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  • 5/28/2018 BADASS Tutorial on Chapter 1 Number System

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    Ms Sandhya Rani Dash

    _____________________________________________________________________

    UNIT 2: NUMBER SYSTEM AND CODES

    Structure

    2.1 Introduction

    2.2 Objectives

    2.3 Binary Numbers

    2.3.1 Binary-to-Decimal conversion

    2.3.2 Decimal-to-Binary Conversion

    2.4 Octal Numbers

    2.4.1 Octal-to-Binary Conversion2.4.2 Binary-to-Octal Conversion

    2.4.3 Octal-to-Decimal Conversion

    2.4.4 Decimal-to-Octal conversion

    2.5 Hexadecimal Numbers2.5.1 Hexadecimal-to-Binary Conversion

    2.5.2 Binary-to-Hexadecimal Conversion

    2.5.3 Hexadecimal-to-Decimal Conversion

    2.5.4 Decimal-to-Hexadecimal Conversion2.5.5 Octal to Hexadecimal Conversion

    2.5.6 Hexadecimal to Octal Conversion

    2.6 Arithmetic Operation

    2.7 1s and 2s Complement2.7.1 Subtraction using 1s complement

    2.7.2 Subtraction using 2s complement

    2.8 9sand 10s Complement

    2.8.1 Subtraction using 9,s complement

    2.8.2 Subtraction using 10s complement

    2.9 Codes

    2.9.1 BCD code

    2.9.2 Excess-3 code

    2.9.3 Gray code

    2.10 Unit Summary

    1.1IntroductionThe study of number system begins with the familiar decimal number system. The

    decimal number system consists of ten unique digits 0 to 9. So, its base/radix is 10

    (since it involves with 10 symbols).The digits 0 to 9 are used to specify any number.Computers and other digital circuits dont process decimal numbers . They work with

    binary numbers which consists of only two digits 0&1.This creates a problem.

    Because to represent larger decimal quantities, a very long binary number is required.

    So to compress this long string of binary numbers, octal & hexadecimal number

    systems are widely used in digital systems .So the knowledge of these number

    systems is very essential for understanding, analyzing and designing of digital

    systems.

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    This unit discuss about the decimal, binary, octal and hexadecimal numbers. The inter

    conversion of these number systems are also explained here. Different types of codes ,

    which are commonly used in digital systems and microprocessors have also been

    discussed in this unit.

    1.2ObjectivesAfter going through this unit you will be able to:

    i) Explain different types of number system(Binary, Decimal, and Octal & Hexadecimal)

    ii) Convert one number system to another

    iii) Appreciate binary arithmetic operation

    iv) Define complements (1s, 2s, 9s & 10s complement)

    v) Apply use of complements in subtraction

    vi) Explain different types of codes and convert a code to binary and vice versa.

    2.3 Binary Numbers

    Binary number system is extensively used in digital electronics since it uses only one

    two symbols 1 & 0. These two digits represent the two voltage levels high or low

    in digital electronics. The base (radix) of the binary number system is 2. It is a

    positional system, i.e., every position is assigned a specific weight. In digital system

    each binary digit is called a bit

    A group of 4 bit is called nibble

    A group of 8 bit is called byte

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    Representation of Decimal and Binary numbers

    The following table represents the decimal number and its corresponding binarynumber

    DecimalNumber

    Binary NumberB3 B2 B1 B0

    0 0 0 0 0

    1 0 0 0 1

    2 0 0 1 0

    3 0 0 1 1

    4 0 1 0 0

    5 0 1 0 1

    6 0 1` 1 0

    7 0 1 1 1

    8 1 0 0 09 1 0 0 1

    2.3.1 Binary to decimal conversion

    Binary numbers can be converted into a decimal number by multiplying by the binary

    numbers 1 or 0 by their positional weight and adding the product.

    For Example

    Convert the binary number 11010 to its decimal equivalent

    Binary

    Number

    Conversation of Binary to

    decimal

    Decimal

    Equivalent

    11010

    1x2 +1x2 +0x2 +1x2 +0x2

    =16+8+0+2+0=26

    26

    2.3.2 Decimal to Binary conversion

    Any decimal numbers can be converted into its equivalent binary numbers by

    successive division by 2 and keeping the track of the reminders. The division is

    continued till the quotient is 0. The number read from bottom to top gives the

    equivalent binary number.

