11 01 1
InputsA B X
Output
InputsA B X
Output
0 00 11 01 1
01 11
a. b.
c. d.
0110
00 01
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2. The truth table for a 2-input NOR gate is
0 00 11 01 1
InputsA B X
Output
0 00 11 01 1
10 00
InputsA B X
Output
0 00 11 01 1
InputsA B X
Output
InputsA B X
Output
0 00 11 01 1
01 11
a. b.
c. d.
0110
00 01
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3. The truth table for a 2-input XOR gate is
0 00 11 01 1
InputsA B X
Output
0 00 11 01 1
10 00
InputsA B X
Output
0 00 11 01 1
InputsA B X
Output
InputsA B X
Output
0 00 11 01 1
01 11
a. b.
c. d.
0110
00 01
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4. The symbol is for which gate
a. OR gate
b. AND gate
c. NOR gate
d. XOR gate
AB X
≥ 1
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5. The symbol is for which gate
a. OR gate
b. AND gate
c. NOR gate
d. XOR gate
AB X
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6. A logic gate that produces a HIGH output only when all of its inputs are HIGH
a. OR gate
b. AND gate
c. NOR gate
d. NAND gate
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7. The expression X = A + B means
a. A OR B
b. A AND B
c. A XOR B
d. A XNOR B
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8. A 2-input gate produces the output shown. (X represents the output)
a. OR gate
b. AND gate
c. NOR gate
d. NAND gate
A
X
B
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9. A 2-input gate produces a HIGH output only when the inputs agree.
a. OR gate
b. AND gate
c. NOR gate
d. XNOR gate
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Answers:
1. c
2. b
3. a
4. a
5. d
6. b
7. c
8. d
9. d
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Logic Gate Characteristics
1. Gate Voltages and Currents
2. Fan – in
3. Fan – out
4. Propagation Delay
5. Power Requirements
6. Noise Margin or Noise Immunity
7. Speed – Power Product (SPP)
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• VIH (min) – High level input voltage. The minimum level required for a logical 1 at an input. Any voltage below this level will not be accepted as a HIGH by the logic circuit
• VIL (max) – The maximum input voltage for logic zero
• VOH (min) – The minimum voltage level at a logic circuit output in the logic 1 state under defined load conditions
Gate Voltages and Currents
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• VOL (max) – Low level output voltage. The maximum voltage level at a logic circuit output in the logical 0 state under defined load conditions
• IIH – High level input current. The current that flows into an input when a specified high level voltage is applied to that input
• IIL – Low level input current. The current that flows into an input when a specified low level voltage is applied to that input
Gate Voltages and Currents
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• IOH – High level output current. The current that flows from an output when a specified high level voltage is obtained at that output
• IOL – Low level output current. The current that flows from an output when a specified low level voltage is obtained at that output
Gate Voltages and Currents
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1. VIH (min) – High level input voltage
2. VIL (max) – Low level input voltage
3. VOH (min) – High level output voltage
4. VOL (max) – Low level output voltage
5. IIH – High level input current
6. IIL – Low level input current
7. IOH – High level output current
8. IOL – Low level output current
Gate Voltages and Currents
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Gate Voltages and Currents
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Fan - in
• Fan-in specifies the number of inputs available on a gate.
• Gate primitives often limit the number of inputs to 4 or 5.
• To build gates with lower fan-in, multiple gates can be
interconnected.
Implementation of a 7-input NAND gate using NAND gates with 4 or fewer inputs.
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Fan - in
Fan-in – the number of inputs to the gate gates with large fan-in are bigger and slower
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Fan out
• Also known as loading factor
• Defined as the maximum number of logic inputs that an output can drive reliably
• A logic circuit that specify to have 10 fan out can drive 10 logic inputs
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The inverter has a fan-out of 3 (i.e., the inverter drives 3 inputs).
Fan out
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Fanout is the number of standard loads that the output can drive.
The number of standard loads is limited by the amount of input current each load requires as compared to the current that the driving gate can deliver.
Fanout, therefore, is generally considered to be the smaller of the following two items:
(max)I
(max)I =fanout or
(max)I
(max)I =fanout
IH
OH
IL
OL
Fan out
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Propagation Delay Every logic gate experiences some delay (though very small) in
propagating signals forward from input to output. This delay is called Gate (Propagation) Delay. Formally, it is the average transition time taken for the output
signal of the gate to change in response to changes in the input signals.
