Post on 31-Mar-2015
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UNIT – 2 Data Representation and Internal Operations of the Computer
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Unit 2 – Data Representation and Internal Operations of the Computer
2.1 Introduction to Number Systems
2.2 Data Representation
2.3 Logic Gates and Circuits
2.4 Boolean Algebra
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Some ancient numerals
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Some ancient numerals
End ShowEgyptian 3rd Century BC
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Some ancient numerals
End ShowCretan 1200-1700BC
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England’s “five-barred gate”
End ShowThe Greek Numeral System
End ShowComputation with Greek Numeral System
Indo-Arabic Indo-Arabic
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Roman Numerals1 I 20 XX2 II 25 XXV3 III 29 XIX4 IV 50 L5 V 75 LXXV6 VI 100 C10X 500 D11XI 1000M16XVI
Now try these:
1. XXXVI2. XL3. XVII4. DCCLVI5. MCMLXIX
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5What is this ?
This is also a symbol
Why do we use this symbol ?
To represent quantity
Five (5)
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0
1
2
3
4
5 10
9
8
7
6
Symbols to represent certain quantities
End ShowHow do you represent this quantity on a spike abacus?
0000
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100
How do you represent this quantity
End ShowHow Represent This Quantity
200
End ShowHow Represent This Quantity
300
End ShowHow Represent This Quantity
400
End ShowHow Represent This Quantity
500
End ShowHow Represent This Quantity
600
End ShowHow Represent This Quantity
700
End ShowHow Represent This Quantity
800
End ShowHow Represent This Quantity
900
End ShowHow Represent This Quantity
900
End ShowHow Represent This Quantity
010
End ShowHow Represent This Quantity
010
End ShowHow Represent This Quantity
110
End ShowHow Represent This Quantity
2100101……… 102
End ShowAbacus
Roman Abacus
Chinese Abacus
JapaneseSoroban (Abacus)
Abacus is a Latin word, related to the Greek “abax”, meaning table. The word `calculus’ originally meant pebble in Latin, used on the counting board.
End Show2.1 – Introduction to the Number System
• A number system defines a set of symbols used to represent quantity.
• Quantifying values and items in relation to each other is helpful us to make sense of our
environment.
• The study of number system is not just limited to computers. We use numbers every day. A computer manipulates and stores numbers inside the computer system.
End ShowWhy we learn it?
These symbols are processed internally by components that can maintain a limited number of discrete states.
To represent these states we have to use number systems.
– Ex: The decimal digits 0,1,2, …..,9 provide 10 discrete symbols (10 digits)
End ShowDecimal Number system – Place Values
Hundreds Tens Ones
We normally use the decimal number system to represent quantities and perform calculations.
3 x 100 4 x 10 2 x 1
342
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Hundreds Tens Ones
3 x 100 4 x 10 3 x 1
343
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Hundreds Tens Ones
3 x 100 4 x 10 4 x 1
344
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Hundreds Tens Ones
3 x 100 4 x 10 5 x 1
345
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Hundreds Tens Ones
3 x 100 4 x 10 6 x 1
346
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Hundreds Tens Ones
3 x 100 4 x 10 7 x 1
347
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Hundreds Tens Ones
3 x 100 4 x 10 8 x 1
348
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Hundreds Tens Ones
3 x 100 4 x 10 9 x 1
349
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Hundreds Tens Ones
3 x 100 0 x 1 5 x 10
350
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Consider this number format 53410
The Base of the Decimal number here is 10
Normally we don’t write base numbers in decimal form
We can use digits 0 - 9
End ShowFace Value and Place value
Weighing Factor The weighing factor is the multiplier value applied to each column position of the number.
For instance, a decimal has a weighing factor of TEN, in that each column on the immediate left indicates an increase in value by a multiple of 10.
i.e.; each column moves to the left increasing by a multiple of 10.
312 = 300 + 10 + 2
= 3 * 100 + 1 * 10 + 2 * 1
= 3 * 102 + 1 * 101 + 2 * 100 Weighing factors
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The decimal Number System: • uses base 10• includes only the symbols 0 through 9
The weighed values for each position is as follows:
104 103 102 101 100 10-1 10-2 10-3
10000 1000 100 10 1 .1 .01 .001
Decimal Number System(cont..)
