02Data Representation

8/13/2019 02Data Representation

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Outline Data Representation

Compliments


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Data Types The data types stored in digital computers may

be classified as being one of the followingcategories:

1. numbers used in arithmetic computations,

2. letters of the alphabet used in data processing, and

3. other discrete symbols used for specific purposes.

All types of data are represented in computers inbinary-coded form.


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Radix representation of

numbers Radix or base: is the total number of

symbols used to represent a value. A

number system of radix r uses a stringconsisting of r distinct symbols torepresent a value.


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Radix representation of

numbers Example: convert the following number to

the radix 10 format. 97654.35

The positions indicate the power of the radix.

Start from the decimal point right to left we get0,1,2,3,4 for the whole numbers.

And from the decimal point left to right

We get -1, -2 for the fractions

= 9x104+ 7x103 + 6x102 + 5x101 + 4x100+ 3x10-1 +

5x10-2


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Binary Numbers Binary numbers are made of binary

digits (bits):

0 and 1

Convert the following to decimal

(1011)2= 1x23+ 0x22+ 1x21 +1x20 =

(11)10


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Example

Use radix representation to convert the binary number (101.01) into

decimal.

The position value is power of 2

1 0 1. 0 1

22 21 20 2-1 2-2

4 + 0 + 1 + 0 + 1/22 = 5.25

(101.01)2(5.25)10

= 1 x 22+ 0 x 2 + 1 + 0 x 2-1+ 1 x 2-2


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Binary Addition

1 1 1 1 0 1

+ 1 0 1 1 1

---------------------

0

1

0

1

1

1111

1 1 00

carries

Example Add (11110)2to

(10111)2

(111101)2+ (10111) 2 = (1010100)2

carry


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Binary Subtraction We can also perform subtraction (with borrows).

Example: subtract (10111) from (1001101)

1 10

0 10 10 0 0 10

1 0 0 1 1 0 1

- 1 0 1 1 1

------------------------

0 1 1 0 1 1 0

borrows

1+1=

2

(1001101)2- (10111)2= (0110110)2


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The Growth of Binary Numbers

n 2n0 20=1

1 21=2

2 22=4

3 23=8

4 24=16

5 25=32

6 26=647 27=12

8

n 2n8 28=256

9 29=512

10 210=1024

11 211=2048

12 212=4096

20 220=1M

30 230=1G

40 240

=1T

Mega

Giga

Tera


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Counting in Octal

0 1 2 3 4 5 6 7

10 11 12 13 14 15 16 17

20 21 22 23 24 25 26 27


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Conversion Between Number Bases

Decimal(base 10)

Octal(base 8)

Binary(base 2)

Hexadecimal

(base16)

We normally convert to base 10because we are naturally used to the

decimal number system.

We can also convert to other number

systems


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Converting an Integer from Decimal toAnother Base

1. Divide the decimal number by the base (e.g. 2)

2. The remainder is the lowest-order digit

3. Repeat the first two steps until no div isorremains.

4. For binary the even number has no remainder0, while the odd has 1

For each digit position:


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Converting an Integer from Decimal toAnother Base

Example for (13)10:Integer

Quotient

13/2 = (12+1) a0 = 1

6/2 = ( 6+0 ) a1 = 0

3/2 = (2+1 ) a2 = 11/2 = (0+1) a3 = 1

Remainder Coefficient

Answer (13)10 = (a3 a2 a1 a0)2 = (1101)2


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Converting a Fraction from DecimaltoAnother Base

1. Multiply decimal number by the base (e.g. 2)2. The integer is the highest-order digit

3. Repeat the first two steps until fraction becomes zero.



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Converting a Fraction from DecimaltoAnother Base

Example for (0.625)10:

Integer

0.625 x 2 = 1 + 0.25 a-1 = 1

0.250 x 2 = 0 + 0.50 a-2 = 0

0.500 x 2 = 1 + 0 a-3 = 1

Fraction Coefficient

Answer (0.625)10 = (0.a-1 a-2 a-3 )2 = (0.101)2


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DECIMAL TO BINARY CONVERSION(INTEGER+FRACTION)

(1) Separate the decimal number into integer and fraction parts.

