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Chapter 2

Chapter 2. Number Systems + Codes. Overview. Objective: To use positional number systems To convert decimals to binary integers To convert binary integers to decimals To represent binary integers in 3 forms: Sign magnitude 1’s complement 2’s complement

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Chapter 2

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  1. Chapter 2 Number Systems + Codes

  2. Overview Objective: To use positional number systems To convert decimals to binary integers To convert binary integers to decimals To represent binary integers in 3 forms: Sign magnitude 1’s complement 2’s complement To reprensent number in octal (base 8) and hexadecimal numbers and convert them to decimal and binary numbers To understand various computer codes

  3. Overview Two types of representations of information: Numeric information Positional number system Conversion between number systems Operation on number systems (add, substract …) Non numeric information Codes Properties of codes

  4. Representation of data in Computer or digital circuits Real life Data 000111000000 : string of 0’s and 1’s • Numeric data • (125.5 : a price • Nonnumeric data • (John : a name ) Overview Correspondance or Representation

  5. I. Representation of numerical information Positional Number Systems (PNS) • PNS – a number is represented by a string of digits. Each digit position is weighted by a power of the base or radix. • Example: Decimal PNS, in base 10 digit position i is weighted by 10i 937.5210=9100 + 310 + 71 + 5.1 + 2.01 + 2.001 + 0.0001 • General Form: Base or Radix = 10 D : dp-1dp-2 dp-3… d2 d1d0. d-1 d-2…d-n D = dp-1.10p-1 + dp-2.10p-2 +…+ d1.101 + d0.100 + d-1.10-1 + d-2.10-2 + … + d-n.10-n p-1 D =  di.10i i=-n di is a decimal digit

  6. I. Representation of numerical information Positional Number Systems (PNS) • General Form: Base or Radix = 2 D : bp-1bp-2 bp-3… b2 b1b0. b-1 b-2…b-n D = bp-1.2p-1 + bp-2.2p-2 +…+ b1.21 + b0.20 + b-1.2-1 + b-2.2-2 + … + b-n.2-n p-1 D = bi.2i i=-n bi is a binary digit

  7. I. Representation of numerical information Summary of PNS • Decimal is natural (we count in base 10) - (Radix or Base 10 ) digits  {0,1,…,8,9} • Binary is used in digital system - (Radix or Base 2 ) digits  {0,1} • Octal is used for representing multibits (group of 3 bits) numbers in digital systems. - (Radix or Base 8=23) digits  {0,1,…,6,7} • Hexadecimal is used for representing multibits (group of 4 bits) numbers in digital systems - (Radix or Base 16=24) digits  {0,1,…9,A,B,…,F}

  8. I. Representation of numerical information Summary of PNS (…) Dec Binary Oct Hex 0 0 0 (000) 0 (0000) 1 01 1 (001) 1 (0001) 2 10 2 (010) 2 (0010) 3 11 3 (011) 3 (0011) 4 100 4 (100) 4 (0100) 5 101 5 ( 101) 5 (0101) 6 110 6 (110) 6 (0110) 7 111 7 (111) 7 (0111) 8 1000 10 (001 000) 8 (1000) 9 1001 11 (001 001) 9 (1001) 10 1010 12 (001 010) A (1010) 11 1011 13 (001 011) B (1011) 12 1100 14 (001 100) C (1100) 13 1101 15 (001 101) D (1101) 14 1110 16 (001 110) E (1110) 15 1111 17 (001 111) F (1111) 16 10000 20 (010 000) 10 (0001 0000)

  9. I. Representation of numerical information ConversionsFrom any base r ---> Decimal Use base 10 arythmetic to expand : p-1 D = di.ri (r is the base and di are the digits) i=0 Two solutions can be used to do this : • Polynomial expansion D = dp-1.rp-1 + dp-2.rp-2 +…+ d1.r1 + d0.r0 Examples : 31016 = 3.162 + 1.161 + 0.160 = 3.256 + 16 + 0 = 78410 1CE816 = 1.163 + 12 .162 + 14. 161+ 8.160 = 740010 436.58= 4.82 + 3. 81 + 6 .80 + 5. 8-1 = 286.62510

  10. I. Representation of numerical information ConversionsFrom any base r ---> Decimal • Iterative multiplication (Expansion of D) D can be written as : ((….((dp-1).r+ dp-2).r+…+d2).r + d1).r+ d0 Examples : 31016 = ((3).16+ 1).16 + 0) = (49).16 + 0 = 78410 Good for programming

  11. I. Representation of numerical information ConversionsFrom Decimal ---> Any base r Successive divisions of D by r yield the digits from the least to the most significant bit. Remember that D can be written in base r as : D = dp-1.rp-1 + dp-2.rp-2 +…+ d1.r+ d0. = ((….((dp-1).r+ dp-2).r+…+d2).r + d1).r+ d0 D/r ----->quotient D1 = (….((dp-1).r+ dp-2).r+…+d2).r + d1 remainder d0 D1/r ----->quotient D2 = (….((dp-1).r+ dp-2).r+… d3).r+d2) remainder d1 D2/r ----->quotient D3 = (….((dp-1).r+ dp-2).r+…+d3) remainder d2 And so on …..