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• This number system uses TEN different symbols to represent values. The set values used in decimal are
0 1 2 3 4 5 6 7 8 9
Lowest value Highest value
Decimal number system (cont..)
When doing a calculation, if the highest digit (9) is exceeded, a carry over transferred to the next column (to the left) occurs.
Lets 17+41718 +119 +220 +321 +4
When 9 is exceeded, we reset (0), and carry a value of 1 to the next column on the left
End ShowBinary number system
TWO symbols are used to represent numerals in the binary number system. These have the values of,
0 1
0 represents low value, and 1 represents high value.
Position x……………
…..5 4 3 2 1
Binary Value 1……………
…1 1 1 1 1
Decimal Value
2 x-1 ……………….
24 23 22 21 20
End ShowMore illustration…..
The binary number system is also known as
the base 2 number system.
The values of the positions are calculated by
raising 2 to some power.
Why is 2 the base in binary numbers?
Because we use 2 digits, the digits 0 and 1.
End ShowBinary Number System (con’t)
• The binary number system is also a positional numbering system.
• Instead of using ten digits, 0 - 9, the binary system uses only two digits, 0 and 1.
Example of a binary number and the position values :
1 0 0 1 1 0 1 26 25 24 23 22 21 20
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4 1
1 x 22 0 x 21 1 x 20 0 x 23
Representation of the binary number 0101
Position x 4 3 2 1
Binary Value 1 0 1 0 1
Decimal Value
2 x-1 23 22 21 20
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• uses base 2
• includes only the digits 0 and 1
The weighted values for each position is as follows:
27 26 25 24 23 22 21 20
128 64 32 16 8 4 2 1
Binary Number System(cont…)
End ShowOctal number system
The octal number system uses EIGHT symbols to represent numbers. The 8 distinct symbols are,
0 1 2 3 4 5 6 7
With 0 having the lowest value and 7 having the highest value
Columns are used in the same way as in the decimal system in that, the left most column is used to represent the greatest value
Octal numbers are represented with the base 8
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3 x 82 1 x 81 4 x 80 0 x 83
Representation of octal number 314
64
64
64
8 1111
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The Octal system is based on the binary system with a 3-bit boundary. The Octal Number System:
• uses base 8 • includes only the symbols 0 through 7
85 84 83 82 81 80
32768 4096 512 64 8 1
The weighted values for each position is as follows:
Octal Number System
End ShowHexadecimal number system
• The hexadecimal number system uses SIXTEEN symbols to represent Numbers. The 16 distinct symbols are, 0 1 2 3 4 5 6 7 8 9 A B C D E F, where A = 10, B = 11, .., F = 15
• With ‘0’ having the lowest value and ‘F’ having the highest value.
• Hexadecimal numbers are represented with the base 16.
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1 x 162 4 x 161 F(15) x 160 0 x 163
Representation of hexadecimal number 34F
256
256
16 111111111111111
16
16
16
256
End ShowThe Hexadecimal Number System:
uses base 16
• includes only the symbols 0 through 9 and the letters A, B, C, D, E, and F to represent 10, 11, 12, 13, 14, and 15 respectively.
163 162 161 160
4096 256 16 1
The weighted values for each position is as follows:
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Binary to Decimal
Multiply each digit by its weighted position, and add each of the weighted values.