(2) Repeatedly divide the integer part by 2 to give a quotient and aremainder and

Remove the remainder. Arrange the sequence of remaindersright to left from the period. (Least significant bit first)

(3) Repeatedly multiply the fraction part by 2 to give an integerand a fraction part

and remove the integer. Arrange the sequence of integers leftto right from the period. (Most significant fraction bit first)


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(Example) (41.6875)10 (?)2

Integer = 41, Fraction = 0.6875

The first procedure produces

41 = 32+8+1

= 1 x 25+ 0 x 24+ 1 x 23+ 0 x 22+ 0 x 2 + 1 = (101001)

Integer remainder

41 /2 1

20 0

10 05 1

2 0

1 1

Overflow Fraction

X by 2 .6875

1 .3750

0 .750

1 .51 0

.Closer tothe point


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Converting an Integer fromDecimal toOctal

1. Divide decimal number by the base (8)

2. The remainder is the lowest-order digit

3. Repeat first two steps until no div isorremains.



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Example for (175)10:

IntegerQuotient

175/8 = 21 + 7/8 a0 = 7

21/8 = 2 + 5/8 a1 = 5

2/8 = 0 + 2/8 a2 = 2

Remainder Coefficient

Answer (175)10 = (a2 a1 a0)2 = (257)8


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1. Multiply decimal number by the base (e.g. 8)2. The integer is the highest-order digit

3. Repeat first two steps until fraction becomeszero.



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Example for (0.3125)10:

Integer

0.3125 x 8 = 2 + 0.5 a-1 = 2

0.5000 x 8 = 4 + 0 a-2 = 4

Fraction Coefficient

Answer (0.3125)10 = (0.24)8

Combine the two (175.3125)10 = (257.24)8

Remainder

of division

Overflow of

multiplication


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Hexadecimal Numbers

Hexadecimal numbers are made of 16 symbols: (0,1,2,3,4,5,6,7,8,9,A, B, C, D, E, F)

Convert a hexadecimal number to decimal (3A9F)16= 3x163+ 10x162 + 9x161 + 15x160 = 1499910

Hexadecimal with fractions: (2D3.5)16= 2x16

2 + 13x161 + 3x160+5x16-1 = 723.312510

Note that eachhexadecimal digit can be represented

with four bits. (1110) 2 = (E)16

Groups of four bits are called a nibble. (1110) 2


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Example

Convert the decimal number(107.00390625)10into hexadecimal number.

(107.00390625)10(6B.01)16

Integer remainder

107 Divide/16

6 11=B

0 6

Overflow FractionX by 16 .

00390625

0 .0625

1 .0000

.Closer to

the period


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One to one comparison

Binary, octal, andhexadecimal similar

Easy to build circuits tooperate on theserepresentations

Possible to convertbetween the threeformats


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Converting between Base 16 andBase 2

Conversion is easy! Determine 4-bit value for each hex digit

Note that there are 24= 16 different values offour bits which means each 16 value is

converted to four binary bits. Easier to read and write in hexadecimal.

Representations are equivalent!

3A9F16= 0011 1010 1001 11112

3 A 9 F


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Converting between Base 16 andBase 8

1. Convert from Base 16 to Base 22. Regroup bits into groups of three starting from right

3. Ignore leading zeros

4. Each group of three bits forms an octal digit (8 isrepresented by 3 binary bits).

352378= 011 101 010 011 1112

5 2 3 73

3A9F16= 0011 1010 1001 11112

3 A 9 F


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ExampleConvert 101011110110011 to

a. octal number

b. hexadecimal number

a. Each 3 bits are converted to octal :

(101) (011) (110) (110) (011)

5 3 6 6 3

101011110110011 = (53663)8

b. Each 4 bits are converted to hexadecimal:

(0101) (0111) (1011) (0011)

5 7 B 3

101011110110011 = (57B3)16

Conversion from binary to hexadecimal is similar except that the bits

divided into groups of four.

Binary Coded Decimal


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Binary Coded Decimal

Binary coded decimal (BCD) represents each decimal digit

with four bits

Ex. 0011 0010 1001 = 32910

This is NOTthe same as 0011001010012

Why use binary coded decimal? Because people think indecimal.

