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

2 Lecture Outline Complement systems Arithmetic operations with binary numbers

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 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 Examples: 9’s Complement E.g.: 9’s complement of : – = : – = : =

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 Examples: 1’s complement E.g.: 1’s complement of : : : : : :

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 Examples: 10’s Complement With addition: (1289, 2389, 5840, 3048) 1289: ( ) + 1 = = : ( ) + 1 = = : ( ) + 1 = = :( ) + 1 = = 6952 With alternative rule 1289: : : :6952

10 Examples: 2’s Complement With addition: ( , , , ) : = : = : = : = With alternative rule: ( , ) : : Carry ignored

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

15 Subtraction Using 2’s Complement X = Y = X – Y = ?Y – X = ? X – Y:Y – X: Res. = -2’s comp. ( ) = Discard the carry No carry

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 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 Subtraction Using 1’s Complement X = Y = X – Y = ?Y – X = ? X – Y:Y – X: ’s complement of = Discard the carry No carry

19 Readings Chapter 1: –Sections 1.6, 1.7