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    For example

    Convert 26 to its binary equivalent

    Decimal

    Number

    Quotient Remainder Binary

    Equivalent

    (26)10

    26/2 13 0-MSB

    (11010)13/2 6 1

    6/2 3 0

    3/2 1 1

    1/2 0 1-LSB

    The decimal fraction is converted to binary fraction by successive multiplication by 2

    and keeping the track of integers generated. The multiplication is continuities till the

    fraction part of the product is 0. The integers read from top to bottom to get the binary

    equivalent.

    For example

    Convert 0.625 to its binary equivalent

    Decimal

    Number

    Generated

    Integer

    Binary

    Equivalent

    0.625

    0.625x 2

    1.250 1-MSB0.1010.250x 2

    0.500 0

    0.500x 2

    1.000 1- LSB

    0.000x 2

    0.000 0

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    Self Check Exercise-1

    A. For Q No. 1 - 5 select the most appropriate alternative from among the four

    alternatives given below each question

    1. The digital systems usually operate on _______ system.

    a) Binary b) decimalc) octal d) hexadecimal

    2. The binary system uses power of ______for positional values.

    a) 2 b) 8 c) 16 d) 10

    3. After counting 0, 1, 10, 11, the next binary number is

    a) 12 b) 100 c) 101 d) 110

    4. The number 1000 is equivalent to decimal number

    a) one thousand b) eight c) four d) sixteen

    5. In binary numbers, shifting the binary point one place to the right

    a) Multiplies by 2 b) divides by 2

    c) Decreases by 2 d) increases by 10

    B. Answer the following questions

    6. Convert the following decimal numbers to their binary equivalents:

    a) 167 b) 72.45 c) 52 d) 163.875

    7. How do you convert a decimal number to an equivalent number in any other basesystem?

    2.4 Octal Numbers

    The base or( radix) of the octal number system is 8.This number system uses 8

    different symbols 0,1,2,3,4,5,6,7. Each significant position in an octal no. system has

    a positional weight. These weights are given in the ascending powers of 8, i.e 80, 8

    1,

    82,.etc. respectively.

    The binary equivalent of an octal no. must be a group of 3 bits word, since the base of

    octal no. system is 8=23.

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    2.4.1Octal to Binary conversion

    Octal numbers can be converted into equivalent binary numbers by replacing each

    octal digit by its 3 bit equivalent binary.

    Example: convert (3702)8to a binary number.

    (3702)8= 3 7 0 2011 111 000 010

    The number (3702)8= (011111000010)2

    = (11111000010)2

    On the most significant digit, leading zeros can be suppressed .i.e. 3 can be written as

    11 instead of 011.

    2.4.2Binary to octal conversion

    To convert from binary to octal back, the reverse procedure is used .i.e. divide the

    whole binary string into groups of 3 bits. Then starting from LSB and moving towards

    MSB, convert each group of 3 bits into the corresponding octal digit.

    For Example

    Convert (10101011111)2to an octal number.

    Solution: 10 101 011 111

    2 5 3 7

    Therefore (10101011111)2= (2537)8

    2.4.3 Octal to decimal conversion

    Any octal number can be converted into its equivalent decimal number by multiplyingeach digit of the octal number by the weight of its position and then adding the

    products.

    Example

    Convert (4257.14)8 into its equivalent decimal number.

    Solution

    (4257.14)8=4 x83+2 x82+5 x 81+7 x80+ 1 x 8-1+4 x 8-2

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    =2048 + 128 + 40 + 0.125 + 0.0625

    = (2216.1875)10

    2.5.4 Decimal to octal conversion

    The octal equivalent of a decimal no. can be obtained by dividing a given decimal

    number by 8 repeatedly until a quotient 0 is obtained. The process is exactly the same

    as the method explained earlier. The following example illustrates the conversion

    from decimal to octal.

    For Example

    Convert (115.1875)10to octal no.

