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Complement in DLD

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shahzad ali
shahzad ali

The document discusses different methods for performing subtraction using addition of complements in digital logic design. It explains that subtraction by borrowing is difficult for digital computers, so subtraction is instead implemented by taking the complement of the numbers being subtracted and then adding them. Two types of complements are described: r's complement and r-1's complement. Examples of performing subtraction using complements in binary and decimal numbers are provided.

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Complements Digital Logic Design BY SHEHZAD ALI
Subtraction using addition Conventional addition (using carry) is easily implemented in digital computers. However; subtraction by borrowing is difficult and inefficient for digital computers. Much more efficient to implement subtraction using ADDITION OF the COMPLEMENTS of numbers.
Types of Complement There are two types of complements: Rs complement e.g. 2s complement for base 2 and 10s complement for base 10 R-1s complement e.g. 1s complement for base 2 and 9s complement for base 10
Complements of Numbers (r-1 )s Complement Given a number N in base r having n digits, the (r- 1)s complement of N is defined as ( r n - 1 ) - N For decimal numbers the base or r = 10 and r- 1= 9, so the 9s complement of N is (10n-1)-N 99999. - N Digit n Digit n-1 Next digit Next digit First digit 9 9 9 9 9 -
2- Find the 9s complement of 546700 and 12389 The 9s complement of 546700 is 999999 - 546700= 453299 and the 9s complement of 12389 is 99999- 12389 = 87610. 9s complement Examples - 5 4 6 7 0 0 9 9 9 9 9 9 4 5 3 2 9 9 - 1 2 3 8 9 9 9 9 9 9 8 7 6 1 0
ls complement For binary numbers, r = 2 and r 1 = 1, r-1s complement is the ls complement. The ls complement of N is (2n - 1) - N. Digit n Digit n-1 Next digit Next digit First digit 1 1 1 1 1 Bit n-1 Bit n-2 . Bit 1 Bit 0 -
ls complement Find r-1 complement for binary number N with four binary digits. r-1 complement for binary means 2-1 complement or 1s complement. n = 4, we have 24 = (10000)2 and 24 - 1 = (1111)2. The ls complement of N is (24 - 1) - N. = (1111) - N
The complement 1s of 1011001 is 0100110 - - 1 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 0 0 1 1 0 The 1s complement of 0001111 is 1110000 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 ls complement
rs Complement Given a number N in base r having n digits, the rs complement of N is defined as rn - N. For decimal numbers the base or r = 10, so the 10s complement of N is 10n-N. 100000. - N Digit n Digit n-1 Next digit Next digit First digit- 1 0 0 0 0 0
10s complement Examples Find the 10s complement of 546700 and 12389 The 10s complement of 546700 is 1000000 - 546700= 453300 and the 10s complement of 12389 is 100000 - 12389 = 87611. Notice that it is the same as 9s complement + 1. - 5 4 6 7 0 0 4 5 3 3 0 0 - 1 2 3 8 9 1 0 0 0 0 0 8 7 6 1 1 1 0 0 0 0 0 0
For binary numbers, r = 2, rs complement is the 2s complement. The 2s complement of N is 2n - N. 2s complement Digit n Digit n-1 Next digit Next digit First digit- 1 0 0 0 0 0
2s complement Example The 2s complement of 1011001 is 0100111 The 2s complement of 0001111 is 1110001 - - 1 0 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
Fast Methods for 2s Complement Method 1: The 2s complement of binary number is obtained by adding 1 to the ls complement value. Example: 1s complement of 101100 is 010011 (invert the 0s and 1s) 2s complement of 101100 is 010011 + 1 = 010100
Fast Methods for 2s Complement Method 2: The 2s complement can be formed by leaving all least significant 0s and the first 1 unchanged, and then replacing ls by 0s and 0s by ls in all other higher significant bits. Example: The 2s complement of 1101100 is 0010100 Leave the two low-order 0s and the first 1 unchanged, and then replacing 1s by 0s and 0s by 1s in the four most significant bits.
Examples Finding the 2s complement of(01100101)2 Method 1 Simply complement each bit and then add 1 to the result. (01100101)2 [N] = 2s complement = 1s complement (10011010)2 +1 =(10011011)2 Method 2 Starting with the least significant bit, copy all the bits up to and including the first 1 bit and then complement the remaining bits. N = 0 1 1 0 0 1 0 1 [N] = 1 0 0 1 1 0 1 1
Example 1 (Decimal unsigned numbers) perform the subtraction 72532 - 13250 = 59282. M > N : Case 1 Do not take complement of sum and discard carry The 10s complement of 13250 is 86750. Therefore: M = 72532 10s complement of N =+86750 Sum= 159282 Discard end carry 105= - 100000 Answer = 59282 no complement
Example 2: Now consider an example with M <N. The subtraction 13250 - 72532 produces negative 59282. Using the procedure with complements, we have M = 13250 10s complement of N = +27468 Sum = 40718 Take 10s complement of Sum = 100000 -40718 The number is : 59282 Place negative sign in front of the number: -59282