2.1.2 Number Base Conversion
Example
The binary value 1011 represents:
1x23 + 0x22 + 1x21 + 1x20
=1x8 + 0x4 + 1x2 + 1x1
=8 + 0 + 2 + 1
=11 (base 10)
End ShowConverting from Binary to Decimal
1 0 0 1 1 0 1 26 25 24 23 22 21 20
20 = 1 24 = 16 21 = 2 25
= 32 22 = 4 26 = 64 23 = 8
1 X 20 = 10 X 21 = 0 1 X 22 = 4 1 X 23 = 8 0 X 24 = 00 X 25 = 01 X 26 = 64
7710
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Division Quotient Remainder Binary Number
47 / 2 23 1 1 2 47
23 --1
Decimal to Binary (Repeated Division By 2)
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Division Quotient Remainder Binary Number
47 / 2 23 1 1
23 / 2 11 1 112 47
2 23 --1
11 --1
Decimal to Binary (Repeated Division By 2)
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Division Quotient Remainder Binary Number
47 / 2 23 1 1
23 / 2 11 1 11
11 / 2 5 1 111
2 47
2 23 --1
2 11 --1
5 --1
Decimal to Binary (Repeated Division By 2)
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Division Quotient Remainder Binary Number
47 / 2 23 1 1
23 / 2 11 1 11
11 / 2 5 1 111
5 / 2 2 1 1111
2 47
2 23 --1
2 11 --1
2 5 --1
2 --1
Decimal to Binary (Repeated Division By 2)
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Division Quotient Remainder Binary Number
47 / 2 23 1 1
23 / 2 11 1 11
11 / 2 5 1 111
5 / 2 2 1 1111
2 / 2 1 0 01111
2 47
2 23 --1
2 11 --1
2 5 --1
2 2 --1
1 --0
Decimal to Binary (Repeated Division By 2)
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Division Quotient Remainder Binary Number
47 / 2 23 1 1
23 / 2 11 1 11
11 / 2 5 1 111
5 / 2 2 1 1111
2 / 2 1 0 01111
1 / 2 0 1 101111
2 47
2 23 --1
2 11 --1
2 5 --1
2 2 --1
2 1 --0
0 --14710 = 1011112
Decimal to Binary (Repeated Division By 2)
End Show• Ex. Lets find the binary equivalent of the
decimal number 254
• 254 / 2 gives 127 with remainder 0
End Show
• Lets find the binary equivalent of the
decimal number 254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
End Show• Lets find the binary equivalent of the
decimal number 254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
End Show• Lets find the binary equivalent of the decimal
number 254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
• 31 / 2 gives 15 with remainder 1
End Show• Lets find the binary equivalent of the decimal
number 254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
• 31 / 2 gives 15 with remainder 1
• 15 / 2 gives 7 with remainder 1
End Show• Lets find the binary equivalent of the decimal number
254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
• 31 / 2 gives 15 with remainder 1
• 15 / 2 gives 7 with remainder 1
• 7 / 2 gives 3 with remainder 1
End Show• Lets find the binary equivalent of the decimal
number 254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
• 31 / 2 gives 15 with remainder 1
• 15 / 2 gives 7 with remainder 1
• 7 / 2 gives 3 with remainder 1
• 3 / 2 gives 1 with remainder 1
End Show• Lets find the binary equivalent of the decimal
number 254
• 254 / 2 gives 127 with remainder 0
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
• 31 / 2 gives 15 with remainder 1
• 15 / 2 gives 7 with remainder 1
• 7 / 2 gives 3 with remainder 1
• 3 / 2 gives 1 with remainder 1
• 1 / 2 gives 0 with remainder 1
111111102
End Show• Lets find the binary equivalent of the decimal number 254
• 254 / 2 gives 127 with remainder 0 LSB
• 127 / 2 gives 63 with remainder 1
• 63 / 2 gives 31 with remainder 1
• 31 / 2 gives 15 with remainder 1
• 15 / 2 gives 7 with remainder 1
• 7 / 2 gives 3 with remainder 1
• 3 / 2 gives 1 with remainder 1
• 1 / 2 gives 0 with remainder 1 MSB
111111102
End ShowBinary to Octal Conversion
It is easy to convert a binary number to an octal. This is accomplished by:
1010111012
001 010 111 110 110 010
1 2 7 6 6 2
Most Significant Bit (MSB)
Least Significant Bit (LSB)
1. Break the binary number into 3-bit sections from the LSB to the MSB.
2. Convert the 3-bit binary number to its octal equivalent.
For example, the binary value 1010111110110010 will be written:
End ShowConversion from Binary to Octal
• Convert 10110111 to octal.
• Each octal digit is represented by 3 binary bits. Split the binary number into groups of 3 bits, starting from the right.
