Number System and Codes PDF

NUMBER SYSTEM AND CODES INTRODUCTION:- The term digital refers to a process that is achieved by using discrete unit. In number system there are different symbols and each symbol has an absolute value and also has place value. RADIX OR BASE:- The radix or base of a...

NUMBER SYSTEM AND CODES INTRODUCTION:- The term digital refers to a process that is achieved by using discrete unit. In number system there are different symbols and each symbol has an absolute value and also has place value. RADIX OR BASE:- The radix or base of a number system is defined as the number of different digits which can occur in each position in the number system. RADIX POINT :- The generalized form of a decimal point is known as radix point. In any positional number system the radix point divides the integer and fractional part. Nr = [ Integer part. Fractional part ] ↑ Radix point NUMBER SYSTEM:- In general a number in a system having base or radix ‘ r ’ can be written as an an-1 an-2 …………… a0. a -1 a -2................ a - m This will be interpreted as Y = an x rn + an-1 x rn-1 + an-2 x rn-2 + ……… + a0 x r0 + a-1 x r -1 + a-2 x r -2 +….......... +a -m x r –m where Y = value of the entire number an = the value of the nth digit r = radix TYPES OF NUMBER SYSTEM:- There are four types of number systems. They are 1. Decimal number system 2. Binary number system 3. Octal number system 4. Hexadecimal number system DECIMAL NUMBER SYSTEM:- The decimal number system contain ten unique symbols 0,1,2,3,4,5,6,7,8 and 9. In decimal system 10 symbols are involved, so the base or radix is 10. It is a positional weighted system. The value attached to the symbol depends on its location with respect to the decimal point. In general,  dn dn-1 dn-2 …………… d0. d -1 d -2................ d - m is given by (dn x 10n) + (dn-1 x 10n-1) + (dn-2 x 10n-2) + … + ( d0 x 100) + ( d-1 x 10 -1) + (d-2 x 10 -2) +…+(d -m x 10 –m) For example:- 9256.26 = 9 x 1000 + 2 x 100 + 5 x 10 + 6 x 1 + 2 x (1/10) + 6 x ( 1/100) = 9 x 103 + 2 x 102 + 5 x 101 + 6 x 100 + 2 x 10-1 + 6 x 10-2 BINARY NUMBER SYSTEM:- The binary number system is a positional weighted system. The base or radix of this number system is 2. It has two independent symbols. The symbols used are 0 and 1. A binary digit is called a bit. The binary point separates the integer and fraction parts. In general, dn dn-1 dn-2 …………… d0. d -1 d -2................ d – k is given by (dn x 2n) + (dn-1 x 2n-1) + (dn-2 x 2n-2) + ….+ ( d0 x 20) + ( d-1 x 2 -1) + (d-2 x 2 -2) +….+(d -k x 2 –k) OCTAL NUMBER SYSTEM:- It is also a positional weighted system. Its base or radix is 8. It has 8 independent symbols 0,1,2,3,4,5,6 and 7. Its base 8 = 23 , every 3- bit group of binary can be represented by an octal digit. HEXADECIMAL NUMBER SYSTEM:- The hexadecimal number system is a positional weighted system. The base or radix of this number system is 16. The symbols used are 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E and F The base 16 = 24 , every 4 – bit group of binary can be represented by an hexadecimal digit. CONVERSION FROM ONE NUMBER SYSTEM TO ANOTHER :- 1. BINARY NUMBER SYSTEM:- (a) Binary to decimal conversion:- In this method, each binary digit of the number is multiplied by its positional weight and the product terms are added to obtain decimal number.  For example: (i) Convert (10101)2 to decimal. Solution : (Positional weight) 24 23 22 21 20 Binary number 10101 = (1 x 24) + (0 x 23) + ( 1x 22 ) + ( 0 x 21 ) + (1 x 20) = 16 + 0+ 4+ 0+ 1 = (21)10 (ii) Convert (111.101)2 to decimal. Solution: (111.101)2 = (1 x 22) + (1 x 21) + ( 1x 20 ) + ( 1 x 2 -1 ) + (0 x 2 -2) + (1 x 2 -3) = 4+ 2+ 1 + 0.5 + 0 + 0.125 = (7.625)10 (b) Binary to Octal conversion:- For conversion binary to octal the binary numbers are divided into groups of 3 bits each, starting at the binary point and proceeding towards left and right. Octal Binary Octal Binary 0 000 4 100 1 001 5 101 2 010 6 110 3 011 7 111 For example: (i) Convert (101111010110.110110011)2 into octal. Solution : Group of 3 bits are 101 111 010 110. 