    Integer Part Quotient Reminder Octal

    Equivalent

    115

    115/8 14 3 MSB

    (163)814/8 1 6

    1/8 0 1 LSB

    Fractional

    Part

    Result Integral Part Octal

    Equivalent

    (0.1875)10

    0.1875x8 1.5000 1

    (0.14)80.5000x8 4.000 4

    Therefore (115.1875)10= (163.14)8

    Self Check Exercise 2

    A. For Q No. 8 -11 select the most appropriate alternative from among the four

    alternatives given below each question

    8. The number 128is equivalent to decimal

    a) 12 b) 20 c) 10 d) 4

    9. The number 10001012is equivalent to octal

    a) 54 b) 45 c) 37 d) 25

    10. The number 178 is equivalent to binary

    a) 11 b) 1110 c) 10000 d) 1111

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    11. Which of the following is NOT an octal number?

    a) 19 b) 77 c) 15 d) 101

    B. Answer the following questions

    12. Convert each of the octal numbers to binary.

    a)158 b) 1738 c) 367.528 d) 1578

    13. Convert the following decimal no. to octal no. then to binary.

    a)59 b) 0.58 c) 64.2 d) 199.3

    2.5 Hexadecimal Numbers

    The hexadecimal number system is the most popular means of direct data entry and

    retrieval in digital system. The base of hexadecimal number system is 16 and uses 16

    symbols, namely, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. The symbol A to F

    represents the decimal 10 to 15 respectively.

    Since its base is 16= 24, every 4 bit binary digit combination can be represented by

    one hexadecimal digit. Each significant position in an hexadecimal number has a

    positional weight. These weights are given in the ascending powers of 16. i.e., 160,

    161,

    162, . etc

    2.5.1 Hexadecimal-to-binary conversion

    The conversion of Hexadecimal to Binary is the reverse process of conversion from

    Binary to Hexadecimal as explained above. i.e. each hexdigit is replaced by its 4-bit

    binary group.

    For example

    Convert (2DAC)16 to Binary

    (2DAC)16 = 2 D A C0010 1101 1010 1100

    = (10110110101100)2

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    2.5.2 Binary to hexadecimal conversion

    To convert a binary no. to a hexadecimal no., divide the whole binary string in to

    groups of 4 bits each and then replace each 4 bit group by the equivalent hexadecimal

    digit

    For example

    Convert (1010100100)2to hexadecimal no.

    (1010100100)2 =0010 1010 0100

    = 2 A 4= (2A4)16

    2.5.3 Hexadecimal to decimal conversion

    To Convert a Hexadecimal number to Decimal multiply each digit in the Hex no. by

    its position weight and take the summation of all these product terms

    For Example

    Convert to (5A6)16 to Decimal

    (5A6)16 =5 x 162 +

    10 x 161+ 6 x 16

    0

    = 1280 + 160 + 6

    = (1446)10

    2.5.4 Decimal to hexadecimal conversion

    A decimal no. can be converted to Hexadecimal no. by successively dividing the

    given decimal no. by 16 till the quotient is 0.

    For Example

    Convert (2598)10to Hexadecimal

    DecimalNumber

    Quotient Remainderin Decimal

    HexadecimalNo.

    HexadecimalEquivalent

    (2598)10

    2898/16 162 6 6

    (A326)16162/16 10 2 2

    10/16 0 10 A

    Therefore (2598)10= (A326)16

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    2.5.5 Octal to hexadecimal conversion

    To convert an octal no. to hexadecimal the easiest method is to first convert the hex

    no. to its binary equivalent and then the binary no. to hexadecimal. It will be clear

    from the example given below.

    For Example:

    Convert (756.402)8to hexadecimal no.

    (756.402)8 = 7 5 6 . 4 0 2

    = 111 101 110 100 000 010

    For the integer part, group the binary bits into 4 from LSB to MSB and for the

    fractional part group the bits from the MSB to LSB

    (756.402)8 = 0001 1110 1110 . 1000 0001 000= 1 E E 8 1 0

    = (1EE.810)16

    2.5.6 Hexadecimal to octal conversion

    To convert a hexadecimal no. to octal, the simple way is to first convert binary no. tooctal.

    For Example: Convert (56)16to Octal no.