Three different propagation delay times associated with a logic gate:
tPHL: output changing from the High level to Low level
tPLH: output changing from the Low level to High level
tPD=(tPLH + tPHL)/2 (average propagation delay)
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Propagation Delay tPHL: output changing from the High level to Low level tPLH: output changing from the Low level to High level tPD=(tPLH + tPHL)/2 (average propagation delay)
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Propagation Delay
In reality, output signals normally lag behind input signals.
Ideally, there will not be any delay.
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Propagation Delay
A glitch in the output of the AND gate caused by propagation delay in the inverter.
Note that we have assumed zero delay for the AND gate.
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Calculation of Circuit Delays Amount of propagation delay per gate depends on:
(i) gate type (AND, OR, NOT, etc) (ii) transistor technology used (TTL,ECL,CMOS etc), (iii) miniaturisation (SSI, MSI, LSI, VLSI)
Propagation delay of logic circuit= longest time it takes for the input signal(s) to propagate to
the output(s).= earliest time for output signal(s) to stabilise, given that
input signals are stable at time 0.
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Calculation of Circuit Delays In general, given a logic gate with delay, t.
If inputs are stable at times t1,t2,..,tn, respectively; then the earliest time in which the output will be stable is:
max(t1, t2, .., tn) + t
To calculate the delays of all outputs of a combinational circuit, repeat above rule for all gates.
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Calculation of Circuit Delays As a simple example, consider the full adder circuit where
all inputs are available at time 0. (Assume each gate has delay t.)
where outputs S and C, experience delays of 2t and 3t, respectively.
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Power Requirements• Every IC need a certain power requirement to
operate
• This power supply comes from the voltage supply that connected to the pin on the chip labeled VCC
• The amount of power require by ICs is determined by the current that it draws from the VCC
• The actual power is ICC x VCC
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ICC(avg) = (ICCH + ICCL)/2
PD(avg) = ICC(avg) X Vcc
Power Requirements
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Noise Margin or Immunity• Stray electric and magnetic fields can induce voltages on
the connecting wires between logic circuits – this unwanted signal called noise
• These cause the input signal to a logic circuit drop below VIH (min) or rise above VIL (max)
• Noise Margin or Noise Immunity refers to the circuit’s ability to tolerate noise without causing spurious changes in the output voltage
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Below given is a diagram showing the range of voltages that can occur at a logic circuit output.
Noise Margin
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• The high state noise margin VNH is defined as
VNH = VOH (min) – VIH (min)
• The low state noise margin VNL is defined as
VNL = VIL (max) – VOL (max)
Noise Margin
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Speed – Power Product
• Product of Propagation Delay and Power of a
Digital IC
SPP = Propagation delay X Power
• Also known as Power – Delay Product.
• Helps measure quality of a logic family.
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Boolean Algebra
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George Boole
My name is George Boole and I lived in England in the 19th century. My work on mathematical logic, algebra, and the binary number system has had a unique influence upon the development of computers. Boolean Algebra is named after me.
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What is Boolean Algebra ?Boolean Algebra is a mathematical technique that provides the ability to algebraically simplify logic expressions. These simplified expressions will result in a logic circuit that is equivalent to the original circuit, yet requires fewer gates.