110 110 011 Convert each group into octal = 5 7 2 6. 6 6 3 The result is (5726.663)8 (ii) Convert (10101111001.0111)2 into octal. Solution : Binary number 10 101 111 001. 011 1 Group of 3 bits are = 010 101 111 001. 011 100 Convert each group into octal = 2 5 7 1. 3 4 The result is (2571.34)8 (c) Binary to Hexadecimal conversion:- For conversion binary to hexadecimal number the binary numbers starting from the binary point, groups are made of 4 bits each, on either side of the binary point.  Hexadecimal Binary Hexadecimal Binary 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 For example: (i) Convert (1011011011)2 into hexadecimal. Solution: Given Binary number 10 1101 1011 Group of 4 bits are 0010 1101 1011 Convert each group into hex = 2 D B The result is (2DB)16 (ii) Convert (01011111011.011111)2 into hexadecimal. Solution: Given Binary number 010 1111 1011. 0111 11 Group of 3 bits are = 0010 1111 1011. 0111 1100 Convert each group into octal = 2 F B. 7 C The result is (2FB.7C)16 2. DECIMAL NUMBER SYSTEM:- (a) Decimal to binary conversion:- In the conversion the integer number are converted to the desired base using successive division by the base or radix. For example: (i) Convert (52)10 into binary. Solution: Divide the given decimal number successively by 2 read the integer part remainder upwards to getequivalent binary number. Multiply the fraction part by 2. Keep the integer in the product as it is and multiply the new fraction in the product by 2. The process is continued and the integer are read in the products from top to bottom. 2 I 52 2 l 26 —0 2 l 13 —0 2l 6 —1 2l 3 —0 2l 1 —1 0 —1 Result of (52)10 is (110100)2 (ii) Convert (105.15)10 into binary. Solution: Integer part Fraction part 2 I 105 0.15 x 2 = 0.30 2l 52 ― 1 0.30 x 2 = 0.60 2l 26 ― 0 0.60 x 2 = 1.20 2l 13 ― 0 0.20 x 2 = 0.40 2l 6 ― 1 0.40 x 2 = 0.80 2l 3 ― 0 0.80 x 2 = 1.60 2l 1 ― 1 0 ― 1 Result of (105.15)10 is (1101001.001001)2 (b) Decimal to octal conversion:- To convert the given decimal integer number to octal, successively divide the given number by 8 till the quotient is 0. To convert the given decimal fractions to octal successively multiply the decimal fraction and the subsequent decimal fractions by 8 till the product is 0 or till the required accuracy is obtained. For example: (i) Convert (378.93)10 into octal. Solution: 8 I 378 0.93 x 8 = 7.44 8 l 47 ― 2 0.44 x 8 = 3.52 8l 5 ― 7 0.52 x 8 = 4.16 0 ― 5 0.16 x 8 = 1.28 Result of (378.93)10 is (572.7341)8 (c) Decimal to hexadecimal conversion:- The decimal to hexadecimal conversion is same as octal. For example: (i) Convert (2598.675)10 into hexadecimal. Solution: Remainder Decimal Hex Hex 16 I 2598 0.675 x 16 = 10.8 A 16 l 162 — 6 6 0.800 x 16 = 12.8 C 16 l 10 — 2 2 0.800 x 16 = 12.8 C 0 — 10 A 0.800 x 16 = 12.8 C Result of (2598.675)10 is (A26.ACCC)16 3. OCTAL NUMBER SYSTEM:- (a) Octal to binary conversion:- To convert a given a octal number to binary, replace each octal digit by its 3- bit binary equivalent.  For example: Convert (367.52)8 into binary. Solution: Given Octal number is 3 6 7. 5 2 Convert each group octal = 011 110 111. 101 010 to binary Result of (367.52)8 is (011110111.101010)2 (b) Octal to decimal conversion:- For conversion octal to decimal number, multiply each digit in the octal number by the weight of its position and add all the product terms For example: - Convert (4057.06) 8 to decimal Solution: (4057.06) 8 = 4 x 83 + 0 x 82 + 5 x 81 + 7 x 80 + 0 x 8 – 1 + 6 x 8- 2 = 2048 + 0 + 40 + 7 + 0 +0.0937 = (2095. 0937)10 Result is (2095.0937)10 (c) Octal to hexadecimal conversion:- For conversion of octal to Hexadecimal, first convert the given octal number to binary and then binary number to hexadecimal. For example :- Convert (756.603)8 to hexadecimal. Solution :- Given octal no. 7 5 6. 6 0 3 Convert each octal digit to binary = 111 101 110. 110 000 011 Group of 4bits are = 0001 1110 1110. 