chapter 2 number systems + codes. overview objective: to use positional number systems to convert...

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

•To reprensent number in octal (base 8) and hexadecimal numbers and convert them to decimal and binary numbers

•To understand various computer codes

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

Overview

Real life Data

•Numeric data

(125.5 : a price

•Nonnumeric data

(John : a name )

Representation of data in

Computer or digital circuits

000111000000 : string of 0’s and 1’s

Correspondance or Representation

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=9100 + 310 + 71 + 5.1 + 2.01 + 2.001 + 0.0001

• General Form: Base or Radix = 10D : 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

I. Representation of numerical information

di is a decimal digit

Positional Number Systems (PNS)

• General Form: Base or Radix = 2D : 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


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}


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)


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


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


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 d2And so on …..



• Example : conversion of 179 from Base 10 -> Base 217910 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 MSBweight Successive divisions by 2

17910 = 101100112



• 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 divisions46710 = 7238


Other ConversionsFrom Binary ---> Octal/Hex

• 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


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


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


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

+ xi ci-1 xi yi si ci

+ yi 0 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


Substraction of binary numbers

• Substraction B 0 10 1

10 1

10

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 1

0 1

0 11

0 10 0 10

X 2110 1 1 0 1 0 0 1 0 - Y 1 09 0 1 1 0 1 1 0 1 1 01 1 1 0 0 1 0 1


Addition of hexadecimal numbers

C 1 1

X 1 9 B 9 16

+ Y C 7 E 6 16

E 1 9 F 16


Addition of two hexadecimal numbers

Representation of negative numbers Signed-magnitude systems

• +99, -57, -13.3Binary 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


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


Representation of negative numbers Radix – Complement representation

• n digit number is substracted from rn

• Example:if r = 10 and n = 4, rn = 104 =10,000The 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


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


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=8010 = 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

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 ……

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.

2’s Complement Addition + SubstractionI. Representation of numerical information

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

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 wrongIf signs of (addends) are same and sign of sum is

different


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


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 + + = +

+ - = -

- - = +


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

II. Representation of non numerical information


• 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}



• 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



• 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

1 -----> 0001 6 -----> 0110

2 -----> 0010 7 -----> 0111

3 -----> 0011 8 -----> 1000

4 -----> 0100 9 -----> 1001

The following codes are unused: 1010, 1011, 1100, 1101, 1110, 1111


chapter 2 number systems + codes. overview objective: to use positional number systems to convert...

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