1. DIGITAL SYSTEMS TCE1111 Representation and Arithmetic Operations with Signed Numbers Week 6 and 7 (Lecture 1 of 2)

2. DIGITAL SYSTEMS TCE1111 2 What we are going to discuss? How to find 1’s and 2’s complement of a given binary number? How to represent negative integers using three ways – sign-magnitude,1’s complement and 2’s complement representation Arithmetic operations with 2’s complement numbers – Addition, Subtraction, Multiplication, Division Overflow Problem and Detection in the addition of signed numbers in 2’ s complement representation.

3. DIGITAL SYSTEMS TCE1111 3 Finding 1’s complement Change all the 1s to 0s and all the 0s to 1s Binary number 0001100100101 1’s complement 1110011011010

4. DIGITAL SYSTEMS TCE1111 4 Example of inverters used to obtain the 1’s complement of a binary number.

5. DIGITAL SYSTEMS TCE1111 5 Finding 2’s complement 2’s complement = 1’s complement +1 Binary number 111000101010001 1’s complement 000111010101110 Add 1 2’s complement 000111010101111

6. DIGITAL SYSTEMS TCE1111 6 Example of obtaining the 2’s complement of a negative binary number.

7. DIGITAL SYSTEMS TCE1111 7 Representation of Signed Integers in Binary – Three forms Sign bit – a 0 is for positive a 1 is for negative Sign-Magnitude form – a negative number has the same magnitude as the corresponding positive number but the sign bit is a 1 rather than a zero Example: +43 in 8-bit S-M form = 0 0101011 -43 in 8-bit S-M form = 1 0101011 1’s complement form- a negative number is the 1’s complement of the corresponding positive number Example: +43 in 8-bit 1’s complement form = 00101011 -43 in 8-bit 1’s complement form = 11010100 2’s complement form- a negative number is the 2’s complement of the corresponding positive number Example: +43 in 8-bit 2’s complement form = 00101011 -43 in 8-bit 2’s complement form = 11010101

8. DIGITAL SYSTEMS TCE1111 8 Determination of decimal value of signed numbers (1)… Sign-magnitude form – convert all but the MSB to decimal and call the decimal value negative if the MSB is 1 Example: 0 1010011 = 64+16+2+1 = +83 since MSB is 0 1 1010011 = 64+16+2+1 = -83 since MSB is 1 1’s complement form – if the bit sign is 0 sum the weighted values of the the binary number, if the bit sign is 1 the sum the weighted value of the binary number assigning the MSB a negative weight and add 1 Example: 01010011 = 64+16+2+1 = + 83 since MSB is 0 10101100 = -128+32+8+4= - 84 +1 since MSB is 1 = - 83

9. DIGITAL SYSTEMS TCE1111 9 Determination of decimal value of signed numbers (2) 2’s complement form - if the bit sign is 0 sum the weighted values of the binary number, if the bit sign is 1 the sum the weighted value of the binary number assigning the MSB a negative weight Example: 01010011 = 64+16+2+1 = + 83 since MSB is 0 10101101 = -128+32+8+4+1= - 83 since MSB is 1

10. DIGITAL SYSTEMS TCE1111 10 Table of 8 bit Two’s Complement numbers 127 01111111 ………. 500000101 400000100 300000011 200000010 100000001 000000000 -111111111 -211111110 -311111101 -411111010 -511111011 ……… -12810000000

11. DIGITAL SYSTEMS TCE1111 11 Range of 2’s complement representation Range of values for n-bit numbers for 2’s complement signed number representation - (2 n-1 ) to +(2 n-1 – 1) Example: For 4 bit 2’s complement representation, range = -8 to +7 For 8 bit 2’s complement representation, range = -128 to +127 For 16 bit 2’s complement representation, range = -32,768 to +32,767

12. DIGITAL SYSTEMS TCE1111 12 Arithmetic operations with 2’s complement numbers – Addition Add the two numbers and discard any final carry bit. Example: Add (34) + (-46) in 8-bit 2’s complement arithmetic 34 = 00100010 -46 = 11010010 -------------- -12= 11110100 ----------------

13. DIGITAL SYSTEMS TCE1111 13 Overflow problem and detection (1)… Problem: When two numbers are added and the number of bits required to represent the sum exceeds the number of bits in the two numbers, an overflow occurs. When it can occur? Overflow may occur only when both numbers are positive or both numbers are negative.

14. DIGITAL SYSTEMS TCE1111 14 Overflow problem and detection (2) Detection of Overflow: –When two operands are having the same sign and the result is having different sign, then there is overflow –If carry into the sign bit is different from carry out of the sign bit, then there is overflow –Example: 01101110 01011011 ------------- 11001001 -------------- Operands sign = +ve (0) Result sign = -ve (1), hence overflow occurs. (or) carry into the sign bit = 1, carry out of sign bit =0, hence overflow occurs.