10 110 111= 2 = 6 = 7= 2678
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Octal Binary (in 3 bits)
0 000
1 001
2 010
3 011
4 100
5 101
6 110
7 111
Relationship between binary and octal
End ShowOctal to Decimal Conversion
To convert from Octal to Decimal, multiply the value in each position by its Octal weight and add each value.
Example, convert this octal 342 to decimal,
we can obtain the decimal value as follows:
3x82 4x81 2x80
3x64 4x8 2x1
192 32 2 192 + 32 + 2 = 226
3 x 82 4 x 81 2 x 80
1
1
8
8
8
8
64
64
64
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Conversion from Octal to Decimal
• Convert 1768 to decimal.
• Each column represents a power of 8, 176 = 1 * 82 + 7 * 81 + 6 * 80
= (1 * 64) + (7 * 8) + (6 * 1) = 64 + 56 + 6
= 126
End ShowDecimal to Octal Conversion
To convert decimal to octal is slightly more difficult. The typical method to convert from decimal to octal is repeated division by 8.
Repeated Division By 8
•For this method, divide the decimal number by 8, and write the remainder on the side as the least significant digit.
•This process is continued by dividing the quotient by 8 and writing the remainder until the quotient is 0.
End ShowDecimal to Octal Conversion
To convert decimal to octal is slightly more difficult. The typical method to convert from decimal to octal is repeated division by 8.
Repeated Division By 8
•For this method, divide the decimal number by 8, and write the remainder on the side as the least significant digit.
•This process is continued by dividing the quotient by 8 and writing the remainder until the quotient is 0.
Division Quotient Remainder 87 / 8 10 7
End ShowDecimal to Octal Conversion
To convert decimal to octal is slightly more difficult. The typical method to convert from decimal to octal is repeated division by 8.
Repeated Division By 8
•For this method, divide the decimal number by 8, and write the remainder on the side as the least significant digit.
•This process is continued by dividing the quotient by 8 and writing the remainder until the quotient is 0.
Division Quotient Remainder 87 / 8 10 7
10 / 8 1 2
End ShowDecimal to Octal Conversion
To convert decimal to octal is slightly more difficult. The typical method to convert from decimal to octal is repeated division by 8.
Repeated Division By 8
•For this method, divide the decimal number by 8, and write the remainder on the side as the least significant digit.
•This process is continued by dividing the quotient by 8 and writing the remainder until the quotient is 0.
Division Quotient Remainder 87 / 8 10 7
10 / 8 1 2
1 / 8 0 18710 = 1278
End ShowBinary to Hexa Conversion
It is easy to convert a binary number to hexa. This is accomplished by:
1.Break the binary number into 4-bit sections from the LSB to the MSB.
2.Convert the 4-bit binary number to its Hexa equivalent.
For example, the binary value 1010111110110010 is written:
1010 1111 1011 0010
A F B 2
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Conversion from binary to hexadecimal
• Convert 10110 to hexadecimal.
• Each hexadecimal digit represents 4 binary bits. Split the binary number into groups of 4 bits, starting from the right.
1 0110 1 6=16 in hexadecimal
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Hexadecimal Binary (in 4 bits)
Hexadecimal Binary (in 4 bits)
0 0000 8 1000
1 0001 9 1001
2 0010 A 1010
3 0011 B 1011
4 0100 C 1100
5 0101 D 1101
6 0110 E 1110
7 0111 F 1111
Relationship between Binary and Hexadecimal
End ShowHexa to Binary Conversion
It is also easy to convert from an integer hexa number to binary. This is accomplished by:
Convert the Hexa number to its 4-bit binary equivalent. Combine the 4-bit sections by removing the spaces.
B 2
1011 0010
This yields the binary number 10110010 or 1011 0010 in our more readable format.
For example the hexa value B2 is written in binary:
End ShowHexadecimal to Decimal Conversion
To convert from Hexa to Decimal, multiply the value in each position by its hexa weight and add each value.