  12. I. Representation of numerical information ConversionsFrom Decimal ---> Any base r • Example : conversion of 179 from Base 10 -> Base 2 17910 179 1 89 1 LSB 2 44 1 4 22 0 8 11 0 16 5 1 32 2 1 64 1 0 128 0 1 MSB weight Successive divisions by 2 17910 = 101100112

  13. I. Representation of numerical information ConversionsFrom Decimal ---> Any base r • Example : conversion of 467 from Base 10 -> Base 8 467 3 1 58 3 LSB 16 8 7 2 448 64 0 7 MSB 467 weight successive divisions 46710 = 7238

  14. Other ConversionsFrom Binary ---> Octal/Hex I. Representation of numerical information • Conversion by substitution • Binary to Oct : Starting from rightmost bit, make groups of 3 bits and convert each group to base8 1000110011102 = 100/011/001/110 = 43168 4 / 3 / 1 / 6 • Binary to Hex : Starting from rightmost bit, make groups of 4 bits and convert each group to base 16 1000110011102 = 1000/1100/1110 = 8CE16 8 / C / E

  15. I. Representation of numerical information Other ConversionsFrom Octal/hex ---> Binary • Conversion by substitution • Octal to Binary : Convert each digit to a 3 bits binary number 5768= 101 111 1102 5 7 6 • Hex to Binary : Convert each digit to a 4 bits binary number 57616 = 0101 0111 01102 5 7 6

  16. I. Representation of numerical information Addition of binary numbers • Example 1 (Addition) Carry 1 01111000 X 190 10111110 + Y 141 10001101 331 1 01001011 Result has more bits than binary addends Carry equals 1 in the last bit position • Example 2 (Addition) Carry 0 01011000 X 173 10101101 + Y 44 00101100 217 11011001 Carry equal to 0 in the last bit position

  17. I. Representation of numerical information Addition of binary numbers • In general, given two binary number X = ( xn-1 … xi … x1 x0 )2 and Y = ( yn-1 … yi … y1 y0)2 the sum (X+Y) bit by bit at position i is given by (ci is the ith position’s carry) ci-1 Addition table + xici-1 xi yi si ci + yi0 0 0 0 0 ci si 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1

  18. Substraction of binary numbers I. Representation of numerical information • Substraction B 0 10 110 110 10 10 X 12129 1 1 1 0 0 1 0 1 - Y 46 0 0 1 0 1 1 1 0 183 1 0 1 1 0 1 1 1 B 0 10 1 0 110 10 0 10 X 2110 1 1 0 1 0 0 1 0 - Y 109 0 1 1 0 1 1 0 1 101 1 1 0 0 1 0 1

  19. I. Representation of numerical information Addition of hexadecimal numbers Addition of two hexadecimal numbers C 1 1 X 1 9 B 9 16 + Y C 7 E 6 16 E 1 9 F 16

  20. I. Representation of numerical information Representation of negative numbers Signed-magnitude systems • +99, -57, -13.3 Binary uses a (sign bit, the most significant bit (MSB)) MSB = 0 (Positive) MSB = 1 (Negative) 0 10101012 = + 8510 1 10101012 = -8510 0 11111112 = + 12710 1 11111112 = -12710 0 00000002 = + 010 1 00000002 = - 010 • Problem: Too much logic needed to - detect and compare sign bit - add or abstract magnitudes - determine the sign of the result

  21. I. Representation of numerical information Representation of negative numbers Complement number system • Difficult to change to complement • But adding or substracting two numbers are easier once they are represented in complement number systems

  22. I. Representation of numerical information Representation of negative numbers Radix – Complement representation • n digit number is substracted from rn • Example: if r = 10 and n = 4, rn = 104 =10,000 The 10’s complement of D=184910 is: (rn – D) = 8151 • Computing 10’s complement: rn – D = (rn - 1 – D) +1 (rn - 1) – D => complement of each digit with rspect to base (r-1) In base 10 complement each digit with respect to 9 and 1 to the result Example: The 10’s complement of 184910 is (104 - 1849)= (9999 – 1849) + 1 = 8 1 5 0 + 1 = 8151