15. DIGITAL SYSTEMS TCE1111 15 Arithmetic operations with 2’s complement numbers – Subtraction-Example (27-12) 2700011011 Two’s Comp00011011 1200001100 Comp11110011 Add 1 1 Two’s comp11110100-12 ADD00011011 11110100 Carry Disregarded 10000111115 Subtraction is addition with the sign of the subtrahend changed Example : (27-12) = 27 + -12

16. DIGITAL SYSTEMS TCE1111 16 Multiplication Two methods: Direct addition and Partial products The sign of the product of a multiplication depends on the signs of the multiplicand and the multiplier according to the following two rules: 1.If the signs are the same, the product is positive 2.If the signs are different, the product is negative

17. DIGITAL SYSTEMS TCE1111 17 Multiplication Direct addition method - add the multiplicand a number of times equals to the multiplier. Disadvantage is this approach becomes very lengthy if the multiplier is a large number. Example:

18. DIGITAL SYSTEMS TCE1111 18 Multiplication Procedure: –Step 1: Determine if the signs of the multiplicand and multiplier are the same or different to determine the sign of the product. (If the signs are the same, the product is positive. If the signs are different, the product is negative) -Step 2: Change any negative number to true (uncomplemented) form by finding 2’s complement -Step 3: Starting with least significant multiplier bit, generate the partial products. When the multiplier bit is 1, the partial product is the same as the multiplicand. When the multiplier bit is 0, the partial product is zero. Shift each successive partial product one bit to the left. -Step 4: Add each successive partial product to the sum of the previous partial products to get the final product. -Step 5: If the sign bit that was determined in step 1 is negative, take the 2’s complement of the product. If positive, leave the product in true form. Attach the sign bit to the product. Partial Product method

19. DIGITAL SYSTEMS TCE1111 19 Multiplication – example

20. DIGITAL SYSTEMS TCE1111 20 Division The numbers in a division are the dividend, the divisor and the quotient. The division operation in computers is accomplished using subtraction. Example: The sign of the quotient depends on the signs of the dividend and the divisor according to the two rules: 1.If the signs are the same, the quotient is positive 2.If the signs are different, the quotient is negative 21 - 7 14 - 7 7 - 7 0

21. DIGITAL SYSTEMS TCE1111 21 Division –Step 1: Determine if the signs of the dividend and divisor are the same or different. This determines the sign of the quotient. (If the signs are the same, the quotient is positive. If the signs are different, the quotient is negative). Initialize the quotient to zero. –Step 2: Subtract the divisor from the dividend using 2’s complement addition to get the first partial remainder and add 1 to the quotient. If this partial remainder is positive, go to step 3. If the partial remainder is zero or negative, the division is complete. –Step 3: Subtract the divisor from the partial remainder and add 1 to the quotient. If the result is positive, repeat for next partial remainder. If the result is zero or negative, the division is complete. –Step 4: Continue to subtract the divisor from the dividend and the partial remainders until there is a zero or negative result. Quotient is the number of times that the divisor is subtracted. When tow binary numbers are dividend, both numbers must be in true ( uncomplemented) form. The basic steps in division process are as follows:

22. DIGITAL SYSTEMS TCE1111 22 Division (1)… Example: Divide 01100100 by 00011001 Solution: Step 1: The signs of both numbers are positive, so the quotient will be positive. The quotient is initially zero: 00000000 Step 2: Subtract the divisor from the dividend using 2’s complement addition by discarding carry if any. 01100100 Dividend + 11100111 2’s complement of divisor ------------------- 01001011 Positive 1 st partial remainder -------------------- Add 1 to the Quotient: 00000000 + 00000001 = 00000001

23. DIGITAL SYSTEMS TCE1111 23 Division (3)… Step 3: Subtract the divisor from the 1 st partial remainder using 2’s complement addition 01001011 1 st partial remainder + 11100111 2’s complement of divisor ------------- 00110010 Positive 2 nd partial remainder -------------- Add 1 to the quotient: 00000001 + 00000001 = 00000010. Step 4: Subtract the divisor from the 2 nd partial remainder using 2’s complement addition. 00110010 2 nd partial remainder + 11100111 2’s complement of divisor --------------- 00011001 Positive 3 rd partial remainder Add 1 to the quotient: 00000010 + 00000001 = 00000011.

24. DIGITAL SYSTEMS TCE1111 24 Division (4) Step 5: Subtract the divisor from the 3 rd partial remainder using 2’s complement addition. 00011001 3 rd partial remainder + 11100111 2’s complement of divisor ------------- 00000000 Zero remainder -------------- Add 1 to the quotient: 00000011 + 00000001 = 00000100 (final quotient). The process is complete.