Using the value from the previous example, B216, we can obtain the decimal value as follows:
B x 161 2 x 160
11 x 16 2 x 1
176 2
176 + 2 = 178
0 x 162 11 x 161 2 x 160 0 x 163
1
1
1616161616161616161616
End Show
Conversion from hexadecimal to decimal
• Convert hexadecimal 17616 to decimal.
• Each column represents a power of 16, 17616= (1 * 162) + (7 * 161) + (6 * 160)
= (1 * 256) + (7 * 16 )+ (6 * 1) = 256 + 112 + 6
= 374
End ShowDecimal to Hexa Decimal Conversion
To convert decimal to hexa is slightly more difficult. The typical
method to convert from decimal to hexa is repeated division by
16.
Repeated Division By 16
For this method, divide the decimal number by 16, and write the
remainder on the right hand side as the least significant digit.
This process is continued by dividing the quotient by 16 and writing
the remainder until the quotient is 0.
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Ex: Convert decimal 2811 to hexa as follows:
Division Quotient Remainder 2811 / 16 175 11 = B
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Division Quotient Remainder 2811 / 16 175 11 = B
175 / 16 10 15 = F
Ex: Convert decimal 2811 to hexa as follows:
End Show
Division Quotient Remainder 2811 / 16 175 11 = B
175 / 16 10 15 = F
10 / 16 0 10 = A
Ex: Convert decimal 2811 to hexa as follows:
End Show
Division Quotient Remainder 2811 / 16 175 11 = B
175 / 16 10 15 = F
10 / 16 0 10 = A
Ex: Convert decimal 2811 to hexa as follows:
281110 = AFB16
End Show
What is BIT (BINARY DIGIT)?
A bit is the smallest element of information used in a computer
A bit holds ONE of TWO possible values,
2.2 Data Representation
A bit which is OFF is also considered to be FALSE or NOT SET; a bit which is ON is also considered to be TRUE or SET
Only one of two values(0 or 1) can be stored in a single bit.
End ShowWith a single bit, you can represent any two distinct items.
Examples
one or zero
true or false
on or off
male or female
right or wrong
End Show
H 01001000
HASCII CODE IS 72
Data Representation
End ShowWhat is Byte ?Bytes are a grouping of 8 bits
A byte is the smallest addressable datum (data item) in the memory by the microprocessor.
End Show
The Binary equivalent of the decimal number 205 is as follows and the Most Significant Bit(MSB) and the Least Significant Bit (LSB)can be defined as below :
1 1 0 0 1 1 0 1
MSB
LSB
b7 b6 b5 b4 b3 b2 b1 b0
Most Significant Bit (MSB) Least Significant Bit (LSB)
1 0 0 1 1 0 0 1
The bits in a byte are filled from LSB (b0) to MSB (b7) respectively as follows:
End ShowSince a byte consists of eight bits, it can represent 28, or 256,
different values. Generally, we use a byte to represent:
Unsigned numeric values in the range 0 to 255
Signed numbers in the range -128 to +127
ASCII(American Standard Code for Information Interchange)
character codes
Other special data types requiring no more than 256
different values.
Many data types have fewer than 256 items so eight bits is
usually adequate.
End ShowCapacity of Various Units
8 bits = 1Byte
(2 10) 1024 Byte = 1 KB (kilobyte)
(210) 1024 KB = 1 MB (Megabyte)
(210) 1024 MB = 1 GB (Gigabyte)
(210 ) 1024 GB = 1 TB (Terabyte)
Unit conversion
1kiloByte = 210 bytes
1MegaByte= 220 bytes
1GigaByte = 230 bytes
1TeraByte = 240 bytes
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2.2.2 Coding Systems
1. BCD (Binary Coded Decimal)
2. ASCII
(American Standard Code for Information Interchange)
End ShowBCD - Binary Corded Decimal
@ Binary Coded Decimal is a numerical code.
@ In this code structure, each of the decimal digits (0-9) is represented by a four-bit binary code (eg: 3 is represented by 0011)
@ Each digit is then represented by it's binary equivalent.
@ 16 unique(different) numbers can be stored in the 4 bit binary code.
@ Thus there are 6 invalid four-bit combinations in the BCD code.