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0 0 X
X Y Z0 0 00 1 01 0 01 1 1
X 1 X X X X 0 X X
Boolean Theorems (1 of 7)
X Y Z0 0 00 1 01 0 01 1 1
X Y Z0 0 00 1 01 0 01 1 1
X Y Z0 0 00 1 01 0 01 1 1
Single Variable - AND FunctionTheorem #1 Theorem #2 Theorem #3 Theorem #4
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X 0 X 1 1 X
X Y Z0 0 00 1 11 0 11 1 1
X X X 1 X X
Boolean Theorems (2 of 7)
X Y Z0 0 00 1 11 0 11 1 1
X Y Z0 0 00 1 11 0 11 1 1
X Y Z0 0 00 1 11 0 11 1 1
Single Variable - OR FunctionTheorem #5 Theorem #6 Theorem #7 Theorem #8
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X X
0 1 0
1 0 1
Single Variable – Invert (NOT) Function
Boolean Theorems (3 of 7)
X XX
Theorem #9
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Summary of Theorems (1,2 & 3 of 7)
AND Function OR Function NOT Function
0 0 X
X 1 X
X X X
0 X X
Theorem #1
Theorem #2
Theorem #3
Theorem #4
X 0 X
1 1 X
X X X
1 X X
Theorem #5
Theorem #6
Theorem #7
Theorem #8
X X
Theorem #9
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Example #1: Boolean Algebra
Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
D C C BA A F 1
SOP – Sum Of Product (Sum of MINTERMS)
Y = AB + BC
POS – Product Of Sum (Product of MAXTERMS)
Y = (A+B) (B+C)
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Example #1: Boolean Algebra
Simplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
D C C BA A F 1
Solution
BA
BA
0 BA
D 0 BA
D C C BA
D C C BA A
1
1
1
1
1
1
F
F
F
F
F
F
; Theorem #3
; Theorem #4
; Theorem #1
; Theorem #5
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Example #2: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
0 B A 1 B A C C B C B BF 2
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Example #2: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
0 B A 1 B A C C B C B BF 2
Solution
B A C BF
B A C B F
0 B A C B F
0 B A B A C B F
0 B A 1 B A C B F
0 B A 1 B A C B C B F
0 B A 1 B A C C B C B BF
2
2
2
2
2
2
2
; Theorem #3 (twice)
; Theorem #7
; Theorem #2
; Theorem #1
; Theorem #5
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Boolean Theorems (4 of 7)
X Y Y X
Commutative Law
Theorem #10A – AND Function
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Boolean Theorems (4 of 7)
X Y Y X
Commutative Law
Theorem #10B – OR Function
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Boolean Theorems (5 of 7)
Z Y)(X Z)(Y X
Associative Law
Theorem #11A – AND Function
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Boolean Theorems (5 of 7)
Z Y) (X Z) (Y X
Associative Law
Theorem #11B – OR Function
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Boolean Theorems (6 of 7)
Z X Y X Z) (Y X
Distributive Law
Theorem #12A – AND Function
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Boolean Theorems (6 of 7)
YZ YW XZ XW Z) Y)(WX(
Distributive Law
Theorem #12B – OR Function
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Example #3: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
T) R)(S (R T RF3
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Example #3: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
T) R)(S (R T RF3
Solution
R S TF
R S 1TF
R S S 1TF
R S S R RTF
T S R S T RT RF
T S R S T R 0 T RF
T S R S T R R R T RF
T RS R T RF
3
3
3
3
3
3
3
3
; Theorem #12B
; Theorem #4
; Theorem #5
; Theorem #12A
; Theorem #8
; Theorem #6
; Theorem #2
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Boolean Theorems (7 of 7)
Y X Y X X
Y X Y X X
Y X Y X X
Y X Y X X
Consensus Theorem
Theorem #13A
Theorem #13B
Theorem #13C
Theorem #13D
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Example #4: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
S Q P S Q P S P F4
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Example #4: Boolean AlgebraSimplify the following Boolean expression and note the Boolean theorem used at each step. Put the answer in SOP form.
S Q P S Q P S P F4
Q P PS F
Q P 1 PS F
Q P Q 1 PS F
S Q P QP S P F
S Q P Q S P F
S Q P S Q S P F
S Q P S Q P S P F
4
4
4
4
4
4
4
; Theorem #12A
; Theorem #13C
; Theorem #12A
; Theorem #12A
; Theorem #6
; Theorem #2
Solution
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Augustus DeMorgan
My name is Augustus DeMorgan. I’m an Englishman born in India in 1806. I was instrumental in the advancement of mathematics and am best known for the logic theorems that bear my name.P.S. George Boolean gets WAY too much credit. He has more theorems, but mine are WAY Cooler!
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DeMorgan’s TheoremsDeMorgan’s Theorems are two additional simplification techniques that can be used to simplify Boolean expressions. Again, the simpler the Boolean expression, the simpler the resulting logic.