1100 0001 1000 Convert 4 bits group to hex. = 1 E E. C 1 8 Result is (1EE.C18)16 (4) HEXADECIMAL NUMBER SYSTEM :- (a) Hexadecimal to binary conversion:- For conversion of hexadecimal to binary, replace hexadecimal digit by its 4 bit binary group. For example: Convert (3A9E.B0D)16 into binary. Solution: Given Hexadecimal number is 3 A 9 E. B 0 D Convert each hexadecimal = 0011 1010 1001 1110. 1011 0000 1101 digit to 4 bit binary Result of (3A9E.B0D)8 is (0011101010011110.101100001101)2  (b) Hexadecimal to decimal conversion:- For conversion of hexadecimal to decimal, multiply each digit in the hexadecimal number by its position weight and add all those product terms. For example: - Convert (A0F9.0EB)16 to decimal Solution: (A0F9.0EB)16 = (10 x 163 )+(0 x 162 )+(15 x 161 ) +( 9 x 160 ) +(0 x 16 – 1) +(14 x 16- 2) +(11 x 16-3) = 40960 + 0 + 240 + 9 + 0 +0.0546 + 0.0026 = (41209.0572)10 Result is (41209.0572)10 (c) Hexadecimal to Octal conversion:- For conversion of hexadecimal to octal, first convert the given hexadecimal number to binary and then binary number to octal. For example :- Convert (B9F.AE)16 to octal. Solution :- Given hexadecimal no.is B 9 F. A E Convert each hex. digit to binary = 1011 1001 1111. 1010 1110 Group of 3 bits are = 101 110 011 111. 101 011 100 Convert 3 bits group to octal. = 5 6 3 7. 5 3 4 Result is (5637.534)8 BINARY ARITHEMATIC OPERATION :- 1. BINARY ADDITION:- The binary addition rules are as follows 0+0=0; 0+1=1; 1+0=1; 1 + 1 = 10 , i.e 0 with a carry of 1 For example :- Add (100101)2 and (1101111)2. Solution :- 100101 + 1101111 10010100 Result is (10010100)2 2. BINARY SUBTRACTION:- The binary subtraction rules are as follows 0-0=0; 1-1=0; 1-0=1; 0 - 1 = 1 , with a borrow of 1 For example :- Substract (111.111)2 from (1010.01)2. Solution :- 1010.010 - 111.111 0 0 1 0.0 1 1 Result is (0010.011)2 3. BINARY MULTIPLICATION:- The binary multiplication rules are as follows 0x0=0; 1x1=1; 1x0=0; 0x1=0 For example :- Multiply (1101)2 by (110)2. Solution :- 1101 x 110 0000 1101 + 1101 1001110 Result is (1001110)2 4. BINARY DIVISION:- The binary division is very simple and similar to decimal number system. The division by ‘0’ is meaningless. So we have only 2 rules 0÷1=0 1÷1=1 For example :- Divide (10110)2 by (110)2. Solution :- 110 ) 101101 ( 111.1 - 110 1010 110 1001 110 110 110 000  Result is (111.1)2 1’s COMPLEMENT REPRESENTATION :- The 1’s complement of a binary number is obtained by changing each 0 to 1 and each 1 to 0. For example :- Find (1100)2 1’s complement. Solution :- Given 1 1 0 0 1’s complement is 0 0 1 1 Result is (0011)2 2’s COMPLEMENT REPRESENTATION :- The 2’s complement of a binary number is a binary number which is obtained by adding 1 to the 1’s complement of a number i.e. 2’s complement = 1’s complement + 1 For example :- Find (1010)2 2’s complement. Solution :- Given 1 0 1 0 1’s complement is 0 1 0 1 + 1 2’s complement 0 1 1 0 Result is (0110)2 SIGNED NUMBER :- In sign – magnitude form, additional bit called the sign bit is placed in front of the number. If the sign bit is 0, the number is positive. If it is a 1, the number is negative. For example:- 0 1 0 1 0 0 1 = +41 ↑ Sign bit 1 1 0 1 0 0 1 = -41 ↑ Sign bit DIGITAL CODES:- In practice the digital electronics requires to handle data which may be numeric, alphabets and specialcharacters. This requires the conversion of the incoming data into binary format before it can be processed. There is various possible ways of doing this and this process is called encoding. To achieve the reverse of it,we use decoders. WEIGHTED AND NON-WEIGHTED CODES:- There are two types of binary codes 1) Weighted binary codes 2) Non- weighted binary codes  In weighted codes, for each position ( or bit) ,there is specific weight attached. For example, in binary number, each bit is assigned particular weight 2n where ‘n’ is the bit number for n = 0,1,2,3,4 the weights are 1,2,4,8,16 respectively. Example :- BCD Non-weighted codes are codes which are not assigned with any weight to each digit position, i.e., each digit position within the number is not assigned fixed value. Example:- Excess – 3 (XS -3) code and Gray codes BINARY CODED DECIMAL (BCD):- BCD is a weighted code. In weighted codes, each successive digit from right to left represents weights equal to some specified value and to get the equivalent decimal number add the products of the weights by the corresponding binary digit. 