End Show
This makes BCD easy to read, but it is not very efficient in
terms of storage space, nor is it as efficiently processed in
hardware.
3 8 6
0011 1000 0110
The number 386 is coded in BCD as follows:
BCD code is 001110000110
The number 59 is coded in BCD as follows
5 9
0101 1001 BCD code is 01011001
End ShowCont… BCD
Decimal BCD
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
Decimal BCD
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
Valid Combinations Invalid Combinations
End ShowCont ….BCD
Converting the Decimal value 546 to BCD
5 = 01014 = 01006 = 0110
Thus 54610 = 0101010001102
Converting the binary value 011100010101 to decimal is
0111 = 70001 = 10101 = 5
Thus 0111000101012 = 71510
End Show
ASCII - The American Standard Code for Information Interchange
• The American National Standards Institute has published an American Standards Code for Information Interchange(ASCII)
• This code is now most widely used by major manufactures. so that their equipment will be compatible with those of other manufactures.
End Show
ASCII is a computer code which uses 128
different coding combinations of a group of
seven bits (27= 128) to represent
• Characters A to Z, both upper and lower case
• Special characters, < , ., ?, : ,etc.,
• Numbers 0 to 9
• Special control codes used in device control
End Show
HEX DEC CHAR
41 65 A
42 66 B
43 67 C
44 68 D
- - -
- - -
- - -
- - -
58 88 X
59 89 Y
5A 90 Z
HEX DEC CHAR
61 97 a
62 98 b
63 99 c
64 100 d
- - -
- - -
- - -
- - -
78 120 x
79 121 y
7A 122 z
HEX DEC CHAR
30 48 0
31 49 1
32 50 2
33 51 3
34 52 4
35 53 5
36 54 6
37 55 7
38 56 8
39 57 9
The tables below gives the ASCII character set :-
End Show
ASCII (Cont……)
Example Code the text string 'Hello.' in ASCII using hexadecimal digits.
H = 48
e = 65
l = 6C
l = 6C
o = 6F
. = 2E
Thus the string is represented by the byte sequence
48 65 6C 6C 6F 2E
Where are we?
2.3 Logic Gates and Circuits
End Show
A logic gate is an elementary building block of a digital circuit. Most logic gates have two inputs and one output.
At any given moment, every terminal is in one of the two binary conditions low (0) or high (1), represented by different voltage levels.
In most logic gates, the low state is approximately zero volts (0 V), while the high state is approximately five volts positive (+5 V).
There are seven basic logic gates: AND, OR, XOR, NOT, NAND, NOR, and XNOR.
End Show
Logic gates are small (several micron) structures which take one or more bits as input, and produce another bit as output
Different logic gates use different techniques to calculate their output
End Show
The most common logic gates used perform the following logic functions:
AND : Output is True(1) if all inputs are True (1)
OR : Output is False (0) if all inputs are False (0)
NOT : Output is the opposite of the single input
i.e., If input is true (1) the output is false (0)
If input is false (0) the output is true (1) End Show
The AND gate has two or more inputs.
The output from the AND gate is 1 if and only if all of the
inputs are 1, otherwise the output from the gate is 0.
The AND gate is drawn as follows
AND - Gate
A
B
F=A.B
+ -A B
A.B
End Show
0
0
End Show
0
0
0
End Show
0
0
0
End Show
1
0
End Show
1
0
0
End Show
0
0
0
End Show
0
1
End Show
0
1
0
End Show
0
0
0
End Show
1
1
End Show
1
1
1
End Show
0
0
1
End Show
Truth table for AND gate
A B A.B
0 0 0
0 1 0
1 0 0
1 1 1
The output from the AND gate is written as A.B
The truth table for a two-input AND gate looks like
00
F=0
01
F=0
10
F=0
11
F=1
End Show
OR-Gate
A
B
F=A+B
+ -
The OR gate has two or more inputs.
The output from the OR gate is 1 if any of the inputs is 1.
The gate output is 0 if and only if all inputs are 0.
