FYBSC IT Digital Electronics Unit I Chapter II Number System and Binary Arithmetic
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The document covers various number systems, including binary, octal, and hexadecimal, alongside binary arithmetic operations such as addition, subtraction, multiplication, and division. It explains different methods for representing negative numbers, including sign-magnitude, 1's complement, and 2's complement, and details how to perform operations using these methods. Furthermore, it describes Binary Coded Decimal (BCD) and Excess-3 arithmetic, demonstrating how to convert between these formats and perform arithmetic operations.
FYBSC IT Digital Electronics Unit I Chapter II Number System and Binary Arithmetic
- 1. NUMBER SYSTEM & BINARY ARITHMETIC UNIT I DIGITAL ELECTRONICS PROF.ARTI GAVAS-PARAB ANNA LEELA COLLEGE OF COMMERCE AND ECONOMICS, SHOBHA JAYARAM SHETTY COLLEGE FOR BMS CHAPTER I1
- 2. UNIT I: CONTENTS Number System: Analog System, digital system, Numbering system, binary number system, octal number system, hexadecimal number system, conversion from one number system to another, floating point numbers, Weighted codes binary coded decimal, non-weighted codes Excess – 3 code, Gray code, Alphanumeric codes – ASCII Code, EBCDIC, ISCII Code, Hollerith Code, Morse Code,Teletypewriter (TTY), Error detection and correction, Universal Product Code, Code conversion. Binary Arithmetic: Binary addition, Binary subtraction, Negative number representation, Subtraction using 1’s complement and 2’s complement, Binary multiplication and division,Arithmetic in octal number system, Arithmetic in hexadecimal number system, BCD and Excess – 3 arithmetic. C H A P T E R I C H A P T E R II
- 3. BINARY ADDITION Binary arithmetic is essential part of all the digital computers and many other digital system. Binary Addition is a key for binary subtraction, multiplication,division.There are four rules of binary addition. In fourth case, a binary addition is creating a sum of (1 + 1 = 10) i.e. 0 is written in the given column and a carry of 1 over to the next column. Example − Addition
- 4. BINARY SUBTRACTION Subtraction and Borrow, these two words will be used very frequently for the binary subtraction.There are four rules of binary subtraction. Example − Subtraction
- 5. BINARY MULTIPLICATION Binary multiplication is similar to decimal multiplication.It is simpler than decimal multiplication because only 0s and 1s are involved. There are four rules of binary multiplication. Example − Multiplication
- 6. BINARY DIVISION Binary division is similar to decimal division. It is called as the long division procedure. Example − Division
- 7. NEGATIVE NUMBER REPRESENTATION Negative numbers can be distinguishable with the help of extra bit or flag called sign bit or sign flag in Binary number representation system for signed numbers. It is not possible to add minus or plus symbol in front of a binary number because a binary number can have only two symbol either 0 or 1 for each position or bit. That’s why we use this extra bit called sign bit or sign flag.The value of sign bit is 1 for negative binary numbers and 0 for positive numbers. There are three possible ways: Sign-Magnitude method, 1’s Complement method,and 2’s complement method.
- 8. SIGNED MAGNITUDE METHOD In this method, number is divided into two parts: Sign bit and Magnitude. If the number is positive then sign bit will be 0 and if number is negative then sign bit will be 1. Magnitude is represented with the binary form of the number to be represented. Example:Let we are using 5 bits register.The representation of -5 to +5 will be as follows: Range of Numbers: For k bits register, MSB will be sign bit and (k-1) bits will be magnitude. Positive largest number that can be stored is (2(k-1)-1) and negative lowest number that can be stored is -(2(k-1)-1). Note that drawback of this system is that 0 has two different representation one is -0 (e.g., 1 0000 in five bit register) and second is +0 (e.g., 0 0000 in five bit register).
- 9. 1’S COMPLEMENT METHOD Positive numbers are represented in the same way as they are represented in sign magnitude method. If the number is negative then it is represented using 1’s complement.First represent the number with positive sign and then take 1’s complement of that number. Example:Let we are using 5 bits register.The representation of -5 and +5 will be as follows: +5 is represented as it is represented in sign magnitude method.-5 is represented using the following steps: (i) +5 = 0 0101 (ii)Take 1’s complement of 0 0101 and that is 1 1010.MSB is 1 which indicates that number is negative. MSB is always 1 in case of negative numbers. Range of Numbers:For k bits register,positive largest number that can be stored is (2(k-1)-1) and negative lowest number that can be stored is -(2(k-1)-1). Note that drawback of this system is that 0 has two different representation one is -0 (e.g.,1 1111 in five bit register) and second is +0 (e.g.,0 0000 in five bit register).