    (56)16= 0101 0110

    = (01010110)2= 01 010 110

    = 1 2 6= (126)8

    Self Check Exercise 3

    A. For Q No. 14 -19, Select the most appropriate alternative from among the

    four alternatives given below each question

    14. The radix/base of the hexadecimal no. system is

    a)10 b)16 c) 8 d) 2

    15. The symbol E represents the decimal

    a)13 b)15 c) 14 d) 12

    16. The number 2D16is equivalent to binary

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    a) 0010 1101 b) 1101 0010 c) 0011 1101 d) 010 0011

    17. Hexadecimal number system is used as a shorthand language for representing

    __________ numbers.

    a) decimal b)binary c) octal d) large

    18.The binary equivalent of A16is

    a)1010 b) 1011 c) 1000 d) 1110

    19. The number 4716is equivalent to octal

    a) 701 b) 137 c) 401 d) 107

    B. Answer the following questions

    20. What are the advantages of aa) Binary no. system b) Hexadecimal no. system

    21. Convert the following hexadecimal numbers to binary.

    a) 4916 b) 32416 c) 4BAC16 d) 3A9.BD16

    2.6 Arithmetic Operation

    Binary arithmetic is much simpler than decimal arithmetic because here only two

    digits 0&1 are involved. Arithmetic operations in a computer are done in its

    arithmetic unit by using binary numbers. The basic arithmetic operations are , addition

    ,subtraction, multiplication ,division.

    These operations are performed by digital logic circuits called the adder, sub tractor ,

    multiplier and divider.

    The arithmetic rules for Addition, Subtraction, Multiplication, and Division of binary

    numbers are given below.

    Addition Subtraction Multiplication Division

    0+0=0 0-0=0 0 x 0=0 0/1=0

    0+1=1 1-0=0 0 x 1=0 1/1=1

    1+0=1 1-1=0 1 x 0=0 0/0=0

    1+1=10 10-1=1 1 x 1=1 1/0=not allowed

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    Self Check exercise 4

    A. For Q No. 22 -26 Select the most appropriate alternative from among the fouralternatives given below each question

    22. The binary addition 1 + 1 + 1 gives

    a) 111 b) 10 c) 110 d) 11

    23. The cumulative addition of the four binary bits ( 1 + 1 + 1 + 1 ) gives

    a) 1111 b) 111 c) 100 d) 1001

    24. The result of binary subtraction (100011) is

    a)111 c) 111 c) 011 d)001

    25. The result of binary multiplication 1112 x 102 is

    a) 1101 b) 0110 c) 1001 d) 1110

    26.The binary division 110002/1002 gives

    a) 110 b) 1100 c) 11 d) 101

    B. Answer the following questions

    27. Subtract the following groups of binary numbers.

    a) 1111011011 b) 10110.1 - 1100.01 c) 101.101100.1

    28. Add the following groups of binary numbers using the binary addition method.

    a) 1010 + 1011 b) 1011.0111 + 1101.101 c) 1110.1011 + 1001.1110

    2.7 1s and 2sComplement

    There are two complement forms used in digital systems to represent signed numbers.These are 1s complement and 2s complement form. These two forms are used to

    represent negative numbers. Most digital computers do subtraction by the 2scomplement method .The advantage of performing subtraction by the complement

    method is reduction in the hardware. Instead of having separate digital circuits foraddition and subtraction, only adding circuits are needed .i.e. subtraction is also

    performed by adders only.

    To take the 1s complement of a binary no., simply change all 1s to 0s and all 0s to

    1s

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    For Example

    The 1s complement of 110010 is 001101

    Similarly the 2s complement of a binary no. can be obtained by adding 1 to its 1scomplement

    For Example

    Find the 2s complement of 10110

    1s complement of 10110= 01001

    For obtaining 2s complement of 10110, you have to add 1 to 01001(1s complement

    of 10110)

    i.e, 1+ 01001= 01010

    2.7.1 Subtraction using 1s complement

    The subtraction using 1s complement can be obtained by the following methods:

    Take the 1s complement of the subtrahend (smaller number). Add this to the minuend (larger number).

    If a carry is generated, it indicates that the answer is positive. Add the carry tothe result. This carry is called the end-around carry (EAC).

    If no carry is generated, then the answer is negative. Take the 1s complementof the original addition and change the sign.

    For Example 1

    Subtract (1011)2from (1110)2using 1s complement method.

    1 1 1 0 (+)

    1s complement- 0 1 0 0____________________

    Carry 1 0 0 1 0

    Add Carry 1____________________

    0 0 1 1

    The answer is (0011)2

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    For Example 2

    Subtract (11000)2from (101)2 using 1s complement method

    1 0 1 (+)

    1s complement 0 0 1 1 1___________________________

    No carry 0 1 1 0 0

    The answer isve, since no carry is obtained. Si the answer is the 1s complement ofthe 01100 and is in opposite sign i.e, -10011

    2.7.2 Subtraction using 2s complement

    2s complement subtraction can be obtained by the following methods.