The OR gate is drawn as follows
A
B A+B
End Show
0
0
End Show
0
0
End Show
0
End Show
0
End Show
1
0
End Show
1
0
End Show
1
End Show
1
End Show
0
1
End Show
0
1
End Show
1
End Show
1
End Show
1
1
End Show
1
1
End Show
1
End Show
1
End Show
A B A+B
0 0 0
0 1 1
1 0 1
1 1 1
The output from the OR gate is written as A+B
The truth table for a two-input OR gate looks like
0
0F=0
0
1F=1
1
0F=1
1
1F=1
Truth table for OR gate
End Show
NOT-Gate
The input to the NOT gate A is inverted
i.e. The binary input state of 0 gives an output of 1 and the binary input state of 1 gives an output of 0.
Ā is known as "NOT A" or alternatively as the complement of A.
A Ā
0 11 0
The NOT gate is unique in that it only has one input. It looks like
AA ĀĀ
End Show
Logic Gates from Transistors• For example, we will build a NOT gate from a
transistor.
Transistor
extrapowersource
Input toNOT gate.
Output fromNOT gate.
End Show
Logic Gates from Transistors• For example, we will build a NOT gate from a
transistor.
TransistorOFF
extrapowersource
NOT 1 0
Input toNOT gateis ON.
Output fromNOT gateis OFF.
End Show
Logic Gates from Transistors• For example, we will build a NOT gate from a
transistor.
TransistorON
extrapowersource
NOT 0 1
Input toNOT gateis OFF.
Output fromNOT gateis ON.
End Show
0
End Show
0
End Show
1
End Show
1
End Show
1
End Show
1
End Show
0
End Show
0
End Show
NAND-Gate
The truth table for a two-input
NAND gate looks like
A B P
0 0 1
0 1 1
1 0 1
1 1 0
BAP .
A
B
BAP .
End Show
NOR-Gate
The truth table for a two-
input NOR gate looks like
A B P
0 0 1
0 1 0
1 0 0
1 1 0
BAP A
B
End Show
XOR-Gate
The truth table for a two-
input XOR gate looks like
A B P
0 0 0
0 1 1
1 0 1
1 1 0
A
B
End Show
A BP
BAP
2.4 Boolean Algebra
A set of rules formulated by the English mathematician George Boole describe certain propositions whose outcome would be either true or false. With regard to digital logic, these rules are used to describe circuits whose state can be either, 1 (true) or 0 (false).
In order to fully understand this, the relation between the AND gate, OR gate and NOT gate operations should be appreciated.
George Boole
End Show
Algebra: variables, values, operations
In Boolean algebra, the values are the symbols 0 and 1If a logic statement is false, it has value 0If a logic statement is true, it has value 1
Operations: AND, OR, NOT.
2.4.1 Basic Laws in Boolean AlgebraEnd Show
Laws of Boolean Algebra
(a) (a) A + B = B + AA + B = B + A
A B A+B
0 0 0
0 1 1
1 0 1
1 1 1
(b) (b) A.B = B.AA.B = B.A
A B A+B
0 0 0
0 1 0
1 0 0
1 1 1
T1 : Commutative LawT1 : Commutative Law
B A B+A
0 0 0
1 0 1
0 1 1
1 1 1
B A B+A
0 0 0
1 0 0
0 1 0
1 1 1
End Show
T2 : Associate Law T2 : Associate Law
(a) (A + B) + C = A + (B + C) (a) (A + B) + C = A + (B + C)
A B C (A+B)
(A+B)+C
0 0 0 0 0
0 0 1 0 1
0 1 0 1 1
0 1 1 1 1
1 0 0 1 1
1 0 1 1 1
1 1 0 1 1
1 1 1 1 1
A B C (B+C)
A+(B+C)
0 0 0 0 0
0 0 1 1 1
0 1 0 1 1
0 1 1 1 1
1 0 0 0 1
1 0 1 1 1
1 1 0 1 1
1 1 1 1 1
End Show
(b) (A.B).C = A.(B.C) (b) (A.B).C = A.(B.C)
Associate Law contdAssociate Law contd……. …….