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DeMorgan’s Theorem #1
0 0 0 10 1 0 11 0 0 11 1 1 0
BA BA A B0 0 1 1 10 1 1 0 11 0 0 1 11 1 0 0 0
BA
A B A B BA
BA Proof
BABA
The truth-tables are equal; therefore, the Boolean equations must be equal.
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DeMorgan’s Theorem #2
Proof
The truth-tables are equal; therefore, the Boolean equations must be equal.
BABA
BA BA
0 0 0 1
0 1 1 0
1 0 1 0
1 1 1 0
BA BA A B
0 0 1 1 1
0 1 1 0 0
1 0 0 1 0
1 1 0 0 0
A B A B BA
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DeMorgan Shortcut
BREAK THE LINE, CHANGE THE SIGNBreak the LINE over the two variables, and change the SIGN directly under the line.
BABA For Theorem #14A, break the line, and change the AND function to an OR function. Be sure to keep the lines over the variables.
BABA For Theorem #14B, break the line, and change the OR function to an AND function. Be sure to keep the lines over the variables.
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DeMorgan’s TheoremsIn other words
NAND = bubbled OR
NOR = bubbled AND
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AND Gate using NOR Gate
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OR Gate using NAND Gates
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Summary
X X 9)
1 X X 8)
X X X 7)
1 1 X 6)
X 0 X 5)
0 X X 4)
X X X 3)
X 1 X 2)
0 0 X 1)
Y X Y X 14B)
Y X YX 14A)
YXYXX 13D)
YXYXX 13C)
YXXYX 13B)
YXYXX 13A)
YZYWXZXWZWYX 12B)
XZXYZYX 12A)
ZYXZY X 11B)
ZXYYZX 11A)
X Y Y X 10B)
X Y Y X 10A)
Commutative Law
Associative Law
Distributive Law
Consensus Theorem
Boolean & DeMorgan’s Theorems
DeMorgan’sTheorem
AND
OR
NOT
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DeMorgan’s: Example #1Example
Simplify the following Boolean expression and note the Boolean or DeMorgan’s theorem used at each step. Put the answer in SOP form.
)ZY()YX(F 1
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DeMorgan’s: Example #1Example
Simplify the following Boolean expression and note the Boolean or DeMorgan’s theorem used at each step. Put the answer in SOP form.
Solution
ZYYXF
)ZY( )YX(F
)ZY( )YX(F
)ZY( )YX(F
)ZY()YX(F
1
1
1
1
1
; Theorem #14A
; Theorem #9 & #14B
; Theorem #9
; Rewritten without AND symbols and parentheses
)ZY()YX(F 1
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DeMorgan’s: Example #2Example
Simplify the output function F2 shown in the logic circuit. Be sure to note the Boolean or DeMorgan’s theorem used at each step. Put the answer in SOP form.
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DeMorgan’s: Example #2Solution
Y XZ XF
)XY()Z X(F
)XY()Z X(F
)XY()ZX(F
)XY()ZX(F
)XY)(ZX(F
2
2
2
2
2
2
; Theorem #14A
; Theorem #9
; Theorem #14B
; Theorem #9
; Rewritten without AND symbols
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Module 11.Introduction2.Number Systems3.Binary Arithmetic4.Logic Functions5.Boolean Algebra6.Minimization
Techniques
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Minimization Techniques
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Minimization Techniques
1.Karnaugh Map
2.Queen Mclusky Method
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Karnaugh Map (K Map)
• 2 Variable K Map
• 3 Variable K Map
• 4 Variable K Map
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2 Variable Karnaugh map.
a b f (a,b)0 00 11 01 1
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K-map for f(a, b) = ab + ab’.
a b f (a,b)0 0 00 1 01 0 11 1 1
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K-map solution for f(a, b) = ab + ab’.
a b f (a,b)0 0 00 1 01 0 11 1 1
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K-map solution for f(a, b) = ab + ab’ + a’b’.
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K-map solution for f(a, b) = ab’ + a’b.