    Take the 2s complement of the subtrahend (smaller number).Add it to the minuend (larger number).Presence of a Carry indicates that the answer is positive. Omit the carry.Absence of a carry indicates that the answer is negative. Take the 2s

    complement of the original addition and change the sign.

    For Example

    Subtract (1011)2from (1110)2using 2s complement method

    1 1 1 0 (+)

    2s complement 0 1 0 1_______________________________

    Carry 1 0 0 1 1

    The carry is ignored. Thus the answer is (0011)2

    For Example 2

    Subtract (101002from (1000)2using 2s complement method

    1 0 0 0 (+)

    2s complement 0 1 1 0___________________________________

    No Carry 1 1 1 0

    No carry is obtained. The answer is the 2s complement of (1110) 2and is in opposite

    sign.

    i.e, -(0010)2

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    2.8 9s and 10s Complement

    9s complement

    The 9s complement of a decimal number can be obtained by subtracting each digit in

    the number from 9,i.e a digit plus its 9s complement equals to 9. The 9s complementof decimal digits 0 to 9 are listed below.

    Decimal No. 9s Complement

    0 0

    1 8

    2 7

    3 6

    4 5

    5 4

    6 3

    7 28 1

    9 0

    For Example

    Find the 9scomplement of each of the following decimal no.s

    a) 15 b)135

    Subtract each digit in the no. from 9 to get the 9s complement

    a) 9915 = 84 (9s complement of 15)

    b) 999135 = 864 (9s complement of 135)

    10s complement

    The 10s complement of a decimal no. i equal to its 9s complement + 1

    For Example

    Convert the following decimal no. in to its 10s complement form

    a) 18 b) 152

    a) 99-18= 81 (9s complement of 18)

    81+1= 82(10s complement of 18)

    b) 999-152= 847 (9s complement of 152)

    847+1 = 848 (10s complement of 152)

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    2.8.1 Subtraction using 9s complement

    Subtraction of a smaller number from a larger number us ing 9s complement

    Steps:

    Take the 9s complement of the smaller number Add the 9s complement to the larger number Carry indicates that the answer is positive. Add the carry to the list significant

    digit

    If there is no carry the answer is negative. Take the 9s complement of theresult and change the sign

    For Example

    Subtract 143 from 363 using 9s complement

    Direction Subtraction 9s complement subtraction

    363 363- 143 9s complement of 143 + 856

    ________ _____________220 Carry 1 219

    Add carry to the result 1+ 219= 220

    The answer is positive. i.e, 220

    For Example

    Subtract 745 from 245 using 9s complement

    Direction Subtraction 9s complement subtraction

    245 245

    - 745 9s complement of 745 + 254________ _____________

    -500 Carry 0 499

    9s complement of 499 = 999-499= 500

    No carry indicates that the answer is negative. i.e, -500

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    2.8.2 Subtraction using 10s complement

    It is analogous to the often used 2s complement method in binary subtraction. In thismethod the minuend is added to the 10s complement of the subtrahend and the carry

    is ignored.

    For Example

    a) Subtract 54 from 88 using 10s complement method

    Direction Subtraction 10s complement subtraction

    88 88- 54 10s complement of 54 + 46

    ________ _____________34 Carry 1 34

    Ignore the carry. The answer is positive i.e, 34

    For Example

    b) Subtract 813 from 432 using 10scomplement method

    Direction Subtraction 10s complement subtraction

    432 432

    - 813 10s complement of 813 + 187________ _____________-381 Carry 0 619

    10s complement of 619 = 381

    Since there is no carry, the answer is negative. i.e, -381

    Self Check Exercise 5

    A. For Q No. 29 -33 select the most appropriate alternative from among the four

    alternatives given below each question

    29. The chief reason why digital computers use complemental subtraction is that it

    a) Simplifies their circuitry b) is a very simple processc) can handle negative numbers easily. d) Avoids direct subtraction

    30. The 1s complement of 10102is

    a) 0110 b) 0101 c) 1110 d) 0011

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    31. The 2s complement of 10002is

    a) 0111 b) 0101 c) 1000 d) 0001

    32. The 9s complement of 19 is

    a) 91 b) 99 c) 80 d) 81

    33.The 10s complement of 46 is

    a) 54 b) 53 c) 45 d) 35

    B. Answer the following questions

    34. How can negative no be represented? Distinguish between 1s complement and2s complement

    35.Distinguish between 9s complement and 10s complement ?