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Complement in DLD

  • 2. Subtraction using addition Conventional addition (using carry) is easily implemented in digital computers. However; subtraction by borrowing is difficult and inefficient for digital computers. Much more efficient to implement subtraction using ADDITION OF the COMPLEMENTS of numbers.
  • 3. Types of Complement There are two types of complements: Rs complement e.g. 2s complement for base 2 and 10s complement for base 10 R-1s complement e.g. 1s complement for base 2 and 9s complement for base 10
  • 4. Complements of Numbers (r-1 )s Complement Given a number N in base r having n digits, the (r- 1)s complement of N is defined as ( r n - 1 ) - N For decimal numbers the base or r = 10 and r- 1= 9, so the 9s complement of N is (10n-1)-N 99999. - N Digit n Digit n-1 Next digit Next digit First digit 9 9 9 9 9 -
  • 5. 2- Find the 9s complement of 546700 and 12389 The 9s complement of 546700 is 999999 - 546700= 453299 and the 9s complement of 12389 is 99999- 12389 = 87610. 9s complement Examples - 5 4 6 7 0 0 9 9 9 9 9 9 4 5 3 2 9 9 - 1 2 3 8 9 9 9 9 9 9 8 7 6 1 0
  • 6. ls complement For binary numbers, r = 2 and r 1 = 1, r-1s complement is the ls complement. The ls complement of N is (2n - 1) - N. Digit n Digit n-1 Next digit Next digit First digit 1 1 1 1 1 Bit n-1 Bit n-2 . Bit 1 Bit 0 -
  • 7. ls complement Find r-1 complement for binary number N with four binary digits. r-1 complement for binary means 2-1 complement or 1s complement. n = 4, we have 24 = (10000)2 and 24 - 1 = (1111)2. The ls complement of N is (24 - 1) - N. = (1111) - N
  • 8. The complement 1s of 1011001 is 0100110 - - 1 0 1 1 0 0 1 1 1 1 1 1 1 1 0 1 0 0 1 1 0 The 1s complement of 0001111 is 1110000 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 ls complement
  • 9. rs Complement Given a number N in base r having n digits, the rs complement of N is defined as rn - N. For decimal numbers the base or r = 10, so the 10s complement of N is 10n-N. 100000. - N Digit n Digit n-1 Next digit Next digit First digit- 1 0 0 0 0 0
  • 10. 10s complement Examples Find the 10s complement of 546700 and 12389 The 10s complement of 546700 is 1000000 - 546700= 453300 and the 10s complement of 12389 is 100000 - 12389 = 87611. Notice that it is the same as 9s complement + 1. - 5 4 6 7 0 0 4 5 3 3 0 0 - 1 2 3 8 9 1 0 0 0 0 0 8 7 6 1 1 1 0 0 0 0 0 0
  • 11. For binary numbers, r = 2, rs complement is the 2s complement. The 2s complement of N is 2n - N. 2s complement Digit n Digit n-1 Next digit Next digit First digit- 1 0 0 0 0 0
  • 12. 2s complement Example The 2s complement of 1011001 is 0100111 The 2s complement of 0001111 is 1110001 - - 1 0 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 1 1 1 1 1 1 1 0 0 0 1 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
  • 13. Fast Methods for 2s Complement Method 1: The 2s complement of binary number is obtained by adding 1 to the ls complement value. Example: 1s complement of 101100 is 010011 (invert the 0s and 1s) 2s complement of 101100 is 010011 + 1 = 010100
  • 14. Fast Methods for 2s Complement Method 2: The 2s complement can be formed by leaving all least significant 0s and the first 1 unchanged, and then replacing ls by 0s and 0s by ls in all other higher significant bits. Example: The 2s complement of 1101100 is 0010100 Leave the two low-order 0s and the first 1 unchanged, and then replacing 1s by 0s and 0s by 1s in the four most significant bits.
  • 15. Examples Finding the 2s complement of(01100101)2 Method 1 Simply complement each bit and then add 1 to the result. (01100101)2 [N] = 2s complement = 1s complement (10011010)2 +1 =(10011011)2 Method 2 Starting with the least significant bit, copy all the bits up to and including the first 1 bit and then complement the remaining bits. N = 0 1 1 0 0 1 0 1 [N] = 1 0 0 1 1 0 1 1
  • 16. Example 1 (Decimal unsigned numbers) perform the subtraction 72532 - 13250 = 59282. M > N : Case 1 Do not take complement of sum and discard carry The 10s complement of 13250 is 86750. Therefore: M = 72532 10s complement of N =+86750 Sum= 159282 Discard end carry 105= - 100000 Answer = 59282 no complement
  • 17. Example 2: Now consider an example with M <N. The subtraction 13250 - 72532 produces negative 59282. Using the procedure with complements, we have M = 13250 10s complement of N = +27468 Sum = 40718 Take 10s complement of Sum = 100000 -40718 The number is : 59282 Place negative sign in front of the number: -59282