1. CPE 201 Digital Design Lecture 3: Digital Systems & Binary Numbers (3)

2. 2 Lecture Outline Complement systems Arithmetic operations with binary numbers

3. 3 Complements Goal: logicalmanipulation –simplify subtraction operation and logical manipulation Complement systems negate a number by taking its complement Two types of complements for each base-r system –Radix complement: r’s complement –Diminished radix complement: (r-1)’s complement Big advantage: allows us to perform subtraction using addition –Thus, only need adder component, no need for separate subtractor component!

4. 4 Diminished Radix Complement The (r-1)’s complement of an n-digit number N = r n – 1 – N Decimal –r = 10  9’s complement of N is 10 n – 1 – N –10 n : a “1” followed by n “0”s –10 n – 1: a number given by n “9”s The 9’s complement of a digital number is obtained by subtracting each digit from 9

5. 5 Examples: 9’s Complement E.g.: 9’s complement of 358433: 999999 – 358433 = 641566 850843:999999 – 850843 = 149156 573852:999999 - 573852 = 426147

6. 6 Diminished Radix Complement – Radix 2 Binary –r = 2  1’s complement of N is 2 n – 1 – N –2 n : a “1” followed by n “0”s –2 n – 1: a number given by n “1”s The 1’s complement of a binary number is obtained by subtracting each digit from 1 Observation: 1-0 = 11-1 = 0 –Bit changes! The 1’s complement of a binary number is obtained by flipping the bits of the number

7. 7 Examples: 1’s complement E.g.: 1’s complement of 11100101: 10010111: 00010001: 10011100: 01110111: 00000000: 00011010 01101000 11101110 01100011 10001000 11111111

8. 8 Radix Complement The r’s complement of an n-digit number N = r n – N This means just adding 1 to the (r-1)’s complement! Decimal –Leave all LSB zeros unchanged –Subtract the first nonzero LSB from 10, subtract the rest from 9 Binary –Leave all LSB zeros and the first “1” unchanged –Flip the rest of the bits

9. 9 Examples: 10’s Complement With addition: (1289, 2389, 5840, 3048) 1289: (9999-1289) + 1 = 8710 + 1 = 8711 2389: (9999-2389) + 1 = 7610 + 1 = 7611 5840: (9999-5840) + 1 = 4159 + 1 = 4160 3048:(9999-3048) + 1 = 6951 + 1 = 6952 With alternative rule 1289:8711 2389: 7611 5840: 4160 3048:6952

10. 10 Examples: 2’s Complement With addition: ( 00010001, 10011101, 01110111, 00000000 ) 00010001: 11101110 + 1 = 11101111 10011101:01100010 + 1 = 01100011 01110111:10001000 + 1 = 10001001 00000000: 11111111 + 1 = 1 00000000 With alternative rule: (1101100, 0110111) 1101100:0010100 0110111:1001001 Carry ignored

11. 11 Subtraction with r’s Complements Subtracting a number-done by adding its complement –Adding its complement results in answer exactly 10 more –Drop the 1  results in subtracting using addition only - - Exemplified with base 10 and 1-digit numbers

12. 12 Subtraction with r’s Complements Compute M – N M – N = M + r’s complement (N) = M + r n – N = (M – N) + r n M>N: = (M – N) + r n M<N: = r n – (N – M) Carry to next digit: IGNORED! The r’s complement of the desired result with negative sign! Take the complement and add a negative sign.

13. 13 Subtraction with r’s Complements To compute M – N: –Take the r’s complement of N –Add that value to M –Two cases can occur: A carry to next digit is obtained (M was greater than N)  ignore the carry to get the result No carry is obtained (M was smaller than N)  take the r’s complement of the resulted value and add a negative sign

14. 14 Subtraction Using 10’s Complement 72532 – 3250 = 72532 + 96750  Result = 69282 3250 – 72532 = 03250 + 27468  Res. = -10’s complement(30718) = - 69282 Discard the carry No carry 692821 30718 Make sure you fill out to the maximum available number of digits! 0

15. 15 Subtraction Using 2’s Complement X = 1010100Y = 1000011 X – Y = ?Y – X = ? X – Y:Y – X: 1010100+1000011+ 01111010101100 100100011101111 Res. = -2’s comp. (1101111) = -0010001 Discard the carry No carry

16. 16 Subtraction with (r-1)’s Complements Compute M – N M – N = M + (r-1)’s complement (N) = M + r n – N – 1 = (M – N – 1) + r n M>N: = (M – N – 1) + r n M<N: = r n – 1 – (N – M) Carry to next digit: IGNORED! Got 1 less than result  add 1 The (r-1)’s complement of the desired result with negative sign! Take the complement and add a negative sign.

17. 17 Subtraction with (r-1)’s Complements To compute M – N: –Take the (r-1)’s complement of N –Add that value to M –Two cases can occur: A carry to next digit is obtained (M was greater than N)  ignore the carry and add 1 to get the result No carry is obtained (M was smaller than N)  take the (r-1)’s complement of the resulted value and add a negative sign

18. 18 Subtraction Using 1’s Complement X = 1010100Y = 1000011 X – Y = ?Y – X = ? X – Y:Y – X: 1010100+1000011+ 01111000101011 10010000+1101110 1 10010001 -1’s complement of 1101110 = -0010001 Discard the carry No carry

19. 19 Readings Chapter 1: –Sections 1.6, 1.7