    2.9 Codes

    We are very comfortable with the decimal number system, but digital systems like

    computers & other digital circuits process the data in the binary format. So various

    binary codes are used to represent the data. A code is a symbolic representation of

    discrete information, which may be present in the form of numbers, letters or physical

    quantities. The information used in every code is represented in binary form. The

    binary code can be classified into two types of codes, i.e. numeric codes and

    alphanumeric codes. Alphanumeric codes represent alphanumeric information, i.e.

    letters of the alphabet, and decimal numbers as a sequence of 0s and 1s. An example

    of alphanumeric cod e is the ASCII code (American Standard Code for Information

    Interchange). Numeric code represents numeric information, i.e. only numbers as a

    series of 0s and 1s.An example of the numeric code is the Binary coded decimal

    (BCD) code which is used to represent decimal digits.

    Codes are also used for error detection and error correction purpose. Since when

    digital signals are transmitted from one location to another location, errors may occur

    due to electrical noise. Various types of codes are discussed below.

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    2.9.1 BCD (or) 8421 code

    A Binary Coded Decimal (BCD) is one in which, decimal digits are encoded by their

    natural binary equivalents-one at a time-into groups of four bits.

    For example (25)10is represented by [0010 0101] using BCD code, rather than(10001)2. It is a positional weight code. Each position of a number represents a

    specific weight.

    These codes are also known as 8421 code or simply BCD code. The number 8421

    indicates the binary weights of the four bits. i.e. the weights of the first (right most)

    position is 20=1, 2

    1=2, 2

    2=4, 2

    3=8.Replacing from left to right, the weights are 8-4-2-

    1,and hence it is called 8421 code.

    The disadvantage of BCD code is that it requires more no. of bits to code a decimal

    number. But the ease of conversion between the BCD code to decimal number and

    vice versa is the main advantage of this code, which makes it useful and popular code

    for input,output operations in digital systems .To convert any decimal number in to

    BCD ,each decimal digit should be replaced by the appropriate 4-bit code.

    The following table summarizes the binary and BCD codes for the decimal numbers

    0 to 15.

    Decimal numbers Binary number Binary Coded

    Decimal(BCD)

    0 0000 0000

    1 0001 0001

    2 0010 0010

    3 0011 0011

    4 0100 0100

    5 0101 0101

    6 0110 0110

    7 0111 0111

    8 1000 1000

    9 1001 1001

    10 1010 0001 0000

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    11 1011 0001 0001

    12 1100 0001 0010

    13 1101 0001 0011

    14 1110 0001 0100

    15 1111 0001 0101

    For Example

    Convert (3906)10to BCD.

    3 9 0 6

    11 1001 00000 0110

    Therefore (3906)10=11100100000110

    Note: Leafing zeros are added to ensure that each digit is represented by four bits .The

    leading zeros can be suppressed on the most significant digit.

    Conversion flow chart

    The flow chart summarizes the conversions that have been discussed so far. There is

    no line directing from octal to BCD ,because that direct conversion was not presented.

    To convert from octal to BCD , convert to binary, then decimal, then BCD.

    2.9.2 Excess-3 code

    The XS-3 code is a non-weight code. This means each position with in a binary

    number isnt assigned affixed value. This code is a self complementing code, which

    HEXADECIMAL BINARY DECIMAL

    OCTAL BCD

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    means 1s complement of the coded number yields 9s complement of the number

    itself. As the name indicates, XS-3 code represents a decimal number ,in binary form,

    as a number greater than 3. An XS-3 code is obtained by adding 3 to a decimal

    number, then convert it into its equivalent BCD code.

    For Example: A decimal 4 can be coded, by adding 3 into4; in order to obtain 7.7 is

    then encoded in its equivalent 4 bit binary code 0111.

    So 0111 is the XS-3 code of the decimal number 4.

    Or it can also be obtained directly in another way by adding binary equivalent of 3 to

    the natural BCD code of the digit.

    i.e. Decimal 4 is coded as 0100+0011=0111 in XS-3 code.