8421 is the most common because 8421 BCD is the most natural amongst the other possible codes. EXCESS THREE(XS-3) CODE:- The Excess-3 code, also called XS-3, is a non- weighted BCD code. This derives it name from the fact that each binary code word is the corresponding 8421 code word plus 0011(3). It is a sequential code. It is a self complementing code. ASCII CODE:- The American Standard Code for Information Interchange (ASCII) pronounced as ‘ASKEE’ is widely used alphanumeric code. This is basically a 7 bit code. The number of different bit patterns that can be created with 7 bits is 27 = 128 , the ASCII can be used to encode both the uppercase and lowercase characters of the alphabet (52 symbols) and some special symbols in addition to the 10 decimal digits. It is used extensively for printers and terminals that interface with small computer systems. The table shown below shows the ASCII groups. The ASCII code LSBs MSBs 000 001 010 011 100 101 110 111 0000 NUL DEL Space 0 @ P P 0001 SOH DC1 ! 1 A Q a q 0010 STX DC2 “ 2 B R b r 0011 ETX DC3 # 3 C S c s 0100 EOT DC4 $ 4 D T d t 0101 ENQ NAK % 5 E U e u 0110 ACK SYN & 6 F V f v 0111 BEL ETB ‘ 7 G W g w 1000 BS CAN ( 8 H X h x 1001 HT EM ) 9 I Y i y 1010 LF SUB * : J Z j z 1011 VT ESC + ; K [ k { 1100 FF FS , < L \ l | 1101 CR GS - = M ] m } 1110 SO RS. > N ^ n ~ 1111 SI US / ? O _ o DLE EBCDIC CODE:- The Extended Binary Coded Decimal Interchange Code (EBCDIC) pronounced as ‘eb – si- dik’ is an 8 bit alphanumeric code. Since 28 = 256 bit patterns can be formed with 8 bits. It is used by most large computers to communicate in alphanumeric data. The table shown below shows the EBCDIC code. The EBCDIC code LSD MSD(Hex) (Hex) 0 1 2 3 4 5 6 7 8 9 A B C D E F 0 NUL DLE DS SP & [ ] \ 0 1 SOH DC1 SOS / a j ~ A J 1 2 STX DC2 FS SYN b k s B K S 2 3 ETX DC3 c l t C L T 3 4 PF RES BYP PN d m u D M U 4 5 HT NL LF RS e n v E N V 5 6 LC BS EOB YC f o w F O W 6 7 DEL IL PRE EOT g p x G P X 7 8 CAN h q y H Q Y 8 9 EM i r z I R Z 9 A SMM CC SM Ø ! I : B VT. $ , # C FF IFS DC4 < * % @ D CR IGS ENQ NAK ( ) _ ‘ E SO IRS ACK + ; > = F SI IUS BEL SUB I ‘ ? ‘  GRAY CODE:- The gray code is a non-weighted code. It is not a BCD code. It is cyclic code because successive words in this differ in one bit position only i.e it is a unit distance code. Gray code is used in instrumentation and data acquisition systems where linear or angular displacement is measured. They are also used in shaft encoders, I/O devices, A/D converters and other peripheral equipment. MORSE CODE :- Morse code is a character-encoding scheme that allows operators to send messages using a series of electrical pulses represented as short or long pulses, dots, and dashes. BAUDOT CODE :- The Baudot code [boˈdo] is an early character encoding for telegraphy invented by Émile Baudot in the 1870s. It was the predecessor to the International Telegraph Alphabet No. 2 (ITA2), the most common teleprinter code in use until the advent of ASCII.It uses 5 bits per character. Hollerith Code : Hollerith code A code for relating alphanumeric characters to holes in a punched card. It was devised by Herman Hollerith in 1888 and enabled the letters of the alphabet and the digits 0–9 to be encoded by a combination of punchings in 12 rows of a card. Unicode :- The Unicode Standard provides a unique number for every character, no matter what platform, device, application or language. It has been adopted by all modern software providers and now allows data to be transported through many different platforms, devices and applications without corruption. Support of Unicode forms the foundation for the representation of languages and symbols in all major operating systems, search engines, browsers, laptops, and smart phones—plus the Internet and World Wide Web (URLs, HTML, XML, CSS, JSON, etc.).