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3 Variable K Map
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3 variable K-map
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K-map solution for Equation ab’c’+ab’c+abc+a’b’c+a’bc+a’bc’
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K-map for Equation f(x,y,z) = x’y’z’+x’y’z+x’yz+xy’z
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K-map for Equation f(x,y,z) = x’y’z’+x’y’z+x’yz+xy’z
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Solution of Equationf(a,b,c) = a’bc+ab’c’+abc+abc’
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4 Variable Karnaugh mapa b c d f (a,b,c,d)0 0 0 00 0 0 10 0 1 00 0 1 10 1 0 00 1 0 10 1 1 00 1 1 11 0 0 01 0 0 11 0 1 01 0 1 11 1 0 01 1 0 11 1 1 01 1 1 1
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f(a,b,c,d) = a’b’c’d’+a’b’cd’+abc’d’+abc’d+abcd+abcd’+ab’c’d’+ab’c’d+ab’cd+ab’cd’
a b c d f (a,b,c,d)0 0 0 0 10 0 0 1 00 0 1 0 10 0 1 1 00 1 0 0 00 1 0 1 00 1 1 0 00 1 1 1 01 0 0 0 11 0 0 1 11 0 1 0 11 0 1 1 11 1 0 0 11 1 0 1 11 1 1 0 11 1 1 1 1
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K-map with don’t caresa b c d f (a,b,c,d)0 0 0 0 00 0 0 1 00 0 1 0 00 0 1 1 10 1 0 0 10 1 0 1 X0 1 1 0 00 1 1 1 11 0 0 0 01 0 0 1 X1 0 1 0 11 0 1 1 01 1 0 0 01 1 0 1 11 1 1 0 01 1 1 1 X
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• Algebraic expressions– f( x, y, z ) = xy+z
• Tabular forms
• Venn diagrams
• Cubical representations
x y z f
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 1
1 0 0 0
1 0 1 1
1 1 0 1
1 1 1 1
x
zy
Representations of Boolean Functions
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Cubical Representation
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Cubical Representation
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x
z
00 10
1101
0-cube in cubic notation
10
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x
z
00 10
1101
0-cube by product terms
x z’
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x
y
z
000 100
101001
010
011
110
0-cube in cubic notation
111
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x
y
z
000
111
100
101001
010
011
110
0-cube by product terms
x y z
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x
y
z
000
111
100
101001
010
011
110
1-cube in cubic notation
1_0
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x
y
z
000
111
100
101001
010
011
110
xz’
1-cube by product terms
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x
y
z
000
111
100
101001
010
011
110_0_
2-cube in cubic notation
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x
y
z
000
111
100
101001
010
011
110Y’
2-cube by product terms
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Cubical Representation of Minterms andImplicants
• f1 = a’b’c’+a’b’c+ab’c+abc+abc’• f2 = a’b’c+ab’c
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Cubical representation of minterms
• f1 = a’b’c’ + a’b’c + ab’c + abc +abc’
• f2 = a’b’c + ab’c
111
f1
c b
a000
001
110
101
α
βγ
δ
f2
001101
α β γ δ
β
β
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• IMPLICANT: An implicant of a function is a product term that is included in the function.
• PRIME IMPLICANT: An implicant is prime if it cannot be included in any other implicants.
• ESSENTIAL PRIME IMPLICANT: A prime implicant is essential if it is the only one that includes a minterm.
Implicants
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Example: f(x,y,z) = xy’ + yz
xy(not I), xyz(I, not PI), xz(PI,not EPI), yz(EPI)
Implicants
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Exact Minimization of Two-Level Logic
• Quine-McClusky (1) generate all primes (2) find a minimum cover
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Quine-McClusky
(1) generate all primes ( utilize AB+AB’=A(B+B’)=A ) f = Sm( 4, 5, 6, 8, 9, 10, 13 ) + d( 0, 7, 15 )
0000 0-00 01-- 0100 -000 -1-1 1000 010- 0101 01-0 0110 100- 1001 10-0 1010 01-1 0111 -101 1101 011- 1111 1-01 -111 11-1
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Quine-McClusky
(2) select a subset of primes f ( x, y, z, w ) = x’z’w’+ y’z’w’ + xy’z’ + xy’w’ + xz’w + x’y+yw => the selected sum for f is f ( x, y, z, w ) = xy’w’ + xz’w + x’y
A subset of implicant is a cover of the function if each minterm for which the function is 1 is included in at least one implicant of the subset.