    For Example

    Convert (463)10into its XS-3 code.

    Given decimal number 4 6 3

    Add 3 to each bit +3 +3 +3

    _________________

    Sum 7 9 6

    Converting the above sum into its BCD code, we have

    Sum 7 9 6

    BCD 0111 1001 0110

    Hence, the XS-3 code for (463)10is 0111 1001 0110

    The following table summarizes the BCD, XS-3 code and 9s complement

    representation for decimal digits.

    Decimal BCD code XS-3 code 9s complement

    0 0000 0011 1100

    1 0001 0100 1011

    2 0010 0101 1010

    3 0011 0110 1001

    4 0100 0111 1000

    5 0101 1000 0111

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    6 0110 1001 0110

    7 0111 1010 0101

    8 1000 1011 0100

    9 1001 1100 0011

    2.9.3 Gray codes

    The gray code isnt a weighted code and isnt suitable for arithmetic operations, but

    finds application in input/output devices and in some types of analog to digital

    converters. Gray code is a cyclic code which has a special property of containing two

    successive code words differ in one bit position only, i.e . it is a unit distance code. It

    is also called as the reflected code.

    Conversion of a binary number to gray code

    Keep the first bit of the gray code is the same as the first bit of the binarynumber.

    The second bit of the gray code equals to the sum of first and second bit of thebinary number. Discard the carry if it is generated from the sum.

    The third bit of the gray cod equals to the sum of second and third bit of thebinary number and so on.

    Example: Convert the binary (110010)2to its gray code.

    Solution:

    1 + 1 + 0 + 0 + 1 + 0 [Binary]

    1 0 1 0 1 1 [Graycode]

    Conversion from gray code to binary

    Conversion of a Gray code into its binary form involves the reverse of the procedure

    as given below

    i) The first binary bit (MSB) is the same as that of the first Gray code bit

    ii) If the second Gray bit is zero, the second binary bit is the inverse of its first binarybit.

    iii) Step 2 is repeated for each successive bit

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    For Example

    Convert [1010111]Gto binary

    Solution:

    1 0 1 0 1 1 1 Gray

    1 1 0 0 1 0 1 Binary

    Therefore [1010111]G = [1100101]2

    Self Check Exercise 6

    A. For Q No. 36 -41 Select the most appropriate alternative from among the four

    alternatives given below each question

    36. BCD code is

    a) non-weighted code b) the same thing as binary numbers

    c) a binary code d) an alphanumeric code

    37. Which of the following 4-bit combinations is/are invalid in the BCD code?

    a) 1010 b) 0010 c) 0101 d) 1000

    38. In XS-3 code each coded number is ________ than in BCD

    a) four larger b) three smaller c) three larger d) much larger

    39. The XS-3 code of 01012 is

    a) 1001 b) 1000 c) 1001 d) 0110

    40. Gray code is

    a) non-weighted code b) weight code

    c) a binary code d) an alphanumeric code

    41. The gray code of 01112is

    a) 1000 b) 1001 c) 0101 d) 0100

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    B. Answer the following questions

    42. What is a BCD code? What are its advantages and disadvantages?

    43. Define the XS-3 code and gray code. Why these are important?

    2.9 Unit Summary

    This unit explains the various number systems, which are widely used in digital

    systems. The general properties of number system, methods of conversion from one

    number system to another and the rules of arithmetic operations are discussed.

    Various complement methods like 1s complement, 2s complement, 9s

    complement& 10s complement are introduced to make the subtraction easier. With

    the application of these methods the subtraction can be achieved by adding the

    complement of the subtrahend to the minuend. Different codes are used in digital

    systems which may be in the form of numbers, letters or physical quantities. Some of

    the more commonly used codes like BCD, XS-3Code, gray code are introduced. The

    knowledge of these number systems & codes is very essential for the effective

    understanding of digital systems

    Answers to Check Your Progress

    1. (a) 2. (a) 3. (b) 4. (b) 5. (a)

    6. a)101001112 b) 1001000.011102 c) 1101002 d) 10100011.1112

    7. A decimal number can be converted to an equivalent no. in any other base system

    with the base b by successively dividing the decimal no by b. In this method, the

    reminders are read from bottom to top. Similarly a decimal fraction can be converted

    to an equivalent no. in any other base system with the base b by successively

    multiplying the decimal no by b. In this method, the integers to the left of the radix

    point are read from top to bottom.