- 10. 2’S COMPLEMENT METHOD 2’s Complement Method: Positive numbers are represented in the same way as they are represented in sign magnitude method. If the number is negative then it is represented using 2’s complement. First represent the number with positive sign and then take 2’s complement of that number. Example: Let we are using 5 bits registers.The representation of -5 and +5 will be as follows: +5 is represented as it is represented in sign magnitude method. -5 is represented using the following steps: (i) +5 = 0 0101 (ii)Take 2’s complement of 0 0101 and that is 1 1011.MSB is 1 which indicates that number is negative. MSB is always 1 in case of negative numbers. Range of Numbers: For k bits register, positive largest number that can be stored is (2(k-1)-1) and negative lowest number that can be stored is -(2(k-1)). The advantage of this system is that 0 has only one representation for -0 and +0. Zero (0) is considered as always positive (sign bit is 0) in 2’s complement representation.Therefore,it is unique or unambiguous representation. 2’s Complement Step 1:Write the absolute value of the given number in binary form. Prefix this number with 0 indicate that it is positive. Step 2:Take the complement of each bit by changing zeroes to ones and ones to zero. Step 3:Add 1 to your result.
- 11. SUBTRACTION USING 1’S COMPLEMENT The steps to be followed in subtraction by 1’s complement are: i)To write down 1’s complement of the subtrahend. ii)To add this with the minuend. iii) If the result of addition has a carry over then it is dropped and an 1 is added in the last bit. iv) If there is no carry over, then 1’s complement of the result of addition is obtained to get the final result and it is negative. Evaluate: (i) 110101 – 100101 Solution: 1’s complement of 10011 is 011010. Hence Minued - 1 1 0 1 0 1 1’s complement of subtrahend - +0 1 1 0 1 0 _________ Carry over - 1 0 0 1 1 1 1 Add 1Becouse carry is 1 + 1 Final Result - 0 1 0 0 0 0 The required difference is 10000
- 12. SUBTRACTION USING 1’S COMPLEMENT: EXAMPLES 1’s complement of 111001 is 000110. Hence Minued - 1 0 1 0 1 1 1’s complement of subtrahend - + 0 0 0 1 1 0 No Carry over - 1 1 0 0 0 1 Take complement of addition 0 0 1 1 1 0 Final Result - - 1 1 1 0 (ii) 101011 – 111001 (iii) 1011.001 – 110.10 1’s complement of 0110.100 is 1001.011 Hence Minued - 1 0 1 1 . 0 0 1 1’s complement of subtrahend - + 1 0 0 1 . 0 1 1 Carry over - 1 0 1 0 0 . 1 0 0 Add 1Becouse carry is 1 + 1 Final Result - 0 1 0 0 . 1 0 1 (iv) 10110.01 – 11010.10 1’s complement of 11010.10 is 00101.01 Minued - 1 0 1 1 0 . 0 1 1’s complement of subtrahend - + 0 0 1 0 1 . 0 1 No Carry over - 1 1 0 1 1 . 1 0 Final Result - – 100.01 Rules: 1. To write down 1’s complement of the subtrahend. 2. To add this with the minuend. 3. If the result of addition has a carry over then it is dropped and an 1 is added in the last bit. 4. If there is no carry over, then 1’s complement of the result of addition is obtained to get the final result and it is negative.
- 13. SUBTRACTION USING 2’S COMPLEMENT The operation is carried out by means of the following steps: (i) At first, 2’s complement of the subtrahend is found. (ii) Then it is added to the minuend. (iii) If the final carry over of the sum is 1, it is dropped and the result is positive. (iv) If there is no carry over, the two’s complement of the sum will be the result and it is negative.
- 14. SUBTRACTION USING 2’S COMPLEMENT: EXAMPLES Rules: 1. At first, 2’s complement of the subtrahend is found. 2. Then it is added to the minuend. 3. If the final carry over of the sum is 1, it is dropped and the result is positive. 4. If there is no carry over, the two’s complement of the sum will be the result and it is negative.
- 15. SUBTRACTION USING 2’S COMPLEMENT: EXAMPLES Rules: 1. At first, 2’s complement of the subtrahend is found. 2. Then it is added to the minuend. 3. If the final carry over of the sum is 1, it is dropped and the result is positive. 4. If there is no carry over, the two’s complement of the sum will be the result and it is negative.
- 16. SUBTRACTION USING 2’S COMPLEMENT: EXAMPLES Rules: 1. At first, 2’s complement of the subtrahend is found. 2. Then it is added to the minuend. 3. If the final carry over of the sum is 1, it is dropped and the result is positive. 4. If there is no carry over, the two’s complement of the sum will be the result and it is negative.
- 17. SUBTRACTION OF LARGER FROM SMALLER USING 2’S COMPLEMENT
- 18. ARITHMETIC IN OCTAL NUMBER SYSTEM: OCTAL ADDITION Following are the characteristics of an octal number system: Uses eight digits, 0,1,2,3,4,5,6,7. Also called base 8 number system. Each position in an octal number represents a 0 power of the base (8). Example: 80 Last position in an octal number represents an x power of the base (8). Example: 8x where x represents the last position - 1. Following octal addition table will help you to handle octal addition. To use this table, simply follow the directions used in this example: Add 68 and 58. Locate 6 in the A column then locate the 5 in the B column. The point in 'sum' area where these two columns intersect is the 'sum' of two numbers. 68 + 58 = 138.