  23. I. Representation of numerical information Representation of negative numbers Radix – Complement representation • Complement of each digit : r –1 – di In base 10 : 10 –1 – di = 9 – di Example : The complement of each digit of 345 is : For 5 : (10 – 1) – 5 = 9 – 5 = 4 For 4 : = 9 – 4 = 5 For 3 : = 9 – 3 = 6 • In base 3 the radix complement of 121 is 101 • In base 2 the radix complement of 1110111 is 0001000

  24. I. Representation of numerical information Representation of negative numbers Two’s Complement • The MSB is the sign bit • Example: 1710 = 000100012 -12710 = 100000012 11101110 01111110 +1 +1 111011112 = -1710 01111111 =12710 • The range is -(2n-1) -> (2n-1-1), For n=8 010 = 000000002 -12810 = 100000002 Note: The weight of the MSB is –2n-1 • The range of positive numbers is 0 to 127 • The range of negative numbers is –1 to -128

  25. I. Representation of numerical information Representation of negative numbers Two’s Complement • Two’s complement of 010 and -12810 010 = 000000002 -12810 = 100000002 11111111 01111111 +1 +1 1000000002 = 010 10000000 = -12810 Note: Ignore carry out of the MSB position Note: -128 does not have a positive counterpart, that is –128 is its own Complement : this can creat problems ……

  26. I. Representation of numerical information 2’s Complement Addition + Substraction +3 0011 -2 0010 -> 1101+1-> 1110 +4 +0100 + -6 0110 -> 1001+1-> + 1010 +7 0111 -8 11000 +6 0110 0110 + -3 0011 -> 1100+1 => 1101 +3 1 0011 = 3 Overflow in the last position Overflow in the last position Rule : Ignore any carry beyond the MSB. The result is correct if the range of the number system Is not exceeded.

  27. I. Representation of numerical information 2’s Complement Addition + Substraction Addition/Substraction rule : Assume that the signed integers are represented in 2s complement : To compute A – B, compute the 2’s complement B’ of B and add B’ to A Examples : 5 0101 5 0101 +2 0010 - 2 + 1110 ( = 2’) 7 0111 3 1 0011 (ignore the leftmost carry

  28. I. Representation of numerical information Overflow • Addition of 2 numbers with different signs can never overflow • Addition of 2 numbers with same sign -3 1101 +5 0101 + -6 1010 + +6 0110 -9 1 0111 = +7 +11 1011 = -5 wrong wrong If signs of (addends) are same and sign of sum is different

  29. I. Representation of numerical information Substraction rules (2’s complement) +4 0100 0100 - +3 -0011 1101 1 1 0001 - 3 1101 1101 - -4 1100 0011 1 1 0001 Same overflow rules apply

  30. I. Representation of numerical information Binary multiplication 11 1011  13 1101 33 1011 11 10000 143 11011 11011 10001111 uses a shift register and adder, and logic control • Signed multiplication +  + = + +  - = - -  - = +

  31. II. Representation of non numerical information Binary codes for decimal numbers Goal: Use binary strings (sequence of 0’s and 1’s) to represent decimal information Definition: • A code is a set of n-bit strings. Each n-bit string represrnts a different number. • A code word is a particular member of a code

  32. II. Representation of non numerical information Binary codes for decimal numbers • A code is a set of n-bit strings. Each n-bit string represrnts a different number. • A code word is a particular member of a code • If n bit is used to represent each code words, the total number of possible code words is 2n Examples: To encode a set consisting of 3 elements {A,B,C} • we need 2 bits to represent each code word. • The possible code words are {00, 01, 10, 11}

  33. II. Representation of non numerical information Binary codes for decimal numbers • We need to assign a code word to each element of the original set. Assignment 1 : 00 -----> A 01 -----> B 10 -----> C Code 11 is unused Another assignment : 10 -----> A 01 -----> B 11 -----> C Code 00 is unused

  34. II. Representation of non numerical information Binary codes for decimal numbers • To encode the 10 decimal digits {0,1,….,8,9} We need 4 bit binary code words. There are 16=24 possible code words. So 6 codes are unused. Assignment 1 : 0 -----> 0000 5 -----> 0101 • -----> 0001 6 -----> 0110 • -----> 0010 7 -----> 0111 • -----> 0011 8 -----> 1000 • -----> 0100 9 -----> 1001 The following codes are unused: 1010, 1011, 1100, 1101, 1110, 1111

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