    8). (c) 9. (b) 10. (d) 11. (a)

    12. a) 00012 b) 001 111 0112 c) 011 110 111.101 0102 d) 001 101 1112

    13 a ) 1110112 b) 0. 100 101 000 1112 c) 001 000 000.001 1002

    d) 011 000 111.010 011 0012

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    14. (a) 15. (c) 16. (a) 17. (a) 18.(a) 19. (d)

    20. The advantages of binary no. system is that, it is very simple, because it consists

    of only two digits, i.e.0&1. So it is useful in digital computers ,since the switching

    circuits used in these computers use two state devices such as transistors ,diodes etc.

    Hexadecimal no. systems are used to compress the long strings of binary numbers.It is

    the most popular means of direct data entry and retrieval in digital systems.

    21. a ) 0100 10012 b) 0011 0010 01002 c) 0100 1011 1010 11002

    d) 0011 1010 1001. 1011 11012

    22. (d) 23. (c) 24. (d) 25. (d) 26. (a)

    27.a) 0011 b) 1010.01 c) 1.001

    28.a) 10101 b) 11001.0001 c) 11000.1001

    29. (a) 30. (b) 31. (c) 32. (c) 33. (a)

    34.Negative numbers can be represented in sign magnitude form ,i.e. MSB representsthe sign bit (0 for positive and 1 for negative) and the remaining bits represent the

    magnitude of the no. Negative numbers can also be represented by the 1s

    complement form, or in 2s complement form

    1s complement can be easily obtained using an inverter. The 2s complement can beobtained by first obtaining the 1s complement and then adding (1) to it

    35. The 9s complement of a decimal no. can be obtained by subtracting each digit of

    that decimal no. from 9. The 10s complement of a decimal no. is obtained by adding

    a 1 to its 9s complement.

    36. (c) 37. (a) 38. (c) 39. (b) 40. (a) 41. (d)

    42. The BCD code uses the binary no. system to specify the decimal numbers 0 to 9.

    In this code each decimal digit, 0 through 9 is coded by 4-bit binary no. It is also

    called as the 8421 code. The main advantage of this code is its ease of conversion to

    and from decimal and the disadvantage of this BCD code is that, arithmetic

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    operations are more complex then they are in pure binary. It is less efficient than the

    pure binary , in the sence that it requires more bits.

    43. A XS-3 code is a BCD code formed by adding 3(0011) to the binary equivalent of

    the decimal no.

    A Gray code is one, in which only one bit changes between successive numbers. It is

    also called! #$%& '%(&!$)* )+'*

    The XS-3 code helps in computing mathematical operations in digital computers.

    The Gray code is useful for its ease of conversion to and from binary. It is used in

    instrumentation and data acquisition systems where linear or angular displacement is

    measured. It is used in I/O devices, A/D converters and other peripheral devices.

    References

    Bignell, J and Donovan, R. (1989) Digital Electronics, Delmar Publishers Inc. New

    York

    Jain, R. P. (2007) Modern Digital Electronics; Third Edition; Tata McGraw-HillPublishing Company Limited; New Delhi

    Kumar, A. A. (2004) Fundamentals of Digital Circuits; Seventh Edition; Eastern

    Economy Editions; Prentice- Hall India Pvt. Ltd; New Delhi

    Mano, M. M. (2006) digital Logic and Computer Design; Second Edition; EasternEconomy Editions; Prentice- Hall India Pvt. Ltd; New Delhi

    Malbert, A. B. and Leach, D. P. (1993) Digital Principles and Applications, Tata Mc

    Graw hill Publishing Company Limited, New Delhi

    Salivahanan, S. and Arivazhagan, S. (2006) Digital Circuits and Design, VikasPublishing House Pvt Ltd, Delhi

    Srinivasan, P. (2007) National Programme Technology Enhanced Learning Video

    Lectures on Introduction to Digital Circuits and System. Lecture I & II

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    Unit End Exercise

    1. State the common features between different number system.

    2. What are meant by 1s and 2s complement of a binary number?

    3. Explain the rules for binary subtraction using 1s and 2s complement methods.

    4. Compare BCD and Binary systems of coding

    5. How do you convert a binary number to a Gray number and vice versa.