- 20. ARITHMETIC IN OCTAL NUMBER SYSTEM: OCTAL SUBTRACTION The subtraction of octal numbers follows the same rules as the subtraction of numbers in any other number system.The only variation is in borrowed number. In the decimal system, you borrow a group of 1010. In the binary system, you borrow a group of 210. In the octal system you borrow a group of 810.
- 21. ARITHMETIC IN HEXADECIMAL NUMBER SYSTEM:ADDITION Following are the characteristics of a hexadecimal number system. Uses 10 digits and 6 letters, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Letters represents numbers starting from 10.A = 10,B = 11, C = 12, D = 13, E = 14, F = 15. Also called base 16 number system. Each position in a hexadecimal number represents a 0 power of the base (16).Example − 160 Last position in a hexadecimal number represents an x power of the base (16).Example − 16x where x represents the last position - 1. Following Hexadecimal addition table will help you greatly to handle Hexadecimal addition. To use this table, simply follow the directions used in this example − Add A16 and 516. Locate A in the X column then locate the 5 in theY column. The point in 'sum' area where these two columns intersect is the sum of two numbers. A16 + 516 = F16.
- 23. ARITHMETIC IN HEXADECIMAL NUMBER SYSTEM: SUBTRACTION The subtraction of hexadecimal numbers follow the same rules as the subtraction of numbers in any other number system. The only variation is in borrowed number. In the decimal system, you borrow a group of 1010. In the binary system, you borrow a group of 210. In the hexadecimal system you borrow a group of 1610. 3F57A – C85E = 32D1C
- 24. BCD AND EXCESS – 3 ARITHMETIC Excess 3 code is a basically a binary code which is made by adding 3 with the equivalent decimal of a binary number and again converting it into binary number. BCD addition and subtraction can be done by this method. Step 1:We have to convert the BCD numbers (which are to be added) into excess 3 forms by adding 0011 with each of the four bit groups them or simply increasing them by 3. Step 2: Now the two numbers are added using the basic laws of binary addition, there is no exception for this method. Step 3: Now which of the four groups have produced a carry we have to add 0011 with them and subtract 0011 from the groups which have not produced a carry during the addition. Step 4:The result which we have obtained after this operation is in Excess 3 form and this is our desired result.
- 25. BCD ADDITION USING EXCESS 3 CODE Example: BCD numbers 356 & 579 to be added using excess 3 code. Step 1: Convert each number into excess code by adding 0011 to each subgroup. 356 → 0011 0101 0110Adding 0011, Excess3 number of 356 is 0110 1000 1001 579 → 0101 0111 1001Add 0011, Excess 3 number of 579 is 1000 1010 1100. Step 2: Simply do binary addition. Carry is produced in second & third subgroups. Step 3: Add 0011 when there is carry. Subtract 0011 when there is no carry. Considering third group where there is no carry.1111 - 0011=1100. Considering other 2 sub groups 0011 + 0011= 0110 0101+0011= 1000. Answer in excess3 is 1100 0110 1000. Step 4:To convert it back to BCD form,subtract 0011 from each subgroup & express in BCD. 1100–0011=1001 -> 9 0110–0011=0011 ->3 1000- 0011= 0101 -> 5 BCD answer is 935. (Verify 356+ 579 = 935). 0110 1000 1001 +1000 1010 1100 ______________ 1111 0011 0101
- 26. BCD SUBTRACTION USING EXCESS 3 CODE Step 1: convert both the number in Excess-3 by adding 0011. Step 2: Follow Normal Binary subtraction. Step 3: Subtract 0011 from each BCD 4- bit binary group in the answer if the subtraction operation of the relevant 4- bit groups required a borrow from the next higher adjacent 4-bit group. Step 4:Add 0011 to the remaining 4-bit groups, if any in the result. Finally, we get the result on excess-3 code. Convert back into BCD. Example: 0001 1000 0101 – 0000 0000 0001 Solution: Step 1: Convert above numbers in XS-3 0100 1011 1000 and 0011 0011 1011. Step 2: Perform Binary Subtraction 0100 1011 1000 - 0011 0011 1011 _____________ 0001 0111 1101 Step 3:The last 4-bit group needed a borrow so we need to subtract 0011 from result of this group.And other two 4-bit groups didn’t require borrow so we need to add 0011 to each of them. 0001 + 0011= 0100 0111 + 0011= 1010 1101 – 0011= 1010 Step 4:To convert it back to BCD form,subtract 0011 from each subgroup & express in BCD. SoThe result will be 0001 0111 0111
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