Digit BCD Code Digit BCD Code

0 0000 5 0101

1 0001 6 0110

2 0010 7 0111

3 0011 8 1000

4 0100 9 1001

3 2 9


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BCD versus other codes

BCD not very efficient

Used in earlycomputers (40s, 50s)

Used to encodenumbers for seven-

segment displays. Easier to read?

(Example)

The decimal 99 is represented

by 1001 1001.


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ASCII Code

American Standard Code for Information Interchange

ASCII is a 7-bit code, frequently used with an 8th

bit forerror detection (more about that in a bit).

Character ASCII (bin) ASCII

(hex)

Decimal Octal

A 1000001 41 65 101

B 1000010 42 66 102

C 1000011 43 67 103

Z

a

1


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Outline

Data Representation

Compliments


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Subtraction using addition

Conventional addition (using carry) is easily

implemented in digital computers.

However; subtraction by borrowing is difficultand inefficient for digital computers.

Much more efficient to implement subtraction

using ADDITION OF the COMPLEMENTS ofnumbers.


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Complements of numbers

(r-1 )sComplement

Given a numberN in base r having n digits,the (r- 1)scomplement ofN is defined as

(rn- 1) - N

For decimal numbers thebase or r =10 and r- 1=9,

so the 9s complement ofN

is (10n-1)-N

99999. - N

Digit

n

Digit

n-1

Next

digit

Next

digit

First

digit

9 9 9 9 9

-


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2- Find the 9s complement of 546700 and 12389

The 9s complement of 546700 is 999999 - 546700=453299

and the 9s complement of 12389 is99999- 12389=87610.

9s complementExamples

5 4 6 7 0- 09 9 9 9 9 9

4 5 3 2 9 9

1 2 3 8- 99 9 9 9 9

8 7 6 1 0


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ls complement

For binary numbers, r =2 and r1=1,

r-1s complement is the ls complement.

The ls complement of N is (2^n- 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

-


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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 ofN is (24- 1) -N. = (1111) -N


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The complement 1s of

1011001 is 0100110

0 1 1 0 0- 1

1 1 1 1 1 1

1 0 0 1 1 0

0 0 1 1 1- 11 1 1 1 1 1

1 1 0 0 0 0

1

1

0

The 1s complement of

0001111 is 1110000

0

1

1

ls complement


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rs Complement

Given a numberN in base r having n digits,

the rscomplement ofN is defined as

rn- N.

For decimal numbers the

base or r =10,

so the 10s complement ofNis 10n-N.

100000. - N

Digit

n

Digit

n-1

Next

digit

Next

digit

First

digit

0 0 0 0 0

-

1


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10s complementExamples

Find the 10s complement of

546700 and 12389

The 10s complement of 546700is 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

0 0 0 0 0 0

4 5 3 3 0 0

1 2 3 8- 91 0 0 0 0 0

8 7 6 1 1

1


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For binary numbers, r =2,

rs complement is the 2s complement.

The 2s complement ofN is 2n -N.

2s complement

Digit

n

Digit

n-1

Next

digit

Next

digit

First

digit

0 0 0 0 0

-

1


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2s complement Example


1011001 is 0100111


0001111 is 1110001

0 1 1 0 0- 10 0 0 0 0 0

1 0 0 1 1 1

0 0 1 1 1

-1

1 1 0 0 0 1

1

0

0

0

1

1

0 0 0 0 0 001


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Fast Methods for 2sComplement

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


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Fast Methods for 2sComplement

Method 2:The 2scomplement can be formed by leaving all least significant 0s

and the first 1 unchanged, and then replacing lsby 0sand 0sby lsin all other higher significant bits.

Example:

The 2s complement of 1101100is

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.

E l


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Examples

Finding the 2s complement of (01100101)2

Method 1 Simply complement each bit andthen add 1 to the result.

(01100101)2[N] = 2s complement = 1s complement (10011010)2+1

=(10011011)2

Method 2Starting with the least significant bit,

copy all the bits up to and including the first 1 bitand then complement the remaining bits.

N = 0 1 1 0 0 1 0 1

[N] = 1 0 0 1 1 0 1 1

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02Data Representation