A B C (A.B) (A.B).C
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 1 1 0 0
1 0 0 0 0
1 0 1 0 0
1 1 0 1 0
1 1 1 1 1
A B C (B.C) A.(B.C)
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 1 1 1 0
1 0 0 0 0
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1
End Show
T3 : Distributive LawT3 : Distributive Law
(a) A.(B + C) = A.B + A.C(a) A.(B + C) = A.B + A.C
A B C (B+C)
A.(B+C)
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 1 0
1 0 0 0 0
1 0 1 1 1
1 1 0 1 1
1 1 1 1 1
A B C AB AC AB+AC
0 0 0 0 0 0
0 0 1 0 0 0
0 1 0 0 0 0
0 1 1 0 0 0
1 0 0 0 0 0
1 0 1 0 1 1
1 1 0 1 0 1
1 1 1 1 1 1
End Show
(b) A + (B.C) = (A + B) (A + C) (b) A + (B.C) = (A + B) (A + C)
A B C BC A+(BC)
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 1 1 1 1
1 0 0 0 1
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
A B C (A+B)
(A+C) (A+B).(A+C)
0 0 0 0 0 0
0 0 1 0 1 0
0 1 0 1 0 0
0 1 1 1 1 1
1 0 0 1 1 1
1 0 1 1 1 1
1 1 0 1 1 1
1 1 1 1 1 1
End Show
T4 : Identity Law T4 : Identity Law
(a) A + A = A (a) A + A = A
(b) A.A = A (b) A.A = A
A A P=A+A
0 0 0
1 1 1
A
0
1
A A P=A.A
0 0 0
1 1 1
A
0
1
End Show
T5 :(a) T5 :(a) ABAAB
A
0
0
1
10
1
0
1
1
0
0
0
AB
0
1
0
0
010
101
111
000
BA BA BAAB B
End Show
ABABA .(b)
A
0
0
1
11
1
0
1
1
1
1
0
A+B
010
101
111
000
BA )).(( BABA BA
End Show
T6 : Redundance Law T6 : Redundance Law
(a) (a) A + A.B = AA + A.B = A
(b) (b) A.(A + B) = AA.(A + B) = A
A B A.B A+A.B
0 0 0 0
0 1 0 0
1 0 0 1
1 1 1 1
A
0
0
1
1
A B A+B
A.(A+B)
0 0 0 0
0 1 1 0
1 0 1 1
1 1 1 1
A
0
0
1
1
End Show
T7 :T7 :
(a) (a) 0 + A = A0 + A = A
(b) (b) 0 . A = 00 . A = 0
0 A 0+A
0 0 0
0 1 1
A
0
1
0 A 0.A
0 0 0
0 1 0
0
0
0
End Show
T8 :T8 :
(a) (a) 1 + A = 11 + A = 1
((b) b) 1 . A = A1 . A = A
1 A 1+A
1 0 1
1 1 1
1
1
1
1 A 1.A
1 0 0
1 1 1
A
0
1
End Show
T9 : T9 : (a)(a) 1 AA
0. AA(b)(b)
A
0 1 1
1 0 1
1
1
1
A AA
A
0 1 0
1 0 0
0
0
0
A AA.
End Show
• T10 :
BAA
BABAA
ABBAA )(
A B
0 0 1 0 0
0 1 1 1 1
1 0 0 0 1
1 1 0 0 1
A B A+B
0 0 0
0 1 1
1 0 1
1 1 1
BAA
)( BAA A B
0 0 1 1 0
0 1 1 1 0
1 0 0 0 0
1 1 0 1 1
A B A.B
0 0 0
0 1 0
1 0 0
1 1 1
BAA
End Show
BABA .
BABA .
(a)
(b)
T11T11 : : De Morgan's TheoremDe Morgan's Theorem
1
1
1
0
A+B
0
0
0
1
10
01
11
00
BA BA A B
0 0 1 1 1
0 1 1 0 0
1 0 0 1 0
1 1 0 0 0
BABA .
A B BA.
A.B
1
0
0
0
0
1
1
1
10
01
11
00BA BA. A B
0 0 1 1 1
0 1 1 0 1
1 0 0 1 1
1 1 0 0 0
A B BA
End Show
End Show